Master Resource Book in JEE Main Physics 9789313195481

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MASTER RES URCE

Book for

JEE Main

Physics Specially Prepared Questions for JEE Main with

Complete Theory 2 Levels Exercises Exams Questions

DB SINGH

ARIHANT PRAKASHAN (Series), MEERUT

MASTER RES URCE

JEE Main

Book for

Arihant Prakashan (Series), Meerut All Rights Reserved © Author No part of this publication may be re-produced, stored in a retrieval system or distributed in any form 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 damages or loss suffered there upon. All disputes subject to Meerut (UP) jurisdiction only.

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MASTER RES URCE

Book for

JEE Main

PREFACE In sync with the recent changes in the test pattern and format of JEE Main (Joint Engineering Entrance), it is my pleasure to introduce Master Resource Book in Physics for JEE Main, for the Students aspiring a seat in a reputed Engineering College. JEE Main is a gateway examination for candidates expecting to seek admission in Bachelor in Engineering (BE), Bachelor of Technology (B.Tech) and Bachelor of Architecture (B.Arch) at Indian Institutes of Information Technology (IIITs), National Institutes of Technology (NITs), Delhi Technological University and other Centrally Funded Technical Institutes (CFTIs). JEE Main is also an examination which is like screening examination for JEE Advanced (The gateway examination to India's most reputed Technical Institutes, Indian Institutes of Technology IITs). Only the top 2.2 lacs students passed in JEE Main will be able to attempt JEE Advanced. Gradually, the number of students aspiring for the seat in the Engineering College has increased rapidly in the last 5 Years or so. This year nearly 10 lacs students appeared for JEE Main and only a few were able to reserve a seat in the college of their choice, so there is a cut throat competition among the aspirants. Thus, it calls for a systematic mastery of all the subjects of the test with paramount importance to problem-solving. Most of the books now in the market have become repetitive with scant respect to the needs of true and effective learning. This book has been designed to fulfill the perceived needs of the students as such. —

This book comprehensively covers all the topics of JEE Main Physics syllabus. The chapters have been sequenced according to the syllabus of class 11th & 12th. Each chapter has essential theoretical discussion of the related concepts with sufficient number of solved examples, practice problems and other solved problems. In each chapter previous years' questions of AIEEE and JEE Main have been included to help students know the difficulty levels and nature of questions asked in competitive exams at this level.

—

All types of questions have been included in this book: Single Correct Answer Types, Multiple Correct Answer Types, Reasoning Types, Matches, Passage-based Questions etc.

—

This is the only book which has its subject matter divided as per class 11th & 12th syllabus. It covers almost all questions of NCERT Textbook & NCERT Exemplar problems.

It is hoped this new effort will immensely benefit the students in their goal to secure a seat in the prestigious engineering college, and would be convenient to teachers in planning their teaching programmes. Suggestions for further improvement are welcome from the students and teachers.

DB Singh

MASTER RES URCE

JEE Main

Book for

CONTENTS 20. Magnetostatics

PART I

21. Electromagnetic Induction and Alternating Current

Chapters from Class 11th Syllabus 1. Units and Measurements 2. Kinematics 3. Vector Analysis

906-944 945-1001

3-40

22. Electromagnetic Waves

1002-1027

41-84

23. Ray Optics and Optical Instruments

1028-1088

85-115

4. Projectile Motion

116-151

24. Wave Optics

1089-1130

5. Circular Motion

152-177

6. Laws of Motion and Friction

178-229

25. Dual Nature of Radiation and Matter

1131-1155

7. Work, Energy and Power

230-267

26. Electronic Devices

1156-1190

8. Centre of Mass

268-306

27. Atoms, Molecules and Nuclei

1191-1247

9. Rotational Motion

307-351

28. Communication Systems

1248-1270

29. Experimental Physics

1271-1308

10. Gravitation

352-396

11. Properties of Solids

397-434

12. Properties of Liquids

435-488

13. Heat and Kinetic Theory of Gases 489-571

JEE Main Solved Papers Solved Papers 2013 (Online & Offline)

1-34

Solved Papers 2014

35-42

Solved Papers 2015

43-52

Solved Papers 2016

53-60

PART II

Solved Papers 2017

1-7

Chapters from Class 12th Syllabus

Solved Papers 2018

1-8

14. Thermodynamics

572-614

15. Oscillations

615-662

16. Waves

663-716

17. Electrostatics

719-803

18. Current Electricity

804-862

19. Magnetic Effect of Current

863-905

Online JEE Main 2019 Solved Papers (April & January Attempt)

1-32

MASTER RES URCE

JEE Main

Book for

SYLLABUS NOTE The syllabus contains two Sections - A & B. Section A pertains to the Theory Part, having 80% weightage, while Section B contains Practical Component (Experimental Skills) having 20% weightage.

SECTION- A UNIT 1 Physics and Measurement

UNIT 5 Rotational Motion

Physics, technology and society, SI units, Fundamental and derived units. Least count, accuracy and precision of measuring instruments, Errors in measurement, Significant figures. Dimensions of Physical quantities, dimensional analysis and its applications.

Centre of mass of a two-particle system, Centre of mass of a rigid body; Basic concepts of rotational motion; moment of a force, torque, angular momentum, conservation of angular momentum and its applications; moment of inertia, radius of gyration. Values of moments of inertia for simple geometrical objects, parallel and perpendicular axes theorems and their applications. Rigid body rotation, equations of rotational motion.

UNIT 2 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, position time graphs, relations for uniformly accelerated motion. Scalars and Vectors, Vector addition and Subtraction, Zero Vector, Scalar and Vector products, Unit Vector, Resolution of a Vector. Relative Velocity, Motion in a plane, Projectile Motion, Uniform Circular Motion.

UNIT 3 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.

UNIT 4 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 nonconservative forces; Elastic and inelastic collisions in one and two dimensions.

UNIT 6 Gravitation The universal law of gravitation. Acceleration due to gravity and its variation with altitude and depth. Kepler's laws of planetary motion. Gravitational potential energy; gravitational potential. Escape velocity. Orbital velocity of a satellite. Geo-stationary satellites.

UNIT 7 Properties of Solids & Liquids Elastic behaviour, Stress-strain relationship, Hooke's. Law, Young's modulus, bulk modulus, modulus of rigidity. Pressure due to a fluid column; Pascal's law and its applications. Viscosity, Stokes' law, terminal velocity, streamline and turbulent flow, Reynolds number. Bernoulli's principle and its applications. Surface energy and surface tension, angle of contact, application of surface tension - drops, bubbles and capillary rise. Heat, temperature, thermal expansion; specific heat capacity, calorimetry; change of state, latent heat. Heat transfer-conduction, convection and radiation, Newton's law of cooling.

MASTER RES URCE

JEE Main

Book for

UNIT 8 Thermodynamics Thermal equilibrium, zeroth law of thermo-dynamics, concept of temperature. Heat, work and internal energy. First law of thermodynamics. Second law of thermodynamics: reversible and irreversible processes. Camot engine and its efficiency.

UNIT 9 Kinetic Theory of Gases Equation of state of a perfect gas, work done on compressing a gas. Kinetic theory of gases - assumptions, concept of pressure. Kinetic energy and temperature: rms speed of gas molecules; Degrees of freedom, Law of equipartition of energy, applications to specific heat capacities of gases; Mean free path, Avogadro's number.

UNIT 10 Oscillations and Waves Periodic motion - period, frequency, displacement as a function of time. Periodic functions. Simple harmonic motion (S.H.M.) and its equation; phase; oscillations of a spring - restoring force and force constant; energy in S.H.M. - kinetic and potential energies; Simple pendulum - derivation of expression for its time period; Free, forced and damped oscillations, resonance. Wave motion Longitudinal and transverse waves, speed of a wave. Displacement relation for a progressive wave. Principle of superposition of waves, reflection of waves, Standing waves in strings and organ pipes, fundamental mode and harmonics, Beats, Doppler effect in sound.

UNIT 11 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.

UNIT 12 Current Electricity Electric current, Drift velocity, Ohm's law, Electrical resistance, Resistances of different materials, V-I characteristics of Ohmic and nonohmic 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, Metre bridge. Potentiometer - principle and its applications.

UNIT 13 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 currentcarrying 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.

MASTER RES URCE

JEE Main

Book for

UNIT 14 Electromagnetic Induction and Alternating Currents

UNIT 17 Dual Nature of Matter and Radiation

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.

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. DavissonGermer experiment.

UNIT 15 Electromagnetic Waves Electromagnetic waves and their characteristics. Transverse nature of electromagnetic waves. Electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays). Applications of e.m. waves.

UNIT 16 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, coherent sources and sustained interference of light. 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.

UNIT 18 Atoms and Nuclei Alpha-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-alpha, beta and gamma 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.

UNIT 19 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 & NOR). Transistor as a switch.

UNIT 20 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)

MASTER RES URCE

JEE Main

Book for

SECTION- B UNIT 21 Experimental Skills Familiarity with the basic approach and observations of the experiments and activities 1. Vernier callipers - its use to measure internal and external diameter and depth of a vessel.

(ii) Internal resistance of a cell. 14. Resistance and figure of merit of a galvanometer by half deflection method.

2. Screw gauge - its use to determine thickness/ diameter of thin sheet/wire.

15. Focal length of (i) Convex mirror (ii) Concave mirror (iii) Convex lens

3. Simple Pendulum - dissipation of energy by plotting a graph between square of amplitude and time.

16. Using parallax method. Plot of angle of deviation vs angle of incidence for a triangular prism.

4. Metre Scale - mass of a given object by principle of moments.

17. Refractive index of a glass slab using a travelling microscope.

5. Young's modulus of elasticity of the material of a metallic wire.

18. Characteristic curves of a p-n junction diode in forward and reverse bias.

6. Surface tension of water by capillary rise and effect of detergents.

19. Characteristic curves of a Zener diode and finding reverse break down voltage.

7. Coefficient of Viscosity of a given viscous liquid by measuring terminal velocity of a given spherical body.

20. Characteristic curves of a transistor and finding current gain and voltage gain.

8. Plotting a cooling curve for the relationship between the temperature of a hot body and time. 9. Speed of sound in air at room temperature using a resonance tube. 10. Specific heat capacity of a given (i) solid and (ii) liquid by method of mixtures. 11. Resistivity of the material of a given wire using metre bridge. 12. Resistance of a given wire using Ohm's law. 13. Potentiometer (i) Comparison of emf of two primary cells.

21. Identification of Diode, LED, Transistor, IC, Resistor, Capacitor from mixed collection of such items. 22. Using multimeter to (i) Identify base of a transistor. (ii) Distinguish between npn and pnp type transistor. (iii) See the unidirectional flow of current in case of a diode and an LED. (iv) Check the correctness or otherwise of a given electronic component (diode, transistor or IC).

Part - I Chapters from Class 11

th

Syllabus

1

Units and Measurements JEE Main MILESTONE

< < < <
0, v < 0, a > 0

acceleration. It passes through two points P and Q separated by a distance with velocities 30 kmh–1 and 40 kmh -2 respectively. The velocity of car midway between P and Q is

other of iron. The thickness of the wooden plate is 4 cm and that of iron plate is 2 cm. A bullet fired goes through the wood first and then penetrates 1 cm into iron. A similar bullet fired with the same velocity from opposite direction goes through iron first and then penetrates 2 cm into wood. If a1 and a2 be the retardation offered to the bullet by wood and iron plates respectively, then (a) t1 + t2 (c) a1 = a2

(b) a2 = 2 a1 (d) data insufficient

line decreases linearly with its displacement s from 20 ms -1 to a value approaching zero at s = 30 m, then acceleration of the particle at s = 15 m is 20

(b) 1 km–1 (d) 35.35 km–1

v (in ms–1)

15. A particle starts from the origin and moves along the X-axis such that the velocity at any instant is given by 4 t 3 - 2 t, where t is in second and velocity is in ms–1. What is the acceleration of the particle when it is 2 m from the origin? (a) 10 ms–2 (c) 22 ms–2

(b) 12 ms–2 (d) 28 ms–2

16. The retardation experienced by a moving motor boat,

dv = - kv 3, dt where k is a constant. If v0 is the magnitude of the velocity at cut-off, the magnitude of the velocity at time t after the cut-off is

O

(a)

2 –2 ms 3

(c) v 0 e

(b) - kt

(d)

v0 2

u (in m)

2v20 kt + 1

(d) -

20 ms–2 3

21. The velocity of a particle moving in a straight line varies with time in such a manner that v versus t graph is velocity is vm and the total time of motion is t0 v

vm t0

v0

30

2 20 (b) - ms–2 (c) ms–2 3 3

after its engine is cut-off, is given by

(a) v 0

(b) x > 0, v < 0, a < 0 (d) x > 0, v > 0, a < 0

20. If the velocity v of a particle moving along a straight

14. A car is moving along a straight road with uniform

(a) 33.3 km–1 (c) 25 2 km–1

18. A lift is coming from 8

(c) 22. 75 m

th

t

p vm 4 (ii) Such motion cannot be realized in practical terms (i) Average velocity of the particle is

58 JEE Main Physics 28. The engine of a train can impart a maximum

(a) Only (i) is correct (b) Only (ii) is correct (c) Both (i) and (ii) are correct (d) Both (i) and (ii) are wrong

22. A police van moving on a highway with a speed of 30 km/h fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km/h . If the muzzle speed of the bullet is 150 m/s, with what speed does the bullet hit the thief’s car? (Note Obtain that speed which is relevant for damaging the thief’s car. [NCERT]

(a) 105 m/s (c) 95 m/s

(b) 100 m/s (d) 110 m/s

with uniform acceleration, then it travels a distance 2s with uniform speed, finally it travels a distance 3s with uniform retardation and comes to rest. If the complete motion of the particle in a straight line then the ratio of its average velocity to maximum velocity in (b) 4/5

(c) 3/5

(a) 108 s (c) 56.6 s

29. The acceleration of a particle increasing linearly with time t is 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) v 0 t + bt3 3

30. The displacement-time graph of a moving particle is shown below. The instantaneous velocity of the particle is negative at the point

(d) 2/5

D

Displacement

acceleration is observed to be a distance a from a fixed point initially. It is at distances b, c, d from the same point after n, 2n, 3n second. The acceleration of the particle is c - 2b + a n2 c + 2b + a (c) 4 n2

c + b+ a 9 n2 c -b+ a (d) n2

Time

(a) C

(b) D

v 30 (in ms–1) 20 10

(b) D > s (d) D £ s

0

26. Three particles start from the origin at the same

–10

time, one with a velocity v1 along x-axis, the second along the y-axis with a velocity v2 and the third along x = y line. The velocity of the third so that the three may always lie on the same line is

–20

3 v1v2 (c) v1 + v2

(d) F

body in 5 s will be 40

2 v1v2 (b) v1 + v2

(c) E

31. In the given v-t graph, the distance travelled by the

constant velocity. At an instant of time the distance of time the distance travelled by it is s and its displacement is D, then (a) D < s (c) D = s

F

E C

(b)

25. A body is moving along a straight line path with

vv (a) 1 2 v1 + v2

1 2 bt 6 1 (d) v 0 t + bt2 3 (b) v 0 t +

Graphical Representation

24. A particle moving in a straight line with uniform

(a)

(b) 191 s (d) time is fixed

(a) v 0 t +

23. A particle starts from rest and travels a distance s

(a) 6/7

acceleration of 1 ms–2 and the brakes can give a maximum retardation of 3 ms–2. The least time during which a train can go from one place to the other place at a distance of 1.2 km is nearly

(d) zero

(a) 20 m

2

3

(b) 40 m

4

5

(c) 80 m

Tme (in s)

(d) 100 m

32. The displacement-time graphs of two moving particles make angles of 30° and 45° with the x-axis. The ratio of the two velocities is

satisfies 0 £ v < v0 .

Displacement

27. In one dimensional motion, instantaneous speed v [NCERT Exemplar]

(a) The displacement in time T must always take non-negative values (b) The displacement x in time T satisfies - v 0 T < x < v 0T (c) The acceleration is always a non-negative number (d) The motion has no turning points

45°

30° Time

(a) 3 : 1

(b) 1 : 1

(c) 1 : 2

(d) 1 : 3

Kinematics 33. A rocket is fired upwards. Its engine explodes fully is 12 s. The height reached by the rocket as calculated from its velocity-time graph is

v (in ms)–1

1200

37. The given graph shows the

x

x0

a

(a)

x

(b)

x

132

t (in s)

(a) 1200 × 66 m 1200 m (c) 12

u

variation of velocity with v0 displacement. Which one of the graph given below correctly represents the variation of acceleration with displacement? a

12

59

(b) 1200 × 132 m

34. v-t graph for a particle is as shown. The distance travelled in the first 4 s is (a) 12 m (b) 16 m (c) 20 m (d) 24 m

a

a

(d) 1200 × 122 m (c)

v (m/s)

(d)

x

8

38. A ball A is thrown up vertically with a speed u and at

4

0

2

4

6

the same instant another ball B is released from a height h. At time t, the speed of A relative to B is t (s)

35. The velocity-time graph of a body is shown in figure. The ratio of the ..... during the v intervals OA and AB is .....

(a) u

(b) 2u

(c) u - gt

(d) (u2 - gt )

39. A body falls freely from rest. It covers as much

C

(a) average velocities : 2 60° 30° O OA 1 A t (b) : AB 3 (c) average accelerations, same as distances covered 1 (d) distances covered : 2

B

distance in the last second of its motion as covered in the first three seconds. The body has fallen for a time of (a) 3 s (c) 7 s

(b) 5 s (d) 9 s

40. Rain is falling vertically with a speed of 30 m/s. A

36. Figure shows the acceleration-time graphs of a particle. Which of the following represents the corresponding velocity-time graphs?

woman rides a bicycle with a speed of 10 m/s in the north to south direction. What is the direction in which she should hold her umbrella? [NCERT] (a) 18° with vertical (c) 28° with vertical

(b) 18° with horizontal (d) 28° with horizontal

41. A ball P is dropped vertically and another ball Q is thrown horizontally with the same velocities from the same height and at the same time. If air resistance is neglected, then

a

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

t

v

v

(a)

(b)

ball P reaches the ground first ball Q reaches the ground first both reach the ground at the same time the respective masses of the two balls will decide the time

42. A particle moves along x-axis as t

t

Which of the following is true?

v

v

(c)

(d)

t

x = 4 ( t - 2) + a ( t - 2)2

t

(a) The initial velocity of particle is 4 (b) The acceleration of particle is 2 a (c) The particle is at origin at t = 0 (d) None of the above

60 JEE Main Physics 43. A ball is thrown vertically upwards. It was observed

50. A body thrown vertically upward with an initial

at a height h twice with a time a interval Dt the initial velocity of the ball is

velocity u reaches maximum height in 6 second. The ratio of the distances travelled by the body in the first second and seventh second is

2

(a) 8 gh + g2 ( Dt )2

æ gDt ö (b) 8 gh + ç ÷ è 2 ø

(c) 1 / 2 8 gh + g2 ( Dt )2

(d) 8 gh + 4 g2 ( Dt )2

(a) 1 : 1

A very small spherical ball slips on this wire the time taken by this ball to slip from A to B is 2 gR g cos q cos q (b) 2 gR g

(c) 2 (d)

A

B

(d) 1 : 11

the balls in regular interval of time. When one ball leaves his hand (speed = 20 ms -1), the position of other ball will be (Take g = 10 ms -2 ) (a) 10 m, 20 m, 10 m (c) 5 m, 15 m, 20 m

(b) 15 m, 20 m, 15 m (d) 5 m, 10 m, 20 m

52. The

velocity-time graph of a particle in one-dimensional motion is shown in figure. Which of the following formulae are correct for describing the motion of the particle over the time interval t1 to t2 .

θ O

R g gR g cos q

(c) 1 : 2

51. A juggler keeps on moving four balls in the air throws

44. A frictionless wire AB is fixed on a sphere of radius R.

(a)

(b) 11 : 1

v C

45. A jet airplane travelling at a speed of 500 km/h ejects its products of combustion at the speed of 1500 km/h relative to the jet plane. What is the speed of the latter with respect to an observer on the ground? [NCERT]

(a) -1000 km / h (c) 100 km / h

(b) 1000 km / h (d) -100 km / h

46. A body is thrown vertically up with a velocity u. It passes three points A, B and C in its upward journey u u u repectively. The ratio of with velocities , and 2 3 4 the separations between points A and B and between AB is B and C, i. e., BC (a) 1

(b) 2

(c)

10 7

(d)

20 7

47. A boy released a ball from the top of a building. It will clear a window 2 m high at a distance 10 m below the top in nearly (a) 1 s

(b) 1.3 s

(c) 0.6 s

(d) 0.13 s

48. A stone is allowed to fall from the top of a tower 100  m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 254 ms -1. The two stones will meet after (a) 4 s

(b) 0.4 s

(c) 0.04 s

(d) 40 s

49. From a balloon rising vertically upwards at 5 m/s a stone is thrown up at 10 m/s relative to the balloon. Its velocity with respect to ground after 2 s is (assume g = 10 m /s2 ) (a) 0 (c) 10 m/s

(b) 20 m/s (d) 5 m/s

O

t1

t

t2

(i) x( t2 ) = x( t1) + v( t1)( t2 - t1) +

1 a( t2 - t1)2 2

(ii) v( t2 ) = v( t1) + a( t2 - t1) é x( t ) - x( t1) ù (iii) vav = ê 2 ú ë ( t2 - t1) û [v( t2 ) - v( t1)] (iv) aav = ( t2 - t1) 1 aav ( t2 - t1)2 2 (vi) x( t2 ) - x( t1) = Area under v-t curve bounded by the t-axis and the dotted line shown (v) x( t2 ) = x( t1) + vav ( t2 - t1) +

(a) (iii) and (vi) (c) (ii), (iii) and (iv)

(b) (iii), (iv) and (vi) (d) (iv) and (vi)

53. Water drops fall from a tap on the floor 5 m below at regular intervals of time, the first drop striking the floor when the fifth drop begins to fall. The height at which the third drop will be, from ground, at the instant when first drop strikes the ground, will be (g = 10 ms–2) (a) 1.25 m (c) 2.73 m

(b) 2.15 m (d) 3.75 m

54. From an elevated point A, a stone is projected vertically upwards. When the stone reaches a distance h below A, its velocity is double of what was at a height above A? The greatest height attained by the stone is (a)

h 3

(b)

2h 2

(c)

h 2

(d)

5h 3

Kinematics 55. A ball is thrown vertically upwards from the top of a tower of height h with velocity v. The ball strikes the ground after v (a) g (c)

é 2 gh ù ê1 + 1 + 2 ú v û ë

v æ 2 gh ö ç1 + 2 ÷ g è v ø

v (b) g

1 /2

(d)

é 2 gh ù ê1 + 1 - 2 ú v û ë

v æ 2 gh ö ç1 - 2 ÷ g è v ø

1 /2

56. A body freely falling from rest has a velocity v after it falls through distance h. The distance it has to fall down further for its velocity to become double is (a) h

(b) 2 h

(c) 3 h

(d) 4 h

57. Two balls A and B are thrown simultaneously from the top of a tower. A is thrown vertically up with a speed of 4 ms–1. B is thrown vertically down with a speed of 4 ms–1. The ball A and B hit the ground with speed v A and vB respectively. Then, (a) v A < v B

(b) v A > v B

(c) v A ³ v B

(d) v A = v B

58. A particle starting from rest falls from a certain height. Assuming that the value of acceleration due to gravity remains the same throughout motion, its displacements in three successive half second intervals are S1, S2 , S3. Then, (a) S1 : S2 : S3 = 1 : 5 : 9 (c) S1 : S2 : S3 = 1 : 1 : 1

(b) S1 : S2 : S3 = 1 : 2 : 3 (d) S1 : S2 : S3 = 1 : 3 : 5

59. A ball thrown upward from the top of a tower with speed v reaches the ground in t1 second. If this ball is thrown downward from the top of the same tower with speed v it reaches the ground in t2 second. In what time the ball shall reach the ground if it is allowed to falls freely under gravity from the top of the tower? (a)

t1 + t2 2

(b)

t1 - t2 2

(c) t1t2

(d) t1 + t2

60. A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 s, the average acceleration during contact is nearly (Take g = 10 ms -2 ) (a) 500 2 ms–2 upwards (c) 1500 5 ms–2 upwards

(b) 1800 ms–2 downwards (d) 1500 2 ms–2 downwards

61. A stone thrown vertically upwards attains a maximum height of 45 m. In what time the velocity of stone become equal to one-half the velocity of throw? (Given g = 10 ms -2 ) (a) 2 s

(b) 1.5 s

(c) 1 s

(d) 0.5 s

62. A body released from a great height falls freely towards the earth. Another body is released from the same height exactly one second later. The separation between the two bodies two second after the release of the second body is (a) 9.8 m

(b) 4.9 m

(c) 24.5 m

(d) 19.6 m

61

63. A particle covers 4 m, 5 m, 6 m and 7 m in 3rd, 4th, 5th and 6th second respectively. The particle starts (a) with an initial non-zero velcoity and moves with uniform acceleration (b) from rest and moves with uniform velocity (c) with an initial velocity and moves with uniform velcoity (d) from rest and moves with uniform acceleration

11 of 36 the height of the tower in the last second of its journey. The height of the tower is (Take g = 10 ms -2 )

64. A balls is released from the top of a tower travels

(a) 11 m (c) 47 m

(b) 36 m (d) 180 m

Relative Motion 65. At a metro station, a girl walks up a stationary escalator in time t1. If she remains stationary on the escalator, then the escalator take her up in time t2 . The time taken by her to walk up on the moving escalator will be [NCERT Exemplar] (a) (t1 + t2 )/2 (c) t1t2 /(t2 + t1 )

(b) t1t2 /(t2 - t1 ) (d) t1 - t2

66. A 120 m long train is moving in a direction with speed 20 m/s. A train B moving with 30 m/s in the opposite direction and 130 m long crosses the first train in a time. (a) 6 s (c) 38 s

(b) 36 s (d) None of these

67. For a body moving with relative speed of the velocity is doubled, then (a) its linear momentum is doubled (b) its linear momentum will be less than doubled (c) its linear momentum will be more than doubled (d) its linear momentum remains unchanged

68. An express train is moving with a velocity v1 its driver

finds another train is moving on the same track in the same direction with velocity v2 . To avoid collision driver applies a retardation a on the train. The minimum time of avoiding collision will be v1 - v2 a (c) None

(a) t =

v21 - v22 2 (d) Both (a) and (b)

(b) t =

69. Rain drops fall vertically at a speed of 20 ms -1. At what angle do they fall on the wind screen of a car moving with a velocity of 15 ms -1, if the wind screen velocity inclined at an angle of 23° to the vertical? ù é -1 æ 4 ö êëcot çè 3 ÷ø » 36° úû (a) 60º (c) 45º

(b) 30º (d) 90º

62 JEE Main Physics 70. A man wants to reach point B on the opposite bank of a river flowing at a speed as shown in figure. What minimum speed relative to water should the man have so that he can reach point B? (a) u 2

(b) u / 2

B

u 45° A

(c) 2u

a quiet day when the water is still and (ii) on a rough day when there is a uniform current so as to help the journey onwards and to impede the journey back. If the speed of the launch on both days was same, the time required for complete journey on the rough day, as compared to the quiet day will be (b) less (d) None of these

72. Two trains travelling on the same track are approaching each other with equal speeds of 40 ms–1. The drivers of the trains begin to decelerate simultaneously when they are just 2 km apart. If the decelerations are both uniform and equal, then the value of deceleration to barely avoid collision should be (a) 0.8 ms–2

(b) 2.1 ms–2 (c) 11.0 ms–2 (d) 13.2 ms–2

73. A 210 m long train is moving due North at a speed of 25 m/s. A small bird is flying due South, a little above the train with speed 5 m/s. The time taken by the bird to cross the train is (a) 6 s

(b) 7 s

(c) 9 s

(d) 10 s

74. A police jeep is chasing with velocity of 45 km/h a theif in another jeep moving with velocity 153 km/h. Police fires a bullet with muzzle velocity of 180 m/s. The velocity with which is will strike of the car of the thief is (a) 150 m/s

(b) 27 m/s

(c) 450 m/s

(d) 250 m/s

75. A boat is sent across a river with a velocity of 8 km/h. If the resultant velocity of boat is 10 km/h, then velocity of river is (a) 10 km/h

(b) 8 km/h

(c) 6 km/h

(d) 4 km/h

76. The distance between two particles moving towards each other is decreasing at the rate of 6 m/s. If these particles travel with same speed and in the same direction then the separatioon increase at the rate of 4 m/s. The particles have speed as (a) 5 m/s 1 m/s (c) 4 m/s; 2 m/s

(b) 4 m/s; 1 m/s (d) 5 m/s; 2 m/s

77. A train is moving towards east and a car is along north, both with same speed. The observed direction of a car to the passenger in the train is

(b) west-north direction (d) None of the above

78. A boat crosses a river from part A to part B which are just on opposite side. The speed of the water is vw and that of boat is vb relative to still water. Assume vb = 2 vw . What is the time taken by the boat? If it has to cross the river directly on the AB line.

(d) u /2

71. A steam boat goes across a lake and comes back (i) on

(a) more (c) same

(a) east-north direction (c) south-east direction

(a)

2D vb 3

(b)

3D 2 vb

(c)

D vb 2

(d)

D 2 vb

79. Two cars A and B are moving with same speed of 45 km/h along same direction. If a third car C coming from the opposite direction with a speed of 36 km/h meets two cars in an interval of 5 minutes. The distance between cars A and B should be (in km) (a) 6.75

(b) 7.25

(c) 5.55

(d) 8.35

80. Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km/h in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m/s2 . If after 50 s, the guard of B just brushes past the driver of A, what was the original distance [NCERT] between them? (a) 1250 m (c) 1450 m

(b) 1350 m (d) None of these

81. On a two lane road, car A is travelling with a speed of 36 km/h. Two cars B and C approach car A in opposite directions with a speed of 54 km/h each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. In this case, the acceleration of car B is required to avoid [NCERT] an accident (a) 1m / s2

(b) 0.1m / s2

(c) 1.9 m / s2

(d) 0.2m / s2

82. A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitoius path 23 km long and reaches the hotel in 28 min. What are the average speed of the taxi and the magnitude of average [NCERT] velocity respectively (in km/h)? (a) 49.3 and 21.43 (c) 21 and 20

(b) 48.3 and 24.43 (d) 21.3 and 49.3

83. A man can swim with a speed of 4 km/h in still water. How long does he take to cross a river 1 km wide, if the river flows steadily 3 km/h and he makes his strokes normal to the river current. How far down the river does he go when he reaches the other bank? [NCERT]

(a) 850 m (c) 650 m

(b) 750 m (d) None of these

Kinematics

Round II Only One Correct Option

63

(Mixed Bag) 8. A particle is dropped vertically from rest from a

1. An automobile travelling with a speed of 60 kmh

-1

can brake to stop with a distance of 20 m. If the car is going twice as fast i. e., 120 kms -1, the stopping distance will be [AIEEE 2004]

height. The time taken by it to fall through successive distances of 1 m each, will then be (a) all being equal to 2 / 9 second

2. Two balls of same size but the density of one is greater

(b) in the ratio of the square roots of the integers 1, 2, 3, … (c) in the ratio of the difference in the square roots of the integers is ( 2 - 1) ( 3 - 2 ) ( 4 - 3 )

than that of the other are dropped from the same height, then which ball will reach the earth first (air resistance is negligible)?

(d) in the ratio of the reciprocal of the square roots of the 1 ö æ 1 1 1 integers is ç , , , ÷ è 1 2 3 4ø

(a) 20 m

(b) 40 m

(c) 60 m

(d) 80 m

(a) Heavy ball (b) Light ball (c) Both simultaneously (d) Will depend upon the density of the balls

9. A man throws balls with the same speed vertically

3. A boat takes two hours to travel 8 km and back in still

water. If the velocity of water 4 kmh -1, the time taken for going ups tream 8 km and coming back is (a) 2 h (b) 2 h 40 min (c) 1 h 20 min (d) cannot be estimated with the information given

4. A particle moving in a straight line covers half the distance with speed of 3 m/s. The other half of the distance is covered in two equal time intervals and with speeds of 4.5 m/s and 7.5 m/s respectively. The average speed of the particle during this motion is (a) 4.0 m/s

(b) 5.0 m/s

(c) 5.5 m/s

(d) 4.8 m/s

5. In a race for 100 m dash, the first and the second runners have a gap of one metre at the mid way stage. Assuming the first runner goes steady, by what percentage should the second runner increases his speed just to win the race. (a) 2% (c) more than 4%

(b) 4% (d) less than 4%

6. Two cars A and B are travelling in the same direction with velocities v A and vB (v A > vB ). When the car A is at a distance s behind car B, the driver of the car A applies the brakes producing a uniform retardation a, there will be no collision when (a) s < (c) s ³

(v A - v B ) 2a

2

(v A - v B ) 2a

2

(b) s = (d) s £

(v A - v B ) 2a

2

(v A - v B ) 2a

2

straight line, where t = time in second. It covers a distance of (b) 6 m

(c) 4 m

(a) At least 0.8 m/s (b) Any speed less than 19.6 m/s (c) Only with speed 19.6 m/s (d) More than 19.6 m/s

10. A motion boat covers a given distance in 6 h moving down stream of a river. It covers the same distance in 10  h moving upstream. The time (in hour) it takes to cover the same distance in still water is (a) 6 h (c) 10 h

(b) 7.5 h (d) 15 h

11. A point initially at rest moves along x-axis. Its acceleration varies with time as a = (6 t + 5) m/s2. If it starts from origin, the distance covered in 2 s is (a) 20 m (c) 16 m

(b) 18 m (d) 25 m

12. From the top of a tower of height 50 m, a ball is thrown vertically upwards with a certain velocity. It hits the ground 10 s after it is thrown up. How much time does it take to cover a distance AB where A and B are two points 20 m and 40 m below the edge of the tower? (g = 10 ms -2 ) (a) 2.0 s (c) 0.5 s

(b) 1.0 s (d) 0.4 s

13. The acceleration of a particle is increasing linearly

7. A bird flies for 4 s with a velocity of |t–2|ms -1 in a (a) 8 m

upwards one after the other at an interval of 2 s. What should be the speed of the throw so that more than two balls are in the sky at any time? (Given g = 9.8 m/s2)

(d) 2 m

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, is 1 2 bt 3 1 (c) v 0t + bt3 3

(a) v 0t +

1 3 bt 6 1 (d) v 0t + bt2 2 (b) v 0t +

64 JEE Main Physics 14. A body of mass m is resting on a wedge of angle q as shown in figure. The wedge is given an acceleration a. What is the value of a that the mass m just falls freely?

19. A graph of x versus t is shown in figure. Choose correct alternatives from below.

[NCERT Exemplar]

x B A

m

C E

θ

(a) g (c) g tan q

(b) g sin q (d) g cot q

15. From the top of a tower, a stone is thrown up and reaches the ground in time t1 = 9 s. A second stone is thrown down with the same speed and reaches the ground in time t2 = 4 s. A third stone is released from rest and reaches the ground in time t3, which is equal to (a) 6.5 s 5 s (c) 36

(b) 6.0 s (d) 65 s

More Than One Correct Option 16. The motion of a body is given by the equation dv ( t) = 6.0 - 3 v ( t), where v( t) is speed in ms–1 and t in dt second. If body was at rest at t = 0

(a) the terminal speed is 2.0 ms–1 (b) the speed varies with the times as v (t ) = 2 (1 - e -3 t ) ms -1 (c) the speed is 1.0 ms–1 when the acceleration is half the initial value (d) the magnitude of the initial acceleration is 6.0 ms–2

17. An elevator ascends with an upward acceleration

of 2.0 ms -2 . At the instant its upward speed is 2.5 ms -1, loose bolt is dropped from the ceiling of the elevator 3.0 m from the floor. If g = 10 ms–2, then (a) the time of flight of the bolt from the ceiling to floor of the elevator is 0.11 s (b) the displacement of the bolt during the free fall relative to the elevator shaft is 0.75 m (c) the distance covered by the bolt during the free fall relative to the elevator shaft is 1.38 m (d) the distance covered by the bolt during the free fall relative to the elevator shaft is 2.52 m

18. A particle of mass m moves on the x-axis as follows : it starts from rest at t = 0 from the point x = 0 and comes to rest at t = 1 at the point x = 1. No other information is available about its motion at intermediate time (0 < t < 1). If a denotes the instantaneous acceleration of the particle, then (a) a cannot remain positive for all t in the interval 0 £ t £ 1 (b) |a | cannot exceed 2 at any points in its path (c) |a | must be ³ 4 at some point or points in its path (d) a must change sign during the motion but no other assertion can be made with the information given

D

t

(a) The particle was released from rest at t = 0 (b) At B, the acceleration a > 0 (c) At C, the velocity and the acceleration vanish (d) The speed at D exceeds that at E

20. The motion of a body falling from rest in a resisting medium is described by the equation dv = A - Bv dt where A and B are constants. Then (a) initial acceleration of the body is A (b) the velocity at which acceleration becomes zero is A/B A (c) the velocity at any time t is (1 - e Bt ) B (d) All of the above are wrong

21. A spring with one end attached to a mass and the other to a rigid support is stretched and released [NCERT Exemplar]

(a) Magnitude of acceleration, when just released is maximum (b) Magnitude of acceleration, when at equilibrium position is maximum (c) Speed is maximum when mass is at equilibrium position (d) Magnitude of displacement is always maximum whenever speed is minimum

22. The displacement (x) of a particle depends on time (t) as

x = a t2 - b t 3

(a) The particle will come to rest after time 2 a / 3 b (b) The particle will return to its starting point after time a / b (c) No net force will act on the particle at t = a /3 b (d) The initial velocity of the particle was zero but its initial acceleration was not zero

23. Suppose a and v denotes the acceleration and velocity respectively of a body in one dimensional motion, then (a) speed must increase when a > 0 (b) speed will increase when v and d are > 0 (c) speed must decreases when a < 0 (d) speed will decrease when v < 0 and a > 0

24. A particle is projected vertically upwards in vacuum with a speed v. (a) The time taken to rise to half its maximum height is half the time taken to reach its maximum height

Kinematics (b) The time taken to rise to three-fourth of its maximum height is half the time taken to reach its maximum height (c) When it rises to half its maximum height, its speed becomes v / 2 (d) When it rises to half its maximum height, its speed becomes v /2

25. A particle is moving with a uniform acceleration along a straight line AB. Its speed at A and B are 2 ms–1 and 14 ms–1 respectively. Then (a) its speed at mid-point of AB is 10 ms–1 (b) its speed at a point P such that AP : PB = 1:5 is 4 ms–1 (c) the time to go from A to mid-point of AB is double of that to go from mid-point to B (d) None of the above

26. A ball is bouncing elastically with a speed 1 m/s between walls of a railway compartment of size 10 m in a direction perpendicular to walls. The train is moving at a constant velocity of 10 m/s parallel to the direction of motion of the ball. As seen from the [NCERT Exemplar] ground. (a) the direction of motion of the ball changes every 10 s (b) speed of ball changes very 10 s (c) average speed of ball over any 20 s interval is fixed (d) the acceleration of ball is the same as from the train

27. The figure shows the velocity (v) of a particle moving on a straight line plotted against time (t). v (ms–1)

5 5

10 15

Comprehension Based Questions Passage When two bodies A and B are moving with velocity v A and v B , then relative velocity of A w.r.t. B is v AB = v A - v B Relative velocity of B w.r.t. A is v BA = v B - v A = v B + ( - v A ) When body C is moving with velocity vC on a body A, which is moving with velocity v A , then velocity of C w.r.t. ground is vC + v A . Suppose two parallel rail tracks run north-south. Train A moves north with a speed of 54 kmh–1 and train B moves south with a speed of 90 kmh–1.

29. Relative velocity of ground w.r.t. B is (a) 25 ms–1 due north (c) 40 ms–1 due north

(b) 25 ms–1 due south (d) 40 ms–1 due south

30. A monkey is moving with a velocity 18 kmh–1 on the roof of train A against the motion of train A. The velocity of monkey as observed by a man standing on the ground is

t (s)

Assertion and Reason

–5

Directions

–10

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

(a) The displacement of the particle in time 2 T is zero (b) The initial and final speeds of the particle are the same (c) The acceleration of the particle remains constant throughout the motion (d) The particle changes its direction of motion at same point

(a) 5 ms–1 towards south (b) 10 ms–1 towards north (c) 10 ms–1 towards south (d) 20 ms–1 towards south

10

0

65

The particle has zero displacement The particle has never turned around The particle has constant acceleration The average speed in the interval 0 to 5 s is the same as the average speed in the interval 5 to 10 s

28. The figure shows the velocity (v) of a particle plotted against time (t). v (ms–1) B

10

Question No. 31 to 35 are Assertion-Reason type. Each of these contains two Statements : Statement I (Assertion), Statement II (Reason). Each of these questions also has four alternative choice, only one of which is correct. You have to select the correct choices from the codes (a), (b), (c) and (d) given below (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 correct explanation of the Assertion (c) If Assertion is true but Reason is false (d) If Assertion is false but the Reason is true

31. Assertion A body is dropped from a height of 40.0 m. 0

–10 A

D T

2T

t (s)

After it falls by half the distance, the acceleration due to gravity ceases to act. The velocity with which it hits the ground is 20 ms–1 (Take g = 10 ms–2). Reason v2 = u2 + 2 as

66 JEE Main Physics 32. Assertion A car moving with a speed of 25 ms–1 takes

34. Assertion The slope of displacement-time graph of a

U turn in 5 s, without changing its speed. The average acceleration during these 5 s is 5 ms–2.

body movng with high velocity is steeper than the slope of displacement-time graph of a body with low velocity. Reason Slope of displacement-time graph = Velocity of the body.

Reason

Acceleration =

Change in velocity Time taken

33. Assertion The average velocity of the object over an interval of time is either smaller than or equal to the average speed of the object over the same interval. Reason Velocity is a vector quantity and speed is a scalar quantity.

35. Assertion A body having non-zero acceleration can have a constant velocity. Reason Acceleration is the rate of change of velocity.

Previous Years’ Questions 36. A ball is dropped from a bridge at a height of 176.4 m

42. A body of mass m is accelerated uniformly from rest

over a river. After 2s, a second ball is thrown straight downwards. What should be the initial velocity of the second ball so that both hit the water [UP SEE 2009] simultaneously?

to a speed v in a time T. The instantaneous power delivered to the body as a function of time is given by

(a) 2.45 ms–1 (c) 14.5 ms–1

(b) 49 ms–1 (d) 24.5 ms–1

37. A scooterist sees a bus 1 km ahead of him moving

with a velocity of 10 ms -1. With what speed the scooterist should move so as to overtake the bus in 100 s.? [Orissa JEE 2008] (a) 10 ms–1 (c) 50 ms–1

(b) 20 ms–1 (d) 30 ms–1

38. A bullet emerges from a barrel of length 1.2 m with a speed of 640 ms–1. Assuming constant acceleration, the approximate time that it spends in the barrel after the gun is fired is [WB JEE 2008] (a) 4 ms (c) 400 ms

(b) 40 ms (d) 1 s

39. A body is fired vertically upwards. At half the

maximum height, the velocity of the body is 10 ms–1. The maximum height raised by the body is (g = 10 ms -2 ). [Orissa JEE 2008] (a) zero (c) 15 m

(b) 10 m (d) 20m

40. The velocity of a particle is v = v0 + gt + ft2 . If its position is x = 0 at t = 0, then its displacement after unit time ( t = 1) is [AIEEE 2007] (a) v 0 - g / 2 + f (c) v 0 + g / 2 + f / 3

(b) v 0 + g / 2 + 3 f (d) v 0 + g + f

41. A proton in a cyclotron changes its velocity from 30 kms–1 due north to 40 kms–1 due east in 20 s. What is the magnitude of average acceleration [BVP 2008] during this time? (a) 2.5 kms–2 (c) 22.5 kms–2

(b) 12.5 kms–2 (d) 32.5 kms–2

[AIEEE 2008] 2

1 mv 2 t 2 T2 mv2 (c) 2 t2 T (a)

2

1 mv t 2 T2 mv2 (d) 2 t T

(b)

43. A body moves with initial velocity 10 ms -1. If it covers a distance of 20 m in 2 s then acceleration of the [Orissa JEE 2011] body is (a) zero (c) 5 ms -2

(b) 10 ms -2 (d) 2 ms -2

44. A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to [UP SEE 2007] 2

(a) x (c) x

x

(b) e (d) log e x

45. A ball which is at rest, is dropped from a height h metre. As it bounces off the floor its speed is 80% of what it was just before touching the ground? The ball [BVP Engg. 2007] will then rise to nearly a height (a) 0.94 h (c) 0.75 h

(b) 0.80 h (d) 0.64 h

46. A particle has an initial velocity of 3 $i + 4 $j and acceleration of 0.4 $i + 0.3 $j. Its speed after 10 s is (a) 10 units (c) 7 units

(b) 7 2 units (d) 8.5 units

47. A particle is moving with velocity v = k ( 4 i$ + x $j ) where k is a constant. The general equation for its [AIEEE 2010] path is (a) y = x2 + constant

(b) y2 = x + constant

(c) xy = constant

(d) y2 = x2 + constant

Kinematics

decelerates at 2 ms -2 . He reaches the ground with a speed of 3 ms -1. At what height, did he fallen out?

48. A particle located at x = 0 at time t = 0, starts moving along the positive x-direction with a velocity v that varies as v = a x . The displacement of the particle varies with time as [AIEEE 2006]

[AIEEE 2005]

(a) 111 m (c) 182 m

(b) t2 (d) t1/2

(a) t3 (c) t

through a distance S, then continues at constant f speed for time t and then decelerates at the rate to 2 come to rest. If the total distance traversed in 15 s, [AIEEE 2005] then

dv declerated at a rate given by = - 2.5 v , where v is dt the instantaneous speed. The time taken by the [AIEEE 2011] object, to come to rest would be

1 2 ft 4 1 (c) s = ft2 6

(a) s =

(b) 4 s (d) 1 s

50. A train accelerated uniformly from rest attains a -1

maximum speed of 40 ms in 20 s. It travels at this speed for 20 s and is brought to rest with uniform retardation in 40 s. The average velocity during this period is [BVP Engg. 2006] (a) (80/3) ms–1 (c) 25 ms–1

friction.

(d) s = ft

[EAMCET 2005]

g (b) (t1 + t2 )2 4 2 æt + t ö (d) 2 g ç 1 2 ÷ è 4 ø

(a) 2g (t1 + t2 )

parachute

opens,

1 2 ft 72

twice in it journey at a height h first after t1 and t2 second. Maximum height reached by the body is

(b) 30 ms–1 (d) 40 ms–1

When

(b) s =

53. A body projected vertically upwards crosses a point

51. A parachutist after alling out falls 50 m without

(b) 293 m (d) 91 m

52. A car starting from rest, accelerates at the rate f

49. An object, moving with a speed of 6.25 m/s, is

(a) 2 s (c) 8 s

67

(c)

it

g (t1 t2 ) 4

Answers Round I 1. 11. 21. 31. 41. 51. 61. 71. 81.

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

2. 12. 22. 32. 42. 52. 62. 72. 82.

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

3. 13. 23. 33. 43. 53. 63. 73. 83.

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

4. 14. 24. 34. 44. 54. 64. 74.

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

5. 15. 25. 35. 45. 55. 65. 75.

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

6. 16. 26. 36. 46. 56. 66. 76.

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

7. 17. 27. 37. 47. 57. 67. 77.

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

8. 18. 28. 38. 48. 58. 68. 78.

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

9. 19. 29. 39. 49. 59. 69. 79.

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

10. 20. 30. 40. 50. 60. 70. 80.

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

Round II 1. 11. 21. 31. 41. 51.

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

2. 12. 22. 32. 42. 52.

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

3. 13. 23. 33. 43. 53.

(b) (b) (b,d) (a) (a) (d)

4. 14. 24. 34. 44.

(a) (b) (b,c) (a) (a)

5. 15. 25. 35. 45.

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

6. 16. 26. 36. 46.

(c) (b,c,d) (b,c,d) (d) (b)

7. 17. 27. 37. 47.

(c) (b,c) (c,d) (b) (d)

8. 18. 28. 38. 48.

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

9. 19. 29. 39. 49.

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

10. 20. 30. 40. 50.

(b) (a,b,c) (b) (c) (c)

the Guidance Round I 1. Horizontal distance covered by the wheel in half revolution pR. A′

6. Initial velocity of the car (u) = 126 km/h = 126 ´

Final

5 m/s 18

5 æ ö m/s÷ çQ 1 km /h = è ø 18

= 35 m/s Final velocity of the car (v) = 0

2h

Distance travelled ( s) = 200 m From equation of motion, v 2 = u 2 + 2as

πR

A

So, the displacement of the point which was initially in contact with ground

a=

or

v 2 - u 2 0 - (35) 2 = 2s 2 ´ 200

= AA¢ = ( pR) 2 + (2 R) 2 = R n2 + 4 = p 2 + 4

[Q R = 1m ]

v = pt

2. Given, Þ

=

x

2

ò 0 dx = pò0 t dt =

2

4´4 pt = =8m 2 2

3. In graph (b), for one value of displacement, there are two timings. As a result of it, for one time, the average velocity is positive and for other time is equivalent negative. Due to it, the average velocity for the two timings (equal to time period) can vanish. 1 u + v1¢ u + u + at1 4. As, v1 = = = u + at1 2 2 2 v ¢ + v ¢ (u + at1) + u + a (t1 + t 2) 1 v2 = 1 2 = = u + at1 + at 2 2 2 2 v ¢ + v ¢ (u + at1 + at 2) + u + a (t1 + t 2 + t3) v3 = 2 3 = 2 2 1 = u + at1 + at 2 + at3 2 1 Then, v1 - v 2 = - a (t1 + t 2) 2 1 v 2 - v3 = - a (t 2 + t3) 2 v1 - v 2 t1 + t 2 \ = v 2 - v3 t 2 + t3

5. x = t + 1 Squaring both sides, we get x = (t + 1) 2 = t 2 + 1 + 2 t Differentiating it w.r.t. t, we get dx = 2t + 2 dt dx Velocity = v = = 2t + 2 dt so increase with time.

-35 ´ 35 49 m/s 2 =400 16

= - 3.06 m/s 2 \Retardation of the car = - 3.06 m/s 2 From equation of motion, v = u + at v - u (0 - 35) or t= = a ( -49 /16) 35 ´ 16 49 5 ´ 16 80 s = = 7 7 =

= 114 . s \Car will stop after 11.4 s

7. Let the particle touches the sphere at the point A. Let PA = l PB =

\ In D OPB, cos a =

l 2

PB r O

α

α

P

\ or

But \

r

r

α

α B

A

PB = r cos a l = r cos a 2 l = 2.4 cos a 1 l = a0t 2 2 æ 2l ö æ 2 ´ 2r cos a ö æ 4r cos a ö t= ç ÷ = ç ÷ = ç ÷ è a0 ø è ø è a0 ø a0

Kinematics 8. Here,

1 2 an n S nth = distance travelled in n second Sn =

For minimum v,

– distance travelled in (n -1) second æ 2n - 1ö =ç ÷a è 2 ø \

or

2 cos q - sin q = 0 tan q = 2,

or 2 sin q = 5 1 cos q = 5

So,

9. According to problem, when s = a, t = p 1 s = ut + ft 2 2

Q

(here, f = acceleration)

fp 2 2

\

a = up +

For

s = b, t = q

8 1 æ 2 ö 2ç ÷+ è 5ø 5 8 = = 3.57 ms–1 5

v min =

\

…(i)

fq 2 2 After solving Eqs. (i) and (ii), we get 2 ( aq - bp) f= pq ( p - q) b = uq +

…(ii)

12. Let a rel = acceleration of ball with respect to groundacceleration of bus with respect to ground. y a Motion of bus

10. Here, v =|t - 2| ms–1

\

g

v = t - 2, when t > 2 s v = 2 - t , when t < 2 s dv a= = 1ms–2 when t > 2 s dt

= - g$i - a$j | a rel| = g 2 + a2

a = -1 ms–2 when t < 2 s a = 1 ms–2 A

dv =0 dq - 8 (2 cos q - sin q) =0 (2 sin q + cos q) 2

or

S nth 2n - 1 2 1 = 2 = - 2 n n Sn n

C

13. From the figure, the relative displacement is a = 1 ms–2

t = 2s

B

7.5 ms–1

In the direction of motion from A to C, bee decelerates but for C to B, bee accelerates. Let

20 ms–1

AC = s1,BC = s2

100 m

uA = 2 ms–1 ,t = 0

\

\

srel = (200 + 100) m = 300 m

uC = 0 at t = 4 s æ u + uC ö s1 = ç A ÷t è 2 ø 1 æ u + uB ö s2 = ç C ÷t è 2 ø 2

v rel = v1 - v 2 = (20 - 7.5) ms–1 = 12.5 ms–1 \

æ0 + 2ö æ2 + 0 ö s = s1 + s2 = ç ÷2 = 4m ÷2+ ç è 2 ø è 2 ø

11. Let the man starts crossing the road at an angle q with the roadside. For safe crossing, the condition is that the man must cross the road by the time truck describes the distance ( 4 + 2 cot q), 4 + 2 cos q 2l sin q So, = 8 v 8 or v= 2 sin q + cos q

69

t= =

srel v rel 300 = 24 s 12.5

14. 40 2 - 30 2 = 2 as, and v 2 - 30 2 = 2a or

s 2

2 (v 2 - 30 2) = 2 as

Comparing, 2 (v 2 - 900) = 1600 - 900 = 700 or or

v 2 = 900 + 350 = 1250 v = 35.35 kmh –1

70 JEE Main Physics dx = 4 t3 - 2 t dt

15.

dx = 4t 3dt - 2 t dt

or

4t 4 2t 2 = t4 -t2 4 2

Integrating,

x=

When

x = 2, t4 -t2 -2 = 0

\

- ( -1) ± 1 + 8 2 2 1± 3 t = =2 2

t2 = or

(Ignoring –ve sign)

18. As lift is coming from 8th to 4th floor, the value of x becomes

d 2x = 12 t 2 - 2 dt 2

Again,

less hence negative, i. e. , x < 0. Velocity is downwards (i. e. , negative). So, v < 0. Before reaching 4th floor lift is retarded, i. e. , acceleration is upwards. Hence, a > 0.

When t 2 = 2, acceleration = 12 ´ 2 - 2 = 22 ms–2

16. As, Þ Þ

19. Let a1 and a2 be the retardations offered to be bullet by wood

dv = - kv3 dt v

òv

0

and iron respectively. For A ® B ® C, v12 - u 2 = 2 a1( 4)

t dv = -kò dt 0 v3

1 v - ò v -3dv = t k v0

0 2 - v12 = 2a1(1)

and

- u 2 = 2 ( 4 a1 + a2)

Adding, we get

-3 + 1 v

or

Dividing Eq. (iii) by Eq. (i), we get 225 x - 15 t 0 = 100 10 - 10 t 0 1 x - 15 ´ 9 2 or = 4 10 - 10 ´ 1 2 45 = 4x - 30 or 4x = 75 75 or x= m = 18.75 m 4

1 v =t k -3 + 1v

For A¢ ® B¢ ® C ¢,

1 é1 1ù ê - ú =t 2k ë v 2 v 02 û 1 1 = 2 kt v 2 v 02 1 1 = + 2 kt v 2 v 02

and

v 22 - u 2 = 2 a2(2)

0

or or or or or

20. Slope of line = -

…(i)

Similarly, in the second case, 20 2 = 2a (30 - 20 t 0)

…(ii)

Again, in the third case, 15 2 = 2a ( x - 5 t 0)

…(iii)

or

…(ii)

4a1 + a2 = 2a1 + 2a2 a2 = 2a1

Þ

decelerated motion is10 - 10 t 0 . Now, in the first case, 10 2 = 2a (10 - 10 t 0)

40 - 40 t 0 = 30 - 20 t 0 20 t 0 = 10 1 t0 = s 2

-u 2 = 2 (2a + 2a2)

Equating Eqs. (i) and (ii) and solving, we get

17. If t 0 is the reaction time, then the distance covered during

or or

0 2 - v 22 = 2a1(2)

Adding, we get

1 1 + 2v 02kt = v2 v 02 v0 v= 2v 02kt + 1

Dividing Eq. (ii) by Eq. (i), 20 2 30 - 20 t 0 = 10 2 10 - 10 t 0

…(i)

2 3

2 Equation of line is (v - 20) = - ( s - 0) 3 2 Þ v = 20 - s 3

…(i)

Velocity at s = 15 m, i. e. , ds 2 v= = 20 - (15) = 10 ms–1 dt s =15 m 3 Differentiate Eq. (i) with respect to time, acceleration dv 2 ds = =dt 3 dt dv 2 ds a= =\ dt s =15 m 3 dt s =15 m =-

20 ms–2 3

21. The displacement of the particle is determined by the area bounded by the curve. This area is p s = vm t 0 4

Kinematics 1 2

The average velocity is

24. As, b - a = un + An2

s p < v > = = vm t0 4

2b - 2a = 2 un + An 2

\

Such motion cannot be relized in practical terms since at the initial and final moments, the acceleration (which is slope of v-t graph) is infinitely large. Hence, both (i) and (ii) are correct. 5 m/ s 18

(Q 1 km/s =

5 m/s) 18

25 m/s 3 Speed of thief’s car (vT ) = 192 km/h 5 m/s = 192 ´ 18 160 m/s = 3 Muzzle speed of bullet (vB) = 150 m/s The bullet is sharing the speed of the police van, therefore effective speed of the bullet vB ¢ = vB + vP 25 475 m/s = 150 + = 3 3 Speed of the bullet with which it hits the thief‘s car = Relative speed of the bullet w.r.t. thief‘s car (vBT ) = vB ¢ - vT æ 475 160 ö =ç ÷ m/s è 3 3 ø =

23. When a particle is moving with uniform acceleration, let v be the velocity of particle at a distance s, 0 +v v then average velocity = = 2 2 s 2s Time taken, t1 = = (v / 2) v When particle moves with uniform velocity, time taken, 2s t2 = v When particle moves with uniform acceleration, time taken, 3s 6s t3 = = (0 + v) / 2 v

or

Total time = t1 + t 2 + t3 2 s 2 s 6 s 10 s = + + = v v v v s + 2 s + 3s 6 v = v av = 10 s / v 10 v av 6 3 = = v 10 5

t = 2n

a b

1 A (2 n) 2 2 Subtracting, Eq. (i) in Eq. (ii), we get c - a = u (2 n) +

Again,

…(ii)

c - a - 2b + 2a = An 2 c -2 b + a A= n2

25. A body is moving on straight line with constant velocity. Between A and B the straight line is the shortest distance. This is the distance travelled. The particle starts at A and reaches B along the straight line. Therefore displacement is also AB . i.e., D = s.

26. Let time interval be chosen as 1 s PA OA v x = = PB OB v y So, P ( x, y) divides AB in the ratio of v x : v y . (0, vy) B

P (x, y)

vy

315 = 105 m/s 3

Therefore, bullet will hit the thief’s car with a speed 105 m/s

\

…(i)

c

=

vBT

t=n

t=0

22. Speed of police van (vP) = 30 km/h = 30 ´

71

A (vy, 0)

O

Using section formula, vx ´ 0 + vy ´ vx v xv y = x= vx + vy vx + vy y=

v xv y + v y ´ 0 vx + vy

=

v = x2 + y 2 = 2

v xv y vx + vy

v xv y v xv y

Now, replace v x by v1 and v y by v 2. 2 v1v 2 v= v1 + v 2

27. The maximum distance covered in time T = v 0 T. Therefore, for the object having one dimensional motion, the displacement x in time T satisfies -v 0 T < x < v 0T. v v 28. As, 1 = and 3 = t1 t2 1 1200 = (t1 + t 2) v , \ 2 vö 1æ 1 4 v2 2 v2 = 1200 = çv + ÷ v = 2è 3ø 2 3 3

72 JEE Main Physics v 2 = 1800 1 1200 = t ´ 1800 2 2400 s t= 1800 2400 = s = 56.6 s 42.43

or \

29. Given,

dv = bt or, dv = bt dt dt v

òv

Þ

t

dv = ò bt dt 0

0

v - v0 =

or

bt 2 2 bt 2 2

or

v = v0 +

or

dx = v 0dt + x

t

ò 0 dx = ò 0

bt 2 dt 2 b t v 0dt + ò t 2 dt 2 0

x = v 0t +

or

bt 3 1 bt 3 = v 0t + 2 3 6

30. Slope is negative at the point E. 31. Area between v-t graph and time axis gives the distance \

32.

D=

1 1 ´ 2 ´ 20 + 15 ´ 3 + 2 ´ ´ 15 ´ 1 2 2

= 80 m . tan 30° 1 = = 1: 3 tan 45° 3 1 2

33. Height reached = ´132 ´1200 m = 66 ´1200 m

37. Given line have positive intercept but negative slope so its equation can be written as V = -mx + v 0

By differentiating w.r.t. time, we get dv dx = - m = - mv dt dt Now substituting the value of v from Eq. (i), we get dv = -m ( -mx + v 0) = m2x - mv 0 dt \ a = m2x - mv 0 The graph between a and x should have positive slope but negative intercept on a-axis. So, graph (a) is correct.

38. At time t

B uA = 0

v A = u - gt upward vB = gt downward

Velocity of A, Velocity of B,

35.

OA tan 60° = AB tan 30° OA tan 30° or = AB tan 60° 1 1 1 = ´ = 3 3 3

40. Velocity of rain falling vertically downward v r = 30 m/s Rain

N

α B –vw O

vw S

α vr vw C

A

Velocity of woman riding a bicycle vw = 10 m/s (north to south) To protect herself from rain, the woman should hold her umbrella in the direction of relative velocity of the rain with respect to the woman, i.e., v rw . The relative velocity of rain with respect to the woman, i.e.,

36. Since acceleration is constant, therefore there is a uniform increase in velocity. So, the v-t graph is a straight line slopping upward to the right. When acceleration becomes zero, velocity is constant. So, v-t graph is a straight line parallel to the time-axis.

h

If we assume that height h is smaller than or equal to maximum height reached by A u =u A A then at every instant v A and vB are in opposite direction \ v AB = v A + vB = u - gt + gt (Speeds in opposite directions get added) =u 1 9 39. As, g (3) 2 = (2n - 1) 2 2 Þ n =5s

34. Distance covered = Area enclosed by v-t graph = Area of triangle 1 = ´ 4 ´ 8 = 16 m 2 1 1 ´ OA ´ AC ´ AB ´ AC 2 2 (i) + =1 OA AB CA CA (ii) tan 60° = and tan 30° = OA AB

…(i) é v0 ù ê where, m = tan q = ú x0 û ë

v rw = v r - vw \

| v rw| = (30) 2 + (10) 2 = 900 + 100 = 1000 m/s = 10 10 m/s

Kinematics If v rw makes an angle a with the vertical, then v 10 tan a = w = v r 30

or

Relative velocity of plane with respect to the observer …(i) v j - v 0 = 500 - 0 = 500 km/h Relative velocity of products of combustion with respect to the jet plane

1 = 0.3333 3

=

…(ii) vg - v j = -1500 km/h (given) (Velocity of ejected gas vg and velocity of v j are in opposite directions)

a = 18°26¢

Hence, woman should hold her umbrella at an angle of 18° 26¢ with the vertical towards south.

Adding Eqs. (i) and (ii), we get (v j - v 0) + (vg - v j ) = 500 - 1500 vg - v 0 = - 1000 km/h

41. Vertical component of velocities of both the balls are same and equal to zero. So, t =

2h . g

Therefore, relative velocity of the ejected gases with respect to the observer is 1000 km/h, -ve sign shows that this velocity is in a direction opposite to the motion of the jet airplane.

Same for both the balls.

42. Given, x = 4 (t - 2) + a (t - 2) 2 At t = 0 , Acceleration

46. Here,

dx v= = 4 + 2a (t - 2) dt v = 4 (1 - a) d 2x a = 2 = 2a dt

and

43. Let the ball be at height h at time t and (t + D t). Then, 1 h = ut - gt 2 2 and

\

1 h = u (t + D t ) - g (t + D t ) 2 2

…(ii)

Dt =

47.

4 u 2 - g 2 ( Dt ) 2 8g

u=

1 8 gh + g 2 ( D t ) 2 2 1 ( g cos q ) t 2 2

AB = 2R cos q 1 2 R cos q = g cos q t 2 2 4R t2 = g R g

velocity v j and ejected gases be moving downwards (-ve direction) with velocity vg while observer be at rest on the ground i.e., v 0 = 0 \ v j = 500 km/h v0 = 0

C=

u2 - u 2 = - 2 gh3 16 u2 ì 8 3 ü 42 5 × í - ý= 2g î 9 4þ 2 g 36

48.

2 ´ 12 2 ´ 10 10 10

æ 2H ö çQ t = ÷ g ø è

= 1.549 s –1.414 s = 0.135 s 1 2 1 As, x = gt and 100 - x = 25 x - g t 2, 2 2 Adding or

25 t = 100 t = 4s

49. Initial velocity of balloon with respect to ground v = 10 + 5 = 15 m/s upward After 2 s its velocity v = u - gt

50.

= 15 - 10 ´ 2 = -5 m/s = 5 m/s (downward) u Time of ascent = = 6 s g Þ u = 60 m/s Distance in first second,

45. Let jet airplane be moving upwards right (+ve direction) with

vg = -1500 km/h

u2 - u 2 = -2 gh2 9

u 2 ì15 8ü u 2 7 × í - ý= 2 g î16 9þ 2 g 144 AB 5 144 20 = ´ = BC 36 7 7

\

h=

t =2

B=

BC =

44. Acceleration of body along AB is g cos q

From D ABC,

u2 - u 2 = - 2 gh1 4

AB =

Substituting Eq. (ii) in Eq. (i), we get,

Distance travelled in time t sec = AB =

A=

…(i)

Equating Eqs. (i) and (ii), we get 2u - g Dt t= 2g

Þ

73

hfirst = 60 -

g (2 ´ 1 - 1) = 55 m 2

Distance in seventh second will be equal to the distance in first second of vertical downward motion. g hseventh = (2 ´ 1 - 1) = 5 m 2 Þ

hfirst / hseventh = 11 : 1

74 JEE Main Physics 51.

v=0

u t=g =2 s

or f t 2 - 2 vt - h = 0 20 m

Þ

15 m

t=

Ground

=

Position of balls

1 2 1 gt = ´ 10 ´ 12 = 5 m 2 2 1 2 1 h2 = gt = ´ 10 ´ 2 2 = 20 m 2 2 From ground, 5 m, 20 m, 15 m (shown in figure) h1 =

not constant and is not uniform. It means acceleration is not constant and is not uniform, therefore relations (a), (b) and (e) are not correct which is for uniform accelerated motion, but relations (c), (d) and (f) are correct, because these relations are true for both uniform or non-uniform accelerated motion.

or

4 v 2 - v 2 = 2 gh ¢

or

3 v 2 = 2 gh ¢

or

3 ´ 2 gh = 2 gh ¢ h¢ = 3 h

or

57. When A returns to the level of top of tower, its downward

velocity is 4 ms -1. This velocity is the same as that of B. So, both A and B hit the ground with the same velocity.

58. As, s¢ µ t 2 s1¢ : s¢2 : s3¢ =

Now,

reaches the ground.

s1 : s2 : s3 = 1 : ( 4 - 1) : (9 - 4)

Hence, the interval of each water drop =

59.

1s = 0.25 s 4

When the 5th drop starts its journey towards ground, the third drop travels in air for t1 = 0.25+0.25 = 0.5 s \ Height (distance) covered by 3rd drop in air is 1 1 h1 = gt12 = ´ 10 ´ (0.5) 2 2 2 = 5 ´ 0.25 = 1.25 m So, third water drop will be at a height of = 5 - 1.25 = 3.75 m

54. Let u be the velocity with which the stone is projected vertically upwards. (v - h) 2 = 4 v h2

\ Now,

2

2

u - 2 g ( -h) = 4 (u - 2 gh) u2 = hmax =

1 2 gt 2 2 h 1 - = -v + gt 2 t2 2

…(i)

h = vt 2 +

and or

h h 1 + = g (t1 + t 2) t1 t 2 2

\

h=

or

1 gt1t 2 2

For fall under gravity from the top of the tower 1 h = gt 2 2 1 1 gt1t 2 = gt 2 \ g 2 Þ

v -h = 2 v h

Given that,

s1 : s2 : s3 = 1 : 3 : 5 1 2 As, h = - vt1 + gt1 2 h 1 or = -v + gt1 t1 2 or

t = 1s

or

1 9 : 1 : or 1 : 4 : 9 4 4

For successive intervals,

1 2 gt 2 1 = ´ 10 ´ t 2 2 1 5 = ´ 10 ´ t 2 2

h=

or

2 gh ù v [v 2 + 2 gh]1/ 2 v é ± = ê1 ± 1 + 2 ú g g g ë v û (2 v) 2 - v 2 = 2 gh ¢

53. By the time 5th water drop starts falling, the first water drop As u = 0 ,

- ( - 2v) ± 4 v 2 + 4 gh 2 v ± v 2 + gh = 2g 2g

56. Now, retain only the positive sign.

52. The slope of the given graph over the time interval t1 to t 2 is

\

1 2 gt 2

h = - vt +

55. As,

t = t1 t 2

60. Average acceleration =

Dv Dt

10 gh 3

=

2 gh ¢ - ( - 2 gh) 2 gh ¢ + 2 gh = Dt Dt

u2 5 h = 2g 3

=

2 ´ 10 ´ 2.5 + 2 ´ 10 ´ 10 ms–2 0.01

…(ii)

Kinematics 50 + 200 5 2 + 10 2 ms–2 = ms–2 0.01 0.01 15 2 = ms–2 = 1500 2 ms–2 0.01 The upward velocity has been taken as positive. Since average acceleration is positive therefore its direction is vertically upward. =

65. Velocity of girl, vg =

t = 1s 15 = 30 - 10 t 10 t = 15 t = 1.5 s 1 2 1 D x = gt - g (t - 1) 2 2 2 1 1 = g [ -(t - 1) 2] = g (2 t - 1) 2 2 1 = ´ 9.8 ´ 5 m = 24.5 m 2

or or or

62. As,

a 2

63. 4 = u + (2 ´ 3 -1) 5a 2 a 5 = u + (2 ´ 4 - 1) 2 7a or 5 =u+ 2 7 a 5 a 2a Subtracting, 1= = =a 2 2 2 5 Again, 4 =u+ 2 5 or u = 4 - = 1.5 ms–1 2 So, the initial velocity is non-zero and acceleration is uniform. 11h 9.8 Clearly, = (2n - 1) 36 2 11 1 9.8 or ´ ´ 9.8 n 2 = (2 n - 1) 36 2 2 11 2 or n 2n - 1 = 36 or 11n 2 = 72 n - 36 or

64.

4 =u+

or

11n 2 - 72 n + 36 = 0

or

11n 2 - 66 n - 6 n + 36 = 0

or Þ

11n (n - 6) - 6 (n - 6) = 0 n = 6 (Rejecting fractional value) 1 h = ´ 10 ´ 6 ´ 6 m = 180 m 2

L t1

Velocity of escalator, v e =

L t1

Effective velocity of girl on escalator = vg + v e L L = + t1 t2

61. Let us solve the problem in terms of relative initial velocity, relative acceleration and relative displacement of the coin with respect to floor of the lift. Given, u = 0 ms–1, a = 9.8 ms–2, s = 4.9 m,t = ? 1 As, 4.9 = 0 ´ t + ´ 9.8 ´ t 2 2 or 4.9 t 2 = 4.9

75

L L L = + t t1 t2

If t is the time taken, then

t=

or

t1t 2 t1 + t 2

66. Total distance = 130 + 120 = 250 m

67.

Relative velocity = 30 - ( -20) = 50 m/s 250 Hence, t= =5s 50 m0v Relativistic momentum = v2 1- 2 c If velocity is doubled, then the relativistics mass also increases. Thus, value of linear momentum will be more than doubled.

68. As the train are moving in the same direction so, the initial relative speed (v1 - v 2) and by applying retardation final relative speed becomes zero. From

v = u - at

Þ

0 = (v1 - v 2) - at æv -v ö t = ç 1 2÷ è a ø

Þ

69. Let the required angle is q. tan (90° - q) =

20 15

20 4 = 15 3 q = 37° q = 37° + 23° = 60°

cos q =

\ Þ \

70. Let v be the speed of boatman in still water. B y v

vb

45°

x A

u

Resultant of v and u should be along AB. Components of v b (absolute velocity of boatman) along x and y directions are, and Further,

v x = u - v sin q v y = v cos q vy tan 45° = vx

76 JEE Main Physics v cos q u - v sin q u u v= = sin q + cos q 2 sin ( q + 45° ) 1=

or

77. v ct = v c - v t

vc

vct

45°

v is minimum at, q + 45° = 90° or q = 45° u v min = 2

and

2 xv x x 2x x x = + + = ,t 2 = v v v v + w v - w v 2 - w2 2 xv 2x = t2 = 2ö æ æ w w2 ö v 2 ç1 - 2 ÷ v ç1 - 2 ÷ è v ø è v ø t1 or t 2 > t1 t2 = w2 1- 2 v t1 =

71. Here, or

or

–vt

78. From figure,

Time taken by the bird to cross the train 210 t= =7s 30

74. Effective speed of the bullet = Speed of bullet + speed of police jeep = 180 m/s + 45 km/h = (180 + 12.5) m/ s = 192.5 m/ s Speed of thief’s jeep = 153 km/h = 42.5 m/s Velocity of bullet w.r.t. thief’s car = 192.5 – 42.5 = 150 m/s

75. Given AB = velocity of boat = 8 km/h AC = Resultant velocity of boat = 10 km/h B

C

θ A

\

BC = Velocity of river = AC 2 - AB2 = (10) 2 - (8) 2 = 6 km/h

76. When two particles moves towards each other, then …(i) v1 + v 2 = 6 When these particles moves in the same direction, then …(ii) v1 - v 2 = 4 By solving Eqs. (i) and (ii), we get, v1 = 5 and v 2 = 1 m/s

vB cos θ θ

vB sin θ

vw

vB sin q = v w v 1 sin q = w = vb 2

velocity. Using,

73. Relative velocity of bird w.r.t. train = 25 + 5 = 30 m/s

B

vB

72. Let us calculate relative deceleration by considering relative v 2 - u 2 = 2 as, 0 2 - 80 2 = 2 ´ a ´ 2000 80 ´ 80 64 or a==ms–2 = -1.6 ms–2 4000 40 1.6 –2 Deceleration of each train is ms , i. e. ,0.8 ms–2 2

vt

v ct = v c + ( - v t ) Velocity of car w.r.t. train (v ct ) is towards west-north.

Þ q = 30° Time taken to cross the river D D 2D t= = = v b cos q v b cos 30° v b 3

79. Distance between the cars A and B remains constant. Let the distance be x. Velocity of C w.r.t. A and B, \

v = 45 + 36 = 81 km/h 5 Distance = 81 ´ = 6.75 km 60

80. Length of each train, l A = lB = 400 m Initial velocities of both trains, uA = uB = 72 km/h 5 m/s = 72 ´ 18

5 ö æ m/s÷ çQ1 km/h = ø è 18

= 20 m/s Distance travelled by train A in 50 s sA = uA ´ t (As for unaccelerated motion distance = Speed ´ Time) sA = 20 ´ 50 = 1000 m Distance travelled by train B in 50 s, 1 sB = uBt + aBt 2 2 (As motion of train B is an accelerated motion) 1 sB = 20 ´ 50 + ´ 1 ´ (50) 2 2 = 1000 + 1250 = 2250 m Original distance between the two trains = sB - sA = 2250 - 1000 = 1250 m

Kinematics 81. Speed of car A(uA) = 36 km/h = 36 ´

Total distance travelled Total time taken 23 = (7 / 15)

The average speed of the taxi = 5 ö æ m/s÷ çQ1 km / h = ø è 18

5 m/s 18

= 10 m/s 1 km

B

345 km/h 7 = 49.3 km/h

=

1 km A

uB

uA uC

C

Speed of car B and car C uB = uC = 54 km/h 5 m/s = 15 m/s = 54 ´ 18 Relative velocity of car B w.r.t. car A uBA = uB - uA = 15 - 10 = 5 m/s Relative velocity of car C w.r.t. car A uCA = uC - uA = 15 - ( -10) = 25 m/s Distance between car A and car B = 1km = 1000 m Time taken by car C to travel distance AC = 1000 m Distance Time (t ) = Relative velocity of car C w.r. t. car A 1000 = s = 40 s 25 Let car B start to accelerate with an acceleration a. Using equation of motion, 1 s = ut + at 2 2 1 or s = uBAt + at 2 2 1 1000 = 5 ´ 40 + a ´ ( 40) 2 2 = 200 + 800 a or or

77

800 a = 800 a = 1m/s 2

Therefore, car B should accelerate with an acceleration 1 m/s 2.

82. Given, shortest distance between the station and the hotel = 10 km \ Displacement of the taxi = 10 km Distance travelled by the taxi = 23 km Time taken by the taxi = 28 min 28 7 h = = 60 15

The magnitude of average velocity Magnitude of the total displacement = Total time taken 10 = ( 7 / 15) 150 km/h 7 = 21.43 km/h

=

No, the average speed of the taxi is not equal to the magnitude of the average velocity of the taxi.

83. Given, speed of man (v m) = 4 km/h A

D

β β O

C

B

Speed of river (v r ) = 3 km/h Width of the river (d) = 1km vr

B

C v

1 km

β

vm A

Time taken by the man to cross the river Width of the river Speed of the man 1km 1 = = h 4 km/ h 4

t=

=

1 ´ 60 = 15 min 4

Distance travelled along the river = v r ´ t 1 3 = 3 ´ = km 4 4 3000 = = 750 m 4

78 JEE Main Physics

Round II 2

12 ö ÷ ´ 20 = 80 m è 60 ø

1. x¢n2x = æç

5. Let v1 and v 2 be the initial speeds of first and second runners

Let a be the retardation in both the cases. Using the relation, v 2 = u 2 + 2 as, when automobile is stopped, v = 0. So,

0 = u 2 + 2 as

or \

s µ u2 s2 = 4 s1 = 4 ´ 20 = 80 m

2. We know that gravity is a universal force with which all bodies are attracted towards the earth. Hence, g is same for both the balls. Also, if t is the time taken by the balls to reach the ground, then from equation of motion. 1 s = ut + gt 2 2 Þ

t=

2 ( s - ut ) g

Since s, u and g are same for both, hence time taken by both the balls is same.

3. Boat covers distance of 16 km in a still water in 2 hours i. e. ,

vB =

16 = 8 kmh –1 2

Now, velocity of water vw = 4 kmh –1 Time taken for going upstream 8 8 t1 = = =2h vB - vw 8 - 4 As water current oppose the motion of boat, therefore time taken for going downstream 8 8 8 t2 = = = h vB + vw 8 + 4 12

respectively. Let t be time by them when the first runner has completed 50 m. During this time, the second runner has covered a distance = 50 - 1 = 49 m. 50 49 …(i) So, t= = v1 v 2 Suppose, the second runner increases his speed to v3 so that he covers the remaining distance ( = 51 m) in time t. 51 49 So t= = v3 v 2 51 or v3 = v2 49 2 ö æ or v3 = ç1 + ÷v è 49 ø 2 v3 -1 = v2 v3 - v 2 = v2

2 49 2 49 2 % increase = ´ 100 = 4.1% 49

or or or

6. For no collision, the speed of car A should be reduced to vB before the cars meet, i. e. , final relative velocity of car A with respect to car B is zero, i. e. , vr = 0 Here, initial relative velocity, ur = v A - vB Relative acceleration, ar = - a - 0 = - a Let relative displacement = sr The equation, v r2 = ur2 + 2 ar sr (0) 2 = (v A - vB) 2 - 2 asr sr =

(water current helps the motion of boat) \

Total time = t1 + t 2 8ö æ = ç2 + ÷ h = 2 h 40 min è 12 ø

4. If t1 and 2 t 2 are the time taken by particle to cover first and second half distance respectively x/2 x t1 = = 3 6 Clearly, x1 = 4.5 t 2 and x2 = 7.5 t 2 x x So, x1 + x2 = = 4.5 t 2 + 7.5 t 2 = 2 2 x t2 = 24 x x x Total time t = t1 + 2 t 2 = + = 6 12 4 So,

average speed = 4 m/s

(v A - vB) 2 2a

For no collision, sr £ s (v A - vB) 2 i. e. , £s 2a

7. Since, v = (t - 2), so v µ t . On plotting a graph between v and t , we get a straight line ABand BC as shown in figure (below). v (ms–1) 2

A

C

1

0

1

2

3

D 4

t

The distance covered in 4 s is equal to the area under the velocity- time graph = Area of D AOB + Area of D BCD 2 ´2 2 ´2 = + =2+2 = 4m 2 2

Kinematics v = 3t2 + 5t + c

1 2

8. As, h = ut + gt 2 Þ

1 = 0 ´ t1 +

Þ

t1 = 2 / g

1 2 gt1 2

Velocity after travelling 1m distance

where c is constant of integration When t = 0 ,v = 0 so c = 0 \

v = 3t2 + 5t

Þ

ds = (3 t 2 + 5 t ) dt

v 2 = u 2 + 2 gh Þ

v 2 = (0) 2 + 2 g ´ 1

Þ

v= 2g

S

1 2 gt 2 2

gt 22 + 2 2 gt 2 - 2 = 0 -2 2 g ± 8 g + 8 g t2 = 2g =

- 2 ±2 g

Taking +ve sign, t 2 = (2 - 2 t ) / g \

t1 2/g 1 and so on. = = t 2 (2 - 2) / g 2 -1

9. Interval of all ball throw = 2 s If we want that minimum three (more than two) ball remain in air then time of flight of first ball must be greater than 4 s t >4s 2u >4s g Þ

u > 19.6 m/s

For u = 19.6 first ball will just about to strike the ground (in air) Second ball will be highest point (in air). Third ball will beat point of projection or at ground (not in air)

10. Let vw be velocity of water and v b be the velocity of motor boat in still water. If xis the distance covered, then as per question x = (v b + vw ) ´ 6 = (v b - vw ) ´ 10 On solving, vw = v b / 4 \ x = (v b + v b / 4) ´ 6 = 7.5 v b Time taken by motor boat to cross the same distance in still water is x 7.5 t= = = 7.5 h xb v b

If t1 and t 2 are the time taken by the ball to reach points A and B respectively, then 1 20 = 45 t1 + ´ 10 ´ t12 2 1 and 40 = - 45 t 2 + ´ 10 ´ t 22 2 On solving, we get, t1 = 9.4 s and t 2 = 9.8 s Time taken to cover the distance AB = (t 2 - t1) = 9.8 – 9.4 = 0.4 s

13. As, Þ Þ At t = 0 ,v = v 0 Þ we get, Again

Þ

v

4

0

0

Integrating it, we have ò dv = ò (6 t + 5) dt

5 ù é s = êt 2 + t 2ú = 8 + 10 = 18 m 2 û0 ë

tower. Taking vertical downward motion of boy from top of tower to ground, we have u = - u, a = g = 10 ms–2, s = 50 m,t = 10 s 1 As s = ut + at 2 2 1 So, 50 = -u ´ 10 + ´ 10 ´ 10 2 2 On solving, u = 45 ms–1

dv = (6 t + 5), dt

dv = (6 t + 5) dt

+ 5 t ) dt

12. Let the body be projected upwards with velocity u from top of

Þ

a=

2

2

\

11. Given, acceleration a = (6 t + 5) m/s 2 Þ

2

ò 0 ds = ò 0 (3 t

1 = 2g ´ t 2 +

ds ö æ ç as v = ÷ è dt ø

Integrating it within the condition of motion is as t changes from 0 to 2 s, S changes from 0 to S, we have

For second 1 m distance

Þ

79

At t = 0 , \

dv = bt dt dv = bt dt v=

bt 2 + K1 2

K1 = v 0 1 v = bt 2 + v 0 2 dx 1 2 = bt + v 0 dt 2 x=

1 bt 3 + v 0t + K2 2 3

x=0 K2 = 0 1 x = bt 3 + v 0t 6

80 JEE Main Physics 14. The horizontal acceleration a of the wedge should be such that in time the wedge moves the horizontal distance BC. The body must fall through a vertical distance AB under gravity. Hence,

or

1 1 BC = at 2 and AB = gt 2 2 2 AB g tan q = = BC a g a= = g cot q tan q

15. Let u be the initial upward velocity of the ball from A and h be the height of the tower.

t = 0 , v = 2 (1 - e-3t )

when

Initially, v = 0, From Eq. (i) acceleration, dv a0 = = 6 - 3 ´ 0 = 6 ms–2 dt a 6 When a = 0 = = 3 then from Eq. (i); 2 2 3 = 6 -3v or 3v = 6 -3 = 3 v = 1ms–1

or

17. Velocity of bolt relative to elevator = 2.5 – 2.5 = 0

Taking the downward motion of the first stone from A to the ground, we have 1 …(i) h = - ut1 + gt12 2 Taking the downward motion of the second stone from A to the ground, we have 1 …(ii) h = ut 2 + gt12 2 Multiplying Eq. (i) t 2 and Eq. (ii) by t1 and adding, we get 1 h (t1 + t 2) = gt1 t 2 (t1 + t 2) 2 1 So, …(iii) h = gt1 t 2 2 For falls under gravity from the top of the tower 1 h = gt32 2

v = 2 (1 - e-3t )

or

Acceleration of bolt relative to elevator, a = 10 - ( - 2) = 12 ms–2 (Q g = 10 m/ s2) 1 2 at 2 1 we have, 3.0 = 0 ´ t + ´ 12 ´ t 2 2 1 or s = 0.707 s = 0.7 s t= 2 1 Displacement = - 2.5 ´ 0.71+ ´ 10 ´ (0.71) 2 2 Using the relation, s = ut +

= - 1.775+2.521 = 0.746 = 0.75 m

…(iv)

=2´

From Eqs. (iii) and (iv),

16. Given, or

18. The body is at rest initially and again comes to rest at t = 1s at

t3 = t1 t 2 = 9 ´ 4 = 6 s dv = 6 -3v dt dv = dt 6 -3v

Integrating it, we have é 1 ù êë - 3 log (6 - 3 v) úû = t + K t = 0,v = 0 1 \ K = - log 6 3 Putting this value in Eq. (ii), we have 1 1 - log (6 - 3 v) = t - log 6 3 3 æ6 -3v ö or log ç ÷ = -3t è 6 ø

(2.5) 2 + 0.75 2 ´ 10

= 0.63+0.75 = 1.38 m

t32 = t1t 2 or

u2 + displacement 2g

Distance covered = 2 ´

position x = 1. …(i)

Thus, firstly acceleration will be positive then negative. Thus a have to change the direction so that body may finally come to rest in the interval 0 £ t £ 1. If we plot v-t graph. The total displacement = 1m = area under v-t graph v B

…(ii)

C

D

vmax

At

or or

6 -3v = e- 3 t 6 v 1 - = e- 3 t 2

A

Now Þ

1/2

1

1 v max × t = s 2 2´s v max = t 2 ´1 v max = = 2 m/s 1

The maximum velocity = 2 m/s

t (s)

Kinematics Now just see the v-t graph During AB a >4 m/s2 For ABC, During BE a 0) then speed will decrease. The speed will increase when v and d are both positive.

24. Maximum height reached, sm =

v2 2g

Time taken to reach the maximum height, Tm = Height s reached in time t is s = ut -

1 2 gt 2

If

t=

T v = 2 2g

then

s=

v ´v 1 æ v ö - g ç ÷ 2 g 2 è2 g ø

21. Maximum restoring force set up in the spring when stretched by distance r is F = - kr and potential energy of stretched 1 spring = kx2. As F µ r and this force is directed towards 2 equilibrium position, hence if mass is left free, it will execute damped SHM due to gravity pull. Magnitude of acceleration in the mass attached to one end of spring when just released is F k a = = r = (maximum) m m k At equilibrium position, r = 0 \a = ´ 0 = 0 m

…(i)

=

2

v2 v2 3 v2 3 = = sm 2g 8g 8g 4

Speed at height s is v 2 = u 2 - 2 gs When then or

s=

sm v 2 = 2 4g

v ¢2 = v 2 - 2 g ´ v¢ =

v 2

v2 v2 = 4g 2

v g

82 JEE Main Physics 25. Here, u = 2 ms–1, v = 14 ms–1

A

C

Since OA = BC, so initial and final speeds are the same.

B

The slope of velocity-time graph represents acceleration. Here, the velocity-time graphs AB is a straight line inclined to time axis hence has equal acceleration throughout. The particle changes its direction of motion after time T.

Distance between A and B = s v 2 - u 2 14 2 - 2 2 194 97 Then acceleration, a = = = = 2s 2s 2s s The speed at mid- point C, s 97 s v 2 = u2 + 2 a = 22 + 2 ´ ´ = 101 s 2 2 v = 101~ - 10 ms–1 1 1 As per question, AP = [ AB] = s 6 6 s When, s= , 6 97 s then, v 2 = 22 + 2 ´ ´ s 6 97 or v =3+2´ = 36.3 ms–1 3 v = 36.3 » 6 ms–1 \

29. vGB = vG + ( -vB) = 0 + ( -25 ms-1) = 25 ms-1 due north

Since velocity at mid- point C is 10 ms–1. \Taking motion from A to C, we have 10 = 2 + a ´ t1 10 - 2 8 or t1 = = a a Taking motion from C to B, we have 14 = 10 + 1 ´ t 2 14 - 10 4 or t2 = = a a t t1 = 2 or t 2 = 1 \ 2 t2

34. As displacement is either smaller or equal to distance but

26. Since the ball is moving with a very small speed in the moving train, the direction of motion of the ball is the same as that of the train. The direction of motion of ball does not change with respect to observer on ground. The speed of the ball as observed by observer on ground before collision with side of train is 10 + 1 = 11 m/s and after elastic collision is10 - 1 = 9 m/s. Since the collision of the ball with side of train is perfectly elastic; the total momentu and kinetic energy are conserved, so average speed of the ball over any 20 seconds interval is constant as observed by observer on ground. Since train is moving with constant velocity, it is an inertial frame,so acceleration of ball is same as from the train.

27. The displacement is the area which the velocity-time graphs encloses with time axis for a given interval of time. Since the area of velocity-time graph for time 0 to 5 s is the same as area of the velocity-time graph for time 5 s to 10 s, hence average speed in these intervals is the same.

28. Displacement = velocity ´ time. In time 0 to 2 the displacement = - Area of D OAB + Area of D OAD + Area of D DBC = 0.

30. v mG = v m + vG = 5 + ( -15 ms-1) = -10 ms–1 towards south = 10 ms–1 towards north 40 = 20 m 2 Using the relation v 2 = u 2 + 2 as = 0 + 2 ´ 10 ´ 20 = 400

32. Here, u = 0 , a = 10 ms–2, s =

v = 20 ms–1

or

change in velocity 25 - ( -25) = = 10 ms–2 time taken 5 Hence, Assertion is wrong but Reason is correct.

33. Acceleration =

never be greater than distance.

35. Since slope of displacement-time graph measures velocity of a moving object.

36. As per definition, acceleration is the rate of chagne of velocity is a =

dv dt

dv =0 \ a =0 dt Therefore, if a body has constant velocity it cannot have non zero acceleration but uniform. 1 2 For first ball, gt = 176.4 2 176.4 ´ 2 t= Þ 10 Þ t = 5.9 s For second ball, t = 3.9s 1 u (3.9)+ g (3.9) 2 = 176.4 2 10 Þ 3.9 u + (3.9) 2 = 176.4 2 Þ u = 25.7 ms–1 If velocity is constant

37.

This value is approximated to 24.5 ms–1.

37. Let u be the velocity of scooterist in order to catch the but in 100 s. Then or

100 u = 2000 2000 u= = 20 ms–1 100

38. Here, u = 0 ,v = 640 ms–1, s = 1.2 m, a = ? and t = ? As, or

v 2 = u 2 + 2 as a=

v 2 - u 2 (640) 2 (640) 2 –1 ms = = 2s 2 ´ 1.2 2.4

Kinematics v - u 640 - 0 = a (640) 2 2.4 2.4 = = 3.75 ´ 10 –3 s~ - 4 ms 640

t=

Also,

\

39. Taking motion of the body from half the maximum height upto the highest point, we have v 2 = u 2 + 2 as

\

0 = 10 2 + 2 ( -10) ´

or

h = 10 m

40. Given, velocity v = v 0 + gt + ft v=

As, So,

t

Q

= 3$i + 4$j + 4$i + 3$j v = 7 $i + 7$j

ò 0 dx = ò 0 v dt = ò 0 (v 0 + gt + ft

2

) dt

Þ

-1/ 2

Þ

x

v dv = - ò kxdx

Þ

dv = 2.5 v dt

0

ò 6.26 v

dv = - 2.5 dt v t

-1 / 2

dv = -2.5 ò dt 0

0

- 2.5 [t ] t0 = [2 v1/ 2]6.25 t =2s

50. The acceleration of train in 20 s is given by

40 - 0 = 2 ms–2 20 [from the formula v = u + at (here, u = 0)]. Now the distance travelled is given by equation of motion, a=

v 2 = u 2 + 2 as

0

v 2 - v 02 kx2 =2 2 æ v 2 - v 02 ö mkx2 mç ÷=2 è 2 ø D K µ x2

dx = ò a dt

2x1/ 2 = at or x µ t 2

or

Þ

displacement and k is a proportionality constant. v dv = - kx i. e. , dx Þ v dx = - kxdx Let for any displacement from 0 to x, the velocity changes from v 0 to v.

Þ

dx dx = a dt = a x or dt x

òx 49. Given,

44. From given information a = - kx, where a is acceleration, x is

Þ

(after integrating)

Integrating it, we have

1 s = ut + at 2 2 1 20 = 10 ´ 2 + a ´ i 2 2 0 =2a Þ a=0

0

y 2 = x2 + C

48. Velocity,

43. Here u = 10 ms , t = 2 s, s = 20 m

v

y dy = xdy

Þ

–1

òv

\ Þ

mv v mv 2 Instantaneous power, P = F × v = ´ t= 2 t T T T

Þ

dx = ky dt dy = kx dt dy dy dt kx = ´ = dx dt dx ky

and

50 = 2.5 kms–2 \Average acceleration = 20 v mv 42. As, acceleration, a = and F = ma = T T v \Velocity acquired, v = at = t T

\

47.

Thus, speed is 7 2 + 7 2 = 7 2 unit Given, v = ky$i + kx$j

2

41. Magnitude of change in velocity = ( 40) 2 + (30) 2 = 50 km/s

Using

u2 æ 80 ö =h´ç ÷ = 0.64 h è100 ø v2

a = 0.4 $i + 0.3 $j v = u + at = 3 $i + 4 $j + (0.4 $i + 0.3$j) 10

and

2

t

v x = v 0t +

h' = h ´

46. Given, u = 3 $i + 4 $j;

h 2

gt f + ft 2 + 2 3 g f When t = 1, then x = v 0 + + 2 3

or

…(ii)

Þ

dx dt

x

…(i)

2

u = 10 ms–1, a = -10 ms–2, v = 0 , s = h / 2 As

v = 2 gh v 2 = u 2 - 2gh u 2 = v 2 + 2gh u 2 = 2gh ¢ v 2 2gh = u 2 2gh ¢

45. Clearly, After rebounce, Þ and

83

So,

[D K is loss in KE]

æ v 2 - u2 ö 40 ´ 40 - 0 = 400 m s1 = ç ÷ a= 2 ´2 è 2 ø

Now distance travelled with constant speed of 40 ms -1 in t = 20 s is s2 = 40 ´ 20 = 800 m

84 JEE Main Physics Again the distance covered in 3rd case is given by 40 ´ 40 - 0 s3 = = 800 m 2 ´1 Therefore, average speed of the train is given by 400 + 800 + 800 200 v av = = = 25 ms–1 20 + 20 + 40 80

51. First 50 m fall is under the effect of gravity only. The velocity –1

acquired, u = 2 gh = 2 ´ 9.8 ´ 50 ms . Taking onward

The velocity of car at A = velocity of car at B = (2 fS)1/ 2 As magnitude of retardation of the car from Bto C is half of that of acceleration from O to A when velocity changes by v, so distance BC = 2 S Distance, AB = 15 S - ( S + 2 S) = 12 S As distance AB is covered with constant velocity in time t So, or

–2

motion of parachutist with retardation 2 ms , we have, u = 10 9.8 ms–1. d = - 2 ms–2,v = 3 ms–1 s=

2

2

(3) - (2 ´ 9.8 ´ 50) v -u = = 243 m 2a 2 ´ ( -2)

Acceleration = f O

A

Retardation = f /2

Constant velocity S

B

144 S 2 = 2 fSt 2 1 2 S= ft 72

53. Time taken by the body to reach the point A is t1 (During

2

\Total height = 50 + 243 = 293 m

52.

or

12 S = vt = (2 fS)1/ 2 ´ t

S

Taking motion of car from 0 to A, Here, u = 0 , a = f , s = S ,v = ? As

v 2 = u 2 + 2 as

So

v 2 = 0 + 2 ´ f ´ S or v = 2 fS

C

upward journey). The body crosses this point again (during downward journey) after t 2, i. e. , the body takes the time (t 2 - t1) to come again at point A. So, the time taken by the body to reach at point B(at maximum height) æt -t ö t = t1 + ç 2 1 ÷ è 2 ø (Q Time of ascending = Time of descending) t +t t= 1 2 2 So, maximum height 2 2 1 1 æt + t ö æt + t ö H = gt 2 = g ç 1 2 ÷ = 2g ç 1 2 ÷ è 4 ø 2 2 è 2 ø

3 Vector Analysis JEE Main MILESTONE < <
45° ) is the inclination with the horizontal of the initial direction of projection, for what value of tan f will the particle strike the plane? (a) tan q = 1 1 (c) sin q = 2

(b) tan q = 2 1 (d) sin q = 2

Let the particle be projected from O with velocity u and strikes the plane at a point P after time t. ON = PN = h, then OP = h 2

P u cos α

h

u θ O

(b) 8h sin a (d) 4h sin a

Solution Velocity before strike u =

Component of acceleration along the inclined plane = g sin a and the perpendicular component = g cos a 1 Using s = ut + at 2, 2 0 = v cos at -

x = u sin at +

...(i) = u sin a

N

1 g cos at 2 2 1 g sin at 2 2

2u 1 æ 2u ö + g sin a ç ÷ èg ø g 2

=

2u 2 sin a 2u 2 sin a + g g

=

4 u 2 sin a g

=4´

45°

2gh

and for horizontal direction,

If the particle strikes the plane horizontally, then its vertical component of velocity at P is zero. h = (u cos q) (t ) Along vertical direction 0 = u sin q - gt

(a) 8h cos a (c) 2h tan a

For vertical direction, we get

Solution Let

1 (u sin q )t 2

tan q = 2

...(i)

u 2 sin 2 ´ 45° = R + 10 g

and

...(ii)

Using Eqs. (i) and (ii) in Eq. (iii),

2

If range is R then,

1 2 gt 2

2gh ´ sin a = 8h sin a g

2

2u ö æ çQ t = ÷ è g ø

Start Practice for

JEE Main Round I

(Topically Divided Problems)

Horizontal Projection

6. Water is flowing from a horizontal pipe fixed at a height of 2 m from the ground. If it falls at a horizontal distance of 3 m, as shown in figure, the speed of water when it leaves the pipe is ( Take, g = 9.8 ms –2 )

1. The horizontal range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an [NCERT Exemplar] angle of 45°, its range will be (a) 60 m

(b) 71 m

(c) 100 m

(d) 141 m

2. A tennis ball rolls off the top of a sister case way with

a horizontal velocity u ms -1. If the steps are b metre wide and h meter high, the ball will hit the edge of the nth step, if 2hu (a) n = 2 gb

2hu2 (b) n = gb2

2hu2 (c) n = gb

hu2 (d) n = 2 gb

2m 3m

(a) 2.4 ms -1 (b) 4.7 ms -1 (c) 7.4 ms -1

7. A stone is just released from the window of a moving train along a horizontal straight track. The stone will hit the ground following (a) straight path (c) parabolic path

3. A bomber plane moves horizontally with a speed of

500 ms -1 and a bomb releases from it, strikes the ground in 10 s. Angle at which it strikes the ground will be ( g = 10 ms -2 ) æ 1ö (a) tan -1 ç ÷ è5ø

æ 1ö (b) tan ç ÷ è5ø

(c) tan -1(1)

(d) tan -1(5 )

4. An aeroplane is flying in a horizontal direction with a velocity 600 kmh -1 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. (a) 3.33 km

(b) 333 km

O

(b) circular path (d) hyperbolic path

A man standing on a hill top projects a stone horizontally with speed v0 as shown in figure. Taking the coordinate system as given in the figure. The coordinates of the point where the stone will hit the hill surface æ 2v2 tan q - 2v20 tan2 q ö ÷ (a) ç 0 , g g ø è

v

æ 2v20

h

2v20

tan q ö ÷ g ø 2

æ 2v2 tan q 2v20 ö ÷ (c) ç 0 , g g ø è B

A

(d) 3330 km

5. The height y and distance x along the horizontal for a body projected in the xy-plane are given by y = 8 t - 5 t2 and x = 6 t. The initial speed of projection is (a) 8 m/s (c) 10 m/s

8.

(b) ç , è g

(c) 33.3 km

(b) 9 m/s (d) (10/3) m/s

(d) 6.2 ms -1

y v0

x

(0, 0)

θ

æ 2v2 tan2 q 2v20 tan q ö ÷ (d) ç 0 , g g è ø

9. A particle moves in the xy-plane with velocity vx = 8 t - 2 and v y = 2. If it passes through the point x = 14 and y = 4 at t = 2 s, find the equation (x-y relation) of the path. (a) x = y2 - y + 2 (b) x = 2 y2 + 2 y - 3 (c) x = 3 y2 + 5 (d) Cannot be found from above data

128 JEE Main Physics 10. The ceiling of a long hall is 25 m high. Then, the maximum horizontal distance that a ball thrown with a speed of 40 m/s can go without hitting the ceiling of the hall, is [NCERT Exemplar] (a) 95.5 m (c) 100 m ^

^

and

a

constant

acceleration

^

a = ( - 100 . i - 0.50 j) ms -2 . When the particle reaches its maximum x -coordinate, what is its y-component a velocity? (a) –2.0 ms -1 (c) –1.5 ms -1

(b) –1.0 ms -1 (d) 1.0 ms -1

12. A projectile can have same range from two angles of projection with same initial speed. If h1 and h2 be the maximum heights, then (a) R =

h1h2

(b) R = 2 h1h2

(c) R = 2 h1h2

(d) R = 4 h1h2

time of flight is 5 s. If = 10 m / s2 , then horizontal component of velocity and the maximum height will be respectively (a) 20 m/s, 62.50 m (c) 80 m/s, 62.5 m

(b) 105.5 m (d) 150.5 m

11. A particle leaves the origin with an initial velocity v = (3.00 i) ms -1

17. An arrow is shot into air. Its range is 200 m and its

(b) 40 m/s, 31.25 m (d) None of these

18. A body of mass m thrown horizontally with velocity v, from the top of tower of height h touches the level ground at distance of 250 m from the foot of the tower. A body of mass 2 m thrown horizontally with v velocity , from the top of tower of height 4h will 2 touch the level ground at a distance x from the foot of tower. The value of x is (a) 250 m (c) 125 m

(b) 500 m (d) 250 2 m

19. A ball is thrown up with a certain velocity at an angle q to the horizontal. The kinetic energy (KE) of the ball varies in the horizontal displacement x as

13. A stone is thrown at an angle q to be the horizontal reaches a maximum height H. Then, the time of flight of stone will be (a) (c)

2H g 2 2 H sin q g

2H g

(b) 2 (d)

14. A bomb is dropped on an enemy post by an aeroplane

400 (a) m 3 1700 m (c) 3

500 (b) m 3 (d) 498 m

15. A body projected with velocity u at projection angle q has horizontal range R. For the same velocity and projection angle, its range on the moon surface will be g moon = g earth / 6) R 36

(a) 36 R

(b)

R (c) 16

(d) 6 R

16. A boy throws a ball with a velocity u at an angle q with the horizontal. At the same instant he starts running with uniform velocity to catch the ball before if hits the ground. To achieve this he should run with a velocity of (a) u cos q (c) u tan q

(b) u sin q (d) u sec q

(b) O

2 H sin q g

flying horizontally with a velocity of 60 kmh -1 and at a height of 490 m. At the time of dropping the bomb, how far the aeroplane should be from the enemy post so that the bomb may directly hit the target?

KE

KE

(a)

O

x

KE

x

KE

(c)

(d) O

O

x

x

20. Two paper screen A and B

are separated by a distance of 100 m. A bullet pierces A and B. The hole in B is 10 cm below the hole in A. If the bullet is travelling horizontally at the time of hitting A. Then, the velocity of the bullet at A is (a) 100 m/s (c) 600 m/s

(b) 200 m/s (d) 700 m/s

21. It was calculated that a shell when fired from a gun with a certain velocity and at an angle of 5p rad should strike a given target. In elevation 36 actual practice, it was found that a hill just prevented the trajectory. At what angle of elevation should the gun be to hit the target? 5p rad 36 7p (c) rad 36 (a)

11 p rad 36 13 p (d) rad 36 (b)

Projectile Motion

129

22. Two projectiles thrown from the same point at angles

29. A projectile shot into air at some angle with the

60° and 30° with the horizontal attain the same height. The ratio of their initial velocities is

horizontal has a range of 200 m. If the time of flight is 5 s, then the horizontal component of the velocity of the projectile at the highest point of trajectory is

(a) 1 (c)

(b) 2 1 (d) 3

3

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

23. A projectile is thrown at angle b with vertical. It reaches a maximum height H. The time taken to reach the highest point of its path is (a)

H g

(b)

2H g

(c)

H 2g

(d)

2H g cos b

(a) 30°

(b)

3Ek 4

(c)

Ek 4

(d) Zero

3 angle q 0 = tan -1 æç ö÷. After 1 s, the particle is moving è4ø

at an angle q to the horizontal, where tan q will be equal to (g = 10 m/s2 ) (b) 2

(c)

1 2

(c) 60°

(a) 100 m (c) 50 m

(d)

1 3

(b) 75 m (d) 25 m

32. A projectile A is thrown at an angle of 30° to the horizontal from point P. At the same time, another projectile B is thrown with velocity v2 upwards from the point Q vertically below the highest point. For B v to collide with A, 2 should be v1

26. When a projectile is projected at a certain angle with

Highest point

the horizontal, its horizontal range is R and time of flight is T1. When the same projectile is throwing with the same speed at some other angle with the horizontal, its horizontal range is R and time of flight is T2 . The product of T1 and T2 is R (a) g 3R (c) g

2R (b) g 4R (d) g

27. A projectile of mass m is thrown with a velocity v making an angle of 45° with the horizontal. The change in momentum from departure to arrival along vertical direction, is (a) 2mv (c) mv

2 mv mv (d) 2

initial velocity and same range. If H is the maximum height attained by one stone thrown at an angle of 30°, then the maximum height attained by the other stone is H 2 (c) 2 H

(b) H (d) 3H

B v2

v1 A

30º

P

(a) 1

Q

(b) 2

(c)

1 2

(d) 4

33. For a projectile thrown into space with a speed v, the horizontal range is

3v2 v2 . × The vertical range is 2g 8g

The angle which the projectile makes with the horizontal initially is (a) 15°

(b)

28. Two stones thrown at different angles have same

(a)

(d) 90°

of a planet at a certain angle with the horizontal surface. The horizontal and vertical displacement x and y vary with time t in second as x = 10 3 t and y = 10 t - t2 . The maximum height attained by the ball is

25. A particle is projected with a velocity of 30 m/s, at an

(a) 1

(b) 45°

31. A ball is projected from a certain point on the surface

kinetic energy E k. What is the kinetic energy at the highest point? Ek 2

30. The kinetic energy of a project at the height point is half of the initial kinetic energy. What is the angle of projection with the horizontal?

24. A cricket ball is hit at 30° with the horizontal with

(a)

40 ms -1 0 ms -1 9.8 ms -1 equal to the velocity of projection of the projectile

(b) 30°

(c) 45°

(d) 60°

34. The velocity of projection of an oblique projectile is ^

^

(6 i + 8 j) ms -1. The horizontal range of the projectile is (a) 4.9 m

(b) 9.6 m

(c) 19.6 m

(d) 14 m

35. A body is projected at an angle q to the horizontal with kinetic energy E k. The potential energy at the highest point of the trajectory is (b) Ek cos2 q

(a) Ek 2

(c) Ek sin q

(d) Ek tan2 q

130 JEE Main Physics 36. Two projectiles A and B are thrown with velocities v v and respectively. They have the same range. If B is 2 thrown at an angle of 15° to the horizontal, A must have been thrown at an angle æ1ö (a) sin -1 ç ÷ è16 ø

æ 1ö (b) sin -1 ç ÷ è 4ø

æ 1ö (c) 2 sin -1 ç ÷ è 4ø

(d)

1 -1 æ 1 ö sin ç ÷ è 8ø 2

37. A particle slides down a frictionless parabolic ( y = x2 ) track ( A - B - C) starting from rest at point A. Point B is at the vertex of parabola and point C is at a height less than that of point A. After C, the particle moves freely in air as a projectile. If the particle [NCERT Exemplar] reaches highest point at P, then A

41. Two particles are simultaneously projected in opposite directions horizontally from a given point in space whose gravity g is uniform. If u1 and u2 be their initial speeds, then the time t after which their velocites are mutually perpendicular is given by (a)

u1u2 g

(b)

u21 + u22 g

(c)

u1 (u1 + u2 ) g

(d)

u2 (u1 + u2 ) g

42. A plane surface is inclined making an angle q with the horizontal. From the bottom of this inclined plane, a bullet is fired with velocity v. The maximum possible range of the bullet on the inclined plane is (a)

v2 g

(b)

v2 g(1 + sin q)

(c)

v2 g(1 - sin q)

(d)

v2 g(1 + sin q)2

y

P

43. A projectile is fired with a velocity v at an angle q with

v0

–x2

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

–x1

the horizontal. The speed of the projectile when its direction of motion makes an angle b with the horizontal is

θ C

B –x0 (x = 0)

(a) v cos q (c) v cos q sec b x

44. A ball is projected up an incline of 30° with a velocity

KE at P = KE at B height at P = height at A total energy at P = total energy at A time of travel from A to B = time of travel from B to P.

38. The horizontal range of an oblique projectile is equal to the distance through which a projectile has to fall freely from rest to acquire a velocity equal to the velocity of projection in magnitude. The angle of projection is (a) 15°

(b) 60°

(c) 45°

(d) 30°

39. A projectile is thrown with velocity v making an angle q with the horizontal. It just crosses the tops of two poles, each of height h, after 1 s and 3 s respectively. The time of flight of the projectile is (a) 1 s (c) 4 s

(b) 3 s (d) 7.8 s

40. Two stones are projected so as to reach the same distance from the point of projection on a horizontal surface. The maximum height reached by one exceeds the other by an amount equal to half the sum of the height attained by them. Then, angle of projection of the stone which attains smaller height is (a) 45° (c) 30°

(b) 60° (d) tan -1 (3 / 4 )

(b) v cos q cos b (d) v cos q tan b

of 30 ms -1 at an angle of 30° with reference to the inclined plane from the bottom of the inclined plane. If g = 10 ms -2 , then the range on the inclined plane is (a) 12 m

(b) 60 m

(c) 120 m

(d) 600 m

45. A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball? [NCERT Exemplar] (a) 40 m (c) 500 m

(b) 45 m (d) 50 m

46. A piece of marble is projected from earth’s surface with velocity of 50 ms -1. 2 s later it just clears a wall 5 m high. What is the angle of projection? (a) 45° (c) 60°

(b) 30° (d) None of these

47. A body is projected with speed v ms -1 at angle q. The kinetic energy at the highest point is half of the initial kinetic energy. The value of q is (a) 30°

(b) 45°

(c) 60°

(d) 90°

48. A ball is projected with velocity u at an angle a with horizontal plane. Its speed when it makes an angle b with the horizontal is (a) u cos a (c) u cos a cos b

u cos b u cos a (d) cos b (b)

Projectile Motion

131

49. The angle of projection of a projectile for which the horizontal range and maximum height are equal to (a) tan -1(2)

(b) tan -1( 4 )

(c) cot -1 (2)

(d) 60°

v0 H θ

50. A particle is projected from horizontal making an -1

angle 60° with initial velocity 40 ms . The time taken by the particle to make angle 45° from horizontal, is (a) 15 s

(b) 2.0 s

(c) 20 s

(d) 1.5 s

51. Two bodies are projected from the same point with equal speeds in such directions that they both strike the same point on a plane whose inclination is b. If a be the angle of projection of the first body with the horizontal the ratio of their times of flight is cos a sin ( a + b ) cos a (c) sin( a - b )

sin( a + b ) cos a sin( a - b ) (d) cos a

(a)

(b)

52. 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. What is the time interval of passing between the two walls? 2h (a) g

(b)

gh g

(c)

h g

h (d) 2 g

53. A projectile is thrown with a velocity of 10 m/s at an angle 60° with horizontal. The interval between the moment when speed is 5 g m/s, is ( g = 10 m/s2 ). (a) 1 s (c) 2 s

√3 H

æ 1 ö (a) tan -1 ç ÷ è 3ø

(b) tan -1 3

æ 2 ö (c) tan -1 ç ÷ è 3ø

æ 3ö (d) tan -1 ç ÷ è 2 ø

57. A particle is projected with speed v at an angle

p q æç0 < q < ö÷ above the horizontal from a height H è 2ø above the ground. If v = speed with which particle hits the ground and t = time taken by particle to reach ground, then (a) (b) (c) (d)

as q increases, v decreases and t increases as q increases, v increases and t increases as q increases, v remains same and t increases as q increases, v remains same and t decreases

58. Two inclined planes are located as shown in figure. A particle is projected from the foot of one frictionless plane along its line with a velocity sufficient to carry it to top after which the particle slides down the other frictionless inclined plane. The total time it will take to reach the point C is

(b) 3 s (d) 4 s 9.8 m

54. A particle is projected from the ground with an initial speed of v at an angle q with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is v (a) 1 + 2 cos2 q 2 v (c) 1 + 3 cos2 q 2

v (b) 1 + cos2 q 2 (d) v cos q

55. A body of mass m is thrown upward at an angle q with

the horizontal with velocity v. While rising up the velocity of the mass after t second will be (a)

( v cos q)2 + ( v sin q)2

(b)

( v cos q - v sin q)2 - gt

(c)

2

22

v + g t - (2v sin q) gt

45°

45°

(a) 2 s

(b) 3 s

(c) 2 2 s

(d) 4 s

59. The equation of motion of a projectile are given by x = 36 t. If and 2 y = 96 t - 9.8 t2 m. The angle of projectile is æ 4ö (a) sin -1 ç ÷ è5ø

æ 3ö (b) sin -1 ç ÷ è5ø

æ 4ö (c) sin -1 ç ÷ è 3ø

æ 3ö (d) sin -1 ç ÷ è 4ø

60. Trajectories of two projectiles are shown in figure. Let T1 and T2 be the time periods and u1 and u2 their speeds of projection. Then, y

(d) v2 + g2t2 - (2v cos q) gt

56. A projectile is thrown at an angle q such that it is just

able to cross a vertical wall as its highest point as shown in the figure. The angle q at which the projectile is thrown is given by

1

(a) T2 > T1

(b) T1 = T2

2 x

(c) u1 > u2

(d) u1 < u2

132 JEE Main Physics 61. A projectile A is thrown at an angle 30° to the horizontal from point P. At the same time another projectile B is thrown with velocity v2 upwards from the point Q vertically below the highest point A would v reach. For B to collide with A the ratio 2 should be v1

point C if both are projected simultaneously? ( g = 10 ms-2 ) B

5 ms–1

10 ms–1

h

v1

60°

v2

C

A 30°

P

3 (a) 2

(b) 2

Q

1 (c) 2

2 (d) 3

62. A fighter plane enters inside the enemy territory, at

time t = 0 with velocity v0 = 250 ms -1 and moves horizontally with constant acceleration a = 20 ms -2 (see figure). An enemy tank at the border, spot the plane and fire shots at an angle q = 60° with the horizontal and with velocity u = 600 ms -1. At what altitude H of the plane it can be hit by the shot?

(a) 10 m (c) 15 m

(b) 30 m (d) 25 m

65. A very broad elevator is going up vertically with a

constant acceleration of 2ms -2 . At the instant when its velocity is 4 ms-1 a ball is projected from the floor of the list with a speed of 4 ms-1 relative to the floor at an elevation of 30°. The time taken by the ball to return the floor is ( g = 10 ms-2 ) (a) 1/2 s (c) 1/4 s

(b) 1/3 s (d) 1 s

66. A projectile is fired at an angle of 30° to the horizontal such that the vertical component of its initial velocity is 80 ms-1. Its time of flight is T. Its T has a magnitude of nearly velocity at t = 4

600 ms–1 H

(a) 200 ms -1 (c) 140 ms -1

θ = 60°

(a) 1500 3 m

(b) 125 m

(c) 1400 m

(d) 2473 m

63. An aircraft, diving at an angle of 53.0° with the vertical releases a projectile at an altitude of 730 m. The projectile hits the ground 5.00 s after being released. What is the speed of the aircraft? (a) 282 ms -1

(b) 202 ms -1 (c) 182 ms -1 (d) 102 ms -1

64. A particle A is projected from the ground with an

initial velocity of 10 ms-1 at an angle of 60° with horizontal. From what height h should an another particle B be projected horizontal with velocity 5 ms-1 so that both the particles collide with velocity 5 ms-1 so that both the particles collide on the ground at

(b) 300 ms -1 (d) 100 ms -1

67. A car is moving rectilinearly on a horizontal path with acceleration a0 . A person sitting inside the car observes that an insect S is crawling up the screen with an acceleration a. If q is the inclination of the screen with the horizontal the acceleration of the insect (a) (b) (c) (d)

parallel to screen is a0 cos q along the horizontal is a0 - a cos q perpendicular to screen is a0 sin q perpendicular to screen is a0 tan q

68. A particle is projected from the ground at an angle of

60° with horizontal with speed u = 20 ms -1. The radius of curvature of the path of the particle, when its velocity makes an angle of 30° with horizontal is ( g = 10 ms -2 ) (a) 10.6 m (c) 15.4 m

(b) 12.8 m (d) 24.2 m

Projectile Motion

Round II Only One Correct Option angle of 45° to the horizontal. The average velocity of stone during its motion from starting point to its maximum height is ( g = 10 ms-2 ) (b) 10 5 ms -1

(c) 20 ms -1

(d) 20 5 ms -1

(Mixed Bag) 7. A car is travelling at a velocity of 10 kmh -1 on a

1. A stone is projected with a velocity 20 2 ms-1 at an

(a) 5 5 ms -1

straight road. The driver of the car throws a parcel with a velocity of 10 2 kmh -1 when the car is passing by a man standing on the side of the road. If the parcel is to reach the man, the direction of throw makes the following angle with direction of the car (a) 135° (c) tan

(b) 45°

-1

æ 1 ö (d) tan ç ÷ è 2ø

( 2) 60°

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 m /s2 ) (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 316 s Actual distance travelled by the ball is more than 49 m

3. A shell is fired from a cannon with a velocity v at angle q with horizontal. At the highest point, it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon. The speed of the other piece just after explosion is (a) 3v cos q 3 (c) v cos q 2

(b) 2v cos q 3 (d) v cos q 2

4. The speed of projection of a projectile is increased by 10%, without changing the angle of projection. The percentage increase in the range will be (a) 10%

(b) 20%

(c) 15%

(d) 5%

5. A projectile is launched with a speed of 10 m/s at an angle 60° with the horizontal from a sloping surface of inclination 30°. The range R is. (Take, g = 10 m / s2 )

8. A particle is dropped from a height h. Another particle was what initially at a horizontal distance d from the first, is simultaneously projected with a horizontal velocity u and two particles just collide on the ground. The three quantities h, d and u are related to (a) d2 =

u2 h 2g

(b) d2 =

9. A body of mass 1 kg is projected with velocity 50 m/s at an angle of 30° with the horizontal. At the highest point of its path a force 10 N starts acting on body for 5 s vertically upward besides gravitational force, what is horizontal range of the body? ( g = 10 m / s2 ) (a) 125 3 m (c) 500 m

(b) 200 3 m (d) 250 3 m

10. If a stone is to hit at a point which is at a distance d away and at a height h above the point from where the stone starts, then what is the value of initial speed u, if the stone is launched at an angle q? u

60°

h θ

R

d

30°

(b) 13.3 m

(c) 9.1 m

magnitude but making different angles with the horizontal. Their ranges are equal. If the angle of p projection of one is and its maximum height is y1, 3 the maximum height of the other will be (a) 3 y1

(b) 2 y1

y (c) 1 2

(a)

g cos q

d 2 ( d tan q - h)

(b)

d cos q

d 2 ( d tan q - h)

(d) 12.6 m

6. Two stones are projected with the same velocity in

y (d) 1 3

2 u2 h g

(d) gd2 = u2 h

(c) d = h

10 m/s

(a) 4.9 m

133

(c)

gd2 h cos2 q

(d)

gd2 ( d - h)

134 JEE Main Physics 11. Figure shows four paths for a kicked football ignoring

Y

the effects of air on the flight rank the paths according to the initial horizontal velocity component highest first

u

θ P

Q

2

(a) mu sin q 1 (c) mu2 sin 2 q 2 1

2

4

3

(a) 1, 2, 3, 4 (c) 3, 4, 1, 2

the slope which is inclined at an angle q with the horizontal. What is the time of flight?

12. After one second the velocity of a projectile makes an

angle of 45° with the horizontal. After another one second it is travelling horizontally. The magnitude of its initial velocity and angle of projection are ( g = 10 ms -2 ) (b) 22.36 ms –1, tan -1(2)

–1

13. A particle of mass m is released from rest and follows a parabolic path as shown. Assuming that the displacement of the mass from, the origin is small, which graph correctly depicts the position of the particle as a function of time? x(t)

(a) 0

(a)

2v2 tan q g

(b)

v2 tan q g

(c)

2v2 sec q g

(d)

2v2 tan q sec q g

17. A body is projected up smooth inclined plane with a

(d) 22.36 ms –1, 60°

(c) 14.62 ms , 60°

V(x) m x

O

velocity v0 from the point A as shown figure. The angle of inclination is 45° and top B of the plane is connected to a well of diameter 40 m. If the body just manages to cross the well, what is the value of v0 ? Length of the inclined plane is 20 2 m, and g = 10 ms -2 . B

(b) 0

A

40 m

t

(a) 20 ms -1

(b) 20 2 ms -1

-1

(d) 40 2 ms -1

(c) 40 ms x(t)

(c) 0

x(t)

t

(d) 0

18. Two projectiles A and B thrown with speeds in the t

14. A particle of mass m is projected with a velocity v at an angle of 60° with horizontal. When the particle is at its maximum height. The magnitude of its angular momentum about the point of projection is 2

(a) zero (c)

3 mv2 16 g

(b)

3 mv 16 g

(d)

3 mv2 3g

C

45°

x(t)

t

(b) mu cos q 1 (d) mu2 cos 2 q 2

16. A projectile is fired with a velocity v at right angle to

(b) 2, 3, 4, 1 (d) 4, 3, 2, 1

(a) 14.62 ms –1, tan -1 (2)

X

2

15. Average torque on a projectile of mass m, initial speed u and angle of projection q, between initial and final position P and Q as shown in figure about the point of projection is

ratio 1 : 2 acquired the same heights. If A is thrown at an angle of 45° with the horizontal, the angle of projection of B will be (a) 0° (c) 30° (e) 15°

(b) 60° (d) 45°

19. A particle is projected with a velocity 200 ms -1 at an angle of 60°. At the highest point, it explodes into three particles of equal masses. One goes vertically upwards with a velocity 100 ms -1, the second particle goes vertically downwards. What is the velocity of third particle? (a) (b) (c) (d)

120 ms -1 making 60° angle with horizontal 200 ms -1 making 30° angle with horizontal 300 ms -1 200 ms -1

Projectile Motion 20. The trajectory of a projectile in vertical plane is

y = ax - bx2 , where a and b are constants and x and y are respectively horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection from the horizontal are b2 , tan -1( b) 4b a2 (c) , tan -1( a ) 4b (a)

a2 , tan -1(2b) b 2a2 (d) , tan -1( a ) b (b)

More Than One Correct Options 21. A particle is projected from a point A with a velocity v at an angle of elevation q. At a certain point B, the particle moves at right angle to its initial direction. Then (a) velocity of particle at B is v sin q (b) velocity of particle at B is v cot q (c) velocity of particle at B is v tan q (d) velocity of flight from A to B is

v g sin q

22. Two projectiles A and B are projected with same speed at angles 15° and 75° respectively to the maximum and have same horizontal range. If h be the maximum height and T total time of flight of a projectile, then (a) hA > hB (c) TA < TB

(b) hA < hB (d) TA > TB

23. Two particles are projected in air with speed v0 at

angles q 1 and q 2 (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then tick the right choices [NCERT Exemplar] (a) angle of projection : q 1 > q2 (b) time of flight : T1 > T2 (c) horizontal range : R1 > R2 (d) total energy : U1 > U2

24. A projectile has the same range R for two angles of projections. If T1 and T2 be the times of flight in the two cases, then (using q as the angle of projection corresponding to T1) (a) T1 T2 µ R

(b) T1 T2 µ R2

(c) T1 /T2 = tan q

(d) T1 /T2 = 1

25. A particle is hurled into air from a point on the horizontal ground at an angle with the vertical. If the air exerts a constant resistive force (a) (b) (c) (d)

the path of projectile will be parabolic path the time of ascent will be equal to time of descent the total energy of the projectile is not conserved at the highest point, the velocity of projectile is horizontal

135

Comprehension Based Questions Passage I A projectile is thrown from the ground with a speed of 2 gh at an angle of 60° to the horizontal from a point on the horizontal ground.

26. The horizontal range of projectile is (a) (c)

(b) 2h 3 (d) 3h/ 2

3h 3h/ 2

27. The time spent by projectile above a height h is (a) 4

h g

(b)

h g

(c)

2h g

(d)

3h g

28. The maximum height attained by projectile is (a) 2h/3 (c) 3h/ 4

(b) 3h (d) 3h/ 2

Passage II Two second after projection, a projectile is travelling in a direction inclined at 30° to the horizontal. After 1 more second, it is travelling horizontally (use g = 10 ms -2 )

29. The initial velocity of its projection is (a) 10 ms -1

(b) 10 3 ms -1

(c) 20 ms -1

(d) 20 3 ms -1

30. The angle of projection of the projectile is (a) 30° (c) 60°

(b) 45° (d) None of these

Assertion and Reason Directions

Question No. 31 to 35 are Assertion-Reason type. Each of these contains two Statements : Statement I (Assertion), Statement II (Reason). Each of these questions also has four alternative choices, only one of which is correct. You have to select the correct choices from the codes (a), (b), (c) and (d) given ahead (a) If both Assertion and Reason are true and Reason is correct explanation of the Assertion (b) If both Assertion and Reason are true but Reason is not correct explanation of the Assertion (c) If Assertion is true but Reason is false (d) If Assertion is false but the Reason is true

31. Assertion If a particle is projected vertices upwards with velocity u, the maximum height attained by the particle is h1. The same particle is projected at angle 30° from horizontal with the same speed u. Now the maximum height is h2 . Thus, h1 = 4 h2 Reason In first case v = 0 at highest point and in second case v ¹ 0 at highest point.

136 JEE Main Physics 32. Assertion At highest point of a projectile dot product

34. Assertion If in a projectile motion, we take air

of velocity and acceleration is zero. Reason At highest point velocity and acceleration are mutually perpendicular.

friction into consideration, then tascent < tdescent . Reason During ascent magnitude of retardation is greater than magnitude of acceleration during descent.

33. Assertion A particle is projected with speed u at an angle q with the horizontal. At any time during motion, speed of particle is v at angle a with the vertical, then v sin a is always constant throughout the motion. Reason In case of projectile motion, magnitude of radial acceleration at top most point is maximum.

35. Assertion In projectile motion if time of flight is 4 s, then maximum height will be 20 m. ( g = 10 m / s2 ). gT Reason Maximum height = . 2

Previous Years’ Questions 36. A projectile is given an initial velocity of ( i$ + 2$j ) m/s, where $i is along the ground and $jis along the vertical. If g = 10 m / s2 , the equation of its trajectory is [JEE Main 2013]

(a) y = x - 5 x2

(b) y = 2x - 5 x2

(c) 4 y = 2x - 5 x2

(d) 4 y = 2x - 25 x2

37. Two cars of masses m1 and m2 are moving in circles of

radii r1 and r2 respectively. Their speeds are such that they make complete circles in the same time t. The ratio of their centripetal acceleration is [AIEEE 2012] (a) m1r1 : m2 r2 (c) r1 : r2

(b) m1 : m2 (d) 1 : 1

38. A boy can throw a stone up to a maximum height of 10 m. The maximum horizontal distance that the boy can throw the same stone up to will be [AIEEE 2012] (a) 20 2 m

(b) 10 m

(c) 10 2 m

(d) 20 m

39. A particle of mass m is projected with a velocity v making an angle of 30° with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at [AIEEE 2011] its maximum height h is (a)

3 mv2 2 g

(b) zero

(c)

mv3 2g

(d)

41. A ball rolls of the top of stair-way with a horizontal velocity of magnitude 1.8 ms -1. The steps are 0.20 m high and 0.20 m wide. Which step will the ball hit first? [Orissa JEE 2011] (a) First (c) Third

42. A projectile is projected with velocity kve vertically

upward direction from the ground into the space (ve is the escape velocity and k < 1). If air resistance is considered to be negligible then the maximum height from the centre of earth to which it can go will [Orissa JEE 2010] be (R = radius of earth) R k +1 R (c) 1 - k2

(a)

v2 g

(b) p

v4 g2

(c) n2

v4 g2

(d) p2

v2 g2

2

with the x-axis with an initial velocity v0 in the X v sin q Y - plane as shown in figure . At a time t < 0 g the angular momentum of the particle is [AIEEE 2010] Y v0

θ

40. A large number of bullets are fired in the all

(a) p

R k -1 R (d) k +1 (b)

2

43. A small particle of mass m in projected at an angle q

3 mv3 16 g

directions with same speed v. What is the maximum area on the ground on which these bullets will [AIEEE 2011] spread?

(b) Second (d) Fourth

X

1 mgv0 t2 cos q i$ 2 (b) - mgv0 t2 cos q $j (c) mgv0 t cos qk$

(a)

1 (d) - mgv0 t2 cos qk$ 2

Projectile Motion

137

44. A point P moves in counter-clockwise direction on a

49. Two particles A and B are projected with same speed

circular path as shown in figure. The movement of P is such that it sweep out a length s = t 3 + 5, where s is in metres and t is in seconds. The radius of the path is 20 m. The acceleration of P when t = 2 s is nearly

so that the ratio of their maximum heights reached is 3 : 1. If the speed of A is doubled without altering other parameters, the ratio of the horizontal ranges [Kerala CET 2008] obtained by A and B is

[AIEEE 2010]

y

(a) 1 : 1 (c) 4 : 1

(b) 2 : 1 (d) 3 : 2

50. A body is projected at such angle that the horizontal range is three times the greatest height. The angle of [BCECE 2008] projeciton is

P(x, y)

(a) 42°8¢ (c) 33°7¢

20 m O

(a) 14 m/s2 (c) 12 m/s2

x

(b) 13 m/s2 (d) 7.2 m/s2

51. A particle is projected with certain velocity at two

45. The maximum range of projectile fired with some initial velocity is found to be 1000 m, in the absence of wind and air resistance. The maximum height reached by the projectile is [Orissa JEE 2009] (a) 250 m

(b) 500 m

(c) 1000 m

(d) 2000 m

46. A particle is projected with velocity v0 along x-axis.

The deceleration on the particle is proportional to the square of the distance from the origin, i. e., a = ax2 , the [MP PET 2009] distance at which the particle stop is 3 v0 2a

æ3 v ö (b) ç 0 ÷ è 2a ø

1/ 3

2 v20 (c) 3a

æ 3 v2 ö (d) ç 0 ÷ è 2a ø

1/ 3

(a)

47. If a body is projected with an angle to the horizontal, then

[EAMCET 2008]

(a) its velocity is always perpendicular to its acceleration (b) its velocity becomes zero at maximum height (c) its velocity makes zero angle with the horizontal at its maximum height (d) the body just before hitting the ground, the direction of velocity coincides with the acceleration

48. A body is thrown upwards from the earth surface

with velocity 5 ms -1 and from a planet surface with velocity 3 ms -1. Both follow the same path. What is the projectile acceleration due to gravity on the planet? Acceleration due to gravity on earth is [Orissa JEE 2008] 10 ms -1. (a) 2 ms -2 (c) 4 ms -2

(b) 3.6 ms -2 (d) 5 ms -2

(b) 53°7¢ (d) 25°8¢

different angles of projections with respect to horizontal plane so as to have same range R on a horizontal plane. If t1 and t2 are the time taken for the two paths, the which one of the following relations is correct? [UP SEE 2008] 2R g g (c) t1 t2 = 2g

(a) t1 t2 =

R g 4R (d) t1 t2 = g

(b) t1 t2 =

52. A particle is projected at 60° to the horizontal with an energy E. The kinetic energy and potential energy at the highest point are [KCET, AIEEE 2007] æ E Eö (a) ç , ÷ è 2 2ø

æ 3E E ö (b) ç , ÷ è 4 4ø

(c) ( E, 0 )

æ E 3E ö (d) ç , ÷ è4 4 ø

53. The maximum height attained by a projectile when thrown at an angle q with the horizontal is found to be half the horizontal range. Then, q is equal to [KCET 2007] -1

(a) tan (2) (c)

p 4

p (b) 6 æ 1ö (d) tan -1 ç ÷ è 2ø

54. A particle is thrown in the upward direction making an angle of 60° with the horizontal direction with a velocity of 147 ms -1. Then, the time after which its inclination with the horizontal is 45°, is [UP SEE 2006]

(a) 15 s (c) 5.49 s

(b) 10.98 s (d) 2.745 s

Answers Round I 1. 11. 21. 31. 41. 51. 61.

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

2. 12. 22. 32. 42. 52. 62.

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

3. 13. 23. 33. 43. 53. 63.

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

2. 12. 22. 32. 42. 52.

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

3. 13. 23. 33. 43. 53.

(a) (d) (a,b) (b) (d) (a)

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

4. 14. 24. 34. 44. 54. 64.

5. 15. 25. 35. 45. 55. 65.

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

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

6. 16. 26. 36. 46. 56. 66.

7. 17. 27. 37. 47. 57. 67.

(c) (b) (b) (c) (b) (c) (c)

8. 18. 28. 38. 48. 58. 68.

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

9. 19. 29. 39. 49. 59.

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

10. 20. 30. 40. 50. 60.

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

8. 18. 28. 38. 48.

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

9. 19. 29. 39. 49.

(d) (c) (d) (d) (c)

10. 20. 30. 40. 50.

(b) (c) (c) (b) (b)

Round II 1. 11. 21. 31. 41. 51.

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

4. 14. 24. 34. 44. 54.

(b) (b) (a,c) (a) (a) (c)

5. 15. 25. 35. 45.

(b) (c) (a.c,d) (c) (a)

6. 16. 26. 36. 46.

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

7. 17. 27. 37. 47.

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

the Guidance Round I 1. Here,

50 = R=

u2 50 50 u 2 sin 2 ´ 15° or = = = 100 g sin 30° 1/ 2 g u 2 sin 2 ´ 45° u 2 = = 100 m g g

\ Angle with which it strikes the ground æ vy ö æ 100 ö q = tan -1ç ÷ = tan -1ç ÷ è 500 ø è vx ø æ 1ö q = tan -1ç ÷ è5ø

1 2

2. nh = gt 2 Þ

…(i)

We have tOB =

Horizontal distance travelled by ball æ 2nh ö nb = ut , nb = u ç ÷ è g ø

…(ii)

Squaring Eq. (ii), we get n 2b 2 =

u 22nh g

1 2

4. From h = gt 2,

æ 2nh ö t= ç ÷ è g ø

\ n=

2u 2h gb 2

3. Horizontal component of velocity v x = 500 ms-1 and vertical components of velocity while striking the ground v y = 0 + 10 ´ 10 = 100 ms-1 u = 500 ms–1

500 ms–1 θ 100 ms–1

2 ´ 1960 2hOA = = 20 s 9.8 g

Horizontal distance AB = vtOB 5ö æ = ç600 ´ ÷ (20) è 18 ø = 3333.33 m = 3.33 km dy dx 5. v y = = 8 -10 t ,v x = = 6 dt dt At t = 0 , \

v y = 8 m/s and v x = 6 m/s v = v x2 + v y2 = 10 m/s

2h 4 = = 0.64 s g (9.8) s 3 Now, v= = = 4.7 m/s t 0.64

6. t =

7. Stone will must follow the parabolic path.

Projectile Motion

or

= sin 33.6° q = 33.6°

\

Horizontal range (R) =

8. Range of the projectile on an inclined plane (down the plane) is, u2 [sin(2a + b) + sin b ] g cos2 b

R= Here,

u = v 0 , a = 0 and b = q

\

R=

139

u 2 sin 2q g

( 40) 2 sin 2 ´ 33.6° 9.8 1600 ´ sin 67.2° = 9.8 1600 ´ 0.9219 = = 150.5 m 9.8

=

2v 02 sin q g cos2 q v2

θ

11. The velocity of the particle at any time t

R

v = v0 + a t The x-component is v x = v ax + axt

θ

tan q g

The y-component is

2v 02 tan 2 q g

When the particle reaches its maximum x-coordinate,

2v 02

Now

x = R cos q =

and

y = - R sin q = -

v y = v oy + ax = ( -0.5 t ) ms-1 v x = 0. i.e., 3 -t =0 Þ t =3 s The y-component of the velocity of this time is

9. v x = 8t - 2 dx = 8t - 2 dt

or or or or

x

v y = - 0.5 ´ 3 = - 1.5 ms-1

t

ò14 dx = ò2 (8t - 2)dt

12. R =

x - 14 = [ 4t 2 - 2t ]t2 = 4t 2 - 2t - 12 x = 4t 2 - 2t + 2

…(i)

u 2 sin q at angle q and 90° - q g

Now,

h1 =

u 2 sin 2 q 2g

and

h2 =

u 2 sin 2 (90° - q) u 2 cos2 q = 2g 2g

Further, v y = 2 or \ or

dy =2 dt y

æ u 2 sin 2q ö 1 R 2 ÷× = h1 h2 = ç g è ø 16 16

t

ò4 dy = ò2 2 dt y - 4 = [2t ]t2 = 2t - 4 or y = 2t y t= 2

or

\ …(ii)

13. H =

Substituting the value of t from Eq. (ii) in Eq. (i), we have x = y2 - y + 2

Let the angle of projection of the ball be q, when maximum height attained by it be 25 m.

or

u 2 sin 2 q 2g

25 =

( 40) 2 sin 2 q 2 ´ 9.8

sin 2 q =

25 ´ 2 ´ 9.8 1600

= 0.3063 or

sin q = 0.5534

2u sin q g

T=

Þ

T2 =

4u 2 sin 2 q T2 8 \ = H g g2

Þ

T=

8H 2H =2 g g

Maximum height attained by the ball H=

u 2 sin 2 q 2g

and

10. Given, initial velocity (u) = 40 m/s Height of the hall (H) = 25 m

R = R h1 h2

14. t =

2h 2 ´ 490 = = 100 = 10 s g 9.8 5ö 500 æ x = vt = ç60 ´ ÷ ms-1 ´ 10 s = m è 18 ø 3

15. R µ

1 \ Rmoon = 6 Rearth g

16. Velocity of a body should be equal to the horizontal component of velocity of ball.

140 JEE Main Physics 17. T =

2 uy g gT = 25 m/s 2

and

uy2

After 1 s, ux will remain as it is uy will decreases by 10 m/s or it will remain 8 m/s vy 8 1 tan q = = = \ v x 24 3

\

uy =

Now,

H=

Further,

R = uxT R ux = = 40 m/s T

\

18. t =

4 5 3 uy sin q0 = 30 ´ = 18 m/s 5

25. Given, ux = u cos q0 = 20 ´ = 24 m/s

2g

(25) 2 = 31.25 m 20

=

26. The two angles of projection are clearly q and (90° - q). 2v sin q g 2v sin(90° - q) T2 = g

2h g

T1 =

Distance from the foot of the tower 2h d = vt = v = 250 m g v 2 height of tower = 4h

When velocity = and

and \

T1T2 =

27. Change in momentum is the product of force and time. Dp = mg ´

v 2( 4h) x= 2 g

Then, distance

\

21.

22.

1 2 (in vertical direction) gt 2 2h 2 ´ 0.1 t= = = 0.141s g 10

h=

20.

Now, in horizontal direction S 100 vx = x = » 700 m/s t 0.141 p 5 p 18 p - 5 p Required angle = = 2 36 36 13 p rad = 36 v 2 sin 2 q As, hmax = 2g In the given problem, hmax is same in both the cases. \ or

23. As, H =

v12 sin 2 60° = v 22 sin 2 30° v1 sin 30° 1 2 1 = = ´ = v 2 sin 60° 2 3 3 v 2 cos2 b or v cos b = 2gH 2g t=

or

24. As, E k ¢ = E k cos2 30° =

v cos b 2gH = g g

t= 3E k 4

2H g

2 sin q g

Dp ö æ çQ F = ÷ è Dt ø

= 2mv sin q = 2mv sin 45° 2mv = = 2 mv 2

2h x=v = 250 m g

19. At the highest point, KE will be minimum but not zero.

2(v) 2(2 sin q cos q) 2R = g ´g g

28. Since, range is given to be same therefore the other angle is (90° - 30° ), i. e. , 60°. H=

v 2 sin 2 30° 1 é v 2 ù = ê ú 2g 4 ë 2g û

v 2 sin 2 60° 3 é v 2 ù = ê ú 2g 4 ë 2g û H¢ 3 4 = ´ = 3 or H ¢ = 3H H 4 1

H¢ =

29. R =

v 2 sin 2q 2v sin q = 200 , T = =5 g g

Dividing,

v 2 ´ 2 sin q cos q g 200 ´ = = 40 g 2v sin q 5 v cos q = 40 ms-1

or

It may be noted here that the horizontal component of the velocity of projection remains the same during the flight of the projectile. 1 30. (KE)H = (KE)i 2 1 1 æ1 ö 1 mv 2 cos2 q = ç mv 2÷ = mv 2 ø 4 2 2 è2 or or or

1 2 1 cos q = 2

cos2 q =

q = 45°

Projectile Motion 31. v y

d d d (y) = (10t ) - (t 2) = 10 - 2t dt dt dt

36. As, R =

At maximum height, v y = 0 10 - 2t = 0

\ or or

2

y = (10 ´ 5 - 5 ´ 5) m = 25 m

or

32. Equating velocities along the vertical,

33. As,

or

v 2 = v1 sin 30° v2 1 = v1 2

or

or

v 2 sin 2q 3v 2 = g 2g

or or

2q = 60°

greater than at P. In the given motion of a particle, the law of conservation of energy is obeyed. Therefore, total energy at P = total energy at A. As vertical distance AB > BP , time of travel from A to B is greater than that from B to P.

38. Using, v 2 - u2 = 2as, we get

q = 30° Let us cross check with the help of data for vertical range. or

2

2

s=

2

v sin q v = 2g 8g 1 sin 2 q = 4 1 sin q = 2

or

or

^

34. Here, v = 6 i + 8 j ms -1 ^

^

Comparing with v = v x i + v y j , we get ux = 6 ms-2 uy = 8 ms-2

and Also,

u

2

= v x2

+

v = 10 ms sin q =

8

θ

36 + 64 = 100 or

10

v y2 -1

6

8 6 and cos q = 10 10

v 2 sin 2q 2v 2 sin q cos q R= = g g 8 6 R = 2 ´ 10 ´ 10 ´ ´ ´ 10 m = 9.6 m 10 10

35. Let v be the velocity of projection and q the angle of projection. Kinetic energy at highest point 1 = mv 2 cos2 q or E k cos2 q 2 Potential energy at highest point = E k - E k cos2 q = E k(1 - cos2 q) = E k sin 2 q

v2 2g

v 2 sin 2q v 2 = g 2g 1 sin 2q = 2

Now,

q = 30°

or ^

v2 æv ö v 2 sin 2q = ç ÷ sin 30° = è2ø 8 1 sin 2q = 8 é 1ù 2q = sin -1ê ú ë8û 1 é 1ù q = sin -1ê ú 2 ë8û

37. Since y = xz , the motion is in two dimensions. Velocity at B is 3 sin 2q = 2

or

v 2 sin 2q g

In the given problem v 2 sin 2q = constant

2t = 10 t =5 s

\

141

or

sin 2q = sin 30°

or

q = 15°

The other possible angle of projection is (90° - 15° ), i. e. ,75°. 1 39. h = v sin qt - gt 2 2 1 2 or gt - v sin qt + h = 0 2 -v sin q t1 + t 2 = 1 g 2 2v sin q or =T t1 + t 2 = g T = (1 + 3) s = 4 s

40.

H + H2 As, H1 - H2 = 1 2 H1 = 3 H2

or \

ì u 2 sin 2 (90° - q) ü u sin q = 2í ý 2g 2g þ î 2

2

tan 2 q = 3 \

tan q = 3

or

q = 60°

Therefore, the other angle is (90° - q) or 30°.

142 JEE Main Physics 41. Since, v1 ^ v 2 \ or

43. As, v ¢ cos b = v cos q

v1 × v1 = 0 $ $ $ (u1 i - gtj ) × ( -u2 i - gt$j ) = 0

(Q horizontal component of velocities are always equal) or BV ¢ = v cos q sec b 2 ´ 30 ´ 30 sin 30° cos 60° 44. R = 10 cos2 30° 1 1 2 ´2 = 180 ´ ´ ´ m = 60 m 2 2 3

g 2 t 2 = u1 u2

\

u1 u2

t=

or

g

42. As, v x = v cos( a - q); v y = v sin( a - q)

45. Horizontal range of a projectile is given by

ax = - g sin q; ay = - g cos q If T is the time of flight, then

v

Y

i

gs

α– θ α θ

nθ g

O

T=

or

Now, or or or or

OA =

v2 [sin(2a - q) - sin q] g cos2 q

Clearly, the range R ( = OA) will be maximum when sin(2a - q) is maximum, i. e. , 1. This would mean p 2a - q = 2 q p or a= + 2 4 Maximum range up the inclined plane,

=

v2 v 2(1 - sin q) (1 - sin q) = 2 g cos q g (1 - sin 2 q) v 2(1 - sin q) v2 = g (1 - sin q) (1 + sin q) g (1 + sin q)

…(i)

v 2 = u 2 + 2as (0) 2 = u 2 + 2( - g )H u2 1 æ u2 ö = ç ÷ 2g 2èg ø 1 = ´ 100 2 = 50 m

H=

or

[using Eq. (i)]

46. Horizontal component = u cos q Vertical component = u sin q g = - 10 ms-2, u = 50 ms-1,h = 5 m, t = 2 s h = uy t +

2

or

u2 g

Using equation of motion,

v [2 sin( a - q) cos a ] g cos2 q v [sin (2a - q) + sin ( - q)] g cos2 q

100 =

When cricketer throws the ball vertically upward, then ball goes upto height H.

2

OA =

Rmax =

\

2v sin( a - q) g cos q

or

u2 g

Rmax = 100 m

Given,

1 g cos q × T 2 2

OB = v cos a ´ T OB cos q = OA OB OA = cos q v sin a × T OA = cos q 2v sin( a - q) 1 OA = v cos a ´ ´ g cos q cos q OA =

Rmax =

B

0 = v sin( a - q) × T -

u 2 sin 2q g

If q = 45°, then R is maximum and is equal to

Y

g cos θ

A

R=

1 2 gt 2

θ

1 ´ 10 ´ 4 2

\

5 = 50 sin q -

or

5 = 50 sin q - 2q 25 1 sin q = = 50 2

or \

47.

q = 30° 1 Given, (KE) highest = (KE) 2 1 1 1 mv 2 cos2 q = × mv 2 2 2 2 1 cos2 q = 2

Þ

cos q =

1 Þ q = 45° 2

Projectile Motion 48. As, v cos b = u cos a

1 2 gt 2 2 gt - 2uyt + 2h = 0 h = uyt -

Further,

(horizontal component of velocities are always equal) u cos a v= \ cos b

or \

t1 =

and

t2 =

49. Given, R = H u 2 sin 2a u 2 sin 2 a = g 2g 2 sin a cos a =

or

2

sin a 2

\

2uy - 4uy2 - 8gh 2g 4uy2 - 8gh g

d

g 2( Dt ) 2 + 2gh 4 ux2 + uy2 = (2 gh) 2

Given,

4h 2 g 2( Dt ) 2 + + 2gh = 4gh 2 4 ( Dt )

\

ux = uy - gt uy - ux t= g 40(sin 60° - sin 30° ) = = 1.5 s 9.8

\

2g

uy2 =

or

50. At 45°, v x = v y or

4uy2 - 8gh

2uy +

Dt = t1 - t 2 =

sin a = 4 or tan a = 4 cos a a = tan -1( 4)

or

143

g2 ( Dt ) 4 - 2gh( Dt ) 2 + 4h 2 = 0 4 ( Dt ) 2 =

2gh ±

2

g /2

51. Let a¢ be the angle of projection of the second body. u

or

4g 2h 2 - 4g 2h 2

Dt = 2

=

4h g

h g

u

53. v 2 = v y2 + v x2 α

β

R=

u2 [sin(2a - b)] g cos b

or

5 g = (uy - gt ) 2 + ux2

or

50 = (5 3 - 10 t ) 2 + (5) 2

\

(5 3 - 10 t ) = ± 5 t1 =

Range of both the bodies is same. Therefore, sin(2a - b) = sin(2a ¢ - b) or 2a ¢ - b = p - (2a - b) p a ¢ = - ( a - b) 2 2u sin( a - b) 2u sin( a ¢ - b) Now, T = and T ¢ = g cos b g cos b T sin( a - b) sin( a - b) = = \ ü T ¢ sin( a ¢ - b) ìp sin í - ( a - b) - bý 2 þ î sin( a - b) sin( a - b) = = ö æa cos a sinç - a ÷ ø è2

5 3 -5 \ t1 - t 2 = 1 s 10 Displacement Average velocity = Time H2 + v av =

…(i)

T /2

H x

R/2

Here,

y v

H = maximum height =

2

v sin 2 q 2g

v 2 sin 2q g 2v sin q T = time of light = g

R = range =

θ ux

or

R2 4

y

52. Let Dt be the time interval. Then,

uy

t2 =

and

54.

5 3 -5 10

2h

2h = (ux) ( Dt ) 2h ux = Dt

and

x

…(i)

Substituting in Eq. (i), we get v v av = 1 + 3 cos2 q 2

144 JEE Main Physics 55. Instantaneous velocity of rising mass after t s will be

1 2 at 2 1 s = 0 ´ t + ( g sin 45° )t 02 2 9.8 2 9.8 2 = t0 2 2

\

s = ut +

v t = v x2 + v y2 where, v x = v cos q = Horizontal component of velocity v y = v sin q - gt = Vertical component of velocity v t = (v cos q) 2 + (v sin q - gt ) 2 v = v 2 + g 2 t 2 - (2v sin q) gt

56.

R /2 3H = = 3 H H (v 02 sin q cos q)/ g or = 3 (v 02 sin 2 q)/2 g

q = tan

g

\ \

t0 = 2 s T = 2t 0 = 4 s dx = 36 m/s dt

vx =

y = 48 t - 4.9 t 2

-1 æ

2 ö ç ÷ è 3ø

v0

θ

t 02 = 4

\

57. From figure, v0 sin

\

59. x = 36 t

2 cot q = 3 2 tan q = 3 or

or

v0 cos θ

\ at t = 0, and

v y = 48 - 4.8 t v x = 36 m/s v y = 18 m/s æ vy ö æ 4ö So, angle of projection q = tan -1 ç ÷ = tan -1 ç ÷ è3ø è vx ø æ 4ö q = sin -1 ç ÷ è5ø

or

60. Maximum height and time of flight depend on the vertical

H

component of initial velocity. Ground

H1 = H2 Þ uy1 = uy 2 T1 = T2

Here

H = ( -v 0 sin q)t +

1 2 gt 2

R=

Range

v x = v 0 cos q

=

v y2 = (v 0 sin q) 2 + 2gHd v = v x2 + v y2 at ground v = v 02 + 2gH

u 2 sin 2q g 2(u sin q) (u cos q) 2uxuy = g g

R2 > R1 ux2 > ux1 or u2 > u1

\

61. Vertical component of velocity of A should be equal to

It means speed is independent of angle of projection. 1 2 Also, gt = H + t v 0 sin q 2 From this where q increase, t increases.

58. The time of ascent = time of descent = t 0

vertical velcotiy of B. v1 sin 30° = v 2 v1 v 1 = v2 \ 2 = 2 v1 2

or or

62. If it is being hit, then

T = total time of flight = 2t 0

d = v 0t +

1 2 at = (u cos q)t 2

9.8 m

(Q acceleration in horizontal direction is zero) g sin 45°

Q 600 ms–1

45°

A

sin 45° = \

9.8 9.8 = BC s

s = 9.8 2

H θ = 60° d

Projectile Motion or

\

t=

u cos q - v 0 a /2

and acceleration of ball relative to lift is 12 ms -2 in negative y- direction or vertically downwards. Hence, time of flight 2uy uy 2 1 T= = = = s 12 6 6 3 u 66. x = cot 30° = 3 uy

1 - 250 2 t= =5 s 10 1 H = (u sin q)t - ´ gt 2 2 600 ´

ux = 80 3 ms-1

\

3 1 = 600 ´ ´ 5 - ´ 10 ´ 25 2 2

T=

63. Since, the projectile is released its initial velocity is the same

At

as the velocity of the plane at the time of release. Take the origin at the point of release. Let x and y( = - 730 m ) be the coordinates of the point on the ground where the projectile hits and let t be the time when it hits. Then, 1 y = - v 0 t cos q - gt 2 2 where, q = 53.0°

64. Horizontal component of velocity of A is10 cos 60° or 5 ms -1 which is equal to the velocity of B in horizontal direction. They will collide at C if time of flight of the particles are equal or t A = tB 2u sin q 2h = g g

67. Acceleration of insect with respect to car a sc is a in the direction shown in figure. Absolute acceleration of insect is a θ

68. Let v be the velocity of particle when it makes 30° with horizontal. Then, y

30

Now,

or and

udy = 4 sin 30° = 2 ms-1

30° v cos 30° x

°

or

x

ux = 4 cos 30° = 2 3 ms

v

g

y

-1

ac

Component of a s along horizontal is a0 - a cos q and perpendicular to screen is a0 sin q.

2

30°

s

a s = a sc + a c

æ 3ö 2(10) 2ç ÷ è 2 ø 2u 2 sin 2 q = = 15 m h= 10 g

u = 4 ms–1

2 ´ 80 = 16 s 10

v = (80 3) 2 + ( 40) 2 » 140 ms-1

\

1 2ö æ çQ h = gtB ÷ è ø 2

65. Components of velocity of ball relative to lift are

=

v y = 80 - 10 ´ 4 = 40 ms-1

This equation gives 1 y + gt 2 2 v0 = t cos q 1 -730 + (9.8) (5) 2 2 = 202 ms-1 = 5 cos 53°

2uy

g T t = = 4 s, v x = 80 3 ms-1 4

H = 2473 m

or

145

g cos 30°

v cos 30° = u cos 60° æ 1ö (20) ç ÷ è 2 ø 20 u cos 60° = ms-1 v= = cos 30° æ 3ö 3 ç ÷ è 2 ø v2 g cos 30° = R 2 æ 20 ö ç ÷ 2 è 2ø v R= = g cos 30° 3 (10) 2 = 15.4 m

146 JEE Main Physics

Round II 1. Shown figure are when projectile is at A, then A

\

u

O

\

θ B

R/2

R 1 u2 1 (20 2) 2 OC = = sin 2q = ´ sin 2 ´ 45° 2 2 g 2 10

dH 2du 1 = =2 ´ H u 10 dH % increase in H = ´ 100 H 2 = ´ 100 = 20% 10

5. At B, S y = 0 y

= 40 m

=

A

(20 2) 2 sin 2 45° = 20 m 2 ´ 10

OA = OC 2 + CA2

\Displacement,

2

= 40 + 20

2

Time of projectile from O to A 1 æ 2u sin q ö u sin q = ç ÷= 2è g ø 2g (20 2) sin 45° =2 s 10 Displacement \ Average velocity = Time

B 30°

\ or Now,

=

=

10 m/s

x

u 2 sin 2 q AC = H = 2g

1 ay t 2 = 0 2 2 uy - 2 (10) 4 t == = s ay -10 ´ 3 /2 3 uyt +

1 2 axt 2 1æ 1 ö æ16 ö = ç10 ´ ÷ ç ÷ = 13.33 m 2è 2ø è 3 ø

AB = R =

6. Given, q1 = p / 3 = 30° Horizontal range is same if q1 + q2 = 90°

40 2 + 20 2 2

\

q2 = 90° - 30° = 60° y1 =

u 2 sin 2 30° 2g

and

y2 =

u 2 sin 2 60° 2g

\

y 2 sin 2 30° æ 1 / 4 ö 1 = =ç ÷ = y1 sin 2 60° è 3 / 4 ø 2

or

y2 =

= 10 5 ms-1

2. As initial velocity is zero. Particle will move in a straight line along anet . a

g

Further,

anet

2h 2 ´ 49 t= = = 10 = 3.16 s g 9.8

2

y1 3

7. Let v1 be the velocity of the car and v 2 be the velocity of the parcel. The parcel is thrown at an angle q from O, it reaches the mass at M. M

3. According to law of conservation of linear momentum at the

v2

highest point. mv cos q = or

m m ( -v cos q) + v1 2 2

v1 = 3 v cos q

u 2 sin 2 q 4. H = 2g 2u sin 2 q dH = du 2g

O

\

v1

cos q = =

So,

v1 10 = v 2 10 2 1 = cos 45° 2

q = 45°

A

Projectile Motion 1 2

u cos q = 2g - g or u cos q = g Squaring and adding Eqs. (i) and (ii), we have

or

8. As, h = gt 2 Distance

d = ut = u

Þ

d 2 = u2 ×

2h g

147 (\ t = 1 s) …(ii)

u 2 = 5 g 2 = 5(10) 2 = 500

2h g

u = 500 = 22.36 ms-1

or

Dividing Eq. (i) by Eq. (ii), we have tan q = 2 or q = tan -1(2).

d

13. Potential v ( x) versus x is parabolic. SHM starts for extreme position and x versus t should be cosine curve.

h

14. Maximum height, H = 9. For 5 s weight of the body is balanced by the given force. Hence, it will move in a straight line as shown.

v 2 sin 2 60° v 2 3 3 v 2 = ´ = 2g 2g 4 8g

Momentum of particle at highest point mv p = mv cos 60° = 2 Angular momentum = pH =

15. Time of flight, t =

2u sin q g

5s

R=

Horizontal range, R =

2

u sin 2 q + (u cos q) (5) g

u 2 sin 2q g

Change in angular momentum,

(50) 2 × sin 60° = + (50 ´ cos 30° ) (5) 10

|dL| = (L f - Li ) about point of projection = (mu sin q) ´

= 250 3 m 1 2

10. h = (u sin q)t - gt 2

u 2 sin 2q g

mu3 sin q sin 2q g Change in angular momentum | t| = Time of flight =

d = (u cos q)t or t = h = u sin q × u=

mv 2 v 2 3 mv 2 ´ = 2 8g 16 g

d u cos q

Torque 2

d 1 d - g× 2 u cos q 2 u cos2 q

dL mu3 sin q 2 sin q × cos q g = ´ T g 2u sin q 1 2 t = mu sin 2 q 2 =

d g cos q 2(d tan q - h)

or

16. We know that the range of projectile projected with velocity u 2 sin 2q 2 ux uy 11. R = = g g \Range µ horizontal initial velocity component (v x) In path 4 range is maximum of football has maximum horizontal velocity component in this path.

12. Time of flight of this particle, T = 4 s. If u is its initial speed and

u sin q = 2g

i. e. ,

Here,

u2 [sin(2q - q0) - sin q0 ] g cos2 q0

u = v , q = (90° + q), q0 = q v2 {sin[2(90° + q)] - q} - sin q} R= g cos2 q0 =

…(i)

After 1 s, the velocity vector particle makes an angle of 45° with horizontal, so vx = vy u cos q = (u sin q) - gt

R=

\

q is the angle of projection, then 2u sin q T=4= g or

u, making an angle q with the horizontal direction up the inclined plane, whose inclination with the horizontal direction is q0 , is

v2 [sin(180° + q) - sin q] g cos2 q0

==

2u 2 v2 sin q tan q sec q 2 = g g cos2 q0

2v 2 tan q (in magnitude) g

148 JEE Main Physics 17. Let v be the velocity acquired by the body at B which will be moving making an angle 45° with the horizontal direction. As v2 the body just crosses the well so = 40 g or

v 2 = 40 g = 40 ´ 10 = 400

or

v = 20 ms-1

s = 20 m, v = 20 ms-1

where particle moves at right angle to its direction, let its velocity be v¢. Then, v ¢x = v sin q = v x = v cos q cos q v¢ = cot q Þ sin q

æ 10 ö 400 = v 02 + 2 ç ÷ ´ 20 2 è 2ø

\

v 02 = 400 + 400 = 800

or

t= =

v = 20 2 ms-1

or

v ¢y = gt v y - v ¢y t= g

Since, or

v 2 = u 2 + 2as

\

where, q = angle of projection \ q = tan -1( a).

21. At initial point, v x = v cos q and v y = v sin q. At second point,

Taking motion of the body from A to B along the inclined plane, we have 10 u = v 0 , a = - g sin 45° = ms-2, 2 As

æ dy ö = a = tan q ç ÷ è dx ø x = 0

and

v sin q - v ¢ cos q v sin q - v cos q × cos q = g g v sin q v (1 - cot q) = g g sin q

22. For qA = 15° and qB = 75° , RA = RB

18. Given, condition h1 = h2 u12 sin 2 45° = u22 sin 2 q

Þ

u2 1 1 1 sin q = 12 sin 2 45° = × = 2 2 4 u2 2

Þ

sin q =

Þ Þ

1 2

But

hA = hB

UA 2 2 uA sin qA uA2 sin 2 qB

Again,

=

19. If a particle is projected with velocity u at an angle q with the v = u cos q = 200 cos 60° = 100 ms-1 If m is the mass of the particle, then its initial momentum at highest point in the horizontal direction = mv = m ´ 100. It means at the highest point, initially the particle has no momentum vertically upwards or downwards. Therefore, after explosion, the final momentum of the particles going upwards and downwards must be zero. Hence, the final momentum after explosion is the momentum of the third particle, in the horizontal direction. If the third particle moves mv¢ with velocity v¢ , then its momentum . According to law of 3 mv ¢ conservation of linear momentum, we have = m ´ 100 or 3 v¢ = 300 ms-1.

20. y = ax - bx2 For height of y to be maximum dy =0 dx or a - 2bx = 0 a or x= 2b y max

v 02

2

sin q i. e. ,h µ sin 2 q 2g ,

\ So,

h1 sin 2 q1 = >1 h2 sin 2 q2 sin 2 q1 > sin 2 q2

q1 > q2 2v 0 sin q Time of flight, T = g or

or \

T µ sin q T1 sin q1 = >1 T2 sin q2

or

T1 > T2

Horizontal range,

R=

\

2

\

23. Height, h =

sin 15° a3 > a2 (d) a1 = a2 , a1 = a3

w = 10N

24. If the surface is smooth, the acceleration of the block m2 will be m

(a) 5 N, zero (c) 5 N, 5 3 N

(b) Zero, N (d) 5 3 N, 5 N

30. A disc of mass 10 g is kept floating horizontally in air by firing bullets, each of mass 5 g, with the same velocity at the same rate of 10 bullets per second. The bullets rebound with the same speed in positive direction. The velocity of each bullet at the time of impact is (a) 196 cms–1 (c) 49 cms–1

m

m2 g 4 m1 + m2 2 m1g (c) m1 + 4 m2

(a)

2 m2 g 4 m1 + m2 2 m1g (d) m1 + m2

31. A ball is travelling with uniform translatory motion.

(b)

This means that

25. If a force of 250 N act on body the momentum acquired is 125 kg-m/s. What is the period for which force acts on the body? (a) 0.5 s

(b) 0.2 s

(c) 0.4 s

(d) 0.25 s

26. A point mass m is moving along inclined plane with acceleration a with respect to smooth triangular block. The triangular block is moving horizontally with acceleration a0 . The value of a is (a) g sin q + a0 cos q (c) g cos q - a0 sin q

(b) g sin q - a0 cos q (d) None of these

27. When a force F acts on a body

F2

of mass m, the acceleration produced in the body is a. If three equal forces 135° 90° F1 = F2 = F3 = F act on the m same body as shown in figure, F 3 the acceleration produced is (a) ( 2 - 1) a

(b) ( 2 + 1) a

(c) 2 a

(d) a

(b) 98 cms–1 (d) 392 cms–1 [NCERT Exemplar]

(a) it is at rest (b) the path can be a straight line or circular and the ball travels with uniform speed (c) all parts of the ball have the same velocity (magnitude and direction) and the velocity is constant (d) the centre of the ball moves with constant velocity and the ball spins about its centre uniformly

32. A satellite in force free space sweeps stationary interplanetary dust at a rate dM / dt = av, where M is the mass, v is the velocity of the satellite and a is a constant. What is the deacceleration of the satellite? (a) -2 av2 / M

(b) -av2 / M

(c) + av2 / M

(d) - av2

33. A metre scale is moving with uniform velocity. This implies F1

[NCERT Exemplar]

(a) the force acting on the scale is zero, but a torque about the centre of mass can act on the scale (b) the force acting on the scale is zero and the torque acting about centre of mass of the scale is also zero (c) the total force acting on it need not be zero but the torque on it is zero (d) neither the force nor the torque need to be zero

200 JEE Main Physics 34. A bird is sitting in a large closed cage which is placed

41. In a rocket of mass 1000 kg fuel is consumed at a rate

on a spring balance. It records a weight of 25 N. The bird (mass m = 0.5 kg) flies upward in the cage with an acceleration of 2 m/s2 . The spring balance will now record a weight of

of 40 kg/s. The velocity of the gases ejected from the rocket is 5 ´ 104 m/s. The thrust on the rocket is

(a) 24 N

(b) 25 N

(c) 26 N

(d) 27 N

velocity u = (3 $i + 4 $j) ms -1 and a final velocity v = ( -3 i$ + 4 $j) ms -1. After behind hit. The change in momentum (final momentum – initial momentum) is [NCERT] (in kg ms -1) (c) - ( 0.9 $i + 1.2 $j )

(b) - ( 0.45 $i + 0.6 $j ) (d) -5 ( $i + $j )

momentum transferred during the hit is [NCERT]

(b) 0.75 kg ms -1 (d) 14 kg ms -1

37. A frictionless inclined plane of length l having inclination q is placed inside a lift which is accelerating downward with an acceleration a ( < g). If a block is allowed to move, down the inclined plane, from rest, then the time taken by the block to slide from top of the inclined plane to the bottom of the inclined plane is (a)

2l g

(b)

2l g-a

(c)

2l g+a

(d)

2l ( g - a ) sin q

inclined surface with angle of inclination (a ). The incline is given an acceleration a to keep the block stationary. Then a is equal to

a

α

(b) g tan a (d) g cosec a

(a) g (c) g / tan a

36. In the above question the magnitude of the (a) zero (c) 1.5 kg ms -1

(b) 5 ´ 10 4 N (d) 2 ´ 10 9 N

42. A block is kept on a frictionless

35. A cricket ball of mass 150 g has an initially

(a) zero

(a) 2 ´ 103 N (c) 2 ´ 106 N

43. A horizontal force F is applied on a block of mass m placed on a rough inclined plane of F inclination q. The normal reaction N is θ

(a) mg cos q (b) mg sin q (c) mg cos q - F cos q (d) mg cos q + F sin q

44. For the system shown in figure, the pulleys are light and frictionless. The tension in the string will be

m m

38. A rocket with a lift-off mass 105 kg is blasted upward

with an initial acceleration of 5 ms–2. If g = 10 ms–2, then the initial thrust of the blast is (a) 1.5 × 102 N (c) 1.5 × 105 N

(b) 1.5 × 103 N (d) 1.5 × 106 N

39. A block of mass 3 kg rests on a horizontal frictionless xy-plane. What would be the acceleration of the block if it is subjected to two forces as shown in figure? (a) 2.5 ms–2 (c) 10 ms–2 along x-axis

y x

O 60° 5N

(b) 5 ms–2 along y-axis (d) 15 ms–2 along y-axis

ends of an unstreatchable massless cord passing over a frictionless pulley. When the masses are released, the pressure on the pulley is (b) 7.5 kgf (d) 15 kgf

(b)

3 mg sin q 2

(d) 2 mg sin q

.

45. An object is kept on a smooth inclined plane of 1 in l.

10 N

40. Two masses of 3 kg and 5 kg are suspended from the

(a) 2 kgf (c) 8 kgf

2 mg sin q 3 1 (c) mg sin q 2 (a)

The horizontal acceleration to be imparted to the inclined plane so that the object is stationary relative to the inclined is (a) g l 2 - 1 (b) g ( l 2 - 1)

(c)

g 2

l -1

(d)

g l2 - 1

46. The monkey B shown in figure is holding on to the tail of the monkey A which is climbing up a rope. The masses of the monkeys A and B are 5 kg and 2 kg respectively. If A can tolerate a tension of 30 N in its tail, what force should it apply on the rope in order to carry the monkey B with it? (Take g = 10 ms–2) (a) 105 N

(b) 108 N

(c) 10.5 N

(d) 100 N

A

B

Laws of Motion and Friction 47. The acceleration of the 500 g block in figure is

30°

54. A cricket ball of mass 150 g collides straight with a

50 g

6g downwards 13 8g downwards (c) 13 (a)

bat with a velocity of 10 ms–1. Batsman hits it straight back with a velocity of 20 ms–1. If ball remains in contact with bat for 0.1s, then average force exerted by the bat on the ball is

7g downwards 13 9g (d) upwards 13 (b)

48. An elevator and its load have a total mass of 800 kg. The elevator is originally moving downwards at 10 ms–1, it slows down to stop with constant acceleration in a distance of 25 m. Find the tension T in the supporting cable while the elevator is being brought to rest. (Take g = 10 ms–2) (b) 1600 N

(c) 9600 N

(d) 6400 N

49. A body with mass 5 kg is acted upon by a force F = ( - 3 $i + 4 $j) N. If its initial velocity at t = 0 is v = (6i$ - 12$j) ms -1, the time at which it will just have [NCERT Exemplar] a velocity along the y-axis is (b) 10 s

(c) 2 s

(d) 15 s

50. A 1000 kg lift is supported by a cable that can support 2000 kg. The shortest distance in which the lift can be stopped when it is descending with a speed of 2.5 ms–1 is [Take g = 10 ms–2] (a) 1 m

(b) 4 ms -1 (d) 8 ms -1

(a) 2 ms -1 (c) 6 ms -1

500 g

(a) never

53. A block is gently placed on a conveyor belt moving horizontally with constant speed. After 4 s the velocity of the block becomes equal to the velocity of belt. If the coefficient of friction between the block and the belt is 0.2, then velocity of the conveyor belt is

100 g

(a) 8000 N

201

(b) 2 m

(c)

5 m 32

(d)

5 m 16

51. A block of weight 5 N is pushed against a vertical wall by a force 12 N. The coefficient of friction between the wall and block is 0.6. The magnitude of the force exerted by the wall on the block is

(a) 15 N (c) 150 N

(b) 45 N (d) 4.5N

55. A block of mass 1 kg is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.5. If g = 10 ms -2 , then the magnitude of the force acting upwards at an angle of 60° from the horizontal that will just start the block moving is (a) 5 N (c) 74.6 N

(b) 5.36 N (d) 10 N

56. 100 g of an iron ball having velocity 10 ms–1 collides with wall at an angle 30° and rebounds with the same angle. If the period of contact between the ball and wall is 0.1s, then the average force experienced by the wall is (a) 10 N (c) 1.0 N

(b) 100 N (d) 0.1 N

57. A heavy uniform chain lies on a horizontal table top. If the coefficient of friction between the chain and the table surface is 0.25 then the maximum fraction of length of the chain, that can hang over one edge of the table is (a) 20% (c) 35%

(b) 25% (d) 15%

58. A block A with mass 100 kg is resting on another 12 N

(a) 12 N

(b) 5 N

(c) 7.2 N

(d) 13 N

52. A blumb bob is hung from the ceiling of a train

block B of mass 200 kg. As shown in figure, a horizontal rope tied to a wall holds it. The coefficient of friction between A and B is 0.2 while coefficient of friction between B and the ground is 0.3. The minimum required force F to start moving B will be

compartment. The train moves on an inclined track of inclination 30° with horizontal. Acceleration of train up the plane is a = 9/2. The angle which the string supporting the bob makes with normal to the ceiling in equilibrium is (a) 30°

æ 3ö æ 2 ö (b) tan -1 ç ÷ (c) tan -1 ç ÷ (d) tan -1 (2) è 3ø è 2 ø

A

B

(a) 900 N (c) 1100 N

F

(b) 100 N (d) 1200 N

202 JEE Main Physics 59. A maximum speed that can be achieved without

66. A smooth inclined plane of length L having

skidding by a car on a circular unbanked road of radius R and coefficient of static friction m is

inclination q with the horizontal is inside a lift which is moving down with retardation a. The time taken by a body to slide down the inclined plane, from rest, will be

(a) m Rg

(b) Rg m

(c) m Rg

(d) m Rg

60. A chain lies on a rough horizontal table. It starts

(a)

(b)

sliding when one-fourth of its length hangs over the edge of the table. The coefficient of static friction between the chain and the surface of the table is

2L ( g + a ) sin q

2L ( g - a ) sin q

(c)

2L g sin q

(d)

2L a sin q

(a)

1 2

(b)

1 3

(c)

1 4

(d)

1 5

61. A fireman of mass 60 kg slides down a pole He is pressing the pole with a force of 600 N. The coefficient of friction between the hands and the pole is 0.5 with what acceleration with the fireman slide down? ( g = 10 m/s2 ) 2

2

(a) 1 m/s (c) 10 m/s2

(b) 2.5 m/s (d) 5 m/s2

62. The minimum velocity (in ms–1) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is (a) 60 ms–1 (b) 30 ms–1

(c) 15 ms–1

(d) 25 ms–1

63. A block of mass 3 kg resting on a horizontal surface. A force F is applied on the block as shown in figure. If 1 what coefficient of friction between the block be 2 3 can be the maximum value of force F so that block does not start moving? (Take g = 10 ms–2)

67. A wooden box of mass 8 kg slides down an inclined plane of inclination 30° to the horizontal with a constant acceleration of 0.4 ms–2. What is the force of friction between the box and inclined plane? (g = 10 ms–2) (a) 36.8 N

(b) 76.8 N

(c) 65.6 N

(d) 97.8 N

68. The coefficient of kinetic friction between a 20 kg box and the floor is 0.40. How much work does a pulling force do on the box in pulling it 8.0 m across the floor at constant speed? The pulling force is directed 37° above the horizontal (a) 343 J (c) 14.4 J

(b) 482 J (d) None of these

69. A car starts from rest to cover a distance s. The coefficient of friction between the road and the tyres is m. The minimum time in which the car can cover the distance is proportional to (a) m

(b) m

(c) 1/m

(d) 1/ m

70. A block of mass m lying on a rough horizontal plane is acted upon by a horizontal force P and another force Q inclined at an angle q to the vertical. The block will remain in equilibrium if the coefficient of friction between it and the surface is

F 60°

(a) 20 N

√3 kg

(b) 10 N

(c) 12 N

(d) 15 N

Q

θ

64. A heavy uniform chain lies on horizontal table top. If the coefficient of friction between the chain and the table surface is 0.25, then the maximum fraction of the length of the chain that can hang over one edge of the table is (a) 20%

(b) 25%

(c) 35%

(d) 15%

65. A block moves down a smooth inclined plane of inclination q. Its velocity on reaching the bottom is v. If it slides down a rough inclined plane of same inclination, its velocity on reaching the bottom is v/n, where n is a number greater than 1. The coefficient of friction is given by 1ö æ (a) m = tan q ç1 - 2 ÷ è n ø 1ö æ (c) m = tan q ç1 - 2 ÷ è n ø

1 /2

1ö æ (b) m = cot q ç1 - 2 ÷ è n ø 1ö æ (d) m = cot q ç1 - 2 ÷ è n ø

P

P + Q sin q mg + Q cos q P + Q cos q (c) mg + Q sin q

(a)

71. A partly hanging uniform chain of length L is resting on a rough horizontal table. l is the maximum possible length that can hang in equilibrium. The coefficient of friction between the chain and table is l L-l l (c) L

(a) 1 /2

P cos q + Q mg - Q sin q P sin q - Q (d) mg - Q cos q (b)

(b)

L l

(d)

lL L+ l

203

Laws of Motion and Friction 72. A box of mass m kg is placed on the rear side of an

(a) 4 kg (c) 9.78 kg

(b) 8 kg (d) It could be any value

73. A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The static coefficient of friction between the block and the slab is 0.60 while the kinetic coefficient of friction is 0.40. The 10 kg block is acted upon by a horizontal force of 100 N. If g = 9.8 ms–2, the resulting acceleration of the slab will be 100 N No friction

(a) 1.47 ms–2 (c) 9.8 ms–2

10 kg 40 kg

C L

mm g M 2mm (c) g M

mm g (M + m) 2mm (d) g (M + m) (b)

(a)

78. A man weighing 60 kg is standing on a trolley weighing 240 kg. The trolley is resting on frictionless horizontal rails. If the man starts walking on the trolley with a velocity of 1 ms -1, then after 4 s, his displacement relative to the ground is (a) 6 m

(b) 4.8 m

74. A body of mass 40 kg resting on a rough horizontal surface is subjected to a force P which is just enough to start the motion of the body. If m s = 0.5, m k = 0.4, g = 10 ms -2 and the force P is continuously applied on the body, then the acceleration of the body is (b) 1 ms–2 (d) 2.4 ms–2

is connected by a cord, passing over a light frictionless pulley to a hanging block of mass 5 kg. The coefficient of kinetic friction between the block and the surface is 0.5. Tension in the cord is (Take g = 9.8 ms–2) (a) 49 N (c) 36.75 N

(b) 36 N (d) 2.45 N

plane with force acting on m1 parallel to the inclined plane. Find the contact force between m2 and m3.

surface of an inclined plane at 45° is 0.5 if g = 9.8 m /s2. The acceleration of the body downwards in m/s2 is 4.9 2

(b) 4.9 2

(c) 19.2 2

support. The tension in the rope at a distance x from the rigid support is æ L - xö (b) ç ÷ Mg è L ø

æ L ö (c) ç ÷ Mg è L - xø

(d)

m3 m1 F

(d) 4.9

76. A rope of length L and mass M is hanging from a right

(a) Mg

(d) 2.4 m

80. Three blocks are placed at rest on a smooth inclined

75. The coefficient of friction between a body and the

(a)

(c) 3.2 m

79. A block of mass 5 kg, resting on a horizontal surface,

(b) 1.69 ms–2 (d) 0.98 ms–2

(a) zero (c) 2 ms–2

B

A

T

open truck accelerating at 4 ms -2 . The coefficient of friction between the box and the surface below it is 0.4. The net acceleration of the box with respect to the truck is zero. The value of m is [Given g = 10 ms -2 ]

x Mg L

77. A plate of mass M is placed on a horizontal frictionless surface (see figure) and a body of mass m is placed on this plate. The coefficient of dynamics friction between this body and the plate is m. If a force 2 mmg is applied to the body of mass m along the horizontal, the acceleration of the plate will be

(a)

m2

θ

( m1 + m2 + m3 ) F m3

(c) F - ( m1 + m2 ) g

(b)

m3F m1 + m2 + m3

(d) None of these

Connected Body Motion 81. Three equal weight 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 weight B and C is (a) zero (b) 13 N (c) 3.3 N (d) 19.6 N

A B

C

204 JEE Main Physics 82. Two blocks are connected by a string as shown in the diagram. The upper block is hung by another string. A force applied on the upper string produces an acceleration of 2 m/s2 in the upward direction in both the blocks. If T and T ¢ be the tensions in the two parts of the string, then ( g = 9.8 m /s2 ) (a) T (b) T (c) T (d) T

= 70.8 N and T ¢ = 47.2 N = 58.8 N and T ¢ = 47.2 N = 70.8 N and T ¢ = 58.8 N = 70.8 N and T ¢ = 0

v sin q

T

v

smooth horizontal plane with the help of a light rope which moves with a velocity v as shown in figure. The horizontal velocity of the block is

(c)

2 kg

4 kg

83. A block is dragged on a

(a) v

F T

F

θ

(b) v sin q v (d) cos q

m

84. A mass of 3 kg descending vertically downward supports a mass of 2 kg by means the end of 5 s, the string breaks. How much higher the 2 kg mass will go further?

87. In the figure, the ball A is released from rest when the spring is at its natural length. For the block B of mass M to leave contact with the ground at same stage, the minimum mass of A must be (a) 2 M (b) M M (c) 2 (d) a function of M and the force constant of the spring

m

M

88. A shell is fired from a cannon with velocity v ms–1 at an angle q with the horizontal direction. At the highest point in its path it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon and the speed in m/s of the piece immediately after the explosion is (a) 3 v cos q (b) 2 v cos q 3v (c) cos q 2 3 v cos q (d) 2

89. A body of weight 2 kg is suspended as shown in figure. The tension T1 in the horizontal string (in kg-wt) is 30°

m m

T

2m

2 kg-wt

(a) 4.9 m (b) 9.8 m (c) 19.6 m (d) 2.45 m

85. Two bodies of masses m1 and m2 are connected by a

light, inextensible string which passes over a frictionless pulley. If the pulley is moving upward with uniform acceleration g, then the tension in the string is 4 m1m2 g m1 + m2 m1m2 (c) g m1 + m2

(a)

m1m2 g 4 m1m2 m - m2 2 (d) 1 g m1 + m2

(b)

(a) 2 / 3

(b) 3/ 2

(c) 2 3

(d) 2

90. Two blocks of masses m and 2 m are connected by a light string passing over a frictionless pulley. As shown in the figure, the mass m is placed on a smooth inclined plane of inclination 30° and 2 m hangs vertically. If the system is released, the blocks move with an acceleration equal to

m 2m

86. In the given arrangement, n number of equal masses are connected by strings of negligible masses. The tension in the string connected to nth mass is mMg nm + M (c) mg

(a)

mMg nmM (d) mng

(b)

30°

g 4 g (c) 2 (a)

(b)

g 3

(d) g

Laws of Motion and Friction

205

91. Refer to the system shown in figure. The ratio of tensions T1 and T2 is T1 T1

m1

T2

m2

T2

m1 m1 + m2 m (c) 1 m2

m2 m1 + m2 m (d) 2 m1

(a)

3 kg

(b)

1 kg C

(a)

weight is suspended by a spring balance. Weights of 1 kg and 5 kg are attached to the opposite ends of a string passing over the pulley and move with acceleration because of gravity. During their motion, the spring balance reads a weight of (a) 6 kg (b) less than 6 kg (c) more than 6 kg (d) may be more or less than 6 kg

T2

(b)

g 6

(c)

g 12

5 kg

6N

5 kg

m3

T3

(a) 3.5 N

(b) 2.5 N

(c) 7 N

(d) 5 N

96. A trolley T (mass 5 kg) on a horizontal smooth surface is pulled by a load L (2 kg) through a uniform rope ABC of length 2 m and mass 1kg. As the load falls from BC = 0 to BC = 2 m, its acceleration (in ms–2) changes from (Take g = 10 ms–2)

(a) 20 N (b) 40 N (c) 10 N (d) 32 N

A

B

C L

94. Refer to the system shown in figure. The acceleration of the masses is

Round II Only One Correct Option acceleration a towards the right along a straight horizontal path. Which of the following represent the surface of the liquid? a

a

a

(a) 20/6 to 20/5 (c) 20/5 to 30/6

(Mixed Bag)

ends of a light inextensible string that goes over a frictionless pulley. The acceleration of the masses and the tension in the string when the masses are released, are respectively [NCERT] (a) 2 m/s2 and 90 N (b) 4 m/s2 and 90 N (c) 2 m/s2 and 60 N

(B)

(b) B

(C)

(c) C

(b) 20/8 to 30/8 (d) None of these

2. Two masses 8 kg and 12 kg are connected at the two

1. A vessel containing water is given a constant

(A)

(d)

7 kg

1 kg

T

(a) A

g 9

contact with each other on a smooth surface. If a force of 6 N is applied on a heavier mass the force on the lighter mass is

by massless strings as shown on a frictionless table in figure. They are pulled with a force T3 = 40 N. If m1 = 10 kg, m2 = 6 kg and m3 = 4 kg, the tension T2 will be m2

g 3

A 5 kg

95. Two block of masses 7 kg and 5 kg are placed in

93. Three blocks of masses m1, m2 and m3 are connected

T1

T2

T1 T1

92. In the figure a smooth pulley of negligible

m1

B

(D)

(d) D

(d) 4 m/s2 and 99 N

206 JEE Main Physics 3. Two elastic blocks P and Q of equal masses m and

8. A circular disc with a groove along its diameter is

connected by a massless spring rest on a smooth horizontal surface, as in figure. A third block R of the same mass M strikes the block P. After the collision, P and Q will

placed horizontally. A block of mass 1 kg is placed as shown. The coefficient of friction between the block 2 and all surfaces of groove in contact is m = , the disc 5 has an acceleration of 25 m/s2 . Find the acceleration of block with respect to disc

R

Q

P

(a) always move in same direction (b) sometimes move in same direction and sometime move in opposite directions (c) always move in opposite directions (d) be at rest with respect to each other

4. Two persons are holding a rope of negligible weight tightly at its ends so that it is horizontal. A 15 kg weight is attached to rope at the mid-point which now no more remains horizontal. The minimum tension required to completely straighten the rope is (a) 15 kg (c) 5 kg

(b) 15/2 kg (d) infinitely large

(a) 19.6 N (b) 25 N (c) 10.6 N (d) 10 N

T

T1

T

6 kg

coefficient of static friction is 0.4. If a force of 2.8 N is applied on the block parallel to floor, the force of friction between the block and floor (Taking g = 10 ms–2) is (b) 8 N

(c) 2 N

(d) zero

7. The pulley and strings shown in figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle q should be

θ

√2 m m

(a) 0º (c) 45º

m

(b) 30º (d) 60º

(a) 10 m/s2 (c) 20 m/s2

4 3 cos θ = — , sin θ = — 5 5

(b) 5 m/s2 (d) 1 m/s2

9. A cylinder roll up an inclined plane, reaches some height and then rolls down (without slipping throughout these motions). The directions of frictional force acting on the cylinder are

10. When forces F1, F2 , F3 are acting on a particle of mass

m 4 kg

6. A block of mass 2 kg is placed on the floor. The

(a) 2.8 N

θ

(a) up the inclined while ascending and down the incline while descending (b) up the incline while ascending as well as descending (c) down the incline while ascending and up the incline while descending (d) down the incline while ascending as well as descending

5. Two bodies of mass 4 kg and 6 kg are attached to the ends of a string passing over a pulley. The 4 kg mass is attached to the table by another string. The tension in this string T1 is

a = 25 m/s2

m such that F2 and F3 are mutually perpendicular, then the particle remains stationary. If the force F1 is now removed, then the acceleration of the particle is (a) F1/ m

(b) FF 2 3/ mF1

(c) ( F2 - F3 ) / m

(d) F2 / m

11. A particle moves in a circular path with decreasing speed. Choose the correct statement. (a) (b) (c) (d)

Angular momentum remains constant Acceleration a is towards the centre Particle moves in a spiral path with decreasing radius The direction of angular momentum remains constant

12. A spring balance, A reads 2 kg with a block m A suspended from it. A balance B reads 5 kg when a beaker filled with liquid is put on the pan of the balance. The two balances B are now so arranged that the hanging mass is inside the liquid as shown in figure. In this situation (a) the balance A will read more than 2 kg (b) the balance B will read more than 5 kg (c) the balance A will read less than 2 kg and B will read more than 5 kg (d) the balance A and B will read 2 kg and 5 kg

Laws of Motion and Friction

207

13. While waiting in a car at a stoplight and 80 kg man

19. A body of mass M is kept on a rough horizontal

and his car are suddenly accelerated to a speed of 5 ms–1 as a result or rear end collision. If the time of impact is 0.4 s, find the average force on the man

surface (friction coefficient m). A person is trying to pull the body by applying a horizontal force but the body is not moving. The force by the surface on the body is F, where

(a) 100 N

(b) 200 N

(c) 500 N

(d) 1000 N

14. A body of mass m is suspended by two strings making angle a and b with the horizontal as shown in figure. Tensions in the two strings are

T2

(b) F = mMgF

(c) Mg £ f £ Mg 1 + m2

(d) Mg ³ f ³ Mg 1 + m2

20. A 5 kg stationary bomb is exploded in three parts having mass 1 : 1 : 3 respectively. Parts having same mass move in perpendicular directions with velocity 39 ms–1, then the velocity of bigger part will be

T1 β

(a) F = Mg

10 ms -1 2 15 (d) ms -1 2

(a) 10 2 ms -1

α

(b)

(c) 13 2 ms -1 mg

21. The upper half of an inclined plane with inclination f

mg cos b = T2 sin ( a + b ) mg sin b (b) T1 = = T2 sin ( a + b ) mg cos b mg cos a (c) T1 = ; T2 = sin ( a + b ) sin ( a + b )

is perfectly smooth, while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if coefficient of friction for the lower half is given by

(a) T1 =

(a) 2 sin f (c) 2 tan f

(d) None of the above

22. An insect crawls up a hemispherical surface very

15. A block of mass 15 kg is resting on a rough inclined plane as shown in figure. The block is tied by a horizontal string which has a tension of 50 N. The coefficient of friction between the surfaces of contact is (g = 10 ms–2) (a) 1/2

(b) 3/4

(c) 2/3

T M

α

(d) 1/4

whirled round in a circle of radius 1.5 m with speed 40 rev/min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N? [NCERT] (b) 300 N

(c) 2000 N

(d) 250 N

17. The mass of a body measured by a physical balance in a lift at rest is found to be m. If the lift is going up with an acceleration a, its mass will be measured as aö æ (a) m ç1 - ÷ è gø

aö æ (b) m ç1 + ÷ è gø

(c) m

(d) zero

a 45° rough inclined plane as it takes to slide down a perfectly smooth 45° incline. The coefficient of kinetic friction between the object and the incline is 1 1 - n2

(b) 1 -

1 n2

(c) 1 -

1 n2

(a) cot a = 3 (c) cosec a = 3

(b) sec a = 3 (d) None of these

23. A mass of 6 kg is suspended by a rope of length 2 m from a ceiling. A force of 50 N is applied in the horizontal direction at the mid-point of the rope. The angle made by the rope, with the vertical, in equilibrium position will be (take g = 10 ms -2 , neglect the mass of the rope) (a) 90º (c) 50º

(b) 60º (d) 40º

24. A rope of mass 0.1 kg is connected at the same height

18. A given object takes n times more time to slide down

(a)

slowly, figure. The coefficient of friction between the insect and the surface is 1/3. If the line joining the centre of the hemispherical surface to the insect makes an angle a with the vertical, the maximum possible value of a is given by

45°

16. A stone of mass 0.25 kg tied to the end of a string is

(a) 200 N

(b) 2cos f (d) tan f

(d)

1 1 - n2

of two opposite walls. It is allowed to hang under its own weight. At the constant point between the rope and the wall, the rope makes an angle q = 10° with respect to horizontal. The tension in the rope at its mid-point between the wall is (a) 2.78 N (c) 2.82 N

(b) 2.56 N (d) 2.71 N

208 JEE Main Physics 25. A 24 kg block resting on a floor has a rope tied to its

32. The motion of a particle of mass m is given by x = 0 for

top. The maximum tension, the rope can withstand without breaking is 310 N. The minimum time in which the block can be lifted a vertical distance of 4.6 m by pulling on the rope is

t < 0 s, x( t) = A sin 4 pt for 0 < t < (1/ 4) s ( A > 0), and x = 0 for t > (1 / 4) s. Which of the following statements is true ? [NCERT Exemplar)

(a) 1.2 s (c) 1.7 s

(b) 1.3 s (d) 2.3 s

(b) The particle is acted upon by on impulse of magnitude 4 p2 A m at t = 0 s and t = (1 / 4 ) s

26. Two small balls of same size and masses m1 and m2 (m1 > m2 ) are tied by a thin weightless thread and dropped from a certain height. Traing upward bouyancy force F into account, the tension T of the thread during the flight after the motion of the ball becomes uniform will be (a) ( m1 - m2 ) g (c) ( m1 + m2 ) g

(a) The force at t = (1 / 8) s on the particle is - 16 p2 Am

(b) ( m1 - m2 ) g / 2 (d) ( m1 + m2 ) g / 2

(c) The particle is not acted upon by any force (d) There is no impulse acting on the particle

33. Mass m1 moves on a slope making an angle q with the

horizontal and is attached to mass m2 by a string passing over a frictionless pulley as shown in figure. The coefficient of friction between m1 and the sloping surface is m.

27. A solid disc of mass M is just held in air horizontal by throwing 40 stones per sec vertically upwards to strike the disc each with a velocity 6 ms–1. If the mass of each stone is 0.05 kg. What is the mass of the disc? (g = 10 ms–2) (a) 1.2 kg

(b) 0.5 kg

(c) 20 kg

m1 m2 B

θ

(d) 3 kg

28. If coefficient of friction between an insect and bowl is m and radius of the bowl is r, the maximum height to which the insect can crawl in the bowl is é 1 ù (a) r ê1 ú êë 1 + m2 úû

(b)

(c) r 1 + m2

(d) r [ 1 + m2 - 1]

r 1 + m2

29. A block of mass m is pushed with a velocity u towards a movable wedge of mass nm and height h, figure. All the surfaces are smooth. The minimum value of u for which the block will reach the top of wedge is 1ö æ (a) 2 gh ç1 - ÷ è nø

1ö æ (b) 2 gh ç1 + ÷ è nø

(c) 3 gh

(d) 2 gh

Which of the following statements are true ? [NCERT Exemplar]

(a) If m2 > m1 sin q, the body will move up the plane (b) If m2 > m1 (sin q + m cos q), the body will move up the plane (c) If m2 < m1 (sin q + m cos q) , the body will move up the plane (d) If m2 < m1 (sin q - m cos q), the body will move down the plane

34. A man of mass M is standing on a board of mass m. The friction coefficient between the board and the floor is m, shown in figure. The maximum force that the man can exert on the rope so that the board does not move is T

T

F

More Than One Correct Option 30. 80 railway wagons all of same mass 5 × 103 kg are pulled by an engine with a force of 4 × 105 N. The tension in the coupling between 30th and 31st wagon from the engine is (a) 25 × 104 N (c) 20 × 104 N

(b) 40 × 104 N (d) 32 × 104 N

31. A gardner waters the plants by a pipe of diameter

(c) 1.27 ´ 10

–4

N

R T f (m + M)g

(b) 1.27 ´ 10 –2 N

(a) m ( m + M ) g m (m + M ) g (b) m +1 m (m + M ) g (c) m -1

(d) 0.127 N

(d) None of the above

1 mm. The water comes out at the rate of 10 cm3 s–1. The reactionary force exerted on the hand of the gardner is (a) zero

T

Laws of Motion and Friction

Comprehension Based Questions A body of mass 10 kg is lying on a rough horizontal surface. The coefficient of friction between the body and horizontal surface is 0.577. When the horizontal surface is inclined gradually, the body just begins to slide at a certain angle a. This is called angle of repose. When angle of inclination is increased further, the body slides down with some acceleration.

35. The minimum force required just to slide the block on the horizontal surface is (b) 100 N

(c) 100 kg

(d) 57.7 kg

36. The minimum force required just to move the body up the incline is (a) 100 N

(b) 57.7 N

(c) 111.5 N

(d) 157.7 N

37. Value of angle of repose in this case is (a) 60º

(b) 57.7º

(c) 5.77º

(d) 30º

Passage II A force that acts on a body for a very short time is called impulsive force. Impulse measures the effect of the force. It is the product of force and time for which the force acts. Impulse is measured by the change in momentum of the body. For a given change in momentum, Fav ´ t = constant. By increasing the tie (t) of impact, we can reduce the average force fav . Read the above passage carefully and answer the following questions (g = 10 ms -2).

38. If the impact lasts for 0.1s, force exerted by the impinging ball on the ground is (a) 45.2 N (c) 42.5 N

impact is (b) 42.5 N

(c) 84.5 N

(d) 45.2 N

40. The loss of energy during impact is Two small balls of same size and and masses and (a) 75 J

(b) 5 J

(c) 57 J

(d) 6.38 J

41. A cricket player lowers his hands while catching a ball, because (a) ball is heavy (b) ball is coming with high speed (c) it increases the time of impact and reduces the impact of force on his hands (d) None of the above

42. A ball of mass 250 g falls from a height of 5 m above the ground, and rebounds to a height of 2.45 m. The impulse on collision is (a) 4.25 kgms–1 (c) 52.4 kgms–1

(b) 45.2 kgms–1 (d) 54.2 kgms–1

Question No. 43 to 51 are Assertion-Reason type. Each of these contains two Statements: Statement I (Assertion), Statement II (Reason). Each of these questions also has four alternative choice, only one of which is correct. You have to select the correct choices from the codes (a), (b), (c) and (d) given below (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 correct explanation of the Assertion (c) If Assertion is true but Reason is false (d) If Assertion is false but the Reason is true

43. Assertion An electric fan continues to rotate for some time after the current is switched off. Reason It is because of inertia of rest.

44. Assertion Force is required to move a body uniformly along a circle. Reason When the motion is uniform, acceleration is zero.

45. Assertion Angle of repose is equal to angle of limiting friction. Reason When a body is just at the point of motion, the force of friction in this stage is called as limiting friction.

46. Assertion A string can never remain horizontal, when loaded at the middle, howsoever large the tension may be. Reason For horizontal spring, angle with vertical, w w = =¥ q = 90° Þ T = 2 cos q 2 cos 90°

47. Assertion Use of ball bearing between two moving

(b) 45.2 kg-wt (d) 42.5 kg-wt

39. The force exerted by the ground on the ball during (a) 90.4 N

Assertion and Reason Directions

Passage I

(a) 57.7 N

209

parts of machine is a common practice. Reason Ball bearings reduce vibration and provide good stability.

48. Assertion The maximum speed with which a vehicle can go round a level curve of diameter 20 m without skidding is 10 ms–1, given m = 0.1. Reason It follows from v £ m rg .

49. Assertion A man is closed cabin falling freely does not experience gravity. Reason Inertial and gravitational equivalence.

mass

have

50. Assertion A cyclist always bends inwards while negotiating a curve. Reason By bending cyclist lower his centre of gravity.

51. Assertion Aeroplane always fly at low altitudes. Reason According to Newton’s third law of motion for every action there is an equal and opposite reaction.

210 JEE Main Physics Previous Years’ Questions 52. A force of (5 + 3 x) N acting on a body of mass 20 kg along the x-axis displaces it from x = 2 m to x = 6 m. The work done by the force is [UP SEE 2009] (a) 20 J

(b) 48 J

(c) 68 J

(d) 86 J

53. Block A of mass m and block B of mass 2m are placed on a fixed triangular wedge by means of a massless, inextensible string and a frictionless at 45° to the horizontal on both the side. If the coefficient of friction between the block A and the wedge is 2/3 and that between the block B and the wedge is 1/3 and both blocks A and B are released from rest, the [UP SEE 2008] acceleration of A will be

m

2m

45°

45°

58. A block of mass m is connected to another block of mass M by a massless spring of constant k. The blocks are kept on a smooth horizontal plane. Initially, the blocks are at rest and the spring is unstretched. Then a constant force F starts acting on the block of mass M to pull it. Find the force on the [AIEEE 2007] block of mass m. MF m+M (M + m) F (c) m

mF M mF (d) m+M

(a)

(b)

59. A player caught a cricket ball of mass 150 g moving at

a rate of 20 ms–1. If the catching process is completed in 0.1s, the force of blow exerted by the ball on the [AIEEE 2006] hands of the player is equal to (a) 3 N (c) 300 N

(b) –30 N (d) 150 N

60. A ball of mass 0.2 kg is thrown vertically upwards by (a) –1 m/s2 (b) 1.2 m/s2

(c) 0.2 m/s2

(d) 0 m/s2

54. A block of mass 5 kg is moving horizontally at a speed

of 1.5 ms–1. A vertically upward force 5 N acts on it for 4 s. What will be the distance of the block from the point where the force starts acting? [BVP Engg. 2008] (a) 2 m

(b) 6 m

(c) 8 m

(d) 10 m

55. A block B is pushed momentarily

B

v

along a horizontal surface with an initial velocity v. If m is the coefficient of sliding friction between B and the surface, block B will come to rest after a time [UP SEE 2007] v (a) gm

gm (b) v

g (c) v

v (d) g

56. An ice cart of mass 60 kg rests on a horizontal snow patch with coefficient of static friction 1/3 Assuming that there is no vertical acceleration, find the magnitude of the maximum horizontal force required to move the ice cart. (g = 9.8 ms–2) [BVP Engg. 2007] (a) 100 N (c) 209 N

applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes upto 2 m height further, find the magnitude of the force [AIEEE 2006] [take g = 10 ms–2] (a) 16 N (c) 22 N

(b) 20 N (d) 4 N

61. A bomb of mass 16 kg at rest explodes into two pieces of masses 4 kg and 12 kg. The velocity of the 12 kg mass is 4 ms–1. The kinetic energy of the other mass is [AIEEE 2006] (a) 288 J (c) 96 J

(b) 192 J (d) 144 J

62. Two masses M and M/2 are joined together by means of light inextensible string passed over a frictionless pulley as shown in figure. When the bigger mass is released, the small one will ascend with an acceleration of [Kerala CET 2005]

(b) 110 N (d) 196 N

57. A uniform metal chain is placed on a rough table such that one end of it hangs down over the edge of the table. When one-third of its length hangs over the edge, the chain starts sliding. Then, the coefficient of [BVP Engg. 2007] static friction is 3 4 2 (c) 3

(a)

1 4 1 (d) 2

(b)

M/2

M

g (a) 3 (c) g

3g 2 g (d) 2 (b)

Laws of Motion and Friction

211

63. A block of mass m is at rest under the action of force F

67. A man drags a block through 10 m on rough surface

against a wall as shown in figure. Which of the following statements is incorrect? [IIT JEE 2005]

(m = 0.5). A force of 3 kN acting at 30° to the horizontal. The work done by applied force is [Orissa JEE 2011]

a

(a) zero

(b) 15 kJ

(c) 5 kJ

which makes an angle 60° with the vertical and it reaches the ground in t1 seconds. Another block is dropped vertically from the some point and reaches the ground in t2 seconds, then the ratio of t1 : t2 is

F

[Kerala CET 2011]

(a) f = mg [where f is the friction force] (b) F = N [where N is the normal force] (c) F will not produce torque (d) N will not produce torque

(a) 1 : 2

(b) 2 : 1

(c) 1 : 3

angle of inclination a as shown in figure. The incline is given an acceleration a to keep to block stationary. [AIEEE 2005] Then, n is equal to

angle 45° with the horizontal and the coefficient of friction is m. The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define N = 10 m, [IIT JEE 2011] then N is mg/√2 m

F1 a

µ mg/√2

α

(b) g

(c) g cosec a

45°

(d) g / tan a

65. A particle of mass m is at rest at the origin at time

g = 0. It is subjected to a force F ( t) = f0 e- bt in the x direction. Its speed v( t) is depicted by which of the following curves? [AIEEE 2012] F0 —– mb

(a) 2

(b) 4

(c) 5

wedge as shown in figure.

M

[Orissa JEE 2011]

M

(b) 53°

37°

v(t)

v(t) t

-2

F0 —– mb

(c)

71. A force vector applied on a mass is represented as $ and acceleration with 1 m/s2 . What F = 6 $i - 8 $j + 10 k

(d) v(t)

will be mass of the body?

v(t) t

(a) 10 2 kg (c) 10 kg

t

66. A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass m and radius R. Assuming pulley to be the perfect uniform circular disc, the acceleration of the mass m. If the string does not slip on the pulley, is [AIEEE 2011]

(a) g g 3

2 g 3 3 (d) g 2 (b)

(b) 2 ms -2 (d) 10 ms -2

(a) 1 ms (c) 0.5 ms -2

t

F0 —– mb

(c)

(d) 6

70. The acceleration of system of two bodies over the

F0 —– mb

(a)

(d) 1 : 2

69. A block is moving on an inclined plane making an

64. A block is kept on a frictionless inclined surface with

(a) g tan a

(d) 10 kJ

68. A block at rest slides down a smooth inclined plane

a

(b) 2 10 kg (d) 20 kg

72. The figure shows the position-time (x-t) graph of one-dimensional motion of a body of mass 0.4 kg. The magnitude of each impulse is [AIEEE 2010] 2 x(m) t 0

(a) 0.2 Ns

2

4

6

(b) 0.4 Ns

8

10 12 14 16

(c) 0.8 Ns

(d) 1.6 Ns

212 JEE Main Physics 73. Two fixed frictionless inclined planes making an

75. A light string passes over a frictionless pulley. To one

angle 30° and 60° with the vertical are shown in the figure. Two blocks A and B are placed on the two planes. What is the relative vertical acceleration of A with respect to B ? [AIEEE 2010]

of its ends a mass of 6 kg is attached. To its other end a mass of 10 kg is attached. The tension in the thread will be [AMU Engg. 2010]

A B

6 kg 60°

30°

10 kg

(a) 4.9 m/s2 in vertical direction (b) 4.9 m/s2 in horizontal direction (c) 9.8 m/s2 in vertical direction (d) zero

(a) 24.5 N

pulled up by men on an inclined plane at angle of 45° as shown in 45° figure. The coefficient of static friction is 0.5. Each man can only apply a maximum force of 500 N. Calculate the number of men required for the block to just start moving up the plane.

(d) 73.5 N

x-axis with a speed of 5.00 ms -1. The magnitude of its momentum is retarded as [AIEEE 2008] (a) 17.565 kg ms -1 (c) 17.57 kg ms -1

(c) 5

(b) 17.56 kg ms -1 (d) 17.6 kg ms -1

77. Rocket pollution is associated with

[J & K CET 2010]

(a) the conservation of angular momentum (b) the conservation of mass (c) the conservation of mechanical energy (d) Newton’s III law of motion

[AMU Engg. 2010]

(b) 15

(c) 79 N

76. A body of mass m = 3.613 kg is moving along. The

74. A block of mass 200 kg is being

(a) 10

(b) 2.45 N

(d) 3

Answers Round I 1. 11. 21. 31. 41. 51. 61. 71. 81. 91.

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

2. 12. 22. 32. 42. 52. 62. 72. 82. 92.

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

2. 12. 22. 32. 42. 52. 62. 72.

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

3. 13. 23. 33. 43. 53. 63. 73. 83. 93.

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

4. 14. 24. 34. 44. 54. 64. 74. 84. 94.

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

5. 15. 25. 35. 45. 55. 65. 75. 85. 95.

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

6. 16. 26. 36. 46. 56. 66. 76. 86. 96.

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

7. 17. 27. 37. 47. 57. 67. 77. 87.

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

8. 18. 28. 38. 48. 58. 68. 78. 88.

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

9. 19. 29. 39. 49. 59. 69. 79. 89.

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

10. 20. 30. 40. 50. 60. 70. 80. 90.

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

8. 18. 28. 38. 48. 58. 68.

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

9. 19. 29. 39. 49. 59. 69.

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

10. 20. 30. 40. 50. 60. 70.

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

Round II 1. 11. 21. 31. 41. 51. 61. 71.

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

3. 13. 23. 33. 43. 53. 63. 73.

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

4. 14. 24. 34. 44. 54. 64. 74.

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

5. 15. 25. 35. 45. 55. 65. 75.

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

6. 16. 26. 36. 46. 56. 66. 76.

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

7. 17. 27. 37. 47. 57. 67. 77.

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

the Guidance Round I 6. Initial mass of the rocket m = 20000 kg

1. When force F is applied on 2m from left, contact force,

When force F is applied on m from right, contact force 2m 2F F2 = F= \ F1 : F2 = 1 : 2 m + 2m 3

T - mg = ma T = mg + ma = m( g + a)

= 29.6 ´ 10 4 N = 2.96 ´ 10 5 N

7. The mass m is not moving with respect to the lift and also has

u = 2g (Hmax )

0.2 m

= 2 ´ 10 ´ 2 = 2 10 m/s

This velocity is supplied to the ball by the hand and initially the hand was at rest it requires this velocity is distance of 0.2 m, it requires this velocity is distance of 0.2 m u2 40 a= = = 100 m/s2 \ 2 s 2 ´ 0.2 So upward force on the ball F = m ( g + a) = 0.2 (10 +100) = 0.2 ´ 110 = 22 N

3. Since, P = (M + m) a Now as in free body diagram of block,

α ma

N

or

no tendency to move. Hence, friction force acting on it is equal to zero. 2 m1m2 2 ´ 4 ´ 2 ´ 10 160 8. T = = = 26.6 » 27 N g = m1 + m2 6 4+2 Total downward thrust on the pulley = 2 T = 2 ´ 27 = 54 N 2 9. Tension in rope, T < Breaking load, mg 3 2 m ( g - a) < mg \ 3 g or a> 3 10. From figure, 8 x = wy x w …(i) = y 8 y

x

mg

mg cos α α + mg α ma sin α

\

= 2 ´ 10 4 ´ 14.8 N

u2 = 2g

co sα

w = mg

= 20000 ´ (9.8 + 5.0)

it reaches up to maximum height Hmax then Hmax = 2 m from

ma

sin

α w

8g

wx = 18 y x 18 = y w

ma cos a = mg sin a sin a a=g = g tan a cos a P = (M + m) g tan a

x

…(ii) y

4. Conservation of momentum in a collision between particles can be understood from both, Newton’s 2nd law and 3rd law. fair

M = 0.05 kg Acceleration g = 9.8 ms , a = 9.5 ms \ Þ

8g

w

5. Here, mass of the body –2

a Rocket

Let initial thrust of the blast be T. \ or

2. Let the ball starts moving with velocity u and

Hmax

T

Initial acceleration a = 5.0 m/s2 in upwards direction

m F F1 = F= 3 m + 2m

–2

mg - fair = ma fair = m ( g - a)

a

= 0.05 (9.8 – 9.5) = 0.015 N mg

Dividing Eq. (i) by Eq. (ii), we have x w y = 8 x 18 y w Þ

w = 18 ´ 8 = 12 g

214 JEE Main Physics 11. Mass of the body m = 5 kg

15. Applying law of conservation of momentum,

Force acting on body F1 = 8 N α

F2 = 6 N Angle between two forces q = 90°

m mö æ v = çm + ÷V è 20 20 ø v 20 v V= ´ = 20 21 21

F

F2 = 6 N

Force perpendicular to force F1 on the body

or

F1 = 8 N

16. Force applied by engine = 6 m

Resultant force acting on the body

When two cars are pulled,

F = F12 + F22 + 2FF 1 2 cos q

(m + m) a = 6 m

= (8) + (6) 2 + 2 ´ 8 ´ 6 ´ cos 90° (Q cos 90° = 0)

= 10 N If resultant force F makes an angle a with force F1, then F 6 tan a = 2 = = 0.75 = tan 36° 33¢ F1 8 a = 36°53¢ Using relation F = ma F 10 Acceleration a = = = 2 m/s2 m 5 \ An acceleration of 2 m/s2 is acting on body at an angle of 36°33¢ from the direction of force F1 = 8 N

12. Here the tension in the cord is given by T

a

mg

–3

r = 5 ms , then x (t ) = 3 t + 4 t 2 + 5 t 3 d 2x (t ) = 8 + 30 t dt 2 t =2s a = 8 + 30 ´ 2 = 68 F = m ´ a = 2 ´ 68 = 136 N a=

14. By drawing the free body diagram of point B. Let the tension in the section BC and BF are T1 and T2 respectively. From Lami’s theorem C

As direction of motion of the body remains unchanged, therefore the direction of force acting on the body is along the direction of motion.

18. The net electromagnetic force = N 2 + f 2

13. Given, x (t) = pt + qt 2 + rt 3 and p = 3 ms–1, q = 4 ms–2,

N = mg , f = m mg Force = mg 1 + m 2

But

19. F = v

dm = 10 ´ 5 N = 50 N dt

20. Mass of the ball m = 0.15 kg Velocity of the ball v = 54 km/h 5 5 ö æ m/s = 54 ´ m/s÷ çQ1 km/h = ø è 18 18 = 15 m/s Let the ball be incident along path PO. Batsman deflects the ball by an angle of 45° along both OQ. N P

Q u cos θ

F T1

120°

T2

120°

120°

10 N = T

u

u

22.5°

\ \ Now

Initial speed, u = 2.0 m/s Final speed, v = 3.5 m/s Time, t = 25 s Force, F =? Using the first equation of motion, v = u + at \ 3.5 = 2.0 + a ´ 25 3.5 - 2.0 or a= m/s2 25 1.5 Acceleration a = m/s-2 25 \Force acting on the body 1.5 4.5 F = ma = 3.0 ´ = N = 0.18 N 25 25

°

T = mg + ma Here upward acceleration = a Mass of sphere = M T = 4 mg Þ 4 mg = mg + ma 3 mg = ma Þ a=3g

17. Mass of the body m = 3.0 kg

22.5

= 64 + 36

2 ma = 6 m or a = 3 ms–2

or

2

T3

45° O

A

T T1 T2 = = sin 120° sin 120° sin120° Þ

T = T1 = T2 = 10 N

u sin θ

θ u cos θ

u

u sin θ

Laws of Motion and Friction 45° = 22.5° 2 The horizontal component of velocity u sin q remains unchanged while vertical component of velocity is just reversed. \ Impulse imparted to the ball = change in linear momentum of the ball = mu cos q - ( - mu cos q) = 2 mu cos q = 2 ´ 0.15 ´ 15 ´ cos 22.5° = 4.5 ´ 0.9239 kg-m/s = 4.16 kg-m/s ÐPON = ÐNOQ =

24.

x1

a1

T1

m1

T1

T1

T2 m2 a2

From force diagram,

becomes zero so upthrust becomes zero.

22. Combined momentum = 2p$i + p$j Magnitude of combined momentum

This must be equal to the momentum of the third part. T = f = 2 mg

2mg

2 mg - mg = ma1 a1 = g

T1 = m1a1

…(i)

T2 = 2 T1

…(ii)

m2g - T2 = m2a1 m2g - 2T2 = m2a1 Total work done by tensions should be zero.

= (2p) 2 + p 2 = 5 p 2 = 5p

\ For B,

x2

m2g

21. Upthrust on the body = Vs g for freely falling body effective g

23. For A,

215

A

\ or

T1x1 - T2x2 = 0 T1x1 = T2x2 T1x1 = 2 T1x2

or

x1 = 2 x2

or mg

From force diagram shown in figure,

…(iii)

2

d x1 2 d 2x2 = dt 2 dt 2

or

\ a1 = 2 a2 After solving Eqs. (i), (iii) and (iv), m2g a2 = 4 m1 + m2 mg

m

2m

m

B

mg mg

2mg

25. Change in momentum = Impulse Þ

2mg

3m

2m

Dp = F ´ Dt Dp Dt = F 125 = = 0.5 s 250

Þ

2mg

2 mg - mg = 3 ma2 g a2 = 3

\

…(iv)

26. For discussion of motion of the point mass m, we assume that observer is situated at the triangular block. The force diagram of point mass m is shown in figure.

For C,

m

a 0

co s

N

θ θ

mg mg

2mg

3m

2mg

m

mg cos θ + ma0 sin θ

F2 = mg

\ \ So,

n

2m

si

m

m

g

mg B

m m

θ ma0

2 mg - mg = 2 ma3 g a3 = 2 a1 > a3 > a2

θ

a0

mg

From force diagram, mg sin q - ma0 cos q = ma \

a = g sin q - a0 cos q

216 JEE Main Physics 27. Resultant acceleration ( a)R = 2a - a

31. In uniform translatory motion, all parts of the ball have the same velocity in magnitude and direction, and this velocity is constant.

a

√2 a

dp æ dm ö 2 =v ç ÷ = av è dt ø dt F a= \ M

32. F =

a a

=

( a)R = ( 2 - 1) a

28. For solving the problem, we assume that observer is situated

av 2 M

33. When a metre scale is moving with uniform velocity, the force acting on the scale is zero and the torque acting about centre of mass of the scale is also zero.

in the frame of pulley (non-inertial reference frame). m1g = w1

34. Reading = Weight of cage + Reaction by bird

m2g = w2

= 20 + 0.5 (10 +2) = 26 N

From force diagram, T

T

w2 m2a0

35. Dp = p2 - p1 = mv - mu a

a

= 0.15 ´ - (3$i - 4$j) - 0.15 (3$i+ 4$j) = - (0.9 $i - 1.2 $j) kg ms -1

36. | Dp| = ( -0.9) 2 + ( -1.2) 2 = 1.5 kg ms–1

m1a0 w1

T - m2a0 - w2 = m2a or

T - m2g - w2 = m2a

or

T - 2 w2 = m2a

37. Effective value of acceleration due to gravity in the lift = g - a (Q a0 = g ) …(i)

From force diagram m1a0 + w1 - T = m1a or

m1a0 + w1 - T = m1a

or

2 w1 - T = m1a

Acceleration down the inclined plane 1 Using, s = ut + at 2, we get 2 1 l = ( g - a) sin qt 2, we get 2 2l ( g - a) sin q

t=

…(ii) (Q a0 = g )

From Eqs. (i) and (ii), T=

4 w1w2 w1 + w2

38. Initial thrust mg + ma = m ( g + a) = 10 5 (10 + 5) N = 15 ´ 10 5 N

29. Various forces acting on the ball are as shown in figure. The three concurrent forces are in equilibrium. Using Lami’s theorem. 10 T1 T2 = = sin 150° sin 120° sin 90° 10 T1 T2 = = sin 30° sin 60° 1 \

T1 = 10 sin 30° = 10 ´ 0.5 = 5 N T2 = 10 sin 60°

and

30. \ or

= 10 ´

= 1.5 ´ 10 6 N 1 2

39. Horizontal component of 10 N is10 cos 60° i. e. ,10 ´ N or 5 N. It is balanced y 5 N force along x-axis. Y-component of 10 N force is 10 sin 60 N. 10 ´

i. e. ,

3 N or 5 3 N 2 ay =

3 =5 3 N 2

2 mnv = Mg Mg v= 2 mn 10 ´ 980 u= cms–1 2 ´ 5 ´ 10 9800 = cms–1 = 98 cms–1 100

5 3N 3 kg

= 5 ms–2

40. T =

2 ´3 ´5 15 kgf kgf = 3+5 4

Pressure on the pulley = 2 T = 2 ´

41. Thrust F = u æç

15 kgf = 7.5 kgf 4

dm ö 4 6 ÷ = 5 ´ 10 ´ 40 = 2 ´ 10 N è dt ø

Laws of Motion and Friction

Suppose A is climbing with acceleration a such that T = 30 N T -2 a = 2 a

42. Let the mass of a block is m. It will remains stationary if forces acting on it are in equilibrium ma cos a = mg sin a a = g tan a

i. e. , Þ

ma

30 - 2 ´ 10 = 2 a a = 5 ms

or or

mg

α

mg α

T 5g T

T ¢ = (30 + 50 + 25) N = 105 N

or sin

T A 5g

–2

T¢ -T -5 g = 5 a T¢ = T + 5 g + 5 a

Again,

co sα

ma α

217

B 2g 2g

500 g - T = 500 a

47.

Here, ma = pseudo force on block, mg = weight

…(i)

T - 100 g sin 30° - T ¢ = 100 a

43. As is clear from figure. F sin θ

T

T a

F

T θ

a θ

mg cos θ

0 sin 3 100g

º

a

mg θ

500 g

R = mg cos q + F sin q T = ma T a m

a m

g

si n

T

T - T ¢ - 50 g = 100 a

…(ii)

T ¢ - 50 g = 50 a From Eqs. (i), (ii) and (iii), 400 g = 650 a 400 g 8 g or a= = 650 13

…(iii)

or

44. Let force in downward to the incline mg sin q - T = ma

Again,

This acceleration is downwards.

48. As the elevator is going down with

T

m

decreasing speed, so acceleration is upward direction. Let it is a

\ or

45.

50 g

mg sin q - T = T 2 T = mg sin q 1 or T = mg sin q 2 1 Here, sin q = l R

o

c ma

m

in gs

T - 800 g = 800 a T = 800 ( g + a)

sθ

a θ

θ

v 2 = u 2 - 2as,

From

ma

θ

v = 10 ms–1

\

a = 2 ms–2

\

T = 800 (10 +2),

\

T = 9600 N

v=0 800 g

49. Here, m = 5 kg, F = ( - 3$i + 4$j) N Initial velocity at t = 0 , u = (6$i - 12$j)

mg cos θ

Let required acceleration of inclined plane be for the object to remain stationary relative to inclined. We have ma cos q = mg sin q 1 a = g tan q = g 2 l -1

46. If A is climbing with constant velocity, then T ¢ = 5 g + T and T = 2 g T¢ = 5 g + 2 g = 7 g

Retardation,

F m æ 3$i 4$j ö ÷ m/s 2 = ç+ 5ø è 5

a=

As final velocity is along Y-axis, its x-component must be zero. From v = u + at , for X-component only, 3$i 0 = 6$i t 5 5 ´6 t= = 10 s 3

218 JEE Main Physics 2000 g = 1000 g + 1000 a

50.

or a=g Direction is upward Now, 0 2 - 2.5 2 = -2 ´ 10 ´ s 2.5 ´ 2.5 625 25 5 or s= = = m= m 20 100 ´ 20 80 16

or

and (ii) frictional force, f2 = mg = 5 N tangentially upward. \Total force exerted by wall on block

56. Change in momentum = 2 ´ 0.1 ´ 10 ´

f

\

30°

Dp 1 = = 10 N Dt 0.1

l 0.25 l æ m ö = = 20% of l ÷l= è m + 1ø (0.25+1) 5

52. g a=

Fav =

57. l ¢ = ç

F = N 2 + fs2 = (12) 2 + (5) 2 = 13 N

58. As, F = FAB + FBG

/2

A

°

fAB B mg

fBG

30° cos

T sin q - mg sin q = ma mg T sin q = mg sin q + 2 T cos q = mg cos q Dividing Eq. (i) by Eq. (ii), we get 2 tan q = 3

= 0.2 ´ 100 ´ 10 +0.3 ´ (300) ´ 10

…(i)

= 200 + 900

…(ii)

= 1100 N

59. In the given condition the required centripetal force is provided by frictional force between the road and tyre. mv 2 = mmg R

53. For block to continue motion on belt, acceleration a = + mg = 0.2 ´ 10 = 2 ms–2 \ Velocity of belt = Velocity of block after 4 s = 2 ´ 4 = 8 ms

–1

m (v - u) 0.15 [20 – (–10)] 0.15 ´ 30 = = = 45 N t 0.1 0.1

55. R + F sin 60° = mg R = mg R

3F 2

F sin 60º F 60º

1

mg

F cos 60° = f = mR

F cos 60º

Ground

= m AB ma g + mBG (mA + mB) g

From diagram,

or

1 2

= 1kg-ms–1

12 N 5N

54. F =

3F = 10 2 20 F= 2+ 3 20 F= N = 5.36 N 3.732 Dp = 2 mu sin 30°

N

30

or

5N

13 N

si n mg

F+

or

51. Wall applies 2 forces of the block (i) normal reaction, R = 12 N,

12 N

æ F 3Fö = 0.5 ç1 ´ 10 ÷ 2 2 ø è

or

\

v = mRg 1 4

3 4

60. Weight of chain on table = mg - mg = mg For maximum possible friction, 3 1 mmg = mg \ 4 4

m=

61. Net downward acceleration Weight - Frictional force = Mass mg - mR = m 60 ´ 10 - 0.5 ´ 600 = 60 300 = = 5 m/s2 60

1 3 F

R

600 N

W

219

Laws of Motion and Friction 62. Using the relation

66. Downward retardation means upward acceleration. mv 2 = mR ,R = mg r mv 2 = m mg r v 2 = mrg

or

g¢ = g + a

v 2 = 0.6 ´ 150 ´ 10 v = 30 ms–1

or or

Now,

t=

2L g ¢ sin q

or

t=

2L ( g + a) sin q

67. ma = m sin q - f

g

8k

63. From acting on block are shown in adjoining figure. N

a

f = mg sin q - ma 1 æ ö = 8 ç10 ´ - 0.4÷ N è ø 2

or

60° f =µN

mg

30°

F cos 60°

60°

F Mg + F sin 60°

= 8 ´ 4.6 N = 36.8 N

As the block does not move, hence

\

f

F cos 60° = f = mN = m (Mg + F sin 60° ) 1 1 æ 3ö F = ç 30 ´ 10 + F ÷ 2 2 3 è 2 ø

68. The work done by the force is F cos 37°, F cos 37° = f = mN

where

F sin 37°

On simplification, we get F = 20 N

F 37°

64. Let length of chain be l and mass m. Let a part x of chain can hang over one edge of table having coefficient of friction. l

f

x

x

mg

In this case, \Pulling force,

F=

mx g l

m ( l - x) g l For equilibrium, F = f , hence mx m m × g = m ( l - x) g = 0.25 ( l - x) g l l l 1 Þ x= 5 x 1 or = = 20% l 5 and friction force, f = mN = m

65. For a smooth plane, v = 2g sin q × s and for a rough plane, v = 2g (sin q - m cos q) s n sin q n= sin q - m cos q sin q n2 = sin q - m cos q

\ or Þ or

2

2

So that,

N

N = mg - F sin 37°, m mg F= (cos 37° + m sin 37° )

Here, m = 0.40 and m = 20 kg \ Hence,

F = 75.4 N W = (75.4 cos 37° ) (8.0) = 482 J

69. Force on the car F = mR ma = m mg a = mg Now from 2nd equation of motion 1 s = ut + at 2 2 1 or s = 0 + at 2 2

or

(Q R = mg )

or

or

t=

2s mg

\

t=

2s mg

or

tµ

1 m

(n - 1) sin q = n m cos q æ n 2 - 1ö 1ö æ m = ç 2 ÷ tan q = tan q ç1 - 2 ÷ è n ø è n ø

F cos 37°

20 kg

(Q u = 0)

220 JEE Main Physics 70. For equilibrium of the block N = mng + Q cos q Q sin q + P = mN

Þ

Q sin q + P = m (mg + Q cos q)

77. Reaction on m is mg. Maximum friction force by m on M is

N

mmg. So, the force on M is mmg forward. m mg Acceleration = M

N P µN θ

P + Q sin θ

78. The trolley shall move backwards to conserve momentum. The backward momentum would be shared by both the trolley and man.

Q

mg

Applying conservation of momentum

mg + Q cos θ

æ Q sin q + P ö m=ç ÷ è mg + Q cos q ø

\

M (L - x) g K Mg (L - x) Tx = L

76. Tx = (mass of rope of length L - x) =

60 ´ 1 (240 + 60) v 60 = 300 v 60 v= 300 1 = ms–1 = 0.2 ms–1 5

or or

71. If m is the mass/length, then Weight of hanging length = m lg Weight of chain on table = m (L - l) g

Speed of man w.r.t. ground = (1 - 0.2) ms–1 = 0.8 ms–1

R = m (L - l) g

Displacement of man = 0.8 ´ 4 m = 3.2 m

f = m sR = m sm (L - l) g Equating, or

79. Refer to the free body diagram of block B

m sm (L - l) g = mlg l ms = L-l

5 g -T = 5 a T = 5 g -5 a Refer to the free body diagram of block A

or

72. Pseudo force on the block = m ´ 4 N (backward) Force of friction = 0.4 ´ m ´ 10 N (forward)

T

Equating, m ´ 4 = 0.4 ´ m ´ 10 = 4 m Clearly, the equation holds good for all values of m.

a a

B

A

T

73. fms = 0.6 ´10 ´ 9.8 N = 58.8 N 5g

Since the applied force is greater than fms therefore the block will be in motion. So, we should consider fk.

T - f = 5a

fk = 0.4 ´ 10 ´ 9.8 N or fk = 4 ´ 9.8 N This would cause acceleration of 40 kg block 4 ´ 9.8 N Acceleration = = 0.98 ms–2 40 kg

or

74. Force, P = fms = m smg (when body is at rest) When the body starts moving with acceleration a, then P - fk = ma or

m smg - m kmg = ma a = (m 0 - m k) g = 0.1 ´ 10 ms

–2

80.

or

a = 0.25 ´ 9.8 ms–2 = 2.45 ms–2

Again,

T = (5 ´ 9.8 – 5 ´ 2.45) N

= (49 - 12.5) N = 36.75 N Net pushing force Acceleration of system a = Total mass F - (m1 + m2 + m3) g sin q or a= (m1 + m2 + m3) Equation of motion for m3

a = (0.5 - 0.4)10

or

5g - 5a - 0.5 ´ 5 ´ g = 5 a 10 a = 2.5g = 2.5 ´ 9.8

or

= 1ms

–2

75. Net acceleration, a = g (sin q - m cos q) = 9.8 ( sin 45° - 0.5 cos45° ) 4.9 m/s2 = 2

or

N - m3 g sin q = m3 a ì F - (m1 + m2 + m3) g sin qü N = m3 g sin q + m3 í ý (m1 + m2 + m3) þ î m3F = m1 + m2 + m3

Laws of Motion and Friction

221

After breaking of string, mass m2 moves under gravity and go further higher through a height h, where final velocity is zero. Hence, (0) 2 - (9.8) 2 = 2 ´ ( -9.8) ´ h or h = 4.9 m

81. Tension between m2 and m3 is given by

85. When the system accelerates upwards, the effective value of m1

acceleration due to gravity is given by

A

f¢ = g + a = g + g = 2 g 2 (m1) (m2) T= (2 g ) m1 + m2

m2

B

Now, m3

C

T=

T=

or

2m1m2 2 ´2 ´2 ´g = ´ 9.8 = 13 N m1 + m2 + m3 2+2+2

4 m1m2 g m1 + m2

86. The system may be represented as follow a

82. FBD of mass 2 kg

FBD of mass 4 kg

T nm

T′

T

8N

4N

T

2 kg

4 kg m2

a

19.6 N

T′

Mg

39.2 N

T - T ¢ = 19.6 = 4

…(i)

T ¢ - 39.2 = 8

…(ii)

From Eq. (ii)

From the force diagram, Mg - T = Ma T = nma From Eqs. (i) and (ii), we get Mg a= nm + M

…(i)

and

T ¢ = 47.2 N and substituting T ¢ in Eq. (i), we get T = 4 + 19.6+ 47.2 = 70.8 N 2

83. From geometry l = x2 + y 2 but y is constant, hence on

…(ii)

The force diagram of nth block is shown in figure.

dl dx differentiating, we have, 2 l = 2x dt dt dl But = v. Hence, horizontal velocity of block, dt dx vx = dt

N

m

Tn

n mg

From the figure,

θ

l

y

θ

x

Þ

lv = x × v x l ×v v = vx = sin q x

or

84. Acceleration of combined system, a=

m1 - m2 3 -2 ´ 9.8 = 1.96 ms–2 ×g = m1 + m2 3+2

Vertically upward velocity of 2 kg mass at the time of breaking of string, v = at = 5 ´ 1.96 = 9.8 ms–2.

Tn = ma =

mMg nm + M

87. For minimum mass of m, mass M breaks off contact when elongation in spring is maximum. At the time of break off, block A is at lowest position and its speed is zero. At an instant t1 mg - kx = ma dv mg - kx v = dx m 0 t æ k ö ò0 v dv = ò0 çè g - m x÷ø dx where x0 is maximum elongation in spring

T = kx mg – kx = mg

mg

222 JEE Main Physics 0 = gx0 -

kx02 2m

T2 - T1 = m2a

91. Net force,

T1 = m1a Dividing Eq. (i) by Eq. (ii) T2 - T1 m2 = T1 m1

and

2 mg k At the time of break off of block B Mg = kx0 Mg = 2 mg M m= 2 x=

(v) vertical = 0 and (v) horizontal = v cos q The initial linear momentum of the system will be mv cos q. Now as force of blasting is internal and force of gravity is vertical. So, linear momentum of the system along horizontal is conserved. p1 + p2 = mv cos q m1v1 + m2v 2 = mv cos q m But it is given that m1 = m2 = and as one part retraces its 2 path, v1 = -v cos q 1 1 \ m ( -v cos q) + mv 2 = mv cos q 2 2

…(ii) a T1

or

T2 m2 m + m1 = +1= 2 T1 m1 m1

or

T1 m1 = T2 m1 + m2

88. In case of projectile motion at the highest point

m2

T2

a m1

T2

92. Spring balance reading in terms of kgf 4 m1m2 4 ´ 5 ´ 1 10 = = m1 + m2 6 3 This is less than 6 kgf.

93. Common acceleration, a = T2

40 ms–2 = 2 ms–2 10 + 6 + 4 40 N

4 kg

40 - T2 = 4 ´ 2

Now,

T2 = ( 40 - 8) N = 32 N

or

94. From diagram, 5 g - T2 = 5 a

…(i)

T2 - T1 = 3 a

…(ii)

T1 - g = a

v 2 = 3 v cos q

or

…(i)

…(iii) T2

T2

T1

89. T sin 30° = 2 kg-wt T sin 30°

30° T 30° T cos 30°

a

T1

3g

B

a

A

5g

T1

B

a

g

Solving Eqs. (i), (ii) and (iii), we get g =9a g a= 9

2 kg-wt

Þ

or

T = 4 kg-wt T1 = T cos 30°

95. Newton’s second law

= 4 cos 30° = 2 3 2 mg - T = 2 ma

90.

…(i)

T - mg sin 30° = ma R

…(ii) T T

m

30

°

30°

30°

in gs

Þ

mg cos 30°

Adding Eq. (i) and (ii), we get mg 2 mg = 3 ma 2 g Þ a= 2

2M

Now

F = ma 6 = (7 + 5) a 1 a = m/s2 2 f ¢ = 5 kg 1 f ¢ = 5 ´ = 2.5 N 2

96. Initially, the weight of load L is the force on the system of mass 8 kg. Acceleration =

2 ´ 10 20 unit = 8 8

Towards the end, force = (2 + 1) ´ 10 N = 30 N 30 units So, acceleration now is 8

Laws of Motion and Friction

223

Round II 1. Due to acceleration in forward direction vessel is in an accelerated frome therefore a pseudo force will be exerted in backward directon. Therefore water will be displaced in backward direction.

2. Masses connected at the two ends of a light m1 = 8 kg, m2 = 12 kg Let T be the tension in the string and masses moves with an acceleration a when masses are released.

T

2T cos q = 2 mg T = mg 2mg cos q = 2 mg 1 cos q = 2

T a

q = 45°

m1

a

For mass m1

8. Making FBD of block with respect to disc.

…(i))

m1 g

Let a be the acceleration of block with respect to disc

m2

For mass m2 m2g - T = m2a Adding Eqs. (i) and (ii), we get \

then in equilibrium, T cos q + T cos q = 2 mg But \

inextensible string are

T - m1g = m1a

7. If T is tension in each part of the string holding mass 2 m,

…(ii)

ma

m2 g

ma cos q

m2g - m1g = (m1 + m2) a (m - m1) a= 2 g (m1 + m2) =

ma sin q

N1

µ

…(iii)

12 - 8 ´ 10 = 2 m/s 2 12 + 8

N2

mg

N1 = mg N2 = ma sin q ma cos q - mN2 - mN1 A= = 10 m/s2 m

Substituting value of a in Eq. (i), we get T = m1g + m1a = m1( g + a) = 8(10 + 2) = 90 N

3. When block R collides with block P, it transfers its momentum

9. As shown in figure, component of weight( mg sin q ) is always

to block P, due to which it moves towards blocks Q. The spring connecting blocks P and Q gets compressed, which will push the block Q outwards. Due to outward motion of the block Q, the spring gets stretched, the block Q is pulled back. The spring gets compressed, it pushes the block P towards left and so on.

down the inclined plane, whether the cylinder is following up or it is rolling down. Therefore, for no slipping, sense of angular acceleration must be the same in both the cases. Therefore, force of friction (f) acts up the inclined plane in both the cases.

4. Let T be the tension in the string. Since

mg sin θ

the system is in equilibrium, therefore from figure. 2T cos q = mg or T = mg / 2 cos q The string will be straight if q = 90° \

f

T

θ

T

θ

10. As F2 and F3 are mutually perpendicular, their mg

T = mg /2 cos 90° = mg /2(0) = ¥

5. For body of mass 6 kg

resultant = F22 + F32 As particle is stationary under F1, F2, F3 therefore, F22 + F32 must be equal and opposite to F1.

T = 6g = 6 ´ 9.8 = 58.8 N For body of mass 4 kg T - T1 = 4g = 4 ´ 9.8 = 39.2 N T1 = T - 39.2 = 58.8 - 39.2 = 19.6 N

6. Minimum force required to move the block = mR = mmg = 0.4 ´ 2 ´ 10 = 8 N Since the force applied is only 2.8 N, the block fails to move and static fraction = applied force = 2.8 N

11. Angular momentum is an axial vector, so its direction is along the axis, perpendicular to the plane of motion which is not changing because of change of speed. Therefore, the direction of angular momentum remains, the same and its magnitude may vary.

12. The reading of balance A will decrease due to the upward thrust caused by buoyancy. The upthrust will be equal to the weight of water displaced. The net downward force due to mass immersed in water will add to effective weight of the system. So, the reading of balance B will increase.

224 JEE Main Physics 13. Impulse = Change in momentum

Þ

17. Mass measured by physical balance remains unaffected due

F ´ t = m (v - u) F ´ 0.4 = 80(5 - 0) 80 ´ 5 F= = 1000 N 0.4

18.

to variation in acceleration due to gravity. 1 1 From s = ut + at 2 = 0 + at 2 2 2

14. Applying Lami’s theorem

or \

T1 T2 mg = = sin(90°+ b) sin(90°+ a) sin[180°- ( a + b)] T1 T mg = 1 = cos b cos a sin( a + b) mg cos b mg cos a ; T2 = T1 = sin( a + b) sin( a + b)

For smooth plane, a = g sin q For rough plane, a, = g (sin q - m cos q) \

\

T cos 45°

µR f= n

45

°

Solving we get, m = 1 -

si

mg cos 45°

g

q = 45° , sin q = cos q = 1 / 2 1 n2

19. Maximum force by surface when friction works

mg

F = f 2 + R 2 = mR 2 + R 2

45°

= R m2 +1

For equilibrium, along the plane

Maximum force = R when there is no friction

mR + T cos 45° = mg sin 45° T mg mR + = 2 2

…(i)

For equilibrium, in direction perpendicular to inclined plane,

Hence, ranging from R to R m 2 + 1 we get

Mg £ f £ mg m 2 + 1

20. As m1 : m2 : m3 = 1: 1: 3

R = T sin 45° = mg cos 45° T mg = = 2 2 m 1 Put in Eq. (i), (T + mg ) = (mg - T) 2 2

and momentum is conserved, \

P12 + P22 + P32 = 3v3 1 ´ 39 2 + 1 ´ 39 2 = 3v3 39 2 = 3v3

m(50 + 15 ´ 10) = (15 ´ 10 - 50) 100 1 m= = 200 2

v3 =

16. Mass of a stone m = 0.25 kg

39 2 = 13 2 ms-1 3

21. For the smooth portion BC,

Radius of the string r = 1.5 m

u = 0 , s = l, g sin f, u = ?

40 2 rev/s = rev/s Frequency n = 40 rev/min = 60 3

From

Centripetal force required for circular motion is obtained from the tension in the string.

For the rough portion CO u = v = 2g sin f. l v = 0, a = g (sin f = m cos f)

T = mrw2 = mr(2pn) 2

[Q w = 2p n ]

= mr 4p 2v 2 2

v 2 - u 2 = 2as v 2 - 0 = 2g sin f ´ l

\Tension in the string = centripetal force

2

æ 22 ö æ2ö = 0.25 ´ 1.5 ´ 4 ´ ç ÷ ´ ç ÷ = 6. 6 N è7ø è3ø Maximum tension which can be withstand by the string Tmax = 200 N

2s g sin q

n 2g (sin q - cos q) = g sin q

when

T

m

2s g (sin q - m cos q)

t¢ =

= nt = n

15. Figure shows free body diagram of the block. R

2s a

t=

s=l From

v 2 - u 2 = 2as 0 - 2gl sin f = 2g (sin f - m cos f) l - sin f = sin f - m cos f m cos f = 2 sinf m = 2 tan f

Laws of Motion and Friction 22. As is from figure.

225

26. As both the balls are of same size, force of buoyancy on each is same. Therefore, in equilibrium, R F

α mg cos α mg

α

F

mg sin α

T

F = mg sin a R = mg cos a F = tan a R 1 m = tan a = 3 cot a = 3

i.e., \

m1g F

23. The three forces acting on the mass

F + F = m1g + m2g g or F = (m1 + m2) 2 Considering the equilibrium of lower ball,

B

at location A have been shown in θ figure. Since the mass is in equilibrium, therefore, the three forces acting on the mass must be represented by the three sides of a S triangle in one order, Hence

or or \

T + F = m1g T = m1g - F

T

T = m1g - (m1 + m2) 50 N

50 6 ´ 10 = SA SB SA 50 5 = = SB 60 6 SA 5 tan q = = SB 6 = 0.8333 = tan 40° q = 40°

In D SBA,

m2g

24. Mass of rope M = 0.1kg, q = 10°

T = (m1 - m2)

A

the bullet on the disc in vertically upward direction. F = nmv = 40 ´ 0.05 ´ 6 = Mg 40 ´ 0.05 ´ 6 M= = 1.2kg 16

6 × 10 N

28. In figure O is the centre of the bowl of radius r. The insect will crawl (from B to A ) till component of its weight (mg ) along the bowl is balanced by the force of limiting friction (f) r R

θ

C B

θ

g m

θ

25. Effective upward force = 310 - mg = 310 - 24 ´ 9.8 = 74.8 N Upward acceleration \ a = 74.8/24 = 3.12ms-2 1 As s = ut + at 2 2 1 4.6 = 0 + ´ 3.12 ´ t 2 2 4.6 2 or t = = 2.95 1.56 t = 2.95 » 1.7s

mg

n si

i. e. , mg

2T sin q = mg mg 0.1 ´ 9.8 T= = = 2.82 N 2 sin q 2sin10°

h

g

T cos θ

y

m

T cos θ

θ

A co s θ

T sin θ

T

T

or

O

f T sin θ

g 2

27. Weight of the disc will be balanced by the force applied by

From figure θ

g 2

or

mg sin q = f = mR = mmg cos q AC m = tan q = OC

or

=

or

m2 =

OA2 - OC 2 r2 - y2 = OC y r2 - y2 y2

m 2y 2 + y 2 = r 2 y=

r m2 +1

h = BC = OB - OC = r - y æ r 1 ö÷ =r= r çç1 ÷ è m2 +1 m 2 + 1ø

226 JEE Main Physics 29. If v is common velocity of the block and movable wedge, then applying the principle of conservation of linear momentum we get, mu + 0 = (m + nm)v mu u v= = m (1 + n) 1 + n This infact, can be taken as velocity of centre of mass of the block and wedge u v CM = v = i. e. , 1+ n Applying the principle of conservation of energy u2 1 1 mu 2 = mgh + m (1 + n) 2 2 (1 + n) 2

At t =

1 1 s, a(t ) = - 16 p 2 A sin 4p ´ = - 16 p 2 A ´ 1 8 8 F = ma (t ) = - 6p 2Am

Impluse = chagne in linear momentum 1 = F ´ t = - 16 p 2 Am ´ = - 4 p 2 Am 4 The impulse (change in linear momentum) at t = 0 is same 1 as at t = sec. 4 Further, as F depends on A which is not constant, therefore, the particle is not acted upon by a constant force.

33. In figure, f is the force of friction. When the body moves up the plane, f acts down the plane.

u2 u 2 = 2gh + 1+ n

or

R

æ 1ö u = 2ghç1 + ÷ è nø

30. Total mass of 80 wagons = 80 ´ 5 ´10 Acceleration,

a=

-3

θ in gs f θ

m1 5

= 4 ´ 10 kg

5

F 4 ´ 10 = ms-2 = M 4 ´ 10 5

Tension in the coupling between 30th and 31st wagon will be due to mass of remaining 50 wagons. Now, mass of remaining 50 wagons. m = 50 ´ 5 ´ 10 3 kg = 25 ´ 10 4 kg

m2g > m1g sin q + m m1 g cos q m2 > m1 (sin q + m cos q) Choice (b) is correct.

(m2g + f ) < m1g sin q

v t

m2g < m1g sin q - f m2g < m1g sin q - m m1 g cos q

Density of water r = 10 3 kg /m3 Cross-sectional area of pipe OA = p (0.5 ´ 10 -3) 2

F=

dv mv Vrv rv v = = = ´ dt t t t At v ö æ ÷ ç\ V = è At ø

(10 ´ 10 -6) 2 ´ 10 3 p ´ (0.5 ´ 10 -3) 2

or m2 < m1 (sin q - m cos q) Choice (d) is correct.

34. As is clear from figure, R + T = (m + M) g R = (m + M) g - T The system will not move till T £ F or (T £ mR) T £ m[(m + M) g = T ] m(m + M) g m (m + M) g T£ \ Fmax = m +1 m +1

= 0.127 N

32. Here, x = 0

for t < 0 s

x(t ) = A sin 4 pt

for t >

x=0 For 0 < t
h1 when its speed becomes v2. Find the work done on the ball by air resistance.

Interpret Work done on the ball by gravity is Wg = - mg(h2 - h1)

7.3 Work-Energy Theorem This theorem is a very important tool that relates the work to kinetic energy. Accordingly, 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 its kinetic energy of the particle. Thus, we can write W = DK = K f - K i We can also write,

K f = Ki + W

Which says that æ Kinetic energy after ö ç ÷ è the net work is doneø ö æ Kinetic energy beforeö æ The net =ç ÷+ç ÷ è the net work done ø è work doneø These statements are known traditionally as the work-kinetic energy theorem for particles. They hold for both positive and negative works. If the net work done on a particle is positive, then the particle’s kinetic energy increases by the amount of the work done. If the net work done is negative, then the particle’s kinetic energy decreases by the amount of the work.

Work done on the ball by air resistance is Wair = ? Q Wg + Wair = DKE 1 Þ -mg (h2 - h1) + Wair = m (v 22 - v12) 2 1 Wair = mg (h2 - h1) + m (v 22 - v12) Þ 2

7.4 Power The time rate of doing work is called power. If an external force is applied to an object (which we assume as a particle), and if the work done by this force is DW in the time interval Dt, then the average power during this interval is defined as DW P= Dt The work done on the object contributes to increasing the energy of the object. The general definition of power is the time rate of energy transfer. The instantaneous power is the limiting value of the average power as Dt approaches zero. DW dW i.e., P = lim = Dt ® 0 D t dt

Work, Energy and Power where we have represented the infinitesimal value of the work done by dW (even though it is not a change and therefore not differential) dW ds ds é ù P= =F = F×v ê as, dt = vú dt dt ë û

Interpret (b) Force required to keep the belt moving = rate of increase of horizontal momentum of sand = mass per second dm ´ velocity change = 2 ´ 0.1 = 0.2 N dt Power = Force ´ velocity = 0.2 ´ 0.1 = 0.02 W

(a) Power is equal to the scalar product of force and velocity. (b) Power is a scalar with dimensions [ML2T -3]. The SI unit of power is Js–1 and is called watt (W) (after James Watt) Practical unit of power is horse power (HP) 1 HP = 746 W Work , any unit of power multiplied by a Time unit of time gives unit of work (or energy), i. e.,

(c) Since, Power =

kilowatt-hour or watt-day are units of work or energy but not of power. 1 kWh = 103 ´ Js–1 ´ (60 ´ 60 s) = 3.6 ´ 106 J (d) The slope of work-time curve gives the instantaneous dW power as P = = tan q [from Fig. (a)] while the area dt dW . under P-t curve gives the work done. Since, P = dt

which means W = ò Pdt = area under P-t curve [as shown in Fig. (b)]. Power

Work

237

7.5 Potential Energy of a Spring 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 = 0denote the position of the block, when the spring is in its natural length. We shall calculate the work done on the block by the spring force as the block moves from x = 0 to x = x1 x = x1

x=0

F

A

A

We have to find the work done during a small interval in which the block moves from x to x + dx. The force in this interval is kx and the displacement is dx. The restoring force and displacement are opposite in direction. So,

dW = F × ds = | F | | ds| cos 180° = - | F | | ds| = - kxdx

during this interval. The total work done as the block is displaced from x = 0 to x = x1 is

θ (a)

Time

dt (b)

Time

W =ò

0

(a) 45 kW (c) 46.9 kW

(b) 45.9 kW (d) None of these

Interpret (c) The work done in accelerating the car is given by 1 1 m(v f2 - v12) = (1200) [(25) 2 - 02 ] 2 2 W = 375 kJ

W = DK =

or

Power =

W 375 = = 46.9 kW t 8

Sample Problem 18 Sand drops vertically at the rate of 2 kgs–1 on to a conveyor belt moving horizontally with a velocity of 0.1 ms–1. The extra power needed to keep the belt moving is (a) 0.05 W (c) 0.06 W

(b) 0.02 W (c) 0.03 W

x

ù 1 é 1 - kxdx = ê - kx 2 ú û0 ë 2 =-

Sample Problem 17 An advertisement claims that a certain 1200 kg car can accelerate from rest to a speed of 25 ms–1 in a time of 8 s. What average power must the motor produce to cause this acceleration? (ignore friction)

x1

1 2 kx1 2

If the block moves from x = x1 to x = x2, the limits of integration are x1 and x2 and the work done is 1 ö æ1 W = ç kx12 - kx22 ÷ = potential energy ø è2 2

Note If the block is displaced from x1 to x 2 and brought back to x = x1 the work done by the spring force is zero. The work done during the return journey is negative of the work during the onward journey. The net work done by the spring force in a round trip is zero.

Sample Problem 19 A block of mass m has a velocity v0 when it just touches a spring. The block moves through a distance l before it stops. The spring constant of spring is k, what is the work done on it by the spring force? kl 2 2 kl 2 (c) 2

(a)

3kl 2 2 3kl 2 (d) 2 (b)

238 JEE Main Physics Interpret (c)

v0

Here, the speed of both the blocks are same. Let the speed is v ms–1. Since block of 2 kg is coming down hence the gravitational potential energy is decreasing while the gravitational potential energy of 1 kg block is increasing.

k

m

So, kinetic energy of both the blocks will increase.

The net force acting on the block by the spring is equal to |Fspring| = kx

Hence,

where, x is the compression in the spring. or

Work done by the spring ò Fspring. ds = ò Fspring | ds| cos180° = -ò

l

0

or

-kl 2 kxdx = 2

\

v2 =

10 = 6.67 15 .

v = 2.58 ms–1

friction in 10 s is equal to (a) –236.8 J (c) –246.9 J

(b) –245.2 J (d) 246 J

Interpret (c) Work done by the force of friction

E = K +U When the forces acting on the system are conservative in nature, the mechanical energy of the system remains constant, K + U = constant DK + DU = 0

There are physical situations, where one or more nonconservative force act on the system but net work done by them is zero, then too the mechanical energy of the system remains constant. If

1.5v 2 = 10

Sample Problem 21 In the above problem work done by

The mechanical energy E of a system is the sum of its kinetic energy K and its potential energy U.

Þ

20 = 10 + 0.5 v 2 + v 2

or

or

7.6 Conservation of Mechanical Energy

1 1 mAv 2 + mBv 2 2 2 1 1 2 ´ 10 ´ 1 = 1 ´ 10 ´ 1 + ´ 1´ v 2 + ´ 2v 2 2 2 mBgh = mA gh +

S Wnet = 0

Mechanical energy, E = constant.

Wf = f ´ s = - 1.96 ´ 126 = –246.9 J

Examples of Conservation of Mechanical Energy 1. Object thrown vertically upwards Energy at the lowest point (at A) is only kinetic energy h = 0, in the middle, energy is both kinetic and potential energy (as h = h1 ) and at the highest point, energy is only potential. ( as v = 0)

Note If only conservative forces are acting on a system of particles and work done by any other external force is zero, then mechanical energy of the system will remain conserved. In this case, some fraction of mechanical energy will be decreasing while the other fraction will increase.

h

Sample Problem 20 In the arrangement shown in figure, string is light and inextensible and friction is absent everywhere. The speed of both the blocks after the block A has ascend a height of 1 m will be (a) 2 ms–1 (b) 2.58 ms–1 –1 (c) 3 ms (c) 3.58 ms–1 Given that, mA = 1 kg and 2 kg.

v=0

B

v1

h1 A

\ A

B

Interpret (b) Since, there is no friction anywhere, so mechanical energy will be conserved.

C

or

v

E = K A = K B + UB = UC 1 E = mv2 2 1 = mv12 + mgh = mgh 2

Work, Energy and Power 2. Freely falling object

(a) KC = U A

At the maximum height, total energy is in the form of potential energy. In the middle, total energy is in the form of both kinetic and potential energy. At the lowest point, total energy is in the form of kinetic energy. \

E = UC = K B + UB = K A

C

UC =0

or

1 mvC2 = mgh = mg (1 - cos q1 ) 2 vC = vmax = 2 g (1 - cos q1 )

\ B

h

(b) U B + KB = U A v1

or or

h1

A

v

KB = U A - U B = mg (h1 - h2 ) 1 mvB2 = mgl (cos q1 - cos q2 ) 2 vB = 2 gl (cos q1 - cos q2 )

\

or

1 1 E = mgh = mv12 + mgh1 = mv2 2 2

(c) If pendulum of length l is released from horizontal position as shown in adjacent figure, then U A = KB

3. Projectile motion

O A

At the highest point, potential energy is maximum and U H = mgh = mg

2

2

u sin q 1 = mu2 sin2 q 2g 2

l

At the highest point, the kinetic energy will be minimum but not zero because at the highest point only vertical component of velocity is zero. u

B

v

or θ

239

H

\

1 1 mu2x = mu2 cos2 q 2 2 1 EH = U H + K H = mu2 = Einitial 2

mgl =

1 mvB2 2

vB = vmean = 2 gl

KH = Hence,

Hence, in projectile motion, mechanical energy is conserved.

1. Does kinetic energy depend on the direction of motion involved? Can it be negative? Does it depend on frame of reference.

2. Can kinetic energy of a system be increased or decreased without applying any external force on the system?

4. Oscillator

3. Out of joule, kilowatt, calorie and electron volt, which one is

O

not the unit of energy?

A

θ2 θ1 l

Check Point

E=U Extreme position

h1

E=U+K

h2

B

C E=K Equilibrium or mean position

4. The protons are brought towards each other. Will the potential energy of the system decrease or increase? If a proton and an electron are brought closer, then?

5. A pump motor is used to deliver water at a certain rate from a given pipe. To obtain n times water from the same pipe in the same time by what amount (a) the force and (b) power of motor should be increased?

240 JEE Main Physics

7.7 Conservative and Non-Conservative Forces

Fig. (a) Let W1, W2, W3 denote the amounts of work done moving a body from A to B along three different paths, 1, 2, 3 respectively. If the force is non-conservative W1 ¹ W2 ¹ W3.

Conservative Forces

1

A force is said to be conservative, if work done by or against the force in moving a body depends only on the initial and final positions of the body and not the nature of path followed between the final and initial positions. This means, work done by or against a conservative force is moving a body over any path between fixed initial and final positions will be the same. For example, gravitational force, electrostatic force etc., are conservative forces. In case of gravitational force, if we take work done in moving the body from A to B, against gravity as negative, the work done in moving the body from B to A, by gravity has to be taken as positive, i. e. ,

A 2 Fig. (b)

Fig. (b) Shown that a particle moving a closed path. A ® 1, B ® 2 ® A. If W1 is work done in moving the particle from A ® 1 ® B and W2 is work done in moving the particle from B ® 2 ® A, then for a non-conservative force | W1 | ¹ | W2 |. \ Net work done along the closed path, A ® B ® A is not zero.

ò F × ds ¹ 0

i. e. ,

WAB = - WBA \ WAB + WBA = 0

In fact, work done is taking the body from A to B is speed in the body in the form of PE. This energy which is spent in moving body from B to A. Thus, over the round trip ( A ® B ® A), net work done is zero.

Non-Conservative Forces A force is said to be non-conservative, if work done by or against the force in moving a body from one position to another, depends on the path followed between these two positions. For example, force of friction and viscous force are non-conservative forces. 2 A

1

B

Sample Problem 22 A particle is taken from point A to point B via the path ACB and then come back to point A via the path BDA. What is the work done by gravity on the body over this closed path, if the motion of the particle is in the vertical plane? (a) mgh 1 (c) mgh 2

(b) –mgh (d) zero

Interpret (d) Here, displacement of the particle is AB, gravity is acting vertically downwards. The vertical component of AB is h upwards. Hence, W( ACB) = -mgh For the path BDA, component of the displacement acting along vertical direction is h (downward) In this case, Total work done

Fig. (a)

B

W(BDA ) = mgh WACB + WBDA = 0

WORKED OUT Examples Example 1

A uniform rope of linear density d and length l is hanging from the edge of a table. The work done in pulling the rope on the table is (a)

dgl 2

(b)

dgl 2 2

(c) dgl 2

(d) d 2gl

Example 5

The work done in time t on a body of mass m which is accelerated from rest to a speed v in time t1 as a function of time t is given by (a)

1 v 2 m t 2 t1

(c)

1 æ mv ö 2 ç ÷ t 2 è t1 ø

(b) m

2

Solution

Here, mass of rope m = l ´ d

For pulling the rope on the table, distance of centre of gravity moved = l /2 l l dgl 2 \Work done = F ´ = ldg ´ = 2 2 2

Solution or

A force of (10 i$ - 3$j + 6 k$ ) N acts on a body of displaces it from A(6 $i - 5 $j + 3k$ ) m to

Example 2 5 kg

and $ B(10 i - 2 $j + 7k$ ) m. The work done is (a) zero

Solution

(b) 55 J

(c) 100 J

(d) 221 J

From second law of motion F = ma = m

m = 5 kg AB = (10 $i - 2$j + 7k$ ) - (6$i - 5$j + 3k$ ) = ( 4$i - 3$j + 4k$ )

= 40 - 9 + 24 = 55 J

Example 3

A ball of mass 5 kg experiences a force F = 2 x 2 + x . Work done in displacing the ball by 2m is (b) 44/3 J

(c) 32/3 J x

x

0

0

(d) 16/3 J

Work done, W = ò F dx = ò (2x2 + x) dx

Solution

2ù2

é 2x x æ16 4 ö 22 J =ê + ú =ç + ÷= 3 2 2ø 3 ë û0 è 3 2

Example 4

A box is dragged across a floor by a rope which makes an angle 45° with the horizontal. The tension in the rope is 100 N while the box is dragged by 10 m. The work done is (a) 607.1 J

Solution

(b) 707.1 J

(c) 1414.2 J (d) 900 J

Work done W = Fs cos q = 100 ´ 10 cos 45° =

100 = 707.1J 2

v t1

Distance travelled from relation s = ut + s =0 +

W = F. s = F . AB = (10 $i - 3$j + 6k$ ) × ( 4$i - 3$j + 4 k$ )

1 v2 2 m t 2 t12

We know, v = u + at ,v = 0 + at1 v a= t1

Given force F = (10 $i - 3$j + 6k$ ) N,

(a) 22/3 J

(d)

v 2 t t1

1 2 at 2

1 vt 2 vt 2 = 2 t1 2t1

W =F ´ s ´

mv vt 2 1 mv 2t 2 = ´ t1 2t1 2 t12

Example 6

A uniform force of 4 N acts on a body of mass 40 kg for a distance of 2.0 m. The kinetic energy acquired by the body is (a) 4 ´ 2 J (c) 4 ´ 4 ´ 2 ´ 10 8 erg

Solution

(b) 4 ´ 4 ´ 2 J (d) 4 ´ 2 ´ 2 erg

KE acquired = Work done F ´ s = 4 ´2J

Example 7

Calculate the KE and PE of the ball half way up , when a ball of mass 0.1 kg is thrown vertically upwards with an initial speed of 20 ms-1. (a) 10 J , 20 J (c) 15 J , 8 J

(b) 10 J ,10 J (d) 8 J ,16 J

Solution

Total energy at the time of projection 1 1 = mv 2 = ´ 0.1(20) 2 = 20 J 2 2 Half way up, PE becomes half the PE at the top 20 i.e., PE = = 10 J 2

\

KE = 20 - 10 = 10 J

242 JEE Main Physics Example 8

A spring is kept compressed by a small cart of mass 150 g. On releasing the cart, it moves with a speed of 0.2 ms-1. The potential energy of the spring is (a) 1 ´ 10 -4 J

Solution

(b) 6 ´ 10 -3 J (c) 4 ´ 10 -4 J

PE of spring = KE of mass = =

(d) 3 ´ 10 -3 J

1 mv 2 2

1 æ 150 ö -3 2 ÷ (0.2) = 3 ´ 10 J ç 2 è1000 ø

The potential energy of a certain spring when stretched through a distance s is 10 J. The amount of work (in J ) That must be done on this spring to stretch it through an additional distance s will be

Solution

(b) 40

(c) 10

Potential energy E1 = 10 =

(d) 20

An obdect of mass 5 kg falls from rest through a vertical distance of 20 m and attains a velocity of 10 ms-1. How much work is done by the resistance of the air on the object? ( g = 10 ms-2)

Solution

(c) 850 J

A single conservative force F(x) acts on a 1.0 kg particle that moves along the x-axis. The potential energy U(x) is given by U(x) = 20 + ( x - 2) 2 where x is in metre. At x = 5.0 m the particle has kinetic energy of 20 J. What is the mechanical energy of the system? (a) 44 J

\

Example 10

(b) - 750 J

Example 12

Solution

1 2 ks 2

1 E 2 = k ( s + s) 2 = 4 ´ 10 = 40 J 2 Amount of work required = E 2 - E1 = 40 - 10 = 30 J

(a) 750 J

= increase in KE of block + increase in elastic potential energy of spring 1 1 So, 45 ´ 9.8 ´ 0.012 = ´ 45 ´ v 2 + ´ 1050 2 2 [(0.075 + 0.024) 2 - (0.075) 2] v = 0.37 ms-1

Example 9

(a) 30

\ Decrease in PE of block

The system is released from rest with the spring initially stretched 75 mm. Calculate the velocity of the block after it has dropped 12 mm. The spring has a stiffness of 1050 Nm -1. Neglect the mass of the small pulley. (d) 2.2 ms-1

When the block descends 12 mm, spring further stretches by 24 mm

45 kg

Mechanical energy = KE + PE = 20 + 29 = 49 J

Example 13

In the above example, the maximum kinetic energy of the particle is (a) 39 J (c) 30 J

Solution

(b) 29 J (d) None of these Maximum kinetic energy at x = 2 m

where PE is minimum and the maximum kinetic energy is Kmax = E - Umin = 49 - 20 = 29 J

When a belt moves horizontally at a constant speed of 1.5 ms-1, gravel is falling on it at 5 kgs -1. Then, the extra power needed to drive the belt is (a) 11.25 W (c) 7.5 W

Solution

Example 11

Solution

(d) 49 J

Example 14

Applying work-energy theorem

(c) 5 ms-1

(c) 48 J

Potential energy at x = 5 m is U = 20 + (5 - 2) 2 = 29 J

(d) - 650 J

Work done ball the forces = Change in kinetic energy 1 or Wmg + Wair = mv 2 2 1 1 \ Wair = mv 2 - Wmg = mv 2 - mgh 2 2 1 2 = ´ 5 ´ (10) - 5 ´ 10 ´ 20 = - 750 J 2

(a) 0.371 ms-1 (b) 0.45 ms-1

(b) 45 J

(b) 37.5 W (d) 0.75 W

Here, v = 1.5 ms-1 dm dm = 5 kg ms-1,F = ´ v = 5 ´ 1.5 = 7.5 N dt dt P = F ´ v = 7.5 ´ 1.5 = 11.25 W

Example 15 A machine gun fires 360 bullets per minute, with a velocity of 600 ms-1. If the power of the gun is 5.4 kW. mass of each bullet is (a) 5 kg

Solution

(b) 0.5 kg (c) 5 g (d) 0.5 g 360 -1 Here, n = = 6 bullets s 60 v = 600 ms-1, m = ?

Power of gun = Power of bullets 1 5.4 ´ 10 3 = (nm) v 2 2 2 ´ 5400 = 6 ´ m(600) 2 2 ´ 5400 kg or m= 6 ´ 600 ´ 600 1 1000 kg = = g =5g 200 200

Start Practice for

JEE Main Round I

(Topically Divided Problems)

Work

6. A 5 kg brick of 20 cm × 10 cm × 8 cm dimensionless

1. Under the action of a force, a 2 kg body moves such that its position x as a function of time t is given by x = t 3 / 3, where x is in metre and t in second. The work done by the force in the first two seconds is (a) 1.6 J (c) 160 J

2. The work done in pulling up a block of wood weighing 2 kN for a length of 10 m on a smooth plane inclined at an angle of 15° with the horizontal is [sin 15° = 0.2588] (b) 5.13 kJ (d) 9.82 kJ

distance h at a constant acceleration g/2. The work done by the string will be Mgh 2

(b)

-Mgh 2

(c)

3Mgh 2

(b) 5 J

(c) 7 J

(d) 9 J

which is inclined at 45° to the horizontal. The coefficient of sliding friction is 0.30. When the block has slide 5 m, the work done on the block by the force of friction is nearly (a) 115 J (b) 75 2 J (c) 321.4 J (d) –321.4 J

8. In a children’s park, there is a slide which has a total

3. A mass M is lowered with the help of a string by a

(a)

(a) 3 J

7. A block of mass 10 kg slides down a rough slope

(b) 16 J (d) 1600 J

(a) 4.36 kJ (c) 8.91 kJ

lying on the largest base. It is now made to stand with length vertical. If g =10 ms -2 , then the amount of work done is

(d)

-3Mgh 2

4. An electron and a proton are moving under the

length of 10 m and a height of 8.0 m. A vertical ladder is provided to reach the top. A boy weighing 200 N climbs up the ladder to the top of the slide and slides down to the ground. The average friction offered by the slide is three-tenth of his weight. The work done by the slide on the boy as he comes down is

influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another. This is because,

[NCERT Exemplar]

(a) the two magnetic forces are equal and opposite, so they produce no net effect (b) the magnetic forces do no work on each particle (c) the magnetic forces do equal and opposite (but non-zero) work on each particle (d) the magnetic forces are necessarily negligible

5. A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes upto 2 m height further. Find the magnitude of force (Consider g = 10 m / s2 ) (a) 22 N (c) 16 N

(b) 4 N (d) 20 N

(a) zero

(b) + 600 J

(c) –600 J

(d) +1600 J

9. A proton is kept at rest. A positively charged particle is released from rest at a distance d in its field. Consider two experiments; one in which the charged particle is also a proton and in another, a positron. In the same time t, the work done on the two moving charged particles is [NCERT Exemplar] (a) same as the same force law is involved in the experiments (b) less for the case of a positron, as the positron moves away more rapidly and the force on it weakness.

244 JEE Main Physics (c) more for the case of a positron, as the positron moves away a larger distance (d) same as the work done by charged particle on the stationary proton

10. A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is [NCERT Exemplar] (a) constant and equal to mg in magnitude (b) constant and greater than mg in magnitude (c) variable but always greater than mg (d) at first greater than mg, and later becomes equal to mg

11. A ball is released from the top of a tower. The ratio of work done by force of gravity in Ist second, 2nd second and 3rd second of the motion of ball is (a) 1 : 2 : 3 (c) 1 : 3 : 5

(b) 1 : 4 : 16 (d) 1 : 9 : 25

12. A plate of mass m, length b and breadth a is initially lying on a horizontal floor with length parallel to the floor and breadth perpendicular to the floor. The work done to erect it on its breadth is é bù êë 2 úû é b - aù (c) mg êë 2 úû (a) mg

é (b) mg a + êë éb+ (d) mg êë 2

bù 2 úû aù úû

(b) zero (d) 36 m

body of mass 7 kg and displaces it from x = 0 m to x = 5m. The work done on the body is x′ joule. If both F and x are measured in SI units, the value of x′ is (a) 135 (b) 235 (c) 335 (d) 935

15. A body of mass 0.5 kg travels in a straight line with

velocity, v = ax 3/ 2 , where a = 5 m -1/ 2 /s. What is the work done by the net force during its displacement from x = 0 to x = 2 m? [NCERT] (c) 20 J

(d) 50 J

16. A uniform chain of length L and mass M overhangs a horizontal table with its two-third part on the table. The friction coefficient between the table and the chain is m. The work done by the friction during the period the chain slips off the table is 1 2 (a) - mMgL (b) - mMgL 4 9 4 6 (d) - mMgL (c) - m MgL 9 7

(b)

350 J 3

(d) zero

18. A force acts on a 30 g particle in such a way that the position of the particle as function of time is given by x = 3 t - 4 t2 + t 3, where x is in metre and t is in second. The work done during the first 4 seconds is (a) 5.28 J (c) 490 mJ

(b) 450 mJ (d) 530 mJ

19. A car weighing 1400 kg is moving at a speed of

54 kmh–1 up a hill when the motor stops. If it is just able to reach the destination which is at a height of 10 m above the point, then the work done against friction (negative of the work done by the friction) is (Take g = 10 ms–2) (a) 10 kJ (c) 17.5 kJ

(b) 15 kJ (d) 25 kJ

(a) Mg

d 4

(b) 3 Mg

d 4

(c) -3 Mg

d 4

(d) Mgd

14. A position-dependent force F = 3x2 – 2x + 7 acts on a

(b) 40 J

500 J 3 750 J (c) 3 (a)

a distance d with constant downward acceleration g / 4 work done by the cord on the block is

moving in one dimension under the action of a force is related to the time t in second by the equation t = x + 3 , the work done by the force (in joule) in first six seconds is

(a) 30 J

of a lake. It is lifted through a height of 5 m in the lake. If g = 10 ms–2, then the work done is

20. A cord is used to lower vertically a block of mass M by

13. The displacement x in metre of a particle of mass m kg

(a) 18 m (c) 9 m/2

17. A 5 kg stone of relative density 3 is resting at the bed

21. Water is drawn from a well in a 5 kg drum of capacity 55 L by two ropes connected to the top of the drum. The linear mass density of each rope is 0.5 kgm–1. The work done in lifting water to the ground from the surface of water in the well 20 m below is (g = 10 ms–2) (a) 1.4 ´ 10 4 J

(b) 1.5 ´ 10 4 J

(c) 9.8 ´ 10 ´ 6 J

(d) 18 J

22. A wire of length L suspended vertically from a rigid support is made to suffer extension l in its length by applying a force F. The work is Fl 2 (c) 2 Fl (a)

(b) Fl (d) Fl

23. A bicyclist comes to a skidding stop in 10 m. During this process, the force on the bicycle due to the road is 200 N and is directly opposed to the motion. The work [NCERT Exemplar] done by the cycle on the road is (a) + 2000 J (c) zero

(b) - 200 J (d) - 20,000 J

Work, Energy and Power

245

24. A uniform chain of length L and mass M is lying on a

30. The relationship between force and position is shown

smooth table and one third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, the work required to pull the hanging part on the the table is

in figure given (in one dimensional case). The work done by the force is displaying a body from x = 1 cm to x = 5 cm is

(b) MgL/3 (d) MgL/18

20

Force (dyne)

(a) MgL (c) MgL/9

25. During inelastic collision between two bodies, which of the following quantities always remain conserved? [NCERT Exemplar]

26. A spring of spring constant 5 ×103 Nm–1 is stretched

(b) 18.75 N-m (d) 6.25 N-m

27. A rod AB of mass 10 kg and length 4 m rests on a horizontal floor with end A fixed so as to rotate it in vertical. Work done on the rod is 100 J. The height to which the end B be raised vertically above the floor is (a) 1.5 m (c) 1.0 m

(b) 2.0 m (d) 2.5 m

3

4

5 6 x (cm)

31.

as a function of its position is shown in figure. What is the net work performed on the brick by the force causing the acceleration as the brick moves from x = 0

to x = 8.0 m ? 20 15 10

other steep meet at A from where two stones are allowed to slide down from rest, one on each track as shown in figure. A I

5 0

1

2

3

(a) 4 J (c) 2 J

5

4

6 7 x (m)

8

(b) 8 J (d) 1 J

32. A 2.0 kg block is dropped from a height of 40 cm onto

II

a spring of spring constant k = 1960 Nm–1. Find the maximum distance the spring is compressed.

θ2 C

Which of the following statement is correct? [NCERT Exemplar]

(a) Both the stones reach the bottom at the same time but not with the same speed (b) Both the stones reach the bottom with the same speed and stone I reaches the bottom earlier than stone II (c) Both the stones reach the bottom with the same speed and stone II reaches the bottom earlier than stone I (d) Both the stones reach the bottom at different times and with different speeds

(a) 0.080 m (b) 0.20 m (c) 0.40 m (d) 0.10 m

33. The graph between the resistive force F acting on a body and the distance covered by the body is shown in the figure. The mass of the body is 2.5 kg and initial velocity is 2 m/s. When the distance covered by the body is 4 m, its kinetic energy would be F (newton)

h θ1

2

(b) 60 erg (d) 700 erg A 10 kg brick moves along an x-axis. Its acceleration

28. Two inclined frictionless tracks, one gradual and the

B

1

(a) 20 erg (c) 70 erg

(ms–2 )

(a) 12.50 N-m (c) 25.00 N-m

0 –10 –20

(a) Total kinetic energy (b) Total mechanical energy (c) Total linear momentum (d) Speed of each body

initially by 5 cm from the unstretched position. Then the work required to stretch it further by another 5 cm is

10

29. A force F = Ay2 + By + C acts on a body in the y-direction. The work done by this force during a displacement from y = – a to y = a is 2 Aa5 2 Aa5 (b) (a) + 2 Ca 3 3 2 Aa5 Ba2 (c) (d) None of these + + Ca 3 2

20 10 0

(a) 50 J (c) 20 J

1

2

3

4

(b) 40 J (d) 10 J

x (cm)

246 JEE Main Physics 34. A stone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is (a) u2 - 2gl

(b) 2gl

(c) u2 - gl

(d) 2 (u2 - gL)

35. The potential energy function for a particle executing

1 2 kx where k is the 2 force constant of the oscillator. For k = 0.5 N/m , the graph of V ( x) versus x is shown in the figure. A particle of total energy E turns back when it reaches x = ± xm . If V and K indicate the PE and KE respectively of the particle at x = ± xm then which of [NCERT Exemplar] the following is correct? linear SHM is given by V ( x) =

V(x)

–xm

(a) V = O, K = E (c) V < E, K = O

39. A body of mass 0.5 kg travels in a straight line with

velocity v = a x 3/ 2 where a = 5m -1/2s -1 . The work done by the net force during its displacement from [NCERT Exemplar] x = 0 to x = 2 m is (a) 1.5 J

(a) 2 ( 2 - 1) ms -1

(b) 2 ( 2 + 1) ms -1

(c) 4.5 ms–1

(d) None of these

41. A 0.5 kg ball is thrown up with an initial speed 14 m/s and reaches a maximum height of 8.0 cm. How much energy is dissipated by air drag acting on the ball during the time of ascent? (a) 19.6 J (c) 10 J

(b) 4.9 J (d) 9.8 J

42. A body is moving unidirectionally under the influence of a source of constant power supplying energy. Which of the diagrams shown in figure correctly shows the displacement-time curve for its [NCERT Exemplar] motion?

xm

x

(b) V = E, K = O (d) V = O, K < O

1 2 k( x + y2 ) 2 1 (d) ky (2x + y ) 2 (b)

d

d

(a)

(b)

t

t

d

d

(d)

(c)

t

t

43. A stone of mass 2 kg is projected upward with kinetic

Energy 37. A ball is projected vertically upwards with a certain initial speed. Another ball of the same mass is projected at an angle of 60° with the vertical with the same initial speed. At highest point of their journey, the ratio of their potential energies will be (a) 1 : 1 (c) 3 : 2

(d) 100 J

that his kinetic energy is doubled, the original speed of the man is

constant k is stretched by a small length x. It is further stretched by another small length y. The work done in the second stretching is 1 2 ky 2 1 (c) k( x + y )2 2

(c) 10 J

40. When a man increases his speed by 2 ms–1, he finds

36. An elastic string of unstretched length L and force

(a)

(b) 50 J

(b) 2 : 1 (d) 4 : 1

38. The kinetic energy K of a particle moving in straight line depends upon the distance s as K = as2

The force acting on the particle is (a) 2 as

(b) 2 mas

(c) 2a

(d) as2

energy of 98 J. The height at which the kinetic energy of the body becomes half its original value, is given by (Take g = 10 ms–2) (a) 5 m (c) 1.5 m

(b) 2.5 m (d) 0.5 m

44. A ball whose kinetic energy is E, is projected at an angle 45° to the horizontal. The kinetic energy of the ball at the highest point of its flight will be E E (a) E (b) (c) (d) zero 2 2

45. A body is falling freely under the action of gravity alone in vacuum. Which of the following quantities [NCERT Exemplar] remain constant during the fall? (a) Kinetic energy (b) Potential energy (c) Total mechanical energy (d) Total linear energy

Work, Energy and Power

247

46. The potential energy as a function of the force

52. Which of the diagrams shown in figure represents

between two atoms in a diatomic molecules is given a b by U( x) = 12 - 6 , where a and b are positive x x constants and x is the distance between the atoms. The position of stable equilibrium for the system of the two atoms is given

variation of total mechanical energy of a pendulum oscillating in air as function of time? [NCERT Exemplar]

(a) x =

a b

(c) x =

(b) x =

E

(a)

47. The potential energy of a particle of mass 5 kg moving in the xy-plane is given by U = ( -7x + 24 y) J, x and y being in metre. If the particle starts from rest from origin, then speed of particle at t = 2 s is (a) 5 ms–1 (c) 17.5 ms–1

(b) 01 ms–1 (d) 10 ms–1

48. A running man has half the kinetic energy of that of a boy of half of his mass. The man speeds up by 1 m/s, so as to have same kinetic energy as that of the boy. The original speed of the man will be (b) 2 - 1 m/s 1 (d) m/s 2

(a) 2 m/s 1 m/s (c) 2 -1

49. If a body looses half of its velocity on penetrating 3 cm in a wooden block, then how much will it penetrate more before coming to rest? (a) 1 cm (c) 3 cm

t

(b)

a b

t

æ2 aö (d) x = 6 ç ÷ è bø

3a b

E

(b) 2 cm (d) 4 cm

50. In the given curved road, if particle is released from A, then M A

E E

(c)

(d) t

t

53. A 50 g bullet moving with a velocity of 10 ms–1 gets embeded into a 950 g stationary body. The loss in kinetic energy of the system will be (a) 95% (b) 100% (c) 5% (d) 50%

54. A car is moving with a speed of 100 kmh–1. If the mass of the car is 950 kg, then its kinetic energy is (a) 0.367 M J (c) 3.67 M J

(b) 3.67 J (d) 367 J

55. A simple pendulum is released from A as shown. If M and l represent the mass of the bob and length of the pendulum respectively, the gain in kinetic energy at B is

A

30°

B

3 Mgl 2 2 (d) mgl 3

Mgl 2 Mgl (c) 2 (a)

(b)

56. Two masses of 1 g and 4 g are moving with equal h

B

(a) kinetic energy at B must be mgh (b) kinetic energy at B may be zero (c) kinetic energy at B must be less than mgh (d) kinetic energy at B must not be equal to zero

51. Two springs have force constants k1 and k2 . These are

extended through the same distance x. If their elastic E energies are E1 and E2 , then 1 is equal to E2 (a) k1 : k2 (c) k1 : k2

(b) k2 : k1 (d) k12 : k22

kinetic energies. The ratio of the magnitudes of their linear momenta is (a) 4 : 1

(b) 2 : 1

(c) 1 : 2

(d) 1 : 16

57. A mass of 5 kg is moving along a circular path of radius 1 m. If the mass moves with 300 revolutions per minute, its kinetic energy would be [NCERT Exemplar]

(a) 250 p2

(b) 100 p2

(c) 5 p2

(d) 0

58. A body of mass 2 kg is thrown up vertically with kinetic energy of 490 J. The height at which the kinetic energy of the body becomes half of its original value is (a) 50 m

(b) 12.25 m

(c) 25 m

(d) 10 m

248 JEE Main Physics 59. In a shotput event an athlete throws the shotput of

mass 10 kg with an inital speed of 1 m s -1 at 45° from a height 1.5 m above ground. Assuming air resistance to be negligible and acceleration due to gravity to be 10 ms -2 , the kinetic energy of the shotput when it just reaches the ground will be [NCERT Exemplar]

(a) 2.5 J (c) 52.5 J

(b) 5.0 J (d) 155.0 J

60. A machine which is 75% efficient uses 12 J of energy in lifting up a 1 kg mass through a certain distance. The mass is then allowed to fall through that distance. The velocity of the ball at the end of its fall is (a) 24 ms -1

(b) 32 ms -1

(c) 18 ms -1

(d) 3 ms–1

61. A body of mass 4 kg is moving with momentum of

8 kg-ms–1. A force of 0.2 N acts on it in the direction of motion of the body for 10 s. The increase in kinetic energy in joule is (a) 10

(b) 8.5

(c) 4.5

(d) 4

62. A body of mass M is dropped from a height h on a sand floor. If the body penetrates x cm into the sand, the average resistance offered by the sand to the body is æ hö (a) Mg ç ÷ è xø

hö æ (b) Mg ç1 + ÷ è xø

(c) Mgh + Mgx

hö æ (d) Mg ç1 - ÷ è xø

66. A bomb of mass 9 kg explodes into 2 pieces of mass 3 kg and 6 kg. The velocity of mass 3 kg is 1.6 m/s, the kinetic energy of mass 6 kg is (a) 3.84 J (c) 1.92 J

67. An engine pumps water continuously through a hole. Speed with which water passes through the hole nozzle is v and k is the mass per unit length of the water jet as it leaves the nozzle. Find the rate at which kinetic energy is being imparted to the water. (a)

(c) 4000 J

(d) 5500 J

64. Given that the position of the body in metre is a function of time as follows x = 2 t4 + 5 t + 4 The mass of the body is 2 kg. What is the increase in its kinetic energy, one second after the start of motion? (a) 168 J (c) 32 J

(b) 169 J (d) 144 J

65. A bomb of mass 3.0 kg explodes in air into two pieces of mass 2.0 kg and 1.0 kg. The smaller mass goes at a speed of 80 m/s. The total energy imparted to the two fragment is (a) 1.07 kJ (c) 2.4 kJ

(b) 2.14 kJ (d) 4.8 kJ

(b)

1 3 kv 2

(c)

v2 2k

(d)

v3 2k

(a) maximum potential energy (b) minimum potential energy (c) minimum kinetic energy (d) maximum kinetic energy

69. A stone is dropped from the top of a tall tower. The ratio of the kinetic energy of the stone at the end of three seconds to the increase in the kinetic energy of the stone during the next three seconds is (a) 1 : 1

(b) 1 : 2

(c) 1 : 3

(d) 1 : 9

70. A rectangular plank of mass m1 and height a is kept on a horizontal surface. Another rectangular plank of mass m2 and height b is placed over the first plank. The gravitational potential energy of the system is (a) [ m1 + m2 ( a + b)] g

machine whose efficiency is 90%, the energy is 5000 J. If the mass is now released, its kinetic energy on hitting the ground shall be (b) 4500 J

1 2 kv 2

68. In the stable equilibrium position, a body has

bö ù é æ m + m2 (b) ê ç 1 a + m2 ÷ ú g 2 2øû ëè

bù bù éæ m éæ m ö ö (c) ê ç 1 + m2 ÷ a + m2 ú g (d) ê ç 1 + m2 ÷ a + m1 ú g è è ø ø 2û 2û ë 2 ë 2

63. A mass of 50 kg is raised through a certain height by a

(a) 5000 J

(b) 9.6 J (d) 2.92 J

Power 71. A

body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to [NCERT] 12 /

(a) t (c) t32/

(b) t (d) t2

72. A 10 m long iron chain of linear mass density

0.8 kg m –1 is hanging freely from a rigid support. If g = 10 ms–2, then the power required to lift the chain upto the point of support in 10 s is (a) 10 W (c) 30 W

(b) 20 W (d) 40 W

73. A 10 HP motor pump out water from a well of depth 20 m and falls a water tank of volume 22380 litre at a height of 10 m from the ground the running time of the motor to fill the empty water tank is ( g = 10 ms -2 ) (a) 5 min (c) 15 min

(b) 10 min (d) 20 min

Work, Energy and Power 74. An engine of power 7500 W makes a train move on a

horizontal surface with constant velocity of 20 ms–1 . The force involved in the problem is (a) 375 N (c) 500 N

(b) 400 N (d) 600 N

75. A one kilowatt motor is used to pump water from a well 10 m deep. The quantity of water pumped out per second is nearly (a) 1 kg (c) 100 kg

(b) 10 kg (d) 1000 kg

accelerated from rest to a velocity of 10 ms–1 in 5 s. If the total mass of the car and its occupants is 1000 kg, then the average horse power developed by the engine is 103 746 105 (c) 746

(b)

(a)

80. A 500 kg car, moving with a velocity of 36 kmh–1 on a straight road unidirectionally, doubles its velocity in one minute. The power delivered by the engine for doubling the velocity is (a) 750 W (c) 1150 W

(b) 1050 W (d) 1250 W

81. The power of a water jet flowing through an orifice of radius r with velocity v is (b) 500 pr2 v2 (d) pr 4 v

(a) zero (c) 500 pr2 v3

76. A car manufacturer claims that his car can be

10 4 746

(d) 8

249

82. A motor of power P0 is used to deliver water at a

certain rate through a given horizontal pipe. To increase the rate of flow of water through the same pipe n times. The power of the motor is increassed to p1. The ratio of p1 to p0 is (a) n : 1

(b) n2 : 1

(c) n3 : 1

(d) n 4 : 1

83. A quarter horse power motor runs at a speed of

77. Which of the diagrams in figure correctly shows the change in kinetic energy of an iron sphere falling freely in a lake having sufficient depth to impart it a terminal velocity? [NCERT Exemplar]

600 rpm. Assuming 40% efficiency, the work done by the motor in one rotation will be (a) 7.46 J (c) 7.46 erg

(b) 7400 J (d) 74.6 J

84. Ten litre of water per second is lifted from well through 20 m and delivered with a velocity of 10 ms–1, then the power of the motor is

(a)

KE

(b)

KE

(a) 1.5 kW

(b) 2.5 kW

(c) 3.5 kW

(d) 4.5 kW

85. A body is moved along a straight line by machine Depth

Depth

delivering a constant power. The distance moved by the body in time t is proportional to (a) t3/ 4

(c) t1/4

(d) t1 /2

86. One man takes 1 minute to raise a box to a height of

(d) KE

(c) KE

(b) t3/2

1 m and another man takes

1 minute to do so. The 2

energy of the two is Depth

Depth

78. A dam is situated at a height of 550 m above sea level and supplies water to a power house which is at a height of 50 m above sea level. 2000 kg of water passes through the turbines per second. What would be the maximum electrical power output of the power house if the whole system were 80% efficient? (a) 8 MW (c) 12.5 MW

(b) 10 MW (d) 16 MW

79. An automobile weighing 1200 kg climbs up a hill that rises 1 m in 20 s. Neglecting frictional effects, the minimum power developed by the engine is 9000 W. If g = 10 ms -2 , then the velocity of the automobile is kmh–1

(a) 36 (c) 72 kmh–1

kmh–1

(b) 54 (d) 90 kmh–1

(a) different (b) same (c) energy of the first is more (d) energy of the second is more

87. The power supplied by a force acting on a particle moving in a straight line is constant. The velocity of the particle varies with the displacement x as (a) x1 /2

(b) x

(c) x2

(d) x1/3

88. A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration a c is varying with time t as ac = k2 rt2 . The power is (a) 2 pmk2 r2t (c)

mk 4 r2t5 3

(b) mk2 r2t (d) zero

250 JEE Main Physics

Round II Only One Correct Option

8. A particle is released from a height s. At certain

1. The bob of a pendulum is released from a horizontal position A as shown in the figure. If the length of the pendulum is 1.5 m, then the speed with which the bob arrives at the lower most point B. Given that it dissipated 5% of its initial energy against air [NCERT] resistance? (a) 6.0 m/s

(b) 6.5 m/s

(c) 4.5 m/s

2. If a man speeds up by 1

(d) 5.3 m/s

ms–1,

his kinetic energy increases by 44%. His original speed in ms–1 is (a) 1

(b) 2

(c) 5

(d) 4

3. A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to (b) x2

(a) x

(c) x0

(d) e x

4. A bullet fired from a gun with a velocity of

104 ms–1goes through a bag full of straw. If the bullet loses half of its kinetic energy in the bag, its velocity when it comes out of the bag will be (a) 7071.06 ms–1 (c) 70.71 ms–1

(b) 707 ms–1 (d) 707.06 ms–1

height its kinetic energy is three times its potential energy. The height and speed of the particle at that instant are respectively s 3 gs , 4 2 s 3 gs (c) , 2 2

constant force. The power P developed by the motor must vary with time t as shown in figure

9. A body of mass 3 kg acted upon by a constant force is 1 2 t , where 3 t is in second. Work done by the force in 2 s displaced by s metre, given by relation s = 8 J 3 5 J (c) 19

19 J 5 3 (d) J 8

(a)

(b)

10. A body of mass 3 kg is under a force which causes a

3 displacement is given by s = t (in m). Find the work 3 done by the force in first 2 seconds.

(b) 3.8 J

(c) t

(d) 24 J

The gun is free to recoil 804 J of recoil energy are released on firing the gun. The speed of bullet (ms–1) is

(b) t

(c) 5.2 J

11. A gun of mass 20 kg has bullet of mass 0.1 kg in it.

(a) 804 ´ 2010

P

(a)

s 3 gs , 4 2 s 3 gs (d) , 4 2 (b)

(a)

(a) 2 J

5. A motor drives a body along a straight line with a

P

(Mixed Bag)

804 2010

(b)

2010 804

(d) 804 ´ 4 ´ 103

12. Power supplied to a particle of mass 2 kg varies with

3 t2 watt. Here t is in second. If the 2 velocity of particle at t = 0 is v = 0, the velocity of particle at time t = 2 s will be

time as P =

P

P

(d)

(c)

t

t

6. A bob of mass m accelerates uniformly from rest to v1

in time t1. As a function of t, the instantaneous power delivered to the body is mv1t (a) t2

mv1t2 (c) t1

mv1t (b) t1

mv2t (d) 2 1 t1

7. Two blocks of mass m each are connected to a spring of spring constant k as shown in figure. The maximum displacement in the block is k m

v 2

(a)

2 mv k

m

2

2

(b)

mv k

v

(c) 2

mv k

(d) 2

k mv2

(a) 1 ms–1 (c) 2 ms–1

(b) 4 ms–1 (d) 2 2 ms -1

13. A body of mass 3 kg is under a force which causes a

displacement in it, given by s = t2 / 3 (in m). Find the work done by the force in 2 s (a) 2 J

(b) 3.8 J

(c) 5.2 J

(d) 2.6 J

14. Power supplied to a particle of mass 2 kg varies with

time as P = t2 /2 watt, where t is in second. If velocity of particle at t = 0 is v = 0, the velocity of particle at t = 2 s will be (a) 1 ms–1 2 (c) 2 ms–1 3

(b) 4 ms–1 (d) 2 2 ms -1

251

Work, Energy and Power 15. Power applied to a particle varies with time as

P = (3 t2 - 2 t + 1) watt, where t is in second. Find the change in its kinetic energy between t = 2 s and t = 4 s. (a) 32 J

(b) 46 J

(c) 61 J

(d) 100 J

16. A car of mass 1000 kg moves at a constant speed of

20  ms–1 up an incline. Assume that the frictional force is 200 N and that sin q = 1/20, where, q is the angle of the incline to the horizontal. The g = 10 ms–2. Find the power developed by the engine? (a) 14 kW (c) 10 kW

(b) 4 kW (d) 28 kW

stretched through a distance s is 10 J. The amount of work (in joule) that must be done on this spring to stretch it through additional distance s will be (a) 30

(b) 40

the arteries at each beat against an average pressure of 10 cm of mercury. Assuming that the pulse frequency is 72 per minute the rate of working of heart in watt, is (Density of mercury = 13.6 g/cc and g = 9.8 ms–2)

(c) 10

(d) 20

22. A bullet when fired at a target with velocity of 100 ms–1 penetrates 1 m into it. If the bullet is fired at a similar target with a thickness 0.5 m, then it will emerge from it with a velocity of 50 m/s 2 (d) 10 m/s

(a) 50 2 m/s

17. The human heart discharges 75 cc of blood through

(a) 11.9 (c) 0.119

21. The potential energy of a certain spring when

(b)

(c) 50 m/s

23. Velocity-time graph of a particle of mass 2 kg moving in a straight line is as shown in figure. Work done by all forces on the particle is 20

(b) 1.19 (d) 119

v (m/s)

18. A particle of mass 2 kg starts moving in a straight line with an initial velocity of 2 ms–1 at a constant acceleration of 2 ms–2. Then rate of change of kinetic energy (a) is four times the velocity at any moment (b) is two times the displacement at any moment (c) is four times the rate of change of velocity at any moment (d) is constant through out

19. The potential energy of a system represent in the first figure. The force acting on the system will be represent by

t(s)

(a) 400 J (c) – 200 J

(b) – 400 J (d) 200 J

24. A particle moves on a rough horizontal ground with

some initial velocity v0 . If 3 th of its kinetic energy is 4 lost due to friction in time t0 , the coefficient of friction between the particle and the ground is (a)

f (x)

2

v0 2 gt 0

(b)

v0 4 gt 0

(c)

3 v0 4 gt 0

(d)

v0 gt 0

25. A box of mass 50 kg is pulled up on an incline 12 m O

long and 2 m high by a constant force of 100 N from rest. It acquires a velocity of 2 ms–1 on reaching the top. Work done against friction (g = 10 ms–2) is

x

a

f (x)

f (x)

(a) 50 J a

(a)

x

a

(b)

f (x)

x

x

(d)

(d) 200 J

half his mass. The man speeds up by 1 ms–1 and then has kinetic energy as that of the boy. What were the original speeds of man and the boy?

f (x) a

(c) 150 J

26. A man running has half the kinetic energy of a boy of

(a) (c)

(b) 100 J

a

x

2 ms -1; 2 2 - 1 ms -1

(b) ( 2 - 1) ms -1, 2( 2 - 1) ms -1 (c) ( 2 + 1) ms -1; 2( 2 + 1) ms -1 (d) None of the above

20. A ball is dropped from a height of 20 cm. Ball rebounds to a height of 10 cm. What is the loss of energy? (a) 25%

(b) 75%

(c) 50%

(d) 100%

27. An engine pumps up 100 kg of water through a height of 10 m in 5 s. Given that the efficiency of the engine is 60%. If g = 10 m/s2 , the power of the engine is (a) 3.3 kW

(b) 0.33 kW

(c) 0.033 kW

(d) 33 kW

252 JEE Main Physics 28. An ideal spring with spring constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is 4 Mg k Mg (c) k

34. Two blocks M1 and M2 having equal mass are to move on a horizontal frictionless surface. M2 is attached to a massless spring as shown in figure. Initially M2 is at rest and M1 is moving toward M2 with speed v and collides head-on with M2 .

2 Mg k Mg (d) 2k

(a)

[NCERT Exemplar]

(b)

M1 = m

M2 = m v

29. A 0.5 kg ball is thrown up with an initial speed

14 ms -1 and reaches a maximum height of 8 m. How much energy is dissipate by air drag acting on the ball during the ascent? (a) 19.6 J (c) 10 J

(b) 4.9 J (d) 9.8 J

30. The kinetic energy k of a particle moving along a circle of radius R depends upon the distance s as k = as2 . The force acting on the particle is (a) 2 a

é s2 ù (b) 2 as ê1 + 2 ú R û ë (d) 2 a

s2 R

(c) 2 as

1/2

More Than One Correct Option 31. A ball moves over a fixed track as shown in figure. From A to B the ball rolls without slipping. Surfaces BC is frictionless. K A , K B and K C are kinetic energies of the ball at A, B and C respectively. Then A

(a) hA > hC : KB > KC (c) hA = hC : KB = KC

Comprehension Based Questions Passage I The stopping distance for vehicle is obtained by dividing their kinetic energy by the stopping force applied. If a car of mass mc and a bus of mass mb, having kinetic energies K c and K b are stopped under the action of the same retarding force in distance xc and xb in time tc and tb respectively, then æm ö t (i) For and K c = K b,xc = xb and c = ç c ÷ tb è mb ø

C

hA

(a) While spring is fully compressed all the KE of M1 is stored as PE of spring (b) While spring is fully compressed the system momentum is not conserved though final momentum is equal to initial momentum (c) If spring is massless, the final state of the M1 is state of rest (d) If the surface on which blocks are moving has friction, then collision cannot be elastic

hC

1/ 2

i.e., bus would take longer time to stop, but they

would cover the same distance before stopping. (b) hA > hC : KC > KA (d) hA < hC : KA > KC

32. A man of mass m, standing at the bottom of the staircaseof height L climbs it and stands at its top. [NCERT Exemplar]

(a) Work done by all forces on man is equal to the rise in potential energy mgL (b) Work done by all forces on man is zero (c) Work done by the gravitational force on man is mgL (d) The reaction force from a step does not do work because the point of application of the force does not move while the force exists

33. If the kinetic energy of a body is directly proportional to time t, the magnitude of the force acting on the body is (a) directly proportional to t (b) inversely proportional to t (c) directly proportional to the speed of the body (d) inversely proportional to the speed of the body

(ii) If pc = pb then

xc mb t and c = 1 = xb mc tb

i.e., stopping distance for car is more than

the bus, though they take the same time. Three cars A, B, C having masses 1000 kg, 2000 kg and 2500 kg are moving with velocities 10 2 ms–1, 10 ms–1 and 8 ms–1 respectively. Exactly same force is applied to stop the cars, A and B. Time taken to stop and the distances travelled before stopping are measured. Similar measurements are made by applying same opposite force on the cars B and C.

35. If car A takes 5 s to stop, the time taken by car A to stop will be (a) 5 s (c) 5 / 2 s

(b) 5 2 s (d) 2.5 s

36. Out of cars B and C, which one stops quickly? (a) B (c) Both take same time

(b) C (d) Cannot be said

Work, Energy and Power 37. Out of cars A and B, which one travels longer distance before stopping.?

38. Out of cars A and B, which one stops first? (a) A (b) B (c) Both stop at the same time (d) Cannot say

39. Out of cars B and C, which one would travel larger distance before stopping?

is equal to the work done on it by the net force. Reason Change in kinetic energy of particle is equal to the work done only in case of a system of one particle.

Passage II In a conservative force field, we can find the radial component of force F from the potential energy function (U) using the relation F = dU . Positive values dr of F mean repulsive forces and vice-versa. We can find

the equilibrium position, where force is zero. We can also calculate ionisation energy, which is the work done to move the particle from a certain position to infinity. Let us consider a particle bound to a certain point at a distance r from the centre of the force. The potential energy function of the particle is given by A B U ( r) = 2 - where A and B are positive constants. r r

40. The nature of equilibrium is (b) stable (d) Cannot be predicted

-3 B2 represents total energy of particle and the 16 A

motion is radial only, then velocity will be zero at 2r 2r r (a) r0 (b) 0 (c) 0 (d) 0 3 5 3

42. The work required to move the particle from equilibrium position to infinity is (a)

B2 4A

(b)

4 B2 A

(c)

of these contains two Statements : Statement I (Assertion), Statement II (Reason). Each of these questions also has four alternative choice, only one of which is correct. You have to select the correct choices from the codes (a), (b), (c) and (d) given below : (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 correct explanation of the Assertion (c) If Assertion is true but Reason is false (d) If Assertion is false but the Reason is true

44. Assertion The change in kinetic energy of a particle

(a) B (b) C (c) Both travel equal distance (d) Cannot say

41. If E =

Assertion and Reason Directions Question No. 44 to 49 are Assertion-Reason type. Each

(a) A (b) B (c) Both travel the same distance (d) Cannot say

(a) neutral (c) unstable

253

4B A

43. The nature of the force is (a) attractive always (b) repulsive always (c) may be attractive or repulsive (d) Cannot predict

(d)

4A B

45. Assertion Power developed in circular motion is always zero. Work done in case of circular motion is zero. Kinetic energy 46. Assertion Stopping distance = Stopping force Reason Work done in stopping a body is equal to change in kinetic energy of the body. Reason

47. Assertion Two springs of force constants k1 and k2

are stretched by the same force. If k1 > k2 , then work done in stretching the first (W1) is less than work done in stretching the second (W2 ). Reason F = k1x1 = k2 x2

x1 k2 = x2 k1 1 k x2 k W1 2 1 1 = 1 = W2 1 k x 2 k2 2 2 2

2

æ k2 ö k2 ç ÷ = k1 è k1 ø

As k1 > k2, W1 < W2

48. Assertion Mass and energy are not conserved separately, but are conserved as a single entity called ‘mass-energy’. Reason This is because one can be obtained at the cost of the other as per Einstein equation. E = mc 2

49. Assertion Energy released when a mass of one microgram disappears in a process is 9 ´ 107 J. 1 Reason It follows from E = mv2 2

254 JEE Main Physics Previous Years’ Questions 50. This question has statement I and statement II. Of the four choices given after the statements, choose the one that best describes the two statements. Statement I A point particle of mass m moving with speed v collides with stationary point particle of mass M. If the maximum energy loss possible is given as æ m ö. æ1 ö f ç mv2 ÷, then f = ç ÷ èM + mø è2 ø Statement II Maximum energy loss occurs when the particles get stuck together as a result of the [JEE Main 2013] collision. (a) Statement I is true, Statement II is true, and Statement II is the correct explanation of Statement I (b) Statement I is true, Statement II is true, but Statement II is not the correct explanation of Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

51. A block of mass 5 kg is resting on a smooth surface at what angle a force of 20 N be acted on the body so that it will acquired a kinetic energy of 40 J after moving 4 m [Orissa JEE 2011] (a) 30° (c) 60°

(b) 45° (d) 120°

56. A block of mass 2 kg is free

F(t)

N to move along the x-axis its is at rest and form t = 0 onwards it is subjected to a 4.5 s t time dependent force F ( t) in O 3s the X-direction. The force F ( t) varies with t as shown in the figure. The kinetic energy of the block after 4.5 [IIT JEE 2010] second is

(a) 4.50 J

(b) 7.50 J

(c) 5.06 J

(d) 14.06 J

57. A variable force given by the two-dimensional vector

F = (3 x2 $i + 4 $j) acts on a particle. The force is in newton and X is in metre. What is the change in the kinetic energy of the particle as it moves from the point with coordinates (2, 3) to (3, 0) (The coordinates are in metres)? [AMU (Med) 2010] (a) –7 J

(b) zero

(c) +7 J

(d) +19 J

58. Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional force are 10% of energy. How much power is generated by the turbine ( g = 10 m/s2 )? [CBSE PMT 2008] (a) 12.3 kW (b) 7.0 kW

(c) 8.1 kW

(d) 10.2 kW

52. A force ( 4 i$ + $j - 2k$ ) N acting on a body maintains its

59. A body of mass m is accelerated uniformly for rest to a

$ ) ms -1. The power exerted is velocity at (2 $i + 2 $j + 3 k

speed v in a time T. The instantaneous power delivered to the body as a function of time is given by

[Kerala CET 2010]

(a) 4 W

(b) 5 W

(c) 2 W

53. A body of mass M is moving with a

(d) 8 W F1

M

F2

uniform speed of 10 m/s on frictionless surface under the influence of two forces F1 and F2 . The net power of the system is [MP PET 2010]

(b) 10 ( F1 + F2 ) M (d) zero

(a) 10 FF 12 M (c) ( F1 + F2 ) M

54. A 2 kg block slides on a horizontal floor with a speed

of 4 ms–1. It strikes an uncompressed spring and compresses it till the block is motionless. The kinetic friction force is 15 N and spring constant is [AIEEE 2007] 10000 ms–1. The spring compresses by (a) 8.5 cm (c) 2.5 cm

1 mv2 (b) t 2 T2 mv2 (d) 2 t T

60. A particle is placed at the origin and a force F = kx is acting on it (where k is a positive constant). If U(0) = 0, the graph of U(x) versus x will be, figure (where U is the potential energy function) [UP SEE 2004] (a)

(b)

U ( x)

U(x)

x

x

(b) 5.5 cm (d) 11.0 cm

55. An engine pumps water through a hose pipe. Water passes through the pipe and leaves to with a velocity of 2 m/s. The mass per unit length of water in the pipe is 100 kg/m. What is the power of the engine? [CBSE PMT 2010]

(a) 800 W

[AIEEE 2005]

1 mv2 2 (a) t 2 T2 mv2 (c) 2 t2 T

(b) 400 W

(c) 200 W

(d) 100 W

(c)

(d)

U(x)

x

U(x)

x

Work, Energy and Power

255

Answers Round I 1. 11. 21. 31. 41. 51. 61. 71. 81.

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

2. 12. 22. 32. 42. 52. 62. 72. 82.

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

3. 13. 23. 33. 43. 53. 63. 73. 83.

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

4. 14. 24. 34. 44. 54. 64. 74. 84.

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

5. 15. 25. 35. 45. 55. 65. 75. 85.

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

6. 16. 26. 36. 46. 56. 66. 76. 86.

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

7. 17. 27. 37. 47. 57. 67. 77. 87.

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

8. 18. 28. 38. 48. 58. 68. 78. 88.

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

9. 19. 29. 39. 49. 59. 69. 79.

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

10. 20. 30. 40. 50. 60. 70. 80.

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

8. 18. 28. 38. 48. 58.

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

9. 19. 29. 39. 49. 59.

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

10. 20. 30. 40. 50. 60.

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

Round II 1. 11. 21. 31. 41. 51.

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

2. 12. 22. 32. 42. 52.

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

3. 13. 23. 33. 43. 53.

(b) (d) (b) (b,d) (a) (d)

4. 14. 24. 34. 44. 54.

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

5. 15. 25. 35. 45. 55.

(a) (b) (b) (b) (a) (a)

6. 16. 26. 36. 46. 56.

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

7. 17. 27. 37. 47. 57.

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

the Guidance Round I 1. v =

dx d æ t 3 ö 2 = ç ÷ =t dt dt è 3 ø

4. As the magnetic forces due to motion of electron and proton

When t = 0, then v = 0, when t = 2, then v = 4 m/s 1 1 W = m[( 4) 2 - (0) 2] = ´ 2 ´ 16 = 16J 2 2

act in a direction perpendicular to the direction of motion, no work is done by these forces. That is why one ignores the magnetic force of one particle on another.

5. Work done by hand = maximum PE of the ball F ´ s = mgh mgh F= s 0.2 ´ 10 ´ 2 F= = 20 N 0.2

2. W = mg sin q ´ S = 2 ´ 10 3 ´ sin 15° ´ 10 = 5.13 kJ s

m

g

sin

θ

θ

3. Tension in the string g ö Mg æ T = M( g - a) = Mç g - ÷ = è 2ø 2 W = Force ´ displacement Mgh =2

6. Initial height of CG = 4 cm Final height of CG = 10 cm Increase in height = 6 cm = 0.06 m Work done = 5 ´ 10 ´ 0.06 = 3 J

7. As, F = mmg cos q or or

F = 0.30 ´ 10 ´ 10 cos 45° 30 N F= 2 30 150 2 ´5 = ´ = 75 2 J W =F´s= 2 2 2

This is negative work because F and s are oppositely directed.

256 JEE Main Physics 8. Force, F =

3 mg 10

15. Mass of the body (m) = 0.5 kg W = - Fs 3 W =mgs 10 3 W =´ 200 ´ 10 J = - 600 J 10

As, or or

9. Force between two protons = force between a proton and a positron. As positron is much lighter than proton, it moves away through much larger distance compared to proton . As work done = force ´ distance, therefore in the same time t, work done in case of positron is more than that in case of proton.

10. When the man gets straight up and stand, reaction of ground on the man = mg. However, when he is squatting on the ground, reaction of ground is more than mg , as the man is to exert some extra force on the ground to stand up.

11. Initial velocity of ball is zero i. e. ,u = 0 \ Displacement of ball in t th second 1 æ 1ö s = gt - g = g çt - ÷ è 2ø 2 æ 1ö s µ çt - ÷ è 2ø or Now,

where,

a = 5 m-1/ 2/s

Velocity of the body at x = 0, v1 = 5 ´ 0 = 0 Velocity of the body at x = 2 m, v 2 = 5 ´ (2)3 / 2 m/s According to work-energy theorem, Work done = Change in kinetic energy 1 1 = mv 22 - mv12 2 2 1 = m(v 22 - v12) 2 1 = ´ 0.5 [5 (23 / 2) - (0) 2] 2 1 = ´ 0.5 ´ 2.5 ´ 23 2 1 = ´ 12.5 ´ 8 = 50 J 2 M 16. dW = - m éê ùú gl dl ëLû 2L

W =ò3 0

1ö æ 1ö æ 1ö æ s1: s2: s3 = ç1 - ÷ : ç2 - ÷ : ç3 - ÷ = 1 : 3 : 5 è 2ø è 2ø è 2ø W = mgs

W µs \ W1: W2: W3 = 1: 3 : 5 a 12. Initial height of CG = 2 b Final height of CG = 2 é b aù Work done = mg = mg êë 2 2 úû

Velocity of the body (v) = ax3 / 2

2L

or

13. Here t = x + 3 x = (t - 3) 2 = t 2 - 6 t + 9 dx v= = 2t - 6 dt at t = 0 s,v = 2 ´ 0 - 6 = - 6 at t = 6 s, v = 2 ´ 6 - 6 = + 6 Initial and final KE are same hence no work is done 1 W = m (v12 - v 22) = 0 2

or

14. This is the case of work done by a variable force 5

W = ò (3x2 - 2x + 7) dx 0

W = | x3 + x2 + 7x|50 or

W = (5 ´ 5 ´ 5 - 5 ´ 5 + 7 ´ 5)

or

W = (125 - 25 + 35) = 135 J

mMg é l 2 ù 3 W =L êë 2 úû 0

mMg 4 L2 -0 2L 9 2 or W = - mMgL 9 Weight in air 17. Relative density = Loss of weight in water 5 ´ 10 \Loss of weight in water = N 3 100 50 ö æ Weight in water = ç50 N ÷N= è 3 3 ø 100 500 Work done = N ´ 5m = J 3 3 dx 18. As, v = = 3 - 8 t + 3 t 2 dt \ v 0 = 3 m/s and v 4 = 19 m/s 1 W = m (v 42 - v 02) 2 (According to work energy theorem) 1 = ´ 0.03 ´ (19 2 - 33) = 5.28 J 2 1 19. Here, 1400 ´10 ´10 + W = ´15 ´15 2 or W = 700 ´ 15 ´ 15 - 1400 ´ 10 ´ 10 = 700(225 - 200)J = 700 ´ 25J = 17.5 kJ or

æ b - aö ÷ ç è 2 ø

mMg l dl L

W =-

Work, Energy and Power 20. When the block moves vertically downward with acceleration

g , then tension in the cord 4 gö 3 æ T = M ç g - ÷ = mg è 4ø 4

Work done by the cord F × s = Fs cos q = Td cos180° d ö æ 3 = ç - Mg ÷ ´ d = -3 Mg ø è 4 4

21. Work done in lifting water and drum = 60 ´10 ´ 20 J = 12000 J Total mass of ropes = 4 ´ 0.5 kg = 20 kg Work done in the case of ropes = 20 ´ 10 ´ 10 = 2000 J Total work done = 14000 J

22. As, dW = Fdl l

W = ò Fdl Y = 0

or or or or or

Fl dl

Yal dl L Yal F= L Ya l W= l dl L ò0

W =ò

l

W=

Ya æ l 2 ö ç ÷ L è2ø 1 1 Yal l = Fl 2 2 L

23. As the road does not move at all, therefore, work done by the cycle on the road must be zero. L 1 24. The weight of hanging part æç ö÷ of chain is æç Mg ö÷. This ø è3 è3ø weight acts at centre of gravity of the hanging part which is at a depth of L /6 from the table. As work done = force ´ distance Mg L MgL W= ´ = \ 3 6 18

25. In inelastic collision between two bodies, total linear 26.

momentum remains conserved. 1 1´ 5 As, W1 = kx12 = ´ 10 3 ´ (5 ´ 10 -2) 2 = 6.25 J 2 2 1 W2 = k ( x1 + x2) 2 2 1 ´ 5 ´ 10 3(5 ´ 10 -2 + 5 ´ 10 -2) 2 = 25 J 2 Net work done = W2 - W1

2

10 ´ 10 ´ h 2 h = 2.0 m

100 =

= 18.75 N - m

h h/2 A

10 × g

28. As both surfaces I and II are frictionless and two stones slide from the same height, therefore, both the stones reach the æ1 ö bottom with same speed ç mv 2 = mgh÷ . As acceleration è2 ø down plane II is larger ( a2 = g sin q2 greater than a1 = g sin q1 ), therefore, stone II reaches the bottom earlier than stone I. +a

29. W = ò Fdy = ò ( Ay 2 + By + C)dy -a

+a

ù ù é Aa3 Ba2 é Aa3 Ba2 ù é Ay3 By 2 ê 3 + 2 + Cy ú = ê 3 + 2 + Caú - ê - 3 + 2 - Caú û û ë û -a ë ë 2Aa3 = + 2Ca 3

30. Work done = Area between the graph and position axis W = 10 ´ 1 + 20 ´ 1 - 20 ´ 1 + 10 ´ 1 = 20 erg

31. According to the graph the acceleration a varies linearly with the coordinate x.We may write a = ax, where a is the slope of the graph 20 a= mg 0 = 2.5 ms-2 8 The force on the brick is in the positive x- direction and according to Newton's second law, its magnitude is given by a a F= = x m m if xf is the final coordinate, the work done by the force is xf a xf W = ò Fdx = ò xdx 0 m 0 a 2 2.5 = ´ (8) 2 = 8 J xf = 2m 2 ´ 10

32. Let m be the mass of the block,h the height from which it is dropped, and x the compression of the spring. Since, energy is conserved, so Final gravitational potential energy = final spring potential energy 1 or mg (h + x) = kx2 2 1 or mg (h + x) + kx2 = 0 2 or kx2 - 2mg (h + x) = 0 kx2 - 2mgx - 2mgh = 0 This is a quadratic equation for x. Its solution is

= 25 - 6.25 = 18.75 J

B

h 27. Work done = mg æç ö÷ è ø

0

W=

257

x= Now,

mg ± (mg ) 2 + 2mghk k

mg = 2 ´ 9.8 = 19.6 N

258 JEE Main Physics and \

33.

hk = 0.40 ´ 1960 = 784 N 19.6 ± (19.6) + 2(19.6)(784) 1960 or = 0.010 m -0.080 m

x=

1 2 1 kx - k ( x + y) 2 2 2 1 2 1 2 or W = kx - k ( x + y 2 + 2xy) 2 2 1 2 1 2 1 2 1 1 = kx - kx - ky - k (2xy) = - kxy - ky 2 2 2 2 2 2 1 = ky( -2x - y) 2 The work done against elastic force is ky Wext = -W = (2x + y) 2 1 37. For first ball mgh1 = mu2 2 u2 h1 i. e. , h1 = u 2g For second ball or

2

Since, x must be positive (a compression), we accept the positive solution and reject the negative solution. Hence x = 0.10 m. 1 1 Initial KE of the body = mv 2 = ´ 25 ´ 4 = 50 J 2 2 Work done against resistive force = Area between F-x graph 1 = ´ 4 ´ 20 = 40 J 2 Final KE = Initial KE – Work done against resistive force = 50 - 40 = 10 J

34. In this case motion of stone is in vertical circle of radius L and

mgh2 = mg

centre at O. The change in velocity is O

Y

W=

v

m

u 2 cos2 q 2g

=

1 1 mu 2 cos2 q = mu 2 cos2 60° 2 2

=

1 1 æ 1ö æ 1ö mv 2 ç ÷ = mv 2 ç ÷ è ø è 4ø 2 2 2

L

2

L u

u2 8g

Dv = v - u = v$j - u$i

h1 u 2 8 g = ´ h2 2 g u 2

Þ

| Dv| = (v) 2 + ( -u) 2 = v 2 + u 2

u cos 60°

According to work-energy theorem, or

W = DK 1 1 WT + Wg = mv 2 - mu 2 2 2

u

\

…(i)

v 2 = u 2 - 2gL | Dv| = v 2 + u 2 = 2 (u 2 - gl)

u sin 60°

Þ

1 2 Differentiating w.r.t. time t , æ ds ö æ dv ö m (2v) ç ÷ = (2a)(2s) ç ÷ è dt ø è dt ø

\

this point is zero.Therefore, kinetic energy K = 0. The total energy E is in the form of potential energy i.e., V = E.

\

W = -DV1 where, W = work done by elastic force of string W = - (Vf - Vi ) = Vi - Vf

h1 4 = h2 1

38. Given, K = as2 or mv 2= as2

35. At x = + xm , the particle turns back. Therefore, its velocity at

36. Elastic force in string is conservative in nature.

h2

60°

WT = work done by the force of tension = 0 Wg = work done by the force of gravity = mgL (path independent) 1 1 From Eq. (i), 0 - mgL = mv 2 - mu 2 2 2 Q

h2 =

Þ or

ds =v dt dv 2m = 4as dt dv m = 2as dt Ft = 2as F = Ft2 + Fr2 F = Ft = 2as

(\Fr = 0)

Work, Energy and Power 39. Here, m = 0.5 kg, v = ax3/ 2, where a = 5 m-1/ 2 s-1,W = ? dv d dx 3 = ( ax3 / 2) = a ´ x1/ 2 dt dt dt 2 3a 1/ 2 3 / 2 x ( ax ) = 2 3 a2 2 A= x 2 3 a2 2 3 a2 2 F = mA = 0.5 ´ x = x 2 4

Acceleration, A =

2

2

0

0

W = ò F. dx = ò =

K2 h2 = K1 h1

Given

K2 =

\

ux = u cos q 2

3 a2 2 3a2 é x3 ù x dx = 4 4 êë 3 úû 0

q = 45°

Given

ux = u cos 45° =

\

=

46. U( x) = (\ K2 = 2 K1)

v12 + 4v1 + 4 = 2v12 v12 - 4v1 - 4

2 1 æ u2 ö E 1 æ u ö mç ÷ = mç ÷ = 2 è2ø 2 2 è 2ø

=0

4 ± 16 + 16 2 4 + 32 v1 = = 2( 2 + 1) ms-1 2 v1 =

41. If there is no air drag then maximum height

\

a b dU - 6 at the stable equilibrium =0 12 dx x x -

12a 6b æ 2a ö + 7 =0 Þ x= ç ÷ 13 èbø x x

1/ 6

¶U $ ¶U $ ij = 7 $i - 24 $j ¶x ¶y F 7 \ ax = X = = 14 . ms-2 along positive x- axis m 5 F 24 ay = Y = m 5

47. F =

= 4.8 ms–2 along negative y-axis

u 2 14 ´ 14 = = 10 m 2 g 2 ´ 9.8

\

v x = axt = 1.4 ´ 2 = 2.8 ms–2

and

v y = 4.8 ´ 2 = 9.6 ms-1

But due to air drag ball reaches up to height 8 m only so, loss of energy

\

H=

= Mg (10 - 8) = 0.5 ´ 9.2 ´ 2 = 9.8 J

v = velocity of boy; V = velocity of man 1 1 é1 ù mV 2 = mv 2 úû 2 2 êë 2

42. Here, P = [ML T ] = constant As, mass M of body is fixed, L2 T -3 = constant L = constant or L µ T 3 / 2 or displacement (d) µ t 3 / 2. T3 1 As, K = mv 2 2 98 ´ 2 v2 = = 98 2 2

98 v = =5 2g 2 ´ 9.8 1 1 K1 = mv 2 = m ´ 2gh 2 2 h=

v = v x2 + v y2 = 10 ms-1

48. Let m = mass of boy, M = mass of man

2 -3

2

æ 1 ö 2 ç\ mu = E ÷ è 2 ø

45. Due to conservation of total energy.

2

1 v12 = 2 (v1 + 2) 2

Þ

u 2

Hence, at the highest point kinetic energy 1 E' = mux2 2

W = 50 J

43.

h2 = 2.5 m

is zero and only horizontal component is left which is

3 (5) 2 [23 - 0 ] 4 ´3

K1 æ v1 ö =ç ÷ K2 è v 2 ø

K1 K1 h2 = = 2 2K1 5

44. At the highest point of its flight, vertical component of velocity

1 40. KE = mv 2 2 Given, v 2 = (v1 + 2) \

\

259

1 æ1 ö M (V + 1) 2 = 1 ç mv 2÷ è2 ø 2 M 1 Putting m = and solving, we get, V = 2 2 -1

49. For first condition, initial velocity = u, final velocity = u / 2, s = 3 cm From

v 2 = u 2 - 2 as 2

Þ

æuö 2 ç ÷ = u - 2 as è2ø

…(i) …(ii)

260 JEE Main Physics Þ

a=

3 u2 8s

55. Vertical height = h = l cos q = l cos 30° Loss of potential energy = mgh = mgl cos 30° =

For second condition, Initial velocity = u / 2, final velocity = 0, s = x From v 2 = u 2 - 2 as

A

u2 - 2 ax 4 8 u2 u2 ´ 8 s = = 1cm = x= 3 4 ´ 2 a 4 ´ 2 ´ 3 u2

h 30°

0=

Þ

l

B

\ Kinetic energy gained =

50. (a) If the surface is smooth, then the kinetic energy at B never

51.

be zero. (b) If the surface is rough, the kinetic energy at B be zero. Because, work done by force of friction is negative. If work done by friction is equal to mgh then, net work done on body will be zero. Hence, net change in kinetic energy is zero. Hence, (b) is correct. (c) If the surface is rough, the kinetic energy at Bmust be lesser than mgh. If surface is smooth, the kinetic energy at B is equal to mgh. (d) The reason is same as in (a) and (b) 1 E = kx2 2

continuously in overcoming resistance due to air. Therefore, total mechanical energy of the pendulum decreases continuously with time. The variation of total mechanical energy E with time t is shown correctly by curve (c).

Þ Initial KE,

Final KE,

1 m1v12 2 1 æ 50 ö 2 = ´ç ÷ ´ 10 = 2.5 J 2 è1000 ø 1 E 2 = (m12 + m22) v 2 2 1 (50 + 950) 1 = ´ = 0.125J 2 1000 2 E2 =

Percentage loss in KE E1 - E 2 2.5 - 0.125 ´ 100 = = 95% E1 2.5

54. Kinetic energy = ´ 950 ´ æç100 ´ 1 2

è

2

5ö ÷ J 18 ø

= 0.3665 ´ 10 6 J = 0.376 MJ

p1 m1 = = p2 m2

\

[Q E k is given to be constant] 1 1 = 4 2

57. Here, m = 5 kg, r =1m 300 rps = 5 rps = 5 ´ 2 p rad s-1 60 1 1 1 KE = mv 2 = m (rw) 2 = ´ 5 (1 ´ 10 p ) 2 = 250 p 2 J 2 2 2 490 Potential energy at the required height = = 245J 2 Again, 245 = 2 ´ 10 ´ h 245 or h= m = 12.25 m 20 w=

58.

59. As the shotput reaches the ground, its KE

60.

m1v1 = (m1 + m2)v m1v1 50 ´ 10 1 v= = = ms-1 m1 + m2 (50 + 950) 2

pµ m

or

53. Applying principal of conservation of linear momentum velocity of the system (v) is

3 mgl 2

56. p = 2mEK

F µk E1 k1 = E 2 k2

52. When a pendulum oscillates in air, it loses energy

3 mgl 2

= PE of shotput when it is thrown + KE given 1 1 = mgh + mv 2 = 10 ´ 10 ´ 1.5 + ´ 10 (1) 2 2 2 = 150 + 5 = 155 J 75 Useful work = ´ 12J = 9 J 100 1 Now, ´1´ v 2 = 9 2 or v = 18 ms-1

61. Initially, 4 u = 8 Þ u = 2 m/ s mv - mu = Ft mv - 8 = 0.2 ´ 10 v = 5 / 2 ms-1 1 Increase in KE = m(v 2 - u 2) 2 2 ù 1 é æ5ö = ´ ê 4 ç ÷ - (2) 2ú = 4.5 J 2 êë è 2 ø úû

Now, or

62. If the body strikes the sand floor with a velocity v, then 1 mv 2 2 With this velocity v, when body passes through the sand floor it comes to rest after travelling a distance x. Let F be the resisting force acting on the body. Net force in downwards direction = Mg - F Mgh =

Work, Energy and Power Work done by all the forces is equal to change in KE 1 (Mg - F) x = 0 - Mv 2 2 (Mg - F) x = -Mgh or Fx = Mgh + Mgx æ hö or F = Mg ç1 + ÷ è xø

63. Because the efficiency of machine is 90%, hence, potential energy gained by the mass 90 = ´ energy spent 100 90 = ´ 5000 = 4500 J 100

64.

When the mass is released now , gain in KE on hitting the ground = Loss of potential energy = 4500 J dx x = 2t4 + 5t + 4 = v = = 8t + 5 dt At t = 0 ,v = 5 m/s At t = 1s,v = 8 ´ 1 + 5 = 13 m/s 1 Increase in KE = m [(13) 2 - (5) 2] = 144 J 2

65. Both fragments will possess the equal linear momentum m1v1 = m2v 2 Þ 1 ´ 80 = 2 ´ v 2 Þ v 2 = 40 m/s 1 1 \ Total energy of system = m1v12 + m2v 22 2 2 1 1 = ´ 1 ´ (80) 2 + ´ 2 ´ ( 40) 2 2 2 = 4800 J = 4.8 kJ

66.

9 kg At rest

v1 = 1.6 m/s

Before explosion

m1

m2

3 kg

6 kg

After explosion

68. Net force on the body is zero and its potential energy should be minimum for stable equilibrium. 1 69. As, E = mg 2t 2 2 1 mg 2 ´ 3 2 E1 9 1 2 = = = 1 E2 mg 2(6 2 - 3 2) 9 ´ 3 3 2 a 70. Height of CG of mass m1 = 2 b Height of CG of mass m2 = a + 2 \ Gravitational potential energy of system a bö ém b æ ù = m1g + m2g ç a + ÷ = 1 + m2 ga + m2g úû è 2 2 ø êë 2 2 bù éæm ö = ê ç 1 + m2÷ a + m2 ú g ø 2û ëè 2

71. Let a body of mass m which is initially at rest undergoes one-dimensional motion under a constant force F with a constant acceleration a. F Acceleration ( a) = …(i) m Using equation of motion, v = u + at F Þ v = 0 + ×t m F v= t Þ m

As there is no external force \

m1v1 + m2v 2 = 0

3 ´ 1.6+6 ´ v 2 = 0 Velocity of 6 kg mass v 2 = 0.8 m/s (numerically) 1 1 its kinetic energy = m2v 22 = ´ 6 ´ (0.8) 2 = 1.92 J 2 2 mass dm 67. K = = length dt 1 KE = mv 2 2 1 d æ dm ö 2 1 æ dm dx ö 2 1 ´ ÷ v = kvv 2 = kv3 Þ ÷v = ç ç 2 è dx dt ø 2 2 dt è dt ø

(Q u = 0) …(ii)

Power delivered (P) = Fv Substituting the value from Eq. (ii), we get F Þ P =F ´ ´t m Þ

P=

F2 t m

Dividing and multiplying by m in RHS,

At the bomb initially was at rest therefore initial momentum of bomb = 0 Final momentum of system = m1v1 + m2v 2

261

P=

F2 ´ mt = a2mt m2

[Using Eq. (i)]

As mass m and acceleration a are constants. \

P µt

72. As, m = 10 ´ 0.8 kg, h = 5 m \

P=

mgh 8 ´ 10 ´ 5 = t 10

= 40 W

73. Volume of water of raise = 22380 L = 22380 ´10 -3 m3 mgh Vrgh = t t Vrgh t= r

P= Þ

t=

22380 ´ 10 -3 ´ 10 3 ´ 10 ´ 10 = 5 min 10 ´ 746

262 JEE Main Physics 74. Power = 7500 W = 7500 Js-1, Velocity v = 20 ms-1 P = Fv or F =

-1

Let rate of flow of motor = ( x) work mgy æy ö Since, power P0 = = = mg ç ÷ èt ø t time y …(i) = x = rate of flow of water = mgx t If rate of flow of water is increased by n times, i. e. ,(nx) Increased power, mgy ¢ æ y¢ö …(ii) P1 = = mg ç ÷ = nmgx èt ø t

P 7500 Js = 375 N = v 20 ms-1

mgh m P or = t t gh m 1000 or = kg = 10 kg t 10 ´ 10 10 - 0 76. As, a = ms-2 = 2 ms-2 5 Q F = ma \ F = 1000 ´ 2 N = 2000 N 0 + 10 Average velocity = ms-1 = 5 ms-1 2 Average power = 2000 ´ 5W = 10 4 W 10 4 Required horse power is 746

75. As, P =

The ratio of powers P1 nmgx = P0 mgx Þ

accelerated due to gravity and retarded due to viscous force. The overall effect is increase in velocity and hence increase in KE till the sphere acquires terminal velocity, which is constant. Hence KE of sphere beyond this depth of lake becomes constant. Choice (b) is most appropriate. dm 78. Given, h = 500 m , = 2000 kgs-1 dt 80 dm \ Power output = ´ gh 100 dt 4 = ´ 2000 ´ 10 ´ 500 W 5 = 8 ´ 10 6 W = 8 MW

79. Minimum force mg sin q. So, minimum power is given by or or

(Q F = mg sin q )

80. Given, u = 10 ms-1, v = 20 ms-1 \

P1 : P0 Þ n : 1

83. Motor makes 600 revolution per minute

77. When an iron sphere falls freely in a lake, its motion is

P = mg sin q v P v= mg sin q 9000 ´ 2 v= = 15 ms-1 1200 ´ 10 ´ 1 18 = 15 ´ = 54 kmh -1 5

82. Power of motor initially = P0

Work done = Increase in kinetic energy 1 = ´ 500 [20 2 - 10 2] 2 500 ´ 30 ´ 10 = 2 500 ´ 30 ´ 10 Power = = 1250 W 2 ´ 60

81. Volume = av = pr 2v Mass = pr 2v ´ 1000 SI units 1 mv 2 1 2 Power of water jet = = ´ pr 2v ´ 1000 ´ v 2 = 500 pr 2v3 t 2

revolution rev = 10 minute sec 1 sec \Time required for one revolution = 10 Energy required for one revolution = Power ´ time 1 1 746 J = ´ 746 ´ = 4 10 40 But work done = 40% of input 746 = 40% ´ 40 40 746 = ´ = 7.46 J 100 40 1 mgh + mv 2 total energy 2 84. As power, P = = t t 1 10 ´ 10 ´ 20 + ´ 10 ´ 10 ´ 10 2 = 1 \

n = 600

= 2000 + 500 = 2500 W = 2.5 kW

85. P = constant Þ Þ Þ Þ Þ

Fv = P Ma ´ v = P P va = M vdv P v´ = ds M v 2 s P ò0 v dv =ò0 M ds

[Q P = force ´ velocity] [Q F = Ma]

vdv ù é Qa= êë ds úû

[Assuming at t = 0 it starts from rest, i. e., from s = 0 ] Þ Þ

v3 P = s 3 M æ 3P ö v=ç ÷ èMø

1/3

´ s1/3

Work, Energy and Power 2

1/3 é ds æ 3P ö ù = ks1/3 êk = ç ÷ ú èMø ú dt êë û

Þ S

ds

æ dx ö ç ÷ = k1t è dt ø dx = k1t dt dx = k2(t )1/ 2 dt

t

ò0 s1/3 = ò0kdt

Þ

s2/3 = kt 2 /3

Þ

æ2 ö s = ç k÷ è3 ø

\

x = k3t 3 / 2

3/ 2

´t

3/ 2

88. Here ac =

86. Energy required = mgh In both, cases, h is the same. Hence, energy given by both is same. [ It is worth noting here that powers of two men will be different as power is the energy expense per unit time and times are different] æ d 2x ö ÷ è dt 2 ø

87. P = Fv = (ma) v = m ç

æ dx ö ç ÷ è dt ø

Since, power is constant.

d dt

2 ö æ çQ k3 = k2÷ è 3 ø

v2 = k2rt r

(Q v = krt )

v = krt

dv dt d(krt ) = = kr dt The work done by centripetal force will be zero. The integral acceleration is at =

So power is delivered to the particle by only tangential force which acts in the same direction of instantaneous velocity.

æ d 2x ö æ dx ö ç 2÷ ç ÷ = k è dt ø è dt ø or

\

(\ k11/ 2 = k2)

dx 1/ 2 µ t µ x1/3 dt

Hence

s µ t3/ 2

Þ

\

Power = Fv t = matkrt = m (kr)(krt )

2

æ dx ö ç ÷ =k è dt ø

= mk2r 2t

Round II 1 2

1. Length of the pendulum = 15 . m 1.5 m

O

2. As, E1 = mv 2 A

1.5 m

B

B

1 m (v + 1) 2 2 1 m [(v + 1) 2 - v 2] (E 2 - E1) 2 44 = = 1 E1 100 2 mv 2 On solving, we get v = 5 ms-1. E2 =

3. Loss of KE = force ´ distance = (ma) x = work done Potential energy of the bob at position A = mgh

As bob moves from position A towards position B its potential energy converted into kinetic energy. 5% of its potential energy is dissipated against air resistance. \

or

263

KE at position B = 95% of its PE at position A 1 95 mv 2 = ´ mgh 2 100 2 ´ 95 ´ gh v= 100 19 = ´ 9.8 ´ 15 . 10 = 5.28 m/s

As

a µ x \ Loss of KE µ x2

1 1 1 4. KE left, mv 2 = æç mu2ö÷ ø è 2

2 2

\ Velocity left, v =

u 10 4 = = 7071.06 ms–1 2 2

5. We know that, P = F × v = F ×

L T

As

F = [MLT –2] = constant

\

L µT2

\

P =F×

or

P µT

L T2 =F× = F ×T T T

264 JEE Main Physics æ v1 ö çQ a = ÷ t1 ø è

6. From v = u + at ,v1 = 0 + at1 F = ma =

mv1 t1

Velocity acquired in t sec = at =

7.

m2 = 0.1 kg v1 = velocity of recoil of gun. v 2 = velocity of bullet As (Q momentum is conserved) m1v1 = m2v 2 0.1 m2 v v1 = v2 = v2 = 2 20 200 m1

v1 t t1

Power = F × v =

m1 = 20 kg

11. Here,

mv1 v1t mv12t ´ = 2 t1 t1 t1

2mv 2 x= k

v 2 = 804 ´ 4 ´ 10 3 ms-1

AC = s v 2 = u 2 + 2g ( s - x)

12. From work energy theorem, DKE = Wnet

…(i)

v 2 = 2g ( s - x) Potential energy at B = mgx \ kinetic energy = 3 ´ potential energy 1 m ´ 2g ( s - x) = 3 ´ mgx 2 or ( s - x) = 3x or s = 4x s or x= 4 From Eq. (i) v 2 = 2g ( s - x)

…(ii)

x=

s 3gs and v = 4 2

1 3

9. Given, s = t 2 ds 2 d 2s 2 = t,a = 2 = v= 3 dt 3 dt 2 F = ma = 3 ´ = 2 N 3 1 2 W =2´ t 3 1 8 W = 2 ´ 2 ´2 = J 3 3

\

At t = 2 s,

t3 3 ds = t 2 dt

\ Þ

a=

A

2

é t3 ù v =ê ú ë 2 û0 2

v = 2 ms-1

B – KE = 3 PE

ò0 3 ´2t ´t

d s d ét ù 2 = ê ú = 2 t m/s dt 2 dt 2 ë 3 û

dt = ò

2

3

2

2

0

0

0

2

0

0

0

3 2 6 t dt = [t 4 ]0 = 24 J 2

2 d 2s ds d 2s ds = ò0 M dt 2 × dt dt dt 2 2

14. From work-energy theorem, DKE = Wnet Kf - Ki = ò P dt 2 æt2ö 1 mv 2 - 0 = ò ç ÷ dt 0 è2ø 2 2

or

1 1 é t3 ù (2)v 2 = ê ú 2 2 ë 3 û0 v =2

\

2

2

2

2 æ2ö æ2 ö 4 ét2ù = ò 3 ç ÷ ç t ÷ dt = ê ú 0 è3ø è3 ø 3 ë 2 û0 4 4 8 W = ´ = = 2.6 J 3 2 3

Þ

2 ms-1 3

dE dt

dE = (3t 2 - 2t + 1)dt E=ò

t =4 s

t =2s

(3t 2 - 2t + 1) dt t =4 s

é 3t 3 2t 2 ù =ê = + 1ú 2 û t =2s ë 3

W = ò F × ds = ò m × ads 2

2

13. As, W = ò Fds = ò Mads = ò M

Now work done by the force

2

2 æ3 ö 1 mv 2 = ò ç t 2÷ dt 0 è2 ø 2

or

15. Given, P = 3t 2 - 2t + 1 =

s=

10. Given,

Kf - Ki = ò Pdt

or

s ö 2g ´ 3s 3gs æ = 2g ç s - ÷ = = è 4ø 4 2 \

2

v 22 10 v 22 = 4 4 ´ 10 4 ´ 10 3

804 =

8. Velocity at B when dropped from A where,

1 1 æ v ö m1v12 = ´ 20 ç 2 ÷ è 200 ø 2 2

Recoil energy of gun =

1 2 1 1 kx = mv 2 + mv 2 = mv 2 2 2 2

= [( 43 - 23) - ( 4 2-2 2) + ( 4 + 2)]

3

or

E = 56 - 12 + 2 = 46 J

Work, Energy and Power P = (mg sin q + F)v

16. As,

= 14000 W = 14kW dW dv P= =P dt dt P = h d g = 10 ´ 13.6 ´ 980 = 1.3328 ´ 10 6dyne/cm2

and \

3 1 th, therefore, KE left is th. Hence, velocity of 4 4 v0 particle reduces from v 0 to = v 0 - mg t 0 2 v or m= 0 2gt 0

24. KE lost is

1 ö æ = ç1000 ´ 10 ´ + 200 ÷ ´ 20 ø è 20

17. As,

25. If W1 = work done by applied force W2 = work done against friction then applying work-energy theorem

dv = pulse frequency ´ blood discharged per pulse dt dv 72 = ´ 75 = 90 cc/s dt 60

W1 - W2 = PE + KE (at the top) 1 F ´ s - W2 = mgh + ´ mv 2 2 1 100 ´ 12 - W2 = 50 ´ 10 ´ 2 + ´ 50 ´ 2 2 2 1200 - W2 = 1100 W2 = 100 J

\ Power of heart = 1.3328 ´ 10 5 ´ 90 erg /s = 1.19 W 1 2

18. As, K = mv 2 dK dv æ dv ö = çm ÷v = mav = 4v = mv × dt dt è dt ø m = 2 kg and a = ms-2

Q

19. As slope of problem graph is positive and constant upto certain distance and then it becomes zero. - dU So, from F = , upto distance dx F = constant (negative) and become zero suddenly.

20. As, K1 = mgh1 and K2 = mgh2 \

%Loss =

K1 - K2 h -h ´ 100 = 1 2 ´ 100 = 50% K1 h1

26. Let mass of boy be m, therefore, mass of man = 2 m, 1 KE of boy 2 1 1 1 \ (2 m) u 2 = ´ mu ¢2 2 2 2 u'2 u' 2 ,u = u = 4 2 When man speeds up to 1 m/s KE of man = KE of boy 1 1 1 (2 m) (u + 1) 2 = mu'2 = m (2 u) 2 2 2 2 (u + 1) 2 = 2 u 2 As,

KE of man =

u +1= 2u

1 2

21. As, U = ks2 = 10 J and

1 æ1 ö U' = k ( s + s) 2 = 4 ç ks2÷ = 40 J ø è2 2

\

W = U' - U = 40 - 10 = 30 J

22. Let v be the velocity with which the bullet will emerge. Now, change in kinetic energy = work done 1 1 For first case, m (100) 2 - m ´ 0 = F 2 2 1 1 For second case, m(100) 2 - mv 2 = F ´ 0.5 2 2 Dividing Eq. (ii) by Eq.(i), we get (100) 2 - (v) 2 0.5 1 or = = 1 2 (100) 2

265

100 v= = 50 2 ms-1 2

23. Initial velocity of particle, vi = 20 ms-1 Final velocity of the particle, v f = 0 According to work-energy theorem Wnet = DKE = Kf - Ki 1 = m(v f2 - vi2) 2 1 = ´ 2(0 2 - 20 2) = -400 J 2

u=

1 2 +1 = 2 - 1 ( 2 - 1) ( 2 + 1)

u = ( 2 + 1) ms-1 u' = 2 u = 2 ( 2 + 1) ms-1

27. Work output of engine = mgh = 100 ´10 ´10 = 10 4 J Efficiency (h) = \

Power = =

output 10 4 10 5 J = ´ 100 = 60 6 input input energy 10 5 / 6 = time 5 10 5 = 3.3 kW 30

28. Let x be the maximum extension of the spring as shown in figure. From conservation of mechanical energy; decreases in gravitational k potential energy = increase in elastic potential energy M 1 Mgx = kx2 \ 2 2 Mg x= \ k M

v =0 x u =0

266 JEE Main Physics 1 2

1 2

29. Energy supplied = mv 2 = (0.5) 142 = 49 J Energy stored = mgh = 0.5 ´ 9.8 ´ 8 = 39.2J \ Energy dissipated = 49 - 39.2 = 9.8 J 1 30. Here K = mv 2 = as2 2 \ mv 2 = 2as2 Differentiating w.r.t. time t dv ds dv 2mv = 4as = 4 asv Þ m = 2as dt dt dt This is tangential force, Ft = 2as mv 2 2as2 = R R \ Force acting on the particle Fc =

Centripetal force

2

æ 2as ö 2 2 F = Ft2 + Fc2 = (2as) 2 + ç ÷ = 2as 1 + s / R è R ø

EB = KB; EC = mghC + KC Using conservation of energy E A = EB = EC , if

KB > KC hA > HC ; KC > KA

and if

hA < hC ; KC < KA

32. When a man of mass m climbs up the staircase of height L, work done by the gravitational force on man is (-) mgL, and work done by muscular force is mgL. If we ignore air resistance and friction, then the work done by all forces on man is equal to - mgL + mgL = zero. Further, force from a step does not do work because the point of application of force does not move while the force exists. 1 mv 2 = a(t ) 2 or v µ t 1/ 2 dv a= µ t -1/ 2 dt 1 1 or and F µ Fµ v t

34. While spring is fully compressed, the entire KE of M1 is not stored as PE of spring as M2 may move. If spring is massless, than as M1 = M2, velocities of M1 and M2 are interchanged on collision.M1 comes to rest, instead of M2. Choice (c) is correct. If surface on which blocks are moving has friction loss of energy is involved . Collision cannot be elastic. Choice (d) is correct.

35. From

t A æ mA ö =ç ÷ tB è mB ø

1/ 2

t A æ 1000 ö =ç ÷ 5 è 2000 ø tA = 5 / 2 s

tC =1 tB

we find that both cars B and C will stop simultaneously.

37. From car A, 1 1 mAv A2 = ´ 1000(10 2) 2 = 10 5 J 2 2 1 1 For car B, KB = mBvB2 = ´ 2000(10) 2 = 10 5 J 2 2 KA =

Thus, cars A and B have same KE. When same stopping force is applied, both would travel the same distance before stopping.

38. As

t A æ mA ö =ç ÷ tB è mB ø

1/ 2

and mA < mB \ t A < tB

i. e. , car A would stop in a shorter time.

39. Linear momentum of car B = mBvB = 2000 ´ 10 = 2 ´ 10 4 kg ms-1 Linear momentum of C = mC vC = 2500 ´ 18 = 2 ´ 10 4 kg ms-1

31. As, E A = mghA + KA

33.

36. When PC = PB;

\

\ xB > xC i. e. , car B will travel large distance before stopping than car C. dU B 2A 40. From = dr r 2 r3 2B 6A d 2U =- 3 + 4 dr 2 r r 2A d 2U 2B4 6AB4 B As , 2 =- 3+ , = r ® r0 = B dr 8A 16A4 8A3 which is positive. \ U is minimum. Hence, the equilibrium is stable.

41. The velocity of particle is zero, where total energy is completely potential. A B 3B2 \ E== U(r) = 2 r 16A r 2r0 2 æ 2A ö 4 B at r= = ç ÷= 3 3è B ø 3 r 2r 2 æ 2A ö 4A r= 0 = ç ÷= Þ 3 3 è B ø 3B, A B A9B2 B.3B 9B2 3B2 3B2 = = = 16A r 2 r 16A2 4A 16A 4A which is the total energy. U(r) =

42. At r = r0 =

=

1 2

A B AB2 2A B2 B2 , U1 = 2 - = ==2 B r 4A 2A 4A r

r = ¥ , U2 = 0 \Work done to move the particle from equilibrium position to infinity, æ B2 ö B2 W = U2 - U1 = 0 - ç ÷= è 4A ø 4A At

1/ 2

PB = PC xC mB 2000 = = = 0.8 xB mC 2500

Thus

Work, Energy and Power 43. Always attractive. 44. Change in kinetic energy = Work done by net force. This

m v2 55. Power = F × v = V æç ö÷ v = è ø

motion only.

(r AV )

t

3

= r AV = 100 (2)3 = 800 W

relationship is valid for particle as well as system of particles.

45. Work done and power developed is zero in uniform circular

56. ò F dt = D p Þ

46. These tells about work energy theorem. 47. As force = kx

1 1 ´ 4 ´ 3 - ´ 1.5 ´ 2 = pf - 0 2 2

Þ

Greater the k greater will be force for constant x.

pf = 6 - 1.5 =

48. The mass may be converted into energy

KE =

E = mc2

as 1 mv 2 2

50. Maximum energy less =

F = 3 x i + 4$j , r = x$i + y k$ dr = dx$i + dy $j

Let,

æ p2 ö ÷ çQKE = 2m ø è

Work done W = ò F dr = ò

Before collision the mass m and after collision the mass is m+M ì M üæ p2 é M ù 1 M ö = mv 2 í = ÷ ý çf = ú ê 2m ë (m + m) û 2 m + Mø î m + Mþ è

51. According to work-energy theorem, W = change in kinetic energy 1 1 FS cos q = mv 2 - mv 2 2 2 Substituting the given values, we get 20 ´ 4 ´ cos q = 40 - 0 40 1 cos q = = 80 2 æ 1ö q = cos-1 ç ÷ = 60° è2ø $ N 52. Here, force F = ( 4 $i + $j - 2 k) $ ms-1 Velocity, v = (2 $i + 2$j+ 3 k) $ (2 $i + 2$j+ 3 k) $ Power P = F × v = ( 4 $i + $j - 2 k) \ = (8 + 2 - 6) = 4 W

53. Q Speed is constant. \ Work done by force = 0 work Power = \ =0 time

=ò

(3 , 0 ) ( 2, 3 )

F = 15 N, k = 1000 Nms-1, x = ? KE spent = Work done against friction + PE of spring 1 1 mv 2 = F ´ x + kx2 2 2 1 1 ´ 2 ´ 4 2 = 15x + ´ 10000 x2 2 2 5000 x2 + 15x - 16 = 0

On solving it, we get x = 0.055 m = 5.5cm

(3 , 0 ) ( 23 )

(3 x2$i + 4$j) (dx $i + dy$j) (3 , 0 )

(3x2 dx + 4 dy) = ( x3 + 4 y)( 2, 3)

= 33 + 4 ´ 0 - (23 + 4 ´ 3)

58.

= 27 + 0 - 20 = + 7 J mgh Power given to turbine = t æ mö Pin = ç ÷ gh = Pin = 15 ´ 10 ´ 60 èt ø Pin = 9000 W Þ Pin = 9 kW As efficiency of turbine is 90% therefore power generat = 90% of 9 kW 90 =9´ 100

Þ

Pout

Þ

Pout = 8.1 kW v T

mv T v Velocity acquired, V = at = t T mv v Instantaneous power, P = F ´ v = ´ t T T

59. Acceleration, a, and F = ma =

=

54. Here, m = 2 kg, v = 4 s-1

\

p2 81 = 2m 4 ´2 ´2

2$

57. Given, p2 p2 2m 2 (m + M)

9 2

KE = 5.06 J

49. The exact reason is governed by E = mc2 and not by E =

267

60. From F = -

mv 2 t T2

dU dx

dU = - Fdx U=ò

U( x )

0

U=

x

x

0

0

dU = - ò Fdx = - ò kxdx

kx2 2

As U(0) = 0 , µ x2 and U is negetive.

8 Centre of Mass JEE Main MILESTONE <
m2 )

(a) 4

m2 T m1

(c) g (d) zero

120°

æ 4ö (a) cos -1 ç ÷ è5ø

æ5ö (b) cos -1 ç ÷ è 4ø

æ 4ö (c) sin -1 ç ÷ è5ø

æ5ö (d) sin -1 ç ÷ è 4ø

17. A bullet of mass M hits a block of mass M ¢. The energy transfer is maximum, when

13. A spherical hollow is made in a

(a) M ¢ = M (c) M ¢ > M

18. A nucleus ruptures into two nuclear parts which (a) 21/3 : 1

14. Two identical blocks A and B, each of mass m resting

m (a) v 2k

(d) 1

T

2

R 7

(c) 2

at an angle q with the horizontal. At the highest point of its trajectory, it collides head-on with a bob connected with a massless string of length l = 10 / 3 m and gets embedded with the bob. After the collision, the string moves to an angle of 120°. What is the angle q ?

10 m 3 50 ms–1

æ m - m2 ö (b) ç 1 ÷ è m1 + m2 ø

(a)

(b) 3

l=

a2

2

A

16. A bullet of mass m is fired with a velocity of 50 ms–1

respectively are tied to the ends of a massless string, which passes over a light and frictionless pulley. The masses are initially at rest and then released. Then acceleration of the centre of mass of the system is æ m - m2 ö (a) ç 1 ÷ g è m1 + m2 ø

v

2v equal masses start moving in opposite direction from a point A in a horizontal circular orbit. Their tangential velocities are v and 2v respectively as shown in the figure between collisions, the particles move with constant speeds. After making how many elastic collisions other than that at A. These two particles will again reach the point A

mv (d) 2k

10 cm, each are placed on a horizontal surface touching one another so that their centres are located at the corners of a square of side 20 cm. What is the distance of their centre of mass from centre of any of the spheres ? (a) 5 cm (c) 20 cm

(b) 10 cm (d) 10 2 cm

20. A bullet of mass 50 g is fired from a gun of mass 2 kg. If the total kinetic energy produced is 2050 J, the kinetic energy of the bullet and the gun respectively are

291

Centre of Mass (a) 200 J, 5 J (c) 5 J, 200 J

(b) 2000 J, 50 J (d) 50 J, 2000 J

21. Two spherical bodies of masses M and 5M in free space with initial separation between their centres equal to 12 R. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is (a) 2.5 R

(b) 4.5 R

(c) 7.5 R

smooth horizontal surface. If the speed of the bullet relative to the gun is v, the recoil-speed of the gun will be m v M

(b)

m M M v (c) v (d) v M+m M+m m

23. A system of three particles having masses m1 = 1 kg,

m2 = 2 kg and m3 = 4 kg respectively is connected by two light springs. The acceleration of the three particles at any instant are 1  ms–2, 2 ms–2 and 0.5 ms–2 respectively directed as shown in the figure. The net external force acting on the system is (a) 1 N

(b) 7 N

(c) 3 N

(d) 6 N

24. A loaded spring gun of mass M fires a shot of mass m with a velocity v at an angle of elevation q . The gun was initially at rest on a horizontal frictionless surface. After firing, the centre of mass of gun-shot system mv (a) moves with a velocity M mv in the horizontal direction (b) moves with a velocity M cos q (c) remains at rest (M - m) v in the horizontal direction (d) moves with velocity (M + m)

25. In the shown figure the magnitude of acceleration of centre of mass of the system is ( g = 10 ms -2 ) µ = 0.2 5 kg

(b) 10 ms -2

(a)

2

æ5ö (b) ç ÷ u è9ø

3

æ5ö (c) ç ÷ u è9ø

4

æ5ö (d) ç ÷ u è9ø

28. Two bodies having masses m1 and m2 and velocities

u1 and u2 collide and form a composite system of m1v1 + m2 v2 = 0 ( m1 ¹ m2 ). The velocity of the composite system is (a) zero

(b) u1 + u2

(c) u1 - u2

(d)

u1 + u2 2

29. Two carts on horizontal straight rails are pushed apart by an explosion of a powder charge Q placed between the carts. Suppose the coefficient of friction between carts and rails are identical. If the 200 kg cart travels a distance of 36 m and stops, the distance covered by the cart weighing 300 kg is (a) 32 m

(b) 24 m

(c) 16 m

(d) 12 m

30. Three identical sphere lie at rest along a line on a smooth horizontal surface. The separation between any two adjacent spheres is L. The first sphere is moved with a velocity u towards the second sphere at time t = 0. The coefficient of restitution for collision between any two blocks is 1/3. Then choose the correct statement. 5L 2u 4L (b) The third sphere will start moving at t = u (c) The centre of mass of the system will have a final speed u/3. (d) The centre of mass of the system will have a final speed u

(a) The third sphere will start moving at t =

and mC are under the action of same constant force for the same time. If m A > mB > mC , the variation of momentum of particles with time for each will be correctly shown as p

p B

(a)

B

(b)

A

C

explodes in mid air. The centre of mass of fragments will move vertically upwards and then downwards vertically downwards in an irregular path in the parabolic path as the unexploded bomb would have travelled

C

A

(c) 2 2 ms -2 (d) 5 ms -2

26. A bomb travelling in a parabolic path under gravity, (a) (b) (c) (d)

5 u 9

31. Three stationary particles A, B, C of masses m A , mB

5 kg

(a) 4 ms -2

distances in a straight line are in geometrical progression with ratio  2 and their coefficients of restitution are each 2/3. If the first ball be started towards the second with velocity u, then the velocity communicated to 5th ball is

(d) 1.5 R

22. A bullet of mass m leaves a gun of mass M kept on a

(a)

27. The masses of five balls at rest and lying at equal

O

(c)

p

O

O

t A, B, C

t

t

p

(d)

A, B, C O

t

292 JEE Main Physics 32. A ball strikes a horizontal floor at an angle q = 45°.

38. A set on n identical cubical blocks lies at rest parallel

The coefficient of restitution between the ball and the floor is e = 1/2. The fraction of its kinetic energy lost in collision is (a) 5/8 (b) 3/8 (c) 3/4 (d) 1/4

to each other along a line on a smooth horizontal surface. The separation between the near surface of any two adjacent block is L. The block at one end is given a speed v towards the next one at time t = 0. All collisions are completely inelastic, then

33. A ball falls freely from a height of 45 m. When the ball is at a height of 25m, it explodes into two equal pieces. One of them moves horizontally with a speed of 10 ms–1. The distance between the two pieces when both strike the ground is (a) 10 m (c) 15 m

(b) 20 m (d) 30 m

More Than One Correct Option 34. A body of mass 2 kg moving with a speed of 3 ms–1 collides with a body of mass 1 kg moving with a speed of 4 ms–1. If the collision is one dimensional and completely inelastic, the speed of composite mass after the collision may be (a)

3 –1 ms 2

(c) 4 ms–1

2 –1 ms 3 10 (d) ms–1 5 (b)

35. A man of mass m is standing at one end of a boat of mass  M and length L. The body walks to the other end, the displacement of the (a) centre of mass of the system is zero m (b) boat is L M+m m (c) man is L M+m m (d) boat is L M

( n - 1) L v n ( n - 1) L (b) the last block starts moving at t = 2v (c) the centre of mass of the system will have a final speed v v (d) the centre of mass of the system will have a final speed n (a) the last block starts moving at t =

Comprehension Based Questions Passage I We sometimes encounter examples where a large force acts for very short duration of time producing an appreciable and finite change in linear momentum of the body. Such forces are know as impulsive forces. As an example, consider two identical cricket balls of mass m and initial speed u, approaching a rigid wall. One ball strikes the wall normally and rebounds with same speed. Another ball strikes the wall making and angle of 30° from normal and is elastically reflected back as shown in figure. Now answer the following questions. u

u u Ball I

36. The velocity of the centre of mass of a two particle system is v and total mass of particles is M. The kinetic energy of the system 1 2 Mv 2 (b) must be equal to or less than Mv2 1 (c) may be equal to or greater than Mv2 2 (d) cannot be exactly calculated as the information given is insufficient (a) may be equal to

37. A ball hits the floor and rebounds after an inelastic collision. In this case choose the correct statement(s). (a) The momentum of the ball just after the collision is the same as that just before the collision (b) The mechanical energy of the ball remains the same in the collision (c) The total energy of the ball and earth is conserved (d) The total momentum of the ball and earth is conserved

30° 30°

u

Ball II

39. What is the magnitude of force on the wall due to first ball ? (a) mu mu (c) 2

(b) 2 mu (d) Data insufficient

40. The ratio of magnitudes of the impulse imparted to the two balls is (a)

2 3

(b)

3 2

(c)

1 2

(d) 2

41. The force exerted on the wall by ball I and ball II is (a) parallel to the wall for both the balls (b) normal to the wall for both the balls (c) normal to the wall for ball number 1 and inclined at 30º to the horizontal for 2nd ball (d) normal to the wall for 1st ball and inclined at 30º to the vertical for 2nd ball

Centre of Mass

44. Assertion The centre of mass of a two particle system

Passage II A tennis ball is dropped from a height h0 on a horizontal marble flooring. The ball rebounds to a height h1, then again falls on the floor, again rebounds and so on.

42. The maximum height of rebound hn after n rebounds will be

(a) ne × h0

(b) n e × h0

(c) e n × h0

(d) e2 n × h0

43. If h0 = 10 m and e = 1/2, then compute the total distance travelled by the ball before it stops bouncing. (a) ¥

(b) 50 m

(c)

293

50 m 3

(d)

25 m 3

Assertion and Reason Directions

(Q. Nos. 44 to 48) Assertion-Reason type. Each of these contains two Statements : Statement I (Assertion), Statement II (Reason). Each of these questions also has four alternative choice, only one of which is correct. You have to select the correct choices from the codes (a), (b), (c) and (d) given below (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 correct explanation of the Assertion (c) If Assertion is true but Reason is false (d) If Assertion is false but the Reason is true

lies on the line joining the two particles, being closer to the heavier particle. Reason This is because product of mass of one particle and its distance from centre of mass is numerically equal to product of mass of other particle and its distance from centre of mass.

45. Assertion A quick collision between two bodies is more violent than slow collision even when initial and final velocities are identical. Reason The rate of change of determines that force is small or large.

momentum

46. Assertion The centre of mass of an electron and proton, when released moves faster towards proton. Reason This is because proton is heavier.

47. Assertion Torque is time rate of change of a parameter, called angular momentum. Reason This is because in linear motion, force represents time rate of change of linear momentum.

48. Assertion Two particles moving in the same direction do not lose all their energy in a completely inelastic collision. Reason Principle of conservation of momentum holds true for all kinds of collisions.

Previous Years’ Questions 49. Which of the following statement(s) is wrong? [UP SEE 2009]

(a) KE of a body is independent of the direction of motion (b) In an elastic collision of two bodies, the momentum and energy of each body is conserved (c) If two protons are brought towards each other, the potential energy of the system decreases (d) A body cannot have energy without momentum

50. The acceleration of the centre of mass of a uniform solid disc rolling down an inclined plane of angle a is [UP SEE 2008]

(a) g sin a (b) 2/3 g sin a (c) 1/2 g sin a (d) 1/3 sin a

collides elastically with block A of mass m and connected to another block B of mass 2m through spring constant k. What is k? If x0 is compression of spring when velocity of A and B is same? [UP SEE 2006] v0

C

(a)

mv 20 x02

(b)

mv 20 2x20

B

A

(c)

3 mv 20 2 x02

(d)

2 mv 20 3 x02

53. A bullet of mass 20 g and moving with 600 ms -1 collides with a block of mass 4 kg hanging with the string. What is velocity of bullet when it comes out of block, if block rises to height 0.2 m after collision? [UP SEE 2006]

51. A 10 kg object collides with stationary 5 kg object and after collision they stick together and move forward with velocity 4 ms -1. What is the velocity with which the 10 kg object hit the second one? [BVP 2007] (a) 4 ms–1 (c) 10 ms–1

52. A block C of mass m is moving with velocity v0 and

(b) 6 ms–1 (d) 12 ms–1

ms–1

(a) 200 (c) 400 ms–1

ms–1

(b) 150 (d) 300 ms–1

54. Consider a rubber ball freely falling from a height h = 4.9 m onto a horizontal elastic plate. Assume that the duration of collision is negligible and the collision with the plate is totally elastic. Then the

294 JEE Main Physics velocity as a function of time the height as function of time will be [AIEEE 2009] v

56. A body A of mass M while falling vertically downwards under gravity breaks into two parts, a body B of mass 1/3 M and a body C of mass 2/3 M. The centre of mass of bodies B and C taken together shifts compared to that the body A towards [AIEEE 2005]

v +v1

v1

(a)

(b)

(a) depends on height of breaking (b) does not shift (c) body C (d) body B

t

O t

O

–v1 v +v1

v +v1

3t1 (c) O t1 2t1 4t1

t

(d) O t1

3t1 2t1

57. A mass m moves with a velocity v and collides

t

inelastically with another identical mass. After v collision the 1st mass moves with velocity in a 3

4t1

–v1

–v1

direction perpendicular to the initial direction of motion. Find the speed of the 2nd mass after collision.

55. Consider a two particle system with particles having masses m1 and m2 . If the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle be moved, so as to keep the centre of mass at the same position?

[AIEEE 2005]

(a) v (b) 3 v 2 v 3 v (d) 3

[AIEEE 2006]

m2 d m1 m (c) 1 d m2

(b)

(a)

(c)

m1 d m1 + m2

(d) d

Answers Round I 1. 11. 21. 31. 41. 51.

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

2. 12. 22. 32. 42. 52.

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

3. 13. 23. 33. 43. 53.

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

4. 14. 24. 34. 44. 54.

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

5. 15. 25. 35. 45. 55.

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

6. 16. 26. 36. 46. 56.

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

7. 17. 27. 37. 47.

(c) (c) (c) (c) (a)

8. 18. 28. 38. 48.

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

9. 19. 29. 39. 49.

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

10. 20. 30. 40. 50.

(a) (c) (a) (c) (c)

2. 12. 22. 32. 42. 52.

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

3. 13. 23. 33. 43. 53.

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

4. 14. 24. 34. 44. 54.

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

5. 15. 25. 35. 45. 55.

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

6. 16. 26. 36. 46. 56.

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

7. 17. 27. 37. 47. 57.

(a) (a) (d) (c, d) (d) (c)

8. 18. 28. 38. 48.

(b) (b) (a) (b, d) (a)

9. 19. 29. 39. 49.

(d) (d) (c) (d) (d)

10. 20. 30. 40. 50.

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

Round II 1. 11. 21. 31. 41. 51.

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

the Guidance Round I 1. Centre of mass is closer to massive part of the body, therefore the bottom piece of bat has larger mass.

7. xCM =

mA xA + mBxB + mC xC + mDxD mA + mB + mC + mD y

2. As shown in figure, centre of mass of respective rods are at their respective mid-points.

D (0, 1)

Hence centre of mass of the system has coordinates ( X CM , YCM), then

C(1, 1)

(0, a)

0,

a 2

X CM =

YCM =

x

1´ 0 + 2 ´1+ 3 ´1+ 4 ´ 0 1+ 2 + 3 + 4 2+3 1 = = = 0.5 m 10 2 mAYA + mBYB + mC YC + mDYD Similarly, YCM = mA + mB + mC + mD

a ,0 2

(a, 0)

a a + m ´ + m ´0 a 2 2 = 3m 3

1´ 0 + 2 ´ 0 + 3 ´1+ 4 ´1 1+ 2 + 3 + 4 7 = = 0.7 m 10

=

a a +m´ 2 2 =a 3m 3

m ´0 + m ´

3. Centre of mass of a bangle lies at the centre of the bangle,

8. From figure, L L L , x2 = + = L 2 2 2 L L L 5L x3 = + + = 2 4 2 4

which is outside the body.

x1 =

4. If speed of man relative to plank be v, then it can be shown easily that speed of man relative to ground 3 M = v v mg = v Mö 4 æ çM + ÷ è 3ø \ Distance covered by man relative to ground must be

\ 3L 4

likely to be at C. This is because lower part of the sphere containing sand is heavier then upper part of the sphere containing air.

6. Here, m1 = 1 kg, v1 = 2 $i m2 = 2 kg, v 2 = 2 cos $i - 2 sin 30 $j m1v1 + m2v 2 m1 + m2

1 ´ 2 $i + 2 (2 cos 30° $i - 2 sin 30° $j) = 1+ 2 $ $ 2 i + 2 3 i - 2 $j æ 2 + 2 3 ö $ 2 $ = =ç ÷ i- j 3 3 è 3 ø

xCM =

M1x1 + M2x2 + M3 x3 M1 + M2 + M3

L 5L M ´ + M ´L + M ´ 2 4 = M+M+M 11 ML 11L = 4 = 3M 12

5. The position of centre of mass of the system shown in figure is

v CM =

B(1, 0)

=

O

m´

A (0, 0)

a a , 2 2

Wall L/4 L/2 L x1

x2

x3

9. The acceleration of centre of mass is aCM = \

F 30 = = 1ms–2 mA + mB 10 + 20

1 aCM t 2 2 1 = ´1´ 22 = 2 m 2

s=

296 JEE Main Physics Since, body of mass m1 moves with acceleration a = 2 ms–2 in upward direction.

10. For the calculation of the position of centre of mass, cut-off mass is taken as negative. The mass of disc is m1 = pr12s

O'

\

f1 = m1g + m1a

O

= p (6) 2s = 36 ps

= 10 ´ 10 + 10 ´ 2 = 120 N \Tension in string = f1 + f2

where s is surface mass density.

= 120 + 80 = 200 N

The mass of cutting portion is 2

m2 = p (1) s = ps xCM =

f1

f2

3 M 13. Mv = v1 + Mv 2 4 4

m1x1 - m2x2 m1 - m2

3 Mv 2 4 4v v2 = 3

Mv =

Taking origin at the centre of disc, x1 = 0 , x2 = 3 cm 36 ps ´ 0 - ps ´ 3 xCM = 36 ps - ps =

T

- 3ps 3 =cm 35 ps 35

(\ v1 = 0)

14. Change in momentum Dp = p2 - p1 = MV - ( -MV ) = 2 MV

15. Given, separation between the nuclei of H and Cl = 1.27Å = 1.27 ´ 10 -10 m

11. The mass of considered element is

Let mass of hydrogen atom = m \Mass of chlorine atom = 35.5 m

dm

CM

O x

Cl

dx

35.5 m

m 1.27 Å

dm = l dx = l 0 xdx \

Let hydrogen atom be at origin i.e., position vector of it, r1 = 0

1

1

ò 0 xdm ò 0 x ( l0xdx) xCM = = 1 ò dm ò 0 l0xdx

\Position vector of chlorine atom r2 = 1.27 ´10 - 10 m Position vector to centre of mass is given by m r + m2 r2 rCM = 1 1 m1 + m2

L

é x3 ù lê ú ë 3 û0

3

L 3 = 2L = = L L2 3 é x2 ù l0 l0 ê ú 2 2 ë û0 l0

12. Since, m2 moves with constant velocity a = 2 ms–1

=

m ´ 0 + 35.5 m ´ 1.27 ´ 10 -10 m+35.5 m

=

35.5 ´ 1.27 ´ 10 -10 36.5

= 1.235 ´ 10 -10 m = 1.24 Å

m1 = 10 kg, m2 =8 kg

From hydrogen atom on the line joining H and Cl atoms.

16. The law of conservation of momentum is applicable on the

m1

process.

17. Change in momentum = Impulse

m2

= Area under force-time graph

v = 2ms–1 f1

\

f2

m1

m2

m1g

m2g

f2 = m2g f2 = 8 ´ 10 = 80 N

\

mv = Area of trapezium

Þ

mv =

3T f0 4 4 mv f0 = 3T

mv = Þ

1æ Tö çT + ÷ f0 2è 2ø

297

Centre of Mass 18. Here, r ( x) = a (1+ bx2) When b ® 0 ,( x) = a = constant i.e., density of rod of length 1 m is constant. In that event, centre of mass of rod would lie at 0.5 m , (i.e., at the centre of rod.) When we try b ®0 in all the four given options, we find 3 (2 + b) 6 choice (a) alone given x = = = 0.5 4 (3 + b) 12

19. As p = mv m=

\

p v

Hence, m-v graph will be rectangular hyperbola. nm Dp Dm 20. F = =v =v t Dt Dt n mvn F = vm = 60 60

21. When a body is equilibrium, net force is zero. Hence, acceleration is zero. v1

22.

1g

3g

E1

E2

v2

As the momentum of both fragments are equal, therefore E1 m2 3 = = E 2 m1 1 According to problem, E1 + E 2 = 6.4 ´ 10 4 J By solving Eqs. (i) and (ii), we get E1 = 4.8 ´ 10 4 J E 2 = 1.6 ´ 10 4 J

and

L v collision is 100% elastic, now first block comes to rest and 2nd block starts moving towards the 3rd block with a velocity L v and takes time = to reach 3rd block and so on. v L Total time = t + t + ¼(n - 1) time = (n - 1) \ v Finally only the last nth block is in motion velocity v, hence final velocity of centre of mass. mv v VCM = = nm n

25. Time taken by first block to reach second block = . Since

26. Since, no external force is present on the system, so conservation principle of momentum is applicable. \

p f = p f = p1 + p 2

\

p1 = - p 2

\

| p1| = |- p 2|

\

p1 = p 2

From this point of view, it is clear that momenta of both particles are equal in magnitude but opposite in direction. Also, fraction is absent. So total mechanical energy of system remains conserved.

27. When spring is massless then according to momentum conservation principle, pi = p f or m1v1 = m2v 2 \ m1v1 = - m2v 2 m1v1 = m2v 2 or p1 = p 2 p2 Q K1 = 1 2 m1 K2 =

23. The resultant force on the system is zero. So, the centre of mass of system has no acceleration.

24. p = px2 + py2 = (2 cos t) 2 + (2 sin t) 2 = 2 p 2 (2) 2 2 = = 2m 2m m

Since kinetic energy does not change with time, both work done and power are zero. Now, as \ or

Power = Fv cos q = 0 F ¹ 0,v ¹ 0 cos q = 0 q = 90°

As direction of p is same that v (Q p = mv), hence angle between F and p is equal to 90°.

K m p22 \ 1 = 2 2 m2 K2 m1

(Q p1 = p2)

28. Since there is no external force acting on the particle, hence y CM =

If m be the mass of the body, then kinetic energy =

(Q pi = 0)

m1y1 + m2y 2 = 0, m1 + m2

æ 3 mö æ mö Hence, ç ÷ ´ (15) + ç ÷ (y ) = 0 è 4 ø 2 è 4ø Þ

y 2 = - 5 cm

29.Velocity or momentum is such that the linear momentum would be conserved.

30. Because in perfectly inelastic collision from colliding bodies stick together and move with common velocities.

31. From conservation of linear momentum \

mv = 3 mv ¢ v v¢ = 3

298 JEE Main Physics 34.

40. If initial velocity of bullet be v, then after collision combined

u 2 = 3m/s

u1 = 4m/s

m1 = 3 kg

velocity of bullet and target is mv v¢ = (M + m)

m2 = 4 kg

m1u1 + m2u2 = (m1 + m2) v

35. Given, m1 = m2 = m, u1 = 4 and u2 = 0

41. From t1 = 0 to 2 t 0 the external force acting on the combined system is m1g + m2g . \ Total change in momentum of the system

(From conservation of linear momentum)

= F ´ t = (m1 + m2) g ´ 2t 0

36. 15 m + 10 m = mv1 + mv 2 and Þ

v 2 - v1 =1 15 - 10

Þ

v 2 - v1 = 5

42. For elastic collision e = 1and velocity of separation is equal to …(i)

velocity of approach. The velocity of the target may be more, equal or less than that of projectile depending on their masses. The maximum velocity of target is double to that of projectile, when porjectile is extremely massive as compared to the target.

…(ii)

Solving Eqs. (i) and (ii), we have v 2 = 15 ms–1,v1 = 10 ms–1

37.

Maximum kinetic energy is transferred from projectile to target when their masses are exactly equal.

43. Let p1 and p2 be the momenta of A and B after collision.

1 m1m2 Loss of kinetic energy = (v1 - v 2) 2 2 m1 + m2

38. v1 = + 3 m/s

=

1 M ´M (v1 - v 2) 2 2 (M + M)

=

M ×M M (v1 - v 2) 2 = (v1 - v 2) 2 4 2 (2 M)

v2 = – 5 m/s m2

m1

As m1 = m2, therefore after collision velocities of masses get interchanged. i. e. ,

velocity of mass m1 = - 5 m/s

and

velocity of mass m2 = + 3 m/s

æ Mö æ M + mö v=ç ÷ × 2 gh = ç1 + ÷ 2 gh è mø è m ø

Þ

v1 æ 1 - e ö =ç ÷ v 2 è1+ e ø

25 = v1 + v 2 v 2 - v1 =1 u1 - u2

v ¢2 or v ¢ = 2 gh 2g

mv = 2 gh (M + m)

\

v v1 = (1 - e) 2 v v 2 = (1 + e) 2 \

h=

and

3 ´ 4 + 4 ´ ( -3) = (3 + 4) v , v = 0

A

p

B

A

Þ

As per definition, mu m (v 2 - v1) v 2 - 0 v 2 e= = = = M = u u M (u2 - u1) 0 - u

A

p2

p1

B

After collision

Then applying impulse = change in linear momentum for the two particles For B For A or

J = p1

…(i)

J = p - p2

…(ii)

p2 = p - J

…(iii)

Coefficient of restitution, e =

momentum must remain conserved. mu + 0 = 0 + mv 2 mu v2 = M

B

Before collision

39. As net horizontal force acting on the system is zero, hence Hence,

J J

p1 - p2 p

=

p1 - p + J p

=

J-p+ J 2J = -1 p p

44. m1v1 - m2v 2 = (m1 + m2) v Þ

2 ´ 3 - 1 ´ 4 = (2 + 1) v 6 - 4 = 3v 2 = 3v 2 v = m/s 3

Centre of Mass

299

y

46. (a) Impulse received by m J = m ( v f - vi ) = m( -2 $i + $j - 3 $i - 2 $j)

12 m/s M 12 m/s

= m( -5 $i - $j)

M

x

135°

and impulse received by M = - J = m(5$i + $j)

M v

(b) mv = m (5 $i + $j) Let v is the velocity of third part. By the conservation of linear momentum,

m $ $ 1 $ $ (5 i + j) = (5 i + j) M 13 (c) e = (relative velocity of separation/relative velocity of approach) in the direction of - $j = 11 / 17 or

v=

47. Retardation due to friction

Þ

Collision is elastic, i. e. , after collision first block comes to rest and the second block acquires the velocity of first block. Or we can understand it is this manner that second block is permanently at rest while only the first block moves. Distance travelled by it will be (5) 2 v2 = =5m 2 a (2) (2.5)

53.

\ Final separation will be ( s - 2) = 3 m

48. Here, m = 0.08 kg, m0 = 0.16 kg According to conservation principle of m momentum, mv1 + mv 2 = (2 m + m0) v CM \

mv1 + mv 2 v CM = 2 m + m0 0.08 ´ 16 1.28 = = 0.16 +0.16 1.32 128 = = 4 ms–1 32

v = 4 2 m/s

52. (a) This is only possible when collision is head-on elastic.

a = mg = (0.25) (10) = 2.5 ms–2

s=

3 m ´ v = m ´ 12 2

v1

(b) When collision is oblique elastic, then in this case, both bodies move perpendicular to each other after collision. (c) Since, in elastic collision, kinetic energy of system remains constant so, this is no possible. (d) The same reason as (b). v -u Acceleration a= t v - v0 or a= t v - v0 or g = \ v =0 t Speed before first bounce v 0 = - 5 ms–1

m0

t=

\ vCM

vB - v A 0 - ( -5) 5 = = = 0.5 s g 10 10

54. For collision between block A and B, e= m

v2

49. When the sphere 1 is released from horizontal position, then from energy conservation, potential energy at height l0 = kinetic energy at bottom 1 or mgl0 = mv 2 or v = 2 gl0 2 Since, all collisions are elastic, so velocity of sphere 1 is transferred to sphere 2, then from 2 to 3 and finally from 3 to 4. Hence, just after collision, the sphere 4 attains a velocity to 2 gl0 .

50. This is an example of elastic oblique collision. When a moving body collides obliquely with another identical body in rest, then during elastic collision, the angle of divergence will be 90°.

51. The momentum of third part will be equal and opposite of the resultant of momentum of rest two equal parts.

\

vB - v A vB - v A vB - v A = = uA - uB 10 - 0 10

vB - v A = 10 e = 10 ´ 0.5 = 5

…(i)

From principle of momentum conservation, mAuA + mBuB = mAv A + mBvB or m ´ 10 + 0 = mv A + mvB \ v A + vB = 10 Adding Eqs. (i) and (ii), we get vB = 7.5 ms–1

…(iii)

Similarly for collision between B and C, vC - vB = 7.5 e = 7.5 ´ 0.5 = 3.75 \ vC - vB = 3.75 ms–1

…(iv)

Adding Eqs. (iii) and (iv), we get 2 vC = 11.25 11.25 vC = = 5.6 ms–1 \ 2

…(ii)

300 JEE Main Physics 55. Since, no force is present along the surface, so momentum

56. As shown in adjoining figure ball is falling from height 2 h and

conservation principle for ball is applicable along the surface of plate.

rebounding to a height h only. It means that velocity of ball just before collision.

v1

θ2 n

v

or

\ \ \ \

θ1

θ1

Plate

mv sin q1 = mv1 sin q2 v sin q1 = v1 sin q2 v cos q2 v1 cos q2 e= 1 = v cos q1 v cos q

u=

4h g

and velocity just after collision.

v1 = cos q2 = ev cos q v1 sin q2 v sin q tan q = = v1 cos q2 ev cos q e tan q =

2 (2 h) = g

v=-

tan q e

\

2h g

-v e= = u

æ tan q ö q2 = tan -1 ç ÷ è e ø

2h 1 g = 4h 2 g

Round II 1. Let the centre of mass be b Then,

4. At the highest point momentum of particle before explosion

(n - 1) mb + ma =0 mn 1 a b=×a = (n - 1) (n - 1)

p = mv cos 60° 1 = m ´ 200 ´ = 100 m horizontally. 2 Now as there is no external force during explosion, hence p = p1 + p 2 + p3 = constant

2. From adjoining figure the component of momentum along x-axis (parallel to the wall of container) remains unchanged even after the collision. mvy m

θ

v mvx

θ

Hence, m

mvx v

mvy

\ Impulse = change in momentum of gas molecule along y-axis, i. e. , in a direction normal to the wall = 2 mv cos q

3. As there is no external force, hence p = p1 + p 2 + p3 = constant Þ

However, since velocities of two fragments, of masses m / 3 each, are 100 ms -1 downward and 100 ms -1 upward.

| p3| = | p1 + p 2| m 5m = (3) 2 + ( 4) 2 = 4 4 [Since v1 and v 2 are mutually perpendicular] 5m m \ p3 = v3 = 2 4 5 Þ v3 = = 2.5 ms–1 2

or

p1 = - p 2 p1 + p 2 = 0 m p3 = × v3 = p = 100 m horizontally 3 v3 = 300 ms-1 horizontally

5. Let mass per unit area of the disc be m. \Mass of the disc (M) = Total area of disc ´ Mass per unit area = pR 2 m. Mass of the portion removed from the disc (M¢ ) 2

2

pR æRö = pç ÷ m = m è2ø 4 =

M 4

The centre of mass of the original disc is O and the centre of mass of the removed part of O1 and let centre of mass of the remaining part be O 2.

Centre of Mass According to the question, figure can be drawn as

7. Here, m1 = u,m2 = Au,u1 = u and u2 = 0 \

O O2

R

x

Þ

R 2

O1

2 m2 u2 (m1 - m2) u1 æ 1- A ö + =ç ÷u (m1 + m2) (m1 + m2) è1 + A ø v1 æ 1 - A ö =ç ÷ u è1+ A ø v1 =

2

\ R 2 The remaining portion of the disc can be considered as a M system of two masses M at O and- M¢ = at O1. 4 If the distance of the centre of mass of the remaining part from the centre O is at a distance x, then R M ´ 0 - M¢ ´ 2 x= M - M¢ OO1 =

Here,

left of centre O.

exchange their velocities. Hence when A collides with B, A transfers its whole velocity to B. When B collides with C , B transfers its whole velocity to C. Hence finally A and B will be at rest and only C will be moving forward with a speed v.

9. Due to presence of contact (frictional) force momenta of blocks A and B separately change but total sum of momenta of A and B taken together is constant because no net external force is acting on the system. pushes the spring forward but 4 kg mass is at rest. Hence, m v + m2v 2 10 ´ 14 + 4 ´ 0 v CM = 1 1 = m1 + m2 10 + 4 140 = = 10 ms–1 14

R to the 6

11. The speed acquired by block, on account of collision of bullet with it, be v 0 ms–1. Since the block rises by 0.1 m, hence 0.1 =

m1x1 + m2x2 + m3 x3 m1 + m2 + m3 1 , √3 C — — 2m 2 2 1m

m A(0, 0)

1m

1m

m B(1, 0)

æ 1ö m ´ 0 ´ m ´ 1 ´ 2m ´ ç ÷ è2ø x= m + m + 2m 2m 1 x= = m 4m 2 m y + m2y 2 + m3y3 y = 11 m1 + m2 + m3 m ´ 0 + m ´ 0 + 2m ´ 3 /2 m + m + 2m 3 = m 4 æ1 3 ö \Centre of mass is ç m, m÷ 2 4 è ø y =

2

8. When two identical balls collide head-on elastically, they

6. The centre of mass is given by x=

Kfinal æ v1 ö æ 1- A ö =ç ÷ =ç ÷ è1+ A ø Kinitial è u ø

10. At the time of applying the impulsive force on block of 10 kg

M R ´ = 4 2 M M4 MR 4 R =´ =8 3M 6 -

Therefore, centre of mass of the remainig part is at

301

v 02 2g

Þ

v 02 = 2 ´ g ´ 0.1

or

v 0 = 2 ms–1

Now as per law of conservation of momentum for collision between bullet and block, mu = mu + Mv 0 M 2 kg Þ v = u - v 0 = 500 ´ 2 ms–1 m 0.01 kg = (500 - 200 2) ms–1 = 220 ms–1 æ m1 - m2 ö ÷ g but a1 è m1 + m2 ø is in downward direction and in the upward direction, i. e. , a 2 = - a1.

12. In the pulley arrangement,| a1| =| a 2| = a = ç

\ Acceleration of centre of mass m a + m2a 2 a CM = 1 1 m1 + m2 é m - m2 ù é m1 - m2 ù m1 ê 1 ú g - m2 ê m + m ú g + m m 2û 2û ë 1 ë 1 = (m1 + m2) 2

é m - m2 ù =ê 1 ú g ë m1 + m2 û

302 JEE Main Physics 13. Let centre of mass of lead sphere after hollowing be at point

15. Let the initially particle x is moving in anti-clockwise direction

O 2, where OO 2 = x .

and y in clockwise direction.

Mass of spherical hollow

As the ratio of velocities of x and y particles are 2

4 æRö pç ÷ M M 3 è2ø + = m= 8 æ 4 3ö p R ç ÷ è3 ø x = OO1 =

and

therefore ratio of their distance covered will be in the ratio of 2 : 1. It means they collide at point B. v

R 2

B

horizontal direction. As bullet of mass m collides with pendulum bob of mass 3 m and two stick together, their common velocity m 50 cos q 25 v¢ = = cos q ms–1 m + 3m 2 Now under this velocity v ¢ pendulum bob goes up to an angle 120°, hence v ¢2 10 é æ 1 ö ù 1- ç f - ÷ = 5 = h = l (1 - cos120° ) = 2g 3 êë è 2 ø úû

B m

A m

Þ

Initial momentum of the system (block C ) = mv After striking with A the block C comes to rest and now both block A and B moves with velocity v. When compression in spring is maximum. By law of conservation of linear momentum, mv = (M + m) V v V= 2

By law of conservation of energy, KE of block C = KE of system + PE of system 1 1 1 mv 2 = (2M) V 2 + kx2 2 2 2 Þ Þ Þ

or

m 2k

v ¢ = 10

æ 4ö q = cos-1 ç ÷ è5ø

17. If M = M¢, then bullet will transfer whole of its velocity (and consequently 100% of its KE) to block and will itself come to rest as per theory of collision.

18. From conservation law of momentum,

1 mv 2 2 x=v

or

Comparing two values of v ¢, we get 25 cos q = 10 2 4 cos q = Þ 5

1 1 1 æv ö mv 2 = (2M) ç ÷ + kx2 è2ø 2 2 2 kx2 =

v ¢2 = 2 ´ 10 ´ 5 = 100 æ 4ö q = cos-1 ç ÷ è5ø

(Q M = m)

2

C

y

16. Velocity of bullet at highest point of its trajectory = 50 cos q in

C

Þ

B C

After first collision at B velocities of particles get interchanged, i. e. ,xwill move with 2 v and particle y with v. Second collision with take place at point C. Again at this point velocities get interchanged and third collision take place at point A. So after two collision these two particles will again reach the point A.

æ M ö R MR M ´0 - ç ÷ ´ è8ø 2 R x= = 16 = 7M M 14 M8 8 R Shift = 14 C m

120°

v

R

14.

x

120°

O1 R 2

\

2v 2v

x

\

A

y

A

x

120°

O2 O

vx 1 = , vy 2

Þ or

4 3 3 pr r ær ö v1 2 m2 3 2 = = = = ç 2÷ v 2 1 m1 4 pr3r è r1 ø 1 3 r2 = (2)1/3 : 1 r1 r1 : r2 = 1 : (2)1/3

Centre of Mass 19. As shown in figure,

26. As there is no net external force, hence motion of centre of

M ´ 0 + M ´ 20 + M ´ 20 + M ´ 0 xCM = = 10 cm 4M Similarly, y CM = 10 cm Hence, distance of centre of mass from centre of any one sphere, say r = (10 - 0) 2 + (10 - 0) 2 = 10 2 cm

mass of fragments should have been as before.

27. We know that velocity of 2nd ball after collision is given by v2 =

u1 (1 + e) m1 (m - m1e) + u2 2 (m1 + m2) (m1 + m2)

In present problem u2 = 0 ,m2 = 2 m1 and e = 2 / 3, hence æ 2ö u ç1 + ÷ m1 è 3ø 5 = u v2 = (m1 + 2 m1) 9

20. Since, there is no external force acting on gun-bullet system, hence pb = pg and

Kb mg 2 kg 4 = = = Kg mb 50 g 1

Kg =

or

Kb 40

As four exactly similar type of collisions are taking place successively, hence velocity communicated to fifth ball 4

Now total energy = Kb + Kg 41 Kb or = , Kb = 2050 40 40 2050 ´ 40 Þ Kb = = 2000 J 41 Kg = 2050 - 2000 = 50 J

and

21. Distance between the centres of spheres = 12 R \ Distance between their surfaces = 12 R - (2 R + R) = 9 R. Since there is no external force, hence centre of mass must remain unchanged and hence Þ

m11 r = m2 r2

Þ

Mx = 5 M (9 R - x ),

æ5ö v5 = ç ÷ u è9ø

28. Since net momentum of the composite system is zero, hence resultant velocity of the composite system should also be zero.

29. Consider the two cart system as a single system. Due to explosion of power total momentum of system remains unchanged, i. e. , p1 + p 2 = 0 or m1v1 = m2v 2, hence v1 m2 = v 2 m1 As coefficient of friction between carts and rails are identical, hence a1 = a2 and at the time of stopping final velocity of cart is zero. Using equation v 2 - u 2 = 2 as, we have s1 v12 m22 = = s2 v 22 m12

where, x = distance covered by smaller body. Þ

x = 7.5 R

22. Speed of the bullet relative to ground v b = v+ v r , where v r is recoil velocity of gun. Now for gun-bullet system applying the conservation law of momentum, we get or

m ( v+ v r ) + Mv r = 0

Þ

vr = -

\

Feq (m1 + m2 + m3)

=

mv mv or v r = m+M m+M

m1a1 + m2a 2 + m3 a3 (m1 + m2 + m3)

Feq = m1a1 + m2a 2 + m3 a3 = 1 ´ 1 + 2 ´ 2 + 4 ´ ( -0.5) = 1+ 4 – 2 = 3 N

24. Since gun-shot system is an isolated closed system, its centre 25.

of mass must remain at rest. 5 g - 5 mg = 4 m/s2 asystem = 5+5 m a + m2a 2 5 ( 4 $i) + 5 ( 4 $j) aCM = 1 1 = m1 + m2 10 =

s2 =

Þ

s1m12 36 ´ (200) 2 = = 16 m (300) 2 m22

30. First sphere will take a time t1, to start motion in second sphere L on colliding with it, where t1 = . u

mv b + Mv r = 0

23. Q a CM =

303

b 42 + 42 = 2 2 m/s2 10

Now speed of second sphere will be v2 =

1ö æ çQ e = ÷ è 3ø

u 2 (1 + e) = u 2 3

Hence, time taken by second sphere to start motion in third L 3L . sphere t 2 = = 2 /3 u 2 u \ Total time

t = t1 + t 2 =

L 3L 5L + = u 2u 2u

31. Change in momentum = Ft and does not depend on mass of the bodies. 1 2

32. Let ball strikes at a speed u the K1 = mu2. Due to collision tangential component of velocity remains unchanged at u sin 45°, but the normal component of velocity 1 change to u sin 45° = u cos 45° 2

304 JEE Main Physics \Final velocity of ball after collision ö æ1 v = (u sin 45° ) + ç u cos 45° ÷ ø è2 2

36. As total mass is M and velocity of centre of mass is v, hence

2

5 æ u ö æ u ö = ç u ÷ +ç ÷ = è 2ø è2 2 ø 8 Hence, final kinetic energy 1 5 K2 = mv 2 = mu 2 2 16 \ Fractional loss in KE

33. Let at the time of explosion velocity of one piece of mass m/2 is (10 $i). If velocity of other be v 2, then from conservation law of momentum (since there is no force in horizontal direction), horizontal component of v 2, must be -10 $i. \Relative velocity of two parts in horizontal direction = 20 ms-1. Time taken by ball to fall through 45 m, 2h 2 ´ 45 = =3s 10 g

and time taken by ball to fall through first 20 m, t¢ =

37. In an inelastic collision neither momentum of ball nor mechanical energy of ball will remain same. However, total energy and total momentum of earth-ball system will remain constant.

38.

1 5 mu 2 - mu 2 K1 - K2 2 3 16 = = = 1 K1 8 mu 2 2

= 20 =

kinetic energy of the system may have any value equal to or 1 greater than Mv 2. However exact value of kinetic energy 2 can be calculated only when values of m1, m2, v1 and v 2 are known to us.

2

2 h¢ 2 ´ 20 = =2s g 10

Hence time taken by ball pieces to fall from 25 m height to ground = t - t ¢ = 3 - 2 = 1s. \ Horizontal distance between the two pieces at the time of striking on ground = 20 ´ 1 = 20 m.

34. If v1 and v 2 are in same direction, then v comp =

=

m1v1 + m2v 2 m1 + m2 2 ´ 3 + 1´ 4 2 +1

10 = ms-1 3 However, if v1 and v 2 are in mutually opposite directions, then 2 ´3 +1 ´ - 4 v comp = 2 +1 2 –1 = ms 3

35. As there is no external force on the system hence displacement of the centre of mass of the system is zero.

L

L

L

v

Since collision is perfectly inelastic, so all the block will stick together one by one and move in a form of combined mass. L Time required to cover distance (d) by first block = . v Now first and second block will stick together and move with v / 2 velocity (by applying conservation of momentum) and L 2L to reach upto block third. combined system will take = v /2 v Now these three blocks will move with velocity v/ 3 and L 3L to reach upto the combined system will take time = v/ 3 v fourth block. L 2L 3L (n - 1) L n (n - 1 ) L So, total time + + +¼ = v v v v 2v

39. We cannot calculate the value of force exerted because time is not known to us.

40. Impulse of ball I, J1 = change in momentum of ball I = 2 mu and impulse of ball II J2 = change in momentum of ballII = 2 mu cos 30°. J1 1 2 Þ = = J2 cos 30° 3

41. Force will be normal to the wall in both situations. 42. Q e = \ Þ

h1 h h hn = 2 = 3 ¼= h0 h1 h2 hn -1 h1 = e2h0 ,h2 = e2h1 = e4h0 and so on hm = e2nh0

43. The total distance travelled by the balls is H = h0 + 2 h1 + 2 h3 + ¼= h0 + 2 e2h0 + 2 e4h0 + 2 e4h0 + ¼ = h0 [1 + 2 e2 (1 + e2 + e4 + ¼)] é æ 1 öù = h0 ê1 + 2 e2 ç ÷ è1 - e2 ø úû ë é 1 + e2 ù = h0 ê 2ú ë 1- e û

Centre of Mass As and

Hence,

h0 = 10 m 1 e= 2

51. As, m1u1 + m2u2 = (m1 + m2) v Þ Þ

1ù é 1+ ê 4 ú = 50 m H = 10 ê 1ú 3 ê 1- ú ë 4û

10 ´ u1 + 5 ´ 0 = (10 + 5) ´ 4 15 ´ 4 u1 = 10 = 6 ms-1

45. In a quick collision, time t is small as F ´ t = constant, therefore, force involved is large in collision is more violent in comparison to slow collision.

46. The position of centre of mass of electron and proton remains at rest, at their motion is due to (internal) forces of electrostatic attraction, which are conservative. No external force, what so ever is acting on the two particles.

47. Angular momentum is rotational analogue of linear momentum, and torque is rotational analogue of force.

52. In an elastic collision, the conservation of linear momentum and conservation of energy hold. Using conservation of linear momentum, we have

Þ

mv 0 = mv + 2 mv v v= 0 3

Using conservation of energy, we have 1 1 1 mv 02 = x02 + (3 m) v 2 2 2 2 where, x0 = compression in the spring.

48. If it is a completely inelastic collision, then m1v1 + m2v 2 = m1v + m2v m v + m1v 2 v= 11 m1 + m2

\

p12 p2 + 2 2 m1 2 m2 As p1 and p 2 both simultaneously cannot be zero, therefore total kinetic energy cannot be lost.

mv 02 = kx02 + (3 m)

Þ

kx02 = mv 02 -

Þ

kx02 =

2 mv 02 3

Þ

k=

2 mv 02 3 x02

KE =

49. If momentum is zero, i. e. , p = 0, then kinetic energy K = p 2/ 2 m = 0 . But potential energy cannot be zero, thus, a body can have energy without momentum.

50. The acceleration of the body which is rolling down an inclined plane of angle a is g sin a g¢ = k2 1+ 2 R

\

or or

a=

2 g sin a 3

mv 02 3

In an inelastic collision between two bodies, only conservation of linear momentum holds.

53. Conservation of linear momentum holds here. According to conservation of linear momentum,

m1v1 = m1v + m2v 2 where v is the velocity of bullet after the collision and v 2 is the velocity of block.

Now, here the body is a uniform solid disc. k2 1 = R2 2 g sin a a= 1 1+ 2 g sin a a= 3 /2

v 02 9

α

where, k = radius of gyration, R = radius of body.

So,

305

\ Here,

0.02 ´ 600 = 0.02 v + 4 v 2 v 2 = 2 gh = 2 ´ 10 ´ 0.2

(as g ¢ = a)

\ Þ \

= 2 ms–1 0.02 ´ 600 = 0.2 v + 4 v 2 0.02 v = 12 - 8 4 v= 0.02 = 200 ms–1

306 JEE Main Physics 1 2

54. h = gt 2 (parabolic) v = - gt and after the collision v = gt (straight line) Collision is perfectly elastic, then ball reaches to same height again and again with same velocity. v

v

56. Since, the acceleration of centre of mass in both the cases is

h

same equal to g, so the centre of mass of the bodies B and C taken together does not shift compared to that of body A.

+v1 O t 1 2t1 3t1 4t1

–ve sign shows that both the particles have to move in opposite directions. md So, 1 is the distance moved by 2nd particle to keep centre m2 of mass at the same position.

t

57. In x-direction t

–v1

mv + 0 = 0 + mv x Þ

55. To keep the centre of mass at the same position, velocity of centre of mass is zero, so m1v1 + m2v 2 =0 m1 + m2 [where v1 and v 2 are velocities of particles 1 and 2 respectively.] Þ Þ

m1

dr1 dr + m2 2 = 0 dt dt

dr dr ù é Q v1 = 1 and v 2 = 2 dt dt ûú ëê

m1dr1 + m2 dr2 = 0

[dr1 and dr2 represent the change in displacement of particles] Let 2nd particle has been displaced by distance x. Þ

m1 ( x) + m2 ( x) = 0

Þ

x=-

m1d m2

Þ

mv = mv x vx = v

In y- direction æ v ö 0 +0 =mç ÷ - mv y è 3ø vy =

v 3

\Velocity of second mass after collision 2

æ v ö 2 v¢ = ç ÷ +v è 3ø =

4 2 v 3

v¢ =

2 v8 3

9 Rotational Motion JEE Main MILESTONE < < <
E2 (c) E1 < E2

(b) E1 = E2 (d) Cannot be said

of angle q. The coefficient of static friction between the plane and cylinder is m s . The condition for the cylinder not to slip is (b) tan q > 3 m s (d) tan q < 3 m s

7. A thin circular ring of mass M and radius r is rotating about its axis with a constant angular velocity w. Two objects each of mass M are attached gently to the opposite ends of a diameter of the ring. The ring will now rotate with an angular velocity w( M - 2m ) M + 2m wM (c) M+m

(a)

wM M + 2m w( M + 2M ) (d) M

(b)

3v 2a

(c)

v M

O

3v 2a

9. A disc of mass M and radius R

(d) zero y

is rolling with angular speed w on a horizontal plane as shown. The magnitude of angular momentum of the disc about O the origin O is

ω M x

3 MR2 w 2 (d) 2 MR2 w (b)

10. A particle performs uniform circular motion with an angular momentum L, if the frequency of particles motion is doubled and its kinetic energy is halved, the angular momentum becomes (a) 4 L (c) 2 L

(b) 0.5 L (d) 0.25 L

11. A uniform rod of length 2 L is placed with one end in contact with the horizontal and is then inclined at an angle a to the horizontal and allowed to fall without stopping at contact point when it becomes horizontal. Its angular velocity will be (a) w =

3 g sin a 2L

(b) w =

2L 3g sin a

(c) w =

6 g sin a L

(d) w =

L g sin a

6. A solid cylinder is rolling down on an inclined plane

(a) tan q ³ 3 m s (c) tan q £ 3 m s

(b)

1 MR2 w 2 (c) MR2 w

P

(b) 100

3v 4a

(a)

3. A thin wire of length l and mass m is bent

a

moving with velocity v on a horizontal smooth plane as shown. It hits a ridge at point O. The angular speed of the block after is hits O is

(b)

2. The moment of inertia of two spheres of equal masses

(a) 80

(Mixed Bag)

12. Two thin discs each of mass M and radius R are placed at either end of a rod of mass m, length l and radius r. Moment of inertia of the system about an axis passing through the centre of rod and perpendicular to its length is 1 mL2 1 + MR 2 + ML2 12 4 4 1 ML2 1 (b) + mR 2 + mL2 12 2 2 2 mR ML2 1 (c) mL2 + + 2 2 12 1 2 mL2 2 (d) + MR + ML 12 2 (a)

M

R

Axis of rotation

m

2r

L

M

R

331

Rotational Motion 13. Four particles each of mass m are lying symmetrically on the rim of a disc of mass M and radius R. Moment of inertia of this system about an axis passing through one of the particles and perpendicular to plane of disc is (a) 16 mR2

(b) 3 ( M + 16 m )

R2 (c) (3M + 12m ) 2

(d) zero

R2 2

are placed along X and Y axis with one end of each at the origin. Moment of inertia of the system about Z-axis is 3 2 ML 2 (c) 2 ML2

2 2 ML 3 (d) None of these (b)

15. If the radius r of earth suddenly changes to x times the present values, the new period of rotation would be (a) dT /dt = (T /r ) ( dr/dt ) (c) dT /dt = ( r/T ) ( dr/dt )

(b) dT /dt = (2T /r ) ( dr/dt ) æ 1 ö æ dr ö (d) dT /dt = ç T /r ÷ ç ÷ è 2 ø è dt ø

16. The curve between log e L and log e p is (L is angular momentum and p is linear momentum) (a)

(b)

(c)

log L

(d) log L

ML2 3

log p

17. A rod of length l is hinged at one end and kept horizontal. It is allowed to fall. The velocity of the other end of the rod is (a) 3 gl

(b) 2 gl

(c) 2 Ml 2

(d) None of these

2 ML2 3

(c)

3 ML2 2

(d)

2 ML2 12

20. A solid sphere of mass M and radius R spins about an

(a)

2 2 p MR 5

(b)

2 pM2 R2 (c) 80 p2 MR2 5

opposite directions along parallel lines separated by a distance d. The vector angular momentum of the two particle system is whatever be the point an out which the angular momentum is taken. (b) different (d) can not be said

(d) 80 pR

21. A ring of radius R is first rotated with an angular velocity w0 and then carefully placed on a rough horizontal surface. The coefficient of friction between the surface and the ring is m. Time after which its angular speed is reduced to half is (a)

w0mR 2g

(b)

2 w0 R mg

(c)

w0 R 2m g

(d)

w0 g 2 mR

22. What is the moment of inertia of solid sphere of density r and radius R about its diameter? (a)

105 5 R r 176

(b)

105 2 R r 176

(c)

176 5 R r 105

(d)

176 2 R r 105

23. If the moment of inertia of a disc about an axis tangential and parallel to its surface be I, then what will be the moment of inertia about the axis tangential but perpendicular to the surface? 6 I 5

(b)

3 I 4

(c)

3 I 2

(d) z

5 I 4

ω

is rotating in the horizontal v plane will constant angular O speed was shown in the figure. At time t = 0 a small insect starts from O and moves with constant speed v with respect to the rod towards the other end it reaches the end of the rod at t = T and stops. The angular speed of system remains w throughout the magnitude of torque ( t) on the system about O as a function of time is best represented by which plot z

z

(a) |r|

(b) |r|

t

18. Two particles, each of mass m and speed v, travel in

(a) same (c) may be same or not

(b)

24. A thin uniform rod pivoted at O log p

log p

(a)

(a)

log L

log L

log p

along X, Y and Z axes in such a way that one end of each rod is at the origin. The moment of inertia of the system about Z-axis is

axis passing through its centre making 600 rpm. Its kinetic energy of rotation is

14. Two uniform thin rods each of mass M and length l

(a)

19. Three rods each of length L and mass M are placed

T

T z

z

(c) |r|

(d) |r|

t T

t

t T

332 JEE Main Physics 25. A rectangular block has a square base measuring a ´ a, and its height is h. It moves on a horizontal surface in a direction perpendicular to one of its edges. The coefficient of friction is m. It will topple if (a) m > h / a 2a (c) m > h

ω(t)

(a)

(b) m > a / h a (d) m > 2h

connected with a massless rod of length 2 R as shown in the figure. What will be the moment of inertia of the system about an axis passing through the centre of one of the sphere and perpendicular to the rod? M

R/2

R/2

2 MR2 5 5 (d) MR2 21

(b)

ω(t)

(d)

B

mg 4

rigid massless rod of length r to constitute a dumb-bell which is free to move in the plane. The moment of inertia of the dumb-bell about an axis perpendicular to the plane passing through the centre of mass is m1m2 r2 m1 + m2 2

(c)

m1mr r m1 - m2

ω0

(d)

ω0

t

t

(b) 2.66 × 10–19 (d) 5.66 × 10–17

moment of inertia about an axis passing through its centre and perpendicular to its axis is minimum, the ratio L / R must be equal to

28. Two particles of masses m1 and m2 are connected by a

(a)

(c)

ω(t)

(a) 2.0 × 10–20 (c) 4.33 × 10–18

(b) mg

(c) 2 mg

t

31. A cylinder of mass M, length L and radius R. If its

and length l is suspended by means of two light inextensible strings as shown in figure. Tension in one string immediately A after the other string is cut is mg 2

ω0

1 part in 5 ´ 1019 per day by the attraction of meteors falling normally on the earth’s surface. Assuming that the density of earth is uniform, the rate of change of the period of rotation of the earth is

27. A uniform rod of mass m

(a)

(b)

30. The mass of the earth is increasing at the rate of

2R

21 (a) MR2 5 5 (c) MR2 2

ω0

t

26. Two spheres each of mass M and radius R / 2 are

M

ω(t)

(b) ( m1 + m2 ) r2 (d) ( m1 - m2 ) r2

29. A circular platform is free to rotate in a horizontal plane about a vertical axis passing through its centre. A tortoise is sitting at the edge of the platform. Now the platform is given an angular velocity w0 . When the tortoise moves along a chord of the platform with a constant velocity (w.r.t. the platform), the angular velocity of the platform will vary with the time t as

(a) 3/2 (c) 2 / 3

(b) 2/3 (d) 3 / 2

32. Four holes of radius R are cut from a thin square plate of side 4 R and mass M. The moment of inertia of the remaining portion about z-axis is p MR2 12 æ4 pö (b) ç - ÷ MR2 è3 4ø æ4 pö (c) ç - ÷ MR2 è3 6ø

y

(a)

x

æ 8 10 p ö 2 (d) ç ÷ MR è 3 16 ø

More Than One Correct Option 33. Consider a bicycle wheel rolling without slipping on a rough level road at a linear speed as shown in figure. Then

C ω

D

O

(a) the speed of the particle A is zero (b) the speed of B, C and D are all equal to v 0 (c) the speed of C is 2 v 0 (d) the speed of B is greater than the speed of O

B

vc

Rotational Motion 34. Choose the correct alternatives

[NCERT Exemplar]

(a) For a general rotation motion, angular momentum L and angular velocity w need not be parallel (b) For a rotational motion about a fixed axis, angular momentum L and angular velocity w are always parallel (c) For a general translational motion, momentum p and velocity v are always parallel. (d) For a general translational motion, acceleration a and velocity v are always parallel

(a) Torque t caused by F about z-axis is along - k (b) Torque t¢ caused by F abut z¢ axis is along - k$ (c) Torque t caused by F about z axis is greater in magnitude than that about z axis (d) Total torque is given be t = t + t ¢

38. With referene to figure of a cube of edge a and mass m, state whether the following are true or false. (O is the centre of the cube). [NCERT Exemplar]

35. The net external torque on a system of particles

G z'

The forces may be acting radially from a point on the axis The forces may be acting on the axis of rotation The forces may be acting parallel to the axis of rotation The torque caused by some forces may be equal and opposite to that caused by other forces

D

36. The figure shows a system consisting of (i) ring of

A

outer surface 3 R rolling clockwise without slipping on a horizontal surface with angular speed w and (ii) an inner disc OP radius 2 R rotating anticlockwise with angular speed w/ 2. The ring and the disc separated by frictionless ball bearing the system is in the xz-plane. The point P on the inner disc is at a distance R from the origin where OP makes an angle of 30° with the horizontal. Then with respect to the horizontal surface. z

ω

x

F Y a

B

x

(a) The moment of inertia of cube about z-axis is Iz = Ix + I y ma2 (b) The moment of inertia of cube about z ¢ is Iz¢ = Iz + 2 mz 2 (c) The moment of inertia of cube about z¢¢ is = Iz + 2 (d) Ix = I y

Comprehension Based Questions

p rad s–1 30 p (c) rad s–1 1800 (a)

11 $ 3 Rw i + Rwk$ 4 4 13 3 (c) The point P has a linear velocity Rw i$ Rw k$ 4 4 (d) The point P has a linear velocity æ 3ö 1 ç3 ÷ Rw i$ + Rw0 k$ 4 ø 4 è (b) The point P has a linear velocity

which is 5 cm long is

F P

p ms -1 120 ´ 60 p (c) ms -1 120

(a)

z

z

p rad s–1 60 p (d) rad s–1 3600 (b)

40. The linear velocity of tip of hours hand of a clock,

37. Figure shows a lamina in xy- plane. Two axes z and z¢

[NCERT Exemplar]

E

39. The angular velocity of minutes hand of a watch is

(a) The point O has a linear velocity 3 Rw $i

pass perpendicular to its plane. A force F acts in the plane of lamina at point P as shown. Which of the following are true? (The point P is closer to z¢-axis than the z-axis.)

C

The three equations of rotational motion are w = w0 + at; q = w0 t + 1 at2 and w2 - w20 = 2 aq , where 2 the symbols have their usual meanings. Also, 2p v = r w; w = = 2 pn are the known standard T relations. Use them to answer the following questions

Rρ 30° 2R

O

Passage I

ωD 3p

z ¢¢

H

about an axis is zero. Which of the following are compatible with it? [NCERT Exemplar] (a) (b) (c) (d)

333

p ms -1 120 ´ 60 ´ 60 p (d) ms -1 60 ´ 60 ´ 60 (b)

41. The spin driver of a washing machine revolving at 15 rps slow down to 5 rps, while making 50 revolutions. Angular acceleration of the driver is (a) - 4 p rads -2

(b) - 4 p rads -2

(c) 8p rads -2

(d) - 8p rads -2

334 JEE Main Physics 42. A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad/s. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis ?

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

44. Assertion The centre of mass of a body will change with the change in shape and size of the body. i=n

(a) 3125 J, 62.5 J-s (b) 72.5 J s and 62.5 J-s (c) 3125 J, 82.5 J-s (d) None of the above

Reason r =

43. Torque of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry and the sphere is free to rotate about an axis passing through its centre. Which of the two will aquire a greater angular speed after a given time ?

å mi i i i =1 i=n

å mi i =1

45. Assertion The velocity of a body at the bottom of an inclined plane of given height is more when it slides down the plane compared to when it rolls down the same plane. Reason In rolling down, a body acquires both, kinetic energy of translation and kinetic energy of rotation.

46. Assertion A ladder is more opt to slip when you are

(a) solid sphere (b) hollow sphere (c) both have some angular speed (d) cannot be said

high on it than when you just begin to climb. Reason At the high up on ladder the torque is large and on climbing up the torque is small.

47. Assertion When ice on polar caps of earth melts,

Assertion and Reason Directions

Question No. 44 to 48 are Assertion-Reason type. Each of these contains two Statements: Statement I (Assertion), Statement II (Reason). Each of these questions also has four alternative choice, only one of which is correct. You have to select the correct choices from the codes (a), (b), (c) and (d) given below (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion

duration of the day increases. 2p Reason L = Iw = I × = constant. T

48. Assertion Moment of inertia of circular ring about a given axis is more than moment of inertia of the circular disc of same mass and same size, about the same axis. Reason The circular ring hollow so its moment of inertia is more than circular disc which is solid.

Previous Years’ Questions 49. A pulley of radius 2m is rotated about its axis by a

force F = (20 t - 5 t2 ) newton (where t is measured in sec) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg-m2 , the number of rotation made by the pulley before its direction of motion of reversed is [AIEEE 2011] (a) less than 3 (b) more than 3 but less than 6 (c) more than 6 but less than 9 (d) more than 9

50. A thin horizontal circular disc is rotating about a vertical axis passing through its centre. An insect is at rest it a point near the rim of disc. The insect now moves along a diameter of the disc to reach its other

end. During the journey of the insect, the angular speed of the disc [AIEEE 2011] (a) continuously of the disc (b) continuously decreases (c) first increases and then decreases (d) remains unchanged

51. A thin uniform rod of length l and mass m is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is w. Its centre of mass rises to a maximum height of [AIEEE 2009] (a)

lw 6g

(b)

l 2 w2 2g

(c)

l 2 w2 6g

(d)

l 2 w2 3g

Rotational Motion 52. Consider a uniform square plate of side a and mass m. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is [AIEEE 2008] (a)

5 ma 2 6

ma 2 12

(b)

(c)

7 ma 2 12

(d)

2 ma 2 3

53. A motor is rotating at a constant angular velocity of 500 rpm. The angular displacement per second is 3p rad 50 50 p (d) rad 3

3 rad 50 p 25 p rad (c) 3

[BVP Engg. 2007]

momentum (L) and kinetic energy (K)? 2

L K L (c) 2 K2

L 2K L (d) 2K

(a)

(b)

60. A disc of mass 2 kg and radius 0.2 m is rotating with

angular velocity 30 rads–1. What is angular velocity, if a mass of 0.25 kg is put on periphery of the disc? [UP SEE 2006]

initial angular speed of 2.00 In a time of 2 s it has rotated through an angle (in radian) of [UP SEE 2007]

(c) 12

(d) 4

55. For the given uniform square lamina ABCD, whose centre is O as shown in figure. D

F

rads–1

(a) 24 (c) 15 rads–1

rads–2.

(b) 10

[UP SEE 2006]

2

rads–1

54. A wheel has angular acceleration of 3.0 rads–2 and an

(a) 6

59. What is moment of inertia in terms of angular

(b)

(a)

335

[AIEEE 2007]

(b) 36 (d) 26 rads–1

61. The moment of inertia of a rod about an axis through

1 ML2 (where, 12 M is the mass and L the length of the rod). The rod is bent in the middle to that the two halves make an angle of 60°. The moment of inertia of the bent rod about the same axis would be [UP SEE 2006] its centre and perpendicular to it is

(a)

1 ML2 48

(b)

1 2 ML 12

(c)

1 ML2 24

(d)

ML2 8 3

62. If the earth is treated as a sphere of radius R and

C

mass M, its angular momentum about the axis of rotation with time period T is [BVP Engg. 2006] O

(a) A

B

E

(a) IAC = 2 IEF

(b) 2 IAC = IEF

(c) IAD = 3 IEF

(d) IAC = IEF [AIEEE 2007]

57. A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the circumference of new disc is a R from the centre of the bigger disc. [AIEEE 2007] The value of a is (b) 1/3 (d) 1/6

58. Four point masses, each of value m, are placed at the corners of a square ABCD of side l. The moment of inertia of the system about an axis passing through A and parallel to BD is [AIEEE 2006] (a) 3 ml 2

(b) 3 ml 2

(c) ml 2

(d) 2 ml 2

MR2T 2p

(c)

2 pMR2 T

(d)

4 pMR2 5T

system as shown in figure. The torque about the point (1, –1) is [AIEEE 2006] Z

(a) constant torque (b) constant force (c) constant linear momentum (d) zero torque

(a) 1/4 (c) 1/2

(b)

63. A force of –F k acts on O, the origin of the coordinate

56. Angular momentum of the particle rotating with a central force is constant due to

pMR3 T

(a) - F( $i + $j) (b) F( i$ + $j) O

(c) - F( $i - $j) (d) F( $i - $j)

Y

X (1, –1)

64. A hoop of radius r and mass m rotating with an angular velocity w0 is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip? [JEE Main 2013] (a)

rw0 4

(b)

rw0 3

(c)

rw0 2

(d) rw0

65. A circular disc of radius R rolls without slipping along the horizontal surface with constant velocity v0 . We consider a point A on the surface of the disc. Then the acceleration of the point A is [UP SEE 2005] (a) constant in magnitude as well as in direction (b) constant in direction (c) constant in magnitude (d) constant

336 JEE Main Physics 66. A solid cylinder of mass 20 kg has length 1 m and

(a) 0.8

(b) 0.4

(c) 0.2

(c) 10 MR2

70. A T shape object with dimensions shown in figure is lying on a smooth A floor. A force F is applied at the point P parallel to AB, such that the object has only the translational motion without F rotation. Find the location of P with [AIEEE 2005] respect to C.

about its diameter with angular velocity of 20 rads–1. Its kinetic energy is

[BVP Engg. 2005]

(b) 100 J

(c) 500 J

(d) 250 J

68. Two discs of the same material and thickness have radii 0.2 m and 0.6 m. Their moments of inertia about the axes will be in the ratio of [BVP Engg. 2005] (a) 1 : 81

(b) 1 : 37

(b)

(d) 20.4

67. A ring of radius 0.5 m and mass 10 kg is rotating

(a) 10 J

40 MR2 9 37 (d) MR2 9

(a) 4 MR2

radius 0.2 m. Then its moment of inertia in kg m 2 about its geometrical axis is [Kerala CET 2005]

(c) 1 : 9

(d) 1 : 3

69. From a circular disc of radius R and mass 9 M, a small disc or radius R/3 is removed. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is [IIT JEE 2005]

4 l 3 2 (d) l 3

(a) l (c)

B O

P D 2l

C

(b)

3 l 2

l

71. An angular ring with inner and outer radii R1 and R2

is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring. F1/ F2 is [AIEEE 2005]

R/3 2R/3

(a)

R1 R2

(b) 1 2

æR ö (c) çç 1 ÷÷ è R2 ø

R

(d)

R2 R1

Answers Round I 1. 11. 21. 31. 41. 51. 61.

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

2. 12. 22. 32. 42. 52. 62.

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

3. 13. 23. 33. 43. 53. 63.

(c) (d) (a) (c) (d) (c) (e)

4. 14. 24. 34. 44. 54. 64.

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

5. 15. 25. 35. 45. 55.

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

6. 16. 26. 36. 46. 56.

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

7. 17. 27. 37. 47. 57.

(c) (b) (c) (c) (b) (b)

8. 18. 28. 38. 48. 58.

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

9. 19. 29. 39. 49. 59.

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

10. 20. 30. 40. 50. 60.

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

Round II 1. (d) 11. (a) 21. (c) 31. (d) 41. (b) 51. (c) 61. (b) 71. (a)

2. (c) 12. (a) 22. (c) 32. (d) 42. (a) 52. (d) 62. (d)

3. (d) 13. (b) 23. (a) 33. (c) 43. (d) 53. (d) 63. (b)

4. (d) 14. (b) 24. (b) 34. (a,c) 44. (a) 54. (b) 64. (c)

5. (c) 15. (b) 25. (a) 35. (a,b,c,d) 45. (d) 55. (d) 65. (a)

6. (c) 16. (b) 26. (a) 36. (a,b) 46. (a) 56. (c) 66. (b)

7. (b) 17. (a) 27. (a) 37. (b,c) 47. (a) 57. (b) 67. (d)

8. (a) 18. (a) 28. (a) 38. (b,d) 48. (b) 58. (b) 68. (a)

9. (c) 19. (b) 29. (c) 39. (c) 49. (b) 59. (b) 69. (a)

10. (d) 20. (c) 30. (a) 40. (b) 50. (c) 60. (a) 70. (b)

the Guidance Round I 1. Moment of inertia of uniform circular disc about diameter = I According to theorem of perpendicular axes, Moment of inertia of disc about its axis = 2 I 1 2ö æ ç where, I = mr ÷ è ø 2 Applying theorem of parallel axes Moment of inertia of disc about the given axis = 2 I + mr 2 = 2 I + 4 I = 6 I

(Here, K is radius of gyrations)

2. About EG, the minimum distance from the axis is the least i. e. , distribution of mass is minimum.

8. Linear momentum of particle before colliding = mv = mv ey Linear momentum of particle after it bounces

3. Applying the principle of conservation of angular momentum, (I1 + I2) w = I1w1 + I1w2 400 600 (6 + I2) ´ 2p = 6 ´ ´ 2p + I2 ´ 0 60 60

= - mv = - mvey Change in linear momentum, Dp = - mv - (mv) = - 2 mv = - 2mv ey Change in angular momentum = DL = r ´ Dp, where r = (yey + aez)

I2 = 3 kg-m 2

which gives,

2 5 Using the theorem of parallel axes, moment of inertia of the sphere about a parallel axis tangential to the sphere is 2 7 I ¢ = I + MR 2 = MR 2 + MR 2 = MR 2 5 5 ö æ 7 7 I' = MK 2 = MR 2, K = ç \ ÷R 5 è 5ø

7. Given, I = MR 2

= (yey + aez) ´ ( - 2mvey ) = 2mv aex

4. Apply parallel axis theorem,

9. Moment of inertia of the system about rod xshown the figure

I = ICM + Mh 2, we get

x

z

2

Þ

=

ML2 7 ML2 æLö +Mç ÷ = è 4ø 12 48 y

5. Here, m1 = m2 = 0.1kg and

r1 = r2 = 10 cm = 0.1m 1 3 I = I1 + I2 = m11 r 2 + m2r22 = m11 r2 2 2 3 = ´ 0.1(0.1) 2 = 1.5 ´ 10 –3 kg-m2 2

æ Ml 2 Ml 2 ö 4 I = Ix + Iy + Iz = 0 + ç + ÷ + Ml 2 = Ml 2 4 ø 3 è 12

(Q m1 = m2)

6. Moment of inertia of the system about axis AX

10. As, m1 = m2 Þ

R12 d 2 = R22 d1

X C

rC l

l

Now, 60° A rB

l

Y

acceleration of the disc is zero. Choice (d) is not true.

= (MA (rA ) 2 + MB(rB) 2 + MC (rC ) 2

12. A raw egg behaves like a spherical shell and a half boiled egg

= M (0) 2 + m ( l) 2 + m ( l cos 60° ) 2 = ml 2 +

5 ml ml = 4 4

1 mR12 I1 2 R2 d = = 12 = 2 1 I2 mR22 R2 d1 2

11. When a disc rotates with uniform angular velocity, angular

= IA + IB + IC

2

pR12xd1 = pR22xd 2

2

behaves like a solid sphere 1ö æ çQ cos 60° = ÷ è 2ø

\

Ir 2 / 3 mr 2 5 = = >1 Ib 2 / 5 mr 2 3

338 JEE Main Physics æ L2

13. Required moment of inertia, I = M ç

è12

+

æ L2 D 2 ö r2ö ÷ ÷ =Mç + 4ø è12 16 ø

ml 2 3 \ Moment of inertia of the system, which rod is bent,

14. Moment of inertia of a uniform rod about one end = 2 ML2 æ M ö (L / 2) = =2´ç ÷ è2ø 3 12

15. Given, mass of bullet (m) = 10 g = 0.01kg Speed of bullet (v) = 500 m/s Width of the door ( l) = 1. 0 m Mass of the door (M) = 12 kg

(L) = mv ´ r

Moment of inertia of the door about the vertical axis at one of its end, Ml 2 12 ´ (1) 2 = = 4 kg-m 2 3 3

But angular momentum, (L) = Iw 2.5 = 4 ´ w 2.5 w= 4

or

With the hole, Ix and Iy both decrease gluing the removed piece at the centre of square plate does not affect Iz. Hence, Iz decreases, overall.

æ2 ö æ 2p ö L = Iw = constant = ç MR 2÷ ç ÷ = constant i. e. , è5 øèT ø

2

\

17. We calculate moment of inertia of the system about AD, a

2

aC

Moment of inertia of each of the spheres A and D about 2 AD = Ma2 5 Moment of inertia of each of the sphere B and C about AD æ2 ö = ç Ma2 + Mb 2÷ è5 ø Using theorem of parallel axis.

I=

t A = 100 ´ 0.75 = 75 Nm counter clockwise, tB = 100 ´ 1.25 = 125 Nm clockwise

23. w1 = 2p rad/day, w2 = 0 and t = 1day w2 - w1 0 - 2 p = t 1 rad rad 2p =2p = day 2 (86400) 2 s2

\

a=

B

b

Da

æ xR ö R22 T1 = ç 1 ÷ ´ 24 h = 24x2h T2 è R1 ø

1 2 1 æ 1ö mr = ´ 16 ç ÷ = 2 kg-m2 è2ø 2 2 2p (n2 - n1) 2 p (2 - 0) p a= = = rads–2 8 2 t p Now, t = Ia = 2 × = p N-m 2

\

I = m11 r 2 + m2r22 = 2 (0.3) 2 + 1(0.3) 2 = 0.27 kg-m2

b

T2 =

21. As,

16. Required moment of inertia,

a

R2 = constant T R12 R22 = T1 T2

Þ

22. As toruqe = force ´ perpendicular distance

w = 0.625 rad/s

A

L ML2 \ K= 12 12

20. As no torque is applied, angular momentum

Þ

1 = 0.01 ´ 500 ´ 2 = 2.5 J-s

\

18. According to the theorem of perpendicular axes, Iz = Ix + Iy .

19. As, I = MK 2 =

As bullet gets embedded exactly at the centre of the door, therefore its distance from the hinged end of the door l 1 (r) = = m 2 2 Angular momentum transferred by the bullet to the door,

(I) =

Total moment of inertia, æ2 ö æ2 ö I = ç Ma2÷ ´ 2 + ç Ma2 + Mb 2÷ ´ 2 è5 ø è5 ø 8 = Ma2 + 2 Mb 2 5

Torque required to stop the earth, t = I a = FR 2 MR 2 ´ a Ia 5 2 F= = MR ´ a = R R 5 2 2p 24 = ´ 6 ´ 10 ´ 6400 ´ 10 3 ´ 5 (86400) 2 = 1.3 ´ 10 22 N

24. By conservation of energy mgh =

1 2 1 1 1 w2 Iw + mv 2 = Iw2 + mr 2w2 = [1 + mr 2] 2 2 2 2 2

é 2 mgh ù w=ê 2 ë I + mr úû

1/ 2

Rotational Motion

339

4.6 t 3 ´ 10 2 ´ 4.6 = 2s t= 6.9 ´ 10 2

25. Here, m = 3 kg, v = 2 ms–1

6.9 ´ 10 2 = 3 ´ 10 2 ´

y

30. Here, r = 0.2 m,M = 10 kg, O

n = 1200 rpm = 20 rps

x

4m

L = Iw = (Mr 2) (2 pn)

r = 4 m,L = ?,T = ?

= 10 ´ (0.2) 2 ´ 2 ´

L = mvr sin 90° = 3 ´ 2 ´ 4 ´ 1 = 24 kg m2s–1 t=

and

dL =0 dt

31. Mass of oxygen molecule (M) = 5.30 ´ 10 -26 kg

26. As the block remains stationary therefore for translatory equilibrium S fx = 0 \ f = N and S fy = 0 \ f = mg For rotational equilibrium S t = 0 By taking the torque of different force about point O t f + t f + t N + t mg = 0 As f and mg passing through point O

$ ´ (2$i - 3$j + 4 k) $ 27. As, t = r ´ F = (3$i + 2$j + 3 k) = - 9 k$ - 12 $j - 4 k$ + 8$i + 6$j + 9$i = 17 $i - 6 $j - 13 k$

28. As, I = 1.2 kg-m2, E r = 1500 J a = 25 rad s–2 w1 = 0 , t = ?

Þ

Er = w=

1 2 Iw 2

Þ

2 Er I

w2 = w1 + a t 50 = 0 + 25 t t =2s 2

29. Here, moment of inertia, I = 3 ´10 kgm

2

Torque, t = 6.9 ´ 10 2 Nm Initial angular speed, w0 = 4.6 rad s–1 Final angular speed, w0 = 0 rad s–1 w0 = w0 + at w - w0 0 - 4.6 4.6 a= = =rad s–2 \ t t t Now, negative sign is for deceleration torque, t =Ia As

Mean speed of the molecule (v) = 500 m/s 2 Given, KE of rotation = ´ KE of translation 3 1 2 2 1 Iw = ´ Mv 2 2 3 2 w= =

2Mv 2 I

2 ´ 5.30 ´ 10 -26 ´ (500) 2 1.94 ´ 10 -46 = 1.35 ´ 10 10 ´ 500 = 6.75 ´ 10 12 rad/s

32. Work done in stopping = change in KE = final KE - initial KE i. e. ,t q = K = constant. As t is same in the two cases, q must be same i. e. , number of revolution must be same. (w - w ) 33. As, t = F ´ r = Ia = I 2 1 t F ´ r ´ t 10 ´ 0.2 ´ 4 = = 20 rad s–1 \ w2 - w1 = I 0.4

34. Maximum height attained (h) =

2 ´ 1500 = = 50 rad s–1 1.2 From,

Moment of inertia I = 1.94 ´ 10 -46 kg-m 2

or

\ tf + tN = 0 As t f ¹ 0 \t N ¹ 0 and torque by friction and normal reaction will be in opposite direction.

Now,

22 ´ 20 = 50.28 kg-m2s–1 7

=

v 2 sin 2 q 2g v 2 sin 2 45° v2 = 2g 4g

At highest point, momentum = mv cos 45° mv = 2 \Angular momentum =

mv v 2 mv3 ´ = 2 4g 4 2 g

35. Given, R = 2 m M = 100 kg Speed of centre of mass (v) = 20 cm/s = 0.20 m/s Work done to stop the hoop = Total kinetic energy of the loop 1 1 W = Mv 2 + Iw2 2 2

340 JEE Main Physics But moment of inertia I = Mr 2 and angular velocity w = \

W=

v R

2ö

æv 1 1 Mv 2 + (Mr 2) ´ ç 2 ÷ 2 2 èR ø

43. As, the rod is highed at one end, its moment of inertia about this end is I =

Total energy in upright position

1 1 = Mv 2 + Mv 2 = Mv 2 2 2

\

= total energy on striking the floor 1 ML2 2 MgL 1 2 0+ = Iw + 0 = w 2 2 2 3

= 100 ´ (0.20) 2 = (100 ´ 0.04) J = 4.0 J

Þ

36. As, Force ´ distance = t = Ia F sin 30° ´

Þ

As, \

75 5 Here, n1 = rps = rps 60 4 For n2 = 0 ,t = 5 s, a = ? w - w1 2 p (n2 - n1) 2p (0 - 5 / 4) a= 2 = = t t 5 p –2 = - = -1.57 rad s 2 ( w / 2) 2 - w2 = 2 a (36 ´ 2p ) 2

2

0 - ( w / 2) = 2a (n ´ 2p )

Similarly,

…(i) …(ii)

39. In rotation per second (rps) w2 - w1 240 - 0 = = 2 rps2 t 2 ´ 60

40. Angular momentum = linear momentum ´ perpendicular distance of line of action of linear momentum from the axis of rotation = mv ´ l

41. As, w = w0 + at = 0 + 2 ´ 6 = 12 rads–1 Also

v = rw

\

v = 2 ´ 12 = 24 ms–1

42. According to conservation of angular momentum Þ \

I1w1 = I2w2 1 1 ì Mü ö æ1 MR 2w = ç MR 2 + í ý R 2÷ w2 è2 2 2 î 4þ ø w2 =

4 w 5

(Q w = 2pn)

10 10 n1 = ´ 2.8 = 4.0 rps 7 7

centre, no torque is exerted i. e. , t = 0. According to the principle of conservation of angular momentum, I ´ w = constant. As mass reduces to half (from 2M to M), moment of inertia I becomen half. Therefore, w must become twices ( = 2 w). 1 1 46. Rotational kinetic energy, E = Iw2 = L ´ 2 pn 2 2 L2 E 2 n1 E µL ´n Þ = ´ \ L1 E1 n2

n = 12

a=

n2 10 = n1 7

45. When the person jumps off the round, radially away from the

Dividing Eq. (i) by Eq. (ii), we get 3 - w2 36 4 = 2 n w 4 \

Lw2 3g or w = 3 L

l2 = 0.7 I1 w2 I1 1 = = w1 I2 0.7

n2 =

w22 - w12 = 2 aq, we get

38. Using,

g =

44. Here, n1 = 2.8 rps,n2 = ?

a = 12 rad s–2

\

37.

2 é 1 æ 1ö ù = 4 ê2 ´ ç ÷ ú a è 4ø ú 2 êë û

1 1 a 24 ´ ´ = 2 2 2

or

ML2 3

or Þ

L2 é E1 /2 ù é n1 ù = ´ L1 êë E1 úû êë 2 n1 úû I I2 = 1 4 L L2 = 4 1 L2 = 4

47. Angular velocity of man relative to platform is v0 r 1 = 0.7+ = 1.2 rad s–1 2

wr = w +

Time taken by the man to complete one revolution, 2p 2p T= = s wr 1.2 Angular displacement of the man w.r.t. ground, vT q = w0T = 0 r 1 æ2 p ö 5 = ç ÷ = p rad 2 è 1.2 ø 6

(as I1 = I)

Rotational Motion 2 5 = 1+ 1 sin 30° sin q2

48. As L = Iw = constant \

2 æ2 p ö MR 2 ´ ç ÷ = constant è 7 ø 5

Þ

R2 = constant T

or

i. e. ,

where R is halved, R 2 becomes 1/4th. Therefore, T becomes 1/4th i. e. , 6 h. é (I w + I w ) 2 ù 1 (I1 + I2) ê 1 1 2 22 ú 2 ë (I1 + I2) û 1 (I1w1 + I2 w2) = 2 I1 + I2

Kf =

49. Here,

Ki =

and Þ

1 (I1w12 + I2 w22) 2

1+

sin q2 =

5 = 0.7143 7

q = 45°

or

52. Here, l = 1m, q = 30°, g = 9.81ms–2,t = ? 2l (l + K 2/ R 2) g sin q

t=

Q

1 2 R 2

For a rupee coin, K 2 =

2 ´ 1(1 + 1 / 2) 6 = = 0.78 s 9.81 sin 30° 9.81

t=

Þ

D K = Kf - Ki I I = - 1 2 ( w1 - w2) 2 2 (I1 + I2)

53. Time taken in reaching bottom of incline is 2l (1 + K 2/ R 2) g sin q

t=

50. Here, mass of man, m = 80 kg

K2 , greater will be the time R2

Mass of platform, M = 200 kg Let R be the radius of platform.

Greater the value of

When man is standing on the rim,

For hollow cylinder (HC), K 2 = R 2 2 For solid sphere (S), K 2 = R2 5

For solid cylinder (SC),

I1 = M (R / 2) 2 + mR 2 2

æRö = ç ÷ (M + 4 m) è2ø I2 = M (R / 2) 2 + m ´ 0 = m (R / 2) 2 As angular momentum is conserved, I1 w2 2 pn2 n2 = = = I2 w1 2 pn1 n1

=

For solid sphere, K 2 =

ö æ ÷ ç 9.8 sin 60° sin q g ÷ çQ a = a= 2 2 ç K ÷ 1+ 1+ 2 ÷ ç 5 è R ø 5 3 -2 a = ´ 9.8 ´ = 6.06 ms 7 2

\

I1 ´ n2 I2

or

(M + 4 m) (R / 2) 2 ´ 12 M (R / 2) 2

(200 + 4 ´ 80) ´ 12 200 520 ´ 12 = 200

2 2 R 5

55. As, it is clear from figure,

=

R

n1 = 31.2 rpm

51. For rolling, t = \

2l (1 + K 2/ R 2) = same g sin q

2l (1 + K12/ R 2) 2l (1 + K22/ R 2) = g sin q1 g sin q2

For sphere,

K12 =

2 2 R , q1 = 30° 5

For hollow cylinder, K22 = R 2, q2 = ?

K 2 = R 2/ 2

54. Here, q = 60°, l = 10 m, a = ?

When man reaches the centre of platform,

n2 =

341

(given in question)

On reaching the bottom of the bowl, loss in PE = mgR, and gain in 1 1 KE = mv 2 + Iw2 2 2 1 1 æ2 ö | DK| = mv 2 + ´ ç mr 2÷ w2 Þ ø 2 2 è5 =

1 1 7 mv 2 + mv 2 = mv 2 2 5 10

As, again in KE = loss in PE 7 \ mv 2 = mgR 10 v=

10 gR 7

342 JEE Main Physics 56. If h is height of the ramp, then in rolling of marble, speed

61. The rolling sphere has rotational as well as translational kinetic energy.

2 gh v= 1 + K 2/ R 2

1 1 mu 2 + Iw2 2 2 1 1 æ 2 2ö 2 2 = mu + ç mr ÷ w ø 2 2 è5

\Kinetic energy =

The speed of the cube to the centre of mass v ¢ = 2 gh v¢ K2 = 1+ 2 v R 2 2 2 For marble sphere, K = R 5

\

v¢ 2 7 = 1+ = = 7: 5 v 5 5

\

1 mv 2 7 = mu 2 + mu 2 2 5 10 Loss in potential energy = Gain in kinetic energy 7 \ mgh = mu 2 10 =

Þ

h=

57. Here, r = 0.5,m = 2 kg Rotational KE = Þ

1 2 1 æ 1 2ö 2 Iw = ´ ç mr ÷ w ø 2 2 è2 1 1 4 = mv 2 = ´ 2 v 2 4 4 v = 8 = 2 2 ms

\

1 2

62. For solid cylinder, q = 30° ,K 2 = R 2 For hollow cylinder, q = ?,K 2 = R 2 Hence, æ 1ö ç1 + ÷ è 2 ø 1+ 1 = sin 30° sin q 2 sin q = = 0.6667 3

–1

58. When a body of mass m slides down an inclined plane, then v = 2 gh

\

When it is in the form of ring, then v ring =

2 gh v 2 gh 2 gh = = = 1+ 1 æ K2 ö 2 2 ç1 + 2 ÷ è R ø 1 2

59. Rotational kinetic energy KR = Iw2

q = 42° 2 gh K2 1+ 2 R

63. As, v =

where K is the radius of gyration.

2

1 MR 1 ´ ´ w2 = Mv 2 (Q v = Rw ) 2 2 4 1 Translational kinetic energy KT = Mv 2 2 1 1 Total kinetic energy = KT + KR = Mv 2 + Mv 2 2 4 3 = Mv 2 4 1 Mv 2 Rotational kinetic energy 4 1 = = \ 3 Total kinetic energy 3 2 Mv 4 1 5g g sin q g sin 30° 5 60. As, a = = g´ = = 2 2 7 2 14 K 1+ 1+ 2 5 R KR =

7 u2 10 g

(R = Radius of sphere)

K2 =1 R2

For ring \

v=

2 gh = gh 1+ 1 2 5

64. Here, m = 8 kg, r = 40 cm = m, w = 12 rad s–1,I = 0.64 kg m2 1 2 1 Iw + mv 2 2 2 1 2 1 2 2 = Iw + mr w 2 2

Total KE =

2

=

1 1 æ2ö ´ 0.64 ´ 15 2 + ´ 8 ´ ç ÷ ´ 15 2 è5ø 2 2

= 216 J

343

Rotational Motion

Round II 1. Moment of inertia of cylinder about an axis through the centre and perpendicular to its axis is æ R 2 L2 ö Ic = M ç + ÷ è 4 12 ø

6. Linear acceleration for rolling, a =

1+

2 æ R 2 L2 ö æ R 2 L2 L2 ö æLö I = Ic + M ç ÷ = M ç + + ÷ =M ç + ÷ è2ø è 4 3ø è 4 12 4 ø

Ih =

2 5

49 MR 2 4

2 3

2. As, Is = MRs2,Ih = MRh2 Is = Ih 2 2 MRs2 = MRh2 5 3

As \

Rs 5 = Rh 3

\

3. Here, pr = l \ r = l / p 1 2 Mr 2 2 1 é æ l ö ù ml 2 \ Moment of inertia of semicircle = êm ç ÷ ú = 2 êë è p ø úû 2 p 2 Moment of inertia of a ring about its diameter =

4. As, w2 = w1 + at \ or From,

a = 2 p rad s–2 w22 - w12

\ For rolling without slipping of a roller down the inclined plane, tan q £ 3 m s .

7. Initial angular momentum of ring L = Iw = MR 2w = (MR 2 + 2 mR 2) w¢ As there is no external torque on the system therefore MR 2w = (MR 2 + 2 mR 2) w¢

= 2 aq

q=

Þ

2

1200 p = 300 p 4p q 300 = = 150 2p 2p

5. From conservation of angular momentum,

Now,

(where, f = force of friction) M But Ra = a \ f = a 2 M 2 M \ f = × g sin q = g sin q 2 3 3 m s = f / N, where N is normal reaction, M g sin q tan q ms = 3 = \ 3 Mg cos q

40 p = 20 p + a ´ 10

Number of rotations completed =

\

θ

K2 1 For cylinder, 2 = 2 R 2 \ acylinder = g sin q 3 For rotation, the torque fR = Ia × (MR 2a) / 2

Final angular momentum of system (Ring + Two particles)

( 40 p ) 2 - (20 p ) 2 = 2 ´ 2 pq Þ

K2 R2

f

Using theorem of parallel axes, moment of inertia of the cylinder about an axis through its edge would be

When L = 6 R ,

g sin q

I1w1 = I2w2 w1 I2 = w2 I1

2

I1 æ I2 ö I ´ç ÷ = 2 I2 è I1 ø I1

As

I1 > I2

\

E1 < E 2

Mw (M + 2 m) a

8. Angular momentum of block w.r.t. a O before collision with O = Mv 2 On collision, the block will rotate about the side passing through O. Now its angular momentum = Iw By law of conservation of angular momentum a Mv = Iw 2 a æ Ma2 Ma2 ö Þ Mv = ç + ÷w Þ 2 è 6 2 ø

1 2 Iw E1 2 1 1 = E 2 1 I w2 2 2 2 =

w¢ =

v M

O

w=

3v 4a

where I is moment of inertia of the block about the axis perpendicular to the plane passing through O.

9. Angular momentum about origin = Itranslation + Irotation = MvR + Ic w = M (Rw) R +

1 3 MR 2w = MR 2w 2 2

344 JEE Main Physics 1 2 w = 2 pn is doubled, w2 becomes 4 times. As E reduces to half, 1 I must have been reduced to th. From L = Iw, L becomes 8 1 1 ´ 2 = times i.e., 0.25 L. 8 4

10. From E = Iw2, we find that when frequency (n) is doubled,

14. According to theorem of parallel axes, moment of inertia of a rod about one of its ends =

\Moment of inertia of two rods about Z-axis = Moment of inertia of 2 rods placed along X and Y-axis

11. By the conservation of energy,

=

Loss in PE of rod = gain of rotational KE 1 1 mg sin a = Iw2 2 2 1 1 ml 2 sin a = w 2 2 3

Þ

l/2 α

3 g sin a l

w=

w=

3g sin a 2L

mL2 . 12 Moment of inertia of each disc about disc about its diameter

12. Moment of inertia of rod about the given axis =

=

MR 2 4

MR 2 ML2 + 4 4

Þ log e L = log e P + log e r If graph is drawn between log e L and log e P then, it will be straight line which will not pass through the origin because of presence of constant in the equation.

17. As the mass is concentrated at the centre of the rod, therefore,

\ For theorem of parallel axes, moment of inertia about the given axis is mL2 æ MR 2 ML2 ö I= +ç + ÷ 12 è 4 4 ø I=

r2 = constant T Differentiating w.r.t. time (t), we get dr dT T ×2 r - r2 dt dt = 0 T2 dr dT or 2 Tr = r2 dt dt dT 2 T dr or = dt r dt

16. As, L = rP

Using theorem of parallel axes, moment of inertia of each disc 2 MR 2 æLö about the given axis = +Mç ÷ è2ø 4 =

Þ

L = Iw = constant p 2 2 æ 2ö = constant ç MR ÷ è5 ø T

or

But in the problem length of the rod 2 L is given \

2 Ml 2 3

15. As no torque is being applied, angular momentum

2

mg

Ml 2 Ml 2 Ml 2 + = 12 4 3

mL2 mR 2 ML2 + + 12 4 4

mg ´ Þ

l 1 2 1 = Iw = 2 2 2 w=

æ ml 2 ö 2 ÷w ç è 3 ø

3g l

l/2

13. According to the theorem of parallel axes, moment of inertia of disc about an axis passing through K and perpendicular to plane of disc, R O

=

R

K

1 3 MR 2 + MR 2 = MR 2 2 2

Total moment of inertia of the system 3 = MR 2 + m (2R) 2 + m ( 2R) 2 + m ( 2R) 2 2 = 3 (M + 16 m)

2

R 2

Velocity of other end of the rod v = wl = 3 gl

18. Let L1, L2 and r1, r2 are the angular momenta and position vectors of the particles at that instant about any arbitrary point O. Angular momentum of the particles, L1 = r1 ´ mv and L 2 = r2 ´ mv It resultant angular momentum of the system is L, then L = L1 + L 2 = r1 ´ mv + ( - r2 + mv) Negative sign shown that both particles are moving in opposite directions.

345

Rotational Motion 5 4 4 2 MR = I 5

23. MI of disc about tangent in a plane = MR 2 = I θ1

M θ1 r1sinθ1 d N

si r2

nθ

2

\

v

P1

r1 θ2

P2

–v

\

r2

24. Angular momentum about an axis

O

\

| L | = |L1 | - |L 2 |

OM - ON = MN = d

(Given) …(ii)

Þ

\

G a/2

ML 3

F = µmg mg

26.

y R/2

R/2

Moment of inertia of the system about yy¢ Iyy¢ = Moment of inertia of sphere P about yy¢ + Moment of inertia of sphere Q about yy¢ Moment of inertia of sphere P about yy¢ 2

=

3 æRö M ç ÷ + M ( x) 2 5 è2ø

=

2 æRö M ç ÷ + M (2R) 2 5 è2ø

=

MR 2 + 4 MR 2 10

2

wR = 0 2 mg

22.

y’

2R

w = w0 - at w - w w0 - w0 / 2 t= 0 = a mg / R

2 2 æ4 ö As, I = MR 2 = ç mR3r ÷ R 2 ø 5 5 è3 8 22 5 = ´ Rr 15 7 176 5 I= Rr Þ 105

Motion

h/2

2

æ 600 ö 2 2 2 = 0.8 p 2 ç ÷ MR = 80 p MR è 60 ø

As

x = vt

R = mg

ML2 ML2 2 ML2 \ I =0 + + = 3 3 3 1 2 1 2 2 20. KE of rotation = Iw = ´ MR (2 pn) 2 2 2 5 1 2 2 = ´ 4 p n MR 2 5

m mgR mg = R mR 2

x

F = mR = m mg . To topple, clockwise moment must be more than the anticlockwise moment h a m mg ´ > mg ´ i. e. , 2 2 or m > a/h

I = I1 + I2 + I3

t I

v

25. As shown in figure normal reaction, R = mg . Frictional force,

The direction of L is perpendicular to the plane of r and v and is inward to the plane of paper, which also remains unchanged with time.

2

O

torque, t = (2 mv 2w) t

|L | = mvd It is constant with time.

19. Moment of inertia of a rod about one end =

v

d Lt = 2 Mv 2tw dt

…(i)

\ r1 sin q1 - r2 sin q2 = d From Eqs. (i) and (ii), we get

21. Angular retardation, a = =

ω

Lt = [I + m (vt ) ] w

where q1 and q2 are the angles between r1, v and r2, v respectively. When particles changes their position with time, their direction of motion ( v) remains unchanged and therefore distances OM = r1 sin q1 and ON = r2 sin q2 remains same.

As

z

2

= mvr1 sin q1 - mvr2 sin q2 = mv(r1 sin q1 - r2 sin q2)

But

3 MR 2 2 3 æ4 ö 6 I¢ = ç I÷ = I 2 è5 ø 5

MI of disce about tangent I to plane I ¢ =

θ2

Moment of inertia of sphere Q about yy¢ is 2

Now, Iyy¢ =

2 æRö Mç ÷ 5 è2ø

2 æRö 21 MR 2 + 4 MR 2 + M ç ÷ = MR 2 10 5 è2ø 5

2

346 JEE Main Physics 27. When one string is cut off, the rod will rotate about the other point A. Let a be the linear acceleration of centre of mass of the rod and a be the angular acceleration of the rod about A. As it clear from figure.

Initial angular momentum, I1 = mR 2 +

MR 2 2

At any time t, let the tortoise reach D moving with velocity v. \

AD = vt AC = R 2 - a2

T A a

…(i) …(ii)

From Eq. (i), T = mg - ma = mg -

m1

torque, therefore

m2 x2

T 1 -1 = = 2 ´ 10 -20 T0 5 ´ 10 19

B

31. Moment of inertia of the cylinder about an axis perpendicular

From Eqs. (i) and (ii),

to the axis of the cylinder and passing through the centre is æ R 2 L2 ö …(i) I =M ç + ÷ è 4 12 ø

m2 r

x1 =

m1 + m2 m1r x2 = m1 + m2 IAB = m1x12 + m2x22 =

m1 m2 r 2 m1 + m2

29. As there is no external torque, angular momentum will remain constant. When the tortoise moves from A to C, figure, moment of inertia of the platform and tortoise decreases. Therefore, angular velocity of the system increases. When the tortoise moves from C to B, moment of inertia increases. Therefore, angular velocity decreases.

R r

If,

I0 w0 = Iw 2 ù é2 2p æ2 2ö æ 2 p ö 2 2 MR ÷ = ê MR + ç MR ÷ ç ú è3 ø è T0 ø ë 5 5 5 ´ 10 19 û T T 1 = 1+ T0 5 ´ 10 19

…(ii)

C x1

A

MR 2 2 As angular momentum is conserved I2 = mr 2 +

…(i)

m1x1 = m2x2

\

OD = r = a2 + [ R 2 - a2 - vt ]2

30. As angular momentum is conserved in the absence of a

3 mg mg = 4 4

x1 + x2 = r

and

\

\ I1w0 = I2 w (t ) This shows that variation of w(t ) with time is non-linear.

l l 3g 3g a = ra = a = = 2 2 2l 4

and

DC = AC - AD = ( R 2 - a2 - vt )

Angular momentum at time t

mg

mg - T = ma t mg ( l / 2) 3 g a= = = I 4 ml 2/ 3

28.

As

D

If r is volume density of the cylinder, then M = ( pR 2L) r = constant M L= \ pR 2r Put in Eq. (i) ö æ R2 M2 I =M ç + ÷ 2 4 4 è 4 12 p R r ø dI For I to be minimum, =0 dR æR dI M2 ö =M ç ÷ =0 2 5 dR è2 3 p R ø R M2 = 2 3 p 2r 2R5

O a C

B

M = mass of platform R = radius of platform m = mass of tortoise moving along the chord AB a = perpendicular distance of O from AB.

or R 6 =

Using Eq. (ii),

R6 =

2 p 2R 4L2r 2 3 p 2r 2

or

R2 =

L2 3 2 2 L or = 3 R2 2

or

L = 3 /2 R

…(ii)

2 M2 3 p 2r 2

Rotational Motion 32. If M mass of the square plate before cutting the holes, then mass of portion of each hole. p M ´ pR 2 = M m= 2 16 16 R \Moment of inertia of remaining portion I = Isquare - 4 Ihole ù é mR 2 M (16 R 2 + 16 R 2) - 4 ê = + m ( 2R) 2ú 12 û ë 2 M ´ 32 R 2 - 10 mR 2 12 8 10 p æ 8 10 p ö 2 = MR 2 MR 2 = ç ÷ MR è 3 16 ø 3 16

=

38. Choice (a) is false, as theorem of perpendicular axes applies only to a plane lamina. Now, Z axis parallel to Z¢ axis and distance between them a 2 a . Therefore, according to the theorem of parallel = = 2 2 axes, 2

ma2 æ a ö Iz ¢ = Iz + mç ÷ = Iz + è 2ø 2 Choice (b) is true. Again, choice (c) is false as Z¢ ¢ axis is not parallel to Z-axis. Choice (d) is true as from symmetry, we find that Ix = Iy .

39. For minutes hand, T = 1h = 60 ´ 60 s 2p 2p = rad s–1 T 60 ´ 60 p = rad s–1 1800

33. From theory of rolling motion without slipping speed of

w=

particle at point of contact A is zero and at the top point C speed is 2 v 0 . Moreover, speed of point O is v 0 but that of B is v 0 2.

34. From the study of theory, we know that for general rotational motion, angular momentum L and angular velocity wneed not be parallel.

35.

Again, for a general translational motion, linear momentum p and linear velocity v are always parallel. This is because p is directed along v only. When net external torque on a system of particles about an axis is zero, i. e. , t = r ´ F = r F sin q t = Zero, where q is angle between r and F, t is unit vector along t, then all the four statements (a), (b), (c), (d) are compatible.

36. As, v 0 = 3 wR$i

347

40. For hour’s hand, T = 12 h = 12 ´ 60 ´ 60 s æ2 p ö v = rw = r ç ÷ è T ø 5 2p ms–1 ´ 100 12 ´ 60 ´ 60 p = ms–1 120 ´ 60 ´ 60 =

41. Here, n0 = 15 rps; n = 5 rps q = 50 revolution = 50 ´ 2 rad From,

60° 30°

3 ωR

P

3 ωR

w2 - w20 = 2 aq a=

w2 - w20 4 p 2 (n 2 - n02) = 2q 2q

a=

4 p 2 (5 2 - 15 2) = - 4 p rad s–2 2 ´5 ´2 p

v = 3 ωR

42. Given, M = 20 kg

For pure rolling, v 0 = 3wR$i wR wR æ ö cos 60° ÷ $i + sin 60 $j v p = ç 3 wR è ø 2 2 =

11Rw $ 3 wR $ i+ j 4 4

37. According to right handed screw rule, the direction of torque $ So choice (a) is false. t caused by F about Z-axis is along k. However choice (b) is true as direction of torque ( t¢ ) caused $ As t = r ´ F and P is closer to Z¢ by F about Z¢ axis is along - k. axis, therefore t caused by F about Z-axis is greater in magnitude than that about Z¢ axis. Choice (c) is true. Choice (d) is false as it is meaningless to add torques about differemt axes

w = 100 rad/s R = 0.25 m Moment of inertia of the solid cylinder about its axis of symmetry. 1 I = MR 2 2 1 = ´ 20 ´ (0.25) 2 2 = 10 ´ 0.0625 = 0.625 kg - m2 Kinetic energy associated with the rotation of the cylinder is given by 1 K = Iw2 2

348 JEE Main Physics =

1 ´ 0.625 ´ (100) 2 2

47. Both, the assertion and reason are true and latter is a correct explanation of the former. Infact, as ice on polar caps of earth melts, mass near the polar axis spreads out, I increases. Therefore, T increases i.e., duration of day increases.

= 0.3125 ´ 10000 = 3125 J Angular momentum, L = Iw

48. In the case of circular ring the mass is concentrated on the rim (at maximum distance from the axis) therefore moment of inertia increase as compared to that in circular disc.

= 0.625 ´ 100 = 62.5 J-s

43. Let M and R be the mass of radius of the solid sphere and

49. To reverse the direction

ò

hollow cylinder. Moment of inertia of the hollow cylinder about its axis of symmetry,

a=

Moment of linear of the solid sphere about its diameter 2 I2 = MR 2 5

or or

t

0

t = I1a1 t = I2a 2 I1a1 = I2a 2 2 MR 2 a1 I2 5 2 = = = 5 a 2 I1 MR 2 a2 =

2t2 -

t3 =0 3

Þ

t3 = 6 t 2

Þ

t =6 s dq =w dt 6 6 æ t3 ö q = ò w dt = ò ç2 t 2 - ÷ dt 0 0 è 3ø

Þ

5 a1 2

t3 3

w is zero at

As

6

…(i) = 2.5 a1 Let after time t , w1 and w2 be the angular speeds of the hollow cylinder and solid sphere respectively. \ and

dw t 40 t - 10 t 2 = = 4t -t2 = I 10 dt

w = ò a dt = 2 t 2 -

Let torque t of triangle magnitude be applied on hollow cylinder and solid sphere. the angular acceleration produced in it are a1 and a 2 respectively.

Therefore,

(work done is zero)

t = (20 t - 5 t 2) 2 = 40 t - 10 t 2

As,

I1 = MR 2

\ and

t dq = 0

w1 = w0 + a1t w2 = w0 + a 2t

…(ii)

= w0 + 2.5 a1t From Eqs. (ii) and (iii), we get

…(iii)

w2 > w1 Therefore, solid sphere will acquire a greater angular speed after a given time.

44. Position vector of centre of mass depends on masses of particles and their location. Therefore, change in shape/size of body do change the centre of mass.

45. In sliding down, the entire potential energy is converted only into linear kinetic energy. In rolling down, a part of same potential energy is converted into kinetic energy of rotation. Therefore, velocity acquired is less.

46. When a person is high up on the ladder. Then a large torque is produced due to his weight about the point of contact between the ladder and the floor whereas when he starts climbing up the torque is small, Due to this reason the ladder is more opt to slip when one is high up on it.

é 2 t3 t 4 ù é 2 1ù =ê - ú = 216 ê - ú = 36 rad ë3 2û ë 3 12 û 0 36 Number of revolution is less than 6. 2p 1 2

50. MI = MR 2 + mx2 where, m = mass of insect and

x = distance of insect from centre

Clearly as the insect moves along the diameter of the disc moment of inertia first decreases then increases. By conservation of angular momentum, angular speed first increases then decreases.

51. If centre of mass rises to a maximum height h, then from loss in KE = gain in PE, we get 1 2 Iw = mgh 2 1 æ ml 2 ö 2 l 2w2 or ÷ w = mgh Þ h = ç 6g 2è 3 ø

52. Moment of inertia of the square plate about an axis passing through the centre and perpendicular to its plane is I=

m ( a2 + a2) ma2 = 12 6

Rotational Motion 53. When the axis passes through one of its corners, we use

349

57. In figure is centre of a circular disc of radius 2 R and mass M.

theorem of parallel axes. 2

\

æa 2ö ma2 ma2 2 + = ma2 I' = I + mç ÷ = 6 2 3 è 2 ø

\

q = wt =

x

500 ´ 2p 50 p rad = 60 3

O

C2

C1

54. Angular accelerations is time derivative of angular speed and angular speed is time derivative of angular displacement. dw By definition a = dt dw = adt

i. e. ,

So, if in time t the angular speed of a body changes from w0 to w w

òw

0

t

dw = ò adt 0

If a is constant w - w0 = at Now, as by definition dq w= dt and dw q= dt

...(i)

C1 is centre of disc of radius R, which is removed. Mass of removed disc, M M1 = p (R) 2r = 4 Mass of remaining disc, M 3M M2 = M - M1 = M - = 4 4 Let its centre of mass be at C 2, where OC 2 = x \

58. As is clear from figure

dq = w0 + at dt

AC = BD = l 2 + l 2 = l 2

i. e. , dq = ( w0 + at )dt So, if in time t angular displacenent is q

Moment of inertia of four point masses about BD 2

2

æl 2ö æl 2ö IBD = m ç ÷ + m ´0 + m ç ÷ + m ´0 è 2 ø è 2 ø

t

ò0 dq = ò0 ( w0 + at) dt q = w0t +

M1 ´ OC1 = M2 ´ OC 2 3M M ´R = x 4 4 R 1 x = = aR Þ a = 3 3

or

Eq. (i) becomes

q

M = p (2R) 2r, where r is mass/area of disc.

1 2 at 2

=

ml 2 ml 2 + = ml 2 2 2 Y

Given, a = 3.0 rads -2, , w0 = 2.0 rads -1 ,t = 2s Hence,

q = 2 ´2 +

1 ´ 3 ´ (2) 2 2

A

or q = 4 + 6 = 10 rad Eqs. (i) and (ii) are similar to first and second equations of linear motion. 1 moment of inertia of lamina about an axis through O 2 and ^ ABCD. 1 and IEF = moment of inertia of lamina about an axis through 2 O and ^ ABCD IAC =

\

IAC = IEF

56. Torque due to central force is zero As

d t = (L) = 0 dt

\

L = constant

l

X

O D

55. From symmetry considerations,

B

l

C

Applying the theorem of parallel axes, IXY = IBD + M ( AO) 2 2

æ l ö 2 = ml 2 + 4 m ç ÷ = 3 ml è 2ø

59. Angular momentum of a rigid body about a fixed axis is given by L = Iw where I is moment of inertia and w is angular velocity about that axis.

350 JEE Main Physics Kinetic energy of body is given by 1 K = Iw2 2 \

K=

Þ

I=

63. Here, F = -Fk$ As

L2 1 (Iw) 2 = 2I 2I

r = ( $i - $j), t = ? t = r ´ F = ( $i - $J) ´ ( -F k$ ) $ = -F ( $i ´ k$ - $j ´ k) = - F ( - $j - $i)

L2 2K

= F( $i + $j)

60. Angular momentum in absence of any external torque

64.

R

remains constant. If no external torque acts on a system of particles, then angular momentum of the system remains constant, i. e. , t = 0 dL =0 \ dt …(i) Þ I1w1 = I2w2 Here, M = 2 kg, m = 0.25 kg, r = 0.2 m w1 = 30 rad s–1 Hence, we get after putting the given values in Eq. (i) 1 1 ´ 2 ´ (0.2) 2 ´ 30 = ´ (2 + 2 ´ 0.25) (0.2) 2 ´ w2 2 2 Þ

1.2 = 0.05 w2 w2 = 24 rad s–1

\

61. Since, rod is bent at the middle, so each part of it will have æLö æMö same length ç ÷ and mass ç ÷ as shown è2ø è2ø

v

From conservation of angular momentum v mr 2w0 = mvr + mr 2 ´ r w0 r Þ v= 2

65. The circular disc of radius R rolls without slipping. Its centre of mass is C. P is point where body is in contact with the surface at any instant. At this instant, each particle of body is moving at right angles to the line which joins the particle with point P with velocity proportional to distance. In other words, the combined translational and rotational motion gives pure rolling and body moves with constant velocity in magnitudes as well as direction.

R C

L/2

M/2

P

60° O

M/2 L/2

Moment of inertia of each part about an axis passing through its one end 2 1 æMö æ L ö = ç ÷ç ÷ 3 è 2 ø è2ø Hence, net moment of inertia about an axis passing through its middle point O is I=

2 2 1 é ML2 ML2 ù ML2 1 æMö æ L ö 1 æMö æ L ö + ç ÷ç ÷ + ç ÷ç ÷ = ê ú= 3 è 2 ø è2ø 3 è 2 ø è2ø 3ë 8 8 û 12

62. Radius of earth = R Mass of earth = M Angular momentum about the axis of rotation is J=

2 2 p 4 pMR 2 MR 2 ´ = T 5 5T

66. Here, m = 20 kg, l = 1m, r = 0.2 m Moment of inertia about its geometrical axis is 1 I = mr 2 2 1 = ´ 20 (0.2) 2 = 0.4 kg m2 2

67. Moment of inertia of a ring about its diameter I=

1 2 mr 2

and kinetic energy is given by 1 E k = Iw2 2 1 1 = ´ mr 2 ´ w2 2 2 1 = ´ 10 ´ 0.5 ´ 0.5 ´ 20 ´ 20 4 = 250 J

Rotational Motion 68. The moment of inertia of a disc about the axis is

l A

1 I = mR 2 2 1 M R2 I1 2 1 1 = I2 1 M R 2 2 2 2

Hence,

=

351

B O

P

F

D 2l

pR12 td ´ R12 [Q M = pR12 td and M1 = M2] pR22 td ´ R22 4

I1 R14 (0.2) 4 æ 0.1ö 1 = 4 = =ç ÷ = 4 è ø I2 R2 (0.6) 81 0.3

\

C

(Q R1 = 0.2 cm,R2 = 0.6 cm)

69. Mass per unit area of disc =

9M pR 2

ml ( l - x) = 2 mlx 2

Mass of removed portion of disc =

9M æRö ´p ç ÷ =M è3ø pR 2

Moment of inertia of removed portion about an axis passing through centre of disc and perpendicular to the plane of disc, using theorem of parallel axis is I1 =

M 2

2

DP = x As P is the centre of mas, therefore

Let

2

1 æRö æ2Rö 2 ç ÷ +M ç ÷ = MR è3ø è 3 ø 2

Þ

l : x = l /3

\

CP = CD + DP = l +

71. Since w is constant, v would also be constant, so, no net force or toruqe is acting on ring. The force experienced by any particle is only along radial direction or we can say it the centripetal force.

When portion of disc would not have been removed, then the moment of inertia of complete disc about the given axis is 1 I2 = MR 2 2 So moment of inertia of the disc with removed portion, about the given axis is 9 1 I = I2 - I1 = MR 2 - MR 2 = 4 MR 2 2 2

70. The object will have translation motion without rotation, when force F is applied at the centre of mass of system. If m is mass per unit length, the mass of AB, m1 = ml at O and mass of OC,m2 = m (2 l). Here, CD = l.

l 4l = 3 3

R2

F1

R1

ω

F2

v

The force experienced by inner part, F1 = mw2R1 and the force experienced by outer part, F2 = mw2R1 \

F1 mR1w2 R1 = = F2 mR2w2 R2

10 Gravitation JEE Main MILESTONE < < < <
1

(c) = 1

times that of modulus of rigidity. Its Poisson’s ratio is (b) 0.2

(c) 0.3

directions on the upper and lower faces of a cube of side 20 cm. The upper face is shifted parallel to itself by 0.25 cm. If the side of the cube were 10 cm, then the displacement would be

litre is subjected to a pressure of 4 ´ 107 Nm –2 . The decrease in its volume is (b) 10 cc

(c) 24 cc

(d) 15 cc

87. The Young’s modulus of the material of a wire is 6 ´ 1012 Nm–2 and there is no transverse strain in it, then its modulus of rigidity will be (a) 3 × 1012 Nm–2 (b) 2 ×1012 Nm–2 (c) 1012 Nm–2 (d) None of the above

Only One Correct Option Brass

modulus of steel and brass wires shown in the figure are a, b and c respectively. The ratio between the increase in length of brass and steel wires would be b2 a (a) 2c

bc (b) 2 2a

ba2 (c) 2c

c (d) 2 2b a

2 kg Steel 4 kg

(d) 16

If its Poisson’s ratio is 0.4, the diameter is reduced by (a) 0.01%

(b) 0.02%

(c) 0.03%

(d) 0.04%

radius r. What force is required to break a wire, of the same material, having twice the length and six times the radius? (a) F (c) 9 F

(b) 3 F (d) 36 F

stretched within the elastic region Force applied to wire 100 N Area of cross-section of wire 10–6 m2 Extension of wire 2 ´ 10–9m Which of the following deductions can be correctly made from this data? 1. The value of Young’s modulus is 1011 Nm–2 2. The strain is 10–3 3. The energy stored in the wire when the load is applied is 10 J (b) 1, 2 are correct (d) 3 only

(Mixed Bag)

Stress

A B C D O

cross-section A is suspended from the roof and mass m2 is suspended from the other end. What is the stress at the mid point of the rod? (a) ( m1 + m2 ) g / A (b) ( m1 - m2 ) g / A é ( m / 2) + m2 ù (c) ê 1 úû g A ë é m + ( m2 / 2) ù (d) ê 1 úû g A ë

4. One litre of a gas is maintained at pressure 72 cm of

2. The figure shows the stress -strain graph of a certain substance. Over which region of the graph is Hooke’s Law obeyed ?

(c) 4

3. One end of a uniform rod of mass m1, uniform area of

1. The ratio of lengths, radii and Young’s

(b) CD (d) OD

(b) 2

(a) 1, 2, 3 are correct (c) 1 only

Round II

(a) BC (c) AB

(a) 1

91. The following data were obtained when a wire was

(b) 0.5 cm (d) 1 cm

86. The compressibility of water is 6 ´ 10-10 N –1m 2 . If one (a) 2.4 cc

qy

90. A force F is required to break a wire of length l and

(d) 0.4

85. Forces of 100 N each are applied in opposite

(a) 0.25 cm (c) 0.75 cm

having equal length. The diameter of rod y is twice the diameter of rod x. If q x and q y are the angles of q twist, then x is equal to

89. The increase in length on stretching a wire is 0.05%.

(d) = 3.2

84. For a given material, the Young’s modulus is 2.4 (a) 0.1

88. Equal torsional torques act on two rods x and y

Strain

mercury. It is compressed isothermally so that its volume becomes 900 cm3. The value of stress and strain will be respectively (a) 0.106 Nm–2 and 0.1 (b) 1.106 Nm–2 and 0.1 (c) 106.62 Nm–2 and 0.1 (d) 10662.4 Nm–2 and 0.1

Properties of Solids

415

5. When a weight w is hung from one end of the wire

12. A load of 4.0 kg is suspended from a ceiling through a

other end being fixed, the elongation produced in it be l. If this wire goes over a pulley and two weights w each are hung at the two ends, the elongation of the wire will be

steel wire of length 2.0 m and radius 2.0 mm. It is found that the length of the wire increases by 0.031 mm as equilibrium is achieved. Taking, g = 3.1 p ms–2, the Young’s modulus of steel is

(a) 4 l

(b) 2 l

(c) l

(d) l / 2

6. When a force is applied on a wire of uniform crosssectional area 3 ´ 10–6 m2 and length 4 m, the increase in length is 1 mm. Energy stored in it will be (Y = 2 ´ 1011 Nm–2) (a) 6250 J (c) 0.075 J

(b) 0.177 J (d) 0.150 J

7. A wire of cross-sectional area A is stretched horizontally between two clamps loaded at a distance 2 l metres from each other. A weight w kg is suspended from the mid point of the wire. The strain produced in the wire, (if the vertical distance through which the mid point of the wire moves down x < l) will be (a) x2 / l2

(b) 2x2 l 2

(c) x2 / 2 l 2

(d) x/2 l

8. The graph shown was obtained

T2

M

(b) 1.8

(c) 1.5

(d) 1.19

14. A uniform cube is subjected to volume compression of each side is decreased by 1%, then the bulk strain is (a) 0.01 (c) 0.02

(b) 0.06 (d) 0.03

(c) 115.5 J

(d) 79.5 J

will be (Y = 2 ´ 1011 Nm–2)

(b) 5 J (d) 250 J 2L

x r is stretched between A and B without the application of any m tension. If Y is the Young modulus of the wire and it is stretched like ACB, then the tension in the wire will be

p2 r 2Y × 2 L2 (c) d2

(d) 17 ´ 103

(c) 6.1

by three wires each 2 m long. These at each end are of copper and middle one is of iron. Determine the ratio of their diameters if each is to have the same tension. Young’s modulus of elasticity for copper and steel are 110 ´ 109 N/m2 and 190 ´ 109 N/m2 respectively. [NCERT]

long wire of cross-sectional area 1mm2 through 1 mm

11. A wire of length 2 L and radius

(b) 1.7

(a) 1 : 1.3

10. The work done in increasing the length of one metre

pr 2Yd 2 (b) 2 L2

(a) 1.2

16. A rigid bar of mass 15 kg is supported symmetrically

the stretching force is increased by 200 kN?

pr 2Yd 3 (a) 2 L2

3.0 ´ 10-5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area 4.0 ´ 10-5 m 2 under a given load. The ratio of the Young’s modulus of steel to that of copper is [NCERT]

(a) 0.61

9. What is the increase in elastic potential energy when

(a) 0.1 J (c) 10 J

13. A steel wire of length 4.7 m and cross-sectional area

is subjected to an external pressure of 107 Nm–2. If the bulk modulus of silver is 17 ´ 1010 Nm–2, the change in density of silver (in kg m–3) is

(a) spring did not obey Hooke’s law (b) amplitude of oscillation was too large (c) clock used needed regulation (d) mass of the pan was not neglected

(b) 636.0 J

(b) 2.0 × 109 Nm–2 (d) 2.0 × 1013 Nm–2

15. A solid block of silver with density 10.5 ´ 103 kg m–3

from the experimental measurements of the period of oscillation T for different masses M placed in the scale pan on the lower end of the spring balance. The most likely reason for the line not passing through the origin is that

(a) 238.5 J

(a) 2.0 × 108 Nm–2 (c) 2.0 × 1011 Nm–2

pr 2Y × 2 L (d) d

(b) 1.3 : 1

(c) 2.3 : 1.3

(d) 2.3 : 1

17. A stress of 1 kg mm2 is applied on a wire. If the modulus of elasticity of the wire is 1010 dyne cm–2, then the percentage increase in the length of the wire will be (a) 0.0098% (b) 0.98%

(c) 9.8%

(d) 98%

18. A rectangular bar 2 cm in breadth and 1 cm in depth and 100 cm in length is supported at its ends and a load of 2 kg is applied at its middle. If Young’s modulus of the material of the bar is 20 ´ 1011 dyne cm–2, the depression in the bar is (a) 0.2450 cm (c) 0.1225 cm

(b) 0.3675 cm (d) 0.9800 cm

19. Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7 ´ 106 Pa. K for copper = 140 ´ 109 Pa. [NCERT]

(a) 5 ´ 10

-7

3

m

(c) 5 ´ 10 -8 m3

(b) 4 ´ 10

-8

3

m

(d) 6 ´ 10 -8 m3

416 JEE Main Physics 20. A wire (Y = 2 ´ 1011 Nm–2) has length 1 m and area of

cross-section 1 mm2. The work required to increase its length by 2 mm is (a) 400 J (c) 4 J

(b) 40 J (d) 0.4 J

21. A steel wire of length 20 cm and uniform

27. Figure shows a 80 cm square brass plate of thickness 0.5 cm. It is fixed at its bottom edge. What tangential force F must be exerted on the upper edge, so that the displacement (x) of this edge in the direction of force is 0.16 mm? The shear modulus of brass is 3.5 ´ 1010 Pa.

cross-section 1 mm2 is tied rigidly at both the ends. The temperature of the wire is altered from 40°C to 20°C. Coefficient of linear expansion of steel is a = 1.1 ´ 10–5 °C–1 and Y for steel is 2.0 ´ 1011 Nm2; the tension in the wire is (a) 2.2 ´ 106 N (c) 8 N

(b) 16 N (d) 44 N

22. If work done in stretching a wire by 1 mm is 2 J, the work necessary for stretching another wire of same material, but double the radius and half length by 1 mm joule is (a) 1/4 (c) 8

(b) 4 (d) 16

23. The Poisson’s ratio of a material is 0.1. If the longitudinal strain of a rod of this material is 10–3, then the percentage change in the volume of the rod will be (a) 0.008% (c) 0.8%

(b) 0.08% (d) 8%

24. The Poisson’s ratio of a material is 0.4. If a force is applied to a wire of this material, there is a decrease of cross-sectional area by 2%. The percentage increase in its length is (a) 3% (c) 1%

(b) 2.5% (d) 0.5%

25. Find the ratio of Young's modulus of wire A to wire B Stress B

(d) 4 ´ 10 -5 N

5

(c) 5 ´ 10 N

28. The twisting couple per unit twist for a solid cylinder of radius 3 cm is 0.1 N-m. The twisting couple per unit twist, for a hollow cylinder of same material with outer and inner radius 5 cm and 4 cm respectively, will be (a) 0.1 N-m (c) 0.91 N-m

(b) 0.455 N-m (d) 1.82 N-m

29. A solid sphere of radius r made of a 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. When a mass m is placed on the piston to compress the liquid, the fractional change in the radius of the sphere (Dr/r) is (a) Ka / mg (c) mg / 3Ka

(b) Ka / 3mg (d) mg / Ka

30. The bulk modulus of a metal is 8 ´ 109 Nm–2 and its density is 11 gcm–2. The density of this metal under a pressure of 20,000 N cm–2 will be (in gcm–3) 440 39

(b)

431 39

(c)

451 39

(d)

30°

Strain

(c) 1 : 3

(d) 1 : 4

is fixed at one end and on other end forces F is applied as shown in figure. Find the shear stress at a plane through the bar making an angle q with the vertical as shown in figure.

26. The stress-strain graph for a metallic wire is shown at two different temperatures, T1 and T2 which temperature is high T1 or T2 ?

FF

θ

Strain T1 T2 Rigid wall

Stress

(a) T1 > T2 (c) T1 = T2

40 39

31. A uniform rectangular bar of area of cross-section A

A

(b) 1 : 1

(b) 3.8 ´ 10 -4 N

(a) 2.8 ´ 10 4 N

(a)

30°

(a) 1 : 1

F

(b) T2 > T1 (d) None of these

F (a) (cos 2 q) 2A F (c) (sin 2 q) 2A

F 2A F (d) cos q 2A (b)

Properties of Solids 32. A uniform rod of length L and area of cross-section A is subjected to tensile load F. If s be Poisson’s ratio and Y be the Young’s modulus of the material of the rod, then find the volumetric strain produced in rod. F (1 + 2 s ) AY (c) Zero (a)

F (1 - 2 s ) AY (d) None of these (b)

33. A wire is suspended from the ceiling and stretched under the action of a weight F suspended from its other end. The force exerted by the ceiling on it is [NCERT Exemplar] equal and opposite to the weight. Tensile stress at any cross section A of the wire is F/A TensiIe stress at any cross section is zero Tensile stress at any cross section A of the wire is 2F/A Tension at any cross section A of the wire is F

34. The wires A and B shown in figure, are A

made of the same material and have radii rA and rB respectively. A block of mass m is connected between them. When a force F is mg/3, one of the wires breaks. (a) (b) (c) (d)

m

B

A will break before B if rA < rB A will break before B if rA = rB F Either A or B will break if rA = 2rB The length of A and B must be known to decide which wire will break

35. A metal wire of length L, area of cross-section A and Young’s modulus Y is stretched by a variable force F such that F is always slightly greater than the elastic forces of resistance in the wire. When the elongation of the wire is l YAl 2 L YAl 2 (b) the work done by F is 2L YAl 2 2L

(d) heat is produced during the elongation

36. The stress-strain graphs for two materials are shown

Stress

in figure. (assume same scale).

Stress

B Al

(YAl = 70 ´ 109 Nm -2 and Ysteel = 200 ´ 109 Nm -2 ) [NCERT Exemplar]

(a) Mass m should be suspended close to wire A to have equal stresses in both the wires (b) Mass m should be suspended close to B to have equal stresses in both the wires (c) Mass m should be suspended at the middle of the wires to have equal stresses in both the wires (d) Mass m should be suspended close to wire A to have equal strain in both wires

38. A metal wire of length L is suspended vertically from a rigid support. When a body of mass M is attached to the lower end of wire, the elongation of the wire is l. (a) The loss in gravitational potential energy of mass M is Mgl (b) The elastic potential energy stored in the wire is Mgl 1 (c) The elastic potential energy stored in the wire is Mgl 2 1 (d) Heat produced is Mgl 2

Passage I

(c) the elastic potential energy stored in the wire is

Strain E Material (i)

mass is suspended at its two A ends by two wires of steel (wire Steel A) and aluminium (wire B) of equal lengths (figure) The m cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2 , respectively.

Comprehension Based Questions

(a) the work done by F is

Ultimate Tension Strength Fracture Point Linear limit

(a) Material (ii) is more elastic than material (i) and hence material (ii) is more brittle (b) Material (i) and (ii) have the same elasticity and the same brittleness (c) Maerial (ii) is elastic over a larger region of strain as compared to (i). (d) Material (ii) is more brittle than material (i)

37. A rod of length l and negligible

More Than One Correct Option

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

417

[NCERT Exemplar] Ultimate Tension Strength Linear Fracture Point limit

Strain E Material (ii)

A boy’s catapult is made of rubber cord 42 cm long and 6 mm in diameter. The boy keeps a stone weighing 0.02 kg on it and stretches the cord by 20 cm. When released, the stone flies off with a velocity of 20 ms -1. Neglect the change in the cross-section of the cord in stretched position.

39. The stress in the rubber cord is (a) 1.8 × 106 Nm–2 (b) 1.4 × 106 Nm–2 (c) 2.4 × 105 Nm–2 (d) 1.8 × 105 Nm–2

40. The strain in the rubber cord is (a) 2.1 (c) 0.96

(b) 1.8 (d) 0.48

418 JEE Main Physics 41. The Young’s modulus of rubber is (a) 2.12 × 106 Nm–2 (c) 3.92 × 106 Nm–2

(b) 2.94 × 106 Nm–2 (d) 1.94 × 106 Nm–2

Passage II A structural steel rod has a radius of 10 mm and length of 1.0 m. A 100 kN force stretches it along its length. Young’s modulus of structural steel is 2 ´ 1011 Nm–2.

42. The elongation in the wire is (a) 1.59 mm (c) 2.38 mm

(b) 3.18 mm (d) 0.79 mm

43. The percentage strain is about (a) 0.16% (c) 0.24%

(b) 0.32% (d) 0.08%

46. Assertion Two identical springs of steel and copper are equally stretched. More work will be done on steel than copper. Reason Steel is more elastic than copper.

47. Assertion Young’s modulus for a perfectly plastic body is zero. Reason For a perfectly plastic body, restoring force is zero.

48. Assertion The bridges are declared unsafe after a long use. Reason The bridges lose their elastic strength with time.

49. Assertion A solid shaft is found to be stronger, than a

Assertion and Reason

hollow shaft of same material. Reason The torque required to produce a given twist in solid cylinder is smaller than that required to twist a hollow cylinder of the same size and material. YA , where Y is 50. Assertion Force constant, k = l Young’s modulus, A is area and l is original length of the given spring. Reason Force constant in case of a given spring is called spring constant.

Direction

51. Assertion The restoring force, F on a stretched string

44. Elastic energy density of stretched wire is (a) 1.26 × 105 Jm–3 (c) 3.79 × 105 Jm–3

(b) 2.53 × 105 Jm–3 (d) 5.06 × 105 Jm–3

45. Stress produced in the steel rod is (a) 1.59 × 108 Nm–2 (c) 4.77 × 108 Nm–2

(b) 3.18 × 108 Nm–2 (d) 6.36 × 108 Nm–2

Question No. 46 to 52 are Assertion-Reason type. Each of these contains two Statements : Statement I (Assertion), Statement II (Reason). Each of these questions also has four alternative choices, only one of which is correct. You have to select the correct choices from the codes (a), (b), (c) and (d) given below (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 correct explanation of the Assertion (c) If Assertion is true but Reason is false (d) If Assertion is false but the Reason is true

for extension x is related to potential energy, U as, dU F=dx 1 Reason F = - kx and U = kx 2 where, k is a spring 2

constant for the given stretched string.

52. Assertion Identical springs of steel and copper are equally stretched. More work will be done on the steel spring. Reason Steel is more elastic than copper.

Previous Years’ Questions 53. A wooden wheel of radius R is made of two semicircular parts (see figure). The two parts are held R together by a ring made on a metal strip of cross- sectional area S and length L. L is slightly less than 2 pR. To fit the ring on the wheel of is heated so that it temperature rises by DT and it just steps over the wheel. As it cools down to surrounding temperature it presses the semicircular parts together if the coefficient of linear expansion of the metal is a and its young’s modulus is Y . The

force that one part of the wheel applies on the other [AIEEE 2012] part is (a) 2 pSy µ DT (c) pSY µ DT

(b) SY µ DT (d) 2 SY µ DT

54. The Poisson’s ratio of the material is 0.5. If a force is applied to a wire of this material, there is a decrease in the cross-sectional area by 4%. The percentage [WB JEE 2009] increase in its length is (a) 1% (c) 2.5%

(b) 2% (d) 4%

Properties of Solids

419

55. Two wires are made of the same material and have

62. If the volume of a block of aluminium is decreased by

the same volume. However, wire 1 has crosssectional area 3A. If length of wire 1 increased by Dx on applying force F, how much force is needed to stretch wire 2 by the same amount ? [AIEEE 2009]

1%, the pressure (stress) on its surface is increased by (Bulk modulus of aluminium = 7.5 ´ 1010 Nm–2)

(a) 4 F

(b) 6 F

(c) 9 F

(d) F

56. When a rod is heated but prevented from expanding, the stress developed is independent of [BVP Engg. 2008] (a) material of the rod (c) length of rod

(b) rise in temperature (d) None of these

57. The Bulk Modulus for an incompressible liquid is [UP SEE 2008]

(a) zero (c) infinity

(b) unity (d) between 0 and 1

58. A metal wire of length L1 and area of cross-section A

is attached to a rigid support. Another metal wire of length L2 and of the same area of the first wire. A body of mass M is then suspended from the free end of the second wire. If Y1 and Y2 are the Young’s modulii of the wires respectively, the effective force constant of the system of two wires is [NSEP 2008] (a)

Y1Y2 A (Y1L2 + Y2 L1 )

Y1Y2 A (c) (Y1L2 + Y2 L1 )

(b)

Y1Y2 A ( L1L2 )1 /2

tensile force is applied on a steel rod of area of cross-section 10–3 m2. The change in temperature required to produce the same elongation if the steel rod is heated is (the modulus of elasticity is 3 ´ 1011 Nm–2 and coefficient of linear expansion of steel is 1.1 ´ 1011° C-1). [EAMCET 2008] (b) 15°C

(c) 10°C

(d) 0°C

60. A load of 1kg weight is attached to one end of a steel

wire of area of cross-section 3 mm2 and Young’s modulus 1011 Nm–2. The other end is suspended vertically from a hook on a wall, then the load is pulled horizontally and released. When the load passes through its lowest position, the fractional [EAMCET 2008] change in length is (g = 10 ms–2) (a) 10–4

(b) 10–3

(c) 103

(d) 104

61. A wire is stretched under a force. If the wire suddenly snaps, the temperature of the wire, (a) remains the same (b) decreases (c) increases (d) first decreases then increases

63. A student performs an experiment to determine the Young’s modulus of a wire, exactly 2 m long, by Searle’s method. In a particular reading, the students measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of ± 0.01 mm. The Young’s modulus obtained from the reading is (Take g = 9.8 ms–2). [IIT JEE 2007]

(a) (2.0 ± 0.3) × 1011 Nm–2 (b) (2.0 ± 0.2) × 1011 Nm–2 (c) (20 ± 0.1) × 1011 Nm–2 (d) (2.0 ± 0.05) × 1011 Nm–2

64. A wire 3 m in length and 1 mm in diameter at 30°C is

(Y Y )1 /2 A (d) 1 2 1 /2 ( L1L2 )

59. There is some change in length when a 33000 N

(a) 20°C

[Kerala CEE 2008]

(a) 7.5 × 1010 Nm–2 (b) 7.5 × 108 Nm–2 (c) 7.5 × 106 Nm–2 (d) 7.5 × 104 Nm–2

[WB JEE 2008]

kept in a low temperature at –170°C and is stretched by hanging a weight of 10 kg at one end. The change in length of the wire is (Y = 2 ´ 1011 Nm–2, g = 10 ms–2 and a = 1.2 ´ 10–5°C–1) [UP SEE 2006] (a) 5.2 mm (c) 52 mm

(b) 2.5 mm (d) 25 mm

65. Two rods of different materials having coefficients of linear expansion a 1 and a 2 and Young’s modulus, Y1 and Y2 respectively are fixed between two rigid massive walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of rods. If a 1: a 2 = 2 : 3, the thermal stress developed in the two rods are equal provided Y1: Y2 equal to [BVP Engg. 2006] (a) 2 : 3 (c) 1 : 2

(b) 4 : 9 (d) 3 : 2

66. The pressure of a medium is changed from 1.010 ´ 105 to 1.165 ´ 105 Pa and change in volume is 10% keeping temperature constant. The Bulk [IIT Screening 2005] modulus of the medium is (a) 204.8 × 105 Pa (b) 102.4 × 105 Pa (c) 51.2 × 105 Pa (d) 1.55 × 105 Pa

Answers Round I 1. 11. 21. 31. 41. 51. 61. 71. 81. 91.

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

2. 12. 22. 32. 42. 52. 62. 72. 82.

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

3. 13. 23. 33. 43. 53. 63. 73. 83.

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

4. 14. 24. 34. 44. 54. 64. 74. 84.

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

5. 15. 25. 35. 45. 55. 65. 75. 85.

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

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

7. 17. 27. 37. 47. 57. 67. 77. 87.

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

8. 18. 28. 38. 48. 58. 68. 78. 88.

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

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

7. 17. 27. 37. 47. 57.

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

8. 18. 28. 38. 48. 58.

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

6. 16. 26. 36. 46. 56. 66. 76. 86.

9. 19. 29. 39. 49. 59. 69. 79. 89.

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

10. 20. 30. 40. 50. 60. 70. 80. 90.

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

10. 20. 30. 40. 50. 60.

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

Round II 1. 11. 21. 31. 41. 51. 61.

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

2. 12. 22. 32. 42. 52. 62.

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

3. 13. 23. 33. 43. 53. 63.

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

4. 14. 24. 34. 44. 54. 64.

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

5. 15. 25. 35. 45. 55. 65.

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

6. 16. 26. 36. 46. 56. 66.

9. 19. 29. 39. 49. 59.

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

the Guidance Round I 1. As, Y = or

Fl aDL DL µ

1 1 ; DL µ 2 a D

DL2 D12 = =4 DL1 D22

then,

or

DL2 = 4DL1 = 4 cm

and

or or or

3. As, Þ

F Fl Y= A = Dl A Dl l Fl ´ 4 Y= pD 2 ´ Dl Dl µ

1 D2

DL2 D12 n 2 = = DL1 D22 1 mgL (Y = Young’s modulus) pr 2l 4mgL 4 mgL = Y= p (2r) 2l p (d) 2l1 Y=

Yµ

Þ

Þ

2. As,

1 d2 1 dµ Y

Þ

So,

d copper µ d iron µ d copper d iron

=

1 Ycopper 1 Yiron Yiron Ycopper

4. If ( A) is the area of cross-section and l is the length of rope, the Al × As the weight of the rope acts at the r mid-point of the rope. mg ( l / 2) So, Y= ´ A Dl mgl Al rgl g rl 2 DL = = = 2 AY 2 AY 2Y

mass of rope, m =

or

DL =

9.8 ´ 1.5 ´ 10 3 ´ 8 2 2 ´ 5 ´ 10 6

= 9.6 ´ 10 -2 m

Properties of Solids 5. According to the figure,

L B

increases in length = BO + OC - BC

L D

DL = 2 BO - 2L = 2 (L2 + x 2 )1/ 2 - 2 L

\

Strain =

1 ´ Young’s modulus ´ (strain) 2 2 1 = ´ Y ´ x2 2 =

O m

13. As, Y µ F

D L x2 / L x2 = = 2 2L 2L 2L

14. F = force developed 2 2

= YA a ( Dq)

or

Y=

ml w A Dl

or

Y=

1 ´ 1 ´ 1 ´ 20 ´ 20 10 -6 ´ 10 -3

= 10 11 ´ 10 -4 ´ 10 -5 ´ 100 = 10 4 N

15. From the question10 6 =

= 4 ´ 10 11 Nm-2

7. As, E =

8.

FL pr 2E

DL =

Clearly,

DL µ L

Maximum stress = 10 8 N/m 2

\ The ratio of

Area of cross-section of steel cable ( A) = pr 2 = 3.14 ´ (1.5 ´ 10 -2) 2 m 2 = 3.14 ´ 2.25 ´ 10 -4 m 2 Maximum force Maximum stress = Area of cross-section

17. As, Y =

Maximum force = Maximum stress ´ Area of cross-section

YA Dl l

or

F=

2.2 ´ 10 11 ´ 2 ´ 10 -6 ´ 0.5 ´ 10 -3 2

= 1.1 ´ 10 2 N

= 7.065 ´ 10 N

18. As, Y =

= 7.1 ´ 10 4 N

10. Let m be the mass of rectangular frame and q be the angle which the tension T in the string make with the horizontal from figure. Then, T sin θ T sin θ T

T θ T cos θ

θ T cos θ m

T is least if sin q has maximum value i.e., sin q = 1 = 90° or q = 90°

11. Strain = fractional change in length Dl l at = = at = 12 ´ 10 -6 ´ 30 = 36 ´ 10 -5 l l

Fl ADl F=

4

2T sin q = mg mg T= 2 sin q 1 Tµ sin q

L is maximum for case (d). d2

or

= 10 8 ´ (3.14 ´ 2.25 ´ 10 -4) N

=

F L A Dl L L Dl µ µ A pd 2 L Dl µ 2 d Y=

Þ

9. Given, radius of steel cable (r) = 1.5 cm = 1.5 ´ 10 -2 cm

or

10 6 m 3 ´ 10 3 ´ 9.8 1000 = = 34.01m 3 ´ 9.8

16. As,

L2dg (8 ´ 10 -2) 2 ´ 1.5 ´ 9.8 As, l = = 9.6 ´ 10 -11 m = 2Y 2 ´ 5 ´ 10 8

LAdg A

L=

\

FL pr 2DL

or

or

FCu YCu 1 = = FFe YEe 3

Þ

Fl (ml w2) l = ADl A Dl

6. As, Y =

1 2

12. Energy stored per unit volume = ´ stress ´ strain

x

é x2 x2 ù DL = 2 L ê1 + ú - 2L = L L2 û ë

or

C

421

Fl 1 Þ Dl µ A Dl A m = Alr, m µ A 1 Dl µ m Dl1 m2 2 = = Dl2 m1 3

Again, \ \

19. As, Y =

F ´ 4 ´1 pD 2Dl

In the given problem, F µ D 2. Since, D is increased by a factor of 4 therefore, F is increased by a factor of 16. So, F will be 100 ´ 16 = 1600 N Fl 20. As, Y = ADl In the given problem, Y , l and Dl are constants. \

F µA

422 JEE Main Physics or

F = pr 2

or

F µ r2

or

F1 r12 1 = = F2 r22 4

(Q area = pr 2)

Fl F 21. As, Y = 2 or Dl = 2 pr Dl pr Y 1 2l Þ Dl µ 2 and Dl ¢ µ ( 2r ) 2 r 1 \ Dl¢ µ 2 r Dl Again, =1 Dl¢ FL 22. As, Y = 2 (Here, l = change in length) pr l FL FL or l = 2 or l µ 2 pr Y r l1 F ´ L ( 4r) 2 = 2 ´ l2 4F ´ 4L r or

23.

24.

25.

28. As, Dl = =

29.

» 0.08 mm Fl As, Y = ADl

or

= 0.71 ´ 10 -3 = 7.1 ´ 10 -4

26. As, L =

r 10 6 100 = = = 33.3 m eg 3 ´ 10 3 ´ 10 3

Dl µ

Þ

30.

F r2

Þ

Dl2 F2 r12 = ´ Dl1 F1 r22

or

Dl2 =2 ´2 ´2 =8 Dl1

or

Dl2 = 8Dl1 = 8 ´ 1 mm = 8 mm

Fl As, Y = ADl or

So, l remain unchanged. Mg ´ 4 ´ l 1 As, Y = Þ Dl µ 2 pD 2 ´ Dl D

F 10 4 ´ 100 A= = Y ´ Breaking strain 7 ´ 10 ´ 0.2

4 ´ 30 ´ 2 ´ 7 22 ´ (3 ´ 10 -3) 2 ´ 1.1 ´ 10 11

= 7.7 ´ 10 -5 m = 0.077 mm

l1 = l2 = l

When D is doubled, Dl becomes one-fourth, i. e. , 1 ´ 2.4 cm, i. e. , 0.6 cm 4 Fl Fl Fl As, Y = or Dl = = 2 ADl AY pr Y 1 In the given problem, Dl = 2 ; when both l and r are double, r Dl is halved. F/A As, Y = Breaking strain

4Fl pD 2Y

YADl l 1 Work done = FDl 2

31. As, Y =

F=

=

1 FA( Dl) 2 YA( Dl) 2 = 2 2l l

=

2 ´ 10 11 ´ 10 -6 ´ 10 -6 = 0.1 J 2 ´1

Fl ADl

where, Y , l and F are constants. 1 Þ Dl µ 2 D Þ

Dl2 D12 1 = = Dl1 D22 16

\

Dl2 =

32. We have, and

1 mm 16 F l Y= ´ A Dl V = Al l=

\

Y=

FV A2Dl

Þ

Dl µ

1 A2

or

Dl µ

1 D4

27. Initial length (circumference) of the ring = 2pr Final length (circumference) of the ring = 2pR Change in length = 2pR - 2pr change in length 2p (R - r) R - r Strain = = = original length 2 pr r F/A F/A Now Young’s modulus, E = = l / L (R - r) / r \

æR - r ö F = AE ç ÷ è r ø

V A

or

Þ

14 DlA DB4 = 16 = 4 = DlB DA æ 1 ö 4 ç ÷ è2ø

(Q Dl1 = 1 mm)

Properties of Solids 33. As, Y =

(mg + mlw2) l pr 2Dl

or

Dl =

or

Dl =

37. As, Y =

=

34. As, Y =

1(10 + 2 ´ 4p 2 ´ 4) 2 p (1 ´ 10 -3) 2 ´ 2 ´ 10 11 (20 + 64 ´ 9.88)7 2 ´ 22 ´ 10 5 4566.24 ´ 10 3 mm = 1 mm 44 ´ 10 5

\

38. As rubber is being more stretched as compare to the iron

2 ´ 10 11 20 Ls Y = = s = LCu YCu 1.1 ´ 10 11 11 l F ´ pR 2 Dl

F, l and Dl are constants.

39.

under the action of same weight. FLS As, YS = ASDLS FL C and YC = A C DL C \

Here,

Þ

1 R µ Y

Þ

RS2 RB2

\

RS 1 = RB 2

or

RS =

2

=

11

YB 10 1 = = YS 2 ´ 10 11 2

40. Net elongation of the rod is 3F

or

l1T2 - lT2 = l2T1 - lT1

or

l(T1 - T2) = l2T1 - l1T2 l T - lT l= 21 12 T1 - T2

or or

l=

l1T2 - l2T1 T2 - T1

3F

2F

(2L /3)

2F (L /3)

æLö æ 2L ö 3F ç ÷ 2F ç ÷ è3ø è3ø + l= AY AY 8FL l= 3AY

Fl We have, Y = ADl Y , l and A are constants. F = constant Þ Dl µ F \ Dl

YC AC DLC æ Y ö æ AC ö æ DLC ö LC F = =ç C÷ ç ÷ç ÷ D Y A L LS è YS ø è AS ø è DLS ø S S S F AC Y DLC 1.1 = 2, = 1, C = AS YS 2 DLS LC 1.1 = ´ 2 ´ 1 = 1.1 LS 2

\

RB 2

Now, l1 - l µ T1 and l2 - l µ T2 l1 - l T1 Dividing, = l2 - l T2

rgL2 1.5 ´ 10 3 ´ 10 ´ 8 ´ 8 = 2Y 2 ´ 5 ´ 10 6 = 96 mm

L µY

Þ

36.

DL =

= 9.6 ´ 10 -2 m = 9.6 ´ 10 -2 ´ 10 3 mm

stress strain = Y DL stress = L Y

Since, cross-section are equal and same tension exists in both wires, therefore, the stresses developed are equal. Also, DL is given to be the same for both the wires.

35. As, Y =

(For the purpose of calculation of mass, the whole of geometrical length L is to be considered.) AlrgL Y= \ 2 A DL

stress strain

or

M = ALr

Now,

or

or

Mg L / 2 ´ A DL L æ ö çLength is taken as because weight acts at ÷ 2 ç ÷ centre of gravity (CG) ø è

m ( g + mlw2) l pr 2Y

Dl =

or

423

41. As, k1 =

Yp (2R) 2 Yp (R) 2 and k2 = L L

Since, k1x1 = k2x2 = w Elastic potential energy of the system 1 1 U = k1x12 + k2x22 2 2 2 2 1 æwö 1 æwö = k1ç ÷ + k2ç ÷ 2 è k1 ø 2 è k2 ø =

1 2ì 1 1ü wí + ý 2 î k1 k2þ

424 JEE Main Physics L L 1 1 + = + k1 k2 4YpR 2 YpR 2

Now,

U=

\

Hence, final length = l + l ¢ = 5a - 4b + 9b - 9a

1 2æ 5L ö 5w 2L = w ç ÷ 2 è 4YpR 2 ø 8pYR 2

l0 = 5 b - 4 a

42. Assume original length of spring = l

E H2 = 1.4p¢ As elasticity of hydrogen and argon are equal

For hydrogen

Þ k1(60) = k2( l - 60) = kl \ mg = k1 = (7.5) According to question, \

Þ \ and

47. As, K =

k1 5.0 ( l - 60) = = k2 7.5 60

or

2 ( l - 60) = 3 60

or

48.

49. As,

mg = Kx K=

k for second case. 2

4 k 6 12 x¢ = = k/2 k 1=

For second case,

…(i)

50. K =

T2 = K( l - l2) T1 l - l1 = T2 ( l - l2)

1 ´ 19.6 ´ 10 2 ´ (5 ´ 10 -2) 2 J = 2.45 J 2

1.5 N = 50 Nm-1 (as mg = kx) 30 ´ 10 -3 m

0.2 ´ 10 m = 0.04 m 50 1 Now, energy stored = ´ 0.20 ´ 10 ´ 0.04 J = 0.04 J 2 0.10 Change in volume, DV = V ´ 100 DV 0.10 …(i) or = = 1 ´ 10 -3 V 100

\

51.

44. As, T1 = K( l - l1)

4 ´ 9.8 or K = 19.6 ´ 10 2 Nm-1 2 ´ 10 -2

\Work done =

…(ii)

Dividing Eq. (ii) by Eq. (i), we get 12 / k x¢ = = 3 cm 4 /k

\

DV = spV pV pV p p As, K = = = ÞT = DV gDT 3aT 3Ka

F k

For first case,

So,

p (Here, K = bulk modulus of elasticity) DV V 1 DV / V = K p DV s= pV

or

x = 12.5 cm

If spring constant is k for the first case, it is

and

1.6p = 1.4p¢ 8 p¢ = p 7

Þ

l = 100 cm kx = k1 ´ 7.5, æ 5k ö kx = ç ÷ ´ 7.5 è3ø

\

43. As, x =

\

mg = k2 = (5.0) kl kl , k2 = k1 = 60 ( l - 60)

and

E Ar = 1.6p

For argon

mg = kx

As,

E = gp

45. Adiabatic elasticity

l=

Bulk modulus of water (K) = 2.2 ´ 10 9 N/m 2

T1l - T1l2 = T2l - T2l1

Pressure on water ( Dp) = ?

(T1 - T2) l = T1l2 - T2l1

Bulk modulus of water

(K) =

or

Dp = K ´

l=

T1l2 - T2l1 (T1 - T2)

l = (5a - 4b) 1 k= b-a

…(i) …(ii)

So, length of wire when tension is 9 N, is given by 9 = kl ¢

(l¢ = change in length) 1 9= ´ dl ¢ = 9 b - 9 a ( b - a)

Dp DV / V DV V

= 2.2 ´ 10 9 ´ 1 ´ 10 -3 = 2.2 ´ 10 6 N-m 2

52. Pressure ( p) = 10 atm = 10 ´ 1.013 ´ 105 Pa (Q 1 atm = 1.013 ´ 10 5 Pa) = 1.013 ´ 10 6 Pa

Properties of Solids Bulk modulus for glass (K) = 37 ´ 10 9 N/m 2

59. As volume, V = pr 2l

æ DV ö Fractional change in volume ç ÷ =? è V ø Bulk modulus

(K) =

p pV = DV / V DV

DV r 2Dl + 2rlDr = V r 2l DV Dl 2 Dr = + V r l

or or

= 2.74 ´ 10 -5 æ DV ö -5 \Fractional change in volume ç ÷ = 2.74 ´10 è V ø

Now, Poisson’s ratio Dr / r Dr / r s === 0.5 Dl Dr -2 r l

60. Let, L be the length of each side of cube, then initial volume = L3 . When each side decrease by 1%.Then

53. As, mg = Kx k=

DV D( pr 2l) = V pr 2l

Þ

DV p 1.013 ´ 10 6 = = V K 37 ´ 10 9 101.3 = ´ 10 -5 37

\

10 N 100 = Nm-1 = 250 Nm-1 -3 4 40 ´ 10 m

Spring constant of combination 250 = Nm-1 = 125 Nm-1 2 1 Energy = ´ 125 ´ ( 40 ´ 10 -3) 2 J = 0.1 J 2

New length, L ¢ = L -

So, 100 m column of water exerts nearly 10 atmospheric pressure, i. e. ,10 ´ 10 5 Pa or 10 6 Pa. 1 DV / V DV é 1ù or = Dp ê ú = V K Dp ëKû

DV ´ 100 = 10 5 ´ 8 ´ 10 -12 ´ 100 = 8 ´ 10 -5 V 1 As, energy stored per unit volume = ´ stress ´ strain 2

or

56.

1 strain stress S 2 = ´ stress ´ Y= = 2 Y strain 2Y

57. As,

Dp = hrg = 200 ´ 10 3 ´ 10 Nm-2 6

3

\ Change in volume, æ 99L ö DV = L3 - ç ÷ è100 ø

\Bulk strain =

Young’s modulus Y G= or

Y = 3G

\

n =3

62. Energy stored per unit volume

coefficient of volume elasticity an elastic coefficient.

64. Work done in stretching the wire 1 ´ force constant ´ x2 2 1 1 For first wire, W1 = ´ kx2 = kx2 2 2 1 2 For second wire, W2 = ´ 2k ´ x = kx2 2 W=

= 10 8 dyne cm -2

\

1 1 Y ( strain) 2 = ´ 1.5 ´ 10 12 ´ (2 ´ 10 -4) 2 2 2

63. Every material has its certain volume. So, it may have

58. As, Dp = 100 atm = 100 ´ 10 6 dyne cm -2 DV 0.01 = = 10 -4 V 100 10 8 K = -4 dyne cm -2 = 10 12 dyne cm-2 10

1 Y 3

= 3 ´ 10 4 Jm-3

Dp 2 ´ 10 6 2 ´ 10 8 K= = = Nm-2 0.1 DV 0.1 V 100 = 2 ´ 10 9 Nm-2

\

DV 3L3 / 100 = = 0.03 V L3

61. For most materials the modulus of rigidity, G is one third of the

= 2 ´ 10 Nm \

3

3 é æ 3 é 3 ù 3L öù = = L2 ê1 - ç1 + K÷ ú = L3 ê ú øû ë100 û 100 ë è 100

=

-2

1 99L = 100 100

æ 99L ö New volume = L ¢3 = ç ÷ , è100 ø

54. 10 m column of water exerts nearly 1 atmosphere pressure.

55. As,

425

æ ö ç Dp ÷ çQ K = ÷ DV ÷ ç è V ø

Hence,

W2 = 2W1

65. Because a liquid at rest begins to move under the effect to tangential force.

426 JEE Main Physics 66. Here, \

Stress = =

and \

1 2

T = m( g + a0) = 10(10 + 2) = 120 N

72. As, energy density = stress ´ strain

T A

=

120 = 60 ´ 10 4 Nm-2 2 ´ 10 -4

stress strain stress strain = Y Y=

=

m (g + a 0 )

73.

4

60 ´ 10 = 30 ´ 10 -7 = 3 ´ 10 -6 2 ´ 10 11

67. Elastic energy stored in the wire is, 1 ´ stress ´ strain ´ volume 2 1 F Dl 1 = ´ ´ ´ Al = FDl 2 A l 2 1 U = ´ 200 ´ 1 ´ 10 -3 = 0.1 J 2 U=

or

uA DB4 = = (2) 4 = 16 uB DA4

Now,

m1

1 1 stress ( stress) 2 stress ´ = µ 4 2Y Y 2 D

(where uA and uB are energy densities) 1 As, energy/volume = ´ stress ´ strain 2 1 1 = Y ´ strain ´ strain = Y ´ strain 2 2 2 1 = ´ 2 ´ 10 10 ´ 0.06 ´ 10 -2 ´ 0.06 ´ 10 -2 2 = 3600 Jm-3

74. Here, force = weight suspended + weight of As,

w1 =

3w 4

68. Breaking strength = tension in the wire = mrw2 (centrefugal force)

w2 =

force stress = = area

\

Þ 4.8 ´ 10 7 ´ 10 -6 = 10 ´ 0.3 ´ w2 48 = 16 0.3 ´ 10

75.

69. To twist the wire through the angle dq, it is necessary to do the dW = tdq 10 p p ´ = rad 60 180 1080 q

q

0

0

hpr 4 q dq hpr 4 q = 2l 4l

\

W = ò dW = ò t dq = ò

or

5.9 ´ 10 11 ´ 10 -5 ´ p (2 ´ 10 -5) 4 p 2 W= 10 -4 ´ 4 ´ 5 ´ 10 -2 ´ (1080) 2

or

W = 1.253 ´ 10 -12 J

70. For cylinder A, For cylinder B,

t=

4

phr q¢ 2l

ph (2r) 4( q - q¢ ) t= 2l

phr 4 q¢ nh (2r) 4( q - q¢ ) = 2l 2l 16 Þ q¢ = q 17 1 71. Energy density = ´ stress ´ strain 2 stress or stress = Ys Y= s 1 1 \Energy density = Ys ´ s = Ys 2 2 2

3 w 4

S

phr 4 As, torque t = q 2l 1 r4

Þ

qµ

Þ

qA r24 = qB r14

work q = 10 ¢ =

w1 +

In the given problem, r 4 q = constant

w = 4 rads-1

and

3L of wire 4

1 2 1 YA YA = ´ ´ 1= 2 2L L

76. As, work done = F ´ extension

1 2

77. As, W = FDl

(where, Dl = entension)

Þ

W=

or

W=

where, (for equal torque) and Þ Þ or

(Q l = 1)

1 Ypr 2Dl ´ Dl 2 l

Ypr 2Dl 2l Fl Y= 2 pr Dl

F=

Ypr 2Dl l

r2 (2r 2) 22 and W µ l l W¢ =8 W

Wµ

W ¢ = 8 ´ 2 J = 16 J

Properties of Solids 1 2

78. As, work done = FDl = =

1 YADl 2 2 l

2 ´ 10 11 ´ 10 -6 (2 ´ 10 -3F) 2 = 0.4 J 2 ´1

79. As, work done, W =

W2 =

85. F 2l

æ pD 2 ö ÷Y 2ç è 4 ø

86.

1 YA / Dl 2 2 l

\

E in calorie =

or

YDl 2 q= 2 2l d JS

or

q= =

81. As,

YmDl 2 2l 2dJ

YmDl 2 mSq = 2l 2dJ

Now,

12 ´ 10 11 ´ 10 -1 ´ 10 -3 ´ 10 -3 2 ´ 2 ´ 2 ´ 9 ´ 10 3 ´ 4.2 ´ 0.1 ´ 10 3 12 ´ 10 5 1 = °C 72 ´ 42 ´ 10 5 252

YA tan qA tan 60° 3 = = = = 3 Þ YA = 3 YB YB tan qB tan 30 1 / 3

82. As, pq = lf; so, f =

0.4 ´ 30° = 0.12° 10

83. Isothermal elasticity q = p, Adiabatic elasticity ( f) = gp \ where \

…(i)

Eq 1 = Ef g g >1 Eq g, then the liquid occupies upper part of the vessel. and vertical acceleration. Consider a container containing liquid is moving up with constant acceleration on an inclined plane as shown in the figure.

θ

p0

a p1

p2 h

p3

h1

2

p4

p5

p6

Y X

φ

p1 ¹ p2

In this case, pressure difference at two points in same horizontal level separated by a distance s is given by p1 - p2 = rs ´ a cos f

p3 = p 4 and p5 = p6 p3 = p4

Further,

p0 + r1hg1 = p0 + r2gh2

Invertical plane, p2 - p1 = r (g + a sin f)h a cos f , tan q = g + a sin f

r1h1 = r2h2 1 hµ r

or or

4. Barometer is used to measure atmospheric pressure while, manometer measures pressure difference, i .e ., gauge pressure.

where q is the angle made by free surface of liquid with horizontal.

7. Variation of pressure within an accelerating closed container.

Vacuum (p = 0) h

p0

p

p0

h

h

1

s

Hg (a) Barometer p0 = hρg

2

(b) Manometer p – p0 = hρg

5. Pressure at two points within a liquid at vertical separation of h when the liquid container is accelerating up are related by expression p2 - p1 = r(g + a )h

a

Here, all the points lying on a particular line making an angle of tan-1(a/g ) with the horizontal have the same pressure. In present situation, point 2 is the least pressure point, if the vessel is completely closed, we can take its pressure to be zero. p1 = p2 + rgh + rsa as p2 = 0 So,

p1 = rgh + rsa

438 JEE Main Physics Sample Problem 2 The two thigh bones (femurs), each of 2

cross-sectional area 10 cm support the upper part of a human body of mass 40 kg. Average pressure sustained by the femurs is (b) 2 ´ 10 5 Nm-2

(a) 10 3 N/m2 (c) 6 ´ 10

-5

(i) Pascal’s law is used in the working of the hydraulic lift which is used to support or lift heavy objects. In hydraulic lift,

N /m

2

(d) 10

-3

Nm

F2 =

-2

where A1, A2 = area of cross-section of smaller and larger piston of hydraulic lift. F1 = force applied on smaller piston.

Interpret (b) Total corss-sectional area of the femurs is

A = 2 ´ 10 cm2 = 2 ´ 10 -4 m2. The force acting on them is F = 40 kgwt = 400 N (taking g = 10 ms–2). This force is acting vertically down and hence, normally on the femurs. Thus, the average pressure is F 400 pav = = = 2 ´ 10 5 Nm–2 A 2 ´ 10 -4

Sample Problem 3 The pressure on a swimmer 10 m below the surface of a lake is (a) 10 atm

(b) 5 atm

(c) 15 atm

(d) 2 atm

Interpret (a) Total pressure, p = pa + rgh, where pa is atmospheric pressure, r is density and g is acceleration due to gravity. p = 1.01 ´ 10 5 pa + 1000 kgm–3 ´ 10 ´ 10

\

= 2.01 ´ 105 pa » 2 atm

Sample Problem 4 The average depth of Indian Ocean is about 3000 m. Bulk modulus of water is 2.2 ´ 10 4 Nm –2, æ DV ö g = 10 ms–2, then fractional compression ç ÷ of water at the è V ø bottom of the Indian Ocean is (a) 1.36%

(b) 20.6%

(c) 13.9%

(d) 0.52%

Interpret (a) The pressure exerted by a 3000 m column of water on the bottom layer –1 –2

= 3 ´ 10 kg m s

7

–2

= 3 ´ 10 Nm

æ DV ö Fractional compression ç ÷ è V ø =

(ii) Hydraulic lift is a force multiplying device which is used in dentist’s chair, car lifts and jacks, many elevators and hydraulic brakes.

Atmospheric Pressure The pressure exerted by atmosphere is called atmospheric pressure. At STP, the value of atmospheric pressure is 1.01 ´ 105 Nm–2 or 1.01 ´ 106 dyne cm–2. (i) The sudden fall in atmospheric pressure produces the possibility of a storm. (ii) Various units of atmospherical pressure are mm or cm of Hg column, torr ( = 1 mm of Hg column). Pressure is also measured in units of atmospheric pressure. The pressure at a depth of 10 m in water is about 2 atmosphere. The unit of pressure used for metrerological purposes is called the bar; one bar is about 105 Nm-2 .

Sample Problem 5 If a room has dimensions 3 m ´ 4 m ´ 5 m. What is the mass of air in the room, if density of air at NTP is 1.3 kgm -3? (a) 78 kg (c) 76 kg

(b) 75 kg (d) 78.5 kg

Density of water is 103 kgm -3 and g = 10 ms-2.

p = hrg = 3000 ´ 1000 ´ 10 7

A2 F1 A1

Stress (3 ´ 10 7 Nm–2) = 1.36 ´ 10 –2 = B (2.2 ´ 10 9)

DV = 1.36% V

12.3 Pascals’ Law and its Applications It states that if gravity effect is neglected, the pressure at every point of liquid in equilibrium or rest is same. Pascal’s law also states that the increase in pressure at one point of the enclosed liquid in equilibrium of rest is transmitted equally to all other points or liquid provided the gravity effect is neglected.

Interpret (a) As we know that density, r = m V

m = rV = 1.3 ´ 3 ´ 4 ´ 5 = 78 kg

or

i. e. , mass of air in the room is 78 kg which is not possible. Since, air is considered as to be weightless. But it is true.

Sample Problem 6 In the above example, what force does water exert on the base of a house tank of base area 1.5 m 2, when it is filled with water upto a height of 1 m? (a) 1.5 ´ 10 4 N (b) 2.5 ´ 10 4 N (c) 3 ´ 10 4 N (d) 3.5 ´ 10 4 N

Interpret (a) We know, p - p0 = hrg = 1 ´ 10 3 ´ 10 = 10 4 Nm–2 F = Dp ´ S = 10 4 ´ 1.5 = 1.5 ´ 10 4 N

Properties of Liquids

(iii) The buoyant force acts at the centre of buoyancy which is the centre of gravity of the liquid displaced by the body when immersed in the liquid.

Sample Problem 7 At a depth of 1000 m is an ocean the force acting on the window of area 20 cm ´ 20 cm of a submarine at this depth, the interior of which is maintained at sea-level atmospheric pressure is [Given, density of seawater is 1.03 ´ 103 kgm -3, g = 10 ms-2] (a) 3.2 ´ 10 8 N (c) 8.3 ´ 10 N

(d) 3.1 ´ 10

–5

(v) Metacentre, is a point where the vertical line passing through the centre of buoyancy intersects the central line.

N

Interpret (b) Given, h = 1000 m, r = 1.03 ´ 103 kgm–3, gauge pressure,

(iv) The line joining the centre of gravity and centre of buoyancy is called central line.

(b) 4.12 ´ 10 5 N

2

pg = rgh pg = 1.03 ´ 10 3 ´ 10 ´ 1000 5

pg = 103 ´ 10 Pa » 103 atm

439

Laws of Floatation When a body of density r B and volume V is immersed in a liquid of density s, the forces acting on the body are (i) The weight of body W = mg = Vr B g acting vertically downwards through the centre of gravity of the body.

The pressure outside the submarine is p = pa + rgh and pressure inside it is pa . Hence, the net pressure acting on the window is gauge pressure pg .

(ii) The upthrust F = Vsg acting vertically upwards through the centre of gravity of the displaced liquid i.e., centre of buoyancy.

Since, area of window is A = 0.4 m2, the force acting on it is F = rg A = 103 ´ 10 5 Pa ´ 0.04 m2 = 4.12 ´ 10 5 N

So, the following three cases are possible.

12.4 Archimedes’ Principle and Buoyancy Whenever a body is immersed in a fluid, the fluid exerts an upward force on the body, which is called the buoyant force. In fact, any body wholly or partially immersed 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, buoyant force = Vi r l g, where Vi is the volume of immersed part of body and r l is the density of fluid. If a body of volume V and density r s is completely immersed in a liquid of density, r l , then its observed weight Wob = Wactual - upthrust = Vr s × g - Vr l × g = V (r s - r l ) g

Case I The density of body is greater than that of liquid (i. e. , r B > s ). In this case, as weight will be more than upthrust, the body will sink. As shown in Fig. (a) (a) ρB > σ

Case II The density of body is equal to the density of liquid (i. e. , r B = s ). In this case, W = F . so, the body will float fully submerged in neutral equilibrium anywhere in the liquid as shown in Fig. (b).

(b) ρB = σ

Case III The density of body is lesser than that of liquid (i. e. , r B < s ). In this case, W < F ,, so the body will move upwards and in equilibrium will float partially immersed in the liquid such that W = Vinsg

(c) ρB < σ

Buoyant force or buoyancy (i) It is an upward force acting on the body immersed in a liquid. (ii) It is equal to the weight of liquid displaced by the immersed part of the body.

[Vin is the volume of body in the liquid] or

Vr B g = Vinrg

or

Vr B = Vins

[as W = mg = r RV g ] …(i)

440 JEE Main Physics Interpret (a) Mass of the man = mass of water displaced

Some Particular Cases

= volume ´ density 1 = 3 ´2 ´ ´ 10 3 kg = 60 kg 100

(i) In liquid, the apparent weight of the body decreases, and this decrease in its weight is equal to the upthrust acting on the body. Hence, apparent weight

Check Point 1

wapp = V ( r S - r L ) g (ii) If object is immersed in water, then

1. One small and one big piece of cork are pushed below the

RD =

weight of body in air Loss in weight in water

RD =

weight of body in air wt. in air – weight in water

So, by weighing a body in air and in water, we can determine the relative density of the body. (iii) The upthrust on a body immersed in a liquid of density r L in a lift moving downward with acceleration a is F = Vr L |g - a | (iv) The upthrust on a body immersed in a liquid of density r L in a lift moving upwards with acceleration a is F = Vr L |g + a |

(vi) The torque of hydrostatic forces per unit width of the wall of a dam is rgH3 t= 6 Here, H = length of wall, r = density of water and g = acceleration due to gravity.

Sample Problem 8 An ice-berg is floating partly immersed in seawater of density 1.03 gcm–3. The density of ice is 0.92 gcm–3. The fraction of the total volume of the ice-berg above the level of seawater is (c) 34 %

(d) 0.8 %

Interpret (b) Let v be the volume of the ice-berg outside the sea water and V be the total volume of ice-berg. According to question, or

0.92 V = 1.03 (V - v) v 1 - 0.92 11 = = V 1.03 103 v 11 ´ 100 = ´ 100 » 11% V 103

Sample Problem 9 A boat having a length of 3 m and breadth 2 m is floating on a lake. The boat sinks by 1 cm, when a man gets on it. The mass of the man is (a) 60 kg

(b) 62 kg

tank. What will happen to the water level, if the stones are unloaded into water?

5. A piece of ice is floating in a vessel containing water. What will be the effect on the level of water in jar, when ice melts and the temperature of water falls from 4ºC to 1ºC ?

The flow of liquids (fluids) is of three types

F=0

(b) 11 %

2. The bags and suitcases are provided with broad handles. Why? 3. Why is it easier to swim in sea water than in river water? 4. A boat carrying a number of large stones is floating in a water

12.5 Flow of Liquids

(v) If lift is falling freely, then

(a) 8.1 %

surface of water. Which will have greater tendency to rise swiftly?

(c) 72 kg

(d) 128 kg

1. Streamline Flow The streamline flow of a liquid is the flow in which each element of the liquid passing through a point travels along the same path and with the same velocity as the preceeding element passing through the same point. Hence, it is a regular flow. The path followed by each element is called streamline. The tangent drawn at any point of streamline gives the direction of the flow of liquid at that point. From figure, velocity at different points may be different. Hence, in the figure v3 v1 = constant, v2 = constant, v1 v3 = constant v2 But v1 ¹ v2 ¹ v3

2. Turbulent Flow A liquid can possess streamlined motion only when its velocity is less than a limiting velocity, called the critical velocity. When the velocity of the liquid becomes greater than the critical velocity for the liquid, the different elements of the liquid move along a zig-zag path. As a result of unsteady motion of the elements of the liquid along zig-zag paths, the liquid gets churned up. Such a motion of the liquid is called turbulent flow.

Properties of Liquids

3. Laminar Flow If a liquid is flowing over a horizontal surface with a steady flow and moves in the from of layers of different velocities which do not mix with each other, then the flow of liquid is called laminar flow. Thus a flow, in which the liquid moves in layers is called a laminar flow.

Reynold’s Number It is a pure number which determines the nature of flow of liquid through a horizontal tube. Value of critical velocity for flow of liquid of density r and coefficient of viscosity h flowing through a horizontal tube of radius r is given by h vc µ rr Reynold’s number (N R) is a unitless and dimensionless number given by rvr NR = h If the value of Reynold’s number (i) lies between 0 to 2000, the flow of liquid is streamline or laminar. (ii) lies between 2000 to 3000, the flow of liquid is unstable and changing from streamline to turbulent flow. (iii) above 3000, the flow of liquid is definitely turbulent.

Equation of Continuity Q Let us consider P stream line flow of an v2 ideal, non-viscous v1 A2 fluid through a tube of varying crossA1 section. Let at two sections, the cross-section areas be A1 and A2 respectively and fluid flow velocities are v1 and v2, then according to equation of continuity

A1v1r1 = A2v2 r 2 where, r1 and r 2 are the respective densities of fluid. Equation of continuity is based on the conservation of mass. If fluid flowing is incompressible, then r1 = r 2 and equation of continuity is simplified as A1v1 = A2v2

Sample

Problem 10 Two syringes of different cross-sections (without needles) filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and longer pistion are 1 cm and 3 cm respectively. If the smaller piston is pushed in through 6 cm, how much does the longer piston move out

(a) 0.37 cm (c) 37 cm

441

(b) 0.67 cm (d) 67 cm

Interpret (b)

Water is considered to be perfectly incompressible. Volume covered by the movement of smaller piston inwards is equal to the volume moved outwards due to longer piston. L1A1 = L2A2 Þ

L2 =

A1 L1 = A2

æ1 ö p ç ´ 10 -2÷ è2 ø

2

æ3 ö p ç ´ 10 -2÷ è2 ø

2

´ 6 ´ 10 -2

= 0.67 ´ 10 –2 m » 0.67 cm

Note Atmospheric pressure is common to both pistons and has been ignored.

Sample Problem 11 In a car, lift compressed air exent a force F1 on a small piston having a radius of 5 cm. This pressure is transmitted to a second piston of radius 15 cm. If the mass of the car be lifted is 1350 kg, the pressure necessary to accomplish this task is (a) 1.9 ´ 10 5 Pa

(b) 3 ´ 10 6 Pa

(c) 6.5 ´ 10 3 Pa

(d) 0.23 ´ 10 3 Pa

Interpret (a) Since, pressure is transmitted undiminished throughout the fluid F1 =

A1 p (5 ´ 10 -2 m) 2 (1350 N ´ 9.8 ms–2) F2 = A2 p (15 ´ 10 -2 m) 2 = 1470 N » 1.5 ´ 10 3 N

The air pressure that will produce this force is p=

F1 1.5 ´ 10 3 N = 1.9 ´ 10 5 Pa = A1 p (5 ´ 10 –2) 2

Note This is almost double the atmospheric pressure.

12.6 Energy of a Flowing Liquid There are three types of energies in a flowing liquid.

Pressure Energy If p is the pressure on the area A of a fluid, and the liquid moves through a distance l due to this pressure, then Pressure energy of liquid = work done = force ´ displacement = pAl The volume of the liquid is Al. Hence, pressure energy per unit volume of liquid pAl = =p Al

442 JEE Main Physics Kinetic Energy If a liquid of mass m and volume V is flowing with velocity 1 v, then the kinetic energy is mv2. 2 \ Kinetic energy per unit volume of liquid 1 æ mö 1 = ç ÷ v2 = rv2 2 èV ø 2

W2 = p2 A2 (v2Dt ) = p2DV or work done on the fluid is - p2DV .

Potential Energy If a liquid of mass m is at a height h from the reference line (h = 0), then its potential energy is mgh. \Potential energy per unit volume of the liquid æ mö = ç ÷ gh = rgh èV ø

Some useful properties for steady or streamline flows can be obtained using the principle of conservation of energy. Consider a fluid moving in a pipe of varying cross-sectional area. Let the pipe be at varying heights as shown. E

D

A2 2

B

p2

1 v2Dt

p1 v1Dt

\ The total work done on the fluid is W1 - W2 = ( p1 - p2 ) DV Part of this work goes into changing the kinetic energy of the fluid, and part goes into changing the gravitational potential energy. If the density of the fluid is r and Dm = rA1v1Dt = r DV is the mass passing through the pipe in time Dt, then change in gravitational potential energy is DU = rg DV (h2 - h1 )

12.7 Bernoulli’s Principle

A1

W1 = p1 A1 (v1Dt ) = p1DV Since the same volume DV passes through both the regions (from the equation of continuity) the work done by the fluid at the other end (DE ) is

Here, r is the density of liquid.

C

In the same interval Dt, the fluid initially at D moves to E, a distance equal to v2Dt. Pressures p1 and p2 act as shown on the plane faces of areas A1 and A2 binding the two regions. The work done on the fluid at left end (BC ) is

h2

h1

An incompressible fluid is flowing through the pipe in a steady flow. Its velocity must change as a consequence of equation of continuity. A force is required to produce this acceleration, which is caused by the fluid surrounding it, the pressure must be different in different regions. Consider the flow at two regions 1 (i. e. , BC ) and 2 (i. e. , DE ). Consider the fluid initially lying between B and D. In a small interval of time (Dt ), this fluid would have moved. Let v1 is the speed of B and v2 at D, then fluid initially at B has moved a distance v1Dt to C (v, Dt is small enough to assume constant cross-section along BC).

The change in its kinetic energy is æ 1ö DK = ç ÷ r DV (v22 - v12 ) è2ø Using the principle of work-energy theorem, we have æ 1ö ( p1 - p2 ) DV = ç ÷ r DV (v22 - v12 ) + rg DV (h2 - h1 ) è2ø Now, we divide each term by DV , we get 1 ( p1 - p2 ) = r (v22 - v12 ) + rg (h2 - h1 ) 2 Re-arranging the above terms, we have æ 1ö æ 1ö p1 + ç ÷ rv12 + rgh1 = p2 + ç ÷ rv22 + rgh2 è2ø è2ø This is Bernoulli’s equation. Since 1 and 2 refer to any two locations along the pipeline, Therefore, the expression in general can be written as æ 1ö p + ç ÷ rv2 + rgh = constant è2ø Bernoulli’s theorem may be stated as we move along a streamline, the sum of the pressure energy ( p), the kinetic æ rv2 ö energy (KE) per unit volume ç ÷ and the potential energy è 2 ø (pE) per unit volume ( rgh) remains constant.

Properties of Liquids

Important Points of Bernolli’s Theorem 1. When a fluid is at rest, i .e ., its velocity is zero everywhere, Bernoulli’s equation becomes p1 + rgh1 = p2 + rgh2 (p1 - p2 ) = rg (h2 - h1)

2. Bernoulli’s equation ideally applies to fluids with zero viscosity or non-viscous fluids.

3. Bernoulli’s equation applies to fluids which must be incompressible, as the elastic energy of the fluid is also not taken into consideration.

4. Bernoulli’s equation does not hold for steady or turbulent flows, because in that situation velocity and pressure are constantly flctuating in time.

5. Bernoulli’s equation for flowing liquid is 1 p + rv 2 + rgh = constant 2 Dividing this equation by rg, we have p v2 + + h = constant rg 2 g In this expression,

v2 p is pressure head. is velocity head and 2g rg

6. Bernoulli’s theorem for unit mass of liquid

Limitations of Bernoulli’s Theorem (i) While deriving the Bernoulli’s equation, it is assumed that velocity of every particle of liquid across any cross-section of tube is uniform. Practically, it is not correct. Infact, the particles of the liquid in the inner most (i. e., central layer) have maximum velocity and those on layer in contact with the tube have least velocity. Therefore, we should take the mean velocity of the liquid. (ii) The viscous drag of the liquid which comes into play when the liquid is in motion has not been taken into account. (iii) While deriving the above equation, it is assumed that there is no loss of energy when liquid is in motion. Infact some KE is converted into heat and is lost. (iv) If the liquid is flowing along a curved path, the energy due to centrifugal force should also be considered.

Sample Problem 12 At what speed, the velocity head of water is equal to pressure head of 40 cm of Hg?

(i) The action of carburetor, paintgun, scent sprayer, atomizer and insect sprayer is based on Bernoulli’s principle. (ii) The action of Bunsen’s burner, gas burner, oil stove and exhaust pump is also based on Bernoulli’s principle. (iii) Motion of a spinning ball (Magnus effect) is based on Bernoulli’s theorem. (iv) Blowing of roofs by wind storms, attraction between two closely parallel moving boats, fluttering of a flag, etc., are also based on Bernoulli’s theorem. (v) Aerofoil or lift on aircraft wing : Aerofoil is shaped to provide an upward dynamic lift. When the aerofoil moves against the wind, the flow speed on top is higher than that below it. There is an upward force resulting in dynamic lift of the wings. It is found that Dynamic lift F = A ( p1 - p2 ) =

1 Ar (v22 - v12 ) 2

Here, A = surface area of aerofoil v1 = velocity below v2 = velocity above and

r = density of air

(b) 2.8 ms-1 (d) 8.4 ms-1

(a) 10.3 ms-1 (c) 5.5 ms-1

Interpret (a) From Bernoulli’s equation, p v2 + + h = constant rg 2 g

p 1 2 + v = constant r 2

Applications Based on Bernoulli’s Principle

443

Here,

p v2 is pressure head. is velocity head and rg 2g

Given that, velocity head = pressure head v2 p = 2 g rg 2p v2 = Þ r Given,

p = 40 cm of Hg = 40 ´ 10 -2 ´ 9.8 ´ 13.6 ´ 10 3 2 ´ 13.6 ´ 10 3 ´ 40 ´ 10 –2 ´ 9.8 10 3 –1 v = 10.32 ms

v2 = Þ

Sample Problem 13 A manometer connected to a closed tap reads 3.5 ´ 10 5 Nm -2 . When the value is opened, the reading of manometer falls to 3 ´ 10 5 Nm -2, then the velocity of flow of water is (a) 100 ms-1

(b) 10 ms-1

(c) 1 ms-1

(d) 10 10 ms-1

Interpret (b) Bernoulli’s theorem for unit mass of liquid is p 1 2 + v = constant r 2 As the liquid starts flowing, it pressure energy decreases

444 JEE Main Physics 1 2 p1 - p2 v = 2 r

12.8 Torricelli’s Theorem

1 2 3.5 ´ 10 5 - 3 ´ 10 5 v = 2 10 3

Þ

2 ´ 0.5 ´ 10 5 10 3

Þ

v2 =

Þ

v 2 = 100

Þ

v = 10 m/s

Sample Problem 14 A cylinder of height 20 m is

It states that the velocity of efflux i. e. , the velocity with which the liquid flows out of an orifice (i. e. , a narrow hole) in a vessel containing liquid is equal to that which a freely falling body would acquire in falling through a vertical distance equal to the depth of orifice below the free surface of liquid in vessel. Quantitatively, velocity of efflux, v = 2 gh, where h is the depth of orifice below the free surface of liquid in vessel.

completely filled with water. The velocity of efflux of water (in ms-1) through a hole on the side wall of the cylinder near its bottom is (ms-1) (a) 10 (c) 30

h v

(b) 20 (d) 40

H

Interpret (b) Let p0 is the atmospheric pressure, r the density of R

liquid and v the velocity at which water is coming out. Applying the Bernoulli’s theorem just inside and outside the hole,

(i) From figure, volume of the liquid coming out per second through an orifice of area of cross-section a at a depth h below the free surface of liquid, in the vessel is

pv 2 pinside + rgh + 0 = poutside + 2 p0 + rgh = p0 +

V = av = a 2 gh.

rv 2 2

(ii) The time after which the liquid strikes the horizontal surface at the base level of liquid is

Þ

v = 2 gh

Þ

v = 2 ´ 10 ´ 20

Þ

v = 20 ms–1

from 2 ´ 10 -2m 2 to 0.01 m2 at pressure 4 ´ 10 4 Pa. What will be the pressure at smaller cross-section? (b) 3.4 ´10 4 Pa

4

4

(c) 2.4 ´ 10 Pa

(d) 4 ´ 10 Pa

Interpret (b) Here, v1 = 2 ms–1; A1 = 2 ´ 10-2 m2; p1 = 4 ´ 10 4 Pa; A2 = 0.01m2; p2 = ? As or

A1v1 = A2v 2 Av v2 = 1 1 A2 2 ´ 10 -2 ´ 2 = 4 ms–1 0.01 1 1 p1 + rv12 = p2 + rv 22 2 2 1 p2 = p1 + r (v12 - v 22) 2 1 p2 = 4 ´ 10 4 + ´ 10 3 (2 2 - 4 2) 2 = 4 ´ 10 4 - 6 ´ 10 3 =

Now, or or

= 3.4 ´ 10 4 Pa

2 (H - h) g

(iii) Horizontal range,

Sample Problem 15 Water is flowing with a speed of 2 ms-1 in a horizontal pipe with cross-sectional area decreasing

(a) 2 ´ 10 4 Pa

t=

R = vt = 2 gh ´

.

2 (H - h) g

= 2 h ( H - h)

Maximum range, Rmax = H , when h = H /2. Regarding this theorem following points are worth noting (a) From above relation, v µ h . This implies that greater is the distance of hole from the free surface of liquid greater will be the velocity of efflux. (b) The range is same for h liquid coming out of H holes at same distance h below the top and above h the bottom, though their A velocities of efflux (v1 and v2) and time taken (t1 and t2) in falling to the ground are different. But

v1t1 = v2t2 = R = 2h ( H - h )

v1 v2 v3

y

x=x'

B

Properties of Liquids (c) If a is the area of orifice at a depth h below the free surface and A is the area of container, the volume of liquid coming out of orifice per second will be

445

A

Check Point 2

h H

a

1. In a stream-lined flow, what is the velocity of the liquid in contact with the containing vessel?

2. Why does the speed of a liquid increase and its pressure (as v = 2 gh )

= av = a 2 gh

If the hole is at the bottom of the tank, time taken to empty the tank is A 2H t= a g

decrease, when the liquid passes through a constriction in a pipe? Explain.

3. Why two ships moving in parallel directions close to each other get attracted?

4. If a small ping pong ball is placed in a vertical jet of air or water, it will rise to a certain height above the nozzle and stay at that level. Explain.

Venturimeter It is a device used to measure the rate of flow of fluids through pipes. In the arrangement shown, the rate of flow of fluid V is given by h

C

A

12.9 Surface Tension 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 is called surface tension.

B A1

A2

v1

v2

V = A1 A2 V = A1 A2

2 gh ( A12 - A22 ) 2 ( p1 - p2 ) r ( A12 - A22 )

Sample Problem 16 The diameter of a pipe at two points, where a venturimeter is connected is 8 cm and 5 cm and the difference of levels in it is 4 cm. The volume of water flowing through the pipe per second is (a) 1889 ccs–1

(b) 1520 ccs–1

ccs–1

(d) 1125 ccs–1

(b) 1321

Interpret (a) Here, r1 = 8/2 = 4.0 cm; Now,

r2 = 5 /2 = 2.5 cm; h = 4 cm A1 = pr12 = p ( 4) 2 = 16 pcm2

and

A2 = pr22 = p (2.5) 2 = 6.25 p cm2

Here,

r = rm

So, the rate of flow of water in venturimeter is given by V = A1 A2

2 gh ( A12 - A22)

= 6.25 p ´ 16 p =

B

E

2 ´ 980 ´ 4 (16 p ) 2 - (6.25 p ) 2

100 p 2 ´ 28 10 (16 p - 6.25 p ) (16 p +6.25 p )

= 1889 ccs–1

A

F

If we draw an imaginary line AB in any direction in a liquid surface, the surface on either side of this line exerts a pulling force on the surface of other side. This force is at right angles to the line AB. The magnitude of such a force per unit length of the line drawn on the surface of the liquid gives the measure of the surface tension. Thus, F Force Surface tension, S = = Length L SI unit of surface tension is N/m or J/m2. It is a scalar and its dimensional formula is [MT –2]. Surface tension is a molecular phenomenon which is due to cohesive force and the root cause of the force is electrical in nature. Surface tension of a liquid depends only on the nature of liquid and is independent of the surface area of film or length of the line considered.Small liquid drops are spherical due to the property of surface tension. Surface tension of a liquid decreases with an increase in temperature. A highly soluble substance like sodium chloride in water, increases the surface tension of water. But the sparingly soluble substance like phenol when dissolved in water, decreases the surface tension of water.

446 JEE Main Physics Force due to Surface Tension If a body of weight w is placed on the liquid surface whose surface tension is S. If F is the minimum force required to pull it away from the water then value of F for different bodies can be calculated from the following table Body

Force

If we establish a relation between surface energy and surface tension, then it is found that the surface tension of liquid is numerically equal to its surface energy W or W = S DA Hence, S= DA i. e. , surface tension may be defined as the amount of work done in increasing the area of the liquid surface by unity against the force of surface tension at constant

Needle (length)

F = 2 lS

Hollow disc (inner radius = r1, outer radius = r2 )

F = 2 p (r1 + r2 ),S = 4 pS

Circular plate or disc (Radius = r)

F = 2 prS

When the surface area of a liquid is increased, work is done against the cohesive force of molecules and this work is stored in the form of additional surface energy.

Square plate

F = 4lS

Increase in surface potential energy,

Square frame (side = l )

F = 8lS

temperature.

DU = Work done (DW ) = S × DA where, DA is the increase in surface area of the liquid.

Sample Problem 17 A rectangular plate of dimensions 6 cm ´ 4 cm and thickness 2 mm is placed with its largest face flat on the surface of water. What will be the downward force on the plate due to surface tension? (Surface tension of water is 7.0 ´ 10 –2 Nm -1) (a) 1.8 ´ 10 –2 N

(b) 1.4 ´ 10 –2 N

(c) 2 ´ 10 -2 N

(d) 2.5 ´ 10 –2 N

Interpret (b) Here,

(i) Work done in blowing a liquid drop If a liquid drop is blown up from a radius r1 to r2, then work done for that is W = S ( A2 - A1 ) = S × 4 p (r22 - r12 ) (ii) Work done in blowing a soap bubble As a soap bubble has two free surfaces, hence, work done in blowing a soap bubble so as to increase its radius from r1 to r2 is given by W = S × 8 p (r22 - r12 )

l = 6 ´ 10 -2 m; b = 4 ´ 10 -2 m; d = 2 ´ 10 -3 m and S = 7.0 ´ 10

–2

(iii) Work done in splitting a bigger drop into n smaller

droplets

–1

Nm

If a liquid drop of radius R is split up into n smaller droplets, all of same size, then radius of each droplet

Force on the plate due to surface tension is F = 2 ( l + b) S = 2 (6 ´ 10 -2 + 4 ´ 10 -2) ´ 7.0 ´ 10 –2

r = R × (n )-1/ 3 and work done W = S × 4 p (nr 2 - R2 ) = S × 4 pR2 (n1/ 3 - 1)

= 1.4 ´ 10 –2 N

(iv) Coalescence of drops If n small liquid drops of radius r each combine together so as to form a single bigger drop of radius R = n1/ 3 × r, then in the process energy is released. Release of energy is given by

14.10 Surface Energy The molecules on the liquid surface experience a net downward force. So, to bring a molecule from the interior of the liquid to the free surface some work is required to be done against the intermolecular forces of attraction. This work done is stored in the surface film of the liquid as its potential energy.

The potential energy per unit area of the surface film is called the surface energy. It may also be defined as the amount of work done in increasing the surface area of the film by unity. Thus, surface energy work done in increasing the surface area = increase in surface area

DU = S × 4 p (nr 2 - R2 ) = S × 4 pr 2 n (1 - n -1/ 3 )

Sample Problem 18 A rectangular film of liquid is extended from 5 cm ´ 3 cm to 6 cm ´ 5 cm. If the work done is 3.0 ´ 10 –4 J. The surface tension of liquid is (a) 0.5 Nm–1 (c) 0.2 Nm–1

(b) 0.1 Nm–1 (d) 2 Nm–1

Interpret (b) Increase in area, DA = 2 (6 ´ 5 – 5 ´ 3) = 2 ´ 15 cm2 Film has 2 free surfaces = 30 ´ 10 -4 m As, work done, W = surface tension ´ increase in area 3.0 ´ 10 –4 = S ´ 2 ´ 30 ´ 10 -4 or

S = 0.1Nm–1

Properties of Liquids Sample Problem 19 Surface tension of a detergent –2

solution is 2.8×10 What is the work done in blowing a bubble of 2 cm diameter? (a) 4 ´ 10 -6 J

(b) 70.3 ´ 10 –6 J

(c) 50.8 ´ 10 –6 J

(d) 60.8 ´ 10 –6 J

When two soap bubbles of same material having different radii r and R ( > r ) are combined to form a double bubble, then p0

pr

–1

S = 2.8 ´ 10 Nm 2 and R = = 1 cm = 0.01m 2 As soap bubble has two free surfaces, \ Work done, W = S ×8 pR 2 = 70.3 ´ 10

–6

J

(i) Oil spreads over the water surface, because the surface tension of oil is smaller than the water. (ii) The surface tension of points and all lubricating oils is low. (iii) The stromy waves at the sea are calmed by pouring oil on the sea water. (iv) The surface tension of antise ptics like dettol is low because they spread faster. (v) The surface tension of soap solution is low, therefore, it can spread over large area.

Surface Tension of Drops and Bubbles Due to the property of surface tension, a drop or bubble tends to contract and so compresses the matter enclosed. This in turn increases the internal pressure which prevents further contraction and equilibrium is achieved. So, in equilibrium, the pressure inside a bubble or drop is greater than outside and difference of pressure between two sides of the liquid surface is called excess pressure. Excess pressure in different cases is given below : p0

p

p

pR

pr

p

R

4S r 4S pR - p0 = R

and

æ 1 1ö pr - pR = 4 S ç - ÷ è r Rø

\

Applications of Surface Tension

p0

R

pr - p0 =

= 2.8 ´ 10 –2 ´ 8 ´ 3.14 ´ (0.01) 2

p0

p0

r

Interpret (b) Given that, –2

447

Radius of the common surface is given, rR . R0 = R-r If two plates are placed in contact with a thin film of liquid in between them to pull them apart, a large force is needed. Excess pressure in this case is

2S , where d is the d

separation between the plates. Force required to separate two plates, each of area A, is 2 A´S given by F = . d

Sample Problem 20 There is an air bubble of radius 1.0 mm in a liquid of surface tension 0.075 Nm–1 and density 103 kgm–3. The bubble is at a depth of 10.0 cm below the free surface. By what amount is the pressure inside the bubble greater than the atmospheric pressure? (a) 1030 Nm–1 (b) 1230 Nm–1 (c) 1130 Nm–1 (d) None of the above

Interpret (c) Here, r = 1.0 mm = 10 -3 m ; p

S = 0.075 Nm–1,

p1i

r = 10 3 kgm–3 , h = 10 cm = 0.10 m

Drop

Air bubble

Excess pressure inside a drop, Dp =

Soap bubble

2S r

Excess pressure inside an air bubble in air, Dp = Excess pressure inside a soap bubble, Dp =

4S r

2S r

The pressure inside the bubble, which is greater than the atmospheric pressure is 2S = + hr g r 2 ´ 0.075 = + 0.10 ´ 10 3 ´ 9.8 10 -3 = 150 + 980 = 1130 Nm–2

448 JEE Main Physics Angle of Contact The angle of contact between a liquid and a solid is defined as the angle enclosed between the tangents to the liquid surface and the solid surface inside the liquid, both the tangents being drawn at the point of contact of the liquid with the solid.

θ?

? θ

12.11 Applications of Capillary Action or Capillarity (i) It is the phenomenon of rise or fall of liquid in a capillary tube. (ii) The root cause of capillarity is the difference of pressure on the two sides of liquid meniscus in the capillary tube. (iii) The height h through which a liquid will rise in a capillary tube of radius r, which wets the sides of the tube, will be given by

h= (i) The angle of contact depends upon (a) the nature of solid and the liquid in contact, (b) the given pair of the solid and the liquid, (c) the impurities. (ii) The angle of contact does not depend upon the inclination of the solid in the liquid. (iii) The value of angle of contact (q) lies between 0° and 180°. For ordinary water and glass, q = 8°. For silver and pure water, q = 90°. For alcohol and clean glass, q = 0°. (iv) The increase in temperature increases the angle of contact. (v) The angle of contact decreases with the addition of impurities in the liquid.

2 S cos q 2 S = rrg Rrg

where, S is the surface tension of liquid, q is the angle of contact, r is the density of liquid and g is the acceleration due to gravity. R is the radius of curvature of liquid meniscus. (iv) If q < 90°, cos q is positive, so h is positive i. e., liquid rises in a capillary tube. (v) If q > 90°, cos q is negative, so h is negative i. e., liquid falls in a capillary tube. (vi) If a capillary tube is of insufficient length as compared to height to which liquid can rise in the capillary tube, then the liquid rises upto the full length of capillary tube but there is no overflowing of the liquid in the form of fountain. It is so because the liquid meniscus adjusts its radius of curvature so, that hR = a constant i. e., hR = h ¢R¢.

(vi) The materials used for water proofing increases the angle of contact as well as surface tension.

(vii) The height of the liquid column in a capillary tube on the surface of moon is six times than that on the earth.

(vii) If a liquid wets the sides of containing vessel, then the value of angle of contact is acute i. e., less than 90°.

(viii) Rise of liquid in a capillary tube does not violate law of conservation of energy.

(viii) If a liquid does not wet the sides of containing vessel, then the value of angle of contact is obtuse i.e., greater than 90°.

(ix) When a capillary tube dipped vertically in a liquid is tilted, length of the liquid (l) in capillary tube increases but vertical height of liquid (h) in the tube above the surface of liquid in trough remains the same.

Shape of Menicus (i) If a liquid wets the sides of the vessel containing liquid, the shape of liquid meniscus is concave upwards. In this case, force of cohesion between liquid molecules is less than force of adhesion between liquid and vessel molecules. (ii) If a liquid does not wet the sides of the vessel containing liquid, the shape of liquid meniscus is convex upwards. In this case, force of cohesion between liquid molecules is greater than the force of adhesion between liquid and vessel molecules. (iii) The shape of liquid meniscus depends upon the molecular forces and is independent of the gravity pull.

h

l=

α

l

h or h = l cos a cos a

(x) If m is the mass of the liquid which rises in a capillary tube of radius r, then

mg = 2 prS or m =

2 prS g

Properties of Liquids

Hot Spot The water meniscus in the tube is along a circle of circumference 2 pr which is in confact with the glass. Due to the surface tension of water, a force equal to S per unit length acts at all points of the circle. If the angle of contact is q, then this force is directed invward at an angle q from the wall of the tube. In accordance with Newton’s third law, the tube exerts an equal and opposite force S per unit length on the circumference of the water meniscus. This force which is directed outward, can be resolved into two components. S cos q per unit length acting vertically upward and S sin q per unit length acting horizontally outward. Considering the entire circumference 2 pr, for each horizontal component T sin q 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 (2 pr ) ( S cos q). It is this force which supports the weight of the water column so raised. Thus, S cos θ

sin θ

S cos θ (2πr)

θ θ

θ

S sin θ

( S cos q) (2 pr ) = Weight of the liquid column = ( pr2rgh) h=

\

2S cos q rrg

Interpret (b)

The excess pressure in a bubble of gas in a 2S , where S is the surface tension of the liquid r gas interface. The radius of the bubble is r. liquid is given by

The pressure outside the bubble p0 equals atmospheric pressure plus the pressure due to 8 cm of water column. That is p0 = (1.01 ´ 10 5 Pa +0.08 m ´ 1000 kg m–3 ´ 9.80 ms–2) p0 = 1.01784 ´ 10 5 Pa The pressure inside the bubble is 2S pi = p0 + r = 1.01784 ´ 10 5 Pa +

(i) If the contact angle q is greater than 90°, the term cos q is negative and hence, h is negative. The expression, then gives the depression of the liquid in the tube (ii) Suppose a capillary tube is held vertically in a liquid which has a concave meniscus, then capillary rise is given by 2S cos q 2S æ as R = r ö = h= ç ÷ è rrg Rrg cos q ø hR =

2S = constant rg

= 1.02 ´ 10 5 Pa

Note This is a 100% increase in pressure from surface level. At a depth of 1 km the increase in pressure is 100 atm. Sub marines are designed to withstand such enormous pressures.

Sample Problem 22 The lower end of a capillary tube is dipped into water and it is seen that water rises through 7.5 cm in the capillary. Given surface tension of water is 7.5 ´ 10 –2 Nm -1 and angle of contact between water and glass capillary tube is zero. What will be the diameter of the capillary tube? (Given, g = 10 ms-2.)

Interpret (c)

(b) 0.3 mm (d) 0.5 mm Given, h = 7.5 cm = 7.5 ´ 10 -2 m, S = 7.5 ´ 10 –2 Nm–1, q = 0 °, 2 r = ? 2 S cos q h= rrg 4 S cos q 2r = hr g

As,

4 ´ 7.5 ´ 10 –2 ´ cos 0° 7.5 ´ 10 –2 ´ 10 3 ´ 10

\

=

Sample Problem 21 The lower end of a capillary tube of

Þ

= 4 ´ 10 –4 m

diameter 2 mm is dipped 8 cm below the surface of water in a beaker. The surface tension of water at temperature of the experiment is 7.3 ´ 10 –2 Nm –1, 1 atmospheric pressure

Þ

= 0.4 mm

= 1.01 ´ 10 5 Pa, desnity of water = 1000 kg/m3, g = 9.8 ms–2, then the pressure inside the bubble is (a) 2.13 ´ 10 3 Pa

(b) 1.02 ´ 10 5 Pa

(c) 5 ´ 10 -5 Pa

(d) 7.3 ´ 10 –3 Pa

(2 ´ 7.3 ´ 10 –2 Pa.m) 10 -3 m

= (1.01784+0.00146) ´ 10 5 Pa

(a) 0.2 mm (c) 0.4 mm

Note The result has following notable features,

or

Capillary Rise

S cos θ

θ

449

Sample Problem 23 Assuming that the density of atmosphere does not change with altitude. How high would the atmosphere extend? (Given, density of the atmosphere at sea level is 1.29 kg /m3). (a) 2 km

(b) 4 km

(c) 8 km

(d) 16 km

450 JEE Main Physics Interpret (c) Pressure = rgh = density ´ gravity ´ height 3

–2

5

Given, r = 1.29 kg/m , g = 9.8 ms , p = 1.01 ´ 10 Pa 1.01 ´ 10 5 = 1.29 ´ 9.8 ´ h

\ Þ

h=

Þ

1.01 ´ 10 5 » 7989 m 1.29 ´ 9.8

h »8m

layer acts in a direction opposite to the relative velocity of flow of fluid. Its unit is poise or dyne cm -2 s in CGS system and poiseuille or deca poiseuille or Newton-s-m -2 in SI system. It is a scalar quantity. 1 poiseuelle = 1 deca poiseucle = 10 poise

Note In reality the density of air decreases with height. So does the

Note In case of a steady flow of a liquid of viscosity hin a capillary tube

value of g. The atmospheric cover extends with decreasing pressure over 100 km. We should also note that the sea level atmospheric pressure is not always 760 mm of Hg. A drop in the Hg level by 10 mm or more is a sign of an approaching storm.

of length L and radius r under a pressure difference p across it, the velocity of flow at a distance x from the axis is given by p (r 2 - x 2 ) v= 4 hL

Check Point 3 1. In summer, cotton dress is preferable. Why? 2. Oil is poured on calm sea waves. Why? 3. Why smearing of glycerine over the glass window prevents rain drops from sticking to it?

4. Why does a small piece of camphor dance about on the water surface?

5. What is the value of surface tension of a liquid at critical temperature?

6. Name the material in whose capillary, water will descend instead of rising?

Sample Problem 28 A Film metal block of area 0.10 m 2 is connected to a 0.010 kg mass via a string that passes over an ideal pully (considered massless and frictionless). A liquid with a film thickness of 0.01 kg 0.30 mm is placed between the block and the table. When released the block moves to the right with a constant speed of 0.085 ms -1. The coefficient of viscosity of the liquid is (a) 45.2 ´ 10 2 Pa-s (c) 3.45 ´ 10

–3

(b) 13.4 ´ 10 –4 Pa -s (d) 1.42 ´ 10 –2 Pa -s

Pa-s

Interpret (c) The metal block moves to the right because of the

12.12 Viscosity The property of a fluid due to which it opposes the relative motion between its different layers is called viscosity (or fluid friction or internal friction) and the force between the layers opposing the relative motion is called viscous force. 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., FµA (ii) Force F is directly proportional to the velocity gradient æ dv ö ç ÷ between the layers. Combining these two, we è dy ø have FµA

dv dv or F = -hA dy dy

where, h is a constant called coefficient of viscosity or simply viscosity of fluid of liquid is equal to the tangential force required to maintain a unit velocity gradient between two parallel layers of liquid each of area unity. The negative sign shows that viscous force on a liquid

tension in the string. The tension T is equal in magnitude to the weight of the suspended mass m. Thus, the shear force F = T = mg = 0.01 kg ´ 9.8 ms–2 = 9.8 ´ 10 -2N Shear stress on the fluid = Strain rate =

F 9.8 ´ 10 –2 = A 0.10

V 0.085 = i 0.030 Stress h= Strain rate h=

(9.8 ´ 10 –2 N) (0.30 ´ 10 –3 m) (0.085 ms–1) (0.10 m2)

h = 3.45 ´ 10 -3 Pa-s

12.13 Poiseuille’s Formula In case of steady flow of a liquid of viscosity h in a capillary tube of length L and radius R under a pressure difference P across it, the volume of liquid flowing per second is given by dQ ppR4 = dt 8 hL This is called Poiseuille’s formula.

Properties of Liquids Poiseuille’s equation can also be written as, p - p2 Dp = Q= 1 X æ 8 hL ö ç 4÷ è pR ø Here,

X=

body will fall with a constant velocity, called terminal velocity vT Upthrust =

3

8 hL pR4

4 πr3ρg 3 Fig (a)

W=

R = electrical resistance

For current flow through a resistance, potential difference is a requirement, similarly for flow of liquid through a pipe, pressure difference is must.

Problems of series and parallel combination of pipes can be solved in the similar manner as is done in case of an electrical circuit. The only difference is, (i) Potential difference (DV ) is replaced by the pressure difference (Dp) L (ii) The electrical resistance, R æç = r ö÷ is replaced by è Aø 8 h L ö and X æç = ÷ è pR4 ø (iii) The electrical current i is replaced by volume flow rate dV Q or . dt

12.14 Stokes’ Law When a spherical body moves through a fluid, the fluid in contact with the body is dragged with it. The fluid exerts a viscous force on the body to oppose its motion. The formula for the viscous force on a sphere moving through a fluid was first derived by the English physicist G. Stokes in 1843. According to him, a spherical body of radius r moving with velocity v experiences a viscous force given by

.

4 6 phrvT = pr3 ( r - s ) g 3 or

vT =

2 r 2 (r - s ) g 9 h

The velocity v of the body as a function of time is shown in Fig. (b).

Terminal Velocity Let a body (density r) of radius r is falling freely in a medium (liquid or gas) of density s. Initially, v = 0 and r > s , so the body will be accelerated downwards. Because of the acceleration, the velocity will increase and hence, viscous force will increase. At a certain instant, when viscous force F will balance the net downward force (weight – upthrust), acceleration will become zero and the

v vT t

O Fig (b)

Sample Problem 25 Water is flowing through a horizontal tube 8 cm in diameter and 4 km in length at the rate of 20 litre/s. Assuming only viscous resistance. The pressure required to maintain the flow in terms of mercury column. (Coefficient of viscosity of water is 0.001 Pa-s) is (a) 69.68 cm (b) 59.68 cm (c) 49.68 cm (d) 39.68 cm

Interpret (b) Here, 2 r = 8 cm = 0.08 m or

r = 0.04 m; l = 4 km = 4000 m; V = 20 litre/s = 20 ´ 10 -3 m3 s–1, h = 0.001Pa -s, p = ?

As,

V=

p pr 4 8 hl

or

p=

8 Vhl pr 4

F = 6 phrv (h = coefficient of viscosity) This law is called Stokes’ law.

πρ3σg F = 6 πηrv

This equation can be compared with the current equation DV through a resistance, i. e. , i = R Here, DV = potential difference and

451

=

8 ´ (20 ´ 10 -3) ´ 0.001 ´ 4000 æ 22 ö 4 ç ÷ ´ (0.04) è7ø

= 7.954 ´ 10 4 Pa \Height of mercury column for pressure difference p will be, p h= rg =

7.954 ´ 10 4 (13.6 ´ 10 3) ´ 9.8

= 0.5968 m = 59.68 cm

452 JEE Main Physics Sample Problem 26 The terminal velocity of a copper ball of radius 2 mm falling through a tank of oil at 20°C is 6.5 cms -1. The viscosity of the oil at 20°C is [Given density of oil is 1.5 ´ 103 kgm -3, density of copper is 8.9 ´ 103 kgm -3].

(a) 0.1 ms–1 (b) 0.2 ms–1 (c) 0.3 ms–1 (d) 0.4 ms–1

(a) 3.3 ´ 10 –1 kgm-1s-1

Interpret (d) Let r be the radius of each of the small rain drop

(b) 6.3 ´ 10 –2 kgm-1s-1

and R be the radius of big rain drop formed.

(c) 9.2 ´ 10

–3

-1 -1

kgm s

(d) 9.9 ´ 10 –1 kgm-1s-1

As, volume of big drop = 8 ´ volume of each small drop \

Interpret (d) Given, v t = 6.5 ´ 10 –2 ms–1, a = 2 ´ 10 -3 m,

4 3 4 p R = 8 ´ p r3 3 3 R = 2r

g = 9.8 ms–2,r = 8.9 ´ 10 3 kgm–3 s = 1.5 ´ 10 3 kgm–3 \

h=

2 a2 (r - s ) g 9 vt

h=

2 (2 ´ 10 –3) ´ 9.8 ´ 7.4 ´ 10 3 kgm–3 ´ 9 6.5 ´ 10 –2

h = 9.9 ´ 10 –1 kgm–1s–1

Sample Problem 27 Eight spherical rain drops of equal size are falling vertically through air with a terminal velocity of 0.10 ms–1. What should be the velocity, if these drops were to combine to form one large spherical drop?

Let terminal velocity of small drop be v1 and of big drop be v 2. As terminal velocity, v=

2 r 2 (r - s ) g 9h

or

v µ r2

\

v 2 R2 = v1 r 2

or

v 2 = v1

R2 æ2r ö = 0.2 ç ÷ è r ø r2

2

= 0.1 ´ 4 = 0.4 ms–1

WORKED OUT Examples Example 1

A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1 cm. What is the pressure exerted by the heel on the horizontal floor? (a) 5.2 ´ 10 6 Nm-2

(b) 4 ´ 10 6 Nm-2

(c) 6.24 ´ 10 6 Nm-2

(d) 5.24 ´ 10 6 Nm-2

Solution

m = 50 kg;

Here,

r = D / 2 = 1 / 2 cm =

1 m 200

Force mg Pressure = = Area pr 2 50 ´ 9.8 = (22 / 7) ´ (1 / 200) 2 = 6.24 ´ 10 6 Nm-2

Example 2

A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the board is 425 cm 2. What maximum pressure would smaller piston have to bear? (a) 5.2 ´ 10 4 Pa

(b) 4.2 ´ 10 5 Pa

(c) 6.92 ´ 10 4 Pa

(d) 5.92 ´ 10 5 Pa

Solution

The maximum force, which the bigger piston can

bear, F = 30000 kgf = 3000 ´ 9.8 N Area of piston , A = 425 cm2 = 425 ´ 10 -4 m2 \Maximum pressure on the bigger piston, F 300 ´ 9.8 p= = A 425 ´ 10 -4 4

= 6.92 ´ 10 Pa Since, the liquid transmits pressare equally,therefore the maximum pressure the shaller piston can bear is 6.92 ´ 10 4 Pa.

The density of ice is 917 kg m -3. What fraction of the volume of a piece of ice will be above water, when floating in fresh water?

Example 3 (a) 0.083 (c) 0.045

(b) 0.053 (d) 0.043

Solution

Here, density of ice, r = 917 kg m-3;

Density of fresh water, r = 1000 kg m-3 Let V be the total volume of the ice and v be the volume of ice above the water. Then volume of the water displaced by the immersed part of ice = (V - v) According to law of floatation, Weight of ice = Weight of the water displaced V ´ 917 ´ g = (V - v) ´ 1000 ´ g 1000 v = 1000 V - 917 V = 83 V v 83 = = 0.083 V 1000

or or

Example 4 A solid floats with1/ 4 th of its volume above the surface of water, the density of the solid is (a) 750 kg m-3

(b) 650 kg m-3

(c) 560 kg m-3

(d) 450 kg m-3

Solution

Let V and r be volume and density of solid respectively and r¢ be the density of water i.e., p¢ = 10 3 kg m-3 Weight of body = Vrg Volume of solid body outside water = V / 4 \Volume of solid body inside water = V - V / 4 = 3V / 4 Weight of water displaced by solid 3V = ´ 10 3 ´ g 4 As solid body is floating, then Weight of body = Weight of water displaced by it 3V Vrg = ´ 10 3 g 4 3 r = ´ 1000 = 750 kg m-3 4

Example 5 A wire ring of 30.0 mm radius resting flat on the surface of the liquid is raised. The pull required is 3.03 gf force, before the film breaks. The surface tension of the liguid is (a) 71.76 dyne cm-1 (c) 75.58 dyne cm

-1

(b) 78.76 dyne cm-1 (d) 70 dyne cm-1

454 JEE Main Physics Solution

Here, r = 30.0 mm = 3 cm; F = 3.00 gf = 3.03 ´ 980 dyne. Since, the liquid is touching the ring, both inside as well as outside therefore, force acting on the ring due to surface tension is given by F" = 2 ( S ´ circumference of ring) = 2( S ´ 2pr) = 4Spr 22 =4´S´ ´ 3 dyne 7 F¢ = F

As,

22 ´ 3 = 3.03 ´ 980 7 3.03 ´ 980 ´ 7 S= 4 ´ 22 ´ 3 =4´S´

= 78.76 dyne cm-1

The work done in blowing a soap bubble of surface tension 0.06 Nm -1 from 2 cm radius to 5 cm radius is

Solution

-1

Here, S = 0.06 Nm ;

= 2 ´ 4pr22 = 2 ´ 4p (0.02) 2 = 32 p ´ 10 -4m2 Final surface area of the bubble = 2 ´ 4pr22 = 2 ´ 4p (0.05) 2 -4

2

= 200 p ´ 10 m

Example 8 A square plate of 10 cm side moves parallel to another plate with a velocity of 10 cm s-1; both plates immersed in water. If the viscous force is 200 dyne and viscosity of water is 0.01 poise, what is their separation distance? (a) 0.05 cm (c) 0.07 cm

(b) 1 cm (d) 7 cm

Here, side of the square plate, I = 10 cm 2

Area of the plate, A = L = 10 2 = 100 cm2 dv = 10 cms-1: F = 200 dyne:

= 200 p ´ 10

-4

- 32 p ´ 10

h = 0.01poise, dx = ? dv As F = hA dx dx =

hAdv 0.01 ´ 100 ´ 10 = = 0.05 cm F 200

Example 9 A rain drop of radius 0.3 mm has a terminal velocity in air 1 ms-1. The viscosity of air is 18 ´ 10 -5 poise. Find the viscous force on the rain drops. (a) 2.05 ´ 10 -7 N (b) 1.018 ´ 10 -7 N

Increase in surface area -4

pr 2 ´ 0.8 ´ 9.8 ´ 10 -2 = 4 200 = 3.92 ´ 10 -2 Nm-1

(b) 0.003168 J (d) 0.004568 J

r1 = 2 cm = 0.02 m; r2 = 5 cm = 0.05 m Since, bubble has two surfaces, initial surface area of the bubble

-4

(c) 1.05 ´ 10 -7 N (d) 2.058 ´ 10 -7 N

2

= 168 p ´ 10 m

\Work done = S × Increase in surface area = 0.06 ´ 168p ´ 10 -4 = 0.003168 J

Example 7

If excess of pressure inside a soap bubble of radius 1 cm is balanced by that due to a column of oil (specific gravity 0.8) 2 mm high, the surface tension of soap bubble is (a) 2.92 ´ 10 -2 Nm-1

(b) 4.92 ´ 10 -2 Nm-1

(c) 5.92 ´ 10 -2 Nm-1

(d) 3.92 ´ 10 -2 Nm-1

Solution

S=

or

Solution

Example 6

(a) 0.004168 J (c) 0.003158 J

Incase of a soap bubble, 4S p= r

Here. r = 1 cm = 10 -2 m;

density of oil, r = 0.8 ´ 10 3 kg m-3 h = 2mm = 2 ´ 10 -3m Pressure due to 2 mm column of oil, p = hrg = (2 ´ 10 -3) (0.8 ´ 10 3) ´ 9.8 = 2 ´ 0.8 ´ 9.8 Pa

Solution

Here, r = 0.3 mm = 0.3 ´ 10 -3 m; v = 1ms-1

h = 18 ´ 10 -5 poise = 18 ´ 10 -6 decapoise viscous force, F = 6 p h rv 22 =6 ´ ´ (18 ´ 10 -6) ´ (0.3 ´ 10 -3) ´ 1 7 = 1.018 ´ 10 -7 N

Example 10

What is the largest average velocity of blood flow in ar rtery of radius 2 ´ 10 -3, if the flow must remain laminar? What is the corresponding flow rate? [Take viscosity of blood to be 2.084 ´ 10 -3 pa-s: Density of blood is 1.06 ´ 103 kgm -3 ] (a) 9.8 ms-1, 2.5 ´ 10 -5m3 s-1 (b) 9.8 ms-1, 3.5 ´ 10 -5m3 s-1 (c) 9.8 ms-1, 1.23 ´ 10 -4m3 s-1 (d) 0.98 ms-1, 1.23 ´ 10 -5m3 s-1

Properties of Liquids Solution

Area of cross-section,

Here, -3

r = 2 ´ 10 m, D = 2r = 2 ´ 2 ´ 10

-3

= 4 ´ 10

-3

m;

h = 2.084 ´ 10 -3 Pa -s; r = 1.06 ´ 10 3 kg m-3 For flow to be laminar, NR = 2000 n=

Now

455

NRh 2.000 ´ (2.084 ´ 10 -3) = 0.98 ms-1 = rD (1.06 ´ 10 3) ´ (4 ´ 10 -3)

22 ´ (0.02) 2 m2 7 Let v be the velocity of the flow of water at the given point. Clearly, V = Av 1 22 or ´ 10 -3 = ´ (0.02) 2 ´ v 3 7 r = pr 2 =

or

v

7 ´ 10 -3 = 0.2639 ms-1 3 ´ 22 ´ (0.02) 2

Volume flowing second = pr 2v c =

22 ´ (2 ´ 10 -3) 2 ´ 0.98 7

= 1.23 ´ 10 -5 m3 s-1

Example 11

Water flows through a horizontal pipe of variable cross-section at the rate of 20 I, per min. What will be the velocity of water at a point where diameter is 4 cm? (a) 0.2639 ms-1 (c) 0.4639 ms-1

(b) 0.5639 ms-1 (d) 0.3639 ms-1

Solution

Volume of the water flowing per second, 20 ´ 1000 3 -1 ms V = 20 L min -1 = 60 ´ (100)3 =

1 ´ 10 -3m3 s-1 3

Radius of the pipe, r=

4 = 2 cm = 0.02 m 2

Example 12

At what speed will the velocity of a stream of water be equal to 20 cm of mercury column? Taking g = 10 ms-2 (a) 6.4 ms-1 (c) 6.4756 ms-1

Solution

(b) 7.3756 ms-1 (d) None of these

Here, velocity head = 20 cm of Hg = 20 ´ 13.6 cm of water.

v2 As velocity head = 2g \

20 ´ 13.6 =

v2 2 ´ 1000

v = 20 ´ 13.6 ´ 2 ´ 1000 = 737.56 cms-1 = 7.3756 ms-1

Start Practice for

JEE Main Round I

(Topically Divided Problems)

Thrust and Pressure

6. A U-tube contains water and methylated spirit

1. Density of ice is r and that of water is s. What will be the decrease in volume when a mass M of ice melts? s -r M 1 æ1 1ö (d) ç - ÷ M èr s ø

M s -r æ1 1ö (c) M ç - ÷ èr s ø

(b)

(a)

(a) 0.221 cm (c) 0.02 cm

2. A 50 kg girl wearing high heel shoes balances on a single heel. If the heel is circular with a diameter 1.0 cm. What is the pressure exerted on the horizontal floor? (a) 6.9 ´ 106 Pa

(b) 6.2 ´ 106 Pa

(c) 9.6 ´ 106 Pa

(d) 9.0 ´ 106 Pa

when it rises from bottom to top of a water tank where the temperature is uniform. If the atmospheric pressure is 10 m of water, the depth of the water in the tank is (b) 40 m

(c) 70 m

(d) 80 m

4. A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. The specific gravity of spirit would be. (a) 0.70

(b) 0.80

(c) 0.90

(d) 0.60

5. A uniform tapering vessel shown in figure is filled with liquid of density 900 kgm–3. The force that acts on the base of the vessel due to liquid is (Take, g = 10 ms -2 ) –3

2

Area = 10 m

0.4 m –3

Area = 2 × 10 m

(a) 3.6 N

(b) 7.2 N

(c) 9.0 N

(b) 2.22 cm (d) None of these

7. A cylindrical vessel is filled with equal amounts of weight of mercury on water. The overall height of the two layers is 29.2 cm, specific gravity of mercury is 13.6. Then the pressure of the liquid at the bottom of the vessel is (a) 29.2 cm of water (c) 4 cm of mercury

3. The surface area of air bubble increases four times

(a) 30 m

separated by mercury. If the 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms? (Specific gravity of mercury = 13.6)

2

(d) 12.0 N

(b) 29.2/13.6 cm of mercury (d) 15.6 cm of mercury

8. The density r of water of bulk modulus B at a depth y in the ocean is related to the density at surface r 0 by the realation æ r 0 gy ö (a) r = r 0 ç1 ÷ B ø è

r gy ö æ (b) r = çr 01 + 0 ÷ è B ø

æ B ö (c) r = r 0 ç1 + ÷ è r 0 hgy ø

æ B ö (d) r = r 0 ç1 ÷ è r 0 gy ø

9. An aquarium tank is in the shape of a cube with one side a 4m tall glass wall. When the tank is half filled and the water is 2 m deep, the water exerts a force F on the wall. What force does the water exerts on the wall when the tank is full and the water is 4 m drop? (a) 1/2 F

(b) F

shows the vertical cross-section of a vessel filled with a liquid of density r. The normal thrust per unit area on the walls of the vessel at point P, as shown will be

(c) 2 F

(d) 4 F

10. Figure

(a) hrg (c) ( H - h) rg

P

θ O

(b) Hrg (d) ( H - h) rg cos q

H Q

h

Properties of Liquids

Relative Density of Substance, Archimedes’ Principle and Laws of Floatation 11. A beaker containing water is balanced on the pan of a common balance. A solid of specific gravity 1 and mass 5 g is tied to the arm of the balance and immersed in water contained in the beaker. The scale pan with the beaker (a) goes down (c) remains unchanged

18. The spring balance A reads 2 kg with a block of mass m suspended from it. A balance B reads 5 kg when a beaker with liquid is put on the pan of the balance. The two balances are now so arranged that the hanging mass is inside the liquid in a beaker as shown in figure.

A

(b) goes up (d) None of these

12. Torricelli’s

barometer used mercury. Pascal duplicated it using French wine of density 984 kg/m3. Determine the height of the wine column for normal atmospheric pressure. (a) 9.5 cm

(b) 5.5 cm

(c) 10.5 cm

than water. A part of block is outside the liquid. When whole of ice has melted, the liquid level will (b) go down (d) first rise then go down

14. A tank 5m high is half filled with water and then is filled to the top with oil of density 0.85 gcm–3. The pressure at the bottom of the tank, due to these liquids is (a) 1.85 g dyne cm -3 (c) 462.5 g dyne cm -3

(b) 89.25 g dyne cm -3 (d) 500 dyne cm -3

15. A balloon of volume 1500 m3 and weighing 1650 kg with all its equipment is filled with He (density 0.2 kg m–3). If the density of air be 1.3 kgm–3, the pull on the rope tied to the balloon will be (a) 300 kg

(b) 1950 kg

(c) 1650 kg

M

B

(d) 11.5 cm

13. An ice block floats in a liquid whose density is less

(a) rise (c) remain same

457

(d) zero

16. A cubic block is floating in a liquid with half of its volume immersed in the liquid. When the whole system accelerates upwards with acceleration of g/3 the fraction of volume immersed in the liquid will be g/3

(a) The balance A will read more than 2 kg (b) The balance B will read less than 5 kg (c) The balance A will read less than 2 kg and B will read more than 5 kg (d) The balance A will read more than 2 kg and B will read less than 5 kg

19. A cylinder of mass m and density r hanging from a string is lowered into a vessel of cross-sectional area A

containing a liquid of density s (< r) until it is fully immersed. The increase in pressure at the bottom of the vessel is (a) Zero

(b)

mg A

(c)

mgr sA

(d)

msg rA

20. A cubical block of wooden edge l and a density r floats in water of density 2 r. The lower surface of cube just touches the free end of a massless spring of force constant k fixed at the bottom of the vessel. The weight w put over the block so that it is completely

immersed in water without wetting the weight is (a) a ( lrg + k ) æ lr g ö (c) a ç + 2k ÷ è 2 ø

(b) a( l2rg + k ) kö æ (d) l ç l2rg + ÷ è 2ø

21. A rectangular plate 2m × 3m is immersed in water in (a)

1 2

(b)

3 8

(c)

2 3

(d)

3 4

17. Two cubes each weighing 22 g exactly are taken. One is of iron ( d = 8 ´ 103 kgm -3) and the other is of marble ( D = 3 ´ 103 kgm -3). They are immersed in alcohol and then weighed again (a) iron cube weighs less (b) iron cube weighs more (c) both have equal weight (d) nothing can be said

such a way that its greatest and least depth are 6 m and 4 m respectively, from the water surface. The total thrust on the plate is (a) 294 ×103 N (c) 100 ×103 N

(b) 294 N (d) 400 ×103 N

22. A block of aluminium of mass 1 kg and volume

3.6 ´ 10-4 m 3 is suspended from a string and then completely immersed in a container of water. The decrese in tension in a container of water. The decrease in tension in the string after immersion is (a) 9.8 N

(b) 6.2 N

(c) 3.6 N

(d) 1.0 N

458 JEE Main Physics 23. A vessel with water is placed on a weighing pan and

29. A hollow cylinder of mass m made heavy at its bottom

it reads 0.8 gcc–1 is sunk into the water with a pin of negligible volume as shown in figure keeping it sunk. The weighing pan will show a reading

is floating vertically in water. It is tilled from its vertical position through an angle q and is left. The

Pin

restoring force acting on it is (a) mg cos q (b) mg sin q ù ù é 1 é 1 (d) mg ê (c) mg ê - 1ú +1 ë cos q û ë cos q úû

30. A hemispherical bowl just floats without sinking in a

Weighing pan

(a) 600 g (c) 642 g

(b) 632 g (d) 640 g

24. A body of density r is dropped from rest at a height h into a lake of density s, where s > r. Neglecting all dissipative forces, calculate the maximum depth to which the body sinks before returning to float on the surface. h (a) s -r hr (c) s -r

hr s hs (d) s -r (b)

25. Two cylinders of same cross-section and length L but made of two material of densities r1 and r2 (in CGS units) are cemented together to form a cylinder of length 2 L. If the combination floats in water with a length L/2 above the surface of water and r1 < r2 , then (a) r1 > 1 (c) r1 > 1 / 2

(b) r1 < 3 / 4 (d) r1 > 3 / 4

26. The density of ice is 0.9 gcc–1 and that of sea water is 1.1 gcc–1. An ice berg of volume V is floating in sea water. The fraction of ice berg above water level is (a) 1/11 (c) 3/11

(b) 2/11 (d) 4/11

27. A solid of density D is floating in a liquid of density d. If v is the volume of solid submerged in the liquid and V is the total volume of the solid, then v / V is equal to d (a) P (c)

D (D + d)

D (b) d D+d (d) D

28. The total weight of a piece of wood is 6 kg. In the

floating state in water its 1 part remains inside the 3 water. On this floating solid, what maximum weight is to be put such that the whole of the piece of wood is to be drowned in the water? (a) 12 kg (b) 10 kg (c) 14 kg (d) 15 kg

liquid of density 1.2 ´ 103 kgm –3. If outer diameter and the density of the bowl are 1 m and 2 ´ 104 kgm –3 respectively, then the inner diameter of the bowl will be (a) 0.94 m (c) 0.98 m

(b) 0.96 m (d) 0.99 m

Surface Tension and Surface Energy 31. A thin metal disc of radius r float on water surface and bends the surface downwards along the perimeter making an angle q with vertical edge of the disc. If the disc displaces a weight of water w and surface tension of water is T, then the weight of metal disc is (a) 2 prT + w (c) 2 prT cos q + w

(b) 2 prT cos q - w (d) w - 2 prT cos q

32. A ring is cut from a platinum tube 8.5 cm internal diameter and 8.7 cm external diameter. It is supported horizontally from a pan of a balance so, that it comes in contact with the water is in glass vessel. If an extra 3.47 g-wt is required to pull it away from water, surface tension of water is (a) 72.07 dyne cm–1 (c) 65.35 dyne cm–1

(b) 70.80 dyne cm–1 (d) 60.00 dyne cm–1

33. What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 4.65 ´ 10-1N/m. The atmospheric pressure is . ´ 105 Pa. Also give the excess pressure inside the 101 drop. (a) 1.01 ´ 105 Pa, 320 Pa 5

(c) 310 Pa, 1.01 ´ 10 Pa

(b) 1.01 ´ 105 Pa, 310 Pa (d) 320 Pa, 1.01 ´ 105 Pa

34. What is the radius of the biggest aluminium coin of thickness, t and density r, which will still be able to float on the water surface of surface tension S? 4S 3S 2S S (a) (b) (c) (d) 3rgt 4rgt rgt rgt

35. 8000 identical water drops are combined to form a big drop then the ratio to the final surface energy to the initial surface energy, if all the drops together is (a) 1 : 10 (c) 1 : 20

(b) 1 : 15 (d) 1 : 25

Properties of Liquids

459

36. A frame made of a metallic wire enclosing a surface

44. What change in surface energy will be noticed when a

area A is covered with a soap film. If the area of the frame of metallic wire is reduced by 50%, the energy of the soap film will be changed by

drop of radius R splits up into 1000 droplets of radius r, surface tension T ?

(a) 100 % (c) 50 %

(b) 75 % (d) 25 %

38. Surface tension of a soap solution is able of 2.0 cm diameter will be (b) 15.2 ´10 -6 pJ (d) 1 ´ 10–4 pJ

In this process, which of the following statements is correct? (a) The sum of the temperatures of the two droplets together is equal to temperature of the original drop (b) The sum of the masses of the two droplets is equal to mass of drop (c) The sum of the radii of the two droplets is equal to the radius of the drop (d) The sum of the surface areas of the two droplets is equal to the surface area of the original drop

40. Work done in splotting a drop of water of 1mm radius water

(b) 895 . ´ 10 -5 J (d) 5.98 ´ 10 -6 J

41. A drop of liquid of diameter 2.8 mm breaks up into 125 identical drops. The change in energy is nearly (S = 75 dyne cm–1) (a) zero (c) 46 erg

(b) 19 erg (d) 74 erg

42. The surface energy of a liquid drop is u. It is sprayed into 1000 equal droplets. Then its surface energy becomes (a) u (c) 100 u

is formed from a given solution. How much work is required to be done to form a bubble of volume 2 V ? (b) 2W (d) 41/3 W

46. What is the ratio of surface energy of 1 small drop and 1 large drop if 1000 drops combined to form 1 large drop? (a) 100 : 1 (c) 10 : 1

39. A drop of water breaks into two droplets of equal size.

(a) 9.8 ´ 10 -5 J (c) 5.89 ´ 10 -5 J

45. Let, W be the work done, when a bubble of volume V

(a) W (c) 21/3 W

(b) 4.35 ´ 10–3 J (d) 4.35 ´ 10–8 J

into 106 droplets is (surface tension of 72 ´ 10-3 J / m 2 )

(d) 36 pR2T

(c) 16 pR T

droplets of equal size. The work done is ( S = 35 ´ 10-2 Nm –1)

(a) 7.6 ´ 10–6 pJ (c) 1.9 ´ 10–6 pJ

(b) 7 pR2T

2

37. A mercury drop of radius 1 cm is broken into 106 (a) 4.35 ´ 10–2 J (c) 4.35 ´ 10–6 J

(a) 4 pR2T

(b) 1000 : 1 (d) 1 : 100

47. A bigger drop of radius R is converted into n smaller drops of radius r, the required energy is (a) ( 4 pr2 n - 4 pR2 ) T 4 ö æ4 (b) ç pr3n - pR3 ÷ T ø è3 3 (c) ( 4 pR2 - 4 pr2 ) nT (d) ( n 4 pr2 - 4 pR2 ) T

Excess of Pressure, Shape of Meniscus and Capillarity 48. The angle of contact at the interface of water-glass is 0° Ethylalcohol-glass is 0°, Mercury-glass is 140° and Methyliodide-glass is 30°. A glass capillary is put in a trough containing one of these four liquids. It is observed that the meniscus is convex. The liquid in [NCERT Exemplar] the trough is (a) water (c) mercury

(b) ethylalcohol (d) methyliodide

49. The diagram shows three soap bubbles A, B and C prepared by blowing the capillary tube fitted with stop cocks S, S1, S2 and S3. With stop cock S closed and stop cocks S1, S2 and S3 opened

(b) 10 u (d) 1000 u

C

43. A water film is made between two straight parallel wires of length 10 cm separated by 5 mm from each other. If the distance between the wires is increased by 2 mm. How much work will be done? Surface tension for water is 72 dyne cm–1. (a) 288 erg (b) 72 erg (c) 144 erg (d) 216 erg

S1 A

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

S3

S S 2 B

B will start collapsing with volumes of A and C increasing C will start collapsing with volume of A and B increasing volume of A, B and C will become equal in equilibrium C and A will both start collapsing with volume of B increasing

460 JEE Main Physics 50. The amount of work done in blowing a soap bubble such that its diameter increases from d to D is (S = surface tension of solution) (a) p( D2 - d2 ) S 2

2

(c) 4 p ( D - d ) S

(d) 8 p ( D2 - d2 ) S

(b) 20 m (d) 30 m

52. In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m/s and 63 m/s respectively. What is the lift on the wing, if its area is 2.5 m2 ? Take the density of air to be 1.3 kg/m 3. (a) 5.1 ´ 102 N 3

(c) 1.6 ´ 10 N

(b) 6.1 ´ 102 N 3

(d) 1.5 ´ 10 N

53. With the increase in temperature, the angle of contact (a) decreases (b) increases (c) remains constant (d) sometimes increases and sometimes decreases

54. Water rises to a height of 10 cm in a capillary tube and mercury falls to a depth of 3.42 cm in the same capillary tube. If the density of mercury and water are 135° and 0° respectively, the ratio of surface tension of water and mercury is (a) 1 : 0.15 (c) 1 : 6.5

(b) 1 : 3 (d) 1.5 : 1

55. Water rises to a height of 16.3 cm in a capillary of height 18 cm above the water level. If the tube is cut at a height of 12 cm in the capillary tube, (a) (b) (c) (d)

(b)

(c)

(d)

(b) 2 p ( D2 - d2 ) S

2 51. If pressure at half the depth of a lake is equal to 3 pressure at the bottom of the lake then what is the depth of the lake? (a) 10 m (c) 60 m

(a)

water will come as a fountain from the capillary tube water will stay at a height of 12 cm in the capillary tube the height of water in the capillary tube will be 10.3 cm water height flow down the sides of the capillary tube

56. Water rises in a capillary tube to a height h. It will rise to a height more than h (a) on the surface of sun (b) in a lift moving down with an acceleration (c) at the poles (d) in a lift moving up with an acceleration

57. If a liquid is placed in a vertical cylindrical vessel and the vessel is rotated about its axis, the liquid will take the shape of figure.

58. By inserting a capillary tube upto a depth l in water, the water rises to a height h. If the lower end of the capillary tube is closed inside water and the capillary is taken out and closed end opened, to what height the water will remain in the tube, when l > h? (a) zero

(b) l + h

(c) 2 h

(d) h

59. Two capillary tubes of radii 0.2 cm and 0.4 cm are dipped in the same liquid. The ratio of heights through which liquid will rise in the tubes is (a) 1 : 2 (c) 1 : 4

(b) 2 : 1 (d) 4 : 1

60. Water rises in a capillary tube to a height h. Choose the false statement regarding rise from the following. (a) On the surface of Jupitor, height will be less than h. (b) In a lift, moving up with constant acceleration, height is less than h. (c) On the surface of the moon, the height is more than h. (d) In a lift moving down with constant acceleration, height is less than h.

Dependence of Surface Tension 61. The surface tension of a liquid at its boiling point (a) becomes zero (b) becomes infinity (c) is equal to the value at room temperature (d) is half to the value at the room temperature

62. When a pinch of salt or any other salt which is soluble in water is added to water, its surface tension (a) increases (b) decreases (c) may increase or decrease depending upon salt (d) None of the above

63. At which of the following temperatures, the value of surface tension of water is minimum? (a) 4°C

(b) 25°C

(c) 50°C

(d) 75°C

64. Two spherical soap bubbles of radii a and b in vacuum coalesce under isothermal conditions. The resulting bubble has a radius given by (a)

( a + b) 2

(c) a2 + b2

(b)

ab a+b

(d) a + b

461

Properties of Liquids 65. When two soap bubbles of radius r1 and r2 ( r2 > r1) coalesce, the radius of curvature of common surface is (a) ( r2 - r1 ) r -r (c) 2 1 r1 r2

(b) ( r2 + r1 ) r r (d) 2 1 r2 - r1

71. Water in a vessel of uniform cross-section escapes through a narrow tube at the base of the vessel. Which graph given below represents the variation of the height h of the liquid with time t? h

h

66. Which graph represent the variation of surface tension with temperature over small temperature ranges for water?

(a)

(b) t

(b)

Surface tension

(a)

Surface tension

t

Temperature

(c)

(d) t

t

Temperature

Surface tension

(d)

Temperature

Temperature

67. A capillary tube of radius R and length L is connected in series with another tube of radius R/2 and length L/4. If the pressure difference across the two tubes taken together is p, then the ratio of pressure difference across the first tube to that across the second tube is (a) 1 : 4 (c) 4 : 1

(b) 1 : 1 (d) 2 : 1

68. The relative velocity of two parallel layers of water is 8 cms–1. If the perpendicular distance between the layers is 0.1 cm, then velocity gradient will be (a) 40 s–1 (c) 60 s–1

(b) 50 s–1 (d) 80 s–1

69. Two water pipes P and Q having diameter 2 ´ 10-2 m and 4 ´ 10-2 m respectively are joined in series with the main supply line of water. The velocity of water flowing in pipe P is (a) 4 times that of Q (b) 2 times that of Q (c) 1/2 times that of Q (d) 1/4 times that of Q

of radius a cm and of length l cm when connected to a pressure head of h cm of water. If a tube of the same length and radius a/2 cm is connected to the same pressure head, the quantity of water flowing through the tube per second will be (a) 16 cm3 (c) 4 cm3

(b) 1 cm3 (d) 8 cm3

73. Under a pressure head, the rate of orderly volume flow of liquid through a capillary tube is Q. If the length of capillary tube were doubled and the diameter of the bore is halved, the rate of flow would become Q 4 Q (c) 8

(b) 16 Q

(a)

(d)

Q 32

Stokes’ Law, Terminal Velocity and Variation of Viscosity 74. The terminal velocity v of a spherical ball of lead of radius R falling through a viscous liquid varies with R such that (a)

v = constant R

(b) vR = constant

(c) v = constant

(d)

v = constant R2

75. The rate of steady volume flow of water through a

70. The rate of flow of liquid through a capillary tube of radius r is V, when the pressure difference across the two ends of the capillary is p. If pressure is increased by 3 p and radius is reduced to r/2, then the rate of flow becomes (a) V/9 (c) V/4

h

72. 16 cm3 of water flows per sec through a capillary tube

Surface tension

(c)

h

(b) 3V/8 (d) V/3

capillary tube of length l and radius r under a pressure difference of p, is V. This tube is connected with another tube of the same length but half the radius in series. Then the rate of steady volume flow through them is (The pressure difference across the combination is p) (a)

V 16

(b)

V 17

(c)

16 V 17

(d)

17 V 16

462 JEE Main Physics 76. A small spherical ball of steel falls through a viscous

82. A marble of mass x and diameter 2 r is gently

medium with terminal velocity v. If a ball of twice the radius of the first one but of the same mass is dropped through the same method, it will fall with a terminal velocity (neglect buoyancy)

released a tall cylinder containing honey. If the marble displaces mass y ( < x) of the liquid, then the terminal velocity is proportional to

v (a) 2 (c) v

v (b) 2 (d) 2 v

pebble is dropped from the top with zero initial velocity. From the plot shown in figure, indicate the one that represents the velocity (v) of the pebble as a function of time (t). [NCERT Exemplar]

(b) W1 < W2 (d) W2 = 32 W1

(b)

84. A spherical ball is dropped in a long column of viscous

(d)

liquid. Which of the following graphs represent the variation of (i) gravitational force with time (ii) viscous force with time (iii) net force acting on the ball with time?

t

t

v

v

(c)

attains its terminal velocity after having fallen 32 m. Then, it covers the rest of the path with terminal velocity only. The work done by air friction during the first 32 m of fall is W1. The work done by air friction during the subsequent 32 m fall is W2 . Then (a) W1 > W2 (c) W1 = W2

v

(a)

(b) (x –y) (x - y ) (d) r

83. A small iron sphere is dropped from a great height. It

77. A tall cylinder is filled with viscous oil. A round

v

(a) (x + y) x+ y (c) r

F t

P

t

Q

78. A rain drop of radius 0.3 mm has a terminal velocity in air = 1 ms–1. The viscous force on it is (a) 101.73 ´ 10–4 dyne (c) 16.95 ´ 10–4 dyne

(b) 101.73 ´ 10–5 dyne (d) 16.95 ´ 10–5 dyne

79. A metallic sphere of mass M falls through glycerine with a terminal velcity v. If we drop a ball of mass 8 M of same metal into a column of glycerine, the terminal velocity of the ball will be (a) 2 v (c) 8 v

(b) 4 v (d) 16 v

80. A rain drop of radius 1.5 mm, experiences a drag force F = (2 ´ 10–5 v) N, while falling through air from a height 2 km, with a velocity v. The terminal velocity of the rain drop will be nearly (use g = 10 ms–2) (a) 200 ms–1 (c) 7 ms–1

(b) 80 ms–1 (d) 3 ms–1

81. Which of the following diagrams (figure) does not represent a streamline flow?

(a)

[NCERT Exemplar]

R t

(a) Q, R, P (c) P, Q, R

(b) R, Q, P (d) R, P, Q

85. A large tank is filled with water to a height H. A small hole is made at the base of the tank if takes T1 H time to decrease the height of water to (h > 1) and if l takes T2 time to take out the rest of water if T1 = T2 then the value of h is (a) 2 (c) 4

(b) 3 (d) 2 2

Liquid Flow 86. An incompressible liquid flows through a horizontal tube as shown in the figure. Then, the velocity v of the fluid is A

(b) v1 = 3 m/s

v2 = 1.5 m/s

A

1.5

A v

(c)

(d)

(a) 3 m/s (c) 1.0 m/s

(b) 1.5 m/s (d) 2.25 m/s

Properties of Liquids

463

87. Water flowing out of the mouth of a tap and falling

92. Three tubes A, B and C are connected to a horizontal

vertically in streamline flow forms a tapering column, i.e., the area of cross-section of the liquid column decreases as it moves down. Which of the following is the most accurate explanation for this?

pipe in which liquid is flowing. The radii of pipe at the joints of A, B and C are 2 cm, 1 cm and 2 cm respectively. The height of liquid A

C B

(a) Falling water tries to reach a terminal velocity and hence, reduces the area of cross-section to balance upward and downward forces (b) As the water moves down, its speed increases and hence, its pressure decreases. It is then compressed by atmosphere (c) The surface tension causes the exposed surface area of the liquid to decrease continuously (d) The mass of water flowing out per second through any cross-section must remain constant. As the water is almost incompressible, so the volume of water flowing out per second must remain constant. As this is equal to velocity ´ area, the area decreases as velocity increases

88. If two ping pong balls are suspended near each other and a fast stream of air is producce within the space of the balls, the balls (a) come nearer to each other. (b) move away from each other. (c) remain in their original positions. (d) move far away.

89. Along a streamline

90. An ideal fluid flows through a pipe of circular cross-section made of two sections with diameters 2.5 cm and 3.75 cm. the ratio of the velocities in the two pipes is [NCERT Exemplar] (b) 3 : 2 (d) 2 : 3

91. Air is streaming past a horizontal air plane wing such that its speed is 120 ms–1 over the upper surface and 90 ms–1 at the lower surface. If the density of air is 1.3 kgm–3, what will be the gross lift on the wing? If the wing is 10 m long and has an average width of 2 m, (a) 81.9 N (c) 81.9 kN

(b) in A and B is equal (d) in A and C is same

93. Figs. (i) and (ii) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is/are incorrect?

Fig. (i)

(a) Fig. (i) (c) both (i) and (ii)

Fig. (ii)

(b) Fig. (ii) (d) None of these

94. A tank is filled with water upto a height H. Water is allowed to come out of a hole P in one of the walls at a depth h below the surface of water (see figure). Express the horizontal distance X in terms of H and h.

[NCERT Exemplar]

(a) the velocity of a fluid particle remains constant (b) the velocity of all fluid particles crossing a given position is constant (c) the velocity of all fluid particles at a given instant is constant (d) the speed of a fluid particle remains constant

(a) 9 : 4 (c) 3 : 2

(a) in A is maximum (c) is same in all three

(b) 8.19 kN (d) 819 kN

h p H

X

(a) X =

h ( H - h)

(c) X = 2 h ( H - h)

h ( H - h) 2 (d) X = 4 ( H - h) (b) X =

95. Water stands at level A in the arrangement shown in the figure. What will happen if a jet of air is gently blown into the horizontal tube in the direction shown in the figure? Jet of air

A

464 JEE Main Physics (a) Water will rise above A in the capillary tube (b) Water will fall below A in the capillary tube (c) There will be no effect on the level of water in the capillary tube (d) Air will emerge from end B in the form of bubbles

96. A cylindrical drum, open at the top, contains 15 L of water. It drains out through a small opening at the bottom. 5 L of water comes out in time t1, the next 5 L in further time t2 and the last 5 L in further time t3. Then (a) t1 < t2 < t3 (b) t1 > t2 > t3 (c) t1 = t2 = t3 (d) t2 > t1 = t3

97. The level of water in a tank is 5 m high. A hole of area 10 cm2 is made in the bottom of the tank. The rate of leakage of water from the hole is (a) 10 -2 m3s -1

98. A fluid flows through a horizontal pipe having two different cross-sections of area A and 2 A. If the pressure at the thin cross-section is p and fluid velocity is v, the velocity and pressure at the thicker cross-section is (take the density of fluid as r) v ,p + 2 v (c) , p + 2

Only One Correct Option

(d) 10 -1 m3s -1

(c) 10 m s

(a)

Round II

(b) 102 m3s -1

3 -1

1 2 rv 2 3 2 rv 8

3 v , p + rv2 4 8 3 2 (d) v , p + rv 4 (b)

(Mixed Bag) 3. A streamline body with relative density r1 falls into

1. There are two identical small holes on the opposite sides of a tank containing a liquid. The tank is open at the top. The difference in height between the two holes is h. As the liquid comes out of the two holes, the tank will experience a net horizontal force proportional to

air from a height h1 on the surface of a liquid of relative density r2 , where r2 > r1. The time of immersion of the body into the liquid will be (a)

2h1 g

(b)

(c)

2h1 r1 ´ g r2

(d)

2h r1 ´ g r2 2h1 r1 ´ g (r2 - r1 )

4. A liquid of density r is filled in a U-tube is accelerated with an acceleration a so that the height of liquid in its two vertical arms are h1 and h2 as shown in the figure. If l is the length of horizontal arm of the tube, the acceleration a is

h

(a) h1 /2

(b) h3 /2

(c) h

(d) h2

2. Water is filled up to a height h in beaker of radius R as shown in the figure. The density of water r the surface tension of water is T and the atmosphere pressure is p0 . Consider a vertical section ABCD of the water column through n 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.

h1 h2 l

g ( h1 - h2 ) g ( h1 - h2 ) towards right (b) towards left 2l 2l g ( h1 - h2 ) g( h1 - h2 ) (c) towards right (d) towards left l l

(a)

2R B

5. A soap film is made by dipping

A C

h

D

(a) |2p 0 Rh + pr2rgh - 2RT | (b) |2p 0 Rh + Rrgh2 - 2RT | (c) |p 0 pR2 + Rrgh2 - 2RT | (d) |p 0 pR2 + Rrgh2 + 2RT |

a circular frame of radius b in soap solution. A bubble is formed by blowing air with speed v in the form of cylinder. The radius of the bubble formed R >> b so that the air is incident normally on the

R b

v

Properties of Liquids surface of bubble. Air stops after striking surface of soap bubble. Density of air is r. The radius R of the bubble when the soap bubble separates from the ring is (surface tension of liquid is S). (a)

S rv2

(b)

4S rv2

(c)

Sb rv

(d)

4 Sb rv2

465

Which of the following graph represents the variation of pressure p along the axis of tube? p

p

(a)

(b)

6. A metal ball immersed in alcohol weighs W1 at 0°C

and W2 at 59°C. The coefficient of cubical expansion of the metal is less than that of alcohol. Assuming that the density of the metal is large compared to that

of alcohol, it can be shown that (a) W1 > W2 (b) W1 < W2 (c) W1 = W2 (d) W1 = 2 W2

x p

p

(c)

(d)

7. A uniform rod of density r is placed in a wide tank containing a liquid s (s > r ). The depth of liquid in the tank is half the length of the rod. The rod is in equilibrium, with its lower end resting on the bottom of the tank. In this position, the rod makes an angle q with the horizontal. Then, sin q is equal to 1 s 1s (b) (a) 2 r 2r (c)

r s

(d)

x

x

x

11. In this figure, an ideal liquid flows through the tube having uniform area of cross-section and is held in vertical plane. Find the ratio of speed of liquid at A and B and also find the pressure difference between these points. A

r s

8. The U-tube has a uniform cross-section as shown in figure. A liquid is filled in the two arms upto heights h1 and

h2 and then the liquid is allowed to move. Neglect viscosity and surface tension. When the level equalize in the two arms, the liquid will

h

h1

B h2

(a) 2lgh 3 (c) lgh 2

æ h -h ö (b) be moving with an acceleration of g ç 1 2 ÷ è h1 + h2 + 2 ø (c) be moving with a velocity of ( h1 - h2 )

g 2( h1 + h2 + h)

(d) exert a net force to the right on the cube

9. A stone of relative density k is released from rest on the surface of a lake. If viscous effects are ignored, the stone sinks in water with an acceleration of (b) g (1 + k ) 1ö æ (d) g ç1 + ÷ è kø

vessel. Water is filled upto a height h and water flows out in t sec. If water is filled to a height 4h, then it will flow out in time (a) 2 t

(b) 4 t

(c) 16 t

rate L. The water travels vertically upwards through the hydrant and then does 90° turn to emerge horizontally at speed v. The pipe and nozzle have uniform cross-section throughout. The force exerted by water on the corner of the hydrant is v v

duct, with cross-section varying as shown in figure. x

(d) 7/4 t

13. A fire hydrant delivers water of density r at a volume

10. A non viscous liquid is flowing through a frictionless

o

(d) zero

12. There is a hole of area A at the bottom of a cylindrical

(a) be at rest

(a) g (1 - k ) æ 1ö (c) g ç1 - ÷ è kø

(b) lgh

(a) zero (b) rvL (c) 2rvL (d) 2 rvL

466 JEE Main Physics 14. A block is submerged in vessel filled with water by a

20. An alloy of Zn and Cu (i.e., brass) weights 16.8 g in air

spring attached to the bottom of the vessel. In equilibrium, the spring is compressed. The vessel now moves downwards with an acceleration a ( < g). The spring length

and 14.7 g in water. If relative density of Cu and Zn are 8.9 and 7.1 respectively then determine the amount of Zn and Cu in the alloy. (a) 2g, 4g (c) 9.345g, 7.455 g

(b) 4g, 2g (d) 0, 3g

21. Two soap bubbles A and B are kept in closed chamber

(a) will become zero (b) will decrease but not zero (c) will increase (d) may increase or decrease or remain constant

15. Calculate the force of attraction between two parallel plates separated by a distance 0.2 mm after a water drop of mass 80 mg is introduced between them. The wetting is assumed to be complete. (surface tension of water is 0.07 Nm–1) (a) 0.14 N (b) 0.28 N (c) 0.42 N (d) 0.56 N

where the air is maintained at pressure 8 N / m 2 .The radius of bubbles A and B are 2 cm and 4 cm respectively surface tension of the soap water used to make bubbles is 0.04 N/m. Find the ratio nB / n A , where n A and nB are the number of moles of air in bubbles A and B respectively [Neglect the effect of gravity] (a) 2 (c) 8

22. A jar shown in figure is filled with a liquid of density r. The jar is placed in vacuum. Cross-section of the jar is circular and base is having a radius R. The force

exerted by the liquid column on the base of the jar is a b

30° 60°

16. A wooden ball of density r is immersed in water of density r 0 to depth h and then released. The height H

F

above the surface of water upto which the ball jump out of water is (a) zero r h (c) 0 r

(b) h ær ö (d) ç 0 - 1÷ h èr ø

17. The bottom of a cylindrical vessel has a circular hole of radius r and at depth h below the water level. If the diameter of the vessel is D, the find then speed with which the water level in the vessel drops. 4 r2 2gh D2 4 D2 (c) 2 2gh r (a)

(b)

4 D2 r2

(d) None of these

18. A canister has a small hole at its bottom. Water penetrates into the canister when its base is at a depth of 40 cm from the surface of water. If surface tension of water is 73.5 dyne/cm, find the radius of the hole. (a) 375 mm (c) 0.0375 mm

(b) 3.75 mm (d) zero

19. A piece of gold weights 50 g in air and 45 g in water. If there is a cavity inside the piece of gold, then find its volume [Density of gold = 19.3 g/cc]. (a) 2.4 cm3

(b) 2.4 m3

(c) 4 .2 m3

(d) 4.2 mm3

(b) 9 (d) 6

c

R

(a) rg ( a + b + c ) pR

2

(b) less than rg ( a + b + c ) pR2 (c) greater than rg ( a + b + c ) pR2 (d) 2rg ( a + b + c ) pR2

23. From a steel wire of density r is suspended a brass block of density r B. The extension of steel wire comes to l. If the brass block is now fully immersed in a liquid of density r L , the extension becomes l ¢. The ratio, l / l ¢ will

be

rB - r rL - r rB - rL (c) rB (a)

rL rB - rL rB (d) rB - rL (b)

24. A glass tube 80 cm long and open at both ends is half immersed in mercury. Then the top of the tube is closed and it is taken out of the mercury. A column of mercury 20 cm long then remains in the tube. The atmospheric pressure (in cm of Hg) is (a) 90 (c) 60

(b) 75 (d) 45

Properties of Liquids

467

25. Equal volumes of two immiscible liquids of densities

31. A vessel whose bottom has round holes with diameter

r and 2r are filled in a vessel as shown in figure. Two small holes are made at depth h/2 and 3 h/2 from the surface of lighter liquid. If v1 and v2 are the velocities of efflux at these two holes, then v1/ v2 is

of 1 mm is filled with water. Assuming that surface tension acts only at holes, then the maximum height to which the water can be filled in vessel without leakage is (Surface tension of water is 75 ´ 10-3 Nm –1 and g = 10 ms–2) (a) 3 cm (c) 3 mm

h v1 h

32. Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine flowing per second at one end is 4.0 ´ 10-3kg/s. What is the pressure difference between the two ends of the tube? (Density of glycerine = 13 . ´ 103 kg/m 3 and viscosity of glycerine = 083 . Pa-s).

v2

1 2 1 (c) 2

1 4 1 d) 2 2 (b)

(a)

26. Two capillaries of radii r1 and r2 , lengths l1 and l2

respectively are in series. A liquid of viscosity h is flowing through the combination under a pressure difference p. What is the rate of volume flow of liquid? (a)

pp 8h

æ l4 l ö ç 4 + 44 ÷ è r1 r2 ø

(d)

pp 8h

æ r14 r24 ö ç + ÷ l2 ø è l1

-1

(b)

l ö 8pp æ l1 ç 4 + 24 ÷ h è r1 r2 ø

(d)

pp æ l1 l ö ç 4 + 24 ÷ 8h è r1 r2 ø

-1

(b) 0.3 cm (d) 3 m

(a) 9.75 ´ 102 Pa (c) 5.75 ´ 102 Pa

33. Two soap bubbles of radii r1 and r2 equal to 4 cm and 5 cm respectively are touching each other over a common surface AB (shown in figure). Its radius will be

-1

A 4cm

(a) there is no change in the size of the bubbles (b) the two bubbles will become of equal size (c) A will become smaller and B will become larger (d) B will become smaller and A will become larger

28. A trough contains mercury to a depth of 3.6 cm. If some amount of mercury is poured in it then height of mercury in the trough will be (a) 3.6 cm (b) 7.2 cm (c) 6 cm (d) None of the above

(b) 4.5 cm (d) 20 cm

34. What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20° C) is . ´ 10-2 N/m? If an air bubble of the same 250 dimension were formed at a depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 101 . ´ 105 Pa.) (b) 2.06 ´ 105 Pa (d) 1.86 ´ 105 Pa

35. Speed of 2 cm radius ball in a viscous liquid is

little water in between them cannot be separated easily because of (b) pressure (d) viscosity

30. An aeroplane of mass 3 ´ 104 kg and total wing area

of 120 m2 is in a level flight at some height. The difference in pressure between the upper and lower surfaces of its wings in kilo pascal is ( g = 10 ms -2 ) (a) 2.5 (c) 10.0

(a) 4 cm (c) 5 cm

(a) 7.06 ´ 105 Pa (c) 1.06 ´ 105 Pa

29. Two pieces of glass plate one upon the other with a

(a) inertia (c) surface tension

5cm B

27. Two soap bubbles A and B are formed at the two open ends of a tube. The bubble A is smaller than bubble B. Valve and air can flow freely between the bubbles, then

(b) 6.75 ´ 102 Pa (d) 6.95 ´ 103 kPa

(b) 5.0 (d) 12.5

20 cms–1. Then the speed of 1 cm radius ball in the same liquid is (a) 7.06 ´ 105 Pa (c) 1.06 ´ 105 Pa

(b) 2.06 ´ 105 Pa (d) 1.86 ´ 105 Pa

36. The work done in increasing the size of a rectangular soap film with dimensions 8 cm ´ 3.75 cm to 10 cm ´ 6 cm is 2 ´ 10-4 J. The surface tension of the film in Nm–1 is (a) 1.65 × 10–2 (c) 6.6 × 10–2

(b) 3.3 × 10–2 (d) 8.25 ×10–2

468 JEE Main Physics 37. The glycerine of density 1.25 ´ 103 kmg –3 is flowing through a conical tube with end radii 0.1 m and 0.04 m respectively. The pressure difference across the ends is 10 Nm–2. The rate of flow of glycerine through the tube is (a) 6.4 ´ 10 –2 m3s –1 (c) 12.8 ´ 10

–2

water as shown in figure Coin h h

(b) 6.4 ´ 10 –4 m3s –1

3 –1

3

3 –1

(d) 12.8 ´ 10 m s

ms

38. A film of water is found between two straight parallel wires of length 10 cm each separated by 0.2 cm. If their separation is increased by 1 mm, while still maintaining their parallelism, how much work will have to be done? (surface tension of water is 7.2 ´ 10–2 Nm –1) (a) 7.22 ´ 10 –6 J

(b) 1.44 ´ 10 –5 J

(c) 2.88 ´ 10 –8 J

(d) 5.76 ´ 10 –5 J

39. Water flows through a vertical tube of variable cross-section. The area of cross-section at A and B are 6 and 3 mm2 respectively. If 12 cc of water enters per second through A, find the pressure difference p A - p B (g = 10 ms -2 ) The separation between cross-section at A and B is 100 cm. (a) 1.6 ´ 105 dyne cm–2 4

44. A wooden block with a coin placed on its top.floats in

–2

(c) 5.9 ´ 10 dyne cm

The distance I and h are shown in the figure. After some time the coin falls into the water. Then [NCERT Exemplar]

(a) I decreases (c) I increases

(b) h decreases (d) h increases

45. A spring balance reads w1 when a ball of mass m is

suspended from it. A weighing machine reads w2 when a beaker of liquid is kept on the pan of balance. When the ball is immersed in liquid, the spring balance reads w3 and the weighing machine reads w4 . The two balances are now so arranged that the suspended mass is inside the liquid in a beaker. Then (a) w3 > w1 (c) w3 < w1 and w 4 > w2

(b) 2.29 ´ 105 dyne cm–2 (d) 3.9 ´ 105 dyne cm–2

Comprehension Based Questions

40. A body of uniform cross-sectional area floats in a liquid of density thrice its value. The portion of exposed height will be (a) 2/3

(b) 5/6

(b) w 4 > w2 (d) w3 > w1 and w 4 < w2

(c) 1/6

(d) 9/10

More Than One Correct Option 41. Streamline flow is more likely for liquids with

Passage I Water of density r at a depth h behind the vertical face of dam whose cross-sectional length is l and cross-sectional area A. It exerts a horizontal resultant force on the dam tending to slide it along its foundation and a torque tending to overturn the dam about the point O.

[NCERT Exemplar ]

(a) high density (c) low density

(b) high viscosity (d) low viscosity

h

42. Two solid spheres A and B of equal volumes but of different densities A dA and dB are connected by a string. They are fully immersed in a fluid of density dF . They get arranged into B an equillibrium state of as shown in the figure with a tension in the string. The arrangement is possible only if (a) dA < dF (c) dA > dF

(b) dB > dF (d) dA + dB = 2dF

43. When an air bubble moves up from the bottom of a lake (a) its acceleration decreases and becomes zero (b) its acceleration increases and becomes constant (c) its velocity increases and becomes constant (d) its velocity decreases and becomes zero

O

46. The pressure energy per unit volume of the water dam is (a) Ahrg

(b) hrg

(c)

1 Ahrg 2

1 (d) rgh2 2

47. The height at which the resultant force would have to act to the same torque is (a)

h 6

(b)

h 3

(c)

h 2

(d)

2h 3

48. Horizontal force on the vertical face of the dam is 1 (a) rghl 2

1 (b) rgh2 l 2

(c) rglh

(d) rglh2

49. Pressure on the vertical face of the dam is (a) rgh

1 (b) rgh 2

(c) rgh2

1 (d) rgh2 2

Properties of Liquids 50. Torque about point O is 1 (a) rglh3 2

1 (b) rglh3 3

1 (c) rglh3 6

(d) rglh3

Passage II A plane is in level flight at a constant speed and each wing has an area of 25 m2. During flight the speed of the air is 216 kmh–1 over the lower wing surface and 252 kmh–1 over the upper wing surface of each wing of aeroplane. Take density of air = 1 kgm -3 and g = 10 ms–2.

51. The mass of the plane is (a) 25 kg

(b) 250 kg

(c) 1750 kg

(d) 3250 kg

52. If a plane is in level flight with a speed of 360 kmh–1 then the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface is (a) 13 %

(b) 9 %

(c) 6.5 %

(d) 4.5 %

53. Pressure difference on each wing of aeroplane is (a) 5

Nm–2

(b) 50

Nm–2

(c) 350

Nm–2

(d) 650

Nm–2

54. Percentage of velocity difference of the upper and lower surface of the wings of aeroplane is (a) 14.3 %

(b) 15.4 %

(c) 16.7 %

(d) 17.4 %

(b) 2500 N

(c) 17500 N

select the correct choices from the codes (a), (b), (c) and (d) given below (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 correct explanation of the Assertion (c) If Assertion is true but Reason is false (d) If Assertion is false but the Reason is true

56. Assertion When height of a tube is less than calculated height of liquid in the tube, the liquid does not overflow. Reason The meniscus of liquid at the top of the tube becomes flat.

57. Assertion The velocity of flow of a liquid is smaller where pressure is larger and vice-versa. Reason This is in accordance with Bernoulli’s theorem.

58. Assertion A hydrogen filled balloon stops rising after it has attained a certain height in the sky. Reason The atmospheric pressure decreases with height and becomes zero when maximum height is attained.

59. Assertion For the flow to be streamline, value of

55. The total upward force on the plane is (a) 250 N

469

(d) 32500 N

Assertion and Reason Direction Question No. 56 to 60 are Assertion-Reason type. Each of these contains two Statements: Statement I (Assertion), Statement II (Reason). Each of these questions also has four alternative choice, only one of which is correct. You have to

critical velocity should be as low as possible. Reason Once the actual velocity of flow of a liquid becomes greater than the critical velocity, the flow becomes turbulent.

60. Assertion A bubble comes from the bottom of a lake to the top. Reason Its radius increases.

Previous Years’ Questions 61. A body floats in a liquid contained in a beaker. If the whole system falls under gravity, then the upthrust [UP SEE 2009] on the body due to liquid is (a) equal to the weight of the body in air (b) equal to the weight of the body in liquid (c) zero (d) equal to the weight of the immersed part of the body

62. A cube made of material having a density of 3

–3

0.9 ´ 10 kgm floats between water and a liquid of density 0.7 ´ 103 kgm –3, which is immiscible with water. What part of the cube is immersed in water? [BVP Engg. 2008]

1 (a) 3

2 (b) 3

3 (c) 4

(d)

3 7

63. A body floats with one-third of its volume consider water and 3/4 of its volume outside another liquid. [BVP Engg. 2008] The density of other liquid is

(a)

9 g/cc 4

(b)

4 g/cc 9

(c)

8 g/cc 3

(d)

3 g/cc 8

64. Bernoulli’s theorem is a consequence of the law of conservation of (a) momentum (c) energy

[UP SEE 2008]

(b) mass (d) angular momentum

65. Two rain drops reach the earth with different terminal velocities having ratio 9 : 4. Then, the ratio [EAMCET 2008] of their volume is (a) 3 : 2 (c) 9 : 4

(b) 4 : 9 (d) 27 : 8

66. The surface tension of soap solution is 0.03 Nm–1. The work done in blowing to form a soap bubble of surface area 40 cm2, in joule is [EAMCET 2008] (a) 1.2 ´ 10 –4 (c) 12 ´ 10 –4

(b) 2.4 ´ 10 –4 (d) 24 ´ 10 –4

470 JEE Main Physics 67. A body weigh 50 g in air and 40 g in water. How much would it weight in a liquid of specific gravity 1.5? [Karnataka CET 2008]

(a) 65 g

(b) 45 g

(c) 30 g

(d) 35 g

68. When the temperature of water rises, the apparent weight of the wood will

Vg (r1 - r2 ) k Vgr1 (c) k

Vg(r1 - r2 ) k Vgr1 (d) k

(a)

(b)

73. A soap bubble is charged to a potential of 16 V. Its

[WB JEE 2008]

radius is, then doubled. The potential of the bubble now will be [BVP Engg. 2007] (a) 16 V (b) 8 V (c) 4 V (d) 2 V

69. The area of cross-section of one limb of an U-tube is

74. A frame made of metallic wire enclosing a surface

(a) increase (b) decrease (c) may increase or decrease (d) remain same twice that of the other. Both the limbs contains mercury at the same level. Water is poured in the wider tube so that mercury level in it goes down by 1 cm. The height of water column is (Density of water = 103 kgm–3, density of mercury = 13.6 ´ 103 kgm –3) [Kerala CET 2008] (a) 13.6 cm (b) 40.8 cm (c) 6.8 cm (d) 54.4 cm

70. A capillary tube (A) is dipped in water. Another identical tube (B) is dipped in a soap-water solution. Which of the following shows the relative nature of the liquid columns in the two tubes? [AIEEE 2008]

area A is covered with a soap film. If the area of the frame of metallic wire is reduced by 50%, the energy of the soap film will be changed by [UP SEE 2007] (a) 100 %

(b) 75 %

(c) 50 %

(d) 25 %

75. The cylindrical tube of a spray pump has a cross-section of 8 cm2, one end of which has 40 fine holes each of area 10–8m2. If the liquid flows inside the tube with a speed of 0.15 m min–1, the speed with which the liquid is ejected through the holes is [Karnataka CET 2007]

(a) 50 ms–1

(b) 5 ms–1

(c) 0.05 ms–1 (d) 0.5 ms–1

76. A boat at anchor is rocked by waves whose crests are A

100 m apart and velocity is 25 ms–1. The boat bounces [UP SEE 2006] up once in every

B

B A

(a)

(b)

(a) 2500 s

(b) 75 s

(c) 4 s

(d) 0.25 s

77. A tank is filled with water of density 1 g per cm3 and B

A

(c)

oil of density 0.9 g cm–3. The height of water layer is 100 cm and of the oil layer is 400 cm. If g = 980 cms–2, then the velocity of efflux from an opening in the bottom of the tank is [UP SEE 2006]

A

B

(d)

(a) 900 ´ 980 cms–1 (c) 92 ´ 980 cms–1

78. A body of density D1 and mass M is moving downward

71. A jar is filled with two non-mixing liquids 1 and 2 having densities r1 and r2 respectively. A solid ball, made of a material of density r 3 is dropped in the jar. It comes to equilibrium in the position shown in the figure. Which of the following is true for r1, r2 and r 3 ? (a) r1 < r3 < r2 (c) r1 > r3 > r2

in glycerine of density D2 . What is the viscous force [Orissa JEE 2006] acting on it?

Liquid 1

ρ1 ρ3

Liquid 2

ρ2

[AIEEE 2008]

(b) r3 < r1 < r2 (d) r1 < r2 < r3

æ D ö (a) Mg ç1 - 2 ÷ è D1 ø

æ D ö (b) Mg ç1 - 1 ÷ è D2 ø

(c) MgD1

(d) MgD2

79. A body of mass 120 kg and density 600 kgm–3 floats in water. What additional mass could be added to the body so that the body will just sink? [Orissa JEE 2006] (a) 20 kg

(b) 80 kg

(c) 100 kg

(d) 120 kg

80. A solid sphere of volume V and density r floats at the

72. A spherical solid ball of volume V is made of a material of density r1. It is falling through a liquid of

density r 2(r 2 > r1 ). Assume that the liquid applied a viscous force on the ball that is proportional to the square of its speeds v, i.e., Fviscous = - kv2 ( k > 0). The

terminal speed of the ball is

(b) 1000 ´ 980 cms–1 (d) 920 ´ 980 cms–1

[AIEEE 2008]

interface of two immiscible liquids of densities r1 and r2 respectively. If r1 < r < r2 , then the ratio of volume of the parts of the sphere in upper and lower liquids [Kerala CET 2006] is (a)

r - r1 r2 - r

(e)

r1 r2 r

(b)

r2 - r r - r1

(c)

r + r1 r + r2

(d)

r + r2 r + r1

Properties of Liquids 81. An incompressible fluid flows steadily through a cylindrical pipe which has radius 2R at a point A and radius R at a point B. Further along the flow of direction if the velocity at point A is v, its velocity at point B will be [BVP Engg. 2006] (a) v/4

(b) 2v

(c) 4v

(d) -

v 2

and the water rises to different height h in them, then we shall have constant [BVP Engg. 2006] (b) h/r

(c) hr2

(d) hr

reaching the surface, its volume becomes (take atmospheric pressure correspond upto 10 m of water) [BVP Engg. 2006]

(b) 4 times

(c) 8 times

(d) 10 times

84. Water rises to a height of 10 cm in a capillary tube and mercury falls to a depth of 3.42 cm in the same capillary tube. If the density of mercury is 13.6 kgm–3 and angle of contact is 135°. The ratio of surface tensions for water and mercury is (angle of contact for water and glass is 8°) [BVP Engg. 2006] (a) 1 : 0.5 (c) 1.5 : 1

(a) more than half filled of r C is less than 0.5 (b) more than half filled of r C is less than 1.0 (c) half filled of r C is less than 0.5 (d) less than half filled if r C is less than 0.5

from a radius of 3 cm to 5cm is nearly (surface tension [IIT JEE 2012] of soap solution = 0.03 Nm -1) (a) 4p mJ

(b) 0.2p mJ

85. Water is flowing through a pipe of constant cross-section. At some point the pipe becomes narrow and the cross-section is halved. The speed of water is [UP SEE 2005]

(a) reduced to zero (b) decreased by factor of 2 (c) increased by a factor of 2 (d) unchanged

A and surface energy E is given by E (a) T = A

(d) 0.4p mJ

(b) T = EA

T (c) E = A

[Orissa JEE 2011]

(d) T =

A E

91. The lower end of a glass capillary tube is dipped in water, water rises to height of 8 cm. The tube is then broken at a height of 6 cm. The height of water column and angle of contact will be [Orissa JEE 2010] 3 4 -1 1 (c) 4 cm, cos 2

3 4 -1 1 (d) 4 cm, cos 2 (b) 6 cm, cos -1

(a) 6 cm, sin -1

internal diameter 8 ´ 10-3 m. The water velocity as it leaves the tap is 0.4 ms -1. The diameter of the water stream at a distance 2 ´ 10-1 m below the tap is close to [AIEEE 2011] (a) 5 ´ 10 -3 (c) 9.6 ´ 10 -3

(b) 7.5 ´ 10 -3 (d) 3.6 ´ 10 -3

93. A uniform long tube is bent into a circle of radius R

[BVP Engg. 2005]

and it lies in a vertical plane. Two liquids of same volume but densities r and d. Fill half tube. The [WB JEE 2010] angle, q is

(a) maximum in solids (b) maximum in liquids (c) maximum in gases (d) same in solid, liquid and gas

δ

87. A thin liquid film formed between a U shaped wire and a light slider supports a weight of 1.5 ´ 10-2 N. The length of the slider is 30 cm and its weight negligible.The surface tension of the [AIEEE 2012] liquid film is (a) 0.0125 Nm-1 (c) 0.05 Nm-1

(c) 2p mJ

92. Water is flowing contineously from a tap having an

(b) 1 : 65 (d) 1 : 3

86. The force of cohesion is

[IIT JEE 2012]

90. The relation between surface tension T.Surface area

83. A bubble rises from bottom of a lake 90 m deep. On

(a) 18 times

water then the correct statement is that the shell is

89. Work done in increasing the size of a soap bubble

82. If we dip capillary tubes of different radii r in water (a) h/r2

471

R

R B

ρ

Film

w

(b) 0.1 Nm-1 (d) 0.025 Nm-1

88. A thin uniform cylindrical shell closed at both ends is partially filled with water . It is floating vertically in water in half-submerged slate. If r c is the relative density of the material of the shell with respect to

ær - dö (a) tan -1 ç ÷ è r + dø

ærö (b) tan -1 ç ÷ è dø

æ dö (c) tan -1 ç ÷ èrø

æ r + dö (d) tan -1 ç ÷ èr - dø

94. If the terminal speed of a sphere of gold (density

= 9.5 kg/ m 3 ) is 0.2 m/s in a viscous liquid (density = 1.5 kg/ m 3 ). Find the terminal speed of sphere of silver (density = 10.5 kg/m 3 ) of the same size in the same liquid [WB JEE 2010] (a) 0.133 m/s (c) 0.2 m/s

(b) 0.1 m/s (d) 0.4 m/s

Answers Round I 1. 11. 21. 31. 41. 51. 61. 71. 81. 91.

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

2. 12. 22. 32. 42. 52. 62. 72. 82. 92.

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

3. 13. 23. 33. 43. 53. 63. 73. 83. 93.

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

4. 14. 24. 34. 44. 54. 64. 74. 84. 94.

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

5. 15. 25. 35. 45. 55. 65. 75. 85. 95.

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

6. 16. 26. 36. 46. 56. 66. 76. 86. 96.

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

7. 17. 27. 37. 47. 57. 67. 77. 87. 97.

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

8. 18. 28. 38. 48. 58. 68. 78. 88. 98.

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

9. 19. 29. 39. 49. 59. 69. 79. 89.

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

10. 20. 30. 40. 50. 60. 70. 80. 90.

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

10. 20. 30. 40. 50. 60. 70. 80. 90.

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

Round II 1. 11. 21. 31. 41. 51. 61. 71. 81. 91.

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

2. 12. 22. 32. 42. 52. 62. 72. 82. 92.

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

3. 13. 23. 33. 43. 53. 63. 73. 83. 93.

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

4. 14. 24. 34. 44. 54. 64. 74. 84. 94.

(c) (c) (c) (c) (a,b) (b) (c) (c) (b) (b)

5. 15. 25. 35. 45. 55. 65. 75. 85.

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

6. 16. 26. 36. 46. 56. 66. 76. 86.

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

7. 17. 27. 37. 47. 57. 67. 77. 87.

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

8. 18. 28. 38. 48. 58. 68. 78. 88.

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

9. 19. 29. 39. 49. 59. 69. 79. 89.

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

the Guidance Round I M r

1. Volume of ice = , volume of water = Change in volume =

M s

M M æ1 1ö - =Mç - ÷ èr s ø r s

2. Given, mass of girl (m) = 50 kg Diameter of circular heel (2r) = 10 . cm \

Radius (r) = 0.5 cm = 5 ´ 10 -3 m

Area of circular heel ( A) = pr 2 = 3.14 ´ (5 ´ 10 -3) 2 m 2 = 78.50 ´ 10 -6 m 2 \ Pressure exerted on the horizontal floor F mg p= = A A 50 ´ 9.8 = 6.24 ´ 10 6 Pa = 78.50 ´ 10 -6

A = 4 pr 2 1/ 2 or r = (A / 4p ) 4 4 Volume V = pr3 = p ( A / 4p )3 / 2 = kA3 / 2 3 3 4p 1 where, ´ = k = constant 3 ( 4 p )3 / 2

3. Surface area,

Using Boyle’s law, we have p1V1 = p2V2 p V (10 + h) kA13./2 or p2 = 1 1 = V2 kA32/ 2 or

æA ö p2 = (10 + h) ç 1 ÷ è A2 ø

3/ 2

As

p2 = 10 of water, so 10 + h 10 = 8

or

80 = 10 + h or h = 70 m

Properties of Liquids Ax ´ 13.6 = (29.2 - x ) ´ 1 x = 2 cm

4. or Water

10 cm

Spirit

12.5 cm

Mercury

Height of water column h1 = 10.0 cm Density of water (r1) = 1 g/cm3 Height of spirit column (h 2) = 12.5 cm Density of spirit (r 2) = ? The mercury column in both arms of the U-tube are at same level, therefore pressure in both arms will be same.

473

\ Height of water column = (29.2 - 2) = 27.2 cm \Pressure of the liquids at the bottom = 27.2 cm of water column + 2 cm of Hg column 27.2 of Hg column + 2 cm of Hg column = 13.6 = 4 cm of Hg column Dp 8. As, Bulk modulus, B = -V0 Dv Dp DV = -V0 Þ B æ Dp ö V = V0 ç1 Þ ÷ è B ø -1

æ Dp ö æ Dp ö r = r 0 ç1 ÷ = p0 ç1 + ÷ ø è è B B ø

\Pressure exerted by water column = Pressure exerted by sprit column

\ Density,

\

where, Dp = p - p0 = hr 0 g Pressure difference between depth and surface of ocean. æ r gy ö ( As,h = y) r = r 0 ç1 + 0 ÷ \ è B ø

or

p1 = p2 h1 r1 g = h 2 r 2 g h1 r1 10 ´ 1 r2 = = = 0.80 g/cm3 h2 12.5 Density of spirit Density of water 0.80 = = 0.80 1

9. Let, b be width of the glass wall. When the tank is half filled

Specific gravity of spirit =

then the average force on the glass wall is F = average pressure ´ area éæ 4ö ù æ4 ö = ê ç ÷ rw g ú ´ ç ´ b÷ ø ëè 2 ø û è2

5. Force on the base of the vessel = pressure ´ area of the base = hrg ´ A = 0.4 ´ 900 ´ 10 ´ 2 ´ 10

When tank is filled up to height 4 m, then -3

F' = ( 4 rw g ) ( 4 ´ b) F¢ 4 ´ 4 = = 4 or F ¢ = 4 F F 2 ´2

= 7.2 N

6. When 15.0 cm of water is poured in each arm then, height of water column (h1) = 10 + 15 = 25 cm Height of spirit column (h 2) = 12.5 + 15 = 27.5 cm Density of water (rw ) = 1g/cm3 3

Density of mercury (r m) = 13.6 g/cm

Let in equilibrium, the difference in the level of mercury in both arms be h cm.

or

hr m g = h1rw g - h2r s g h1 rw - h2 r s h= rm =

= (1 + h). Since, the liquid exerts equal pressure in all direction at one level, hence the pressure at P = (H - h) rg .

11. Effective weight of solid of specific gravity 1 when immersed

Density of spirit (r s ) = 0.80 g/cm3

\

10. Depth of point P below the free surface of water in the vessel

25 ´ 1 - 27.5 ´ 0.80 13.6

= 0.221 cm Therefore, mercury will rise in the arm containing spirit by 0.221 cm.

7. Let, A be the area of cross-section of the cylindrical vessel and

in water will be zero.

12. Atmospheric pressure ( p) = 1013 . ´ 10 5 Pa Density of French wire (r) = 984 kg/m3 Let h be the height of the wine column for normal atmospheric pressure. For normal atmospheric pressure ( p) = hrg \

h=

p 1.013 ´ 10 5 = = 10.5 m 984 ´ 9.8 rg

13. Ice is lighter than water. When ice melts, the volume occupied by water is less than that of ice. Due to which the level of water goes, down.

14. Pressure at the bottom p = (h1d1 + h2 d 2) g

x cm be the height of mercury in vessel. The height of water in the vessel = (29.2 ´ x ) cm.

= [250 ´ 1 + 250 ´ 0.85] g = 250 [1.85] g

As per question

= 462.5 g dyne/cm

474 JEE Main Physics 15. Pull on the rope = effective weight

22. Here, mass of block = m = 1kg

= [1650 + (1500 ´ 0.2) - 1500 ´ 1.3] kgf

Volume of the block, V = 3.6 ´ 10 -4m3 Tension in the string, T = mg = mg - Vr water g

= 1650 +300 –1950 = 0 r 16. Fraction of volume immersed in the liquid, Vin = æç ö÷ V i. e. , It

\ Decrease in the tension of string

ès ø

depends upon the densities of the block and liquid so, there will be no change in it if system moves upward of downward with constant velocity i. e. , uniform acceleration.

T - T ¢ = mg - [mg - vr water g ] = Vr water g

= 3.6 ´ 10 -4 m3 ´ 10 3 kgm-3 ´ 10 ms-2 = 3.6 N

23. The upward thrust (i. e., buoyancy force) acts on the body and an equal and opposite force acts on the water so the weight will be the sum of the two = 600 + 40 = 640 g

17. Since, density of iron is more than that of marble, the volume of iron is less than that of marble for the given mass. The upthrust of water on iron will be less than that on marble. Due to which iron cube will weight more.

24. The speed of the body just before entering the liquid is v = 2gh. The buoyant force B of the lake (i. e. , upward thrust of liquid on the body) is greater than the weight of the body w, since s > r. If V is the volume of the body and a is the acceleration of the body inside the liquid, then

18. The effective weight of the block in liquid will become less than 2 kg due to buoyancy of liquid. As a result of which A will read less than 2 kg.

19.

B - w = ma

As, the body immersed in liquid has some effective weight acting downwards so the reading of B will be more than 5 kg. m Volume of cylinder = r æ mö Upthrust on cylinder = ç ÷ sg èrø

or or or

Since, the density of block is half than that of water, hence half of its volume is immersed in water.

(s - r) g = ra (s - r) g a= r

Using the relation, v 2 = u 2 + 2as, we have

From Newton’s third law, the downward force exerted by æ mö cylinder on the liquid is = ç ÷ sg èrø msg \ Increase in pressure = rA

20. Initially the position of wooden block is as shown in figure.

sVg - rVg = rVa

0 = ( 2gh) 2 - 2g or

s=

(s - r) s r

hr s -r

25. Mass of the cylinders = AL (r1 + r2). As cylinders float with length L /2 outside the water, therefore length of cylinder inside the water = 3 L /2. When cylinders are floating, then, weight of cylinder = weight of water displaced by cylinder.

w l 2

So,

l 2 l

or As

AL (r1 + r 2) g = A(3L / 2) ´ 1 ´ g r1 + r 2 = 3 / 2 r1 < r 2, so, r1 < 3 / 4

26. Let, v be the volume of ice-berg outside the sea water while (a)

floating. Therefore, volume of ice-berg inside the sea water = (V - v). As ice-berg is floating, so weight of ice-berg = weight of sea water displaced by ice-berg.

(b)

When weight, w is put on the block, the remaining half of the volume of block is immersed in water, figure (b). Therefore,

i. e. ,

w = additional upthrust + spring force kö l æ ælö = l ´ l ´ ´ 2r ´ g + k ç ÷ = l ç l 2rg + ÷ è è2ø 2 2ø

21. Given, size of the plate = 2m× 5m and We know that area of the plate A = 2 ´ 3 = 6 m2

= 10 3 ´ 9.8 ´ 6 ´ 5 = 294 ´ 10 3 N

or

1.1v = 1.1V - 0 / 9 V

or

v / V = 0.2/1.1 = 2 /11

27. As, solid is floating in liquid, so, weight of solid body = weight

Greatest and least depths of the plate are 6m and 4m. and depth of centre of the plate 6+ 4 x= = 5m 2 \Total thrust on the plate r = rw g A x

V ´ 0.9 ´ g = (V - v) ´ 1.1 ´ g

28.

of liquid displaced by immersed part of the body i. e. , VDg = v dg or v /V = D/d V Given, 6 g = ´ 10 3 ´ g 3 and

(6 + m) g = V ´ 10 3 ´ g

Dividing Eq. (ii) by Eq. (i), we get or

m = 18 - 6 = 12 kg

...(i) …(ii)

Properties of Liquids 29. Let, l be the length of the cylinder in water when it is in the vertical position and A be the cross-sectional area of the cylinder. As cylinder is floating So, weight of cylinder = upward thrust or mg = Alrg or m = Alr When the cylinder is tilted through an angle q, then length l of cylinder in water = cos q l Weight of water displaced = Arg cos q lArg Restoring force = \ - lArg cos q é 1 ù é 1 ù - 1ú = lArg ê - 1ú = mg ê cos cos q q ë û ë û

30. Let D1 be the inner diameter of the hemispherical bowl and D2 be the outer diameter of the bowl. As, bowl is just floating so 3 4 æ 1ö p ç ÷ ´ 1.2 ´ 10 3 3 è2ø 3 3 4 é æ 1ö æ D ö ù = p ê ç ÷ - ç 1 ÷ ú ´ (2 ´ 10 4) 3 êë è 2 ø è 2 ø úû 1.2 ´ 10 3 or = 1 - D13 2 ´ 10 4 1/ 3 1/ 3 æ18.8 ö æ 1.2 ö Þ D1 = ç1 ÷ ÷ =ç è 20 ø è 20 ø D1 = 0.98 m

On solving,

31. As, weight of metal disc = total upward force T

θ T

θ

r

= w + 2 prT cos q

32. Force on the ring due to surface tension of water = ( pD1 + pD2) S = mg mg 3.47 ´ 980 S= = p (D1 + D2) (22/7) ´ (8.5+8.7) = 72.07 dyne cm–1

= 1.01 ´ 10 5 + 3.10 ´ 10 2 = 1.01 ´ 10 5 + 0.00310 ´ 10 5 = 1.01310 ´ 10 5 Pa Excess pressure inside the drop ( Dp) =

2S 2 ´ 4.65 ´ 10 -1 = R 3 ´ 10 -3

= 310 . ´ 10 2 = 310 Pa

34. Let R be the radius of the biggest aluminium coin which will be supported on the surface of water due to surface tension. mg = S ´ 2 pR

Then, 2

pR t rg = S ´ 2 pR

or

R = 2 S /rgt

or

35. As volume remains constant i. e. ,R3 = 8000 r3 or R = 20 r Now,

Surface energy of one big drop Surface energy of 8000 small drops =

4 pR 2T 1 R2 (20 r) 2 = = = 2 2 8000 ´ 4pr T 8000 r 8000 r 2 20

36. As, surface energy = surface tension ´ surface area i. e. , E = S ´2 A New surface energy, E1 = S ´ 2 ( A /2) = S ´ A E - E1 % decrease in surface energy = ´ 100 E 2 SA - SA = ´ 100 = 50% 2 SA drop, then according to question, 4 3 4 pR = 10 6 ´ pr3 3 3 R or r= = 0.01R 100 = 0.01 ´ 10 –2 m = 10 –4 m \Work done = surface tension ´ increase in area = 35 ´ 10 –2 ´ [(10 6 ´ 4 p ´ (10 -4) 2 - 4p ´ (10 -3) 2] = 4.35 ´ 10 –2 J

38. As, work done = surface tension ´ surface area = 1.9 ´ 10 –2 ´ ( 4 pR 2) ´ 2

33. Given, radius of drop (R) = 3.00 mm = 3 ´ 10

2 ´ 4.65 ´ 10 -1 3 ´ 10 -3

37. If r is the radius of smaller droplet and R is the radius of bigger

= upthrust force + force due of surface tension = weight of displaced water + T cos q (2 pr)

So,

= 1.01 ´ 10 5 +

475

-3

= 1.9 ´ 10 –2 ´ 4 ´ p (1 ´ 10 -2) 2 ´ 2

m

Surface tension of mercury (S) = 4.65 ´ 10

-1

= 15.2 ´ 10 –6 pJ N/m

5

Atmospheric pressure (p0 ) = 1.01 ´ 10 Pa Pressure inside the drop = Atmospheric pressure + Excess inside the liquid drop 2S = p0 + R

39. When two drops are splitted, the law of conservation of mass is obeyed.

40. Work done in splotting a water drop of radius R into n drops of equal size = 4pR 2T (n1/3 - 1) = 4 p ´ (10 -3) 2 ´ 72 ´ 10 -3 ´ 10 6 /3 - 1 = 4 p ´ 10 -6 ´ 72 ´ 10 -3 ´ 99 = 8.95 ´ 10 –5 J

476 JEE Main Physics 41. Here, R = 2.8 /2 = 1.4 mm = 0.14 cm 4 3 4 (equality of volume) pR = 125 ´ pr3 3 3 R 0.14 or r= = = 0.028 cm 5 5 \Change in energy = surface tension ´ increase in area

50. Change in surface area = 2 ´ 4 p [(D /2) 2 - (d /2) 2] = 2 p (D 2 - d 2)

Now,

\Work done = surface tension ´ change in area = 2pS (D 2 - d 2) h 2

51. Pressure at half the depth = p0 + dg

= 75 ´ (125 ´ 4 ´ 4 pr 2 - 4 pR 2) = 74 erg

Pressure at the bottom = p0 + hdg

42. Given ; u = S ´ 4 pR 2; when droplet is splitted into

According to given condition, 2 h p0 + dg = ( p0 + hdg ) 2 3 3h 3 p0 + dg = 2p0 + 2 hdg Þ 2

1000 droplets each of radius r, then 4 3 4 pR = 1000 ´ pr3 or r = R /10 3 3 \ Surface energy of all droplets = S ´ 1000 ´ 4 pr 2 = S ´ 1000 ´ 4 p (R /10) 2

Þ

= 10 ( S 4pR 2) = 10 u

43. As, work done = surface tension × increase in area = 72 ´ [10 ´ 0.7 - 10 ´ 0.5] ´ 2 = 288 erg

be at the same height h and speeds of air on the upper and lower surfaces of the wings be v1 and v 2. Speed of air on the upper surface of the wing v1 = 70 m/s

surface area 4 4 Rö æ = S (1000 ´ 4 pr 2 - 4 pR 2) ç100 ´ pr3 = R 2 or r = ÷ è 3 3 10 ø æ ö R2 = S ´ 4 p ç1000 ´ - R 2÷ = 36 pR 2S 100 è ø

Speed of air on the lower surface of the wings v 2 = 63 m/s Density of the air r = 1.3 kg/m3 Area A = 2.5 m 2 According to Bernoulli’s theorem, 1 1 p1 + rv12 + rgh = p2 + rv 2 + rgh 2 2 1 2 or p2 - P1 = r(v1 - v 22) 2

45. Let R and R ¢ be the radius of bubble of volume V and 2 V respectively. Then 4 3 4 pR = V and pR ¢3 = 2 V 3 3 R ¢3 So, = 2 or R ¢ = (2)1/3R R3 As W = S ´ ( 4 pR 2) 2 W ¢ = S ´ ( 4 pR ¢2) 2

or

W ¢ R ¢2 = 2 = 2 2/3 = ( 4)1/3 W R W ¢ = ( 4)1/3 W

4 3 4 pR = 1000 ´ pr3 3 3 Þ R = 10 r Surface energy of small drop E1 = S ´ 4 pr 2

\Lifting force acting on the wings, 1 F = ( p2 - p1) ´ A = r(v12 - v 22) ´ A 2 Force ù é êëQ Pressure = Area úû 1 = ´ 1.3 ´ [( 70) 2 - (63) 2] ´ 2.5 2 1 = ´ 1.3 [4900 - 3969] ´ 2.5 2 1 = ´ 1.3 ´ 931 ´ 2.5 = 1.51 ´ 10 3 N 2

46. As,

Surface energy of large drop E 2 = S ´ 4 p (10 r) 2 \

E1 /E 2 = 1/100

47. As, work done = surface tension ´ increase in surface area = T (n 4pr 2 - 4 pR 2)

53. With the increase in temperature, the surface tension of liquid 54.

decreases and angle of contact also decreases. 2S cos q As, (height raised = h) h= r rg or

48. The meniscus of liquid in a capillary tube will be convex upwards if angle of contact is obtuse. It is so when one end of glass capillary tube is immersed in a trough of mercury.

49. As excess pressure, p µ1/ r, therefore, pressure inside C is highest and pressure inside B is lowest. The pressure inside A is in between. Therefore, C starts collapsing with volume of A and B increasing.

2 r0 2 ´ 10 5 = 20 m = 3 dg 10 ´ 10

52. Let the lower and upper surface of the wings of the aeroplane

44. Increase in surface energy = surface tension ´ increase in

and

h=

\

hr rg hr or S µ 2 cos q cos q h cos q2 r1 = 1´ ´ h2 cos q1 r 2

S= Sw S Hg

10 cos 135° 1 ´ ´ ( -3.42) cos 0° 13.6 10 0.707 1 = ´ = 3.42 13.6 6.5

=

Properties of Liquids 55. There will be no over flowing of liquid in a tube of insufficient height but there will be adjustment of the radius of curvature of meniscus so that hR = a finite constant.

56. When lift is accelerated downwards, the observed weight of body in a lift decreases. Hence, to counter balance the upward pull due to surface tension on the liquid meniscus, the height through which the liquid rises must increase.

57. For the given angular velocity of rotation, the centrifugal force F µ r; Therefore, more liquid will be accumulated near the wall of tube and the liquid meniscus will become concave upwards.

58. Due to surface tension, water rises in the capillary tube upto a height, h with concave meniscus on both the sides. Therefore, the total height of water column in the capillary tube = h + h = 2 h.

59. Height, h µ1/R h1 /h2 = R2 /R1 = 0.4 /0.2 = 2

So,

70. As, V =

ppr 4 p (3p + p) (r /2) 4 and V ¢ = 8 hl 8 hl V¢ 1 = 4 ´ (1 / 2) 4 = V 4 V V¢ = 4

\ or

71. Let at a time t dV be the decrease in volume of water in vessel in time dt. Therefore rate of decrease of water in vessel = rate of water flowing out of narrow tube dV p ( p1 - p2) r 4 = 8 hl dt

So,

p1 = p2 = hrg

But, \

dV p (hrg ) r 4 ( prgr 4) = = ´ (h ´ A) 8 hl 8 hl ´ A dt

where

h ´ A = volume of water in vessel at a time t = V

\

æ p rgr 4 ö dV = - ç ÷ ´ V dt = - lV dt è 8 hlA ø

or

dV = -l dt V

62. When a highly soluble salt (like sodium chloride) is dissolved in water, the surface tension of water increases.

63. Surface tension of water decreases with rise in temperature. 64. Since, the bubbles coalesce in vacuum and there is no change in temperature, hence its surface energy does not change. This means that the surface area remains unchanged. Hence, 4 p a2 + 4 p b 2 = 4 p R 2 R = a2 + b 2

or

65. The excess of pressure inside the first bubble of radius r1 is p1 = 4 S /r1; and in the second bubble of radius r2 is, p2 = 4 S /r2. 4S 4S 4S = \Excess pressure p = r r1 r2 r2 1 r2 - r1 Þ = r r1 r2 r=

Þ

r1 r2 r2 - r1

r1

Integrating it within the limits as time changes 0 to t, volume changes from V0 to V. V or log e = -lt V0 or V = V0 e- lt where, V0 = initial volume of water in vessel = Ah0 Therefore,

V=

4

p 1 prR pp (R /2) 1 or 1 = = 2 8 hL 8h (L /2) p2 4

68. The velocity gradient,

DV 8 = = 80 s–1 Dr 0.1

69. Using theorem of continuity, we have pDp2v p

=

pDQ2 vQ

ppr 4 ,i. e. ,V µ r 4 8 hl

Þ

V ¢ ( a /2) 4 1 = = V 16 a4

or

V¢ =

V 16 = = 1 cm3 16 16

73. As,

Q=

ppr 4 8 hl

and

Q1 =

pr (r /2) 4 Q = 8 h (2l) 32

74. Terminal velocity, v =

2

æ 4 ´ 10 -2 ö æD ö v p = ç Q ÷ vQ = ç ÷ ´ vQ = 4 vQ è DP ø è 2 ´ 10 -2 ø

h = h0 e- lt

Thus, the variation of h and t will be represented by exponential curve as given by (a).

67. Volume of liquid flowing per second through each of the two 4

h ´ A = h0 Ae- lt

or

66. As, Tc = T0(1- at), i. e. ,surface tension decreases with increase

tubes in series will be the same. So,

prgr 4 = l = constant 8 hlA

where,

72. As, V = in temperature.

477

or

2R 2 (r - r 0) g 9h

v 2 (r - r 0) g = = constant 9h R2

478 JEE Main Physics 75. Rate of flow of liquid V =

p R

83. Work done against air friction is the average gain in kinetic energy before attaining the terminal velocity 1 2 0 + mv ter 1 2 2 W1 = = mv ter 4 2 Work done against air friction after attaining terminal velocity is 1 2 W2 = mv max 2

8 hl pr 2 For another tube liquid resistance 8 hl 8 hl R¢ = = 4 ×16 = 16 R 4 pr r æ ö pç ÷ è2ø

where liquid resistance, R =

For series combination p p p V = = = Vnew = R + R ¢ R + 16 R 17 R 17

76.

2 r 2rg Given, v= 9h 4 3 4 Mass = pr r = p (2 r)3r1 3 3 or r1 = r /8 Terminal velocity of second ball is 2 (2 r) 2 (r /8) g v = v1 = 8h 2

\

W2 > W1

84. Gravitational force remains constant on the falling spherical …(i)

77. When a round pebble is dropped from the top of a tall cylinder, filled with viscous oil the pebble acquires terminal velocity (i.e., constant velocity) after some time.

ball. It is represented by straight line P. The viscous force (F = 6 phrv) increases as the velocity increases with time. Hence, it is represented by curve Q. Net force = gravitational force – viscous force. As viscous force increases, net force decreases and finally becomes zero. Then the body falls with a constant terminal velocity. It is thus represented by curve R.

85. t =

A a

2 [ H1 - H2 ] g T1 =

A a

and

T2 =

ù A T é H - 0ú ê a g ë h û

78. From Stokes’ law, F = 6 phrv = 6 ´ 3.14 ´ (18 ´ 10 –5) ´ 0.03 ´ 100

According to problem, T1 = T2

= 101.73 ´ 10 –4 dyne

\

4 3

4 3

79. As, M = pr3r and 8 M = pR3r, 3

R = 8r

So, Þ Now v µ r 2 so,

R = 2r

F = 2 ´ 10 -5 v =

4 3 pr rg 3

4 22 2 ´ 10 -5v = ´ ´ (1.5 ´ 10 –3)3 ´ 10 3 ´ 10 3 7

On solving,

v = 7.07 ms–1 » 7 ms–1

81. In a streamline flow the two streams cannot cross each other. 82. If v is the terminal velocity, then equation of force,

or

86. If the liquid is incompressible, then mass of liquid entering

Þ

AV1 = AV2 + 1.5 A. v A ´ 3 = A ´ 1.5 A. v v = 1m/s

v1 = 4 v

80. When terminal velocity v is reaching, then

or

H H -0 Þ H =2 Þn = 4 N n

M = M1 + M2 2

Þ

H-

through left end should be equal to mass of liquid coming out from the right end.

3

v1 æ 2 r ö =ç ÷ =4 v è r ø or

2 é Hù ê Hú g ë hû

Now,

xg - yg = 6 p hrv ( x - y) g v= r 6 ph vµ

( x - y) r

87. According to equation of continuity, av = constant. As v increases, a decreases.

88. When air stream is produced in between two suspended balls, the pressure there becomes less than the pressure on the opposite faces of the balls. Due to which the balls are pushed towards each other.

89. Along a streamline, the velocity of every fluid particle while crossing a given position is the same.

90. According to equation of continuity 2

a1v1 = a2v 2 or 1 2

æd ö v1 a2 pd 22 / 4 æ 3.75 ö 9 = = = ç 2÷ = ç ÷= . 2 è 2.50 ø 4 è d1 ø v 2 a1 pd1 / 4 1 2

91. As, p1 + rv12 = p2 + rv 22 or

p1 - p2 =

(from Bernoulli’s equation)

1 r (v 22 - v12) 2

Properties of Liquids 1 ´ 1.3 ´ (120 2 - 90 2) 2 = 4.095 ´ 10 3 Nm–2

479

95. When, air is blown in the horizontal tube, the pressure of air

=

decreases in the tube. Due to which the water will rise above the tube A.

Gross lift on the wing = ( p1 - p2) ´ area = 4.095 ´ 10 3 ´ 10 ´ 2 = 81.9 ´ 10 3 N

96. If h is the initial height of liquid in drum above the small opening, then velocity of efflux, v = 2 gh. As the water

92. Since, the tubes A and C are connected to a tube of same area of cross-section, and the liquid flowing there will have same velocity, hence, the height of liquid in A and C will be same. Since, tube B is connected to a tube of smaller area of cross-section, therefore the liquid is flowing faster in this tube and pressure there is less according to Bernoulli’s theorem.

drains out, h decreases, hence v decreases. This reduces the rate of drainage of water. Due to which, as the drainage continues, a longer time is required to drain out the same volume of water. So, clearly t1 < t 2 < 3.

97. As, velocity of efflux, v = 2 gh; Volume of liquid flowing out per sec = v ´ A = 2 gh ´ A

93. Fig. (a) is incorrect. From equation of continuity, the speed of liquid is larger at smaller area. According to Bernoulli’s theorem due to larger speed the pressure will be lower at smaller area and therefore, height of liquid column will also be at lesser height, while in Fig. (a) height of liquid column at narrow area in higher.

= 2 ´ 10 ´ 5 ´ (10 ´ 10 -4) = 10 -2 m3 s–1

98. As, Av = 2 Av ¢ or v ¢ = v /2 For a horizontal pipe, according to Bernoulli’s theorem p+

94. Vertical distance covered by water before striking ground = (H - h). Time taken is, t = 2 (H - g ) × g ; Horizontal velocity of water coming out of hole at P , u = 2 gh \ Horizontal range = ut = 2 gh ´ 2 (H - g )/ g = 2 h (H - h)

1 2 1 æv ö rv = p ¢ + r ç ÷ 2 2 è2ø

2

or

p¢ = p +

1 2 æ 1ö rv ç1 - ÷ è 4ø 2

Þ

p¢ = p +

3 2 rv 8

Round II 1. Here, v1 = 2g (h + x); v 2 = 2gx

2. As, net force = Averge pressure ´ Area - T ´ 2 R hö æ = ç p0 + rg ÷ (2 Rh) - T 2R è 2ø

x

= |2p0Rh + Rrgh 2 - 2 RT|

v2 h

3. If V is the volume of the body, its weight = V r1 g . Velocity v1

Let, a = area of cross-section of each hole r = density of the liquid The momentum of the liquid flowing out per second through lower hole = mass ´ velocity = av1 r ´ v1 = a r v12 The force exerted on the lower hole towards left

gained by body when it falls from a height h1 = 2gh1. The weight of liquid displaced by the body as body starts immersing into the liquid = V r 2 g . The net retarding force on the body when it starts going in the liquid, F = V (r 2 - r1) g é V (r 2 - r1) g ù F =ê \Retardation, a = ú V r1 ë V r1 û The time of immersion of the body is that time in which the velocity of the body becomes zero. Using the relation v = u + at , we have v = 0 , u = 2gh1 ,

= a r v12

a=

Similarly, the force exerted on the upper hole towards right = a r v 22

we have

ær -r ö 0 = 2gh1 = ç 2 1 ÷ g è r1 ø

or

t=

Net force on the tank, F = a r(v12 - v 22) = a r[2g (h + x) - 2gx] = 2argh Þ

F µh

ær -r ö v(r 2 - r1) g = -ç 2 1 ÷ g è r1 ø V r1

2h1 æ r1 ö ´ç ÷ è r 2 - r1 ø g

480 JEE Main Physics 4. Pressure on left end of horizontal tube, p1 = p0 + h1 rg Pressure on right end of horizontal tube, p2 = p0 + h2 rg As p1 > p2, so acceleration should be towards right hand side. If A is the area of cross-section of the tube in the horizontal portion of U-tube, then

Upthrust of liquid on rod = A l s g acting upwards through the mid-point of AD. For rotational equilibrium of rod, net torque about point A should be zero. So, L l (LA r g ) cos q = ( lA sg ) cos q 2 2 or

p1 A - p2 A = ( lAr) a or or

(h1 - h2)r g A = lAra g (h1 - h2) towards right a= l

5. In figure, total force on the ring due to surface tension of soap film = (2pb) ´ 2S sin q Mass of air entering per second the bubble = volume ´ density = ( Av)r = pb 2 ´ vr Momentum of air entering per sec, pb 2v r ´ v = p 2b 2v 2r The soap bubble will separate from the ring, when force of surface tension of ring equal to the force b or 2 pb ´ 2S ´ = pb 2v 2r R 4S or R= r v2

6. Let V0 , Vt = volume of the metal ball at 0° C and t° C

then height of liquid in each arm of U-tube =

(h1 + h2) h1 - h2 of the liquid has = 2 2 been transferred from left arm to right arm of U-tube æh -h ö = ç 1 2 ÷ Ar è 2 ø consider that a height, h1 -

where, A = area of cross-section of tube andr = density of liquid. æn -n ö The decrease in height of this liquid = ç 1 2 ÷ è 2 ø 2

æh -h ö Loss in potential energy of this liquid = ç 1 2 ÷ Arg è 2 ø The mass of the entire liquid in U-tube

W1 = W0 - V0 r 0 g

Using law of conservation of energy, we have

= V0(1 + g mt ) ´ = V0 r 0

r0 g (1 + g at )

(1 + g ml) g (1 + g a l)

Weight of the rod = ALr g acting vertically downwards at C. B

A

s σ θ?

v = (h1 - h2)

g 2(h1 + h2 + h)

m mg ´g = r h

where r ¢ is the density of water. Therefore, k = r /r ¢. Thus, the buoyant force acting upwards is mg /k whereas the weight mg of the stone acts vertically downwards. Therefore, the net force in the downward direction = mg - mg /k æ 1ö = mg ç1 - ÷ è kø If a is the acceleration of the sinking stone, then æ 1ö ma = mg ç1 - ÷ è kø

D

C

or

water displaced by it = r ¢Vg = r ¢ ´

A = area of cross-section of the rod.

L 2

2

1 æh -h ö (h1 + h2 + h)r Av 2 = ç 1 2 ÷ A r g è 2 ø 2

9. If m is the mass of the stone and V its volume, the weight of the

As g m < g a , hence upthrust at t° C is less than at 0° C. It means upthrust has been decreased with increase in temperature. Due to which W2 > W1. L As, AB = L , AC = ; AD = l (say) 2

h=

h1 + h1 . We may 2

W2 = Wt - Vt r t g

Upthrust at t ° C = Vtr t g

Let

1 s 2 r

8. When, there is equal level of liquid in two arms of U-tube,

r0 Vt = V0(1 + g mt ) and r t = (1 + g a l)

where,

sin q =

= (h1 + h2 + h)rA If this liquid moves with velocity v, then its 1 KE = (h1 + h2 + h)r Av 2 2

respectively, r 0 r 2 = density of alcohol at 0° C and t ° C respectively. Then

7.

or

l2 r = L2 s

or

æ 1ö a = g ç1 - ÷ è kø

Properties of Liquids 10. As we know according to equation of continuity, when

kx F

kx' F'

cross-section of duct decreases, the velocity of flow of liquid increases and in accordance with Bernoulli’s theorem, in a horizontal pipe, the place where speed of liquid is maximum, the value of pressure is minimum. Hence the 2 nd graph correctly represents the variation of pressure.

11. As area of cross-section is uniform therefore according to equation of continuity speed of liquid is same at all points i. e. , v A = vB But during motion of liquid from A to B the potential energy decreases. \According to Bernoulli’s theorem, 1 2 p v + gh + = constant 2 r p p gh + A = B Þ r r Þ

pB - pA = rgh

12. Volume of water in the vessel of base area A' and height h is V = A' h. Averege velocity of out flowing water when height of water changes from h to 0 is v=

2gh + 0 2gh = 2 2

\ V = Av t When vessel is filled to height 4 h, then volume in vessel = 4V = 4Avt = 4A

2gh ´t 2

If t is the time taken for the out flowing liquid and v1 is the averege velocity of out flowing liquid, then

a

w

w

When the vessel moves downwards with accleration a ( < g ) the effective downward acceleration = g - a. Now upthrust is reduced say it becomes F' F where F ¢ = ( g - a) g In figure, then w - kx¢ - F ¢ = ma wa æ g - aö or w - kx¢ - ç ÷F = è g ø g a wa or (w - F) - kx¢ + F = g g a wa or kx - kx¢ + F = g g a or x¢ = x + (F - w) gk Hence, the spring length will increase.

15. Let A be the circular area over which the liquid wets the plate and d be the distance between two plates. Mass of liquid drop, m = Adr. If S is the force of surface tension of water, then excess of pressure inside the liquid film in excess of atmospheric pressure is given by

4 V = Av1 t1 or

t1 =

4V 4 A 2gh ´ t ´ 2 = 2t = Av1 2 ´ A ´ 2g ´ 4h

13. In time Dt , momentum of water entering the hydrant p1 = (rLDt )v$j Momentum of water while leaving the hydrant in time Dt is p = (rLDt )v( - $i) 2

Change in momentum in time Dt is D p = p 2 - p1 = r L t v( - $i - $j) | D p| = r L D t v ( -1) 2 + ( -1) 2 = 2 r LD t v | Dp | Force exerted by water, F = = 2rLv Dt

14. Let k be the spring constant of spring and its gets compressed by length x in equilibrium position. Let m be the mass of the block and F be the upward thrust of water on block. When the block is at rest, w = kx + F or …(i) w - F = kx

481

d

p=

S S 2S = = r d /2 d

Force of attraction between the plates, 2S F= A d 2S 2 Sm ´ Ard = F= rd 2 rd 2 –6 2 ´ 0.07 ´ (80 ´ 10 ) Þ = 0.28 N 10 3 ´ ( 4 ´ 10 -8)

Fù é êë\ p = A úû

16. Let V be the volume of wooden ball. The mass of ball is m = V r. Upward acceleration, upward thrust - weight of ball a= mass of ball V r 0 g - Vrg (r 0 - r) g = = Vr r If v ¢ is the velocity of ball on reaching the surface after being released at depth h is 1/ 2 é æ p -rö ù v = 2as = ê2ç 0 ÷ ghú ë è r ø û

482 JEE Main Physics If h' is the vertical distance reached by ball above the surface of water, then v 2 2(r 0 - r) 1 h' = = gh ´ 2g r 2g ær -rö æ r0 ö =ç 0 ÷h = ç - 1÷h è r ø èr ø

17. The velocity of efflux = 2gh The rate of flow of liquid out of hole = Av = pr 2 2gh By using equation of continuity ( Av) container = ( Av) hole p

D2 v = pr 2 2gh 4 v=

4r 2 2gh D2

\Speed with which water level falls =

4r 2 2gh D2

18. As the water tries to enter the hole, it forms a liquid surface through the hole with its concave surface downward. Due to which it can withstand the pressure of the liquid upto which the canister is lowered 2s 2T \In equilibrium, = hr g Þ h = rrg r Putting the given values, we get 2 ´ 73.5 r= = 0.00375 cm 40 ´ 1 ´ 980

Also, the total mass of alloy m1 + m2 = 16.8 g Solving Eqs. (i) and (ii), we get

21.

m1 = 9.345 g and m2 = 7.455 g 4S Excess pressure inside the soap bubble = . So, the pressure r 4S inside the soap bubble = patm+ r From ideal gas equation, pV = nRT pAVA nA = pBVB nB æ 4Sö 4 3 ç8 + ÷ è rA ø 3 prA nA …(i) Þ = æ 4 S ö 4 pr 2 nB ç8 + ÷ 3 B è rB ø Substituting, we get, S = 0.04 N/m, rA = 2 cm, rB = 4 cm in Eq. (i) nA 1 = nB 6 nB \ =6 nA

22. When jar is placed in vacuum, the liquid level rises up to the top of jar. The force exerted by liquid on the base of jar = force due to vertical column of liquid of height ( a + b + c) + vertical downward. a

= 0.0375 mm

19. Let Vc is the volume of cavity and V is the actual volume of gold piece [excluding volume of cavity] 50 V= = 2.6 cm3 \ 19.3

F

Now, loss in wt. of gold in water = Thrust due to water Þ Þ Þ

c

R

50 g - 45 g = [V + Vc ] rw g

Component of thrust F acting on the portion BC of jar

5 = (2.6 + Vc ) ´ 1

= ( a + b + c) rg ´ pR 2 + F sin 60°

Vc = 2.4 cm3

= greater than ( a + b + c) r g ´ pR 2

20. Let m1g and m2g be the mass of Cu and Zn respectively in alloy,

23. Let V be the volume of the brass block weight of brass block = V rBg - V rLg . If A is the area of cross-section of steel wire, then V rBg L (V rBg - V rLg ) L Y= ´ = ´ A l l¢ l rB or = l¢ rB - rL

m Volume of Cu = 1 cc \ 8.9 m and Volume of Zn = 2 cc 7.1 m ù ém \Total volume of alloy = ê 1 + 2 ú cc 8.9 7.1 ë û Now, loss of wt. in water = thrust due to water m ù ém (16.8 - 14.7) g = ê 1 + 2 ú ´ 1g Þ ë 8.9 7.1û m m Þ 2.1 = 1 + 2 8.9 71 .

b

30° 60°

24. Let p0 = atmospheric pressure. Then, or ...(i)

or

p1V1 = p2V2 V p2 = p1 1 V2 æ 40 ö 2 p2 = p0 ç ÷ = p0 è 60 ø 3

Properties of Liquids Now, or or

p2 + (20 cm of Hg) = p0 2 p0 + (20 cm of Hg) = p0 3 p0 = 20 cm of Hg 3 p0 = 60 cm of Hg

32. Given, length of the tube ( l) = 1.5 m Radius of the tube (r) = 10 . cm = 1 ´ 10 -2 m Mass of glycerine flowing per second = 4 ´ 10 -3 kg/s Density of glycerine (r) = 1.3 ´ 10 3 kg/m3 Viscosity of glycerine (h) = 0.83 Pa-s m Volume of glycerine flowing per second (V ) = r

40 cm p1

60 cm

Mass ù é êëQ Density = Volume úû

p2

=

20 cm (a)

(b)

4 4 ´ 10 -3 3 m /s = ´ 10 -6 m3 /s 3 1.3 1.3 ´ 10

According to Poiseuille’s formula, the rate of flow of liquid through a tube

26. The rate of flow of liquid (V ) through capillary tube is æ pr 4 ö p pressure difference p pr V= = pç ÷= = 8hhl resistance è 8hl ø R

V=

2

where,

8hhl T= pr 4

When two tubes are in series,

p p é l1 l2 ù = = ê + ú é ù 8 h ë r14 r24 û 8 h l1 l2 + ê ú p ë r 4 r24 û p

p pr 4 8 hr

where, p is the pressure difference between the two ends of the tube. 8hrV or p= pr 4

total resistance, R = R1 + R2 p Rate of flow of liquid, V ' = R1 + R2

=

. ´ 4 ´ 10 -6 8 ´ 0.83 ´ 15 = 975.37 Pa . ´ (1 ´ 10 -2) 2 ´ 1.3 314

= 9.75 ´ 10 2 Pa -1

27. Excess of pressure inside the bubble, p = 4S / r. So, smaller is the radius r, the larger is the excess of pressure p. It means, the pressure of air is more in bubble A to bubble B.

28. Let A be the area of cross-section of through and r be the

33. Excess pressure or

4S 4S 4S = r1 r2 r

1 1 1 1 1 1 or r = 20 cm = - = - = r r1 r2 4 5 20

34. Given, surface tension of soap solution ( S) = 2.5 ´ 10 -2 N/m Density of soap solution (r) = 12 . ´ 10 3 kg/m3 Radius of soap bubble (r) = 5.00 mm = 5.0 ´ 10 -3 m

density of mercury. Initial mass of mercury in trough = A ´ 3.6 ´ r

Atmospheric pressure ( p0) = 101 . ´ 10 5 Pa

Final mass of mercury in trough = Ah' r = ( A ´ 3.6 ´ r) ´ 2

Excess pressure inside the soap bubble =

h ¢ = 7.2 cm

or

29. The force of surface tension pulls the plates towards each other.

30. In level flight of aeroplane, mg = pA or

31. As, or

=

483

mg 3 ´ 10 4 ´ 10 = = 2.5 kPa A 120 Pa

2S r 2S h= rrg

hr g =

2 ´ 75 ´ 10 -3 = 0.03 m = 3 cm = æ1 -3 ö 3 ç ´ 10 ÷ ´ 10 ´ 10 ø è2

=

4 ´ 2.5 ´ 10 -2 = 20 Pa 5.0 ´ 10 -3

Excess pressure inside the air bubble =

4S r

2S 2 ´ 2.5 ´ 10 -2 = R 5.0 ´ 10 -3 = 10 Pa

\ Pressure inside the air bubble = Atmospheric pressure + Pressure due to 40 cm of soap solution column + Excess pressure inside the bubble = (101 . ´ 10 5) + (0.40 ´ 12 . ´ 10 3 ´ 9.8) + 10 = (101 . ´ 10 5) + 4.704 ´ 10 3 + 10 = 101 . ´ 10 5 + 0.04704 ´ 10 5 + 0.00010 ´ 10 5 = 105714 . ´ 10 5 Pa = 106 . ´ 10 5 Pa

484 JEE Main Physics 35. Terminal velocity, v =

2r 2(r - r 0) g 9h

v µr

i. e. ,

2.

v1 r12 = v r2

Þ

2

36. Change in surface energy = 2 ´10 -4 J DA = 10 ´ 6 - 8 ´ 3.75 = 30 cm2 = 30 ´ 10 -4 m2 \ Work done, W = T ´ 2 ´ (change in area) Now, change in surface energy = Work done 2 ´ 10 -4 = T ´ 2 ´ 30 ´ 10 -4

Þ or

37. v = a1a2

46. Pressure energy per unit volume of the dam is equal to

2( p1 - p2) r(r14 - r24)

= 6.4 ´ 10 -4m3 s-1

38. As, work done = surface tension ´ increase in area W = surface tension ´ [0.10 ´ 0.006 - 0.10 ´ 0.005] ´ 2 = 7.2 ´ 10 -2 ´ 0.10 ´ 0.001 ´ 2 = 1.44 ´ 10 -5 J

and

V 12 ´ 10 -6 = 2 ms-1 = 200 cms-1 = A1 6 ´ 10 -6 V 12 ´ 10 -6 = 4 ms-1 = 400 cms-1 = v2 = A2 3 ´ 10 -6 r 2 2 (v 2 - v1 ) 2 1 = 1 ´ 1000(100) + (16 ´ 10 4 - 4 ´ 10 4) 2

Now, pA - pB = rg (h2 - h1) +

= 10 5 + 6 ´ 10 4 = 1.6 ´ 10 5 dyne cm-2

40. Let x be the portion of exposed height of the body of length l, area of cross-section A. As, the body is floating, so A l r g = A( l - x)3 rg or or or

l = 3l - 3x x = 2l/3 x 2 = l 3

41. Streamline flow more likely for liquids with high viscosity and low density.

water falls in water, upthrust on the block decreases. Due to it, l decreases as well as h decreases.

2( p1 - p2) r[( pr12) 2 - ( pr22) 2]

22 2 ´ 10 = ´ (0.1) 2 ´ (0.04) 2 7 (125 . ´ 10 3)[(0.1) 4 - (0.04) 4 ]

39. As, v1 =

44. When a coin placed on the top of a wooden box floating in

45. The effective weight of ball in liquid w3 becomes less then w1

= pr12 ´ pr22

Þ

upward thrust of water acting on bubble which is greater than the weight of air bubble. So, upward acceleration is maximum. As bubbles moves a viscous force act on it which increase with the increases in speed and finally a stage comes when upward thrust becomes equal to weight of bubble and viscous force. Then the bubble moves with constant velocity and zero acceleration.

T = 3.3 ´ 10 -2 Nm-1 2( p1 - p2) r( a12 - a22)

= pr12r22

capillary tube will change in such a way so that vertical component of the surface tension forces just balance the weight of liquid column.

43. An air bubble moves up from the bottom of take due to

æ 1ö = 20 ç ÷ = 5 cms-1 è2ø

As,

42. The angle of contact at the free liquid surface inside the

due to buoyancy of liquid. As, the ball immersed in liquid has some effective weight acting vertically downwards, so, w4 > w2. hydrostatic pressure (i. e. , pressure due to a column of liquid p) = hrg

47. Let, H be the height above O at which the total force F would have to act to produce the given torque. Then H ´F = t t or H= F rglh3 / 6 h H= = 2 3 (r glh / 2)

48. Consider a strip of dam of thickness, dy at a depth, y as shown in figure. Pressure at depth, y is p = r gy The force against the shaded strip in the figure is dF = p ´ l dy = r gyl dy h

Total force,

h æy2ö 1 F = ò r gy l dy = rgl ç ÷ = r glh 2 0 è 2 ø0 2

49. Pressure due ot water at the bottom end of vertical face = hrg Pressure due to water at the top end of vertical face = 0 \ Average pressure on the vertical face of dam hr g + 0 1 = = hr g 2 2

50. Torque of the force dF about an axis through O is dt = dF ´ (h - y) = (rgyldy) ´ (h - y) = rgyl (h - y) dy Total torque about the point O is h

t = ò rgly (h - y) dy 0

h

é hy 2 y3 ù rglh3 = r gl ê - ú = 3 û0 6 ë 2

Properties of Liquids 51. If, m is the mass of the aeroplane, then, mg = 32500 or

32500 32500 m= = = 3250 kg g 10

52. If v1 and v 2 are the speeds of air on the lower and upper surface of the wings of aeroplane and p1, p2 are the pressures there, then Assume difference, 1 p1 - p2 = r (v 22 - v12) 2 æv + v ö or Dp = r ç 2 1 ÷ (v 2 - v1) = rv av (v 2 - v1) è 2 ø Here, Þ

v av = 360 kmh

-1

= 100 ms

62. Let

= 0.65 = 6.5%

53. Here, A = 25 ´ 2 = 50 m2 v1 = 216 kmh -1 = 60 ms-1 v 2 = 252 kmh -1 = 70 ms-1 Pressure difference on each wing of aeroplane 1 Dp = p1 - p2 = r (v 22 - v12) 2 1 2 = ´ 1(70 - 60 2) = 650 Nm-2 2

54. Percentage of velocity difference on the upper and lower surface of the wings of aeroplane is v -v 70 - 60 10 = 2 1= = = 0.154 = 15.4% v av (70 + 60) / 2 15

55. Upward force on the aeroplane = Dp ´ A = 650 ´ 50 = 32500 N

56. It can be shown that R ´ h = constant, where R is radius of curvature of the meniscus of liquid in the tube. When height of tube is less, the meniscus becomes flat i. e. , R = ¥ . That is why liquid does not overflow.

l3 ´ 0.9 ´ 10 3 ´ g = ( l 2 ´ x) ´ 1000 g + l 2 ( l - x) ´ 0.7 ´ 10 3 g

or

floatation in water Vrg =

\

rl 8 = rw 3 8 8 r l = rw = g /cc 3 3

…(i)

…(ii)

(Q rw = 1g / cc)

64. According to the Bernoulli’s theorem, the total energy (pressure energy, potential energy and kinetic energy) of an incompressible and non-viscous fluid in steady flow through a pipe remains constant throughout the flow 1 r + rgh + rv 2 = constant i. e. , 2 So, it is clear that Bernoulli's theorem is a consequence of the law of conservation of energy.

65. As terminal velocity,

2 (r - s ) 2 r g 9 h v µ r2

v= i. e. , Þ

v1 æ r1 ö =ç ÷ v 2 è r2 ø

2

2

or

61. Upthrust is independent of all factors of the body such as its mass, size, density etc, except the volume of the body inside the fluid. Fraction of volume immersed in the liquid ærö Vin = ç ÷ V i.e., it depends upon the densities of the block ès ø and liquid. So, there will be no change in it if system moves upward or downward with constant velocity or some acceleration. Therefore, the upthrust on the body due to liquid is equal to the weight of the body in air.

2 V ´ rw g 3

2 (\ V is immersed in water of density r w ) 3 Similarly, in a liquid. 1 Vrg = Vr l g 4 From Eqs.(i) and (ii), we have 2 1 Vrw g = Vr l g Þ 3 4 rl æ 2 ö æ 4 ö Þ = ç ÷ ´ç ÷ rw è 3 ø è 1 ø

60. The shape of liquid drop is spherical due to surface tension of liquid.

l ´ 0.9 = x + ( l - x) ´ 0.7 x 2 0.3 = 0.2 l or = l 3

63. Let V be volume of body and r its density, then by law of

59. The moment when actual velocity of flow of liquid exceeds critical velocity, the flow becomes turbulent. Hence for the flow to be streamline, the limiting value of critical velocity should be as large as possible.

l = side of the tube

x = side of cube immersed in water, l - x = side of cube immersed in liquid. According to law of floatation,

-1

Dp v 2 - v1 mg / A 3250 ´ 10 / 50 = 2 = = 2 v av rv av rv av 1 ´ (100) 2

485

or

9 æ r1 ö =ç ÷ 4 è r2 ø r1 3 = r2 2 4 3 3 pr (Volume)1 3 1 æ r1 ö = =ç ÷ (Volume) 2 4 pr3 è r2 ø 2 3 3 27 æ3ö ç ÷ = è2ø 8

486 JEE Main Physics 66. As, work done = surface tension ´ change in area = 0.03 ´ 2 ´ ( 40 ´ 10 -4) J

72. When falling drop attains the terminal velocity v, then the acceleration of the drop is zero. In this situation mg = FB + FV

= 2.4 ´ 10 -4 J

Density of liquid r1 = specific gravity of liquid ´ density of water = 1.5 ´ 1 = 1.5 gcc-1 If V is the volume of the body, then weight of body in water = weight of body air- upthrust of liquid = 50 g - Vr1 g = (50 - 10 ´ 1.5) g = 35g

68. When a piece of wood is floating in water, then the weight of wood is balanced by the upthrust of water, i.e., weight of wood is equal to weight of the water displaced by the immersed part of the body. When temperature of water is raised, the wood will remain float in water. Due to it, the apparent weight of wood will remain same.

69. A and C be the initial levels of Hg in limb I and limb II of U-tube when mercury is in equilibrium state. Let, h be the height of water column added in limb I of U-tube. Due to it, the mercury level gets depressed to level B in limb I and gets raised to level D in limb II, I

II

v=

or

73. Potential on bubble, Þ Þ

1 q 4pe 0 r V r 1 Vµ \ 1 = 2 V2 r1 r 16 2 or V2 = 8 V = V2 1

mg

V=

74. As, surface energy = surface tension ´ surface area Þ

E = S ´ 2A æ Aö \New surface energy, E1 = S ´ 2 ç ÷ = S ´ A è2ø Now % decrease in surface energy, E - E1 2SA - SA ´ 100 = ´ 100% = 50% E 2SA a1v1 = a2v 2

C

Þ

æ 0.15 ö ( 40 ´ 10 -8) v1 = (8 ´ 10 -4) ´ ç ÷ è 60 ø

On solving,

1 cm E

B

Vg (r1 - r 2) k

75. Using equation of continuity, 2 cm

A

v

kv 2 = V g (r1 - r 2)

or

D h

FB

Vr1 g = Vr 2g + kv 2

or

67. Density of water, rw = 1gcc-1

FV

v1 = 5 ms–1

76. The distance between the two consecutive crests in transverse wave motion is called wavelength. The boat bounces up, i.e., it travels from crest to the consecutive crest along wave motion.

When, AB = 1cm and CD = x (say) Then, or

2a ´1 = a ´ x x = 2 cm

If point E of limb II, is in level with B of limb I, then or

hr 0 g = (2 + 1) r HG g 3 ´ 13.6 h = 3r HG / rw = = 40.8 cm 1

70. Surface tension of soap solution is less than that of water. 2S cos q , rrg

As,

h=

So,

h µS

Hence, the height of liquid raised in the capillary tube is less for soap solution and more for pure water, i. e., option (c) is correct.

71. Since, liquid 1 is over the liquid 2, so r1 < r2. If r3 is greater than r 2 or r3 < r 2, the ball would not have been partially inside liquid 2 but would have sunk totally. Therefore, r 2 < r3 and r1 < r3 Hence, r1 < r3 < r 2.

As, wavelength = distance between two consecutive crests So, l = 100 m Velocity of wave, v = 25 ms–1 Hence, time in one bounce of boat l 100 t= = =4s v 25

77. Pressure at the bottom of tank must equal pressure due to water of height, h. Let, dw and d o be the densities of water and oil, then the pressure at the bottom of the tank = hwdw g + hod o g Let this pressure be equivalent to pressure due to water of height h . Then hdw g = hw g + hod o g hd 400 ´ 0.9 h = hw + o o = 100 + dw 1 = 100 + 360 = 460 According to Torricelli’s theorem, v = 2gh = 2 ´ 980 ´ 460 = 920 ´ 980 cms–1

487

Properties of Liquids 78. Volume of the body, V = m / D1.

84. From the formula, rise of liquid in capillary tube is

Mass of the liquid displaced by body, mD2 m¢ = VD2 = D1

2S cos q rgr S cos q S cos q or h µ hµ r r h=

Þ

\ Viscous force = effective weight of the body = mg - m¢g = mg -

79. Volume of the body, V =

Angle of contact for water and glass q1 = 80° Angle of contact for mercury and glass q2 = 135° h1 S1 cos q1 ´ r 2 Hence, = h2 r1 ´ S 2 cos q2

æ D ö mD2 g = mg ç1 - 2 ÷ è D1 ø D1

120 = 0.2 m3 600

Weight of water displaced = 0.2 ´ 10 3 ´ g = 200 kg -wt Mass of water displaced = 20 kg

S1 h1 r1 cos q2 = S 2 h2 r 2 cos q2

or

Therefore, additional mass which can be added to body, so that body can just sink

(given)

=

= 200 –120 = 80 kg

0.1 1 cos135° 1 ´ ´ = 0.324 13.6 cos 8° 65

80. Let V be the total volume of a solid sphere, V1 is the volume of

85. From the equation of continuity, the amount of mass that

the part of the sphere immersed in a liquid of density r1 and V2 is the volume of the part of the sphere immersed in a liquid of density r 2 . Then V = V1 + V2 As sphere is floating

flows past any cross-section of a pipe has to be the same as the amount of mass that flows past any other cross-section,

So,

m1 = m2

i. e. ,

r1A1v1 = r 2A2v 2 A1

Vrg = V1 r1g + V2 r 2g (V1 + V2) rg = V1 r1g + V2 r 2g

or or

A2

V1(r - r1) g = V2(r 2 - r) g V1 r 2 - r = V2 r - r1

v1

and

the volume of liquid flowing per second at B Þ v ´ p (2R) 2 = v ¢ ´ pR 2

Þ

v¢ = 4 v

82. If the height h is the rise of liquid in capillary tube of radius r,

p1 = 90 + 10 = 100 m of water Let its volume be V1 Let its surface, the pressure will be, p2 = 10 m of water Now, on reaching the surface its volume be V2 Now according to Boyle's law p1V1 = p2V2 \

100 ´ V = 10 ´ V2 100 ´ V V2 = = 10 V 10

v 2 = 2v1

86. We know that the force of cohesion (attraction between molecules) is maximum in solids.

87. Here, 2TL = mg …(i)

Since, we dip capillary tube of different radii in water and water rises to different height in them, then equation becomes 2S cos q hr = = constant rg

83. Pressure on the bottom of the lake is given by

r1 = r 2 A A2 = 1 2

Given,

81. Since, the volume of liquid flowing per second at A is equal to

then expression of height is given by 2S cos q h= rgr

v2

. ´ 10 -2 1.5 15 mg = = 0.025 N / m = 2L 2 ´ 30 ´ 10 -2 600 V As, net force, mg + r c ´ Vl ´ g = 0 ´ r c ´ g 2 V0 m Þ Vl = 2 rc r Þ

88.

T=

c

V Vl < 0 2

So,

89. Here, W = T ´ DA (Q work = surface tension ´ area) = 0.03 [2 ´ 4p ´ (5 2 - 3 2) 10 -4 ] = 24p (16) ´ 10 -6 = 0.384 p ´ 10 -3 J = 0.4 p mJ

90. As,surface tension =

Surface energy Area

or T =

E A

v

488 JEE Main Physics 91. When a capillary tube is broken at height of 6 cm, the height of water column will be 6 cm 2S cos q h As h= = = constant rrg cos q 8 6 (from question) = \ cos 0° cos q 6 cos 0° 3 = Þ cos q = 8 4 3 q = cos-1 4

92. Given, diameter = 8 ´10 -3 m and

v1 = 0.4 m/ s

\

v 2 = v12 + 2gh 2

= (0.4) + 2 ´ 10 ´ 0.2 = 2 m/ s Now, A1v1 = A2v 2 Þ Þ

æ 8 ´ 10 pç 2 è

A

θ D

h1 h2

R θ

θ

h3

E C P

Þ

dgh1 + rgh2 = rgh3

or dgR (sin q + cos q) + rgR (1 - cos q) = rg (1 - sin q) R or

d (sin q + cos q) + r (1 - cos q) = r (1 - sin q) (r + d) sin q = (r - d) cos q

or

ær - dö tan q = ç ÷ è r + dø

or

ær - dö q = tan -1ç ÷ è r + dø

or

-3 ö 2

æd2ö ÷ ´ 0.4 = p ´ ç ÷ ´ 2 è 4ø ø d = 3.6 ´ 10 -3 m

93. Vertical height of the liquid in portion AC, h1 = DO + OE = R sin q + R cos q = R (sin q + cos q) Vertical height of the liquid in portion CP, h2 = R - R cos q = R (1 - cos q) Vertical height of the liquid in portion PB,

94. As, terminal speed, v = where,

2 r 2g (r - s ) 9 h

r = density of the substance s = density of the liquid

If h and r are constants then, v µ (r - s ) v 2 r Ag - r Li Þ = v1 r Au - r Li or

v 2 10.5 - 1.5 = v1 19.5 - 1.5 9 1 = 18 2 0.2 v = 0.1m/s v2 = 1 = 2 2 =

h3 = R - R sin q = R (1 - sin q) In equilibrium, the pressure due to liquid on the both sides must be equal at the lowest point P

B

or

Heat and 13

Kinetic Theory of Gases

JEE Main MILESTONE < < < < < < < <
F

And if g S = g L, F ¢ = F or

Similarly, if the temperature is decreased the length and hence, the time period gets decreased. A pendulum clock in this case runs fast and it gains the time.

or

l – laDq T¢ l¢ = = T l l 1 » 1 - aDq 2 1 æ T ¢ = T ç1 – aDqö÷ è ø 2 DT = T – T ¢ =

1 T a Dq 2

Here, DT is the temperature difference. However, at lower temperature scale reading will be more or true value will be less.

or

¢ < wapp wapp

and

¢ = wapp wapp

5. Effect of temperature on the time period of a

pendulum The time period of a simple pendulum is given by

Time lost in time t (by a pendulum clock whose actual time period is T and the changed time period at some higher temperature is T ¢ ) is DT ö Dt = æç ÷t è T¢ ø

true value = scale reading (1 + aDT )

æ ö 1 ç ÷ 1 g D + T è ø L

æ 1 + g S DT ö F¢ = F ç ÷ è 1 + g L DT ø

Now, if vice-versa.

l + laDq T¢ = = (1 + aDq)1/ 2 T l 1 T ¢ » T æç1 + aDqö÷ è ø 2 1 DT = T ¢ – T = TaDq 2

6. At some higher temperature a scale will expand and scale reading will be lesser than true values, so that

ö (VS + DVS ) æ 1 ×ç ÷ VS è 1 + g L DT ø

æ V + g S VS DT ö =ç S ÷ VS è ø or

Here, we put Dl = l aDq in place of l aDT , so as to avoid the confusion with change in time period. Thus,

and time gained in time t is the same, i.e., DT ö Dt = æç ÷t è T¢ ø

F µ VS r L F ¢ VS¢ r L¢ = × F VS r L =

T µ l

l + Dl T¢ l¢ = = T l l

F = upthrust = VS r L g

where

l g

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.

or

This expression can also be written as,

\

or

or

r r¢ = 1 + gDT

\

T = 2p

493

7. When a rod whose ends are rigidly fixed such as to prevent from expansion or contraction undergoes a change in temperature, thermal stresses are developed in the rod. This is because, if the temperature is increased, the rod has a tendency to expand but since, it is fixed at two ends, the rod exerts a force on supports.

494 JEE Main Physics l, α

Let q be the temperature at which the clock is correct. Time lost per day = 1/2 a(rise in temperature ) ´ 86400

Thermal strain = So,

Dl = a × DT l

thermal stress = (g ) (thermal strain) = YaDT

or

force on supports F = A (stress) = YA aDT

Here, Y = Young’s modulus of elasticity of the rod. F = YAaDT

Sample Problem 4 A surveyor’s 30 m steel tape is correct at a temperature of 20°C. The distance between two points, as measured by this tape on a day when the temperature is 35°C, is 26 m. What is the true distance between the points? (a steel = 1.2 ´ 10 –5 / ° C ) (a) 26.00476 m (c) 25.6658 m

(b) 27.00468 m (d) None of these

Interpret (a) Let temperature above the correct temperature be Þ

q = 35 - 20 = 15° C (Using the relation) Correct length = measured length (1+ aq)

True distance between the points = 26 (1 + 1.2 ´ 10–5 ´ 1.5) True distance = 26.00476 m Þ

Sample Problem 5 A steel ring of 3.000 inch inside diameter at 20°C is to be heated and slipped over a brass shaft measuring 3.002 inch in diameter at 20°C. To what temperature should the ring be heated? ( asteel = 1.1 ´ 10–5 °C-1) (a) 70.6°C (c) 80.6°C

(b) 75.6°C (d) 78.6°C

Interpret (b) Let qbe the temperature to which the ring must be heated. Final diameter of ring should be 3.002 inch. Þ Þ Þ

3.002 = 3 [1 + a ( q - 20)] 3.002 - 3 q= + 20 3a q = 75.6° C

Sample Problem 6 A pendulum clock loses 12 s a day, if the temperature is 40°C and goes fast by 4 s a day if the temperature is 20°C. Find the temperature at which the clock will show correct time and the coefficient of linear expansion of the metal of the pendulum clock. (a) 120°C, 1.85 ´ 10–5 ºC–1 (b) 28°C, 1.85 ´ 10–6 ºC–1 (c) 25 °C, 1.85 ´ 10–5 ºC–1 (d) 27°C, 1.85 ´ 10–6 ºC–1

Interpret (c) A pendulum clock keeps proper time at temperature q1 and if temperature is increased to q2 > ( q1), then due to linear expansion length of pendulum and hence its time period will increase Fractional change in time period DT 1 = a Dq T 2

Þ 12 = 1/2 a ( 40 - q) ´ 86400 Time gained per day = 1/2 a (drop in temperature) ´ 86400 1 4 = a ( q - 20) ´ 86400 2

…(i)

…(ii)

On adding Eqs. (i) and (ii), we get 32 = 86400 a ( 40 - 20) Þ a = 1.85 ´ 10 –5° C–1 On dividing Eqs. (i) and (ii), we get 12 ( q - 20) = 4 ( 40 - q) Þ q = 25° C Clock shows correct time at 25°C

13.5 Calorimetry Calorimetry means measurement of heat. When a body at higher temperature is brought in contact with another body at lower temperature, the heat lost by the hot body is equal to the heat gained by the colder body and provided no heat is allowed to escape to the surrounding. A device in which heat measurement can be made is called a calorimeter. 1 calorie is the quantity of heat required to raise the temperature of 1 g of water by 1°C.

Water Equivalent It is the quantity of water whose thermal capacity is same as the heat capacity of the body. It is denoted by W. W = ms = Heat capacity of the body

Principle of Calorimetry When two bodies at different temperatures are placed in contact with each other or mixed with each other (liquid-in-liquid, solid-in-liquid), the heat will pass from the body at higher temperature to the body at lower temperature until both bodies reach a common temperature. This state is called as thermal equilibrium. At this state, Heat lost by one body = Heat gained by the other body Let two bodies of masses m1 and m2, specific heats s1 and s 2 and at temperatures q1 and q2 are brought in contact with each other. Assuming q1 > q2, heat will flow from body 1 to body 2. If q is the common temperature of two bodies at the state of thermal equilibrium, then (assuming no heat is gained or lost from or to the surroundings)

Heat and Kinetic Theory of Gases Heat lost by body 1 = Heat gained by body 2 m1s1 (q1 - q) = m2s 2 (q - q2 )

(q2 < q < q1 )

495

measured in terms of calg–1 or kcalkg–1. It is given by Q = mL, where L is the latent heat.

Latent Heat of Fusion

13.6 Specific Heat The specific heat (s) of a substance is the quantity of heat in calorie required to raise the temperature of 1 g of that substance by 1°C. Its unit is cal g–1C–1. The heat lost by a body or gained from a body depends upon the difference in the temperature. The heat lost or gained by a body Q = msDq

It is the quantity of heat required to change the unit mass of a solid substance to the liquid state at its melting point. For ice, latent heat of fusion is 80 calg–1.

Latent Heat of Vaporisation It is the quantity of heat required to convert unit mass of a liquid to gaseous state at the boiling point of the liquid. For water, latent heat of vaporisation is 540 calg–1.

m = mass of the body, s = specific heat, Dq = rise or fall in the temperature of body. –1

–1

Specific heat for ice s ice = 0.5 cal g C

for water s water = 1 cal g–1C–1 for steam s steam = 0.47 cal g–1C–1

Heat Capacity The heat capacity of a body is the quantity of heat required by the body to raise its temperature by 1°C. It is also known as thermal capacity. Heat capacity = ms (mass ´ specific heat)

13.7 Phase Changes and Latent Heat Normally, matter exists in three states : solid, liquid and gas. The conversion of one of these states of matter to another is called the change of state. There are two common changes of states (i) The change of state from solid to liquid is called melting and from liquid to solid is called fusion. Both the solid and liquid states of the substance coexist in thermal equilibrium during the change states from solid to liquid. (ii) The change of state from liquid to gas (or vapour) is called vaporisation. The temperature at which the liquid and the vapour states of the substance coexist is called its boiling point. The change of state from solid state to vapour state without passing through the liquid state is called sublimation.

Latent Heat The latent heat is the amount of heat that has to be supplied to (or taken from) the body during the change of state while temperature remaining constant. It is

Sample Problem 7 Calculate the heat of fusion of ice from the following data for ice at 0ºC added to water. Mass of calorimeter = 60 g, mass of calorimeter + water = 460 g, mass of calorimeter + water + ice = 618 g, initial temperature of water = 38°C, temperature of the mixture = 5°C. The specific heat of calorimeter = 0.10 ca lg -1 °C -1. (a) 73.85 calg -1 (b) 78.35 calg -1 (c) 88.7 calg -1 (d) 84.3 calg -1

Interpret (b) Mass of water = 460 - 60 = 400 g Mass of ice = 618 - 460 = 158 g Heat lost by water = Heat gained by ice to melt + Heat gained by (water + calorimeter) to reach 5ºC Þ 400 ´ 1 ´ (38 - 5) = 158 ´ L + 158 ´ 1 ´ 5 + 60 ´ 0.1 ´ 5 (where L is the latent heat of fusion of ice) L = 78.35 calg –1

Þ

Sample Problem 8 What will be the temperature, when 150 g of ice at 0°C is mixed with 300 g of water at 50°C? Specific heat of water = 1ca lg -1° C -1. Latent heat of fusion of ice = 80 ca lg -1. (a) 6.0°C (c) 6.7°C

(b) 5.6°C (d) 17.6°C

Interpret (c) Let us assume that T > 0° C Heat lost by water = Heat gained by ice to melt + heat gained by water formed from ice 300 ´ 1 ´ (50 - T) = 150 ´ 80 + 150 ´ 1 ´ (T - 0) Þ

T = 6.7° C

Hence, our assumption that T > 0°C is correct. For example water at 1 atm 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 atm is 539 cal/g.

496 JEE Main Physics Figure shows how the temperature varies when we add heat continuously to a specimen of ice with an initial temperature below 0°C. Suppose we have taken 1 g of ice at –20° C specific heat of ice is 0.53 cal/g-°C. T(°C)

d

b

0

c

–20 a Q1

Q2

Q3

5 g of water at 30°C. Find the temperature of the mixture in equilibrium.

Interpret 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 or m1cwater | Dt1| = m2cwater | Dt 2| \ (10) (70 – t ) = 5 (t – 30) \ t = 36.67° C

e

100

Sample Problem 10 10 g of water at 70°C is mixed with

Q (cal)

Q4

In the figure

a to b Temperature of ice increases until it reaches its melting point 0°C. Q1 = mcice [0 – (–20)]

Sample Problem 11 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 Lv = 539 cal / g and c water = 1cal / g -° C )

Interpret Let Q be the heat required to convert 200 g of water

w= (1) (0.53) (20) = 10.6 cal

at 40°C into 100°C, then Q = mcDT

b to c Temperature remains constant until all the ice has

= (200) (1.0) (100 – 40) = 12,000 cal

melted. Q2 = mLf = (1) (80) = 80 cal

c to d Temperature of water again rises until it reaches

d to e Temperature is again constant until all the water is

Now, suppose m0 mass of steam converts into water to liberate this much amount of heat, then Q 12000 = = 22.26 g m0 = 539 L

transformed into the vapour phase.

Since, it is less than 30 g, the temperature of the mixture is 100°C

its boiling point 100°C. Q3 = mcwater [100 – 0] = (1) (1.0) (100) = 100 cal

Q4 = mLv = (1) (539) = 539 cal Thus, the net heat required to convert 1 g of ice at – 20° C into steam at 100°C is

Mass of steam in the mixture = 30 – 22.26 = 7.74 g and mass of water in the mixture = 200 + 22.26 = 222.26 g

Q = Q1 + Q2 + Q3 + Q4 = 729.6 cal

Sample Problem 9 How much heat is required to convert 8.0 g of ice at –15° C to steam at 100°C? (Given c ice = 0.53 cal/g-°C, Lf = 80 cal / g and Lv = 539 cal/g, and c water = 1cal/g-°C) ice

ice

water

0°C

–15°C Q1

water

0°C Q2

100°C Q3

13.8 Heat Transfer There are three different ways in which heat can be transferred; conduction, convection and radiation.

steam

Conduction

100°C

It is a process by which the heat is transferred in solid. In conduction, molecules vibrate about a fixed location and transfer the heat by collision.

Q4

Interpret Q1 = mcice (Tf – Ti ) = (8.0) (0.53) [0 – (–15)] = 63.6 cal Q 2 = mLf = (8) (80) = 640 cal Q3 = mcwater (Tf – Ti ) = (8.0) (1.0) [100 – 0] = 800 cal Q 4 = mLv = (8.0) (539) = 4312 cal \ Net heat required Q = Q1 + Q 2 + Q3 + Q 4 = 5815.6 cal

When a metallic rod is put in a flame, the other end of rod will soon be so hot that you cannot hold it by your hands. It means heat transfer take place by conduction from hot end of rod through its different parts of the other ends.

Thermal Conductivity In solids, heat is transferred through conduction. We will study conduction of heat through a solid bar.

Heat and Kinetic Theory of Gases Regarding conduction following points are worth noting (i) The amount of heat flowing in a rod of surface area A in time t is

DQ Dq = - KA Dt Dx

(ii) The ratio of thermal and electrical conductivities is the same for the metals at a particular temperature and is proportional to the absolute temperature of the metal. If T is the absolute temperature, then

K µT s

DT1 = DT2 = DT3 =¼ 1 1 1 1 = = + +¼ Rp R1 R2 R3

(vi) Heat current H=

dQ DT = dt R

H2

T1

T l1

T =

l2

T2

K1T1l2 + K2T2l1 K1l2 + K2l1

(iv) The thermoelectric conductivity or diffusivity is defined as the ratio of the coefficients of thermal conductivity. So, m Thermal capacity per unit volume = æç ö÷ c = rc, where r èV ø is the density of substance. K Diffusivity D = rc (v) The hindrance offered by a body to the flow of heat is called its thermal resistance. Temperature difference (DT ) R= Heat current ( H ) =

DT l = H KA

where l is length of rod, A the area and DT the temperature difference across its ends. If different rods are connected in series, then heat flowing per second is same. i. e.,

H1 = H2 = H3 =¼

\

Rs = R1 + R2 + R3 + ¼

l ö æ ç where R = ÷ è sAø

We find the following similarities in heat flow through a rod and current flow through a resistance. Current flow through a resistance

dQ dt

dq dt = rate of charge flow Electric current i =

= rate of heat flow H=

DT TD = R R

i =

R=

l KA

R=

K = thermal conductivity K2

K1

l ö æ ç where R = ÷ è KAø

Current flow through a resistance dq DV i= = dt R

Heat current H =

(iii) Let two rods of thermal conductivities K1, K2 lengths l1, l2 and cross-sectional area A are connected in series. In steady state the temperatures of ends of rod are T1 and T2 and the temperature of junction is T. Then H1

i. e.,

Heat flow through a conducting rod

K = constant sT

or

If different rods are connected in parallel, then temperature difference is same, i. e., \

Here, K = coefficient of heat conduction Dq = temperature gradient between faces of a rod Dx DQ In the above relation, negative sign is used to make Dt a positive quantity since, is negative.

497

DV PD = R R l sA

s = electrical conductivity

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 resistances in series and parallel. This analogy is of great importance in solving complicated problems of heat conduction. (vii) In series combination of rods of different materials, equivalent conductivity K1

K2

l1

K3

l2

l3

L

l1 + l2 + l3 l l l = 1 + 2 + 3 Ks K1 K2 K3 If lengths of rods are equal, then 1 1 1 + + K1 K2 K3 1 = 3 Ks (viii) In parallel combination of slabs of different materials, equivalent conductivity K 1 A 1 + K 2 A 2 + K 3 A3 Kp = A1 + A 2 + A3 If areas of slabs are equal, then

498 JEE Main Physics Which collection of the answer is correct ?

l A1 K1 A2 K2 A3 K3

(i) (a) 3 kW -1

(ii) 6W

(iii) 5 Cm-1

(iv) 8°C

(b) 5.9 kW -1

60 W

-50 Cm-1 87.5°C

(c) 15.9 kW -1

6.3 W

-50 Cm-1 67.5°C

(d) 15.9 kW -1

6.3 W

-50 Cm-1 87.5°C

Interpret (d) (i) Thermal resistance K1 + K2 + K3 3

Kp =

(ix) Ingen Hauz’s experiment If a number of identical rods state of different metals are coated with wax and one of their ends is put in boiling water, then in steady state the square of length of the bar over which wax melts is directly proportional to the thermal conductivity of the metal, i. e., K = constant l2 (x) When the atmospheric pressure falls below 0°C (say-T°C), the cold air above water extracts heat from the water. As a result, the water begins to freeze into the ice layers. Let at any time the thickness of ice is x and further layer of ice of thickness dx is formed in time dt. If r is density of ice and L be the latent heat of fusion, then

or

R=

l l = KA K ( pr 2)

R=

(2) ( 401) ( p ) (10 -2) 2

= 15.9 kW –1 (ii) Thermal current, H =

DT Dq 100 = = R R 15.9

H = 6.3 W 0 - 100 (iii) Temperature gradient = = -50 km–1 2 or

= -50° Cm–1 (iv) Let q°C, be the temperature at 25 cm from the hot end, then 100°C

°C

0°C

0.25 m 2.0 m

at – T°C Air

or

( q - 100) = ( temperature gradient) ´ (distance) q - 100 = ( -50) (0.25) q = 87.5° C

Ice

Sample Problem 13 Two metal cubes with 3 cm edges of copper and aluminium are arranged as shown in figure. Find

x

dx

at 0°C Water

at 4°C

time taken by ice to grow to a thickness x is rL x rL 2 x dx = x t= Kq ò 0 2 Kq If the thickness is from x1 to x2, then time taken rL t= ( x22 - x12 ) 2 KT Here, K = coefficient of thermal conductivity of ice.

Sample Problem 12 A copper rod 2m long has a circular cross-section of radius 1 cm. One end is kept at 100°C and the other at 0°C and the surface is insulated so that negligible heat is lost through the surface. Find (i) the thermal resistance of the bar (ii) the thermal current H dT and (iii) the temperature gradient dx (iv) the temperature 25 cm from the hot end. Thermal conductivity of copper is 401 Wm-1K -1

100°C

Al Cu

20°C

(i) the total thermal current from one reservoir to the other. (ii) the ratio of the thermal currents carried by the copper cube to that carried by the aluminium cube. Thermal conductivity of copper is 401 Wm-1K -1 and that of aluminium is 237 Wm-1K -1. (a) 0.08 kW -1, 1.75 (c) 0.25 kW -1, 1.32

(b) 0.01 kW -1, 1.05 (d) 0.02 kW -1, 1.02

Interpret (a) Thermal resistance of aluminium cube, R1 = or

R1 =

(3.0 ´ 10 –2) = 0.14 kW –1 (237) (3.0 ´ 10 –2) 2

and thermal resistance of copper cube R2 =

l , KA

l , KA

Heat and Kinetic Theory of Gases R2 =

or

(3.0 ´ 10 –2) = 0.08 kW –1 ( 401) (3.0 ´ 10 –2) 2

Interpret (a) Let q be the temperature of inner surface of box.

As these two resistances are in parallel, their equivalent resistance will be RR (0.14) (0.08) R= 1 2 = = 0.05 kW –1 R1 + R2 (0.14) (0.08) Temperature difference Thermal resistance (100 - 20) = = 1.6 ´ 10 3 W 0.05

\ Thermal current, H =

Heat transfer per second through A + Heat produced by source per second = Heat transfer per second through B æ dQ ö æ dQ ö Þ ç ÷ + 36 = ç ÷ è dt ø A è dt ø B

Sample Problem 14 Water is boiled in flat bottom kettle

KA (100 - q ) KA ( q - 4) + 36 = d d

Þ Þ

KA ( q - 4 - 100 + q ) = 36 ´ d

A 100°C

In parallel, thermal current distributes in inverse ratio of resistances HCu RAl R1 0.14 Hence, = = = = 1.75 HAl RCu R2 0.08

B 4°C Source

Now, d = 8 cm, A = 12 cm2,K = 0.5 cals–1° C–1cm–1

2

placed on a stove. The area of the bottom is 3000 cm and the thickness is 2 mm. If the amount of steam produced is 1 gmin -1, calculate the difference of temperature between the inner and outer surfaces of the bottom. K for the material of kettle is 0.5 cal°C -1s -1cm -1 (a) 2.1 ´ 10 –3 °C

(b) 3.1 ´ 10 –3 °C

(c) 1.2 ´ 10 –3 °C

(d) 2.5 ´ 10 –3 °C

Interpret (c) Mass of steam produced = dm = 1 gs–1 dt

60

dQ dm Heat transferred per second = =L dt dt dQ 1 = 540 ´ cal° C–1s–1cm–1 Þ dt 60

Þ Þ Þ Þ

q = temperature difference d = thickness = 2 m = 0.2 cm dQ K Aq = dt d dm K Aq L = dt d 0.5 ´ 3000 ´ q 9= 0.2

Þ Þ

2 q - 104 =

36 ´ 8 12 ´ 0.5

q = 76° C

Convection It is a process by which heat is transferred in fluids (liquids and gases). In convection, transfer of heat takes place by transport of matter (in form of motion of particles). When a liquid in a container is heated, the molecules at the lower layers are heated up and their densities decrease. As a result the molecules rise up and heavier ones come down and hence a continuous movement of molecules takes place giving rise to convection currents. In this manner the whole of liquid gets heated.

Radiation In radiation, heat is transferred from one body to other or to the surroundings even in the absence of any medium in the intervening space. Heat energy of the sun is transmitted to earth through radiations.

q = 1.2 ´ 10 –3° C

Sample Problem 15 A closed cubical box made of perfectly insulating material has walls of thickness 8 cm and the only way for the heat to enter or leave the box is through the solid cylindrical metallic plugs each of cross-sectional area 12 cm 2 and length 8 cm fixed in the opposite walls of the box as shown in figure. The outer surface A is kept at 100°C while the outer surface B of other plug is kept at 4°C. K if the material of the plugs is 0.5 cals-1 ° C -1cm -1. A source of energy generating 36 cals-1 is enclosed inside the box. Find the equilibrium of the inner surface of the box, assuming that it is same at all points on the inner surface. (a) 76°C

499

(b) 86°C

(c) 66°C

(d) 56°C

13.9 Heat Transfer through Radiation Radiation is only a mode of transfer of energy by transverse electromagnetic waves. While studying heat radiations (Radiant energy) we are concerned with thermal radiations which form the infrared region of electromagnetic waves. All bodies emit heat to the surroundings at all temperatures and at all times. When the temperature of a body remains constant, it emits as much heat to the surroundings as it gains from them. The body is then in a state of dynamic (thermal) equilibrium.

500 JEE Main Physics Absorption, Reflection and Transmission

where, Q represents the energy of thermal radiation.

When radiations are incident on a surface, then three things happen¾a part of the radiation is absorbed, some is reflected back, and remaining is transmitted.

Absorptivity or absorptive power, a =

Incident

Absorption

Reflected

Transmitted

Qincident = Qabsorbed + Qreflected + Qtransmitted

Qabsorbed Qincident

Qreflected Qincident Q Transmissivity, t = transmitted Qincident Qs Qv Qt + + = a+ r+t =1 Q Q Q Reflectivity,

r=

For a perfect black body,

a = 1, r = t = 0

For a perfect reflector,

a = t = 0, r = 1

For a perfect transmitter,

a = r = 0, t = 1

Some Common Terms and Points The thermal radiation emitted by a body comprises of all the wavelengths; intensities of radiation corresponding to different wavelengths are different.

Emissive power (e) For a given surface it is defined as the

Absorptive power (a) It is defined as the ratio of the radiant

It is the total amount of energy radiated by a body per second per unit area of surface. 1 DQ e= A Dt

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 (al ) 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 l. It may have different values for different wavelengths for a given surface. The spectral absorptive power al is related to absorptive power a through the relation ¥

radiant energy emitted per second per unit area of the surface.

Spectral emissive power (el ) It is emissive power for a particular wavelength l. Thus, ¥

e = ò el dl 0

Emissivity ( e ) Emissivity of a body at a given temperature is defined as the ratio of the total emissive power of the body (e) to the total emissive power of a perfect black body (E) at that temperature, e e= i. e. , E

a = ò al dl 0

Perfectly Black Body A body which can absorb all radiations of each wavelength at any temperature, which are incident on its and emits the full radiation spectrum on being heated is known as a black body. Such a body neither reflects nor transmits any part of the incident heat radiation and hence, appears black irrespective of the colour of the incident radiation. This implies that a perfectly black body has unit absorptance. Also, a perfectly black body when heated emits radiation of all possible wavelengths at that temperature.

P

Q

Cavity approximating an ideal black-body. Radiation entering the cavity has little chance of leaving before it is completely absorbed.

Heat and Kinetic Theory of Gases Materials like black velvet or lamp black come close to being ideal black bodies but the best practical realization of an ideal black body is a small hole leading into a cavity, as this absorbs 98% of the radiation incident on them.

Ferry’s black body Ferry suggested and designed a perfectly black body on the principle that any space which is almost wholly closed having a small hole is capable of emitting and absorbing full radiation spectrum. Ferry’s black body consists of a double walled hollow sphere having a small opening O on one side and a conical projection P just opposite to it. The inner wall of enclosure is painted with lamp black. The heat radiations entering the hollow sphere through O get completely absorbed due to multiple reflections. The chance of heat radiations getting and through O is reduced by conical projection and the lamp black coating. Therefore, all the radiations are absorbed completely and hence, the absorptance of the enclosure is 100% approximately.

501

If Q is the total energy radiated by the ordinary body, then Q e= = esT 4 A ´t Q = A esT 4t

Þ

Net Heat Loss from the Surface of a Body The rate at which a body radiates energy is determined by the temperature of the body and its surroundings. When a body is hotter than its surroundings, the rate of emission exceeds the rate of absorption; there is net loss of energy, and the body cools down, unless it is heated by some other means. When a body is cooler than its surroundings, the rate of absorption is greater than the rate of emission, and its temperature rises. At thermal equilibrium the two rates are equal. Walls

T1

Kirchhoff’s Law At any temperature and for particular wavelength, the ratio of the emissive power to the absorptive power of all the bodies is same and is equal to the emissive power of a perfectly black body. e (constant) i. e. , =E a Now,

E =1

So,

e=a

T2

Hence, for a body at a temperature of T1, surrounded by walls at a temperature T2 (as in figure), the net rate of loss (or gain) of energy per second by radiation is Hnet = AesT14 - AesT24

(for perfectly black body)

Kirchhoff’s law signifies that good absorbers are always good emitters.

Stefan’s Law The energy emitted per second per unit area of a black body (emissive power = 1) is proportional to the fourth power of the absolute temperature.

Hnet = Aes (T14 - T24 )

Newton’s Law of Cooling According to this law, if the temperature T of the body is not very different from that of the surroundings T0, then dT is proportional to the temperature rate of cooling dt difference between them. To prove it let us assume that T = T0 + DT

4

i. e. ,

E = sT

Here,

s = Stefan’s constant

So that

= 5.67 ´ 10–8 Jm–2s–1K –4

æ 4 DT ö » T04 ç1 + ÷ è T0 ø

For any other body, e = e sT 4 Here, e = emissivity of body (e = 1for a black body)

æ DT ö T 4 = (T0 + DT )4 = T04 ç1 + ÷ è T0 ø

\

(T 4 - T04 ) = 4 T03 (DT )

or

(T 4 - T04 ) µ DT

4

(from binomial expansion)

…(i) (T0 = constant)

502 JEE Main Physics Rate at which heat is emitted, dQ1 = seAT 4 dt

l

Rate at which radiation is absorbed, dQ2 = seAT04 dt Net rate of heat loss,

l

H = seA [ T 4 - T04]

[Q ee = a]

dT seA 4 =[ T - T04] dt ms

Rate of cooling,

where, m is mass of the body and s its specific heat capacity. Negative sign is there because temperature is falling with time. dT Now, = - K DT dt seA ´ 4 T03 where, K = ms Solving above equation, we get T ( t ) = T0 + (T1 - T0 ) e- kt where T1 is the temperature of body at t = 0.

Note Approximate solution for Newton’s law of cooling is T1 - T2 T +T = K éê 1 2 - T0 ùú , where t is the time in which temperature of t ë 2 û body changes fromT1 toT2 .

Wien’s Displacement Law According to this law, the wavelength (l m ) of maximum intensity of emission of black body radiation is inversely proportional to absolute temperature of the black body, i. e. , 1 lm µ T or

l mT = b constant

where b is Wien’s constant and has value 2.89 ´ 10–3 m- K and l m is the wavelength corresponding to maximum intensity (energy constant) of radiation emitted by body at temperature T. l

The thermal radiation emitted by a body at any temperature consists of all wavelengths from small to large values. The intensities of all wavelengths are different. Eλ

l

The figure above shows the experimental curves for radiation emitted by a black body versus wavelength for different temperatures. The most significant feature of the curves obtained is that they are universal i. e. , black-body radiation curves obtained depend only on the temperature and not on the shape, size or material of the black body. As the temperature of the body increases, the wavelength at which the spectral intensity (El ) is maximum shifts towards left.

Note 1. Diathermanous A surface or a medium which transmits most of the radiation (t = 1) is called diathermanous, also the substances, which allow heat radiation to pass through them are called diathermanous e.g., dry air, rock salt etc. 2. Adiathermanous A surface or a medium which does not transmit radiation at all (t = 0) is known as a opaque or adiathermanous medium. Moreover, the substances which absorb heat radiation and get themselves heated are called adiathermanous, e . g ., water, wood and solid.. 3. Solar constant The amount of heat received from the sun by one square centimetre area of a surface placed normally to the sun rays at mean distance of the earth from the sun is known as solar constant. It is denoted by S. 3

r S = æç ö÷ sT 4 èRø Here, r is the radius of the sun and R the mean distance of the earth from the centre of the sun. Value of solar constant is 1.937 calcm–2 min–1.

Sample Problem 16 The emissivity of tungsten is approximately 0.35. A tungsten sphere 1 cm in radius 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 the supports is neglected? s = 5.67 ´ 10 –8 SI units. (a) 2119.8 W (b) 2019.8 W (c) 2219.8 W (d) 1919.8 W

Interpret (b) Net heat lost by sphere per second Hnet = es A (T 4 - T04) where, T = temperature of sphere = 3000 K T0 = temperature of surrounding = 300 K A = 4 pr 2 = 4 p (0.01) 2

Area,

T1 > T2 > T3

To maintain constant temperature,

T1

Power input required = net heat loss from the surface

T2

Pinput = es A (T 4 - T04) T3

λm1 λm2 λm3

λ

= 0.35 ´ 5.67 ´ 10 –8 ´ 4 p (0.01) 2 ´ (3000 4 - 300 4) Pinput = 2019.8 W

Heat and Kinetic Theory of Gases Sample Problem 17 The rate at which the radiant energy reaches the surface of the earth from the sun is about 1.4 kWm -2. The distance from the earth to the sun is about 1.5 ´ 1011 m, and the radius of the sun is about 0.7 ´ 109 m. What is the rate of radiation of energy per unit area from the sun’s surface? (b) 6.43 ´ 10 6 Wm-2

(a) 6.43 ´ 10 7 Wm-2 7

(c) 5.43 ´ 10 Wm

(d) 6.43 ´ 10 -7 Wm-2

-2

Sample Problem 19 A body cools down from 60°C to 55°C in 30 s. Using Newton’s law of cooling, calculate the approximate time taken by same body to cool down from 55°C to 50°C. Assume that the temperature of surroundings is 45°C. (a) 41.28 s (c) 51.28 s q1 - q2 =K t

é q1 + q2 ù êë 2 - q0 úû 60 - 55 é 60 + 55 ù =K ê - 45ú 30 2 ë û

= 1.5 ´ 10 11 m

R = radius of the sun = 0.7 ´ 10 9 m

Let power of the sun, P = energy radiated from the surface of the sun per second. Hence, in every one second, P joule of energy are radiated from the surface of the sun and this energy passes through a big sphere of radius D centred at the sun. Hence, at the circumference of this big sphere (i. e. , near the surface of the earth), the energy crossing through a unit area per second =

P p = area of big sphere 4 pD2

P = 1.4 ´ 10 3 Wm–2 4 pD 2 diu ra sD

D

P = 4 p (1.5 ´ 10 11) 2 ´ 1.4 ´ 10 3 W

Þ Þ

P = 3.96 ´ 10

26

Similarly, for 2nd case, 55 - 50 =K t

…(i)

é 55 + 50 ù - 45ú êë 2 û

…(ii)

Dividing Eq. (i) by Eq. (ii), we get, t = 51.28 s

Sample Problem 20 A black body at 227°C radiates heat at a rate of 7 cal/cm 2s. At a temperature of 727°C, the rate of heat radiated in the same units will be (a) 112 (c) 101

(b) 105 (d) 89

Interpret (a) According to Stefan’s law E = sT 4

sphere o big f

Sun

(b) 55.28 s (d) 60.28 s

Interpret (c) According to Newton’s law of cooling

Interpret (a) Let D = distance from the sun to the earth Let

503

W

\

7 = s (227 + 273) 4 = s ´ (500) 4

and

x = s (727 + 273) 4 = s ´ (1000) 4

Hence,

x (1000) 4 = = 16 7 (500) 4

Þ

x = 16 ´ 7 = 112 cal/cm 2s

Sample Problem 21 Which of the following is vm -T graph for perfectly black body? n m is the frequency of radiations with maximum intensity and T is the absolute temperature.

Rate of radiation of energy per second per unit area of the sun’s surface is given by

νm(Hz)

B D

P P = area of big sphere 4 pR 2

C

= 6.43 ´ 10 7 Wm–2

Sample Problem 18 In the above problem, if the sun radiates as an ideal black body, what is the temperature of its surface? (a) 6803 K (c) 5803 K

(b) 5603 K (d) 5503 K

Interpret (c) If the sun is an ideal black body, e = 1 Þ

E = sT 4

Þ

æEö T=ç ÷ ès ø

1/ 4

æ 6.43 ´ 10 7 ö =ç ÷ è 5.67 ´ 10 –8 ø

(a) A

= 5803 K

(b) B

(c) C

(d) D

Interpret (c) According to Wein’s displacement law l mT = b = Wein’s constant If n m is the frequency corresponding to wavelength l m then æ C ö ç ÷T = b è nm ø or

1/ 4

A T(K)

O

nm =

C T b

i. e. , nm µ T \ n m - T graph is straight line shown by the curve C.

504 JEE Main Physics V/T

13.10 Ideal Gas or Perfect Gas

V/T m = constant p = constant

An ideal gas or perfect gas is that gas which strictly obeys the gas laws such as Boyle’s law, Charles, law, Gay-Lussac’s law etc.

Boyle’s Law

1/V

(d)

According to it for a given mass of ideal gas at constant temperature (called isothermal process), the volume of a gas is inversely proportional to its pressure i. e. , 1 (if m and T = constant) V µ p Graphical forms of such law are shown in figure m = constant T = constant

According to it for a given mass of an ideal gas at constant volume (called isochoric process), pressure of a gas is directly proportional to its absolute temperature i. e. , This is shown graphically p m = constant V = constant

V

p/T

(a)

T(in K)

m = constant T = constant

m = constant T = constant

m = constant V = constant

1/V

(b) 1/p

pV

(if m and V = constant)

p µT

m = constant T = constant

(a)

1/T

(e)

Gay-Lussac’s Law or Pressure Law

p

p

m = constant p = constant

(b)

p or T

Avogadro’s Law (c)

(d)

p or V

According to it at same temperature and pressure equal volumes of all the gases contain equal number of molecules, i. e. , N1 = N 2

1/V

pV m = constant T = constant

13.11 Equation of State of a Perfect Gas p

In practice, the gases do not obey the gas laws at all values of temperature and pressure. It is because of the intermolecular forces between the gas molecules.

(e)

Charles’ Law According to it for a given mass of an ideal gas at constant pressure (called isobaric process), volume of a gas is directly proportional to its absolute temperature i. e. , V µT

(if m and p = consant)

Graphical forms of such law are shown in figure V

1/T

V/T m = constant p = constant

m = constant p = constant

m = constant p = constant

An ideal gas is one whose molecules are free from intermolecular attraction and obeys gas laws at all values of temperature and pressure. Ideal gas equation is a form of combined effect of above first four laws. Thus, the equation is given by m pV = nRT = RT M m Here, n = number of moles of the gas = M m = total mass of the gas M = molecular mass

(a)

T(in K)

(b)

V or T

(c)

V

R = universal gas constant = 8.31 Jmol–1 K –1 = 2.0 cal mol–1K –1

Heat and Kinetic Theory of Gases The above first four laws can be obtained by this ideal gas equation. For example, for a given mass of a gas pV = constant (at constant temperature) V = constant (at constant pressure) T

Þ

(Boyle’s law)

Eq. (ii) can now be rewritten as T p

…(vi)

Putting this value is Eq. (i), we get pT n = constant pn T

Þ

n-1 p n

…(vii) …(viii)

= constant

Work done by a gas which is compressing from state 1 to state 2 is given by 2

1

…(ix)

p dV

From Eq. (i), we have p=

For the purpose of calculations, it is convenient to place the ideal gas in the form pf V f pV i i = Ti Tf

If the temperature is constrained to be constant, this becomes

C Vn

…(x)

Putting this value in equation (ix), we get 2

W = - ò V - n dV On integration it leads to

…(xii) 1

Using equation pV n = C , we have æ V 1- n ö W = - ç pV n ÷ 1- n ø è

which is referred to as Boyle’s law.

which is referred to as Charles’ law.

2

æ V - n+1 ö W =C ç ÷ è - n + 1ø

pV i i = pf V f If the pressure is constant, then the ideal gas law takes the form Tf Vi V f or V f = Vi = Ti T f Ti

…(xi)

1

where the subscripts i and f refer to the initial and final states of some process.

æ pV ö W = -ç ÷ è 1- n ø Þ

W =-

2

…(xiii) 1

n

…(xiv) 1

( p2V2 - p1V1 ) 1- n

…(xv)

Ideal gas law also follows the equation

Work Done on Compressing a Gas The expansion and compression of ideal gases follow the expression.

…(xvi)

p2V2 = mRT2

…(xvii)

W =-

mR (T2 - T1 ) 1- n

…(xviii)

Note Similar expressions are obtained by similar methods for work done during expansion of gas but starting from

We obtain,

…(iii)

p1V1 = mRT1 Work done now becomes

…(i)

where n is number of moles of the gas. Ideal gases also follow the combined gas law pV …(ii) = constant T pV , we get T pV pV n ¸ = constant T

…(v)

W = -ò

13.12 Ideal Gas Law with Constraints

Dividing Eq. (i) by

TV n-1 = constant

(Charles’ law)

and if p, V and T are constants then n = constant for all gases. Since, equal number of moles contain equal number of molecules, thus at constants p, V and T all gases will contain equal number of molecules which is nothing but Avogadro’s law.

pV = constant

…(iv)

V = constant ×

p = constant (at constant volume) T (Gay-Lussac’s law)

n

T = constant pV

pV n ×

Þ

505

2

W=

ò1 pdV

W=

p 2V 2 - p1V1 n -1

W=

mR (T2 - T1) n -1

506 JEE Main Physics Sample Problem 22 Two moles of an ideal gas is

Sample Problem 23 A closed container of volume 0.02 m3

contained in a cylinder fitted with a frictionless movable piston, exposed to the atmosphere, at an initial temperature T0. The gas is slowly heated so that its volume becomes four times the initial value. The work done by the gas is

contains a mixture of neon and argon gases at a temperature of 27°C and pressure of1 ´ 10 5 Nm 2. The total mass of the mixture is 28 g. If the gram molar weights of neon and argon are 20 and 40 respectively, find the masses of the individual gases in the container, assuming them to be ideal. (R = 8.31 Jmol -1K -1).

(a) zero (c) 4 RT0

(b) 2 RT0 (d) 6 RT0

Interpret (d) Given that gas is slowly heated, which means it remains in equilibrium (more or less) with the atmosphere, i. e. , the process takes place at constant pressure.

Piston Gas

From the equation of ideal gas law pV = nRT For infinitesimal change pdV = nR dT or pDV = nR DT Also, pDV = Work done by the gas = DW \ Also \ Given,

DW = nRDT DV µ DT DT µ DV µ V2 - V1 V2 = 4 V1

\ DT µ 4 V1 - V1 µ 3 V1 µ 3 T0 Also given m = 2 moles The expression for work done becomes DW = nRDT DW = 2 R 3T0 = 6 RT0

Note 1. STP or NTP refers to standard (normal) temperature of 273K and 1 atm pressure of 1.01 × 105 Pa. 2. Whatever be the process, in equilibrium state, an ideal gas satisfies the equation pV = nRT. 3. In terms of density, the ideal gas equation may be expressed as p = constant rT 4. In terms of number of molecules (n) per unit volume of a gas, the ideal gas equation may be expressed as p = nkT 5. If n1 mole of a non-reactive gas in thermodynamical state ( p1, V1, T1) be mixed with n2 mole of another non-reactive gas at ( p2 , V2 , T2 ) and the resultant gas mixture is at a state ( p, V , T ) then p1V1 p2V2 pV + = T1 T2 T

(a) 24 g (c) 26 g

(b) 25 g (d) 27 g

Interpret (a) Let in the given container mass of neon be m and mass of argon be (28 - m) g, so that m nNe = 20 28 - m and nA = 40 m (28 - m) 28 + m n = nNe + nA = + = 20 40 40 and using ideal gas equation for the mixture, we have pV 1 ´ 10 5 ´ 0.02 = = 0.8 RT 8.314 ´ 300 Comparing Eqs.(i) and (ii), we get 28 + m = 0.8 40 n=

Þ \ and

…(i)

...(ii)

m= 4g mNe = 4 g mA = 28 - 4 = 24 g

Sample Problem 24 During an experiment, an ideal gas is found to obey an additional law Vp 2 = constant. The gas is initially at temperature T and volume V. What will be the temperature of the gas when it expands to a volume 2V? (a) 3 T

(b) 1/2 T

(c) 2 T

(d) 3 T

Interpret (c) Here it is given that Vp2 = constant K (say). Hence, we may write the gas equation as, pV = nRT or or

K × V = nRT V nR T V = K

Þ

V1 T1 = V2 T2

\

T2 = T1 =T

V2 V1 2V = 2T V

Heat and Kinetic Theory of Gases

13.13 Kinetic Theory of Gases

vrms

The kinetic theory of gases correlates the macroscopic properties of gases e. g. , pressure, temperature etc., to the microscopic properties of gas molecules e. g. , speed, momentum, kinetic energy of molecules etc. The kinetic theory of gases is based on the following assumptions

1/ 2

y

v

(i) A gas consists of a large number of tiny, identical, spherical and electrically neutral, stable elastic particles called molecules.

m

d

vx

(ii) The space occupied by the molecules of a gas is extremely small as compared to the volume of the gas. (iii) The molecules of a gas are in a state of continuous, random motion with all possible speeds ranging from zero to infinity in different possible directions. The speed distribution is in accordance with Maxwell’s distribution law of molecular speeds and has been shown in figure.

é v2 + v22 + ¼+ vN2 ù =v =ê 1 ú N û ë

507

z

d

x

d

A cubical box with sides of length d containing an ideal gas. The molecule shown moves with velocity v.

Thus, pressure exerted by a gas p = where r =

1 mN 2 1 2 v = rv , 3 V 3

mN = density of given gas. V

Number of molecules (n)

T1

Now T2 > T1

or Also

(vmp)

Molecular speed (v)

(iv) Each molecule behaves as an independent entity. There is no force of attraction among the molecules. Thus, gas molecules have no potential energy but possess only kinetic energy which is directly proportional to temperature of the gas. (v) The pressure of a gas is due to elastic collision of gas molecules with the walls of the container. (vi) The dynamics of the particles is governed by Newton’s laws of motion. (vii) The time of contact of a moving molecule with the container walls at the time of collision is negligible as compared to the time between two successive collisions with the same wall of the container.

Concept of Pressure On the basis of these assumptions we can do mathematical calculations to find expression for pressure 1 exerted by a gas. Accordingly, we find that pV = mN (v ) 2 3 where, m = mass of 1 gas molecule and N = total number of gas molecules, v is as root mean square velocity

1 æNö 2 2 N æ1 2ö mç ÷v = ç mv ÷ ø 3 èV ø 3 V è2 2 pV = N KE 3 1 2 2 æ 1 2ö p = rv = ç rv ÷ ø 3 3 è2 p=

1 2 rv = average kinetic energy of the gas per unit 2 volume. 2 p = ´ average kinetic energy per unit volume \ 3 2 p= E 3 Now,

Kinetic Energy and Temperature According to kinetic theory of gases, pV =

1 mNv 2 3

but according to equation of state for an ideal gas pV = nRT 1 1 3 RT 3 mN v 2 = nRT to mv 2 = = kT , 3 3 2 n 2 where k is the Boltzmann’s constant. Its value is 1.38 ´ 10–23 J mol–1 K –1. \ Mean translational kinetic energy of a gas molecule 3 = kT i. e. , the mean translational kinetic energy of a gas 2 molecule depends only on its temperature and is independent of its nature or mass etc.

508 JEE Main Physics On this basis, we can define absolute zero temperature as the temperature at which translational kinetic energy of a gas molecule becomes zero i. e. , at which the molecular motion ceases altogether.

Critical temperature, pressure and volume Gases cannot 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 Waals’ constants a and b are as under a Vc = 3 b, pc = 27 b2 8a Tc = 27 Rb

and Further,

for all gases.

RMS Speed of Gas Molecules Root mean square (rms) speed. It is defined as the square root of the mean of squares of the speeds of different molecules i. e. , vrms = v = (v12 + v22 + ¼+ vN2 ) /N According to kinetic theory of gases it is observed that 3p 3 pV vrms = v = = M r

(a) 7330 K (c) 7530 K

(b) 7730 K (d) 7430 K

Interpret (b) Kinetic energy gained by an electron when accelerated by a potential difference of 1 V is1 eV = 1.6 ´ 10 –19 J. 3 2

According to kinetic theory of gases, kinetic energy = kT 3 kT = 1 eV = 1.6 ´ 10 19 J 2 2 ´ 1.6 ´ 10 –19 T= 3k

As Þ

where, M is the molar mass of gas while m is the mass of a single gas molecule.

Average speed It is the arithmetic mean of the speeds of molecules in a gas. Thus, v + v2 + ¼+ vN vav = v = 1 N On the basis of kinetic theory it is observed that 8p 8 kT 8 RT = = vav = pr pm pM

Most probable speed It is the speed possessed by maximum number of gas molecules in a given gas. On kinetic theory basis it is found that 2 RT 2 kT 2p = = M m r

2 ´ 1.6 ´ 10 –19 = 7730 K 3 ´ 1.38 ´ 10 –28

Van der Waals’ Gas Equation The gases actually found in nature are called real gases. They do not obey gas laws. A real gas behaves as ideal gas most closely at low pressure and high temperature. Equation of state for real gases is given by van der Waals’ equation with two corrections in ideal gas (i) volume correction (ii) pressure correction. van der Waals’ gas equation for 1 mole of gas is given by a ö æ ç p + 2 ÷ (V - b) = RT è V ø For n moles

3 RT 3 kT = = M m

Thus, we find that for a given gas 8 vmp : vav : v = 2 : : 3. p

average translation kinetic energy of a molecule in a gas becomes equal to the kinetic energy of a electron accelerated from rest through a potential difference of one volt? (k = 1.38 ´ 10 –23 JK -1)

=

RTc 8 = is called critical coefficient and is same pcVc 3

vmp =

Sample Problem 25 At what temperature does the

æ an2 ö ç p + 2 ÷ (V - nb) = nRT V ø è

Here, a and b are constants called van der Waals’ constants.

13.14 Degree of Freedom (f) The term degree of freedom refers to the number of possible independent ways in which a system can have energy. y f=2

f=1

x (a)

(b) f=3

vmp < vav < v and (c)

Heat and Kinetic Theory of Gases 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. In Fig. (b), the projectile 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.

Vibrational Energy The forces between different atoms of a gas molecule 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.

freedom, 3 translational, 2 rotational and 2 vibrational. Thus, 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 NH3) 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 vibrational energy greater than that of a diatomic gas. But at high enough temperatures it is also significant. So, it has 8 degrees of freedom 3 rotational, 3 translational and 2 vibrational. Thus,

Degree of Freedom of Monoatomic Gas

z

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

509

y

x

(all translational)

It can also rotate but due to its small moment of inertia, rotational kinetic energy is neglected. (3 translational + 3 rotation)

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 H2) cannot only move bodily but also rotate about any one 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 z

(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 y

x

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 potential energy). Thus, at high temperatures a diatomic molecule has 7 degrees of

(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 andf = 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.

510 JEE Main Physics

13.15 Internal Energy of an Ideal Gas

13.16 Law of Equipartition of Energy

Suppose a gas is contained in a closed vessel as shown in figure. If the container as a whole is moving with some speed, then this motion is called the ordered motion of the gas. 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 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.

An ideal gas is just like an ideal father. As an ideal father distributes whole of its 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 1 freedom energy of one mole of an ideal gas is RT , where 2 T is the absolute temperature of the gas. Thus, if f be the number of degrees of freedom, the internal energy of f 1 mole of the gas will be RT or internal energy of n moles 2 n of the gas will be fRT . Thus, 2 n …(i) U = fRT 2

Disordered motion Ordered motion

For a monoatomic gas, f = 3. 3 Therefore, U = RT 2 (for 1 mole of a monoatomic gas)

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 ) = K T + K R at room temperatures.

and for non-linear polyatomic gas at low temperatures, f = 6, so 6 (for 1 mole) U = RT = 3RT 2

Note From Eq. (i) we can see that internal energy of an ideal gas depends only on its temperature and which is directly proportional to its absolute temperature T. In an isothermal processT = constant. Therefore, the internal energy of the gas does not change or dU = 0.

Internal Energy (U)

Kinetic Energy

Potential Energy

For a dia and linear polyatomic gas at low temperatures, f = 5, so, 5 (for 1 mole) U = RT 2

Check Point 2 1. Although the rms speed of gas molecules is of the order of the

Due to intermolecular forces

Due to Translational Rotational Vibrational interatomic KE KE KE Forces (vibrational)

Later in the next article we will see that K T (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. or

U µT

speed of sound in that gas, yet on opening a bottle of ammonia in one corner of a room, its smell takes time in reaching the other corner. Explain why? 2. The ratio of vapour densities on two gases at the same temperature is 8 : 9. Compare the rms velocities of their molecules. 3. Can the temperature of a gas be increased keeping its pressure and volume constant? 4. On driving the scooter for a long time, the air pressure in the tyres slightly increases. Why?

Heat and Kinetic Theory of Gases

Molar Specific Heat of the Gases Consider a container containing m gram of gas of molecular mass M. If n is the number of moles of gas in container, DQ is the heat supplied and rise in temperature is DT , then 1 DQ c= m DT m Further, or m = nM n= M 1 DQ so c= nM DT Thus, molar specific heat 1 æ DQ ö C = Mc = ç ÷ 2 è DT ø We can write this relation as [Molar specific heat (C) of the mass] = [Molar mass (M) of the gas] ´ [gram specific heat (c) of the gas] Molar specific heat has two kinds

Sample Problem 26 A flask contains argon and chlorine in the ratio of 2 : 1 by mass. the temperature of the mixture is 27° C. Root mean square speed vrms of the molecules of the two gases is [Given, atomic mass of argon = 39.9 u, molecular mass [NCERT] of chlorine = 70.9 u] (a) 0.22 (c) 1.33

(b) 2.2 (d) 3.3

Interpret (c) The average kinetic energy (per molecule) of any 3 (ideal) gas is always equal to k BT . It depends only on temperature 2 and is independent of the nature of the gas. Since argon and chlorine both have the same temperature in the flask, the ratio of average kinetic energy (per molecule) of the two gases is 1 : 1. 1 2 Now, mv rms = average kinetic energy per molecule 2 3 = kBT 2 2 (v rms ) Ar (m) Cl (M) Cl = = 2 (v rms) Cl (M) Ar (M) Al

=

(i) Specific heat at constant volume (CV ) When heat is supplied to gas at constant volume the entire heat supplied just increases the internal energy of gas and does nothing else. 1 DQ ö CV = æç ÷ n è DT øV Thus,

CV =

1 æ DU ö ç ÷ n è DT ø

(ii) Specific heat at constant pressure (C p ) When heat is supplied to the gas at constant pressure a part of it increases the internal energy of the gas and remaining does an external work. So, specific heat at constant pressure 1 DQ ö C p = æç ÷ n è DT ø p

At constant pressure to increase in the internal energy of the gas by the same amount (as in case of heat supplied at constant volume), more amount of heat has to be supplied. Hence, we conclude that C p > CV

Note l

l

l

The relation betweenC p andCV is given by Mayer’s relation which is C p - CV = R Here R is gas constant. C p and CV in terms of degrees of freedom f can be written as f CV = R 2 f f and C p = CV + R = R + R = æç + 1ö÷ R è2 ø 2 Ratio of specific heats C p and CV is C 2 g = p = 1+ CV f

511

70.9 = 1.77 39.9

where M denotes the molecular mass of the gas. Taking square root (v rms) Ar = 1.33 (v rms) Cl

Sample Problem 27 A vessel contains two non-reactive gases neon (monoatomic) and oxygen (diatomic). The ratio of their partial pressure is 3 : 2. The ratio of number of molecules is [Given, atomic mass of Ne = 20.2 u, molecular mass of [NCERT] O 2 = 32 u] (a)

2 3

(b)

3 2

(c)

4 3

(d)

3 4

Interpret (b) Each gas (assumed ideal) obeys gas laws. Since V and T are common to the two gases, we have and

p1V = m1 RT p2V = m 2 RT p1 m1 = p2 m 2

Here 1 and 2 refer to neon and oxygen respectivley. p1 3 = p2 2 Given By definition and

m1 3 = m2 2 N m1 = 1 NA N m2 = 2 NA

where N1 and N2 are the number of molecules of 1 and 2, and NA is Avogadro’s number. N1 m1 3 = = N2 m 2 2

512 JEE Main Physics

Hot Spot

Specific Heat Capacity of Monoatomic, Diatomic and Polyatomic Gases

The molecule of a monoatomic gas has only three translational degrees of freedom. Thus, the average energy of a molecule at 3 temperature T is kBT . The total internal energy of a mole of such 2 gas is 3 3 U = kBT ´ N A = RT 2 2

Monoatomic Gases The molar specific heat at constant volume CV is C V (monoatomic gas) =

dU 3 = RT dT 2

From Mayer’s formula C p - CV = R where, C p is molar specific heat at constant pressure. Thus 5 Cp = R 2

Ratio of specific heat g =

Cp 5 = CV 3

Diatomic Gases A diatomic molecule has 5 degree of freedom, 3 translational and 2 rotational. Using the law of equipartition of energy the total internal energy of a mole of such a gas is 5 5 U = kBT ´ N A = RT 2 2

The molar specific heats are then given by 5 7 C V (rigid diatomic) = R,C p = R 2 2 7 g (rigid diatomic) = 5

If the diatomic molecule is not rigid but has in addition a vibrational mode 5 7 U = æç kBT + kBT ö÷ N A = RT è2 ø 2 7 9 9 C V = R ,C p = R , g = R 2 2 7

Polyatomic Gases A polyatomic molecule has 3 translational, 3 rotational degrees of freedom and a certain number (f ) of vibrational modes. From law of equipartition of energy, one mole of such a gas has ö æ3 3 U = çç kBT + kBT + f kBT ÷÷ N A 2 2 ø è C V = (3 + f )R , C p = (4 + f )R (4 + f ) g= (3 + f )

Note Specific heat of lighter elements is higher than heaveir elements and vice versa. Specific heat of the same substance in different states (solid, liquid and vapour) is different. For example, specific heat of water is 1 cal g–1° C–1 and that of ice is 0.5 cal g–1° C–1.

Table Values of f, U, C V , C p and g for Different Gases Nature of gas f

U=

Cp f dU f RT CV = = R Cp = CV + R Y = CV dT 2 2

Monoatomic 3

3 RT 2

3 R 2

5 R 2

1.67

Di and linear 5 polyatomic

5 RT 2

5 R 2

7 R 2

1.4

Non-linear polyatomic

3RT

3R

4R

1.33

6

Sample Problem 28 A cylinder of fixed capacity 44.8 litres contains helium gas at standard temperature and pressure. The amount of heat needed to raise the temperrature of the gas in the cylinder by 15°C is [Given, R = 8.31 J mol -1K -1] (a) 45 J (b) 374 J (c) 273 J (d) 432 J

Interpret (b) From ideal gas law pV = mRT 1 mol of any (ideal) gas at standard temperature (273 K) and pressure (1 atm = 1.01 ´ 10 5 Pa) occupies a volume of 22.4 litres. This universal volume is called molar volume. Thus, the cylinder in this example contains 2 mol of helium. Further, since helium is monoatomic, its predicted (and observed) molar specific heat at 3 constant volume CV = R, and molar specific heat at constant 2 pressure. 3 5 Cp = R + R = R 2 2 Since, volume of the cylinder is fixed, the heat required is determined by CV . \ Heat required = Number of moles ´ Molar specific heat ´ rise in temperature = 2 ´ 1.5 R ´ 15 = 45 R = 45 ´ 8.31 = 374 J

Heat and Kinetic Theory of Gases Sample Problem 29 A sphere of aluminium of 0.047 kg placed for sufficient time in a vessel containing boiling water, so that the sphere is at 100°C. It is then immediately transferred to 0.14 kg copper calorimeter containing 0.25 kg of water at 20°C. The temperature of water rises and attains a steady state at 23°C. The specific heat capacity of aluminium is [NCERT] (a) 0.911 kJ kg -1K -1

(b) 211 kJ kg -1K -1

(c) 423 kJ kg -1K -1

(d) 143 kJ kg -1K -1

Interpret (a) At a steady state, heat given by an aluminium sphere will be equal to the heat absorbed by the water and calorimeter. Mass of aluminium sphere (mi ) = 0.047 kg Initial temperatue of aluminium sphere = 100° C Final temperatuee = 23° C Change in temperature ( DT) = 100° C - 23° C = 77° C Let the specific heat capacity of aluminium be S Al, The amount of heat lost by the aluminium sphere = m1S AlDT = 0.047 ´ S Al ´ 77 Mass of water (M2) = 0.25 kg Mass of calorimeter (m3) = 0.14 kg Initial temperature of water and calorimeter = 20° C Final temperature of the mixture = 23° C Change in temperature ( DT2) = 23° C - 20° C = 3° C Specific heat capacity of water ( Sw ) = 4.18 ´ 10 3 Jkg –1 K –1 Specific heat capacity of copper calorimeter = 0.386 ´ 10 3 Jkg –1 K –1 The amount of heat gained by water and calorimeter = m2Sw DT2 + m3 S Cu DT2 = 0.25 ´ 4.18 ´ 10 3 + 0.14 ´ 386 ´ 10 3 ´ (23 - 20) In the steady state heat lost by the aluminium sphere = heat gained by water + calorimeter So, 0.047 kg ´ S Al ´ 77° C = 0.25 ´ 4.18 ´ 10 3 + 0.14 ´ 0.386 ´ 3 S Al = 0.911kJ kg –1 K –1

13.17 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 collision against one another. Therefore, path of a single gas molecule consists of a series of short zig-zig paths of different lengths. The mean free path of a gas molecule is the average distance between two successive collisions. Mathematically it is expressed as 1 l= 2 pd 2nV where,

d = diameter of molecules,

nV = number of molecules per unit volume.

513

The number of molecules per unit volume can be determined from Avogadro’s number and the ideal gas law leading to nN A nN A N A p nV = = = nRT V RT p \

l=

RT 2 pd 2N A p

13.18 Avogadro’s Number A mole (abbreviated mol) of a pure substance is a mass of the material in grams that is numerically equal to the molecular mass in atomic mass unit (amu). A mole of any material will contain Avogadro’s number of molecules. For example, carbon has an atomic mass of exactly 12.0 atomic mass units a mole of carbon is therefore 12 grams. One mole of an ideal gas will occupy a volume of 22.4 litres at STP. Avogadro’s number N A = 6.0221367 ´ 1023 mol

Standard Temperature and Pressure (STP) STP is used widely as a standard reference point for expression of the properties and processes of ideal gases. The standard temperature is the freezing point of water and the standard pressure is one standard atmosphere. These can be quantified as follows : Standard temperature 0° C = 273.15 K Standard pressure = 1atmosphere = 760 mm of Hg = 101.3 kPa Standard volume of 1 mole of an ideal gas at STP = 22.4 litres.

Sample Problem 30 A vessel contains a mixture of 7 g of nitrogen and11g of carbon dioxide at temperature T = 300 K. If the pressure of the mixture is 1 atm (1 ´ 10 5 N / m 2), its density is (gas constant R = 2513 J / mol K) (a) 0.72 kg /m3

(b) 1.44 kg /m3

3

(d) 5.16 kg /m3

(c) 2.88 kg /m

Interpret (b) The expression for density of a mixture of gases is given by, r mix =

pMmix RT

where, Mmix is the mass of mixture of non-reactive gases, p is pressure, R is gas constant and T is temperature. n1M1 + n2 M2 Mmix = n1 + n2

514 JEE Main Physics Given, Mass of nitrogen gas, mN = 7 g = 7 ´ 10 -3 kg

Interpret (b) We have mass of ice m = 3 kg Specific heat capacity of water, S water = 4186 Jkg –1 K –1

Molecular weight of nitrogen gas, mN = 28 ´ 10 -3 kg Molecular weight of carbon dioxide, MCO2 = 44 ´ 10 -3 kg

Mmix =

Mmix

mCO2 mN MCO2 MN + MCO2 MN = mN mCO2 + MN MCO2

nNMN nN + nCO2

Mmix =

18 ´ 10 1 1 + 4 4

= m S iceDT1 = (3 kg) (2100 Jkg –1 K –1) [0 - ( -12)] C = 75600 J Q 2 = heat required to melt ice at 0°C to water at 0°C = mLfice = (3) (3.35 ´ 10 5 Jkg –1)

= 36 ´ 10 -3 kg

= 1005000 J Q3 = heat required to convert water at 0°C to water at 100°C = mSw DT2 = (3 kg) (4186 Jkg –1 K –1) (100° C) = 1255800 J Q 4 = heat required to convert water at 100°C to steam at 100°C

Sample Problem 31 When 0.15 kg of ice at 0° C is mixed with 0.30 kg of water at 50° C in a container the resulting temperature is 6.7°C. The heat of fusion of ice is (Swater = 4186 J kg -1K -1) [NCERT] 3

(a) 1.45 ´ 10 J kg

-1

(c) 5.23 ´ 10 6 J kg -1

Lsteam = 2.256 ´ 10 6 Jkg –1

Latent heat of steam,

Q1 = heat required to convert ice at –12°C to ice at 0°C

(1 ´ 10 5) (36 ´ 10 -3) = 1.44 kg /m3 r= 25 ´ 300 3

\

Lice = 3.35 ´ 10 5 Jkg –1

Latent heat of fusion of ice,

Q = heat required to convert 3 kg of ice at –12°C to steam at 100°C

mN + mCO2 (7 + 11) ´ 10 -3 = = mN mCO2 æ 7 11 ö + + ÷ ç è 28 44 ø MN M CO2 -3

S ice = 2100 Jkg –1 K –1

Specific heat capacity of ice,

Mass of carbon dioxide, mCO2 = 11 g = 11 ´ 10 –3 kg

5

(b) 3.34 ´ 10 J kg

-1

(d) 6.23 ´ 10 7 J kg -1

Interpret (b) Heat lost by water = msw (Q f - Qi )w = (0.30) (4186) (50 - 6.7) = 54376.14 J Heat required to melt ice = m2 Lf = (0.15) Lf Heat required to raise temperature 0 ice water to final temperature = m1 sw (Q f - Qi )i

= mLsteam = (3 kg) (2.256 ´ 10 6) = 6768000 J So, Q = Q1 + Q 2 + Q3 + Q 4 = 75600 J+1005000 J+1255800 J + 6768000 J = 9.1 ´ 10 6 J

Sample Problem 33 The temperature of the steel-copper junction in the steady state of the system as shown in the figure is [Given length of the steel rod = 15 cm, length of the copper rod = 10 cm, temperature of furnace = 300° C, temperature of other end = 0° C, the area of cross-section of the steel rod is twice that of the copper rod, thermal conductivity of steel = 50.2 Js-1m -1 K -1 and of copper = 385 Js-1 m -1 K -1] [NCERT]

= (0.15 kg) (4186 Jkg –1 K –1) (6.7 - 0)

Furnace Steel 300°C

= 4206.93 J Heat lost = heat gained

Ice box 0°C

Insulating material copper

54376.14 J = (0.15 kg) Lf + 4206.93 J

(a) 4.4°C (c) 44.4°C

Lf = 3.34 ´ 10 5 Jkg –1

(b) 40°C (d) 52°C

Sample Problem 32 Heat required to convert 3 kg of ice

Interpret (c) The insulating material around the rod reduces

at -12° C kept in a calorimeter to steam at 100° C at atmospheric pressure. [Given specific heat capacity of ice = 2100 J kg -1 K -1, specific heat capacity of water

heat loss from the sides of the rods. Therefore, heat flows only along the length of the rods. Consider any cross-section of the rod in the steady state, heat flowing into the element must equal the heat flowing out of it, otherwise there would be a net gain or loss of heat by the element and its temperature would not be steady. Thus, in the steady state, rate of heat flowing across a cross-seciton of the rod is the same at every point along the length of the combined steel-copper rod. Let T be the temperature of the steel copper junction in the stedy state. Then k1A1 (300 - T) k2A2 (T - 0) = L1 L2

= 4186 J kg -1 K -1, latent heat of fusion of ice = 3.35 ´ 10 5 J kg -1, and latent heat of steam = 2.256 ´ 106 J kg -1]

[NCERT]

(a) 1250 J

(b) 9.1 ´ 10 6 J

(c) 9.1 ´ 10 –6 J

(d) 1.250 ´ 10 3 J

515

Heat and Kinetic Theory of Gases where 1 and 2 refer to the steel and copper rod. For A1 = 2 A2, L1 = 15 cm, L2 = 10 cm, k1 = 50.2 Js–1 m–1 K –1, k2 = 385 Js–1 m–1 K –1, we have 50.2 ´ 2 ´ (300 - T) 385 T = 15 10 which gives T = 44.4° C

Sample Problem 34 A pan filled with hot food cools from 94° C to 86° C in 2 min, when the room temperature is at 20° C, how long will it take to cool from 71° C to 69° C? [NCERT Exemplar]

(a) 14 s

(b) 3 s

(c) 42 s

(d) 13 s

Interpret (c) The average temperature of 94°C and 86°C is 90°C, which is 70°C above the room temperature, under these conditions the pan cools 8°C in 2 minutes, we have Change in temperature = kD T Time 8° C …(i) = K (70° C) 2 min The average of 69°C and 71 °C is 70°C, which is 50°C above room temperature K is the same for this situation is for the original 2° C …(ii) = K (50° C) Time Dividing Eqs. (i) and (ii), we get 8° C/2 min K (70° C) = ,T = 0.7 min = 42 s 2° C/time K (50° C)

Sample Problem 35 A blacksmith fixes iron ring on the rim of the wooden wheel of a bullock cart. The diameter of the rim and the iron ring are 5.243 m and 5.231m respectively at 27° C. The temperature to which the ring be heated so as to fit [NCERT] the rim of the wheel is (a) 100°C (c) 218°C

(b) 50°C (d) 420°C

Interpret (c) Given, T = 27° C, LT1 = 5.231m,LT2 = 5.243 m So,

LT2 = LT1 = [1 + a1 (T2 - T1)]

5.243 = 5.231[1+1.20 ´ 10 –5 (T2 - 27° C)] Þ

T2 = 218° C

Sample Problem 36 In the arrangement shown in the figure gas is k thermally insulated. An ideal gas is filled in the cylinder having pressure p 0 (> m, S atmospheric pressure p a ). The spring of force p0 constant k is initially unstretched. The piston of mass m and area S is frictionless. In equilibrium, the piston rises up a distance x0, then the decrease in internal energy of the gas is given by 1 (a) pa Sx0 + kx02 + mgx0 2 4 2 (c) 2 pa Sx0 + x02 + mgx0 5 3

4 (b) pa Sx0 + x02 + 2 mgx0 5 (d) pa Sx0 + 2 kx02 + 2 mgx0

Interpret (a) Equilibrium of piston gives pS = pa S + mg + kx0 mg kx0 p = pa + + S S (p = final pressure of gas)

paS

pS mg

kx0

Work done by the gas is equal to work done against atmospheric pressure + elastic potential energy stored in the spring + increase in gravitational potential energy of the piston. 1 1 = pa DV + kx02 + mgx0 = pa Sx0 + kx02 + mgx0 2 2 This is also the decrease in internal energy of the gas, because the gas is thermally insulated and this work is done at the expense of internal energy of the gas.

Sample

Problem 37 Carbon monoxide is carried around a closed p2 cycle abc, in which bc is an isothermal process, as shown. The gas absorbs 7000 J of heat as its p1 temperature is increased from 300 K to 1000 K is going from a to b. The O quantity of heat ejected by the gas. process ca is (a) 4200 J (c) 9800 J

b

a V1

c V2

(b) 5000 J (d) 3500 J

Interpret (c) Given that gas absorbs 7000 J of energy, hence ( DQ) ab = mCV DT \ + 7000 = mCV (1000 - 300) For the process ca Ta = 300 K Tc = Tb = 1000 K ( DQ) ca = mC p DT = mC p (300 - 1000) Also

= - mC p ´ 700 C p - CV = R

\

C p = R + CV ( DQ) ca = -m (CV + R) 700 For carbon monoxide, 2 2 7 g = 1+ = 1+ = n 5 5 5R R R = = CV = g -1 7 -1 2 5

Hence, we have

or

( DQ) ab = mCV 700 5R ´ 700 = 7000 ( DQ) ab = m 2 20 mR = =4 5

\ ( DQ) ca = - (7000 + 4 ´ 700) = -9800 J Negative sign shows that heat is ejected.

V

WORKED OUT Examples Example 1

The coefficient of volume expansion of glycerine is 49 ´ 10 -5° C -1 . What is the fractional change in density for a 30° C rise in temperature? (a) 0.0155 (c) 0.0255

Solution

(b) 0.0145 (d) 0.0355

Here, g = 49 ´ 19 -5 ° C -1 DT = 30° C V ¢ = V + DV = V (1 + g DT)

\

V ¢ = V (1 + 49 ´ 10 -5 ´ 30) = 1.0147 V

As

r=

r - 0.9855 r m = 0.0145 , r¢ = r V

Example 2

How much should the temperature of a brass rod be increased so as to increase its length by 1%? Given a for brass is 0.00002°C -1. (a) 300°C (c) 500° C

Solution

(b) 400°C (d) 550° C

Here, DT = ?,

DL 1 = L 100

a = 0.00002°C-1 As \ or

DL = aLDT DL a DT = L DL 1 DT = = La 100 ´ 0.00002 DT =

10 5 = 500°C 2 ´ 10 2

If some heat is lost to the surroundings, value of s so, obtained will be less than the actual value of s.

Example 4 A geyser heats water glowing at the rate of 3.0 Lmin-1 from 27°C to 77°C. If the geyser operates on a gas burner, what is the rate of combustion of fuel, if its heat of combustion is 4.0 ´ 10 4 Jg -1? (a) 25.75 gmin -1 (c) 15.75 gmin -1

Solution

Example 3

In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cc of water at 27°C. The final temperature is 40ºC. Calculate the specific heat of the metal. If heat losses to the surroundings are not negligible, is our answer greater or smaller than the actual value of specific heat of the metal? (a) 0.02 (c) 0.01

Solution

Fall in temperature of metal DT = 150 - 40 = 110°C If c is specific heat of the metal, then heat lost by the metal, …(i) DQ = mcDT = 200 s ´ 110 Volume of water150 cc Mass of water, m¢ =150 g Water equivalent of calorimeter w = 0.025 kg = 25g Rise in temperature of water in calorimeter DT ¢ = 40 - 27 = 13°C Heat gained by water and calorimeter DQ ¢ = (m¢ + w) DT ¢ = (150 + 25) ´ 13 …(ii) DQ¢ = 175 ´13 As DQ = DQ ¢ \ From Eqs. (i) and (ii) 200 ´ s ´ 100 = 175 ´ 13 175 ´ 13 s= » 0.1 200 ´ 110

(b) 0.2 (d) 0.1 Here mass of metal, m = 0.20 kg = 200 g

(b) 10.75 gmin -1 (d) 35.75 gmin -1

Here, volume of water heated = 3.0 Lmin -1

Mass of water heated, m = 3000 gmin -1 Rise of temperature, DT = 77 - 27 = 50° C Specific heat of water, s = 4.2 Jg -1 ° C- 1 Amount of heat used, DQ = msDT = 3000 ´ 4.2 ´ 50 = 63 ´ 10 4 Jmin -1 Heat of combustion = 4 ´ 10 4 Jg -1 Rate of combustion of fuel =

63 ´ 10 4 4 ´ 10 4

= 15.75 gmin -1

Heat and Kinetic Theory of Gases Example 5

A 19 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 min assuming 50% of power is used up in heating the machine itself or lost to the surrounding, specific heat of aluminium is 0.91 Jg -1 ° C -1 ? (a) 105°C (c) 106°C

Solution

(b) 103°C (d) 108°C

Here, P = 10kW = 10 4 W 3

m = 8.0 kg = 8 ´ 10 g

Mass,

Rise in temperature DT = ? t = 2.5 min = 2.5 ´ 60 = 150 s

Time,

Specific heat, s = 0.91 Jg

-1

°C

or F =

Y ´ a ´ Dl 0.0024 = 0.91 ´ 10 11 ´ 3.142 ´ 10 -6 ´ l2 1.8 = 3.81 ´ 10 2 N

Example 7

From the following data, find the magnitude of Joule's mechanical equivalent of heat : C p for hydrogen = 3.409 cal g -1 C -1; Cv for hydrogen = 2.409 cal g -1° C -1 and molecular weight of hydrogen = 2. (a) J = 2.11 J cal -1

(b) J = 1.11 J cal -1

(c) J = 3.11 J cal -1

(d) J = 4.11 J cal -1

Solution

Here, J = ? C p = 3.409 calg -1° C-1

-1

CV = 2.409 calg -1° C-1, M = 2

Total energy P ´ t = 10 4 ´ 150

R = 8.31 J mol-1K -1

= 15 ´ 10 5 J As 50% of energy is lost. \ Energy available, DQ =

1 ´ 15 ´ 10 5 2

= 7.5 ´ 10 5 J DQ = msDT

As DT =

\

7.5 ´ 10 5 DQ = 103°C = ms 8 ´ 10 3 ´ 0.91

Example 6

A brass wire 1.8 m long at 27ºC is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of - 39° C, what is the tension developed in the wire, if the diameter is 2.0 mm? Coefficient of linear expansion of brass = 2.0 ´ 10 -5 °C -1 and Young's modulus of brass = 0.91 ´ 1011 pascal (Nm -2). (a) 3.81 ´ 10 1 N

(b) 3.81 ´ 10 2 N

(c) 3.81 ´ 10 3 N

(d) 3.81 ´ 10 4 N

Solution

r R = J MJ 8.31 \ 3.409 - 2.409 = 2J 4155 . 1= J As

An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12°C. To what valume does if grow when it reaches the surface, which is at temperature of 3°C? Given, 1 atm = 1.01 ´ 10 5Pa

Example 8

(a) 5.275 ´ 10 -6m3

(b) 6.275 ´ 10 -6m3

(c) 3.275 ´ 10 -6m3

(d) 4.275 ´ 10 -6m3

Solution

T1 = 12 ° C = 12 + 273 = 285 K P1 = 1atm+ h1 r g

If l2 is length of the wire at - 39° C, then

Therefore, decrease in length, Dl = l1 - l2 = 1.8 - 1.7976 = 0.0024 m Also,

Y = 0.91 ´ 10 11 Pa (Nm-1)

Diameter of wire, d = 2.0 min = 2.0 ´ 10 -3 m Therefore, area of cross-section of wire, p d2 p a= = ´ (2.0 ´ 10 -3) 2 = 3.142 ´ 10 -6 m2 4 4 Now, Young's modulus of the material of the wire is given by F/a F/l Y= = Dl / l2 a ´ Dl

When the air bubble is at 40 m depth, then V1 = 1 cm3 = 1.0 ´ 10 - 6 m3

a = 2.0 ´ 10 -5 ° C-1

= 1.8 (1 - 1.32 ´ 10 3 ) = 1.7976 m

C p - CV =

J = 411 . J cal-1

Here, l1 = 1.8 m; DT = ( -39) - 27 = - 66° C

l2 = l1 (1+ aDT) = 1.8 (1+ 2.0 ´ 10 -5 ´ ( -66)

517

= 1.01 ´ 10 5 + 40 ´ 10 3 ´ 9.8 = 493000 Pa When the air bubble reaches at the surface of lake, then V2 = ?, T2 = 35° C = 35 + 273 = 308 K P2 = 1atm = 1.01 ´ 10 5 Pa Now, or \

p1V1 PV = 2 2 T1 T2 V2 = V2 =

p1V1T2 T2 p2 (493000) ´ 1.0 ´ 10 -6 ´ 308 285 ´ 1.01 ´ 10 5

= 5.275 ´ 10 - 6 m3

518 JEE Main Physics Example 9.

Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m3 at a temperature of 27°C and 1 atm pressure. (Boltzmann constant = 1.38 ´ 10 -23 JK -1) (a) 5.10 ´ 10 26

(b) 4.10 ´ 10 26

(c) 6.10 ´ 10 26

(d) 2.10 ´ 10 26

Solution

(i) Average speed, c1 + c2 + c3 + c4 4 2+ 4+ 6+ 8 = = 5 kms-1 4 (ii) Root mean square speed, cav =

c=

Here, V = 25.0 m3 =

T = 27 + 273 = 300 K k = 1.38 ´ 10 -23 JK -1 pV = nRT = n (Nk)T = (nN)kT = N ¢ kT where nN = N ¢ = total number of air molcules in the given gas pV N¢ = kT Now,

=

(1.01 ´ 10 5) ´25 (1.38 ´ 10 -23) ´ 300

= 6.10 ´ 10

What is the mean kinetic energy of one gram molecule of hydrogen at STP. Given density of hydrogen at STP is 0.09 kgm -3 (a) 3403.4 J (c) 3203.4 J

(b) 4403.4 J (d) 2403.4 J

Here, r = 0.09 kgm-3

At STP, pressure p = 1.01 ´ 10 5 Pa

26

(a) 5 kms -1, 5.48 kms -1 (b) 4 kms -1, 3.48 kms -1 (c) 5 kms -1 , 8.48 kms -1 (d) 4 kms -1, 2.48 kms -1 Here, c1 = 2 kms

22 + 42 + 62 + 82 = 5.48 kms-1 4

Example 11

Solution

Example 10 Four molecules of a gas have speed 2, 4, 6, 8 kms-1 respectively. Calculate (i) average speed and (ii) root mean square speed.

Solution

c12 + c22 + c32 + c42 4

According to kinetic theory of gases, 1 p = rc 2 3 or

c=

3 ´ 1.01 ´ 10 5 3p = = 1837.5 ms-1 0.09 r

Volume occupied by one mole of hydrogen at STP = 22.4 L = 22.4 ´ 10 -3 m3 \ Mass of hydrogen, -1

; c2 = 4 kms

-1

c3 = 6 kms-1 and c4 = 8 kms-1

M = Volume ´ Density = 22.4 ´ 10 -3 ´ 0.09 = 2.016 ´ 10 -3 kg

Start Practice for

JEE Main Round I

(Topically Divided Problems)

Thermometry and Calorimetry

7. If the ratio of densities of two substances is 5 : 6 and

1. Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the [NCERT] relation between TA and TB ? (a)

TA 4 = TB 7

(b)

TA 3 = TB 7

(c)

TA 7 = TB 3

(d)

TA 7 = TB 4

2. A faulty thermometer has its fixed points marked 5 and 95. When this thermometer reads 68, the correct temperature in celsius is (a) 68°C

(b) 70°C

(c) 66°C

(d) 72°C

3. The Fahrenheit and Kelvin scales of temperature will give the same reading at (a) – 40

(b) 313

(c) 574.25

(d) 732.75

4. An amount of water of mass 20 g at 0°C is mixed with 40

that of the specific heats is 3 : 5. Then, the ratio between heat capacities per unit volume is (a) 1 : 1 (c) 1 : 2

(b) 2 : 1 (d) 1 : 3

8. Heat capacity of a substance is infinite. It means (a) heat is given out (b) heat is taken in (c) no change in temperature whether heat is taken in or given out (d) All of the above

9. A cylinder containing an ideal gas is in vertical position and has a piston of mass M that is able to move up or down without friction. If the temperature is increased. [NCERT Exemplar]

g of water at 10°C, final temperature of the mixture is (a) 5°C (c) 20°C

(b) 0°C (d) 6.66°C Temperature (°A)

5. The graph between two 180 temperature scales A and B is shown in figure. ∆tA =150° Between upper fixed point and lower point ∆tB =100° there are 150 equal division on scale A and 0 Temperature (°B)100 100 on scale B. The relationship for conversion between the two scales is [NCERT Exemplar] given by t - 180 t B (a) A = 100 150 t B - 180 t A (c) = 150 100

t - 30 t B (b) A = 150 100 t B - 40 t A (d) = 100 180

6. One gram of ice is mixed with one gram of steam. At thermal equilibrium the temperature of mixture is (a) 0°C

(b) 100°C

(c) 55°C

(d) 80°C

(a) both p and V of the gas will change (b) only p will increases according to Charles' law (c) V will change but not p (d) p will change but not V

10. Water falls from a height of 500 m. What is the rise in temperature of water at the bottom if whole energy is used up in heating water ? (a) 0.96°C (c) 1.16°C

(b) 1.02°C (d) 0.23°C

11. 540 g of ice at 0°C is mixed with 540 g of water at 80°C. The final temperature of the mixture is (a) 0°C (c) 80°C

(b) 53°C (d) less than 0°C

520 JEE Main Physics 12. Which one of the following would raise the temperature of 20 g of water at 30°C most when mixed with it? (a) 20 g of water at 40°C (c) 10 g of water at 50°C

(b) 40 g of water at 35°C (d) 4 g of water at 80°C

13. A metal sphere of radius r and specific heat c is rotated about an axis passing through its centre at a speed of n rotations per second. It is suddenly stopped and 50% of its energy is used in increasing its temperature. Then the rise in temperature of the sphere is (a) (c)

2 p2 n2 r2 5 c

(b)

1 p2 n2 10 r2 c 2

14. Volume versus temperature graphs for a mass of an ideal gas are shown in figure at two different values of constant pressure. What can be inferred about relation between P1 and P2 ? [NCERT Exemplar]

p2

30

p1

20 10 100 200 300 400 500

(a) p1 > p2 (c) p1 < p2

T (K)

(b) p1 = p2 (d) data is insufficient

15. A sphere, a cube and a thin circular plate, all of same material and same mass are initialy heated to same high temperature. [NCERT Exemplar] (a) Plate will cool fasted and cube the slowest (b) Sphere will cool fasted and cube the slowest (c) Plate will cool fasted and sphere the slowest (d) Cube will cool fastest and plate the slowest

16. When the room temperature becomes equal to the dew point the relative humidity of the room is (a) 100% (c) 70%

(b) 0% (d) 85%

17. An aluminium sphere is dipped into water. Which of the following is true?

[NCERT Exemplar]

(a) increases as its effective length increases even though its centre of mass still remains at the centre of the bob (b) decreases as its effective length increases even though its centre of mass still remains at the centre of the bob (c) increases as its effective length increases due to shifting of centre of mass below the centre of the bob (d) decreases as its effective length increases remains same but the centre of mass shifts above the centre of the bob

Thermal Expansion of Solids and Liquids at 0°C so that the length of the steel rod is 5 cm longer than the copper rod at any temperature ? a (Steel) = 1.1 ´ 10–5 ° C–1 a (Copper) = 1.7 ´ 10–5 ° C (a) 14.17 cm; 9.17 cm (c) 28.34 cm; 18.34 cm

(b) 9.17 cm, 14.17 cm (d) 14.17 cm, 18.34 cm

20. When a liquid in a glass vessel is heated, its apparent

expansion is 10.30 ´ 10–4 C–1. When the same liquid is heated in a metal vessel, its apparent expansion is 10.06 ´ 10–4 °C–1. If the coefficient of linear expansion of glass = 9 ´ 106 °C–1, what is the coefficient of linear expansion of metal?

V (L) 40

pendulum

19. What should be the lengths of a steel and copper rod

é prn ù (d) 5 ê ú ë14 c û

7 2 2 pr n c 8

18. As the temperature is increased, the time period of a

[NCERT Exemplar]

(a) Buoyancy will be less in water at 0° C than that is water at 4° C (b) Buoyancy will be more in water at 0° C than that is water at 4° C (c) Buoyancy in water at 0° C will be same as that in water at 4° C (d) Buoyancy may be more or less in water at 4°C depending on the radius of the sphere

(a) 51 ´ 10 -6 ° C–1 (c) 25 ´ 10

-6

°C

–1

(b) 17 ´ 10 -6 ° C–1 (d) 43 ´ 10 –6 ° C–1

21. A steel wire of uniform area 2 mm2 is heated upto 50°C and is stretched by tying its ends rigidly. The change in tension when the temperature falls from 50°C to 30°C is (Take Y = 2 ´ 1011 Nm -2 , a = 1.1 ´ 10–5 ° C–1) (a) 1.5 ´ 1010 N

(b) 5 N

(c) 88 N

(d) 2.5 ´ 1010 N

22. Density of substance at 0°C is 10 g/cc and at 100°C, its density is 9.7 g/cc. The coefficient of linear expansion of the substance is (a) 1.03 × 10–4°C -1 (c) 19.7 ´ 10 –3°C -1

(b) 3 ´ 10 -4 °C -1 (d) 10 -3 °C -1

23. A rectangular block is heated from 0°C to 100°C. The percentage increase in its length is 0.2%. What is the percentage increase in its volume? (a) 0.6%

(b) 0.10%

(c) 0.2%

(d) 0.4%

24. A cubic vessel (with faces horizontal + vertical) contains an ideal gas at NTP. The vessel is being carried by a rocket which is moving at a speed of 500 ms -1 in vertical direction. The pressure of the gas inside the vessel as observed by us on the ground [NCERT Exemplar]

Heat and Kinetic Theory of Gases (a) remains the same because 500 ms -1 is very much smaller than Vrms of the gas (b) remains the same because motion of the vessel as a whole does not affect the relative motion of the gas molecules and the walls (c) will increase by a factor equal to ( v2rms + (500 )2 / v2rms where v rms was the original mean square velocity of the gas (d) will be different on the top wall and bottom wall of the vessel

25. A metal rod having linear expansion coefficient -5

2 ´ 10 °C–1 has a length of 1 m at 20°C. The temperature at which it is shortened by 1 mm is (a) –20°C (c) –30°C

(b) –15°C (d) –25°C

26. A bimetallic strip is made of aluminium and steel (a Al > a steel ). On heating, the strip will [NCERT Exemplar]

(a) remain straight (b) get twisted (c) will bend with aluminium on concave side (d) will bend with steel on concave side

27. A bimetallic is made of two strips A and B having coefficients of linear expansion a A and a B . If a A < a B , then on heating, the strip will (a) bend with A on outer side (b) bend with B on outer side (c) not bend at all (d) None of the above

28. A clock with an iron pendulum keeps correct time at 15°C. What will be the error in second per day, if the room temperature is 20°C? (The coefficient of linear expansion of iron is 0.000012°C–1.) (a) 2.6 s (c) 1.3 s

(b) 6.2 s (d) 3.1 s

29. A

uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly (a) its speed of rotation increases (b) its speed of rotation decreases (c) its speed of rotation remains same (d) its speed increases because its moment of inertia increases

30. A uniform metal rod is used as a bar pendulum. If the room temperature rises by 10°C and coefficient of linear expansion of the metal of the rod is 2 ´ 106 °C–1 the period of pendulum will increase by (a) 1 ´ 10 -3 %

(b) -1 ´ 10 -3 %

(c) 2 ´ 10 -3 %

(d) -2 ´ 10 -3 %

521

31. A vertical column 50 cm long at 50°C balances another column of same liquid 60 cm along at 100°C. The coefficient of absolute expansion of the liquid is (a) 0.005/°C (c) 0.002/°C

(b) 0.0005/°C (d) 0.0002/°C

32. A bar of iron is 10 cm at 20°C. At 19°C it will be (a of iron = 11 ´ 10-6 /°C)

(a) 11 ´ 10 -6 cm longer (c) 11 ´ 10 -5 cm shorter

(b) 11 ´ 10 -6 cm shorter (d) 11 ´ 10 -5 cm longer

33. The radius of a metal sphere at room temperature T

is R, and the coefficient of liner expansion of the metal is a . The sphere is heated a little by a temperature DT so that its new temperature is T + DT. The increase in the volume of the sphere is approximately [NCERT Exemplar] (a) 2p Ra DT

(b) p R2 a DT

(c) 4 p R3a DT /3

(d) 4 pR3aDT

34. The volume of a metal sphere increases by 0.24% when its temperature is raised by 40°C. The coefficient of linear expansion of the metal is …°C. (a) 2 ´ 10 –5 per°C (c) 2.1 ´ 10 –5 per°C

(b) 6 ´ 10 -5 per°C (d) 1.2 ´ 10 –5 per°C

Thermal Conduction and Convection 35. A wall has two layers A and B, made of two different materials. The thermal conductivity of material A is twice that of B. If the two layers have same thickness and under thermal equilibrium, the temperature difference across the wall is 48°C, the temperature difference across layer B is (a) 40°C

(b) 32°C

(c) 16°C

36. Two

plates of same thickness, of coefficients of thermal conductivity K 1 Q 1 and K 2 and areas of cross section A1 and A2 are connected as shown in figure. The common coefficient conductivity K will be (a) K1 A1 + K2 A2 (c)

K1 A1 + K2 A2 A1 + A2

(d) 24°C K1 A1 K2 A2

Q2

of

thermal

K1 A1 K2 A2 K A + K2 A1 (d) 1 2 K1 + K2 (b)

37. Ice starts forming in a lake with water at 0°C, when

the atmospheric temperature is –10°C. If time taken for 1 cm of ice to be formed is 7 h, the time taken for the thickness of ice to increase from 1 cm to 2 cm is (a) 7 h (b) less than 7 h (c) more than 7 h but less than 14 h (d) more than 14 h

522 JEE Main Physics 38. When a bimetallic strip is heated, it

44. Three rods of material X and three rods of material Y

(a) does not bend at all (b) gets twisted in the form of an helix (c) bend in the form of an arc with the more expandable metal outside (d) bends in the form of an arc with the more expandable metal inside

are connected as shown in figure. All are identical in length and cross-sectional area. If end A is maintained at 60°C, end E at 10°C, thermal conductivity of X is 0.92 cals–1 cm–1°C–1 and that of Y is 0.46 cals–1 cm–1°C–1, then find the temperatures of junctions B, C, D. C

39. Four rods of different radii r and length l are used to connect two reservoirs of heat at different temperatures. Which one will conduct heat fastest? (a) r = 2 cm, l = 0.5 m (c) r = 2 cm, l = 2 m

x A

(b) r =1 cm, l = 0.5 m (d) r = 1 cm, l = 1 m

y

x x

y

y

40. Two rods of equal length and area of cross-section are kept parallel and lagged between temperatures 20°C and 80°C. The ratio of the effective thermal conductivity to that of the first rod is é æ K1 ö 3 ù ÷= ú ê the ratio ç è K2 ø 4 û ë (a) 7 : 4 (c) 4 : 7

(b) 7 : 6 (d) 7 : 8

41. Two rods of same length and material transfer a given amount of heat in 12 s, when they are joined end to end (i.e., in series). But when they are joined in parallel, they will transfer same heat under same conditions in (a) 24 s (c) 48 s

nine times that of steel. In the composite cylindrical bar shown in figure, what will be the temperature at the junction of copper and steel? Copper

(a) 75°C (c) 33°C

Steel

0°C

6 cm

18 cm

(b) 67°C (d) 25°C

43. Five rods of same dimensions are arranged as shown in figure. They have thermal conductivities K 1, K 2 , K 3, K 4 and K 5 . When points A and B are maintained at different temperatures, no heat would flow through central rod, if (a) K1K4 = K2 K3 (b) K1 = K4 and K2 = K3 K K (c) 1 = 2 K4 K3 (d) K1 K2 = K3 K4

(b) 30°C, 20°C, 20°C (d) 20°C, 20°C, 20°C

45. A cylindrical rod with one end in

θ1 a steam chamber and the other R2 end in ice results in melting of θ 0.1 g of ice per second. If the rod R1 is replaced by another with half θ2 the length and double the radius of the first and if the thermal conductivity of the material of the second rod is 1/4 that of the first, the rate at which ice melts in gs–1 will be

(b) 1.6

(c) 0.2

(d) 0.1

46. Consider two insulating sheets with thermal

42. The coefficient of thermal conductivity of copper is

100°C

D

(a) 20°C, 30°C, 20°C (c) 20°C, 20°C, 30°C

(a) 3.2

(b) 3 s (d) 1.5 s

E

B

C K1 A

q1R2 + q2 R1 R1 + R2 ( q1 + q2 ) R1R2 (b) R21 + R22 q R + q2 R2 (c) 1 1 R1 + R2 q1q2 R1R2 (d) ( q1 + q2 ) ( R1 + R2 ) (a)

θ1 R2 θ R1

θ2

47. Two rods P and Q have equal lengths. Their thermal K2 B

K5 K4

K3

resistances R1 and R2 as shown in figure. The temperature q is

conductivities are K 1 and K 2 and cross-sectional areas are A1 and A2 . When the temperature at ends of each rod are T1 and T2 respectively, the rate of flow of heat through P and Q will be equal, if (a)

A1 K2 = A2 K1

(b)

A1 K2 T2 = ´ A2 K1 T1

(c)

A1 = A2

(d)

A1 æ K2 ö =ç ÷ A2 è K1 ø

D

2

K1 K2

48. If l is length, A is the area of cross-section and K is thermal conductivity, then the thermal resistance of the block is given by (a) K l A

(b) 1/ KlA

(c) l + KA

(d) l / KA

Heat and Kinetic Theory of Gases 49. The amount of heat conducted out per second through a window, when inside temperature is 10°C and outside temperature is –10°C, is 1000 J. Same heat will be conducted in through the window, when outside temperature is –23°C and inside temperature is (a) 23°C (c) 270 K

(b) 230 K (d) 296 K

523

56. The rate of radiation of a black body at 0°C is E watt. The rate of radiation of this body at 273°C will be (a) 16 E

(b) 8 E

(c) 4 E

(d) E

57. Two circular discs A and B with equal radii are blackened. They are heated to same temperature and are cooled under identical conditions. What inference do you draw from their cooling curves, shown below? A

50. Two identical square rods of metal are welded end to end as shown in figure (i) 20 cal of heat flows through it in 4 minutes. If the rods are welded as shown in figure (ii) the same amount of heat will flow through the rods in [NCERT Exemplar]

R B

(θ–θ0)

0°C 0°C

(a) 1 min

100°C

100°C

(b) 2 min

(c) 4 min

(d) 16 min

51. The ratio of thermal conductivity of two rods is 5 : 4. The ratio of their cross-sectional areas is 1 : 1 and they have the same thermal resistances. The ratio of their lengths, must will be (a) 4 : 5

(b) 9 : 1

(c) 1 : 9

(d) 5 : 4

52. In heat transfer which method is based on gravitation (a) Natural convection (c) Radiation

(b) Conduction (d) Stirrling of liquid

53. If a liquid is heated in weightlessness the heat is transmitted through (a) (b) (c) (d)

conduction convection radiation neither because the liquid cannot be heated in weightlessness

Thermal Radiation ; Stefan’s Law, Wien’s Law and Newton’s Law of Cooling 54. The wavelength of maximum intensity of radiation emitted by a star is 289.8 nm and the radiation intensity for the star is (Stefan’s constant = 5.67 ´ 10–8 Wm –2K –2 , Wien’s constant b = 2878 mK). (a) 5.67 ´ 10 8 Wm–2

(b) 5.67 ´ 10 –12 Wm–2

(c) 10.67 ´ 107 Wm–2

(d) 10.67 ´ 1014 Wm–2

55. A polished metal plate with a rough black spot on it is heated to about 1400 K and quickly taken to a dark room. The spot will appear (a) darker than plate (c) equally bright

(b) brighter than plate (d) equally dark

(a) A and B have same specific heats (b) specific heat of A is less (c) specific heat of B is less (d) nothing can be said

58. The temperature of a black body is increased by 50%, then the percentage of increase of radiation is approximately (a) 100%

(b) 25%

(c) 400%

(d) 500%

59. A body cools from 80°C to 50°C in 5 min. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surroundings is 20°C. [NCERT] (a) 9 min

(b) 7 min

(c) 8 min

60. The

(d) 10 min

frequency ( nm ) corresponding to which energy emitted by a black vm body is maximum may vary with temperature T of the body as shown in figure. Which of the curves represents correct variation? (a) A

y

(b) B

(c) C

D A C B

x T

(d) D

61. If temperature of a black body increases from 7°C to 287°C, then the rate of energy radiation increases by (a) (287 /7) 4

(b) 16

(c) 4

(d) 2

62. A black body is at a temperature of 2880 K. The energy of radiation emitted by this object with wavelength between 499 nm and 500 nm is U1 and between 999 nm and 1000 nm is U2 . The Wien constant = 2.88 ´ 106 nm K. Then (a) U1 = 0 (c) U1 > U2

(b) U3 = 0 (d) U2 > U1

63. If wavelength of maximum intensity of radiation

emitted by sun and moon are 0.5 ´ 10–6 m and 10–4 m respectively, the ratio of their temperatures is (a) 1 : 100

(b) 1 : 200

(c) 200 : 1

(d) 400 : 1

524 JEE Main Physics 64. The maximum energy in the thermal radiation from

a hot source occurs at l = 11 ´ 10–5 cm. If temperature of another source is n times, for which wavelength of maximum energy is 5.5 ´ 10–5 cm, then n is (a) 2

(b) 4

(c)

1 2

(d) 1

65. A black body radiates at two temperatures T1 and T2 ,

such that T1 < T2 . The frequency corresponding to maximum intensity is (a) less at T1 (c) equal in the two cases

(b) more at T1 (d) cannot say

66. An object is cooled from 75°C to 65°C in 2 min in a room at 30°C. The time taken to cool another identical object from 55°C to 45°C in the same room, in minutes is (a) 4

(b) 5

(c) 6

(d) 7

67. A black body at 1373°C emits maximum energy corresponding to a wavelength of 1.78 micron. The temperature of moon for which l m = 14 micron would be (a) 62.6°C

(b) –58.9°C

(c) 63.7°C

(d) 64.2°C

68. A planet is at an average distance d from the sun and its average surface temperature is T. Assume that the planet receives energy only from the sun, and loses energy only through radiation from its surface. Neglect atmospheric effects. If T µ d- n ,the value of n is (a) 2

(b) 1

(c) 1/2

(d) 1/4

69. The rectangular surface of area 8 cm ´ 4 cm of a black body at a temperature of 127°C emits energy at the rate of E per second. If the length and breadth of the surface are each reduced to half of its initial value, and the temperature is raised to 327°C, the rate of emission of energy will become 3 E 8 9 (c) E 16

81 E 16 81 (d) E 64

(b)

(a)

70. The plots for intensity versus wavelength for three black bodies at temperatures T1, T2 , T3 respectively are shown in figure. Their temperatures are such that I

T1

T3

T2

the room temperature is 20°C, then its temperature in next 5 min will be (a) 38°C

(b) 33.3°C

(c) 30°C

(d) 36°C

72. A black body radiates heat energy at the rate of

2 ´ 105 Js–1m–2 at a temperature 127°C. The temperature of black body, at which the rate of heat radiation is 32 ´ 105 Js–1m–2 is (a) 273°C

(b) 527°C

(c) 873°C

(d) 927°C

73. A liquid is filled in a container which is kept in a room whose temperature is 20°C. When temperature of liquid is 80°C, it emits heat at the rate of 45 cals–1. When temperature of liquid falls to 40°C, its rate of heat loss will be (a) 15 cals–1 (c) 45 cals–1

(b) 30 cals–1 (d) 60 cals–1

74. The maximum wavelength of radiation emitted at 2000 K is 4 mm. What will be the maximum wavelength emitted at 2400 K? (a) 3.3 mm (c) 1 m

(b) 0.66 mm (d) 1 mm

75. Two bodies A and B are placed in an evacuated vessel maintained at a temperature of 27°C. The temperature of A is 327°C and that of B is 227°C. The ratio of heat loss from A and B is about (a) 2 : 1 (c) 1 : 2

(b) 4 : 1 (d) 1 : 4

76. The reflectance and emittance of a perfectly black body are respectively (a) 0, 1 (c) 0.5, 0.5

(b) 1, 0 (d) 0, 0

77. The rate of emission of radiation of a black body at temperature 27°C is E1. If its temperature is increased to 327°C, the rate of emission of radiation is E2 . The relation between E1 and E2 is (a) E2 = 24 E1 (c) E2 = 8 E1

(b) E2 = 16 E1 (d) E2 = 4 E1

78. The rates of heat radiation from two patches of skin each of area A, on a patient’s chest differ by 2%. If the patch of the lower temperature is at 300 K and emissivity of both the patches is assumed to be unity, the temperature of other patch would be (a) 306 K (c) 308.5 K

(b) 312 K (d) 301.5 K

79. The rays of sun are focussed on a piece of ice through a lens of diameter 5 cm, as a result of which 10 g ice melts in 10 min. The amount of heat received from sun, per unit area per min is

λ

(a) T1 > T2 > T3 (c) T2 > T3 > T1

71. A metallic sphere cools from 50°C to 40°C in 300 s. If

(b) T1 > T3 > T2 (d) T3 > T2 > T1

(a) 4 cal cm–2 min–1 (c) 4 Jm–2 min

(b) 40 cal cm–2 min–1 (d) 400 cal cm–2 min–1

Heat and Kinetic Theory of Gases

525

80. Solar radiation emitted by sun resembles that

87. If a given mass of gas occupies a volume of 100 cc at

emitted by a black body at a temperature of 6000 K. Maximum intensity is emitted at a wavelength of about 4800Å. If the sun were cooled down from 6000 K to 3000 K, then the peak intensity would occur at a wavelength of

1 atm pressure and temperature of 100°C (373.15 K). What will be its volume at 4 atm pressure; the temperature being the same?

(a) 4800 Å (c) 2400 Å

(b) 9600 Å (d) 19200 Å

observed that the wavelength corresponding of maximum energy changes from 0.26 mm to 0.13 mm to a body at the respective temperature. Then ratio of E the emissivities 2 is E1 (b) 4/1

(c) 1/4

(d) 1/16

82. The energy emitted per second by a black body at 27°C is 10 J. If the temperature of black body is increased to 327°C, the energy emitted per second will be (a) 20 J

(b) 40 J

(c) 80 J

(d) 160 J

83. A black body at a temperature of 327°C radiates 4 cal cm–2s–1. At a temperature of 927°C, the rate of heat radiated per unit area in cal cm–2s–1 will be (a) 16

(b) 32

(c) 64

(b) 400 cc (d) 104 cc

88. 1 mole of H2 gas is contained in a box of volume

81. When the temperature of a black body increases, it is

(a) 16/1

(a) 100 cc (c) 25 cc

(d) 128

84. The temperature of coffee in a cup with time is most likely given by the curve in figure.

V = 100 . m 3 at T = 300K. The gas is heated to a temperature of T = 3000K and the gas gets converted to a gas of hydrogen atoms The final pressure would be (considering all gases to be ideal) [NCERT Exemplar] (a) same as the pressure initially (b) 2 times the pressure initially (c) 10 times the pressure initially (d) 20 times the pressure initially

89. Two gases A and B having the same temperature T, same pressure p and same volume V are mixed. If the mixture is at same temperature T and occupies a volume V, the pressure of the mixture is (a) 2 p (c) p/2

(b) p (d) 4 p

90. When a gas filled in a closed vessel is heated through 1°C, its pressure increases by 0.4%. The initial temperature of the gas was (a) 250 K (c) 250°C

(b) 2500K (d) 25°C

91. A vessel of volume V contains a mixture of 1 mole of (a) θ

(b)

θ

t(time)

(c)

t(time)

θ

θ

(d) t(time)

t(time)

85. A solid cube and a solid sphere have equal surface areas. Both are at the same temperature of 120°C. Then (a) (b) (c) (d)

both of them will cool down at the same rate the cube will cool down faster than the sphere the sphere will cool down faster than the cube whichever of the two is heavier will cool down faster

86. A surface at temperature T0 K receives power P by

radiation from a small sphere at temperature T > T0 and at a distance d. If both T and d are doubled, the power received by the surface will become (a) P (c) 4 P

(b) 2 P (d) 16 P

hydrogen and 1 mole of oxygen (both considered as ideal). Let f1 (v) dv, denote the fraction of molecules with speed between v and (v + dv) with f2 (v) dv, similarly for oxygen. Then [NCERT Exemplar] (a) f1 ( v ) + f2 ( v ) = f ( v ) obeys the Maxwell's distribution law (b) f1 ( v ), f2 ( v ) will obey the Maxwell's distribution law separately (c) Neither f1 ( v ), nor f2 ( v ) will obey the Maxwell's distribution law (d) f2 ( v ) and f1 ( v ) will be the same

92. An inflated rubber balloon contains one mole of an ideal gas, has a pressure p, volume V and temperature T. If the temperature rises to 1.1 T, and the volume is increased to 1.05 V. the final pressure [NCERT Exemplar] will be (a) 1.1 p (c) less than p

(b) p (d) between p and 1.1 p

93. The air density at Mount Everest is less than that at the sea level. It is found by mountainers that for one trip lasting few hours, the extra oxygen needed by them corresponds to 30000 cc at sea level (pressure 1 atm, temperature 27°C). Assuming that the temperature around Mount Everest is –73°C and

526 JEE Main Physics that the pressure cylinder has capacity of 5.2 L, the pressure at which oxygen be filled (at site) in the cylinder is (a) 3.86 atm (c) 5.77 atm

(b) 5.00 atm (d) 1 atm

is subjected to pressure and temperature variation. The experiment is performed at high pressures as well as high temperature. The result obtained are pV with shown in the figure. The correct variation of RT p will be exhibited by 4 3

0.0

100 200 300 400 500 600 p (in atm)

(b) curve (3) (c) curve (2)

(d) curve (1)

95. How much should the pressure be increased in order to decrease the volume of a gas by 5% at a constant temperature? (b) 5.26%

(c) 10%

(d) 4.26%

96. An ideal gas is found to obey an additional law 2

pV = constant. The gas is initially at temperature T and volume V. Then it expands to a volume 2 V, the temperature becomes (a) T / 2

(b) 2 T

(c) 2 T / 2

(d) 4 T

97. The rms velocity of gas molecules is 300 ms–1. The rms velocity of molecules of gas with twice the molecular weight and half the absolute temperature is (a) 300 ms–1 (c) 75 ms–1

(b) 600 ms–1 (d) 150 ms–1

velocity, then (a) c s < c

(b) c s = c

(b) 1.5 ms–1 (d) 1.5 kms–1

102. At a certain temperature, the ratio of the rms velocity of H2 molecules to O2 molecule is (b) 1 : 4 (d) 16 : 1

(d) None of these

molecules, each of mass 2 m, of gas B are contained in the same vessel which is maintained at a temperature T. The mean square velocity of molecules of B type is denoted by V2 and the mean V square velocity of A type is denoted by V1, then 1 is V2 (b) 1

(c) 1/3

(a) 0.14 g (c) 0.14 kg

(b) 0.02 g (d) 0.014 kg

104. RMS velocity of a particle is c at pressure p. If pressure is increased two times, then rms velocity becomes (a) 0.5 c

(b) c

(c) 2 c

(d) 3 c

105. If the molecular weight of two gases are M1 and M2 , then at a temperature the ratio of rms velocity c1 and c2 will  be æM ö (a) ç 1 ÷ è M2 ø

1 /2

æM ö (b) ç 2 ÷ è M1 ø

æ M - M2 ö (c) ç 1 ÷ è M1 + M2 ø

1 /2

1 /2

æ M + M2 ö (d) ç 1 ÷ è M1 - M2 ø

1 /2

sample of helium is 5/7th that of the molecules in a sample of hydrogen. If the temperature of the hydrogen as is 0°C, that of helium sample is about (a) 0°C

1 /2

99. N molecules, each of mass m, of gas A and 2 N

(a) 2

(a) 1.5 mm s–1 (c) 1.5 cms–1

106. The root mean square velocity of the molecules in a

98. If cs is the velocity of sound in air and c is the rms ægö (c) c s = c ç ÷ è 3ø

mass 5 ´ 10-17 kg in their Brownian motion in air at NTP ( k = 1.38 ´ 10–23 JK –1 )

gauge pressure of 15 atm and a temperature of 27°C. After some oxygen is withdrawn from the cylinder the gauge pressure drops to 11 atm and its temperature drops to 17°C. The mass of oxygen taken out of the cylinder ( R = 8.31Jmol –1 K -1) . molecular mas of O2 = 32 u) is [NCERT]

1

0.5

(a) 5%

101. Calculate the rms speed of smoke particles each of

103. An oxygen cylinder of volume 30 L has an initial

1.0

(a) curve (4)

(b) F2 (d) Cl2

(a) 1 : 1 (c) 4 : 1

2

2.0 pV RT 1.5

of a certain diatomic gas is found to be 1930 ms–1. The gas is (a) H2 (c) O2

94. A fixed amount of nitrogen gas (1 mole) is taken and

2.5

100. At room temperature, the rms speed of the molecules

(d) 2/3

(b) 4 K

(c) 273°C

(d) 100°C

107. The average translatory energy and rms speed of molecules in a sample of oxygen gas at 300 K are 6.21 ´ 10–21 J and 484 ms–1 respectively. The corresponding values at 600 K are nearly (assuming ideal gas behaviour) (a) 12.42 ´ 10 –21 J, 968 ms –1 (b) 7.78 ´ 10 –21 J, 684 ms –1 (c) 6.21 ´ 10 –21 J, 968 ms –1 (d) 12.42 ´ 10 –21 J, 684 ms –1

Heat and Kinetic Theory of Gases 108. The average energy and the rms speed of molecules in a sample of oxygen gas at 400 K are 7.21 ´ 10-21 J and 524 ms–1 respectively. The corresponding values at 800 K are nearly (a) 14.42 ´ 10 –21 J, 1048 ms –1 (b) 10.18 ´ 10 –21 J, 741 ms –1 (c) 7.21 ´ 10 –21 J, 1048 ms –1 (d) 14.42 ´ 10

–21

J, 741 ms

–1

109. The average kinetic energy of a gas molecule at 27°C is 6.21 ´ 10–21 J. Its average kinetic energy at 127°C will be (a) 12.2 ´ 10 –21 J

(b) 8.28 ´ 10 –21 J

(c) 10.35 ´ 10 –21 J

(d) 11.35 ´ 10 –21 J

(b)

nR 2

(c)

nRT 2

(d) 2 nRT

111. For a gas, if the ratio of specific heats at constant pressure and constant volume is g, then the value of degrees of freedom is (a)

g +1 g -1

(b)

g -1 g +1

(c)

( g - 1) 2

(d)

2 g -1

112. The value of molar specific heat at constant pressure for one mole of triatomic gas (triangular arrangement) at temperature T K is (R = universal gas constant) (a) 3 R

(b)

2 R 7

(c)

5 R 2

(d) 4 R

113. The diameter of a gas molecule is 2.4 ´ 10–10 m. The mean free path of gas molecule at NTP is ´ 10–23

(k = 1.38 (a) 1.46 ´ 10 –7 m (c) 1.46 ´ 10

–6

m

JK–1)

(b) 2.46 ´ 10

–6

m

(d) 2.46 ´ 10

–7

m

Round II Only One Correct Option 1. A thin copper wire of length l increase in length by 1%, when heated from 0°C to 100°C. If a thin copper plate of area 2 l ´ l is heated from 0°C to 100°C, the percentage increase in its area would be (a) 1% (c) 3%

(b) 4% (d) 2%

(a) Ne

(b) O3

(c) N2

(d) NH3

115. There is a rough black spot on a polished metallic plate. It is heated upto 1400 K. Approximately and then at once taken in a dark room which of the following statements is true? (a) In comparision with the plate the spot will shine more (b) In comparision with the plate the spot will appear were black (c) The spot and the plate will be equally bright (d) The plate and the black spot can not be seen in the dark room

velocity of

for 1 mole of polyatomic gas having n number of degrees of freedom at temperature T K is (R = universal gas constant) nR 2T

114. The value of g for gas X is 1.33, the X is

116. The thermal radiation from a hot body travels with a

110. The value of molar specific heat at constant volume

(a)

527

(a) 330 ms–1

(b) 2 ´ 10 8 ms –1

(c) 3 ´ 10 8 ms -1

(d) 230 ´ 10 8 ms –1

117. Assuming the sun to have a spherical outer surface of radius r radiating like a black body at temperature t° C, the power received by a unit surface (normal to the incident rays) at a distance R from the centre of the sun is (s is stefan’s constant) (a) 4 pr2st 4 (c)

16 p2 r2st 4 R2

(b)

r2s (t + 273) 4 4 pR 2

(d)

r2s (1 + 273) 4 R2

118. The temperature of sun is 5500 K and it emits maximum intensity radiation in the yellow region (5.5 ´ 10–7 m). The maximum radiation from a furnace occurs at wavelength 11 ´ 10-7 m. The temperature of furnace is (a) 2550 K

(b) 2750 K

(c) 2650 K

(d) 2850 K

119. The temperature of a liquid drops from 365 K to 361 K in 2 minutes. Find the time during which temperature of the liquid drops from 344 K to 342 K. Temperature of room is 292 K (a) 84 s

(b) 72 s

(c) 66 s

(d) 60 s

(Mixed Bag) 2. A steel ball of mass 0.1 kg falls freely from a height of

of 10 m and bounces to a height of 5.4 m from the ground. If the dissipated energy in this process is absorbed by the ball, the rise in its temperature is (a) 0.01°C (c) 1.1°C

(b) 0.1°C (d) 1°C

528 JEE Main Physics 3. The triple point of neon and carbon dioxide are

12. The ends of 2 different materials with their thermal

24.57 K and 216.55 K respectively. These temperatures on the celsius and fahrenheit scales are respectively [NCERT]

conductivities, radii of cross-section and length all in the ratio of 1 : 2 maintained at temperature difference. If the rate of the flow of heat in the longer rod is 4 cals–1, that in the shorter rod in cals–1 will be

(a) - 415.44° F and - 69.88° F (b) 415.44° F and 69.88° F (c) - 315.44° F and 69.88° F (d) - 69.88° F and 415.44° F

(a) 1

100°C, its density is 9.7 g/cc. The coefficient of linear expansion of the substance is (b) 10–2 °C–1 (d) 102 °C–1

5. The rate of cooling at 600 K, if surrounding temperature is 300 K is R. The rate of cooling at 900 K is (a)

16 R 3

(b) 2 R

(c) 3 R

(d)

2 R 3

6. The temperature of a piece of metal is increased from 27°C to 84°C. The rate at which energy is radiated is increased to (a) four times (c) six times

(b) two times (d) eight times

7. Two identical square rods of metal are welded end to

(a) 1.44 ´ 10 -2 cm

(b) 2.44 ´ 10 -3 cm

(c) 1.44 ´ 10 -2 mm

(d) 2.44 ´ 10 -3 mm

14. A body takes 10 min to cool from 60°C to 50°C. If the temperature of surroundings is 25°C and 527°C respectively. The ratio of energy radiated by P and Q is (a) 48°C

(b) 46°C

cross-section have been joined as shown in figure. Each rod is of same length. The left and right ends are kept at 0°C and 90°C respectively. The temperature of the junction of the three rods will be 2 1 0°C 3

(c) 1 min

(d) 4 min

(c) 37°C

(d) 30°C

(c) 1 : 16

(d) 1 : 8

10. The amount of heat required to convert 10 g of ice at –10°C into steam at 100°C is (in calories) (a) 6400

(b) 5400

(c) 7200

(d) 7250

11. A lead bullet of 10 g travelling at 300

ms–1

strikes against a block of wood comes to rest. Assuming 50% of heat is absorbed by the bullet, the increase in its temperature is (Specific heat of lead = 150 JkgK –1) (a) 100°C

(b) 125°C

(c) 30°C

90°C

(d) 20°C

heated it bends into an arc with brass on the convex and iron on the concave side of the arc. This happens because

material have their diameters in the ratio 1 : 2 and lengths in the ratio 2 : 1. If the temperature difference between their ends is same, the ratio of heat conducted respectively by A and B per second is (b) 1 : 4

(b) 60°C

16. A bimetallic strip consists of brass and iron when it is

9. Two cylindrical conductors A and B of same metallic

(a) 1 : 2

90°C

(b)

37°C. The temperature of the mixture is (b) 27°C

(d) 42.85°C

(a)

8. 22 g of CO2 at 27°C is mixed with 16 g of oxygen at (a) 32°C

(c) 49°C

15. Three rods made of same material and having same

(a) 45°C

(b) 12 min

(d) 6

hole is 4.24 cm at 27.0°C. What is the change in the diameter of the hole when the sheet is heated to 227°C? Coefficient of linear expansion of copper is [NCERT] . ´ 10-5 / ° C. 170

end as shown in figure, Q cal of heat flows through this combination in 4 min. If the rods were welded as shown in figure, the same amount of heat will flow through the combination in

(a) 16 min

(c) 8

13. A hole is drilled in a copper sheet. The diameter of the

4. The density of a substance at 0°C is 10 g/cc and at (a) 10–4 °C–1 (c) 10–3 °C–1

(b) 2

(c) 150°C

(d) 200°C

(a) brass has a higher specific heat capacity than iron (b) density of brass is more than that of iron (c) it is easier to bend an iron strip than a brass strip of the same size (d) brass has a higher coefficient of linear expansion than iron

17. The efficiency of a Carnot engine is 50% and temperature of sink is 500 K. If temperature of source is kept constant and its efficiency raised to 60%, then the required temperature of sink will be (a) 100 K

(b) 600 K

(c) 400 K

(d) 500 K

18. 70 cal of heat are required to raise the temperature of 2 mole of an ideal gas at constant pressure from 30°C to 35°C. The amount of heat required to raise the temperature of the same sample of the gas through the same range at constant volume is nearly (Gas constant = 1.99 cal K–1-mol–1) (a) 30 cal

(b) 50 cal

(c) 70 cal

(d) 90 cal

Heat and Kinetic Theory of Gases 19. A flask of volume 103 cc is completely filled with mercury at 0°C. The coefficient of cubical expansion of mercury is 1.80 ´ 10–6 ° C–1 and that of glass is 1.4 ´ 10–6 ° C–1. If the flask is now placed in boiling water at 100°C, how much mercury will overflow? (a) 7 cc (c) 21 cc

(b) 1.4 cc (d) 28 cc

when determined using two different vessels A and B are l1 and l2 , respectively. If the coefficient of linear expansion of the vessel A is a, the coefficient of linear expansion of vessel B is g -g (b) 1 2 2a g -g (d) 1 2 + a 3

maximum heat, when their ends are maintained at a constant temperature difference ? (b) l = 1m, r = 0.1 m (d) l = 0.1m, r = 0.3 m

22. Two spheres made of same substance have diameters in the ratio 1 : 2. Their thermal capacities are in the ratio of (a) 1 : 2

(b) 1 : 8

(c) 1 : 4

(d) 2 : 1

23. A child running at a temperature of 101°F is given an antipyrin (i.e., medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98°F in 20 min, what is the average rate of extra evaporation caused by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of child is 30 kg. The specific heat of the human body is approximately the same as that of water and latent heat of evaporation of water at that [NCERT] temperature is about 580 cal/g. (a) 4.31 g/min (c) 2.31 g/min

(b) 4.31 g/s (d) 2.31 g/s

24. A wheel is 80.3 cm in circumference. An iron tyre measures 80.0 cm around its inner face. If the coefficient of linear expansion for iron is 1.2 ´ 0-5 ° C-1, the temperature of the tyre must be raised by (a) 105°C (c) 312°C

(b) 417°C (d) 223°C

25. The temperature gradient in the earth’s crust is

32°C km–1 and the mean conductivity of earth is 0.008 cals–1cm–1°C -1. Considering earth to be a sphere of radius 6000 km loss of heat by earth everyday is about (a) 1030 cal (c) 1020 cal

(b) 1040 cal (d) 1018 cal

w2 at 50°C. The coefficient of cubical expansion of metal is less than that of water. Then (a) w1 < w2 (c) w1 = w2

(b) w1 > w2 (d) data is not sufficient

90.0 cm, when both are at 10°C, the calibration temperature, for the tape. What would be tape read for the length of the rod when both are at 30°C? Given, a for steel 1.2 ´ 10–5 ° C–1 and a for copper is

1.7 ´ 10–5 ° C–1 (a) 90.01 cm (c) 90.22 cm

(b) 89.90 cm (d) 89.80 cm

28. A cylinder of radius r and thermal conductivity K 1 is

21. Which of the following cylindrical rods will conduct (a) l = 1 m, r = 0.2 m (c) l = 10 m, r = 0.1 m

26. A metal ball immersed in water weighs w1 at 0°C and

27. A steel tape measures the length of a copper rod as

20. The coefficiency of apparent expansion of a liquid

a g1g2 (a) g1 + g2 g -g +a (c) 1 2 3a

529

surrounded by a cylindrical shell of linear radius r and outer radius 2 r, whose thermal conductivity is K 2 . There is no loss of heat across cylindrical surfaces, when the ends of the combined system are maintained at temperatures T1 and T2 . The effective thermal conductivity of the system, in the steady state is K1K2 K1 + K2 K + 3 K2 (c) 1 4 (a)

(b) K1 + K2 (d)

3 K1 + K2 4

29. The power radiated by a black body is P, and it radiates maximum energy around the wavelength l 0 . If the temperature of black body is now changed so that it radiates maximum energy around a wavelength l 0 / 4, the power radiated by it will increase by a factor of 4 3 64 (c) 27

(a)

16 9 256 (d) 81 (b)

30. A solid whose volume does not change with temperature floats in liquid. For two different temperatures t1 and t2 , the fractions f1 and f2 of volume of solid remain submerged. What is the coefficient of volume expansion of liquid? f1 - f2 f2t1 - f1t2 f +f (c) 1 2 f2t1 - f1t2 (a)

f1 - f2 f1t1 - f2t2 f +f (d) 1 2 f1t1 - f2t2 (b)

31. A vessel of volume 4 L contains a mixture of 8 g of oxygen, 14 g of nitrogen and 22 g of carbon dioxide at 27°C. The pressure exerted by the mixture is (a) 5.79 ´ 105 Nm–2 (b) 6.79 ´ 105 Nm–2 (c) 7.79 ´ 103 Nm–2 (d) 7.79 ´ 105 Nm–2

530 JEE Main Physics 32. 22 g of carbon dioxide at 27°C is mixed in a closed container with 16 g of oxygen at 37°C. If both gases are considered as ideal gases, then the temperature of the mixture is (a) 24.2°C (c) 31.5°C

33. Two chambers containing m1 and m2 gram of a gas at

pressures p1 and p2 respectively are put in communication with each other, temperature remaining constant. The common pressure reached will be p1p2 ( m1 + m2 ) p2 m1 + p1m2 m m ( p + p2 ) (c) 1 2 1 p2 m1 + p1m2

p1p2 m1 p2 m1 + p1m2 m1m2 p2 (d) p2 m1 + p1m2 (b)

34. At room temperature (27°C) the rms speed of the molecules of a certain diatomic gas is found to be 1920 ms–1.The gas is (a) Cl2 (c) N2

container of 10 L capacity at 27°C. The pressure exerted by the mixture in terms of atmospheric pressure is (R = 0.082 L atm K–1 mol–1) (b) 2.5 atm (d) 8.7 atm

36. Inside a cylinder closed at both ends is a movable piston. On one side of the piston is a mass m of a gas, and on the other side a mass 2 m of the same gas. What fraction of the volume of the cylinder will be occupied by the larger mass of the gas when the piston is in equilibrium? The temperature is the same throughout. 1 (b) 3

1 (c) 2

1 (d) 4

37. Two containers of equal volume contain the same gas at the pressures p1 and p2 and absolute temperatures T1 and T2 respectively. On joining the vessels, the gas reaches a common pressure p and a common temperature T. The ratio p/T is equal to p T + p2T2 (a) 1 2 T1 ´ T2 (c)

1 é p1T2 + p2T1 ù ú ê 2ë T1T2 û

p T + p2T1 (b) 1 2 T1 + T2 (d)

p1T2 - p2T1 T1 ´ T2

38. Two moles of monoatomic gas is mixed with three moles of a diatomic gas. The molar specific heat of the mixture at constant volume is (a) 1.55 R (c) 1.63 R

(b) 2.10 R (d) 2.20 R

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

kinetic energy of the atoms increases potential energy of the atoms increases total energy of the atoms increases the potential energy curve is asymmetric about the equilibrium distance between neighbouring atoms

41. An iron tyre is to be fitted on a wooden wheel 1 m in diameter. The diameter of tyre is 6 mm smaller than that of wheel. The tyre should be heated so that its temperature increases by a minimum of (the coefficient of cubical expansion of iron is 3.6 ´ 10–5 / ° C (b) 334°C

(c) 500°C

(d) 1000°C

42. A glass flask of volume one litre at 0°C is filled, level

35. 8 g of O2 , 14 g of N2 and 22 g of CO2 is mixed in a

2 (a) 3

(b) 6 ´ 10 -6 /°C (d) 27 ´ 10 -6 /°C

40. Solids expand on heating because

(a) 167°C

(b) O2 (d) H2

(a) 1.4 atm (c) 3.7 atm

glass vessel is 153 ´ 10-6 /°C and in a steel vessel is 144 ´ 106 /°C. If a for steel is 12 ´ 10-6 /°C, then that of glass is (a) 9 ´ 10 -6 /°C (c) 36 ´ 10 -6 /°C

(b) 28.5°C (d) 33.5°C

(a)

39. The coefficient of apparent expansion of mercury in a

full of mercury at this temperature. The flask and mercury are now heated to 100°C. How much mercury will spill out, if coefficient of volume expansion of mercury is 1.82 ´ 10–4 / ° C and linear expansion of glass is 0.1 ´ 10–4 / ° C respectively? (a) 21.2 cc (c) 1.52 cc

(b) 15.2 cc (d) 2.12 cc

43. A steel scale measures the length of a copper wire as 80.0 cm, when both are at 20°C (the calibration temperature for scale). What would be the scale read for the length of the wire when both are at 40°C? (Given a steel = 11 ´ 10–6 per° C and a copper = 17 ´ 10–6 per°C) (a) 80.0096 cm (c) 1 cm

(b) 80.0272 cm (d) 25.2 cm

44. When the temperature of a rod increases from t to t + Dt, its moment of inertia increases from I to I + DI . If a be the coefficient of linear expansion of DI is the rod, the then the value of I (a) 2 a Dt a Dt (c) 2 Dt (e) 2a

(b) a Dt Dt (d) a

45. Two metal strips that constitute a thermostat must necessarily differ in their (a) mass (b) length (c) resistivity (d) coefficient of linear expansion

Heat and Kinetic Theory of Gases 46. A metal ball immersed in alcohol weighs w1 at 0°C

and w2 at 59°C. The coefficient of cubical expansion of the metal is less than that of alcohol. Assuming that the density of metal is large compared to that of alcohol, it can be shown that (a) w1 > w2 (c) w1 < w2

(b) w1 = w2 (d) w2 = ( w1/ 2)

49 ´ 10-5 /K. What is the fractional change in its density for a 30°C rise in temperature? [NCERT] (a) 1.45 ´ 10 -3

(b) 2.45 ´ 10 -3

(c) 2.45 ´ 10 -2

(d) 1.45 ´ 10 -2

48. A piece of metal weighs 46 g in air. When it is immersed in the liquid of specific gravity 1.24 at 27°C it weighs 30 g. When the temperature of liquid is raised to 42°C the metal piece weighs 30.5 g, specific gravity of the liquid at 42°C is 1.20, then the linear expansion of the metal will be (a) 3.316 ´ 10 –5 /° C

(b) 2.316 ´ 10 –5 /° C

(c) 4.316 ´ 10 –5 /° C

(d) None of these

49. It is known that wax contracts on solidification. If molten wax is taken in a large vessel and it is allowed to cool slowly, then (a) it will start solidifying from the top to downward (b) it will starts solidifying from the bottom to upward (c) it will start solidifying from the middle, upward and downward at equal rates (d) the whole mass will solidify simultaneously

50. A substance of mass m kg requires a power input of P watts to remain in the molten state at its melting point. When the power is turned off, the sample completely solidifies in time t sec. What is the latent heat of fusion of the substance? Pm t m (c) Pt

Pt m t (d) Pm

(a)

(b)

contained in a calorimeter of water equivalent to 0.02 kg at 15°C till the temperature of the calorimeter and its contents rises to 80°C. The mass of the steam condensed in kg is (b) 0.065

(c) 0.260

coil of 1 kW at 27°C. The lid of the container is open and energy dissipates at rate of 160 J/s. In how much time temperature will rise from 27°C to 77°C ? [Given specific heat of water is 4.2 kJ/kg] (b) 6 min 2 s (d) 14 min

54. A lead bullet at 27°C just melts when stopped by an obstacle. Assuming that 25% of heat is absorbed by the obstacle, then the velocity of the bullet at the time of striking (MP of lead = 327°C, specific heat of lead = 0.03 cal/g°C, latent heat of fusion of lead = 6 cal/g and J = 4.2 joule/cal) (a) 410 m/s (c) 307.5 m/s

(b) 1230 m/s (d) None of these

55. We have seen that a gamma-ray dose of 3 Gy is lethal to half the people exposed to it. If the equivalent energy were absorbed as heat, what rise in body temperature would result? (a) 300 mK (c) 455 mK

(b) 700 mK (d) 390 mK

56. 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 is 23°C. The temperature when A and C are mixed, is (a) 18.2°C (c) 20.2°C

(b) 22°C (d) 25.2°C

57. In an industrial process 10 kg of water per hour is to be heated from 20°C to 80°C. To do this steam at 150°C is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at 90°C. How many kg of steam is required per hour? (Specific heat of steam = 1 calorie per g°C, Latent heat of vaporisation = 540 cal/g) (a) 1 g

51. Steam at 100°C is passed into 1.1 kg of water

(a) 0.130

53. Water of volume 2L in a container is heated with a

(a) 8 min 20 s (c) 7 min

47. The coefficient of volume expansion of glycerine is

(b) 1 kg

(b) 6 kg

(c) 4 kg

(d) 2 kg

(d) 10 kg

are maintained at different temperatures t1 and t2 . The liquid columns in the two arms have heights l1 and l2 respectively. The coefficient of volume expansion of the liquid is equal to t1

52. 2 kg of ice at –20°C is mixed with 5 kg of water at

(a) 7 kg

(c) 10 g

58. In a vertical U-tube containing a liquid, the two arms

(d) 0.135

20°C in an insulating vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heats of water and ice are 1 kcal/kg per°C and 0.5 kcal/kg/°C while the latent heat of fusion of ice is 80 kcal/kg.

531

t2 l1

l2

(a)

l1 - l2 l2t1 + l1t2

(b)

l1 - l2 l1t1 - l2t2

(c)

l1 + l2 l2t1 + l1t2

(d)

l1 + l2 l1t1 + l2t2

532 JEE Main Physics 59. The coefficient of linear expansion of crystal in one

66. A closed compartment containing gas is moving with

direction is a1 and that in every direction perpendicular to it is a2 . The coefficient of cubical expansion is

some acceleration in horizontal direction. Neglect effect of gravity. Then, the pressure in the compartment is

(a) a1 + a2 (c) a2 + 2 a2

(b) 2 a1 + a2 (d) None of these

60. Three rods of equal length l are

(a) same everywhere (c) lower in rear side

(b) lower in front side (d) lower in upper side

R

67. A room is maintained at 20°C by a heater of

joined to form an equilateral triangle PQR. O is the mid point of PQ. Distance OR remains same for small change in temperature. Coefficient of P O linear expansion for PR and RQ is same, i. e., a 2 but that for PQ is a 1. Then

resistance 20 W connected to 200 V mains. The temperature is uniform throughout the room and heat is transmitted through a glass window of area 1 m 2 and thickness 0.2 cm. What will be the temperature outside? Given that thermal conductivity K for glass is 0.2 cal/m/°C sec and J = 4.2 J /cal.

(a) a2 = 3 a1 (c) a1 = 3 a2

Q

(b) a2 = 4 a1 (d) a1 = 4 a2

61. An electric kettle takes 4 A current at 220 V. How much time will it take to boil 1 kg of water from temperature 20°C? The temperature of boiling water is 100°C (a) 12.6 min (c) 6.3 min

(b) 4.2 min (d) 8.4 min

62. 10 g of ice at –20°C is droped into a calorimeter containing 10 g of water at 10°C; the specific heat of water is twice that of ice. When equilibrium is reached, the calorimeter will contain (a) 20 g of water (b) 20 g of ice (c) 10 g ice and 10 g water (d) 5 g ice and 15 g water

63. A copper block of mass 2.5 kg is heated in furnace to a temperature of 500°C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 039 . J/g-K; heat of fusion of water = 335 J/g) [NCERT] (a) 25 kg (c) 9 kg

(b) 15 kg (d) 13 kg

64. Steam is passed into 22 g of water at 20°C. The mass of water that will be present when the water acquires a temperature of 90°C (Latent heat of steam is 540 cal/g) is (a) 24.8 g (c) 36.6 g

(b) 24 g (d) 30 g

65. Ice starts forming in lake with water at 0°C and when the atmospheric temperature is –10°C. If the time taken for 1 cm of ice be 7 h, then the time taken for the thickness of ice to change from 1 cm to 2 cm is (a) 7 h (c) Less than 7 h

(b) 14 h (d) More than 7 h

(a) 15.24°C (c) 24.15°C

(b) 15.00°C (d) None of these

68. There is formation of layer of snow x cm thick on water, when the temperature of air is - q °C (less than freezing point). The thickness of layer increases from x to y in the time t, then the value of t is given by ( x + y ) ( x - y ) rL 2 kq ( x + y ) ( x - y ) rL (c) kq

( x - y ) rL 2 kq ( x - y ) rLk (d) 2q

(a)

(b)

69. A composite metal bar of uniform section is made up of length 25 cm of copper, 10 cm of nickel and 15 cm of aluminium. Each part being in perfect thermal contact with the adjoining part. The copper end of the composite rod is maintained at 100°C and the aluminium end at 0°C. The whole rod is covered with belt so that no heat loss occurs at the side. If K Cu = 2 K Al and K Al = 3 K Ni , then what will be the temperatures of Cu-Ni and Ni-Al junctions repectively Cu

Ni

Al 0°C

100°C

(a) 23.33°C and 78.8°C (c) 50°C and 30°C

(b) 83.33°C and 20°C (d) 30°C and 50°C

70. Three rods of identical area of cross-section and made from the same metal form the sides of an isosceles triangle ABC right angled at B. The points A and B are maintaned temperatures T and 2T respectively. In the steady state the temperature of the point C is TC . Assuming that only heat conduction T takes place, C is equal to T 1 ( 2 + 1) 1 (c) 2 ( 2 - 1) (a)

3 ( 2 + 1) 1 (d) 3 (2 - 1)

(b)

Heat and Kinetic Theory of Gases 71. The only possibility of heat flow in a

77. Two metallic spheres S1 and S2 are made of the same

themros flask is through its cork which is 75 cm2 in area and 5 cm thick its thermal conductivity is 0.075 cal/cm sec°C. The outside temperatue is 40°C and latent heat of ice is 80 cal g -1. Time taken by 500 g of ice at 0°C in the flask to melt into water at 0°C is (a) 2.47 h (c) 7.42 h

material and have identical surface finish. The mass of S1 is three times 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

(b) 4.27 h (d) 4.82 h

72. 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 g/s be the rate of melting of ice in two cases respectively. The ratio of q1/ q2 is (a) 1/2

(b) 2/1

(c) 4/1

(d) 1/4

73. A solid cube and a solid sphere of the same material have equal surface area. Both are at the same temperature 120°C, then (a) (b) (c) (d)

both the cube and the sphere cool down at the same rate the cube cools down faster than the sphere the sphere cools down faster than the cube whichever is having more mass will cool down faster

74. A black body is at a temperature of 2880 K. The energy of radiation emitted by this object with wavelength between 499 nm and 500 nm is U1, between 999 and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien’s constant b = 2.88 ´ 106 nmK. Then (a) U1 = 0

(b) U3 = 0

(c) U1 > U2

(d) U2 > U1

75. A black metal foil is warmed by radiation from a small sphere at temperature T and at a distance d. It is found that the power received by the foil is P. If both the temperature and distance are doubled, the power received by the foil will be (a) 16 P

(b) 4 P

(c) 2 P

(d) P

76. Three rods of same dimensions

R

are arranged as shown in figure. They have thermal K1 K2 conductivities K 1, K 2 and K 3. The points P and Q are maintained at different P Q K3 temperatures for the heat to flow at the same rate along PRQ and PQ, then which of the following options is correct? 1 (a) K3 = ( K1 + K2 ) 2 K1K2 (c) K3 = K1 + K2

(b) K3 = K1 + K2 (d) K3 = 2 ( K1 + K2 )

533

(a) 1/3

(b) (1 / 3)1 /3

(c) 1 / 3

(d) 3 /1

78. Three discs A, B and C having radii 2 m, 4 m and 6 m respectively are coated with carbon black on their other surfaces. The wavelengths corresponding to maximum intensity are 300 nm, 400 nm and 500 nm respectively. The power radiated by them are Qa , Qb and Qc respectively (a) Qa is maximum (c) Qc is maximum

(b) Qb is maximum (d) Qa = Qb = Qc

79. The total energy radiated from a black body source is collected for one minute and is used to heat a quantity of water. The temperature of water is found to increase from 20°C to 20.5°C. If the absolute temperature of the black body is doubled and the experiment is repeated with the same quantity of water at 20°C, the temperature of water will be (a) 21°C (c) 24°C

(b) 22°C (d) 28°C

80. A solid sphere and a hollow sphere of the same material and size are heated to the same temperature and allowed to cool in the same surroundings. If the temperature difference between each sphere and its surroundings is T, then (a) (b) (c) (d)

the hollow sphere will cool at a faster rate for all values of T the solid sphere will cool at a faster rate for all values of T both spheres will cool at the same rate for all values of T both spheres will cool at the same rate only for small values of T

81. A solid copper cube of edges 1 cm is suspended in an evacuated enclosure. Its temperature is found to fall from 100°C to 99°C in 100 s. Another solid copper cube of edges 2 cm, with similar surface nature, is suspended in a similar manner. The time required for this cube to cool from 100°C to 99°C will be approximately (a) 25 s (c) 200 s

(b) 50 s (d) 400 s

82. A body initially at 80°C cools to 64°C in 5 min and to 52°C in 10 min. The temperature of the body after 15 min will be (a) 42.7°C (c) 47°C

(b) 35°C (d) 40°C

534 JEE Main Physics 83. A 5 cm thick ice block is there on the surface of water in a lake. The temperature of air is –10°C; how much time it will take to double the thickness of the block ? (L = 80 cal/g, K ice = 0.004 erg/s-K, dice = 0.92 g cm –3) (b) 191 h

(c) 19.1 h

(d) 1.91 h

84. Four identical rods of same material are joined end to end to form a square. If the temperature difference between the ends of a diagonal is 100°C then the temperature difference between the ends of other diagonal will be (a) 0°C 100 (b) °C; where l is the length of each rod l 100 °C (c) 2l

85. 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 (b) 1.6

(c) 0.2

(d) 0.1

86. One end of a copper rod of length 1.0 m and area of

cross-section 10-3 m 2 is immersed in boiling water and the other end in ice. If the coefficient of thermal conductivity of copper is 92 cal/m-s-°C and the latent heat of ice is 8 ´ 104 cal/kg, then the amount of ice which will melt in one min is -3

(b) 8 ´ 10 kg

-3

(d) 5.4 ´ 10 -3 kg

(a) 9.2 ´ 10 kg (c) 6.9 ´ 10 kg

-3

wall area of 1 m2 and a wall thickness of 5.0 cm. The thermal conductivity of the ice box is K = 0.01 joule/metre-°C. It is filled with ice at 0°C along with eatables on a day when the temperature is 30°C. The latent heat of fusion of ice is 334 ´ 103 joule/kg. The amount of ice melted in one day is (1 day = 86.400 s) (b) 7760 g

(c) 11520 g

88. Five rods of same dimensions are aranged as shown in the K1 figure. They have thermal conductivities K 1, K 2 , K 3, K 4 A and K 5 . When points A and B K3 are maintained at different temperatures, no heat flows through the central rod if (a) K1 = K4 and K2 = K3 (c) K1K2 = K3 K4

(b) K1K4 = K2 K3 K2 K (d) 1 = K4 K3

(b) proportional to (d) proportional to 1 / r

90. A solid copper sphere (density r and specific heat capacity c) of radius r at an initial temperature 200 K is supended inside a chamber whose walls are at almost 0 K. The time required (in ms) for the temperature of the sphere to 100 K is (a)

72 rrc 7 s

(b)

7 rrc 72 s

(c)

27 rrc 7 s

(d)

7 rrc 27 s

at 0°C and the other end B in water at 100°C. If a point P on the rod is maintained at 400°C, then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporationof water is 540 cal/g and latent heat of melting of ice is 80 cal/g. If the point P is at a distance of lx from the ice end A, find the value of l (Neglect any heat loss to the surroundings). (a) 9

(b) 2

(c) 6

(d) 1

92. A sphere and a cube of same material and same volume. One heated upto same temperature and allowed to cool in the same surroundings. The ratio of the amounts of radiation emitted will be 4p :1 3 2 /3 1 æ4pö (d) ç ÷ :1 2è 3 ø

(a) 1: 1 æpö (c) ç ÷ è6ø

(b)

1 /3

:1

93. Four rods of identical cross-sectional area and made

87. An ice box used for keeping eatable cold has a total

(a) 776 g

(a) independent of r (c) proportional to r2

91. A metal rod AB of length 10 x has its one end A in ice

(d) 100°C

(a) 3.2

cooling is

(d) 1552 g

(a)

2 +1 T 2

2 T 2 +1 (d) None of these

(b)

(c) 0

94. The graph AB shown in figure is a plot of temperature body in degree celsius and degree fahrenheit. Then,

C K2 K5

B K4

D

from the same metal form the sides of square. The temperature of two diagonally opposite points are T and 2 T respectively in the steady state. Assuming that only heat conduction takes place, what will be the temperature difference between other two points?

B

100°C

Centigrade

(a) 1 h

89. A hot metallic sphere of radius r radiates heat. Its

A

32°F

(a) slope of line AB is 9/5 (c) slope of line AB is 1/9

212°F Fahrenheit

(b) slope of line AB is 5/9 (d) slope of line AB is 3/9

Heat and Kinetic Theory of Gases 95. The graph shows the variation of temperature ( T) of

Y T°C

one kilogram of a material with the heat ( H ) supplied to it. At O, the substance in the solid state. From the graph, we can conclude that

535

A

90 B

D

80

T

D(H4, T2)

(a) 40 H

O

(a) T2 is the melting point of the solid (b) BC represents the change of state from solid to liquid (c) ( H2 - H1 ) represents the latent heat of fusion of the substance (d) ( H3 - H1 ) represents the latent heat of vaporization of the liquid

converted to steam at 100°C. Which of the following curves represents the phenomenon qualitatively

210

E

B 60

C

30 A

(b) CD

Time

(c) ED

(d) EF

(b)

Density

(a)

Heat supplied

(c)

the state of mattter denotes

Temperature

(d)

Density

Temperature

97. The portion AB of the indicator diagram representing p

F D

Density

Heat supplied

240

Density

Heat supplied

(d)

(d) 20

represented by the

Temperature

Temperature

Heat supplied

(c)

(c) 100

100. The variation of density of water with temperature is

Temperature

Temperature

(a) BC

(b)

(b) 80

energy is supplied at a constant rate. Then temperature versus time graph is as shown in the figure. The substance is in liquid state for the portion (of the graph)

96. A block of ice at –110°C is slowly heated and

(a)

X

99. A solid substance is at 30°C. To this substance heat

B(H2, T1)

α

4 min Time (min.)

Temperature (T°C)

β

A(H1, T1)

2 min

γ

C(H3, T2)

Temperature

A

Temperature

101. If a graph is plotted taking the temperature in Fahrenheit along Y -axis and the corresponding temperature in celsius along the X-axis, it will be straight line

C B D V

(a) the liquid state of matter (b) gaseous state of matter (c) change from liquid to gaseous state (d) change from gaseous state of liquid state

98. The figure given below shows the cooling curve of pure wax material after heating. It cools from A to B and solidifies along BD. If L and C are respective values of latent heat and the specific heat of the liquid wax, the ratio L / C is

(a) having a +ve intercept on Y-axis (b) having a +ve intercept on X-axis (c) passing through the origin (d) having a –ve intercepts on both the axis

102. Which of the curves in figure represents, the relation between Celsius and Fahrenheit temperatures? (a) 1 (c) 3

(b) 2 (d) 4

°C 2

3

4

1

°F

536 JEE Main Physics 107. Two substances Aand B of equal mass m are heated

at uniform rate of 6 cal s -1 under similar conditions. A graph between temperature and time is shown in figure. Ratio of heat absorbed H A / H B by them for complete fusion is

Time

(b) 2.52 ´ 103 K

(c) 2.52 ´ 10 4 K

(d) 3.52 ´ 103 K

105. The graph signifies

t

(b) 4/9

2 3 4

5 6

(c) 8/5

(b) 50°C

Temperature (T)

(d) 5/8

(c)

50°C

Temperature (T)

(d) 50°C

Temperature (T)

Temperature

7

Density (D)

(a)

50°C

Temperature (T)

109. Which curve shows the rise of temperature with the amount of heat supplied, for a piece of ice? B

A C

106. Which of the substances A, B or C has the highest specific heat? The temperature versus time graph is shown A B C

Time (t)

(a) A (b) B (c) C (d) All have equal specific heat

(a) A

D 400 300 200 Amount of heat supplied

(b) B

(c) C

110. A solid material is supplied with heat at constant rate and the temperature of the material changes as shown. From the graph, the false conclusion drawn is

(d)D Temperature

Temperature in K

500

adiabatic expansion of a gas isothermal expansion of a gas change of state from liquid to solid cooling of a heated soil

Temperature (T)

20

dependence of density of water correctly?

Time

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

B

40

108. Which one of the figures gives the temperature

of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at –20°C? (Atomic mass of [NCERT] Ar = 39.9 u, of H = 4.0 u). (a) 3.52 ´ 103 K

60

(a) 9/4

Density (D)

104. At what temperature is the root mean square speed

A

80

1

Density (D)

(a) Its specific heat capacity is greater in the solid state than in the liquid state (b) Its specific heat capacity is greater in the liquid state than in the solid state (c) Its latent heat of vaporization is greater than its latent heat of fusion (d) Its latent heat of vaporization is smaller than its latent of fusion

100

Density (D)

Temperature

matter, at a uniform rate. Its temperature is plotted against time, as shown. Which of the following conclusions can be drawn?

Temperature (°C)

103. Heat is supplied to a certain homogeneous sample of

O

E C D A B CD = 2AB Heat input

(a) AB and CD of the graph represent phase changes (b) AB represents the change of state from solid to liquid (c) Latent heat of fusion is twice the latent heat of vaporization (d) CD represents change of state from liquid to vapour (e) Latent heat of vaporization is twice the latent heat of fusion

Heat and Kinetic Theory of Gases

537

115. Following graph shows the correct variation in

pressure of 1 atm. The rate of heating is constant. Which one of the following graphs represents the variation of temperature with time?

intensity of heat radiations by black body and frequency at a fixed temperature

(b)

Temperature

3500 K 2500 K 1500 K ν

Time

(d)

Eλ

Temperature

Temperature

Time

(c)

Visible Infrared

(a)

Ultraviolet

(a)

Temperature

Eλ

Ultraviolet

111. Liquid oxygen at 50 K is heated to 300 K at constant

Visible Infrared 1500 K 2500 K 3500 K

(b)

Time

Time

ν

112. The graph shown in the adjacent diagram, represents the variation of temperature ( T) of two bodies, x and y having same surface area, with time ( t) due to the emission of radiation. Find the correct relation between the emissivity ( e) and absorptivity (a) of the two bodies

Eλ

Infrared Visible Ultraviolet 3500 K

(c)

2500 K 1500 K

T

ν y

Eλ

x t

(a) ex > e y and ax < a y (c) ex > e y and ax > a y

wavelength for three black bodies at temperatures T1, T2 I T1 and T3 respectively are as shown. Their temperatures are such that (a) T1 > T2 > T3 (c) T2 > T3 > T1

2500 K 3500 K ν

T3 T2

λ

(b) T1 > T3 > T2 (d) T3 > T2 > T1

114. The adjoining diagram shows the spectral energy density distribution El of a black body at two different temperatures. If the areas under the curves are in the ratio 16 : 1, the value of temperature T is TK Eλ

2000 K

116. A brass boiler has a base area 0.15 m2 and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. The temperature of the part of the flame in contact with the boiler will be. (Thermal conductivity of brass = 109 J/s-m-K, Heat of [NCERT] vapourization of water = 2256 ´ 103 J/kg) (a) 158°C

(b) 208°C

(c) 238°C

(d) 264° C

117. A body cools in a surrounding which is at a constant

temperature of q 0 . Assume that it obeys Newton’s law of cooling. Its temperature q is plotted against time t. Tangents are drawn to the curve at the points P (q = q 2 ) and Q(q = q 1). These tangents meet the time axis at angles of f2 and f1, as shown θ

λ

(a) 32000 K (c) 8000 K

1500 K

(d)

(b) ex < e y and ax > a y (d) ex < e y and ax < a y

113. The plots of intensity versus

Infrared Visible Ultraviolet

(b) 16000 K (d) 4000 K

θ2

P

θ1 θ0

Q φ2

φ1

t

538 JEE Main Physics tan f2 q - q0 = 1 tan f1 q2 - q0 tan f1 q1 (c) = tan f2 q2

tan f2 q - q0 = 2 tan f1 q1 - q0 tan f1 q2 (d) = tan f2 q1

(a)

121. The energy distribution E with

(b)

118. Shown below are the black body radiation curves at temperatures T1 and T2 ( T2 > T1). Which of the following plots is correct? T2

I

T2

I

T1

(a)

T1

(b)

λ

(a) shift towards left and become higher (b) rise high but will not shift (c) shift towards right and become higher (d) shift towards left and the curve will become broader

x = 0 to x = l. If its thermal resistance per unit length is uniform which of the following graph is correct? T

T2

I

T1

(c)

T

T1

(d)

(a)

T2

(b) O

λ

Intensity

27°C and 327°C is shown in the figure. Let A1 and A2 be the areas under the two curves respectively. The A value of 2 is A1

T

x

T

(c)

(d) O

O

x

x

123. Radius of a conductor increases uniformly from left and to right end as shown in figure.

2 327°C

1

27°C

T1

Wavelength

(a) 1 : 16 (c) 2 : 1

H

(b)

x

H

(c)

(d) O

Time

O

x

H

Time

(d) Time

H

O

Temperature

Temperature

Time

Material of the conductor is isotropic and its curved surface is thermally insulated from surrounding. Its ends are maintained at temperatures T1 and T2 ( T1 > T2 ). If in steady state, heat flow rate is equal to H, then which of the following graphs is correct?

(a)

Temperature

Temperature

higher than the room temperature and allowed to cool in a room free from air currents. Which of the following curves correctly represents the rate of cooling?

(b)

T2 x

(b) 4 : 1 (d) 16 : 1

120. A block of metal is heated to a temperature much

(c)

O

x

λ

119. The spectrum of a black body at two temperatures

(a)

λ

122. Heat is flowing through a conductor of length l from

λ I

E

the wavelength ( l) for the black body radiation at temperature T kelvin is shown in the figure. As the temperature is O increased the maxima will

x

O

x

Heat and Kinetic Theory of Gases 124. Which of the following graphs correctly represents R

the relation between ln E and ln T, where E is the amount of radiation emitted per unit time from unit area of a body and T is the absolute temperature? ln E

B

(θ – θ 0)

(b) ln T

ln T ln E

ln E

(c) ln T

ln T

O

125. A hollow copper sphere S and a hollow copper cube C, both of negligible thin walls of same area, are filled with water at 90°C and allowed to cool in the same environment. The graph that correctly represents their coolings is T

T

(a)

(b)

S

C S

t

t

T S C, S

(d) C t

t

126. Which of the following is the nm - T graph for a perfectly black body (n m = maximum frequency of radiation) ? νm

ratio 1 : 2. Both are at same temperature. Ratio of heat radiation energy emitted per second by them is (a) 1 : 2

(b) 1 : 4

(c) 1 : 8

(d) 1 : 16

More Than One Correct Option 129. If a and l are coefficiehnts of linear, superficial and volume expansion respectively, then b 1 = a 2 g 3 (c) = a 1

(a)

130. Mark the correct options

C

T

(a) A and B have same specific heats (b) Specific heat of A is less (c) Specific heat of B is less (d) Nothing can be said

128. Two spheres made of same material have radii in the

(d)

b 2 = g 3 b g (d) = a b (b)

[NCERT Exemplar]

(a) A system X is in thermal equilibrium with Y but not with Z. System Y and Z may be in thermal equilibrium with each other. (b) A system X is in thermal equilibrium with Y but not with Z. System Y and Z are not in thermal equilibrium with each other. (c) A system X is neither in thermal equilibrium with Y nor with Z. The systems Y and Z must be in thermal equilibrium with each other. (d) A system X is neither in thermal equilibrium with Y nor with Z. The system Y and Z may be in thermal equilibrium with each other.

131. ‘Gulab Jamuns’ (assumed to be spherical) are to be

D B C A T

(a) A (c) C

A

ln E

(a)

(c)

539

(b) B (d) D

127. Two circular discs A and B with equal radii are blackened. They are heated to same temperature and are cooled under identical conditions. What interference do you draw from their coolings curves?

heated in an oven. They are available in two sizes, one twice bigger (n radius) than the other. Pizzas (assumed to be discs) are also to be heated in oven. They are also in two sizes, one twice big (in radius) than the other. All four are put together to be heated to oven temperature. Choose the correct option from [NCERT Exemplar] the following (a) Both size gulab jamuns will get heated in the same time (b) Smaller gulab jamuns are heated before bigger ones (c) Smaller pizzas are heated before bigger ones (d) Bigger pizzas are heated before smaller ones

540 JEE Main Physics 132. A spherical body of radius r radiates power P and its rate of cooling is R. (c) R µ r2

(b) P µ r2

(a) P µ r

1 (d) R µ r

133. Refer to the plot of temperature versus time showing

Temperature (°C)

the changes in the state of ice on heating (not to scale) Which of the following is correct? [NCERT Exemplar] E 100

C A

B tm

0

137. Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface areas of the two bodies are the same. The two bodies emit total radiant power at the same rate. The wavelength l B corresponding to maximum spectral radiancy in the radiation from B is shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A, by 1.00 mm. If the temperature of A is 5802 K (a) The temperature of B is 1934 K (b) l B = 1.5 mm (c) The temperature of B is 11604 K (d) The temperature of B is 2901 K

D

138. ABCDEFGH is hollow cube made of an insulator.

time (min)

(a) The region AB represents ice and water in thermal equilibrium (b) At B water starts boiling (c) At C all the water gets converted into steam (d) C to D represents water and steam in equilibrium at boiling point

Face ABCD has positive charge on it. Inside the cube, we have ionized hydrogen. B

C +P

A

D

134. Under which of the following conditions, the law pV = RT is not obeyed by a real gas? (a) High pressure and high temperature (b) Low pressure and low temperature (c) Low pressure and high temperature (d) High pressure and low temperature

E

pV

[NCERT Exemplar]

[NCERT Exemplar]

pV

(a)

H

The usual kinetic theory expression for pressure

135. Which of the following graphs do/does not represent the behaviour of an ideal gas?

G

F

(b)

(a) will be valid (b) will not be valid since the ions would experience forces orger than due to collisions with the walls (c) will not be valid since collisions with walls would not be elastic (d) will not be valid because isotropy is lost

139. In a diatomic molecule, the rotational energy at a V

V

pV

pV

(c)

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

(d) V

given temperature

V

136. A glass full of hot milk is poured on the table. It begins to cool gradually Which of the follwong is correct? (a) The rate of cooling is constant till milk attains the temperature of the surrounding (b) The temperature of milk falls off exponentially with time (c) While cooling, there is a flow of heat from milk to the surrounding as well as from surrounding to the milk but the net flow of heat is from milk to the surrounding and that is why it cools (d) All three phenomenon, conduction, convection and radiation are responsible for the loss of heat milk to the surroundings.

[NCERT Exemplar]

obeys Maxwell's distribution have the same value for all molecules equals the translational kinetic energy for each molecule is (2/3)rd the translational kinetic energy for each molecule

Comprehension Based Questions Passage I When two substances at different temperatures are mixed together, exchange of heat occurs between them till they acquire a common temperature. In thermal equilibrium, heat gained by one substance is equal to heat lost by the other substance. This is called principle of calorimetry.

140. 10 g of ice at 0°C and water at 100°C are mixed together. The resultant temperature would be (a) 10°C

(b) 5°C

(c) 50°C

(d) 40°C

Heat and Kinetic Theory of Gases 141. Equal masses of ice at 0°C is put in 10 g of water at 80°C. The final temperature would be (a) 10°C (c) 40°C

(b) 0°C (d) 80°C

142. An ice block of mass m at 0°C is put in water of mass 2 m at 60°C. The final temperature would be (a) 60°C (c) 30°C

(b) 0°C (d) 13.3°C

Passage II The latent heat of fusion of ice is 80 calg–1 and latent heat of steam is 540 calg–1. Change of state occurs only at melting point or boiling point of the substance. There is no change in temperature during the entire change of state. For rise in temperature ( DT) heat required DQ = mcDT, where c is specific heat of the substance.

143. Heat required to convert 1 g of ice at – 5°C to water at 0°C is (specific heat of ice = 0.5 calg–1°C–1) (a) 80 cal (c) 77.5 cal

(b) 82.5 cal (d) 802.5 cal

144. Heat released when 10 g of steam at 100°C cools to water at 100°C is (a) 540 cal (c) 5400 cal

(b) 54 cal (d) 54000 cal

145. Heat required to melt 10 g of ice at 0°C to water at 0°C is (a) 800 cal (c) 8 cal

(b) 80 cal (d) None of these

146. SI unit of latent heat is (a) J kg–1 (c) kg J–1

(b) J kg–1 K–1 (d) kg J–1 K–1

Passage III A cylinder is containing nitrogen at 2 atm and temperature 17°C. The radius of a nitrogen molecule to be roughly 1.0A. Molecular mass of nitrogen = 28.0 u, Boltzmann constant, k = 1.38 ´ 10–23 JK –1.

147. The mean free path of nitrogen molecule is

541

150. Time taken for collision of nitrogen molecules is (a) 4 ´ 10 -13 s (c) 6 ´ 10

-13

s

(b) 5 ´ 10 –13 s (d) 7 ´ 10 -13 s

Assertion and Reason Directions Question No. 151 to 161 are Assertion-Reason type. Each of these contains two Statements: Statement I (Assertion), Statement II (Reason). Each of these questions also has four alternative choice, only one of which is correct. You have to select the correct choices from the codes (a), (b), (c) and (d) given below (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 correct explanation of the Assertion (c) If Assertion is true but Reason is false (d) If Assertion is false but the Reason is true

151. Assertion The SI unit of Stefan’s constant is Wm–2 K–4. Reason This follows from Stefan’s Law, E = aT4 E Wm -2 \ a= 4 = T K4

152. Assertion The rate of loss of heat of a body at 300 K is R. At 900 K, the rate of loss becomes 81 R. Reason This is as per Newton’s law of cooling.

153. Assertion When temperature difference across the two sides of a wall is increased, its thermal conductivity increases. Reason Thermal conductivity depends on nature of material of the wall.

154. Assertion Cooking in a pressure cooker is faster. Reason

Because steam does not leak out.

155. Assertion For higher temperatures, the peak emission wavelength of a black body shifts to lower wavelengths. Reason Peak emission wavelength of a black body is proportional to the fourth power of temperatures.

(a) 1.1 ´ 10 –7 m

(b) 2.1 ´ 10 –7 m

156. Assertion Two bodies at different temperature, if

(c) 3.1 ´ 10 –7 m

(d) 0.8 ´ 10 –7 m

brought in thermal contact do not necessary settle to the mean temperature. Reason The two bodies may have different thermal capacities.

148. Root mean square velocity of nitrogen molecules (a) 4 ´ 102 ms –1 2

(c) 6 ´ 10 ms

–1

(b) 5 ´ 102 ms –1 (d) 7 ´ 102 ms –1

149. Collision frequency of the nitrogen molecules is (a) 4 ´ 10 9 s –1

(b) 5 ´ 10 9 s –1

(c) 6 ´ 10 9 s –1

(d) 8 ´ 10 9 s –1

157. Assertion When

small temperature difference between a liquid and its surrounding is doubled, the rate of loss of heat of the liquid becomes twice. Reason This is as per Newton’s law of cooling.

542 JEE Main Physics 158. Assertion When temperature of a black body is halved, wavelength corresponding to which energy radiated is maximum becomes twice. Reason This is as per Wien’s law.

If the temperature is doubled and oxygen gas dissociates into atomic oxygen, the rms speed would be 2 c. Reason c µ

159. Assertion When speed of sound in a gas is c, then crms = Reason

c=

T M

161. Assertion The number of molecules in 1 cc of water is

3 ´c g

1 ´ 1022 . 3 Reason The number of molecules per gram mole of water is equal to Avogadro’s number ( = 6.023 ´ 1023 g –1 mol –1). nearly equal to

g p r

160. Assertion The root mean speed (rms) of oxygen molecules at a certain absolute temperature T is c.

Previous Years’ Questions 162. Statement

I The temperature dependence of resistance is usually given as R = R0 (1 + Dt). The resistance of a wire changes from 100 W to 150 W when its temperature is increased from 27°C to 227°C. This implies that a = 2.5 ´ 10–3° C–1. Statement II R = R0 (1 + a Dt) is valid only when the change in the temperature is small and [AIEEE 2009] DR = ( R - R0 ) T3 > T1 (d) T1 > T3 > T2

placed on table. Box A contains one mole of nitrogen at temperature T0 , while box B contains 1 mole of helium at temperature (7/3) T0 . The boxes are then put into thermal contact with each other, and heat flows between them until the gases reach a common final temperature (Ignore the heat capacity of boxes) then, the final temperature of gases, Tf , in terms of [Kerala CET 2008] T0 is 7 (b) Tf = T0 3 5 (d) Tf = T0 3

temperature constant. What is the effect on kinetic energy of molecules? [UP SEE 2006] (a) Increase (c) No change

179. Two balloons are filled, one with pure helium gas and the other by air, respectively. If the pressure and temperature of these balloons are same then the number of molecules per unit volume is [UP SEE 2006] (a) more in the helium filled balloon (b) same in both balloons (c) more in air filled balloon (d) in the ratio of 1 : 4

(b) 8.3 J mol–1K–1 (d) 2 cal mol–1K–1

[UP SEE 2006]

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

174. The value of a metal sphere increase by 0.24% when its temperature is raised by 40°C. The coefficient of linear expansion of the metal is...°C–1. [BVP Engg. 2007]

(b) 6 ´ 10 -5 (d) 1.2 ´ 10 –5

does not change increase decreases may either increase or decrease depending on the process used

181. The average energy for molecules in one degree of freedom is 3 (a) kT 2

composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively are T2 and T1 ( T2 > T1). The rate of heat transfer through the æ A ( T2 - T1) K ö slab, in a steady state is ç ÷ f , with f è ø x equal to

[UP SEE 2007]

neutral temperature is (a) zero (c) negative

[BVP Engg. 2005]

(b) maximum (d) minimum but not zero

liquids A, B and C are 12°C and 28°C respectively. The temperature when A and B are mixed is 16°C and when B and C are mixed is 23°C. The temperature when A and C are mixed is [Kerala CET 2005]

1 2

(c)

2 3

(b) 22°C

(c) 20.2°C

(d) 24.2°C

184. A black body has maximum wavelength l m at

T1

2K

K

(d) kT

183. The temperature of equal masses of three different

(a) 18.2°C

(b)

3 (c) kT 4

4x

x

(a) 1

[BVP Engg. 2006]

kT (b) 2

182. The thermoelectric power for a thermocouple at the

175. The temperature of the two outer surface of a

T2

(b) Decrease (d) Can’t be determined

180. When you make ice cubes, the entropy of water

pV for one mole of an ideal gas is nearly 173. The value of T equal to [BVP Engg. 2007]

(a) 2 ´ 10 -5 (c) 18 ´ 10 -5

(b) visible region (d) infrared region

178. Pressure of an ideal gas is increased by keeping

172. Two rigid boxes containing different ideal gases are

(a) 2 J mol–1K–1 (c) 4.2 J mol–1K–1

[BVP Engg. 2006]

(a) ultraviolet region (c) gamma region

λ

3 (a) Tf = T0 7 3 (c) Tf = T0 2

3 RT 2

(d)

177. Thermal radiations are electromagnetic waves belonging to

(a) T3 > T2 > T1 (c) T1 > T2 > T3

543

2000 K. Its corresponding wavelength at 3000 K will be [Kerala CET 2005]

(d)

1 3

(a)

3 lm 2

(b)

2 lm 3

(c)

16 lm 81

(d)

81 lm 16

544 JEE Main Physics 185. A body with area A at maintained temperature T and emissivity e = 0.6 is kept inside a spherical black body. What will be the maximum energy radiated per second? [IIT JEE 2005] (a) 0.60 s AT

4

(b) 0.80 s AT

4

(d) 0.40 s AT 4

(c) 1.00 s AT 4

186. Calorie is defined as the amount of heat required to raise temperature of 1 g of water by 1°C and it is defined under which of the following conditions ? [IIT JEE 2005]

(a) From 14.5 °C to 15.5 °C at 760 mm of Hg (b) From 98.5 °C to 99.5 °C at 760 mm of Hg (c) From 13.5 °C to 14.5 °C at 76 mm of Hg (d) From 3.5 °C to 4.5 °C at 76 mm of Hg

187. In which of the following process, convection does not take place primarily ?

maintained at temperatures 2 T and 3 T respectively. The temperature of the middle ( i. e., second) plate under steady state condition is [IIT JEE 2012] æ 65 ö (a) ç ÷ è 2ø æ 97 ö (c) ç ÷ è2ø

ends to the other end under steady state. The variation of temperature q along the length x of the bar from its hot end is best described by which of the [AIEEE 2009] following figures? (a) θ

(c)

ær ö (c) ln ç 2 ÷ è r1 ø

r1

θ

r2

(d)

T2

r2 - r1 r12r

t

loge (θ – θ0)

loge (θ – θ0)

(b)

O

t

t

loge (θ – θ0)

loge (θ – θ0) O

(d)

O

x

(d) x

at a temperature of T K evaluate the total radiant power incident on the earth at a distance r from the [AIEEE 2006] sun. (a) pr20 R2sT 4 / r2

(b) r20 R2sT 4 / 4 pr2

(c) R2sT 4 / r2

(d) 4 pr20 R2sT 4 / r2

where r0 is the radius of the earth and s is stefan’s constant.

193. Two rigid boxes containing different ideal gases are placed on a table box A contains one mole of nitrogen at temperature T0 while box B contains one mole of 7 helium at temperature æç ö÷ T0 . The boxes are then è3 ø put into thermal contact with 0each other and heat flows between them untill the gases reach a common final temperature (ignore the heat capacity of boxes). Then, the final temperature of the gases in terms of [AIEEE 2006] T0 is (a) Tf =

(c)

(b)

192. Assuming the sun to be a spherical body of radius R T1

and q 0 is temperature of surroundings, then according to Newton’s law of cooling the correct graph between log e (q - q 0 ) and t is [AIEEE 2012[

O

(d) (97)1/4 T

x

189. A liquid in a beaker has temperature q ( t) at time t

(a)

T

T

1/4

x

concentric spheres of radii r1 and r2 kept at temperatures T1 and T2 , respectively. The radial rate of flow of heat in a substance between the two concentric spheres is proportional to [AIEEE 2005] r12r ( r2 - r1 )

æ 97 ö (b) ç ÷ è4ø

191. A long metallic bar is carrying heat from one of its

188. Figure shows a system of two

(b)

1/4

T

[IIT JEE 2005]

(a) Sea and land breeze (b) Boiling of water (c) Warming of glass of bulb due to filament (d) Heating air around a furnace

(a) ( r2 - r1 )

1/4

7 3 5 T0 (b) Tf = T0 (c) Tf = T0 3 2 2

(d) Tf =

3 T0 7

194. Three perfect gases at absolute temperature T1, T2

t

190. Three very large plates of some area are kept parallel and close to each other. They are considered as ideal black surfaces and have very high thermal conductivity. The first and third plates are

and T3 are mixed. The masses of molecules are m1, m2 and m3 and the number of molecules are n1, n2 and n3 respectively. Assuming no loss of energy the final temperature of the mixture is [AIEEE 2011] (a)

(T1 + T2 + T3 ) 3

(b)

n1T1 + n2T2 + nT 33 ( n1 + n2 + n3 )

(c)

2 n1T12 + n2T22 + nT 33 n1T1 + n2T2 + nT 33

(d)

n21T12 + n22T22 + n32T32 ( n1T1 + n2T2 + nT 3 3)

Heat and Kinetic Theory of Gases 195. One kg of a diatomic gas is at a pressure of

8 ´ 106 N / m 2 . The density of the gas is 4 kg/m 3. What is the energy of the gas due to its thermal motion? [AIEEE 2009] (a) 3 ´ 10 4 J (c) 6 ´ 10 4 J

197. If a piece of metal is heated to temperature q and then allowed to cool in a room which is at temperature q 0 , the graph between the temperature T of the metal and time t will be closed to [JEE Main 2013]

(b) 5 ´ 10 4 J (d) 7 ´ 10 4 J

T

T

(b) θ0

(a)

196. Two thermally insulated vessels 1 and 2 are filled with air at temperatures ( T1, T2 ) volumes ( V1, V2 ) and pressures ( p1, p2 ) respectively of the value joining the two vessels is opened the temperature inside the vessel at equilibrium will be [AIEEE 2008, 04] (a) T1 + T2 T T ( p V + p2V2 ) (c) 1 2 1 1 p1VT 1 2 + p2VT 2 1

545

O

(b) (T1 + T2 ) / 2 T T ( p V + p2V2 ) (d) 1 2 1 1 p1VT 1 1 + p2VT 2 2

O

t

t

T

T

(c) θ0

(d) θ0

O

O

t

t

Answers Round I 1. 11. 21. 31. 41. 51. 61. 71. 81. 91. 101. 111.

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

2. 12. 22. 32. 42. 52. 62. 72. 82. 92. 102. 112.

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

3. 13. 23. 33. 43. 53. 63. 73. 83. 93. 103. 113.

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

4. 14. 24. 34. 44. 54. 64. 74. 84. 94. 104. 114.

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

5. 15. 25. 35. 45. 55. 65. 75. 85. 95. 105. 115.

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

6. 16. 26. 36. 46. 56. 66. 76. 86. 96. 106. 116.

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

7. 17. 27. 37. 47. 57. 67. 77. 87. 97. 107. 117.

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

8. 18. 28. 38. 48. 58. 68. 78. 88. 98. 108. 118.

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

9. 19. 29. 39. 49. 59. 69. 79. 89. 99. 109. 119.

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

10. 20. 30. 40. 50. 60. 70. 80. 90. 100. 110.

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

Round II 1. (d) 11. (c) 21. (d) 31. (d) 41. (c) 51. (a) 61. (c) 71. (a) 81. (c) 91. (a) 101. (a) 111. (a) 121. (a) 131. (b,d) 141. (b) 151. (a) 161. (a) 171. (c) 181. (b) 191. (c)

2. (b) 12. (a) 22. (b) 32. (c) 42. (b) 52. (b) 62. (c) 72. (c) 82. (a) 92. (c) 102. (a) 112. (c) 122. (c) 132. (b,d) 142. (d) 152. (c) 162. (c) 172. (d) 182. (b) 192. (a)

3. (a) 13. (a) 23. (a) 33. (a) 43. (a) 53. (a) 63. (b) 73. (b) 83. (c) 93. (c) 103. (b,c) 113. (a) 123. (b) 133. (a,d) 143. (b) 153. (d) 163. (b) 173. (d) 183. (c) 193. (b)

4. (a) 14. (d) 24. (c) 34. (d) 44. (a) 54. (a) 64. (a) 74. (d) 84. (a) 94. (d) 104. (b) 114. (d) 124. (d) 134. (a,b,d) 144. (c) 154. (c) 164. (a) 174. (a) 184. (b) 194. (b)

5. (a) 15. (b) 25. (d) 35. (c) 45. (d) 55. (b) 65. (b) 75. (b) 85. (c) 95. (c) 105. (c) 115. (c) 125. (c) 135. (a,b,c) 145. (a) 155. (c) 165. (c) 175. (d) 185. (a) 195. (b)

6. (b) 16. (d) 26. (a) 36. (a) 46. (c) 56. (c) 66. (b) 76. (c) 86. (c) 96. (a) 106. (c) 116. (c) 126. (b) 136. (a,b,d) 146. (a) 156. (a) 166. (a) 176. (b) 186. (a) 196. (c)

7. (c) 17. (c) 27. (a) 37. (c) 47. (d) 57. (b) 67. (a) 77. (b) 87. (d) 97. (a) 107. (c) 117. (b) 127. (b) 137. (a,b) 147. (a) 157. (a) 167. (c) 177. (d) 187. (c) 197. (c)

8. (a) 18. (b) 28. (c) 38. (b) 48. (b) 58. (a) 68. (a) 78. (b) 88. (b) 98. (d) 108. (a) 118. (a) 128. (b) 138. (b,d) 148. (b) 158. (a) 168. (b) 178. (c) 188. (d)

9. (d) 19. (b) 29. (d) 39. (a) 49. (b) 59. (a) 69. (b) 79. (d) 89. (d) 99. (b) 109. (c) 119. (d) 129. (b,c) 139. (a,d) 149. (b) 159. (b) 169. (b) 179. (b) 189. (a)

10. (d) 20. (d) 30. (a) 40. (d) 50. (b) 60. (c,d) 70. (b) 80. (a) 90. (b) 100. (a) 110. (c) 120. (b) 130. (c) 140. (a) 150. (a) 160. (a) 170. (c) 180. (c) 190. (c)

the Guidance Round I 1. Given, triple point of water on scale A = 200 A

m = c ´r V c1 r1 3 5 Desired ratio = = ´ =1 :2 c2 r 2 5 6

7. Heat capacity/volume = c ´

Triple point of water on scale B = 350 B We know that triple point of water on absolute scale = 27316 . K \

200 A = 350B = 27316 . K 27316 27316 . . 1A = K and 1B = K \ 200 350 If TA and TB are the triple point of water on two scales A and B, then 27316 . 27316 . TA = TB 200 350 4 TA 200 4 \ = = or TA = TB 7 TB 350 7

2.

68 - 5 63 7 C x - lower fixed point = = = = 100 upper fixed point – lower point 95 - 5 90 10 700 ÞC= = 70° 10

practically no change in temprature whether heat is taken in or given out.

9. As piston is of fixed mass M, and it is able to move up or down without friction, therefore, when temperature is increased, piston moves out, increasing the volume V. The pressure p on the gas remains the same, because of fixed mass. mgh 10. Heat absorbed by water = Heat produced mc DT = J gh 980 ´ 500 ´ 100 900 DT = = = 1.16° C = Jc 420 4.2 ´ 10 7 ´ 1

11. From the principle of calorimetry, Þ

3. Let, F = K = X As

8. Infinite thermal capacity implies that there would be

F - 32 K - 273 x - 32 x - 273 = = \ 9 5 9 5 9x - 2457 = 5x - 160 4x - 2457 + 160 = 0 2297 x= = 574.25° 4

4. Let, q be the temperature of the mixture. Heat gained by water at 0°C = Heat lost by water at 10°C c1 m1 ( q - 0) = cm2 (10 - q) 400 q= = 6.66° C 60

5. From graph, we note that for scale A, the lowest fixed point is higher than 0° A and the higest point is 180°A. For scale B, the lowest point is 0°B and the highest point 100°B t - 30 t -0 t Therefore, the relation A = B = B is correct. 150 100 100

6. Heat required to melt 1 g of ice at 0°C to water at 0°C

Þ

m1s1Dq1 = m2s2Dq2 sw 540 ´ sw ´ (80 - x ) = 540 ´ ´ ( x - 0) 2 160° x= C 3

12. Let m gram of water, whose temperature is q ( > 30° C) be added to 20 g of water at 30°C. Then m ´ 1 ( q - q0) = 20 ´ 1 ( q0 - 30) (m + 20) q0 = 60 + mq 600 + mq q0 = 20 + m For q0 to be maximum m should be small and q should be large. 50 (KE of rotation) = cm q 13. Here, 100 1 æ 1 2ö 1 æ 2 2ö 2 ç Iw ÷ = cm q ç Ir ÷ (2 pn) = cmq ø è 2 2 4 è5 ø 2 p 2n 2r 2 Þ q= 5 c

= 1 ´ 80 cal Heat required to raise temperature of 1 g of water from 0°C to 100°C = 1 ´ 1 ´ 100 = 100 cal Total heat required for maximum temperature of100° C = 80 + 100 = 180 cal

14. When pressure of an ideal gas is constant, Charles’ law is

As one gram of steam gives 540 cal of heat when it is converted to water at 100°C therefore, temperature of the mixture would be 100°C.

15. Rate of cooling

obeyed i.e., V 1 V µ T or = constant = T p From the slope of curve shown in figure 3; p1 > p2. Dq Aes (T 4 - T04) Dq Þ µA = t mc t

Since area of plate is largest so it will cool fastest and sphere the slowest.

Heat and Kinetic Theory of Gases f F where, f and F are vapour pressures at closed point and air temperature. temperature at dew point RH = ´100 Þ Air temperature x Þ RH = t ´ 100 = 100% x where, dew point = air temperature

16. Relative humidity (RH) = ´100%

17. Let V be the volume of the sphere and r be the density of water. Buoyancy (F) on sphere due to water is F = V r g or F µ r Since, r 0°C > r 4°C, so, F0°C > F4°C, simple pendulum increases even though its centre of mass still remains at the centre of the bob. As time period, T = 2 p l/g or T µ l. So, T increases as temperature increases.

\

l0( s) a( c) 1.7 ´ 10 –5 = 1.545 = = l0( c) a( s) 1.1 ´ 10 –5

24. The pressure of the gas inside the vessel, as observed by us, on the ground remains the same. This is because motion of the vessel as a whole does not effect the relative motion of gas molecules and the walls of the vessel.

Þ

t 2 = t1 +

28. Here,

DT = 20 - 15 = 5°C a = 0.00012°C–1 = 12 ´ 10 -6°C–1 1 a ( DT) ´ 86400 s 2 1 = ´ 12 ´ 10 -6 ´ 5 ´ 86400 s = 2.590 s 2 ~ - 2.6 s

Time lost per day =

g ag = 10.30 ´ 10 –4°C–1 a a = 9 ´ 10 -60 °C-1, a m = ?

Now,

g r = g ag + g glass = g am + g m

10.30 ´ 10 –4 + 3 ´ 9 ´ 10 -6 = 10.06 ´ 10 –4 + g m (Q g g = 3 ´ a a )

\

g m = (10.30 +0.27 - 10.06)10 am =

–4

= 0.51 ´ 10 –4

1 0.50 ´ 10 -4 gm = 3 3

29. When a metallic rod is heated it expands. Its moment of inertia (I) about a perpendicular bisector increases. According to law of conservation of angular momentum, its angular speed ( w) decreases, since w µ1/ l (according to law of conservation of angular momentum)

30. From, T = 2p

= 0.17 ´ 10 –4 = 17 ´ 10 -6°C–1 F L AY F = a ( DT) AY = 1.1 ´ 10 –5 ´ (50 - 30) ´ 2 ´ 10 -6 ´ 2 ´ 10 11

21. As, DL = aL ( DT) =

= 88 N

22. Here, and

r 0 = 10 g / cc r100 = 9.7 g /cc, a = ?

l , we get g

DT 1 Dl a DT = = T 2 l 2 1 = ´ 2 ´ 10 -6 ´ 10 = 10 -3% 2

= 1.7 ´ 10 –5°C–1

\

-10 3 = -30°C 1.0 ´ 2 ´ 10 –5

27. As aB > aA , therefore strip B will appear on outer side.

g am = 10.06 ´ 10 –4°C–1

\

l2 - l1 l1a

= 20 +

l0( s) = 1.545 ´ 9.17 cm = 14.17 cm

20. Here,

l2 = l1 [1 + a (t 2 - t1)]

Al Steel

l0( s) = 1.545 l0( c) l0( s) - l0( c) = 5 0.545 l0( c) = 5 5 l0( c) = = 9.17 cm 0.545

and

V = L3 DV ´ 100 DL ´ 100 =3 = 3 ´ 0.2% = 0.6% V L

aluminium strip will expand more than that of steel strip. Due to it, aluminium strip will bend more on convex side and steel strip on concave side.

a (copper) = 1.7 ´ 10 –5°C–1

Þ

23. As,

g 3.09 ´ 10 –4 = = 1.03 ´ 10 –4° C–1 3 3

26. Since, a Al > a steel, so in bimetallic strip on heating,

19. Here, a (steel) = 1.1´10 –5°C–1

\ Also,

r 0 = r100 (1 + g ´ 100) r -r 10 - 9.7 g = 0 100 = = 3.09 ´ 10 –4 r100 ´ 100 9.7 ´ 100 a=

25. From,

18. With increases in temperature, the effective length (l) of

As,

From,

547

31. As,

h1 r 2 (1 + g q1) = = h2 r1 (1 + g q2)

Þ

50 1 + g ´ 50 = 60 1 + g ´ 100

Þ

g = 0.005/° C

é p0 ù êQ r = 1 + g q ú ë û

548 JEE Main Physics 32. As,

L = L0 (1 + a Dq)

38. A bimetallic strip on being heated bends in form of an arc with -6

L1 1 + a ( Dq1) 10 1 + 11 ´ 10 ´ 20 = = = L2 1 + a ( Dq2) L2 1 + 11 ´ 10 -6 ´ 19

more expandable metal ( A) outside (as shown)

Þ L2 = 9.99989 Length is shorter by » 10 - 9.99989 = 0.00011 = 11 ´ 10

cm

4 pR3 . 3

αA

39.

Increase in volume of sphere with rise in temperature D T is 4 D V = g V D T = 3a ´ p R3 D T = 4 p R3 a D T 3 DV 0.24 34. Here, g= = = 6 ´ 10 -5 /°C V ´ DT 100 ´ 40 g a = = 2 ´ 10 -5 /°C 3 If q is temperature of junction, (dT) A = qA - q, (dT)B = ( q - qB) æ dQ ö æ dQ ö As, ÷ ÷ =ç ç è dt ø A è dt ø KA A

2 æ DQ ö r \ç ÷ µ , which is maximum in case (a). è Dt ø l

40. For parallel combination of two rods of equal length and equal area of cross-section.

2 KB( qA - q) = KB ( q - qB) 2 qA - 2 q = q - qB 2 qA + qB = 3 q qA - qB = 48° qA = 48 + qB

As

Hence,

41. As, Dt =

A (dT)B (dT) A = KB (dx) A (dx)B

…(i)

2 ( 48 + qB) + qB = 3 q 96 + 3 qB = 3 q 96 = 3 ( q - qB) q - qB = 96 /3 = 32° C

36. As is clear from figure. dQ dQ1 dQ 2 = + dt dt dt K ( A1 + A2) dT dT dT = K1A1 + K2A2 dx dx dx K1A1 + K2A2 K= A1 + A2

37. Growth of ice in a pond is conduction process governed by the relation, t =

K + K2 K= 1 = 2 K 7 = K1 6

4 K1 3 = 7 K1 2 6

K1 +

DQ ( Dx ) KA ( DT)

when two rods of same length are joined in parallel, A ® 2 1 times and ( Dx) = times. 2 1 1 \ D t becomes times, i. e. , ´ 12 s = 3 s 4 4

42. Let the temperature of junction be q. æ DQ ö æ DQ ö =ç ÷ ÷ ç è Dt ø copper è Dt ø steel

Put in Eq. (i)

\

αA

αB

DQ æ DT ö 2 DT = KA ç ÷ = K ( pr ) è Dx ø ( l) Dt

35. Here, KA = 2 KB (dx) A = d (dx)B,

\

αB

A

33. Let V be the volume of sphere of radius R at temperature. then V=

B

B

A –5

Þ Þ 9 K2 or or or

The ratio of times for thickness of ice from 0 to y is 1 : 3

(100 - q) K2 A ( q - 0) = 18 6

(100 - q) = K2q 3 3 q = 900 - 9 q 12 q = 900 q = 75°C

43. The equivalent electrical circuit, figure in this case is of Wheatstone bridge. No current would flow through central rod CD when the bridge is balanced. The condition for P R balanced Wheatstone bridge is = (in terms of resistances) Q S

rL y 2 Kq 2

\ Time taken to increase the thickness from 1 cm to 2 cm is equal to 3 ´ 7 = 21 h.

K1A

1 / K1 1 / K3 = 1 / K2 1 / K4 or

K2 K4 = K1 K3

or

K1K4 = K2 K3

Heat and Kinetic Theory of Gases æ DQ ö æ DQ ö ÷ =ç ÷ è D t øP è D t øQ

44. Let L be the length of each rod.

47. As, ç

Temperature of A = 60° C temperature of E = 10° C Let q1, q2, q3 be respective temperture of B, C , D.

Þ

If Q1, Q 2, Q3 , Q 4 , Q5 , Q 6 are the amounts of heat following/sec respectively from A to B; B to C ; B to D; C to D; D to E and C to E, then using figure.

or

C (θ2) Q2 Q1 A(60°) y

B (θ1)

x

x

Q6

x Q4 y

E (10°)

0.46 A (60 - q1) 0.92 A ( q1 - q2) , Q2 = L L 0.46 A ( q1 - q3) 0.92 A ( q2 - q3) Q3 = , Q4 = L L 0.46 A ( q3 - 10) 0.92 A ( q2 - 10) Q5 = , Q6 = L L

or Again, Again,

…(i) …(ii) …(iii)

46. For the two sheets, shown in figure, rate of heat transfer is

\

Þ or

q1R2 - qR2 = qR1 - q2R1 q R + q2R1 q= 1 2 R1 + R2

ts 4 = = 1min 4 4 Series resistance, Rs = R1 + R2 and parallel resistance, RR Rp = 1 2 R1 + R2

Þ

Mass of ice melted/s = 2 ´ 0.1 g = 0.2 g s -1

dQ1 dQ 2 = dt dt dT1 dT2 = R1 R2 q1 - q q - q2 = R1 R

(Q Dx = l )

t µR t p Rp R /2 1 = = = t s Rs 2 R 4

Þ

q1 = 30° C, q2 = 20° C, q3 = 20° C dQ dT 45. As = KA , therefore, when dt dx 1 1 dt ® , A ® (2) 2 = 4, K ® 2 4 dQ becomes twice; m would become twice. dt

same, i. e. ,

DT Dx l = ( DQ / Dt ) KA KA

Þ

Q1 = Q 2 + Q3 0.46 A (60 - q1) 0.92 A ( q1 - q2) 0.46 A ( q1 - q3) = + L L L

q1 + 2 q2 - 4 q3 = 10° Solving Eqs. (i), (ii) and (iii), we get

K1A1 = K2A2 A1 K2 = A2 K1

must be same. Initial temperature difference = 10 - ( -10) = 20° C = 20 K Outside temperature = -23° C = -23 + 273 = 250 K Inside temperature = 250 + 20 = 270 K Q kADq Dq Dq (R = Thermal resistance) 50. As, = = = t l ( l / kA) R

Q1 =

60 - q1 = 2 ( q1 - q2) + q1 - q3 4 q1 = 2 q2 - q3 = 60° Q 2 = Q 4 + Q 6 gives q1 - 3 q2 - q3 = 10° Q5 = Q3 + Q 4 given

(T1 - T2) (T - T ) = K2A2 1 2 l l

49. For the same heat to be conducted, temperature difference

Q5 D (θ ) 3

Now

K1A1

or

48. Rth =

y

Q3

549

tp =

51. Given A1 = A2 and

K1 5 = K2 4 R1 = R2 l1 l = 2 k1A k2A

Q Þ

l1 k1 5 = = l2 k2 4

Þ

52. Natural convection arises due to difference of density at two places and is a consequence of gravity.

53. Convection is not possible in weigtlessness so the liquid will be heated through conduction.

54. Here, lm = 289.8 nm = 289.8 ´10 –9 m s = 5.67 ´ 10 –8 Wm–2 K –4 b = 2889 m mK = 2889 ´ 10 -6 mK If T is temperature of star, then according to Wien’s law l mT = b T=

2889 ´ 10 –6 b = 10 4 = l m 289.8 ´ 10 –9

From Stefan’s law, E = sT 4 = 5.67 ´ 10 –8 (10 4) 4 = 5.67 ´ 10 8 Wm–2

550 JEE Main Physics 55. On heating, black spot absorbs maximum radiation. Therefore, when plate is taken to a dark room, the spot will emit more radiations than the rest of the plate. Hence, it will appear brighter than the plate. æT ö E 56. As, 2 = ç 2 ÷ E1 è T1 ø

4

60. According to Wien’s displacement law, lm µ

i. e. , vm µ T Hence, curve A represents the correct variation.

61. Given, T1 = 7° C = 7 + 273 = 280 K 4

\

E 2 æ 727 + 273 ö =ç ÷ = 16 or E 2 = 16 E 7 è 0 + 273 ø

T2 = 287° C = 287 + 273 = 560 K 4

57. Cooling rate (R) is rate of fall of temperature, according to Newton’s law. It varies inversely as specific heat of the liquid. For A, rate of cooling is larger. Therefore, specific heat of A is smaller.

58. When temperature of a black body is increased by 50% 150 2 T1 = T1 100 3 According to Stefan’s law T2 =

\

Percentage increase in radiation (E 2 - E1) (81 - 16) ´ 100 = ´ 100 » 400% E1 16

b 2.88 ´ 10 6 = = 10 3 nm 2880 T From the shape of the E versus graph U2 > U1 lm =

63. As,

T1 l m2 10 -4 = = = 200 T2 l m1 0.5 ´ 10 -6

64. As,

lm T1 11 ´ 10 -5 =2 =n = 1 = l m2 5.5 ´ 10 –5 T2

65. According to Wein’s law, as T increases, lm decreases and v m increase. \ Therefore,

59. Given, initial temperature T1 = 80°C

In second condition, Initial temperature T1¢ = 60°C Final temperature T2¢ = 30°C Now,

or

t¢ = ? (60 - 30) 2 æ 60 + 30 ö = - 20 ÷ ç ø 15 è 2 t¢ 30 2 = ( 45 - 20) t ¢ 15 30 ´ 15 t¢ = 2 ´ 25 = 9 min

T1 < T2 v m1 < v m2

66. According to Newton’s law of cooling,

Final temperature T2 = 50 °C Temperature of the surroundings T0 = 20 °C t1 = 5 min According to Newton’s law of cooling. dT é T + T2 ù Rate of cooling, =kê 1 - T0 ú dt ë 2 û (80 - 50) ù é 80 + 50 =kê - 20 ú 5 2 û ë 30 = k(65 - 20) 5 6 = k ´ 45 6 2 k= = 45 15

4 E 2 æ T2 ö æ 560 ö 4 =ç ÷ =ç ÷ = 2 = 16 è 280 ø E1 è T1 ø

62. From Wien’s law lmT = b

4

4 E 2 æ T2 ö 81 æ3ö =ç ÷ =ç ÷ = è2ø E1 è T1 ø 16

or

c 1 1 or µ vm T T

dq µ ( q - q0) dt dq = K ( q - q0) dt

-

where, K is a constant of proportionality. 75 - 65 1 æ 75 + 65 ö Now, =K ç - 30 ÷ = K ´ 40 Þ K = è ø 2 2 8

…(i)

In case of second identical object 55 - 45 ö æ 55 + 45 =K ç - 30 ÷ ø è 2 t 10 1 = ´ 20 t 8 80 t= = 4 min Þ 20 T2 l m1 1.78 67. As, = = 14 T1 l m2 T2 =

1.78 1.78 ´ T1 = (1373 + 273) 14 14

= 209.3 K = 309.3 - 273 = 63.7° C

68. Let R = radius of planet P = power radiated by the sun P Energy received by planet = ´ 4 pR 2 4 pd 2 Energy radiated by planet = ( 4 pR) 2sT 4

Heat and Kinetic Theory of Gases For thermal equilibrium, 4 pR 2(sT 4) =

75. As,

P ´ 4 pR 2 4 pd 2

E1 T14 - T04 (327 + 273) 4 - (27 + 273) 4 = = E 2 T24 - T04 (227 + 273) 4 - (27 + 273) 4

1 T µ 2 d 4

T µ d -1/ 2 \ n =

or

1 2

4 E 2 E ¢ 1 (327 + 273) 4 1 æ 600 ö 81 = ´ç = = ÷ = 4 E1 E 64 4 (127 + 273) 4 è 400 ø

E¢ =

or

81 E 64

77. Here,

1 1 or T µ T lm

and T2 = 327° C = (327+273) K = 600 K According to Stefan’s law,

Þ

( l m)1 < ( l m) 2 < ( l m)3

\ T1 > T2 > T3

q1 - q2 ö æq -q = K ç 1 2 - q0 ÷ ø è t t 50 - 40 æ 50 + 40 ö =K ç - 20 ÷ è ø 300 2 1 K= 25 ´ 30 q 40 - q ö Kq æ 40 + q = =K ç - 20 ÷ = ø 2 1500 è 2 300

Now,

300 q = 60000 - 1500 q 60000 q= = 33.3° C 1800

or

T2 æ E 2 ö =ç ÷ T1 è E1 ø

or \

1/ 4

æ 32 ´ 10 5 ö =ç ÷ è 2 ´ 10 5 ø

1/ 4

=2

T2 = 2 T1 = 2 ´ (127 + 273) = 800 K = 800 - 273 = 527° C

73. According to Newton’s law of cooling, \

Rate of heat loss = Temperature difference R2 40 - 20 1 = = R1 80 - 20 3 R2 =

74. As, Þ Þ

l m2 l m1

R1 45 = = 15 cals–1 3 3

T = 1 (Wein’s displacement law) T2 l m2 2000 = l m4 2400 l m2 = 4 ´

20 = 3.3 mm 24

E 2 = 16 E1

78. According to Stefan’s law, E1 = sT 4 and E 2 = s (T + DT) 4 \

E 2 - E1 s [(T + DT) 4 - T 4 ] DT = =4 E1 T sT 4

Now,

E 2 - E1 2 DT = =4 E1 100 300

300 ´ 2 = 1.5 K 4 ´ 100 \Temperature of other patch = T + DT

\

DT =

= 300 + 1.5 = 301.5 K 10 79. Mass of ice melted/min = = 1g 10 Quantity of heat used = 1 ´ 80 cal Area of lens = pr 2 = 3.14 (2.5) 2 = 19.625 cm2

4

æ T2 ö E ç ÷ = 2 è T1 ø E1

72. As,

1215 = 2.23 544

4 E 2 æ T2 ö æ 600 ö =ç ÷ =ç ÷ = 16 è 300 ø E1 è T1 ø

71. According to Newton’s law of cooling

or

=

4

lm µ

Þ

(600) 4 - (300) 4 10 8 (1296 - 81) = (500) 4 - (300) 4 10 8 (625 - 81)

T1 = 27° C = (27 + 273) = 300 K

70. According to Wein’s displacement law

From figure,

=

76. For a perfectly black body, a = e = 1and r = 0

69. As energy emitted µ AT 4 \

551

\Amount of heat received/min/cm 2 from sun 80 = » 4 cal cm–2 min –1 19.625

80. Here, T1 = 6000 K, l1 = 4800 Å T2 = 3000 K, l 2 = ? According to Wein’s law l 2 T1 \ = l1 T2 Þ

l2 =

6000 T1 ´ l1 = ´ 4800 = 9600 Å 3000 T2

81. According to Wein’s displacement law ( l m)1 T2 0.26 = = =2 ( l y ) 2 T1 0.13 4

4 E1 æ T1 ö 1 æ 1ö =ç ÷ =ç ÷ = è2ø E 2 è T2 ø 16

\ 4

4

æT ö 327 + 273 ö 82. As, E 2 = E1 = ç 2 ÷ = 10 æç ÷ = 160 J ø è è T1 ø

27 + 273

552 JEE Main Physics 83. Here, T1 = 327° C = (327 + 273) K = 600 K and T2 = 927° C = (927+273) K = 1200 K According to Stefan’s law E µT4

92. Here, p1 = p , V1 = V , T1 = T T2 = 1.1 T , V2 = 1.05 V , p2 = ? p2V2 p1 V1 From = T2 T1

4

4 E 2 æ T2 ö æ1200 ö =ç ÷ =ç ÷ = 16 è 600 ø E1 è T1 ø

p2 = p1

E 2 = 16 E1 = 16 ´ 4 = 64 cal cm–2 s–1

=p´

84. Fall of temperature with time follows an exponential curve as shown in option (c) figure. This is as per Newton’s law of cooling.

4

93. Here,

88. Applying standard gas equation, p2V2 p1V1 = T2 T1 p2 V1 T2 = × p1 V2 T1 T2 3000 = = 10 and every molecule of H2 splits into T1 300 hydrogen atoms, doubling the number, therefore volume available to given number of entities becomes half i.e., p 1 V2 = V1 therefore, 2 = 2 ´ 10 = 20 . p1 2 As

or

p1 = p V2 = V p1V1 = p2V2 pV p ´2V p2 = 1 1 = =2p V V

90. Here, p1 = p and T1 = T 0.4 p2 = p + p = 1.004 p 100 As \ or

T2 = 200 K; V2 = 5.2 L p V T 1 ´ 30 ´ 200 p2 = 1 1 2 = = 3.86 atm T1V2 300 ´ 5.2

\

pV pV = constant. Thus, the variation of RT RT and p will be horizontal straight line (1). 5 95 V 95. New volume, V1 = V V= 100 100 pV pV 100 New pressure, p1 = = = p V1 (95 /100) V 95

94. As, pV = n RT, so

\% increase in pressure æ p - pö æ p1 ö =ç 1 ÷ ´ 100 = ç - 1÷ ´ 100 è p ø èp ø æ100 ö =ç - 1÷ ´ 100 = 5.26 è 95 ø

96. As pV 2 = a constant, therefore, when V becomes 2 V , p2 becomes half i. e. , p 2 =

89. Here, initially p1 = p,V1 = V + V = 2 V Finally, As

As

91. As the vessel contains 1 mole of hydrogen and 1 mole of exygen, therefore, as per Maxwell's law of speed distribution, f1 (v) and f2 (v) will obey the law separately.

1 1 times. or p = 2 2

p = a constant, so T µ p. T

Thus, T becomes T / 2.

97. As, c =

3 pV 3 RT = M M

and the new rms speed, c1 =

T2 = (T + 1) p2 T2 = p1 T1 1.004 p T + 1 1 = = 1+ p T T 1 T= = 250 K 0.004

p1 = 1 atm; T1 = 300 K; V1 = 30000 cc = 30 L;

4

2 T µ = 2 2 = 4 times, i. e. , 4 P d 2 22 pV 1 ´ 100 87. As, V1 = = = 25 cc p1 4 Pµ

V T p ´ 1.1 = 1.1 1.05V T 1.05

p2 = 1.05 p

85. When both have same area and same temperature, they will cool at the same rate. 1 86. As P µ T 4 and P µ 2, therefore, d

V1 T2 × V2 T1

=

98. As, cs =

3 R (T /2) 1 3 RT = (2 M) 2 M c 300 = = 150 ms–1 2 2

3 pV 3p gp and c = = M r r g cs = 3 c

or

cs = c

g 3

Heat and Kinetic Theory of Gases 1M 2 3 pV 3 RT c or c2 = = M M 3V 3 ´ RT 3 RT 2 For gas A, V1 = = M M 3 RT 2 For gas B, ; V2 = M

99. As, p =

So,

V12 =1 V22

Þ

V1 =1 V2

3 kT 3 RT 100. As, m = 2 = 2 c Nc 3 ´ 8.31 ´ 300 = 3.3 ´ 10 –27 kg = 6.023 ´ 10 23 ´ (1930) 2 Mass of H2 molecule = 1.66 ´ 10 –27 ´ 2 = 3.32 ´ 10 –27 kg Thus, the gas is hydrogen.

105. As, Mc2 = RT or c = æç 1 2

3 RT ö ÷ è M ø 1 cµ M

3 2

Þ

c1 æ M2 ö =ç ÷ c2 è M1 ø

So,

106. As,

æ cHe ö rH = ç ÷= è cH ø r He ( cHe) t = ( cHe) 0

1/ 2

1/ 2

1 1 = 4 2 T T0

( cHe) t ( cHe) 0 = ( cH) 0 ( cH) 0

\

T 5 = T0 7

T » 2 T0 = 2 ´ 273 = 546° K = 273° C

or

c2 T = 2 c1 T1

107. As,

T2 2T =c = c 2 = 484 2 = 684 ms–1 T1 T

3 ´ 1.38 ´ 10 –23 ´ 273 3 kT 101. As, c = = m 5 ´ 10 -17

or

= 15 ´ 10 -3 ms–1 = 1.5 cms–1

\

3 RT c MO 16 ; so H = = =4 M cO MH 1

E 2 c22 ( c 2) 2 = = =2 E1 c12 c2

or

E 2 = 2 E1 = 2 ´ 6.21 ´ 10 –21 J

102. As c =

103. Absolute pressure p1 = (15 + 1) atm [ Q Absolute pressure = Gauge pressure + 1 atm] = 16 ´ 1013 . ´ 10 5 Pa V1 = 30 L = 30 ´ 10 -3 m3 T1 = 27315 . + 27 = 300.15 K Using ideal gas equation or

pV = nRT pV n= RT =

´ 10 5 ´ 30 ´ 10 -3 p1V1 16 ´ 1013 . = RT1 8.314 ´ 300.15

= 19.48 Final

p2 = (11 + 1) = 12 atm = 12 ´ 1013 . ´ 10 5 Pa V2 = 30 L = 30 ´ 10 -3m3 T2 = 27315 . + 17 = 290.15 K

Number of moles =

´ 10 5 ´ 30 ´ 10 -3 p2V2 12 ´ 1013 . = RT2 8.314 ´ 300.15

= 1512 . Hence, moles removed = 19.48 - 1512 . = 4.36 Mass removed = 4.36 ´ 32 g = 0.1396 kg

104. RMS velocity does not change with pressure, till temperature remains constant.

553

c1

= 12.42 ´ 10 –21 J

108. Average energy E µ T \

E 800 800 = =2 E 400 400

or

E 800 = 2 ´ E 400 = 2 ´ 7.21 ´ 10 –21 = 14.42 ´ 10 –21 J

RMS velocity, \

cµ T

c800 = c400

800 = 524 ´ 2 » 741 ms–1 400

109. As, KE µ T So,

E127 = E 27 ´ (27 + 273) / (27 + 273) = 6.21 ´ 10 –21 ´ 400 / 300 = 8.28 ´ 10 –21 J

110. According to law of equipartition of energy, average kinetic energy per molecule per degree of freedom at temperature T 1 is kT. The average kinetic energy per molecule of 2 n polyatomic gas molecule = kT (n = number of mole) 2 The average kinetic energy per mol of polyatomic gas n n E = kT ´ N = RT 2 2 d æn ö n CV = ç RT ÷ = R dT è 2 ø 2

554 JEE Main Physics 111. As, g = 1+

2 2 2 or = g - 1 or n = n g -1 n

112. Here, n = 6,

116. The thermal radiation from a hot body travels with a velocity of light in vaccum, i. e. ,3 ´ 10 8 ms–1.

117. Power radiated by sun at t° C = s (t + 273) 4 4pr 2

æ 6ö æ nö C p = ç1 + ÷ R = ç1 + ÷ R = 4 R è 2ø è 2ø

Power received by a unit surface =

113. At NTP, T = 273 K, p = 1.01´105 Nm–2 =

d = 2.4 ´ 10 –10 m kT l= 2 pd 2p

Here,

= 1.46 ´ 10 2 n

2 n

114. g = 1+ or 1.33+ or

r 2s (t + 273) 4 R2

118. lm1T1 = lm2T2 5.5 ´ 10 –7 ´ 5500 = 11 ´ 10 -7 T

Þ

(1.38 ´ 10 –23) ´ 273 = 1.414 ´ 3.14 ´ (2.4 ´ 10 –10) 2 ´ 1.01 ´ 10 5 –7

s (t + 273) 4 4 pr 2 4 pR 2

T = 550 ´ 5 K = 2750 K

m

119. As,

4 2 = 1+ 3 n

On solving n = 6. It is triatomic gas molecule having triangular structure, i. e. , O3 .

115. The black spot on heating absorbs radiation and emits it then

365 - 361 é 365 + 361 ù =Kê - 293ú 2 2 ë û

Þ Again,

in the dark room while the polished shining part reflects radiation and absorbs nothing and so does not emit radiations and becomes invisible in the dark.

344 - 342 1 = t 35 14 t= 10

(from Newton’s law of cooling) 1 K= 35 é 344 + 342 ù 10 - 293ú = êë 2 û 7 14 min = ´ 60 = 84 s 10

Round II 1 ö 1. As, l = l0 æç1+ ÷ è

\ or Þ or

2. Here,

100 ø 1 ö æ 2 l 2 = 2 l02 ç1 + ÷ è 100 ø

2

2 100 2 DS = S ´ 100 DS 2 = = 2% S 100

2 l 2 - 2 l02 = 2 l02 ´

m = 0.1kg,h1 = 10 m,h2 = 5.4 m c = 460 J kg -1°C–1, g = 10 ms–2, q = ?

Energy dissipated, Q = mg (h1 - h2) = 0.1 ´ 10 (10 – 5.4) = 4.6 J From Q = cmq Q 4.6 q= = = 0.1°C cm 460 ´ 0.1

3. Triple point of neon, (T1) = 24.57 K Triple point of CO2, (T2) = 216.55 K On celsius scale, °C = K - 27315 . Triple point of neon, t1° C = 24.57 - 27315 . = - 248.58°C Triple point of CO2, t 2° C = 216.55 - 27315 . = - 56.60°C

On Fahrenheit scale, K - 27315 . F - 32 = 5 9 9 + 32 5 9 Triple point of neon, F1 = (K1 - 27315 . ) ´ + 32 5 9 = (24.57 - 27315 . ) ´ + 32 5 9 = - 248.58 ´ + 32 = - 415.44°F 5 9 Triple point of CO2, F2 = (K2 - 27315 . ) ´ + 32 5 9 = (216.55 - 27315 . ) ´ + 32 5 9 = - 56.6 ´ + 32 = - 69.88°F 5 F = (K1 - 27315 . )´

or

4. As r = r0 (1- g DT) \

\

9.7 = 10 (1 - g ´ 100) 9.7 = 1 - g ´ 100 10 9.7 0.3 g ´ 100 = 1 = = 3 ´ 10 -2 10 10 g = 3 ´ 10 -4 1 a = g = 10 -4° C–1 3

Heat and Kinetic Theory of Gases 5. Rate of cooling is proportional to (T 4 - T04), as per Stefan’s law.

= R¢ =

4

4

4

v = 300 ms–1, q = ?, C = 150 J - kg –1 K –1 Q=

4

9 -3 3 (3 - 1) 80 16 = = = 3 6 4 - 3 4 3 4 (2 4 - 1) 15 16 R 3

4

6. As,

11. Here, m = 10g = 10 –2 kg

R ¢ (900) 4 - (300) 4 = R (600) 4 - (300) 4

\

4 4 E 2 æ T2 ö æ 273 + 84 ö æ 357 ö =ç ÷ =ç ÷ =ç ÷ = 2.0 è 273 + 27 ø è 300 ø E1 è T1 ø

DQ æ DT ö 7. From, = KA ç ÷ è Dx ø Dt

Q = cm q Q 225 q= = = 150° C cm 150 ´ 10 -2 K 1 r 1 12. Here, 1 = , 1 = K2 2 r2 2

=

0.5 ´ (27+273)+0.5 (37+273) 0.5+0.5

= 305 K = 305 - 273 = 32°C

9. Here, \ Þ

D1 1 = D2 2

dx1 1 dQ 2 dQ1 = 4 cals–1, =? = , dx2 2 dt dt dQ 2 / dt dQ1 / dt

dx1 2 = dx2 1 dQ1 æ dT ö dQ 2 æ dT ö dQ1 / dt ç = KA1 ÷: ç = KA2 ÷= dt è dx1 ø dt è dx2 ø dQ 2 /dt A dx 1 1 1 = 1× 2= ´ = dx1 A2 4 2 8

10. 10 g of ice at –10°C to ice at 0°C Q1 = cm Dq = 0.5 ´ 100 = 50 cal 10 g of ice 0°C to water at 0°C Q 2 = mL = 10 ´ 80 = 800 cal 10 g of water at 0°C to water at 100°C Q3 = cm Dq = 1 ´ 10 ´ 100 = 1000 cal 10 g water at 100°C to steam at 100°C Q 4 = mL = 10 ´ 540 = 5400 cal Total heat required, Q + Q1 + Q 2 + Q3 + Q 4 = 50+800+1000+5400 = 7250 cal

=

K2 A2 dT / dx2 K1 A1 dT / dx1

=

K2 A2 dx1 K1 A1 dx2

1 =4 2 dQ1 dQ 2 / dt 4 = = = 1 cals–1 dt 4 4 =2´4´

13. Given, diameter of the hole (d1) = 4.24 cm Initial temperature T1 = 27 + 273 = 300 K Final temperature T2 = 227 + 273 = 500 K

Coefficient of linear expansion ( a) = 170 . ´ 10 -5 /°C Coefficient of superficial expansion (b) = 2a = 3.40 ´ 10 -5 /°C Initial area of hole at 27°C ( A1) = pr 2 = =

A1 D12 1 = = A2 D22 4

A1 1 = A2 4

\

In arrangement (b), A is doubled and Dx is halved. 1/ 2 1 Dt ® ® time \ 2 4 1 ´ 4 min = 1min i. e. , 4 16 g of O2 is half mole of O2, i. e. ,n2 = 0.5 nT +n T T= 11 22 \ n1 + n2

50 æ 1 1 -2 2ö 2 ç mv ÷ = ´ 10 (300) = 225 J ø 4 100 è 2

From

DQ Dx Dt = KA ( DT)

8. 22 g of CO2 is half mole of CO2, i. e. ,n1 = 0.5

555

p ( 4.24) 2 = 4.494p cm 2 4

pd12 4

Area of hole at 227°C ( A2) = A1(1 + b × Dt ) = 4.494 p [1 + 3.40 ´ 10 -5 ´ (227 - 27)] = 4.494p [1 + 3.40 ´ 10 -5 ´ 200 ] = 4.495p ´ 10068 . = 4.525p cm 2 If diameter of hole becomes d 2 at 227°C, then A2 = 4.525p = or

pd 22 4

d 22 = 4.525 ´ 4

or d 2 = 4.2544 cm \Change in diameter ( Dd) = d 2 - d1 = 4.2544 - 4.24 = 0.0144 cm = 144 . ´ 10 -2 cm

pd 22 4

556 JEE Main Physics g a = g r - g = (180° - 4.0) 10 -6

14. According to Newton’s law q1 - q2 ö æ q + q2 =K ç 1 - q0 ÷ ø è 2 t \

60 - 50 =K 10

æ 60 + 50 ö - 25÷ ç è ø 2

Let q be the temperature after another 10 min 50 - q ö æ q + 50 \ =K ç - 25÷ ø è 2 10

Vt = V0 (1 + 1.40 ´ 10 –6 ´ 10 2) = (10 3 + 1.4) cc …(i)

…(ii)

Dividing Eq. (i) by Eq. (ii), we get 10 30 ´ 2 = 50 - q q \

q = 42.85° C

15. Let the temperature of junction be Q. In equilibrium, rate of flow of heat through rod 1 = sum of rate of flow of heat through rods 2 and 3. æ dQ ö æ dQ ö æ dQ ö ç ÷ =ç ÷+ç ÷ è dt ø 1 è dt ø è dt ø 3 KA

( q - 0) KA (90° - q) KA (90° - q) = + l l l

Þ or

q = 2 (90° - q) 3 q = 180° 180° q= = 60° 3

Þ

16. Two strips of equal lengths but of different materials (different coefficient of linear expansion) when joined together, is called bimetallic strip, and can be used in thermostat to break or make electrical contact. This strip has the characteristic property of bending on heating due to unequal linear expansion of the two metals. The strip will bend with metal of greater a on outer side i.e., convex side. T 17. As, h = 1- 2 T1 \

50 500 or T1 = 1000 K =1= 100 T1

Again,

60 T = 1- 2 100 1000 T2 = 400 K

or

18. As, dQ = C pm DT

20. Using g r = g a + g , we get g r = g1 + 3 a = g 2 + 3 b g -g b= 1 2+a 3

\

21. As, As

70 = C p ´ 2 (35 - 30) CV = C p - R = 7 - 1.99 = 5.01 cal mol–1°C–1 dQ ¢ = CVm DT = 5.01 ´ 2 ´ (35 – 30) = 50.1 cal

19. Here, V0 = 103 cc g r = 180 ´ 10 -6°C–1 g = 40 ´ 10 -6°C–1,t = 100°C

DQ r2 æ DT ö 2 æ DT ö = KA ç ÷ = K pr ç ÷µ è Dx ø è l ø l Dt

r2 is maximum for (d), it is correct choice. l

22. Thermal capacity = Mass ´ specific heat Due to same material both spheres will have same specific heat. Also mass = Volume (V ) ´ density (r) \Ratio of thermal capacity 4 3 3 3 pr m1 V1 r 3 1 æ r1 ö 1 æ 1ö = = = =ç ÷ =ç ÷ = è2ø m2 V2 r 3 pr3 è r2 ø 8 2 4

23. Given, mass of the child (m) = 30 kg Time taken (t ) = 20 min Fall in temperature = (101 - 98)°F 5 5 DT = 3°F = 3 ´ ° C = °C 9 3 Specific heat of human body ( s) = 4.2 ´ 10 3 J/kg-°C Latent heat of evaporation (L) = 580 cal/g = 580 ´ 10 3 cal/kg = (580 ´ 10 3 ´ 4.2) J/kg Heat given by body during fall in temperature Q1 = msDT Let m¢ be the mass of sweat evaporates from the human body. Heat taken in evaporation

\

Q 2 = m¢ L Q1 = Q 2 msDT = m¢ L

or

m¢ =

But

\

\

\Volume of mercury that will overflow = Vt - V0 = 14 cc

msDT 30 ´ 4.2 ´ 10 3 ´ 5 / 3 = L 580 ´ 4.2 ´ 10 3

10 = 0.0862 kg 116 0.0862 \ Rate of evaporation of sweat = 20 =

= 0.00431 kg/min = 4.31 g/min

Heat and Kinetic Theory of Gases 24. Here, Dl = 80.3 – 80.0 = 0.3 cm l = 80 cm, a = 1.2 ´ 10 –6 °C–1 Dl Rise in temperature DT = la DT =

DT ö 2 ÷ Dt , where, A = 4 pr è Dx ø 22 æ 32 ö (6 ´ 10 8) 2 ´ ç 5 ÷ ´ 86400 è10 ø 7

= 10 18 cal

26. Apprent weight (wa ) = Actual weight (w) - upthrust (F) where, upthrust = weight of water displaced = V rw g F50 V50 r50 g 1 + g m ´ 50 Now, = = F0 V0 r 0 g 1 + gw ´ 50 As g m < gw , therefore, F50 < F0 . Hence, w2 > w1

P = P0 ´

=

90 (1 + 1.7 ´ 10 –5 ´ 20) 1 (1+1.2 ´ 10 –5 ´ 20)

=

90 ´ 1.00034 = 90.01 cm 1.00024

DQ ö æ DQ ö æ DQ ö +ç =ç ÷ ÷ ÷ è Dt ø inner è Dt ø outer è Dt ø total

28. As, æç

K1pr 2(T2 - T1) K2p [(2 r) 2 - r 2] (T2 - T1) + l l K p (2 r) 2 (T2 - T1) = l pr 2 (T2 - T1) K p 4 r 2(T2 - T1) or (K1 + 3 K2) = l l K1 + 3 K2 or K= 4

29. Let, T0 be the initial temperature of the black body \ l 0T0 = b (Wien’s law) Power radiated, P0 = CT04 , where C is constant. If T is new temperature of black body, then 3 l0 T = b = l 0T0 4

4

256 81

P 256 = P0 81

or

30. Let V be the volume of solid; d be its density and m be its mass; if g coefficient of volume expansion of liquid, then d0 Density at temperature t1 is d1 = 1 + g t1 d0 Density at temperature t 2 is, d 2 = 1+ g t 2 According to Archimedes’s principle, f1Vd1 = m = f2Vd 2 d1 f2 d 0 (1 + g t 2) or = = d 2 f1 1 + g t1) d 0 or

27. As the steel tape is calibrated at 10°C, therefore, adjacent centimetre marks on the steel tape will be separated by a distance of lt = l10 (1 + a s DT) = (1 + a s 20) cm Length of copper rod at 30°C = 90 (1 + a c 20) cm Therefore, number of centimetres read on the tape will be 90 (1 + a c 20)) = 1(1 + a s 20)

4 T0 3

æ 4ö Power radiated, P = CT 4 = CT04 ç ÷ è3ø

0.3 = 312.5°C 80 ´ 1.2 ´ 10 -5

25. As, DQ = KA æç

= 0.008 ´ 4 ´

T=

or

557

f1 + f1g t 2 = f2 + f2g t1 f1 - f2 = g ( f2 t1 - f1 t 2) (f - f ) g= 1 2 f2 t1 - f1 t 2

31. As, p = p1 + p2 + p3 æ n RT ö æ n RT ö æ n RT ö =ç ÷ ÷ +ç ÷ +ç è V ø N è V ø CO è V øO 2 2 2 RT V (0.25+0.5+0.5) (8.31) ´ 300 = 4 ´ 10 -3 = 7.79 ´ 10 5 Nm–2 = (nO2 + nN2 + nCO2 )

22 1 = ; 44 2 molar specific heat of CO2 at constant volume CV1 = 3 R 16 1 For oxygen, number of moles (n2) = = ; 32 2 5R molar specific heat of O2 at constant volume CV2 = . 2 Let T K be the temperature of mixture. Heat lost by O2 = Heat gained by CO2. n2CV2 DT2 = n1CV1 DT1 1 æ5 ö 1 ç R ÷ (310 - T) = ´ (3 R) (T - 300) ø è 2 2 2

32. For carbon dioxide, number of mole (n1) =

or or

1550 - 5 T = 6 T - 1800 T = 304.54 K = 31.5° C

33. According to Boyle’s law, pV = k (a constant) or or

p

m pm = k or r = r k r=

p k

k ö æ ç where, = k = constant ÷ ø è m

558 JEE Main Physics p1 k p1 m1 m km V1 = = 1 = 1 k p1 r1 p1/ k km2 V2 = p2 r1 =

So, and Similarly,

æm m ö Total volume = V1 + V2 = k ç 1 + 2 ÷ è p1 p1 ø Let p be the common pressure and r be the common density of mixture. Then m + m2 m1 + m2 r= 1 = æ m1 m2 ö V1 + V2 kç + ÷ è p1 p2 ø \

34. As, crms

p = kr =

m1 + m2 p1p2 (m1 + m2) = m1 m2 (m1p2 + m2p1) + p1 p2

3 RT = M M=

or

3 RT 3 ´ 8.31 ´ 300 = 2 (1920) 2 crms

38. Molar specific heat of the mixture at constant volume is CV =

n1CV1 + n2CV2 (n1 + n2)

36. When the piston is in equilibrium, the pressure is same on both the sides of the piston. It is given that temperature and weight of gas on the two sides of piston not change. From ideal gas equation, pV = n RT , we have V µ mass of the gas. V1 m1 So, = V2 m2 V1 m or +1= 1 +1 V2 m2 V1 + V2 m1 + m2 or = V2 m2 V2 m2 2m 2 or = = = V1 + V2 m1 + m2 m + 2 m 3

37. For a closed system, the total number of moles remains constant. So, \ \

p1V = n1RT1 and p2V = n2RT2 p (2 V ) = (n1 + n2) RT p (n1 + n2) 1 æp p ö R = ç 1 + 2÷ = T 2 2 è T1 T2 ø =

1 æ p1T2 + p2T1 ö ç ÷ 2 è T1T2 ø

2+3

= 2.1 R

39. g real = g app. + g vessel So

( g app. + g versel) glass = ( g app. + g versel) steel

Þ 153 ´ 10 -6 + ( g versel) glass = (144 ´ 10 -6 + g vessel) steel ( g vessel) steel = 3a = 3 ´ (12 ´ 10 -6)

Further,

= 36 ´ 10 -6 /°C Þ 153 ´ 10 -6 + ( g vessel) glass = 144 ´ 10 - -6 + 36 ´ 10 -6 ( g vessel) glass = 3a = 27 ´ 10 -6 /°C

Þ

a = 9 ´ 10 -6 /°C

Þ

40. The expansion of solids can be well understood by potential energy curve for two adjacent atoms in crystalline solid as a function of their intermolecular separation (r). U

= 2 ´ 10 -3 kg = 2 g Since, M = 2 for the hydrogen molecule. Hence, the gas is hydrogen. n RT + n2RT + n3RT RT 35. As, p = 1 = (n1 + n2 + n3) V V æ 8 14 22 ö 0.082 ´ 300 =ç + + = 3.69 atm ÷´ è16 28 44 ø 10

=

æ5 ö æ3 ö 2 ç R÷ + 3 ç R÷ è2 ø è2 ø

r P3

E

P2

C r0

P1

A

BT

FT 3 D T2

1

r1

r2

T3 > T2 > T1 At ordinary temperature Each molecule of the solid vibrates about its equilibrium position P1 between A and B (r0 is the equilibrium distance of it from some other molecules) At high temperature Amplitude of vibration increases (C « D and E « F). Due to asymmetry in the curve, the equilibrium positions (P2 and P3) of the molecule is displaced. Hence its distance from other molecules increases (r2 > r1 > r0). Thus, on raising the temperature, the average equilibrium distance between the molecules increase and the solid as a whole expands.

41. Initial diameter of tyre = (1000 - 6) mm = 994 mm Initial radius of tyre R =

994 = 497 mm 2

and change in diameter DD = 6 mm 6 DR = = 3 mm 2 After increasing temperature by Dq tyre will fit on wheel Increment in the length (circumference) of the iron tyre g g [As a = ] DL = L ´ a ´ Dq = L ´ ´ Dq 3 3 ægö 2 pDR = 2 pR ç ÷ Dq è3ø

Heat and Kinetic Theory of Gases Þ

Dq =

Þ

3 ´3 3 DR = g R 3.6 ´ 10 -5 ´ 497

As Þ or

Dq @ 500°C

42. Due to volume expansion of both liquid and vessel change in volume of liquid relative to container is given by DV = V0[ gL - gg ]Dq. V0 = 1000 cc, ag = 0.1 ´ 10 -4 /°C

Given \

gg = 3ag = 3 ´ 0.1 ´ 10 -4 /°C = 0.3 ´ 10 -4 /°C

\

DV = 1000[1.82 ´ 10 -4 - 0.3 ´ 10 -4 ] ´ 100 = 152

43. With temperature rise (same 20°C for both), steel scale and copper wire both expand. Hence length of copper wire w.r.t. steel scale or apparent length of copper after rise in temperature

Þ

Lapp = L'cu - L'steel = [L0(1 + a cuDq) - L0(1 + a s Dq)] Lapp = L0( a cu - a s ) Dq

= 80(17 ´ 10 -6 - 11 ´ 10 -6) ´ 20

p59 < p0 , w2 > w1 w1 < w2

47. Given, coefficient of volume expansion ( g) = 49 ´ 10 -5/K Rise in temperature ( Dt ) = 30°C Let initial volume of glycerine be V0 . \ Volume of glycerine when temperature is increased by 30°C V = V0[1 + gDt ] = V0[1 + 49 ´ 10 -5 ´ 30 ] = V0[1 + 0.01470 ] = 10147 . V0 V0 1 \ = V 10147 . If mass of glycerine is m, then m initial density of glycerine (r 0) = V0 and final density of glycerine (r) =

= 0.0096 cm \Length of the wire read = 80.0096 cm

\

44. Moment of inertia of a rod, 1 ...(i) ML2 12 where M is the mass of the rod and L is the length of the rod 1 \ Dl = 2MLDL (\M is a constant) ...(ii) 12

Fractional change in density =

...(iii)

DL = LaDt DL or = aDt L DL in Eq. (iii), we get Substituting the value of L Dl = 2aDt l As

45. Thermostat is used in electric apparatus like refrigerator, iron, etc., for automatic cut-off. Therefore, for metallic strips to bend on heating their coefficient of linear expansion should be different.

46. As the coefficient of cubical expansion of metal is less as compared to the coefficient of cubical expansion of liquid, we may neglect the expansion of metal ball. So when the ball is immersed in alchohol at 0°C, it displaces some volume V of alchohol at 0°C and has weight w1. \ w1 = w0 - v p 0 g where, w0 = weight of ball in air Similarly, w2 = w0 - v p0 g where, and

p 0 = density of alchohol at 0°C p59 = density of alchohol at 59°C

…(i)

m V

r 1 m / V V0 = = = r 0 m / V0 V 10147 .

l=

Divide Eq. (ii) by (i), we get Dl DL =2 l L

559

…(ii)

Dr r - r 0 r = = -1 r0 r0 r0

Substituting value from Eq. (ii), we get 1 Fractional change in density = -1 10147 . = - 0.0145 = - 145 . ´ 10 -2 Negative sign shows that the density of glycerine decreases with rise in temperature.

48. Loss of weight at 27°C is = 46 - 30 = 16 = V1 ´ 1. 24 r1 ´ g Loss of weight at 42°C is

...(i)

= 46 - 30.5 = 15.5 = V2 ´ 12 . r1 ´ g 16 V 1.24 Now dividing Eq. (i) by (ii), we get = 1 ´ 15.5 V2 1.2

...(ii)

But

V2 = 1 + 3a(t 2 - t1) V1 =

Þ Þ

15.5 ´ 1.24 = 1.001042 16 ´1.2

3a( 42° - 27° ) = 0.001042 a = 2.316 ´ 10 -5 /°C

49. Substances are classified into two categories (i) Water like substances which expand on solidification (ii) CO2 like (Wax, Ghee etc. ) substances which contract on solidification. Their behaviour regarding solidification is opposite. Melting point of with rise of pressure but that of wax etc increases with increase in pressure. Similarly ice starts forming from top to downwards whereas wax starts its formation from bottom to upwards.

560 JEE Main Physics mL . This must be t the heat supplied for keeping the substance in molten state per sec. Pt mL = P or L = \ m t

50. Heat lost in t sec = mL or heat lost per sec =

51. Heat is lost by steam in two stages (i) for change of state from steam at 100°C to water at 100°C is m ´ 540 (ii) to change water at 100°C to water at 80°C is m ´ 1 ´ (100 - 80), where m is the mass of steam condensed. Total heat lost by steam is m ´ 540 + m ´ 20 = 560 m(cals.) Heat gained by calorimeter and its contents is = (1.1 + 0.02) ´ (80 - 15) = 1.12 ´ 65 cal Using principle of calorimetery, Heat gained = heat lost \ 560m = 112 . ´ 65 m = 0.130 g

52. Initially ice will absorb heat to raise its temperature to 0°C then its melting takes place If mi = Initial mass of ice, mi ¢ = Mass of ice that melts and mw = Initial mass of water By law of mixture, Heat gained by ice = Heat lost by water Þ mi ´ c ´ (20) + mi ¢ ´ L = mw cw (20) Þ 2 ´ 0.5(20) + m1¢ ´ 80 = 5 ´ 1 ´ 20 Þ m1¢ = 1 kg So final mass of water = Initial mass of water + Mass of ice that melts = 5 + 1 = 6 kg

53. Heat gained by the water = (Heat supplied by the coil) - (Heat dissipated to environment) mcDq = PCoil t - PLosst

Þ Þ Þ

2 ´ 4.2 ´ 10 3 ´ (77 - 27) = 1000t - 160t t=

Now the bullet will melt if Q 2 ³ Q1 3 mv 2 ´ 10 -3 ³ 15 m ´ 4.2 i. e. , 8 Þ v min = 410 m/s

55. We can relate an absorbed energy Q and the resulting temperature increase DT with relation Q = cmDT. In that equation, m is the mass of the material absorbing the energy and c is the specific heat of that material. An absorbed dose of 3 Gy corresponds to an absorbed energy per unit mass of 3 J/kg. Let us assume that c the specific heat of human body, is the same as that of water, 4180 J/kg K. Then we find that Q/m 3 DT = = = 7.2 ´ 10 -4K » 700 mK c 4180 Obviously the damage done by ionizing radiation has nothing to do with thermal heating. The harmful effects arise because the radiation damages DNA and thus interferes with the normal functioning of tissues in which it is absorbed.

56. Heat gain = Heat lost Þ and Þ Þ

54. If mass of the bullet is m g then total heat required for bullet to just melt down Q1 = mcDq + mL = m ´ 0.03(327 - 27) + m ´ 6 = 15 m cal = (15m ´ 4.2)J Now when bullet is stopped by the obstacle, the loss in its 1 mechanical energy = (m ´ 10 -3)v 2J 2 (As mg = m ´ 10 -3 kg) As 25% of this energy is absorbed by the obstacle, The energy absorbed by the bullet 75 1 3 Q2 = ´ mv 2 ´ 10 -3 = mv 2 ´ 10 -3 J 100 2 8

CB(23 - 19) = CC (28 - 23) CB 5 = CC 4 C A 15 = CC 16

…(i)

If q is the temperature when A and C are mixed, then Þ

C A ( q - 12) = CC (28 - q) C A 28 - q = CC q - 12

On solving Eqs. (i) and (ii), q = 20.2°C

57. Suppose m kg steam is required per hour Heat is released by steam in following three steps (i) When 150°C steam ¾® 100°C steam Q1

Q1 = mcsteam Dq = m ´ 1150 ( - 100) = 50 m cal

5

4.2 ´ 10 = 500 s = 8 min 20 s 840

C A (16 - 12) = CB(19 - 16) CA 3 = CB 4

(ii) When 100°C steam ¾® 100°C water Q2

Q 2 = mLv = m ´ 540 = 540 m cal (iii) When 100°C water ¾® 90°C water Q2

Q3 = mcw Dq = m ´ 1 ´ (100 - 90) = 10 m cal Hence total heat given by the steam Q = Q1 + Q 2 + Q3 = 600 mcal Heat taken by 10 kg water Q ¢ = mcw Dq = 10 ´ 10 3 ´ 1 ´ (80 - 20) = 600 ´ 10 3 cal Hence Þ Þ

Q = Q' 600m = 600 ´ 10 3 m = 10 3 g = 1 kg

…(ii)

Heat and Kinetic Theory of Gases 58. Suppose, height of liquid in each arm before rising the temperature is l.

l

l

l1

l2

= 335 ´ 10 3 J/kg

With temperature rise height of liquid in each arm increases i. e., l1 > l and l2 > l l1 l2 Also, l= = 1 + gt1 1 + gt 2 Þ

l1 + gl1t 2 = l2 + gl2t1 l -l g= 1 2 l2t1 - l1t 2

Þ

63. Given, mass of copper block (m) = 2.5 kg Change in temperature ( Dt ) = 500°C Specific heat ( s) = 0.39 J/g-K = 390 J/kg-K Latent heat of fusion of water (L) = 335 J/g

t1 t2

.

Heat energy absorbed by copper block Q1 = msDt = 2.5 ´ 390 ´ 500 J Let m¢ kg of ice be melted. Heat energy required to melt ice, Q 2 = m¢ L But heat energy absorbed by copper block = Heat energy utilized by ice in melting Q1 = Q 2 msDt = m¢ L msDt m¢ = L 2.5 ´ 390 ´ 500 = 335 ´ 10 3

59. V = V0(1 + gDq) 3

L = L0(1 + = L30(1 +

a1Dq)L20(1 +

a 2Dq)

2

a1Dq)(1 + a 2Dq) 2

or

L30 = V0 and L3 = V

Since,

Hence, 1 + gDq = (1 + a1Dq)(1 + a 2Dq) 2 @ (1 + a1Dq)(1 + a 2Dq) @ (1 + a1Dq + 2a 2Dq) g = a1 + 2a 2

Þ

561

l 60. (OR) 2 = (PR) 2 - (PO) 2 = l 2 - æç ö÷

2

kg = 1455 . » 15 . kg

64. Let m g of steam get condensed into water (By heat loss). This happens in following two steps

è2ø 100ºC Steam

2

él ù = [ l (1 + a 2t )]2 - ê (1 + a1t ú ë2 û l2 l2 2 2 2 2 l - = l (1 + a 2 t + 2a 2 t ) - (1 + a12t 2 + 2a1t ) 4 4 Neglecting a 22 t 2 and a12t 2 0 = l 2(2a 2t ) 2a 2 =

Þ

Þ

[(H2 = m × 1 × (100 – 90)] 90ºC Water

2

l (2a 2t ) 4

2a1 Þ a1 = 4 a 2 4

61. P ´ t = mcDq Þ

100ºC (H1 = m × 540) Water

mcDq 4200mDq 4200 ´ m ´ Dq = = P P VI {QC water = 4200 J / kg ´°C } 4200 ´ 1 ´ (100 - 20) t= = 381 s » 6.3 min 220 ´ 4 t=

62. Heat given by water Q1 = 10 ´ 10 = 100 cal Heat taken by ice to melt Q 2 = 10 ´ 0.5 ´ [0 - ( -20)] + 10 ´ 80 = 900 cal As Q1 < Q 2, so ice will not completely melt and final temperature = 0°C As heat given by water in cooling up to 0°C is only just sufficient to increase the temperature of ice from -20° C to 0°C, hence mixture in equilibrium will consist of 10 g ice and 10 g water at 0°C.

Heat gained by water (20°C) to raise its temperature upto 90° = 22 ´ 1 ´ (90 - 20) Hence, in equilibrium; heat lost = heat gain Þ m ´ 540 + m ´ 1 ´ (100 - 90) = 22 ´ 1 ´ (90 - 20) Þ m = 2.8 g The net mass of the water present in the mixture = 22 + 2.8 = 24.8 g Pl 2 65. t = ( x2 - x12) 2Kq Þ t µ ( x22 - x12) Þ

( x2 - x2) t = 22 12 t ' ( x'2 - x'1 )

Þ

7 (12 - 0 2) = t' (2 2 - 12)

Þ

t ¢ = 21h

562 JEE Main Physics 66. The pressure on the rear side would be more due to fictitious force (acting in the opposite direction of acceleration) on the rear face. Consequently, the pressure in the front side would be lowered.

Þ

71. mL =

KADqt Dx

Þ

q = 15.24°C rL 2 ( x2 - x12) 2kq rL 2 rL( x + y)( x - y) t= ( x - y 2) = 2kq 2kq

68. Since, t = \

K( 2T - Tc ) A K(Tc - T) A = a 2a Tc 3 = T 1+ 2

Þ

V 2 t (200) 2 ´ t 67. Heat developed by the heater H = . = R J 20 ´ 4.2 0.2 ´ 1 ´ (20 - q)t Heat conducted by the glass H = 0.002 (200) 2 ´ t 0.2 ´ (20 - q) t Hence, = 20 ´ 4.2 0.002 Þ

Rate of flow of heat in path BCA will be same æQ ö æQ ö i. e. ç ÷ =ç ÷ è t ø BC è t ø CA

500 ´ 80 =

0.0075 ´ 75 ´ ( 40 - 0)t 5

t = 8.9 ´ 10 3 s = 2.47 h

Þ

72. Initially the rods are placed in vessels as shown below RP = R/ 2

69. If suppose KNi = K

R

Þ KAl = 3K and KCu = 6K Since all metal bars are connected in series. æQ ö æQ ö æQ ö æQ ö So, ç ÷ =ç ÷ =ç ÷ =ç ÷ è t ø Combination è t ø Cu è t ø Al è t ø Ni

Q

25 cm

10 cm

15 cm

Cu

Ni

Al

Hence, if Þ Þ Þ Similar if Þ Þ

θ1

θ2

mL æQ ö = q2 L ç ÷ = è t ø2 t

Þ Q

0ºC

æQ ö æQ ö =ç ÷ ç ÷ è t ø Combination è t ø Cu Keq A (100 - 0)

KCu A(100 - q1) lCombination lCu 2KA(100 - 0) 6KA(100 - q1) = (25 + 10 + 15) 25 =

q1 = 83.33°C æQ ö æQ ö =ç ÷ ç ÷ è t ø Combination è t ø Al 2KA(100 - 0) 3 KA( q2 - 0) = 50 15 q2 = 20 °C

70. QTB > TA Þ Heat will flow B to A via two paths (i) B to A (ii) and

R 100ºC l

(100 - 0) = 2R q1 4 From Eqs. (i) and (ii), = q2 1

Req =2R

R 0ºC l

…(ii)

73. Rate of cooling of a body A Area Dq Aes(T 4 - T04) µ = Þ Rµ m Volume t mc 1 For the same surface area R µ Volume R=

Volume of cube < Volume of sphere Q Þ RCube > RSphere i. e., cube, cools down with faster rate.

74. Wien's displacement law is lmT = b Þ

lm =

b 2.88 ´ 10 6 = = 1000 nm t 2880

Energy distribution with wavelength will be as follows

along BCA as shown.

Eλ

(T )A

U2

a√2 a

U1

499 500

√2T B

...(i)

Finally when rods are joined end to end as shown

Keq = 2 K

100ºC

Q ( q1 - q2) = t R Q mL (100 - 0) æ ö = q1L = ç ÷ = è t ø1 t R /2

Þ

1 1 1 9 + + = 6K 3K K 6K

0ºC

a

C(TC)

U3

1499 1500

Þ

100ºC

3 1 1 1 = + + Keq KCu KAl KNi =

0ºC

l

999 1000

and

R

100ºC

λ (nm)

From the graph it is clear that U2 > U1.

Heat and Kinetic Theory of Gases 75. Energy received per second i. e., power P µ (T 4 - T04) P µT4

Þ

Also energy received per sec (P) µ Pµ

Þ

(Q T0 mhollow . Hence hollow sphere will cool fast.

Þ

K3

77. Rate of cooling (R) =

Þ

80. Rate of cooling

l

It is given that H1 = H2 KA( q1 - q2) K3 A( q1 - q2) = Þ 2l l K KK Þ K3 = = 1 2 2 K1 + K2

Þ

16 q - 20 = 1 0.5 q = 28°C

81. Rate of cooling

H1 H2

if m be mass of water taken and S be its specific heat capacity, then Q1 = ms(20.5 - 20) and Q 2 = ms( q - 20) q°C = Final temperature of water Q 2 q - 20 Þ = Q1 0.5

Þ

2

K1

Q 2 = 16 Q1

Þ

4

2 æT ö æd ö = çç 1 ÷÷ ´ ç 2 ÷ è d1 ø P2 è T2 ø

P1

Þ

563

1

5 min

2

10 min

52ºC

3

15 min

θ=?

64ºC

For first process, (80 - 64) é 80 + 64 ù = Kê - q0 ú 5 2 ë û For second process, (80 - 52) ù é 80 + 52 = Kê - q0 ú 10 2 û ë

...(i)

...(ii)

For third process, (80 - q) é 80 + q ù ...(iii) = Kê - q0 ú 15 ë 2 û 1 On solving Eqs. (i) and (ii), we get K = and q0 = 24° C 15 Putting these values in Eq. (iii), we get q0 = 42.7°C

564 JEE Main Physics 83. t =

Ql mLl = KA( q1 - q2) KA( q1 - q2) VrLl = KA( q1 - q2)

88. For no current flow between C and D æQ ö æQ ö ç ÷ =ç ÷ è t ø AC è t ø CB K1A( qA - qC ) K2A( qC - qB) = l l qA - qC K2 = qC - qB K1

Þ

5 + 10 5 ´ A ´ 0.92 ´ 80 ´ 2 .h = = 191 0.004 ´ A ´ 10 ´ 3600

Þ

84. Suppose temperature difference between A and B is 100°C and qA > qB

æQ ö æQ ö ç ÷ =ç ÷ è t ø AD è t ø DB

Also, C H/2 A H

Þ

H H/2

H/2

Heat current will flow from A to B via path ACB and ADB. Since, all the rods are identical, so ( Dq) AC = ( Dq) AD Dq [because heat current H = , here R = same for all] R Þ qA - qC = qA - qD Þ qC = qD i. e., temperature difference between C and D will be zero. Q KADq 85. = t l mL K( pr 2) Dq Þ = t l

Þ

1 Since for second rod K becomes th r becomes double and 4 length becomes half, so rate of melting will be twice i. e æ mö æ mö ç ÷ = 2 ç ÷ = 2 ´ 0.1 = 0.2 g/s è t ø2 è t ø1

87.

10 -3 ´ 92 ´ (100 - 0) ´ 60 1 ´ 8 ´ 10 4 -3

kg

Heat transferred in one day (86400 s) Q = 6 ´ 86400 = 518400 J Now, Q = mL m=

91.

Q 518400 = L 334 ´ 10 3

kg = 1552 g = 1552 .

[As J = 4.2]

P(400°C)

0°C

dQ KA 0.01 ´ 1 = dq = ´ 30 = 6 J/s l dt 0.05

Þ

Þ

s 4 pr 2 100 (200 4 - 0 4) = 4 dt 3 pr rcJ 3 rrcJ rrc 4.2 dt = ´ 10 -6 s = ´ 10 -6 . 48s s 48 7 rrc 7 rrc = ms » ms 80 s 72 s

Ice A

ice so, KA( q1 - q2)t =m´L l

= 6.9 ´ 10

K1K4 = K2 K3

dq Aes(T 4 - T04) 89. Rate of cooling Rc = = dt mc dq A r 2 dq 1 µ Þ µ µ 3 Þ dt V r dt r dT sA 4 90. = (T - T04) [ In the given problem fall in temperature dt mcJ of body dT = (200 - 100) = 100K , temperature of surrounding T0 = 0K, Initial temperature of body T = 200 K]

2 æ m ö Kr Rate of melting of ice ç ÷ µ èt ø l

86. Heat transferred in one minute is utilised in melting the

...(ii)

It is given that qC = qD , hence from Eqs. (i) and (ii), K2 K4 we get, = K1 K3

D

m=

K3 A( qA - qD ) K4 A( qD - qB) = l l qA - qD K4 = qD - qB K3

Þ

H/2 B

...(i)

QA λx

QB (10 – λ)x

B Water 100°C

Heat received by end A, for melting of ice KA( 400 - 0)t QA = = mLice lx Heat received by end B, for vaporisation of water KA( 400 - 100)t QB = = mLvap (10 - l) x 400 L lx Dividing both equations, = ice 300 Lvap (10 - l) x 4 (10 - l) 80 Þ = 3 l 540 Þ l =9

...(i)

...(ii)

Heat and Kinetic Theory of Gases 92. Q = sAt(T 4 - T04)

98. Let the quantity of heat supplied per minute be Q. Then, quantity of heat supplied in 2 min = mC(90 - 80) In 4 min, heat supplied = 2mC(90 - 80) L \ 2mC(90 - 80) = mL Þ = 20 C

If T ,T0 , s and t are same for both bodies Q sphere Asphere 4pr 2 = = Q cube Acube 6 a2 But according to problem, volume of sphere = volume of cube 4 3 Þ pr = a3 3 1/3

æ4 ö a = ç p÷ r è3 ø

Þ

99. In the given graph CD represents liquid state. 100. Density of water is maximum at 4°C and is less on either side of this temperature. C F - 32 9 or F = C + 32 101. We know that, = 100 180 5 Y

Substituting the value of a in Eq. (i), we get Q sphere 4pr 2 4 pr 2 = = 2 Q cube 6 a2 ìï æ 4 ö 1/3 üï 6í ç p ÷ rý ïî è 3 ø ïþ =

4 pr 2 æ4 ö 6ç p ÷ è3 ø

2/3

r2

æpö =ç ÷ è6ø

F

O

1/3

:1

93. Temperature difference between C and D is zero. R

R

A √2T

B T

C

X

Equation of straight line is, y = mx + c Hence, m = (9 / 5), positive and c = 32 positive. The graph is shown in figure. C F - 32 = Þ 5 9

102.

C

20 æ5ö C =ç ÷F è9ø 3

Hence, graph between °C and ° F will be a straight line with positive slope and negative intercept.

103. The horizontal parts of the curve, where the system absorbs R

R D

94. Relation between Celsius and Fahrenheit scale of temperature is

565

C F - 32 5 160 = ÞC = F 5 9 9 9

Equating above equation with standard equation of line 5 y = mx + c, we get slope of the line AB is m = 9

heat at constant temperature must depict changes of state. Here, the latent heats are proportional to lengths of the horizontal parts. In the sloping parts, specific heat capacity is inversely proportional to the slopes.

104. Root mean square speed of organ atom 3RTAr MAr

(v rms) Ar =

Root mean square speed of helium atom (v rms) He =

95. Since in the region AB temperature is constant, therefore at this temperature phase of the material changes from solid to liquid and (H2 - H1) heat will be absorb by the material. This heat is known as the heat of melting of the solid.

Dividing Eq. (i) by Eq. (ii)

Similarly in the region CD temperature is constant, therefore at this temperature phase of the material changes from liquid to gas and (H4 - H3) heat will be absorbed by the material. This heat as known as the heat is vaporisation of the liquid.

Given

96. Initially, on heating temperature rises from -10°C to 0°C. Then ice melts and temperature does not rise. After the whole ice has melted, temperature begins to rise until it reaches 100°C. Then it becomes constant, as at the boiling point will not rise.

97. The volume of matter in portion AB of the curve is almost constant and pressure is decreasing. These are the characteristics of liquid state.

…(i)

3RTHe MHe

(v rms) Ar MHe 3RTAr = ´ (v rms) He MAr 3RTHe

Þ Þ \

(v rms) Ar = (v rms) He æ T ö æM ö 1 = ç Ar ÷ × ç He ÷ è THe ø è MAr ø MHe THe = MAr TAr æM ö TAr = THe ç Ar ÷ è MHe ø æ 39.9 ö = 253.15ç ÷ è 4 ø = 2523.675 K= 2.52 ´ 10 3 K

…(ii)

566 JEE Main Physics 106. Substances having more specific heat take longer time to get heated to a higher temperature and longer time to get cooled. T A B

1 Þ v m µ T. As the T temperature of body increases, frequency corresponding to maximum energy in radiation (v m) increases this is shown in graph ( c).

115. According to Wien's

law, l m µ

116. Given, base area of boiler ( A) = 0.15 m 2

C

Thickness (d) = 10 . cm = 1 ´ 10 -2 m t

tA tB tC

If we draw a line parallel to the time axis, then it cuts the given graphs at three different points. Corresponding points on the times axis shows that tC > tB > t A CC > CB > C A

Þ

107. From given curve, Melting point for A = 60°C and melting point for B = 20°C Time taken by A for fusion = (6 - 2) = 4 min Time taken by B for fusion = (6.5 - 4) = 2.5 min HA 6 ´ 4 ´ 60 8 Then, = = HB 6 ´ 2.5 ´ 60 5

6.0 kg /s = 0.1 kg /s 60 Thermal conductivity of brass (K) = 109 J/s-m-K

Rate of water boils = 6.0 kg/min =

Latent heat of vapourization of water (L) = 2256 ´ 10 3 J/kg Let q1 be the temperature of the part of the boiler in contact with the stove. Rate of heat energy supplied = Rate of heat energy utilized in vaporization KADq = mL d KA( q1 - q2) = mL d 109 ´ 0.15( q1 - 100) = 0.1 ´ 2256 ´ 10 3 1 ´ 10 -2 1635 ( q1 - 100) = 2256 ´ 10 2

108. Anomalous density of water is given by (a). It has maximum

q1 =

density at 4°C. When ice is formed it floats.

111. Initially liquid oxygen will gain the temperature up to its boiling temperature, then it change its state to gas. After this again its temperature will increase.

112. Rate of cooling æç -

dT ö ÷ µ emissivity ( e) è dt ø

æ dT ö æ dT ö From graph, ç ÷ Þ ex > ey ÷ > çè dt ø x è dt ø y Further emissivity ( e) µ Absorptive power ( a) Þ ax > ay (Q good absorbers are good emitters). 1 113. According to Wien's law, lm µ and from the figure T ( l m)1 < ( l m)3 < ( l m) 2, therefore T1 > T3 > T2 Ar 16 114. [Given] = A2000 1 Area under el - l curve reperesents the emissive power of body and emissive power µ T 4 (Hence, area under el - l curve) µ T AT æ T ö =ç ÷ A2000 è 2000 ø

4

Þ

16 æ T ö =ç ÷ 1 è 2000 ø

4

Þ

T = 4000 K

Þ

4

225600 + 100 1635

= 137.98 + 100 = 237.98°C » 238°C dq 117. For q - t plot, rate of cooling = = slope of the curve. dt dq At P, = tan f2 = k ( q2 - q0), dt where, k = constant. dq At Q, = tan f1 = k ( q1 - q0) dt tan f 2 q2 - q0 = Þ tan f1 q1 - q0

118.According to Wien's displacement law, lm µ Þ

1 T

l m2 < l m1

(QT1 < T2)

Therefore I - l graph for T2 has lesser wavelength ( l m) and so curve for T2 will shift towards left side.

119. Area under given curve represents emissive power and emissive power µ T 4 Þ

A µT4 4

16 A2 T24 (273 + 327) 4 æ 600 ö =ç = = ÷ = è 300 ø 1 A1 T14 (273 + 27) 4

Heat and Kinetic Theory of Gases

Negative sign shows that temperature decreases i. e. , the body cools, c is the specific heat of the material and q0 is the surrounding temperature. dq 1 µ Þ dt c dq ö æ i. e., rate of cooling çR = ÷ is inversely proportional to è dt ø the specific heat of material. For A, rate of cooling is large. Therefore, specific heat of A is smaller.

120. According to Newton's law of cooling, θ θi θ0 t

Rateof cooling µTemperature differernce dq µ ( q - q0) Þ dt dq Þ = a( q - q0) dt q t dq Þ òqi ( q - q0) = - aò0 dt Þ

4

(a = constant)

q = q0 + ( qi - q0) e- at

This relation tells us that, temperature of the body varies exponentially with time from qi to q0 . 1 121. According to Wien's displacement law, lm µ . Hence, T if temperature increases l m decreases i. e. , peak of the E - l curve shift towards left. dQ dq 122. = - KA dt dx dQ , K and A are constants for all points Q dt Þ dq µ - dx , i. e., temperature will decrease linearly with x.

123. Since the curved surface of the conductor is thermally insulated, therefore in steady state, the rate of flow of heat at every section will be the same. Hence, the curve between H and x will be straight line parallel to x-axis.

124. According to Stefan's law, E = sT 4 In E = In s + 4 In T Þ Þ ln E = 4 In T + In s On comparing this equation with y = mx + c, we find that graph between In E and In T will be a straight line, having positive slope (m = 4) and intercept on In E axis equal to In s. dq eAs 3 125. = 4q0 Dq dt mc eAs 3 For given sphere and cube 4q0 Dq is constant, so for both mc dq rate of fall temperature = constant. dt 1 126. Wien's law lm µ or n m µ T T n m increases with temperature. So, the graph will be straight line.

127. When a body cools by radiation, the rate of cooling is given by

dq eAs 4 =( q - q40) dt mc

567

128. As,

2 E1 A1 æ T1 ö 1 4 pr12 æ 1ö = ×ç ÷ = ´1 = ç ÷ = 2 è2ø E 2 A2 è T2 ø 4 4 pr2

129. As, b = 2 a and g = 3 a \

b 2a 2 g 3a 3 = = and = = g 3a 3 a a 1

130. If Tx = Ty and Tx ¹ Tz, then Ty ¹ Tz If Tx ¹ Ty and Tx ¹ Tz, then Ty ¹ Tz but Ty may be equal to Tz.

131. With the given amount of heat, the heating effect is more on a body of smaller surface area than that of larger surface area.

132. According to Stefan’s law, E = aT 4 æ dT ö Power radiated, P = ( 4 pr 2) sT 4 = ms ç ÷ è dt ø = where,

3 3 dT pr rs 4 dt

dT = R = rate of cooling dt

P µ r2 4 Also, from 4 rr 2T 4 = pr3r s (R) 3

\

3 sT 4 rrs 1 Rµ r

R=

133. In the given graph, the region AB represents no change in temperature with time. It means ice and water are in thermal equilibrium. The region BC shows the change in temperature with time. The region CD represents a constant temperature (100°C) with time. It means, water and steam are in thermal equilibrium at boiling point.

134. A real gas can behave as an ideal gas under low pressure and high temperature, then all the gas laws are obeyed.

135. For an ideal gas, pV = a constant, when temperature is constant. Thus, the variation between pV and V is a straight line parallel to V-axis. Hence, graphs (a), (b) and (c) are wrong.

136. The heat from hot milk spread on the table is transferred to the surrounding by conduction, convection and radiation, and the temperature of milk falls off exponentially with time according to Newton's law of cooling.

568 JEE Main Physics 143. As, DQ = cm DT + mL = 0.5 ´1(5)+1´ 80 = 82.5 cal

137. According to Stefan's law, E = eAsT 4 Þ

E A = eA AsTA4

and

EB = eB AsTB4

Q \ Þ

144. As, DQ = mL = 10 ´ 540 = 5400 cal 145. As, DQ = mL = 10 ´ 80 = 800 cal 146. SI unit of latent heat is Jkg –1

E A = EB eATA4 = eBTB4 TB =

1 æ eA 4 ö 4 ç TA ÷ è eB ø

1 4ö 4

æ1 = ç ´ (5802) ÷ è 81 ø

Þ TB = 1934 And, from Wien's law l A ´ TA = lB ´ TB l A TB Þ = lB TA Þ

147. Here, p = 2 atm = 2 ´1.013 ´105 Nm–2 T = 17 + 273 = 290 K, s = 2 r = 2 ´ 1 Å = 2 ´ 10 –10 m l=

= 1.11 ´ 10 –7 m

148. For nitrogen molecule, M = 28 g = 28 ´ 10 -3 kg

lB - l A TA - TB = lB TA

Þ

1 5802 - 1934 3868 = = lB 5802 5802

Þ

lB = 1.5 mm

kT (1.38 ´ 10 –23) ´ 290 = 2 ms 2p 1.414 ´ 3.14 ´ (2 ´ 10 –10) 2 ´ 2.026 ´ 10 5

v rms =

= 5.1 ´ 10 2 ms–1

149. As, collision frequency = number of collisions per second

138. As face ABCD has positive charge on it and the gas consists of ionized hydrogen, therefore, isotropy is lost, The usual expression for pressure on the basis of kinetic theory will not be valid as ions would also experience forces, other than the forces due to collisions with the walls of the container.

=

140. As, Q mix

mL mwQw - l i 10 ´ 100 - 10 ´ 80 Cw 1 = = 10 + 10 mi + mw 1000 - 800 20 200 = = 10°C 20

=

141. 10 g water heat taken by ice to melt at 0°C is Q1 = mL = 10 ´ 80 = 800 cal Heat given by water or cool upto 0°C is Q 2 = ms Dq = 10 ´ 1 (80 - 0) = 800 cal Hence, heat given by water is just sufficient to melt the whole ice and final temperature of mixture is 0ºC. 2 m ´ 60 ´ (1 - m) ´ 80 142. Temperature of mixture = (2 m + m) = 13.3° C

v rms 5.1 ´ 10 2 = l 1.1 ´ 10 –7

= 4.58 ´ 10 9 s–1 » 5 ´ 10 9 s–1

150. Time taken for collision of nitrogen molecules t1 =

139. In a diatomic molecule, the rotational energy at a given temperature obeys Maxwell's energy distribution law. As each such atom has three translational degrees of freedom and two rotational degrees of freedom, therefore, at a given 2 temperature, rotational energy is rd the translational KE of 3 each molecule. Energy associated with each molecule per æ 1 ö degree of freedom is fixed ç = kB T ÷ . è 2 ø

3 RT 3 ´ 8.31 ´ 290 = M 28 ´ 10 –3

s 2 ´ 10 -10 = 4 ´ 10 -13 s = v rms 5.1 ´ 10 2

152. Stefan’s law applies here and not the Newton’s law of cooling. According to Stefan’s law, 4

4 E 2 æ T2 ö æ 900 ö =ç ÷ =ç ÷ = 81 è 300 ø E1 è T1 ø

E2 = 81 \ E 2 = 81R R

153.Thermal conductivity of the wall depends only on nature of material of the wall; and not on temperature difference across its two sides.

154. On increasing pressure, boiling point of water increases. Therefore, cooking is faster. 1 T

155. lm µ as per Wien’s displacement law. 156.When two bodies at temperatures T1 and T2 are brought in thermal contact, they do settle to the mean temperature æ T1 + T2 ö ç ÷ . They will do so, in case the two bodies were of è 2 ø same mass and material i. e. , same thermal capacities. In other words, the two bodies may have different thermal capacities that is why they do not settle to the mean temperature, when brought together.

Heat and Kinetic Theory of Gases 157.According to Newton’s law of cooling dQ µ ( q - q0) dt 1 158.According to Wien’s law, lm µ when T is halved, lm T becomes twice. gp c 3 Þ rms = r c g

3p and c = r

159.We know, crms =

T 160. As, crms µ ; When T is doubled and M has become half, M the, crms will become two times.

161.Density of water = 1g/cc \ Mass of 1 cc of water = volume×density = 1 ´ 1 = 1g of water In 1 g mole (or 18 g) of water, the total number of molecules = 6.023 ´ 10 23 \Number of molecules of water in 1 g

(m1 + m 2) RT = p (V1 + V2) Using Eqs. (i) and (ii), we get æ p1V1 p2V2 ö ( p V + p2V2) (V1 + V2) + ç ÷ RT = 1 1 è RT1 RT2 ø (V1 + V2) or

æ p1V1 p2V2 ö + ç ÷ T = ( p1V1 + p2V2) è T1 T2 ø

167. When we increase the temperature of a liquid, the liquid will expand. So, the volume of the liquid will increase and hence, the density of the liquid will decrease.

168. An ideal gas is a gas which satisfying the assumptions of the

q1 - q2 æ q1 + q2 ö µç - q0 ÷ ø è 2 t where, q0 = temperature of surroundings 59 - 49 æ 50 + 49 ö µç - 30 ÷ \ ø è t1 2

Average kinetic energy of diatomic gas due to thermal motion, is m 5 5 5 E = (nR) T = pV = ´ p ´ J 2 2 2 5 ´ (8 ´ 10 4) ´ 1 = = 5 ´ 10 4 J 2´4

=

c12 + c22 + ¼+ c62 6 2

2

2

2 +5 +3 +6 +3 +5 6

t2 =

Þ

39 ´ 5 = 10 s 19

therefore, K =

x+ x Sxi 2 K1K2 = = x x Sxi / Ki K1 + K2 + K1 K2

p2V2 = m 2R T2

b T 1 Tµ lm

lm = \

As l m increases, hence T decreases Þ T1 > T2 > T3 . nT +n T 172. As T = 1 1 2 2 n1 + n2

p1V1 = m1R T1 …(i)

As no work is done in removing the partition, total energy remains conserved. Therefore, 3 3 ( p1V1 + p2V2) = p (V1 + V2) 2 2

…(ii)

170. As, the two portions of the slab are connected in series,

2

166. According to standard gas equation

…(i)

Dividing Eq. (ii) by Eq. (i), we get t 2 39 = t1 19

171. According to Wien’s displacement law,

= 3 2 = 4.242 unit

and

40 - 39 æ 40 + 39 ö µç - 30 ÷ ø è t2 2

and

molecules is, n = 5

2

( p1V1 + p2V2) T1T2 ( p1V1T2 + p2V2T1)

169. From Newton’s law of cooling

164. For diatomic gas, number of degrees of freedom per

2

T=

On solving, we get

kinetic energy.

6.023 ´ 10 23 1 = ´ 10 23 18 3 5 7 1´ 1´ 3 + 5 m1 g1 m 2g 2 æ 5 ö æ 7 ö + 1 1÷ ç ÷ ç g -1 g 2 -1 è 3 ø è 5 ø 3 = = 1.5 = 1 = m1 m ö æ ö 2 æ + 2 g1 - 1 g 2 - 1 ç 1 ÷ + ç 1 ÷ ÷ ç ÷ ç ç 5 - 1÷ ç 7 - 1÷ è3 ø è5 ø

165. As, crms =

…(ii)

For mixture of two gases

=

163. g mix

p1V1 + p2V2 V1 + V2

p=

\

569

\

Tf =

173. The value of

æ7 ö 1(T0) + 1 ç T0 ÷ è3 ø 1+ 1

=

10 T0 5 T0 = 3 ´2 3

pV for one mole of an ideal gas T = gas constant = 2 cal mol–1 K –1

570 JEE Main Physics In this equation, N = number of moles of the gas p = pressure of the gas V = volume of the gas R = universal gas constant and T = temperature of the gas from Eq. (i), N p we have = = constant all gases will contain, V RT

174. As, DV = V g Dt Þ

Þ

0.24 = 100 ´ g ´ 40 0.24 g= = 0.00006 = 6 ´ 10 –5 100 ´ 40 g a= 3 a = 2 ´ 10 -5° C–1

175. Let the temperature of common interface be T°C.

So, at constant pressure and temperature equal, numbers of molecule per unit volume.

Rate of heat flow

\ and

Q KA DT H= = l t 2 KA (T - T1) æQ ö H1 = ç ÷ = è t ø1 4x

Note This result is nothing but Avogadro’s laws.

180. The entropy function gives us a numerical measure of the

KA (T2 - T) æQ ö H2 = ç ÷ = è t ø1 x

In steady state the rate of heat flow should be same in whole system i.e., Þ Þ Þ Þ

H1 = H2 2 KA (T - T1) KA (T2 - T) = 4x x

182. We know that the thermo electromotive force is maximum at neutral temperature. Hence, the thermoelectric power will also maximum at neutral temperature.

T - T1 = T2 - T 2 T - T1 = 2 T2 - 2 T 2T + T T= 2 1 3

183. Let m1 = m2 = m3 = m …(i)

Hence, heat flow from composite slab is 2 T2 + T1 ö KA é KA (T2 - T) ù KA æ (T2 - T1) …(ii) H=ê ÷= úû = x çèT2 3 ø 3x x ë [From Eq. (i)] Accordingly,

é A (T2 - T1) K ù H=ê úû f x ë

Let s1, s2, s3 be the respective specific heats of the three liquids, When A and B are mixed, temperature of mixture = 16°C As heat gained by A = heat lost by B \

ms1 (16 - 12) = ms2 (19 - 16) 4 s1 = 3 s2

176. According to kinetic theory of gas, the mean kinetic energy of 1 molecules per degree of freedom is given by kT 2 and for a gram mole, the kinetic energy NkT RT Rö æ = = ç since, k = ÷ è 2 2 Nø

177. We know that thermal radiations consists of larger wavelength as compared to gamma rays and wavelength in visible regions and so, thermal radiations belong to infrared region.

178. Kinetic energy of ideal gas depends only on its temperature. Hence, it remains constant whether its pressure is increased or decreased.

179. Ideal gas equation can be written as ...(i)

…(i)

When B and C are mixed, temperature of mixture = 23°C. As heat gained by B = heat lost by C, ms2 (23 - 19) = ms3 (28 - 23) …(ii) 4 s2 = 5 s3 3 15 From Eqs.(i) and (ii), s1 = s2 = s3 4 16 When A and C are mixed, suppose temperature of mixture = t heat gained by A = heat lost by C

\

By comparing Eqs. (ii) and (iii), we get 1 f= Þ 3

pV = n RT

irreversibility of a given process, i. e. , it is a measure of disorder of a system. During formation of ice cubes orderedness increases i.e., disorderness decreases, hence entropy decreases. 1 181.Average energy per molecule per degree of freedom = kT 2

ms1 (t - 12) = ms3 (28 - t ) 15 s3 (t - 12) = s3 (28 - t ) 16 15 t - 180 = 448 - 16 t 628 t= = 20.2° C 31

184. Here, lm1 = lm ,T1 = 2000 K l m2 = ?, T2 = 3000 K According to Wein’s displacement law, l m2T2 = l m1T1 2000 2 T l m2 = l m1 1 = l m ´ = lm 3000 3 T2

Heat and Kinetic Theory of Gases sA (3T) 4 - sA (T ¢) 4 = sT (T ¢) 4 - sA (2 T) 4

185. Energy radiated per second by a body at temperature T K as per Stefan’s law is 4

E = esAT = 0.6 sAT

(3 T) 4 - (T ¢) 4 = (T ¢) 4 - (2 T) 4

4

(2 T ¢) 4 = (16 + 81) T 4

186. One calorie is defined as the amount of heat required to raise

æ 97 ö T¢ = ç ÷ è2ø

the temperature of 1 g of water from 14.5°C to 15.5°C at atmospheric pressure.

187. In convection process, the heat is transferred by the bodily motion of the heated particles. It is not so in case of warming of glass bulb due to filament heating. In fact, warming of glass bulb is due to radiation.

dx

r2

r1

dx 1 = 4 pKx2 4 pK

r2

é 1 1ù r2 - r1 ê - ú= r r pK (rr 4 ë 1 2û 1 2)

0

O

or log ( q - q0) = - kt + C So, graph is straight line.

7 T0 3

3 5 æ7 ö 1 ´ R ´ ç T0 - Tf ÷ = 1 ´ R ´ (Tf - T0) è3 ø 2 2 3 By solving, we get Tf = T0 2 f f f f 194. As, n1k1T1 + n2kT2 + n3kT3 = (n1 + n2 + n3) kT 2 2 2 2 n1T1 + n2T2 + n3T3 Þ T= n1 + n2 + n3 Þ

As Therefore,

t dq = -k ò dt 0 q - q0

Temperature =

f 2

loge (θ – θ0)

òq

1 mole He

195. For 1 kg gas energy E = rT

rr 12 r2 - r1

dq = -K ( q - q0) dt (Newton’s law of cooling) q

Box B

1 mole N 2 Temperature = T0

189. As,

Þ

R 2sT 4 r2

Heat lost by the Helium gas = Heat gained by the Nitrogen gas ö æ7 Þ mB ´ (CV ) He ´ ç T0 - Tf ÷ = m A ´ (CV ) N2 ´ (Tf - T0) ø è3 Box A

Rate of heat flow = H T -T T -T = 1 2 = 1 2 ´ 4pK (rr 1 2) R r2 - r1 µ

R 2sT 4 r2

193. When two gases are mixed together, then x

ò dR = R R=ò

hot end to cold end. Also the temperature decreases more rapidly near hot end and goes down to slow towards cold end.

\Radiation power incident to earth = pr02 ´

1 ù é ú êFrom R = KA ú ê ë where, K ® thermal conductivity û

Þ

T

r1

Let us consider an element (spherical shell) of thickness dx and radius x as shown in figure. Let us first find the equivalent thermal resistance between inner and outer sphere. dx Resistance of shell = dR = K ´ 4 px2

As,

1/ 4

191. In steady state the temperature decreases exponentially from

192. Solar constant =

188. To measure the radial rate of heat flow, we have to go for integration technique as here the area of the surface through which heat will flow is not constant.

571

p = rrT rT = p / r 5 8 ´ 10 4 ´ 2 4

\

E=

or

E = 5 ´ 10 4 J

[f = 5, for diatomic gas]

196. The number of moles of the system temains same t

190. In steady state energy absorbed by middle plate is equal to

\

p1V1 p2V2 p (V1 + V2) + = RT1 RT2 RT

Þ

T=

energy released by middle plate

p (V1 + V2) T1T2 p1V1T2 + p2V2T1

According to Boyle’s law \ 3T

T

2T

p1V1 + p2V2 = p (V1 + V2) ( p V + p2V2) T1T2 T= 11 ( p1V1T2 + p2V2T1)

197. According to Newton’s cooling law, options (c) is correct answer.

14 Thermodynamics JEE Main MILESTONE <
Wisothermal > Wadiabatic > Wisochoric

1. For this process Charles’ law is obeyed. V V Hence, V µT Þ 1 = 2 T1 T2 2. Specific heat of a gas during an isobaric process Q æf ö C p = ç + 1÷ R = è2 ø nDT

Important Points 1. For a reversible process, the first law of thermodynamics gives the change in the internal energy of the system. dU = dQ - dW Replacing work with a change in volume gives dU = dQ - pdV Since the process is isochoric, dV = 0 the above equation becomes dU = dQ

3. Work done in a isobaric process W = p(Vf - Vi ) = nR (T f - Ti ) = nRDT 4. From first law of thermodynamics DQ = DU + DW Here,

DW = nRDT

2. Also for specific heat capacity at constant volume CV =

DU = nCV DT \

Integrating both sides yields

DQ = nCV DT + nRDT

b

Q = m ò C V dT

= nC p DT

a

5. Bulk modulus of an isobaric process is zero Dp Kisobaric = =0 (As Dp = 0) DV V 6. p-V curve is a straight line parallel to volume axis, dp slope of p-V curve for an isobaric process = 0. dV (i) Graph 1 incidates isobaric p expansion. The heat is 1 given to the gas. The 2 volume and temperature of the gas both will rise. The gas expands during V positive work.

where C V is the specific heat capacity at constant volume, a is initial temperature and b is final temperature, hence Q = mC V DT On pressure-volume diagram an isochoric process appears as a straight vertical line. p

p –ve work

+ve work

Isobaric compression

Isobaric expansion V

V

3. Work done on gas in some process p

p 1

(ii) Graph 2 indicates isobaric compression. In this process, heat is taken out of the gas. The temperature falls and the gas contracts causing negative work.

1

2

2 V

DV = 0 but dV ¹ 0 Work done = + ve

Indication Diagram of p-V Curves

V

DW can be zero but dW ¹ 0

p

Isothermal

We can determine the work done from area under p-V curve. p

dU dQ = mC V dT dT

(compression) Adiabatic O

Isochoric

V

Isobaric Isothermal Adiabatic (expansion) V

Dp p For isothermal process, = - × For adiabatic process, DV V Dp gp =i. e. , it means that at a particular point, slope DV V (value) of adiabatic curve is more than that for isotherm or

For clockwise For anti-clockwise

DW = + ve DW = - ve

4. Work done is least for monoatomic gas expansion p Isothermal Polyatomic Monoatomic V1

V2

V

adiabatic process

Thermodynamics p

14.8 Efficiency of a Cycle

T1 A Adiabatic compression

In a cyclic process, DU = 0 and

(p1V1)Q1 Isothermal expansion B

T2 p4

(from first law of thermodynamics)

=

Wtotal | Q | – | Q–ve | ´ 100 = +ve ´ 100 | Q+ve | | Q+ve |

ì Q ü = í1 – –ve ý ´ 100 Q +ve þ î Thus,

h=

Wtotal ´ 100 | Q+ve |

ì Q ü = í1 – –ve ý ´ 100 Q+ve þ î

a

14.9 Carnot Cycle The most efficient heat engine cycle is the Carnot cycle consisting of two isothermal processes and two adiabatic process. In order to approach the Carnot efficiency, the processes involved in the heat engine cycle must be reversible and involve no change in entropy. This means that the carnot cycle is an idealization, since no real engine processes are reversible and all real physical processes involve some increase in entropy.

b

Carnot efficiency, h =

p3 V 3 T2 C Heat out

Q2

c

d

V

TH - TC ´ 100% TH

The temperature in the Carnot efficiency expression must be expressed in kelvins. The efficiency of a heat engine cycle is given by W QH - QC h= = QH QH For the ideal case of the Carnot cycle, this efficiency can be written as T - TC h= H TH Using these two expressions together Q T 1- C = 1- C QH TH QC QH = TC TH

Note (i) There cannot be a cycle whose efficiency is 100%. Hence, h is always less than100%. Thus, WTotal ¹ Q + ve (ii) It is just like a shopkeeper. He takes some money from you. (Suppose he takes ` 100/- from you). In lieu of this he provides services to you (suppose he provides services of worth ` 80/-). Then, the efficiency of the shopkeeper is 80%. There cannot be a shopkeeper whose efficiency is 100%. Otherwise what will he save?

V4

D Isothermal compression

First we see what is the meaning of efficiency of a cycle. Suppose 100 J of heat is supplied to a system (in our case it is an ideal gas) and the system does 60 J of work. Then, efficiency of the cycle is 60%. Thus, efficiency (h) of a cycle can be defined as

(p 2 V 2 )

Heat in

Qnet = Wnet

æ Work done by the working substance ö ç ÷ (an ideal gas in our case) during a cycle ÷ ´ 100 h=ç ç Heat supplied to the gas during the cycle ÷ ç ÷ è ø

579

or

QH QC =0 TH TC

If we take Q to represent heat added to the system, then heat taken from the system will have a negative value. For the Carnot cycle Q S i =0 i Ti Which can be generalized as an integral around a reversible cycle dQ (Clausius theorem) ò T =0 For any part of the heat engine cycle, this can be used to define a change in entropy S for the system 2 dQ S2 - S1 = ò 1 T Or in differential form at any point in the cycle dQ dS = T

580 JEE Main Physics For any irreversible process, the efficiency is less than that of the Carnot cycle. This can be associated with less heat flow to the system and/or more heat flow out of the system. The inevitable result is dQ ò T £ 0 (Clausius inequality)

14.10 Carnot’s Engine A heat engine is a device which transforms heat into mechanical work continuously. Carnot designed a theoretical engine which is free from all defects of practical engine. This engine cannot be realised in practice. It consists of a cylinder having perfectly insulating walls and perfectly conducting base. Heat can enter or leave the cylinder through its conducting base. It is fitted with a piston in the cylinder. The working substance draws heat from the heat source but the temperature of the source remains constant (T1 ). The engine also consists a heat sink of infinite thermal capacity, the heat may be rejected by the working substance to the sink whose temperature also remains constant (T2 ). The engine has an insulating stand which thermally isolates the working substance from the surrounding. An ideal gas is used as the working substance.

Sample Problem 11 Carnot’s engine takes in a 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? (a) 2.70 ´ 10 5 cal , 70.70%

Sample Problem 12

A gasoline engine takes in 5 moles of air at 20°C and 1 atm, and compresses it adiabatically to 1/10 th of the original volume. Find the final temperature and pressure. Assume air to be diatomic. The work done and change in internal energy is (a) 46 kJ , - 46 kJ (b) 36 kJ , - 36 kJ (c) -46 kJ , 46 kJ (d) 36 kJ , - 46 kJ

Interpret (c) Let p1 = 1 atm, n = 5 moles, T1 = 293 K V2 =

T1V1g -1 = T2V2g -1

Using, Þ

V1 10

æV ö T2 = T1ç 1 ÷ è V2 ø

Work done =

g -1

= 293(10) 0.4 = 736 K

nR(T1 - T2) 5 ´ 8.3 ´ (293 - 736) = = - 46 kJ g -1 0.4

DU = DQ - W = 0 - W = 46 kJ

Sample Problem 13

How much work is done by ideal gas in expanding isothermally from an initial volume of 3 L at 20 atm to a final volume of 24 L? (a) 1.36 ´ 10 5 J

(b) 1.26 ´ 10 4 J

(c) 1.36 ´ 10 4 J

(d) 2.36 ´ 10 5 J

Interpret (b) In isothermal process at temperature T V2 V1 V W = 2.303( p1V1) log10 2 V1 W = 2.303 nRT log10

= 2.303(20 ´ 2) log10

(b) 2.70 ´ 10 5 cal, 72.72%

(c) 2.70 ´ 10 5 cal , 80.70% (d) 3.70 ´ 10 5 cal , 70.70%

(Using p1V1 = nRT) 24 L-atm 3

= [2.303 ´ 60 ´ log10 8] (101) J = 1.26 ´ 10 4 J

Interpret (b) Given, Q = 10 6 cal and As, \

Check Point

T1 = (827 + 273) = 1100 K T2 = (27 + 273) = 300 K Q1 Q 2 = T1 T2

1. A piece of metal is hammered. Does its internal energy

T æ 300 ö 6 Q 2 = 2 Q1 = ç ÷ (10 ) è1100 ø T1

will the temperature of the gas change, if the volume of the cylinder is gradually increased?

increase?

2. A cylinder filled with gas is placed in heat-proof jacket. How

= 2.720 ´ 10 5 cal

3. If hot air rises, why is it cooler at the top of a mountain than

Efficiency of the engine æ T ö h = ç1 - 2 ÷ ´ 100 è T1 ø

4. On removing the valve, the air escaping from a cycle tube

or

300 ö æ h = ç1 ÷ ´ 100 = 72.72% è 1100 ø

near sea level? becomes cool. Why?

5. Why is it impossible for a ship to use the internal energy of sea water to operate its engine?

Thermodynamics

581

Table 16.1 Different Thermodynamic Processes at a Glance S. No.

Change or Name of Process

1.

Definition

p = constant

V = constant

dQ

(i) For solids dQ = mC p dT

(i) For solids dQ = mC VdT

(ii) For gases dQ = mC p dT = nC p dT

(ii) For gases, dQ = nC VdT

Isobaric

Isochoric

Isothermal

Adiabatic

T = constant

(a) Q = constant (b) Entropy, S = constant

2.

dQ = dW

Zero

Zero

- dW

(iii) For change in state dQ = mL 3.

dU

(i) dQ - pdV

dQ

(ii) dQ - nRdT 4.

dW

(i) pdV

Zero

(ii) nRdT

5.

V = constant T V1 V2 = T1 T2

Equation of state or

6.

p-V graph

or

p = constant T p1 p2 = T1 T2

(i) 2. 303 nRT log10

V2 V1

(i)

R(T2 - T1 ) (1 - g )

(ii) 2.303 p1V1 log10

V2 V1

(ii)

p2 V2 - p1V1 (1 - g )

(iii) 2.303 p1V1 log10

p1 p2 (i) pV g = constant (ii) TV g - 1 = constant (iii) p1 - g T g = constant

pV = constant or p1V1 = p2 V2

p

p

p

Isotherm pV = constant

Adiabatic

pi

p p

Hyperbola pf

Vi Vf Isobaric expansion

V

Isobaric expansion

¥

Zero

V

V Vi Vf Isothermal expansion

V

p V

Asiabatic expansion

gp V

7.

Slope of p-V curve

8.

Law

Charle’s law

Gay-Lussac’s law

Boyle’s law

Poisson’s law

9.

Form of first law

dQ = dU + dW = nC p dT + pd V

dQ = dU = nC VdT

dQ = dW = pdV

(i) dW = - dU (ii) dU = - dW

10.

Bulk modulus

Zero

Infinity

11.

Result of maximum work

Maximum

Zero

-

- p Less from isobaric process but greater from adiabatic process

-

-gp Minimum but not zero

WORKED OUT Examples Example 1

A cylinder containing an ideal gas and closed by a movable piston is submerged in an ice-water mixture. The piston is quickly pushed down from position (1) to position (2) (process AB).

Example 2 A fixed mass of gas is taken through a process A ® B ® C ® A. Here A ® B is isobaric B ® C is adiabatic and C ® A is isothermal 105

A

B

1 p(N/m2)

C

2

V

4

1

The pressure at C is given by (g = 1.5) The piston is held at position (2) until the gas is again at 0°C (process BC). Then the piston is slowly raised back to position (1) (process CA). Which one of the following p-V diagrams correctly represent the processes. AB, BC and CA and the cycle ABCA p

p A

( a)

B

V2

A C

V1

( b) V1

V2

(c)

( d)

C

V2

V1

V

pB = pC For isothermal process CA,

…(i)

pAVA = pC VC From Eqs. (i) and (ii), we get

…(ii)

C

g = 1.5, VC = 43 = 64 m3

V1

\

A

V2

V

In an adiabatic process, heat is neither allowed to enter nor allowed to escape the system, the process AB is adiabatic compression because piston is pushed very quickly from position 1 to position 2. The process BC is isochoric because in this case volume remains constant, whereas process CA is an isothermal expansion because temperature remains constant. These are shown on the p-V diagram correctly in option (d).

Note Slope of adiabatic curve is more in magnitude in comparison to be

1

é V g ù g -1 é 4g ù g -1 VC = ê B ú =ê ú ë VA û ë 1û For

Solution

slope of the isothermal curve.

(a) For adiabatic process BC (pressure constant)

V

B

B

10 5 N / m2 32 (d) 10 5 N /m2 (b)

1

p A

Solution

B

V

p

C

10 5 N / m2 64 (c) zero (a)

pC =

pAVA 10 5 = N / m2 VC 64

Example 3

An ideal gas is taken through a cyclic thermodynamic process through four steps. The amount of heat involved in these steps are Q1 = 5960 J , Q 2 = - 5585 J , Q3 = - 2980 J and Q 4 = 3645 J respectively. The efficiency of the cycle is (a) 100% (c) 28%

Solution

(b) 10.82% (d) 15%

From the given problem, DQ = Q1 + Q 2 + Q3 + Q 4 = 5960 - 5585 - 2980 + 3645 DQ = 9605 - 8565 = 1040 J

Thermodynamics Efficiency of a cycle is defined as h=

Network DW DQ = = Input heat Q1 + Q 4 Q1 + Q 4

DQ = 1040 J

Putting

Q 2 + Q 4 = 5960 + 3645 = 9605 J 1040 h= = 0.1082 = 10.82% 9605

and \

Example 4

The efficiency of an ideal gas with adiabatic exponent g for the shown cyclic process would be

Example 6 A Carnot's engine, with its cold body at 17°C has 50% efficiency. If the temperature of its hot body is now increased by 145° C, the efficiency becomes (a) 55% (c) 40%

Solution

(b) 60% (d) 45% 1-

T2 = 0.5 or T1 = 2T2 = 2 (17 + 273) = 580 K T1

Temperature of hot body is increased by 145°C or 145 K, \

T1¢ = (580 + 145) = 725 K

and

T 2 = (17 + 273) = 290 K æ 290 ö h = ç1 ÷ ´ 100 = 60% è 725 ø

V 2V0

\

C

Example 7 V0

A

B T0

(a)

2T0

(2 ln 2 - 1) (1 - 1 ln 2) (b) g / ( g - 1) g / ( g - 1)

Solution

(c)

T

(2 ln 2 + 1) (2 ln 2 - 1) (d) g /( g - 1) g /( g + 1)

As, WBC = pDV = nRDT = - nRT0 WCA = + 2nRT0 ln 2 nRg T0 DQBC = nC p DT = g -1

and Also,

Efficiency =

\

Work (2 ln 2 - 1) = Input heat g /( g - 1)

A steam engine delivers 5.4 ´ 10 8 J of work per

minute and takes 36 . ´ 109 J of heat per minute from the boiler. What is the efficiency of the engine? How much heat is wasted per minute? (a) 31 . ´ 10 9 J/min

(b) 2 ´ 10 9 J/min

(c) 4 ´ 10 9 J/min

(d) 6 ´ 10 9 J/min

Solution Here, Q1 =

We know that h % = \

W ´ 100 Q1

h% =

5.4 ´ 10 8 J ´ 100 3.6 ´ 10 9 J

=

3 ´ 100 = 15% 20

Also using the relation, Q 1 = W + Q 2, we get Q 2 = Q1 - W

Pressure

Temperature

(a)

increase

increase

= 36 ´ 10 8 - 5.4 ´ 10 8

(b)

decrease

decrease

= 30.6 ´ 10 8 J / min

(c)

increase

decrease

(d)

decrease

increase

= 3.06 ´ 10 9 J / min

Solution

= 31 . ´ 10 9 J / min

According to first law of thermodynamics

DQ = DU + DW For an adiabatic process, DQ = 0

Tµ

1 V

Example 8

The volume of system produced by 1g of water at 100°C is 1650 cm3. What is the change in internal energy during the change of state? Given, J = 4.2 ´ 107 erg cal -1, g = 981 cms-2. Latent heat of steam = 50 calg -1.

\ DU = - W In adiabatic process, 1 pµ g V and

Heat absorbed per minute

Q 2 = Heat rejected per minute

Example 5

During an adiabatic expansion, the increase in volume is associated with which of the following possiblities w.r.t. pressure and temperature?

583

g -1

g > 1, because volume increases. then, p and T will decrease.

(a) 2.0 ´ 10 9 erg

(b) 2.0 ´ 10 11 erg

(c) 2.0 ´ 10 10 erg

(d) 2.0 ´ 10 12 erg

Solution \

Here, mass of water = 1 g

Initial volume of water, V1 = 1 cm3 Volume of steam, V2 = 1650 cm3

Change of internal energy, dU = ?

584 JEE Main Physics As the state of water is changing, \

Solution

= 540 ´ 4.2 ´ 10 7 erg

Let the original volume V1 = V \Final volume V2 = V / 2 Initial pressure p1 = 0.76 m of Hg column.

= 22.68 ´ 10 9 erg

Let p2 be the final pressure after compressions

dQ = mL = 1 ´ 540 cal

As the change in adiabatic,

Taking p = 1atm, = 76 ´ 13.6 ´ 981 dyne cm

p1V1g = p2V2g

\

-2

g

dW = pdV = p (V2 - V1)

1.4 æV ö æ V ö p2 = p1 ç 1 ÷ = p1 ç ÷ èV / 2ø è V2 ø

or

= 76 ´ 13.6 ´ 981 (1650 - 1) = 76 ´ 13.6 ´ 981 ´ 1649 erg

p2 = 0.76 ´ (2)1.4

= 1.67 ´ 10 9 erg As

dQ = dU + dW

\

dU = dQ - dW = 22.68 ´ 10 9 - 1.67 ´ 10 9 dU = 21.01 ´ 10 9 erg

Example 9

Find the work required to compress adiabatically 1g of air initially at NTP to half its volume. Density of air at NTP = 1.29 ´ 10 -3 gcm -3 and g = 1.4. (b) - 62.75 J (d) - 82.75 J

(a) 62.75 J (c) 82.75 J

Solution

Here, T1 = 0 + 273 = 273 K; p1 = 1.013 ´ 10 6 dyne cm -2; g = 1.4

and density of air at NTP, r = 1.29 ´ 10 -3 gcm-3 Mass 1 \ = = 775.2cm3 V1 = Density 1.29 ´ 10 -3 V 775.2 and = 387.6 cm3 V2 = 1 = 2 2 æV ö p1V1g = p2V2g or p2 = p1 ´ ç 1 ÷ è V2 ø

Now,

æ V ö or p2 = 1.013 ´ 10 6 ´ ç 1 ÷ è V1 / 2 ø

g

1× 4

= 1.013 ´ 10 6 ´ (2)1.4

= 1.013 ´ 10 6 ´ 2.639 = 2.673 ´ 10 6 dyne cm -2 Work done during adiabatic change, p V - p2V2 W= 11 g -1 =

1 th of its 4 original volume. What is the rise in temperature, the original temperature being 27°C and g = 1.5?

Example 11

(a) 300°C (c) 400°C

Solution

6

7.853 ´ 10 8 - 10.36 ´ 10 8 = - 6.27 ´ 10 8 erg 0.4 = - 62.7 J =

Let the initial volume, V1 = V

Final volume, V2 = V/ 4 Here, initial temperature, T1 = 27° C = 273 + 27 = 300 K and g = 1.5 Let T2 K be the final temperature after compression. Since the change is adiabatic, \ T1V1g - 1 = T2V2g - 1 or

æV ö T2 = T1 ç 1 ÷ è V2 ø

g -1

é V ù = 300 ê ë V/ 4 úû

1.5 - 1

= 300 ( 4)1/ 2 = 300 ´ 2 = 600 K T2 = 600 - 273 = 327° C \Rise in temperature = 317° C - 27° C = 300° C or 300 K

Example 12 A gram molecule of a gas at 127°C expands isothermally until its volume is doubled. How much will be the amount of heat energy absorbed? (a) 400 cal (c) 500 cal

(b) 548 cal (d) 580 cal

Here, temperature of the gas, T = 273 + 127 = 400 K Let initial volume of the gas, V1 = V \ Final volume of the gas, V2 = 2V In an isothermal expansion, V Work done (W ) = 2.3026 RT log10 2 V1

A certain gas at atmospheric pressure is compressed adiabatically so that its volume becomes half of its original volume. What will be the resulting pressure in Nm -2? Take, g = 1.4, for air. 1.4

(b) 0.76 ´ (2) 76 (d) (2)1.4

2V V = 2.3026 ´ 8.3 ´ 400 ´ 0.3010 W = 2.30 ´ 10 3 J = 2.3026 ´ 8.2 ´ 400 ´ log10

Example 10

(a) 76 ´ (2) 0.76 (c) (2)1.4

(b) 350°C (d) 450°C

Solution

6

1.013 ´ 10 ´ 775.2 - 2.673 ´ 10 ´ 387.6 1.4 - 1

1.4

A gas is suddenly compressed to

or

If H is the amount of heat absorbed H=

W 2.30 ´ 10 3 = = 548 cal J 4.2

Thermodynamics Example 13

A cylinder containing one gram molecule of the gas was compressed adiabatically until its temperature rose from 27°C to 97°C. The heat produced in the gas (g = 1.5) is (a) 250.6 cal (b) 276.7 cal

Solution

(c) 298.5 cal (d) 320.6 cal

T1 = 27° C = 273 + 27 = 300 K Final temperature, T2 = 97° C = 273 + 97 = 370 K When a gas is compressed adiabatically, work done on the gas is given by R W= (T2 - T1) (1 - g) =

(a) 25% (c) 40% Here, steam point,

Ice point, As \

T1 = 100° C = 100 + 273 = 373 K T2 = 0° C = 0 + 273 = 273 K 273 100 h =1= 373 373 273 100 h =1373 373 100 = ´ 100% = 26.81% 2373

A Carnot engine intakes steam at 200°C and after doing work, exhausts it to a sink at 100°C. The percentage of heat which is utilised for doing work is

W 11.62 ´ 10 2 = = 276.7 cal J 4.2

(a) 21%

Example 14

5 moles of an ideal gas is carried by a quasi-static isothermal process at 500 K to twice its volume as shown in figure.

Solution

(b) 25%

(d) 30%

T2 = 100°C = 100 + 273 = 373 K T 273 100 h =1- 2 =1= T1 473 473 h=

pB

(c) 27%

Here, T1 = 200°C = 200 + 273 = 473 K

A

pA

(b) 35% (d) 50%

Example 16

\Heat produced, H=

The efficiency of a Carnot’s engine working between steam point and ice point is

8.3 ´ (370 - 300) 1 - 1.5

W = - 11.62 ´ 10 2J

or

Example 15

Solution

Here, initial temperature,

585

B

100 ´ 100% = 21.14% 473

This is the percentage of heat which is utilized for doing work. O

VA V

(i) How much work was done by the gas along the path AB? p (ii) Calculate the pressure ratio B pA Given, R = 8.31J mol -1 K -1 (a) 14,401.3 J, 2 1 (c) 14,401.3 J, 2

Solution

(a) 15 ´ 10 8 J

(b) 15 ´ 10 4 J

(c) 5 ´ 10 5 J

(d) 2 ´ 10 4 J

Solution (b) 34,28.9 J, 2 1 (d) 3428.9 J, 2

T1 = 227° C = 227 + 273 = 500 K

(i) Here, n = 5; T = 500K; VB = 2 VA ;

Wiso = n ´ 2.303 ´ RT log

VB VA

= 5 ´ 2.303 ´ 8.31 ´ 500 log

\ 2VA VA

= 5 ´ 2.303 ´ 8.31 ´ 500 log2 = 5 ´ 2.303 ´ 8.31 ´ 500 ´ 0.3010 = 14,401.3 J (ii) For an isothermal change, pA VB = pBVB pB VA V 1 \ = = A = PA VB 2VA 2

Here, Q1 = 6 ´ 10 5 cal

T2 = 127° C = 127 + 273 = 400 K Work done/cycle, W = ? Q 2 T2 As = Q1 T1

R = 8.31 mol-1 K -1 Now,

A Carnot engine absorbs 6 ´ 10 5 cal at 227°C. The work done per cycle by the engine if its sink is maintained at 127°C is

Example 17

VB

Q2 =

T2 400 ´ Q1 = ´ 6 ´ 10 5 T1 500

= 4.8 ´ 10 5 cal As

W = Q1 - Q 2 = 6 ´ 10 5 - 4.8 ´ 10 5 W = 1.2 ´ 10 5 cal = 1.2 ´ 10 5 ´ 4.2 J W = 5.04 ´ 10 5 J

Start Practice for

JEE Main Round I

(Topically Divided Problems)

Thermodynamic Processes

6. If

1. The ratio of the slopes of p-V graphs of adiabatic and isothermal is (a)

g -1 g

(b) g - 1

(c) g /1

(d) g

2. An ideal gas at a pressure 1 atm and temperature of 27°C is compressed adiabatically until its pressure becomes 8 times, the initial pressure. Then, the final 3ö æ temperature is ç g = ÷ è 2ø (a) 627°C

(b) 527°C

(c) 427°C

(d) 327°C

3. A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of heat insulator, and the piston is insulated by having a pile of sand on it/ By what factor does the pressure of the gas increase, if the gas is compressed to half its original [NCERT Exemplar) volume? (a) 1.40

(b) 1.60

(c) 2.64

(d) 1.94

4. p-V diagram of an ideal gas is as shown in figure. Work done by the gas in the process ABCD is 2p0

7. An ideal gas is taken from state A to state B following three different paths as shown in p-V diagram. Which one of the following is true? y A

C

p

B

D

x

O V

(a) Work done is maximum along AB (b) Work done is minimum along AB (c) Work done along ACB = work done along ADB (d) Work done along ADB is minimum.

of a gas. How does the work done in the process changes with time?

p p0

(a) 0.25 kg (b) 2.25 kg (c) 0.05 kg (d) 0.20 kg

8. Figure shows a thermodynamic process on one mole

D

C

an average person jogs, hse produces 14.5 ´ 103 cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires [NCERT Exemplar] 580 ´ 103 cal for evaporation) is

B

A

y B

V0

p

3V0

2V0 V

(a) 4 pV0

(b) 2p 0V0

(c) 3p 0V0

(d) p 0V0

5. If for hydrogen C p - CV = m and for the nitrogen

C p - CV = n, where C p , CV refer to specific heats per unit mass respectively at constant pressure and constant volume, the relation between m and n is (a) m = 14 n (c) m = 7 n

(b) n = 7 n (d) n = 14 n

A O

x V

(a) decreases continuously (b) increases continuously (c) remains constant (d) first increases and then decreases

Thermodynamics 9. During an adiabatic process, the pressure p of a fixed mass of an ideal gas changes by Dp and its volume V changes byDV. If g = C p / CV , then DV / V is given by Dp p Dp (c) gp

(a) -

(b) -g (d)

587

14. In the indicator diagram, net amount of work done will be

Dp p

p 1

Dp g2 p

10. Figure shows four p-V diagrams. Which of these curves represent isothermal and adiabatic process?

2

V

(a) positive

(b) zero

(c) infinity

(d) negative

15. A cyclic process is shown in figure. Work done during

y

isobaric expansion is A

D C V

(b) A and C

(c) A and B

(d) B and D

102

state B along ACB and is brought back to A along BDA as shown in figure. Net work done during one complete cycle is given by area C p1

D A

O

(a) ACBDA (c) AVV 1 2 BDA

V1

(a) 1600 J

1

3 V (Vm–3)

2

(b) 100 J

(c) 400 J

(d) 600 J

16. In a p-V diagram for an ideal gas (where p is along

X

V2

(a) 1 (c) C p / CV

(b) 2 (d) CV / C p

17. An ideal gas undergoes cyclic process ABCDA as

V

(b) ACB p2 p1 A (D) BD Ap1p2 B

shown in given p-V diagram. The amount of work done by the gas is [NCERT Exemplar] p

12. A gas at pressure p is adiabatically compressed so that its density becomes twice that of initial value. Given that g = C p /CV = 7/5, what will be the final pressure of the gas?

D

2p0 p0

7 p 5 (d) p

(a) 2p

C

y-axis and V is along x-axis), the value of the ratio “slope of adiabatic curve/slope of the isothermal curve” at any point will be (where symbols have their usual meanings).

B

p

D

O

11. A thermodynamic system is taken from state A to

p2

B

p

x

O

(a) D and C

A

2 × 102

B

(Nm–2)

p

C B

A

(b)

(c) 2.63 p

13. Two isothermal curves are shown in figure at temperature T1 and T2 . Which of the following relations is correct?

V

3V0

V0

(b) - 2p 0V0 (d) + 4 p 0V0

(a) 6 p 0V0 (c) + 2 p 0V0

18. In the following p-V diagram figure two adiabates cut two isothermals at T1 and T2 . The value of Vb/ Vc is

p A

B

p

T1

T1 T2

D

V

(a) T1 > T2 (c) T1 = T2

(b) T1 < T2 1 (d) T1 = T2 2

Va

C

Vd

Vb

T2

Vc

V

(a) = Va / Vd (c) > Va / Vd

(b) < Va / Vd (d) Cannot say

588 JEE Main Physics 19. In figure a certain mass of gas traces three paths 1, 2,

23. An ideal monoatomic gas is taken around the cycle

3 from state A to state B. If work done by the gas along three paths are W1, W2 , W3 respectively, then

ABCD as shown in p versus V diagram. Work done during the cycle is y

B 1

p

(V, 2p) D

(2V, 2p) C

A (V, p)

B (2V, p)

p

2 3 A

x

V

(a) W1 < W2 < W3 (c) W1 > W2 > W3

V

(a) pV (c) 2 pV

(b) W1 = W2 = W3 (d) Cannot say

24. By what percentage should the pressure of the given

20. Work done by the system in closed path ABCA, is

mass of gas be increased so to decrease its volume by 10% at a constant temperature?

y p2 A

(a) 5% (c) 12.5%

p

(b) 7.2% (d) 11.1%

25. One mole of an ideal gas expands adiabatically from

p1

C

B O

(b) 0.5 pV (d) 3 pV

V1

x

V2 V

(b) (V1 - V2 )( p1 - p2 ) ( p + p1 )(V2 - V1 ) (d) 2 2

(a) zero ( p - p1 )(V2 - V1 ) (c) 2 2

an initial temperature T1 to a final temperature T2 . The work done by the gas would be (a) (C p - CV )(T1 - T2 ) (c) CV (T1 - T2 )

(b) C p (T1 - T2 ) (d) (C p - CV )(T1 + T2 )

26. A gas at pressure 6 × 105 Nm–2 and volume 1 m3 and

21. Figure shows four thermodynamic process to which a

its pressure falls to 4 × 105 Nm–2. When its volume is 3m3.Given that the indicator diagram is a straight line, work done by the system is

gas sample may be subjected. The isobaric and isothermal curves are

27. A thermodynamic system goes from state (i) (p, V) to

.

y IV

V

(b) II and IV (d) II and III

1

2

Out of the following diagrams (figure), which represents the [NCERT Exemplar] T-p diagram?

22. In the indicator diagram, Ta , Tb, Tc , Td represents

T

V

T

temperatures of gas at A, B, C, D respectively. Which of the following is correct relation?

2

2

(a)

(b) 1

1

A p

(d) pV, pV.

p

ideal gas shown in figure Constant p= V

II

(a) IV and III (c) I and III

(c) pV, zero

28. Consider p-V diagram for an

III

(d) 10 × 105 J

(2 p, V) and (ii) (p,V) to (p, 2V). Work done in the two cases is (a) zero, zero (b) zero, pV

p

I

(a) 6 × 105 J (b) 3 × 105 J (c) 4 × 105 J

B

p

p T

D

T

C V

(a) Ta = Tb = Tc = Td (c) Ta = Tb and Tc = Td

(c)

2

1

(d)

1

2

(b) Ta ¹ Tb ¹ Tc ¹ Td (d) None of these p

p

Thermodynamics 29. The pressure inside a tyre is 4 atm at 27°C. If the tyre bursts suddenly, new temperature will be (g = 7/5) (a) 300 (4)7/2 (c) 300 (2)7/2

(b) 300 (4)2/7 (d) 300 (4)–2/7

30. A monoatomic 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, L2 are the lengths of the gas column before and after expansion respectively, then T1/ T2 is given by (a) ( L1 / L2 ) 2 /3

(b) ( L1 / L2 )

(c) L1 / L2

(d) ( L2 / L1 ) 2 /3

31. For adiabatic expansion of a perfect monoatomic gas, when volume increases by 24%, what is the percentage decrease in pressure? (a) 24% (c) 48%

(b) 30% (d) 71%

32. 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 adiabatic. Then, (a) W2 > W1 (b) W2 > W3 (c) W1 > W2 (d) W1 > W3

> W3 > W1 > W3 > W2

volume of 3 L under adiabatic conditions. If g = 1.40, the work done is (31.4 = 4.6555) (b) 60.7 J (d) 100.8 J

(b) 20 J

(c) 60 J

(d) 40 J

38. In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (a) 15.6 J

(b) 11.2 J

(c) 14.9 J

(d) 16.9 J

39. During an isothermal expansion, a confined ideal gas does –150 J of work against its surrounding. This implies that (a) 150 J of heat has been added of the gas (b) 150 J heat has been removed from the gas (c) 300 J of heat has been added to the gas (d) No heat is transferred because the process is isothermal

Laws of Thermodynamics and Internal Energy

(b) II (d) Cannot say

system and at the same time 105 J of mechanical work was done on the system. The increase in its internal energy is (a) 20 cal (c) 404 cal

(b) 303 cal (d) 425 cal

(a) 40 J

II p

(b) 70 J

(c) 150 J

(d) 110 J

42. A gas is expanded from volume V0 to 2 V0 under three

III

different processes shown in figure.

V

35. An ideal gas is heated at constant pressure and absorbs amount of heat Q. If the adiabatic exponent is g then the fraction of heat absorbed in raising the internal energy and performing the work, is 1 g 2 (c) 1 g

(a) 30 J

it acquires internal energy of 40 J, then the amount of internal work done is

I

diagrams. In which case, work done is minimum

(a) 1 -

of a gas is changed in such a manner that the gas molecules gives out 20 J of heat and 10 J of work is done on the gas. If the internal energy of gas 40 J then the final internal energy will be

41. If the heat of 110 J is added to a gaseous system and

34. As shown in figure three p-V

(a) I (c) III

37. In a thermodynamic process pressure of a fixed mass

40. In a certain process, 400 cal of heat are supplied to a

33. A litre of dry air at STP is allowed to expand to a (a) 48 J (c) 90.5 J

1 g 2 (d) 1 + g (b) 1 +

Process 1 is isobaric process, process 2 is isothermal and process 3 is adiabatic. Let DU1, DU2 and DU3 be the change in internal energy of the gas in these three processes. Then, 1

p0

2

p

3

36. A thermodynamical system is changed from state ( p1V1) to ( p2 V2 ) by two different process. The quantity which will remain same is (a) DQ (c) DQ + W

589

(b) DW (d) DQ - DW

V0

(a) DU1 > DU2 > DU3 (c) DU2 < DU1 > DU3

V

2V2

(b) DU1 < DU2 < DU3 (d) DU2 < DU3 < DU1

590 JEE Main Physics 43. In a thermodynamic process, pressure of a fixed mass

48. If amount of heat given to a system be 50 J and work

of gas is changed in such a manner that the gas releases 20 J of heat and 8 J of work is done on the gas. If internal energy of the gas was 30 J, then the final internal energy will be

done on the system be 15 J, then change in internal energy of the system is

(a) 42 J

(b) 18 J

(c) 12 J

(d) 60 J

44. What is the nature of change in internal energy in the following three thermodynamic processes shown in figure?

p

p

(a) 35 J

(b) 50 J

change in heat energy DQ is equal to (a) 0.5 W

(b) W

(d) 2 W

50. 5 mole of an ideal gas with (g = 7/5) initially at STP are compressed adiabatically so that its temperature becomes 400°C. The increase in the internal energy of gas in kJ is (b) 41.55 (d) 50.55

respect of usual quantities represented by DQ, DU and DW?

(ii)

(a) DU and DW are path dependent (b) DQ and DU are path dependent (c) DU does not depend on path (d) DQ does not depend upon path

p

52. When an ideal monoatomic gas is heated at constant

V (iii)

(a) DU is positive in all the three cases (b) DU is negative in all the three cases (c) DU is positive for (i), negative for (ii), zero for (iii) (d) DU = 0, in all the cases

45. Figure shows two processes a and b for a given sample of a gas, if DQ1, DQ2 are the amounts of heat absorbed by the system in the two cases and DU1, DU2 are changes in internal energies respectively, then y p

(c) 1.5 W

51. Which one of the following statements is true in

V

(I)

(d) 15 J

49. In an isothermal change of an ideal gas, DU = 0. The

(a) 21.55 (c) 65.55 V

(c) 65 J

pressure, fraction of heat energy supplied which increases the internal energy of gas is (a) 2/5 (c) 3/7

(b) 3/5 (d) 3/4

53. Consider two containers A and B containing identical gases at the same pressure, volume and temperature. The gas in container A is compressed to half of its original volume isothermally while the gas in container B is compressed to half of its original value adiabatically. The ratio of final pressure of gas in B to that of gas in A is [NCERT Exemplar]

a

æ 1ö (b) ç ÷ è 2ø

(a) 2g - 1 b

O

2

æ 1 ö (c) ç ÷ è1 - g ø

x

g -1

2

æ 1 ö (d) ç ÷ è g - 1ø

V

(a) DQ1 = DQ2 ; DU1 = DU2 (c) DQ1 < DQ2 ; DU1 < DU2

(b) DQ1 > DQ2 ; DU1 > DU2 (d) DQ1 > DQ2 ; DU1 = DU2 .

46. 1 cm3 of water at its boiling point absorbs 540 cal of heat to become steam with a volume = 1.013 ´ 105 Nm–2 and the mechanical equivalent of heat = 4.19 Jcal–1. The energy spend in this process in overcoming intermolecular forces is (a) 540 cal

(b) 40 cal

(c) 500 cal

(d) zero

47. During adiabatic expansion of 10 moles of a gas, the internal energy decreases by 50 J. Work done during the process is (a) + 50 J (c) zero

(b) – 50 J (d) Cannot say

54. Three copper blocks of masses M1, M2 and M 3 kg

respectively are brought into thermal contact till they reach equilibrium. Before contact, they were at T1, T2 , T3 ( T1 > T2 > T3). Assuming there is no heat loss to the surroundings, the equilibrium temprature [NCERT Exemplar] T is (s is specific heat of copper) T1 + T2 + T3 3 M1T1 + M2T2 + M3T3 (b) T = M1 + M2 + M3 M1T1 + M2T2 + M3T3 (c) T = 3( M1 + M2 + M3 ) M1T1s + M2T2 s + M3T3s (d) T = M1 + M2 + M3 (a) T =

Thermodynamics 55. In which of processess does the internal energy of the system remains constant (a) isobaric (c) adiabatic

(b) isothermal (d) isochoric

56. A thermodynamic system is taken through the cycle PQRSP process. The net work done by the system is p S

200 kPa

64. In a refrigerator, the low temperature coil of evaporator is at – 23°C and the compressed gas in the condenser has a temperature of 77°C. How much electrical energy is spent in freezing 1 kg of water already at 0°C ? (a) 134400 J (c) 80000 J

65. A refrigerator absorbs 2000 cal of heat from ice trays.

R

(b) –20 J

(a) 2100 J (c) 8400 J

Q

P 100 cc

300 cc

(c) 400 J

V

(d) –374 J

Carnot Engine and Refrigerator

and 500 K , (ii) T K and 900 K. The value of T is

800 K and 500 K is (c) 0.375

(d) 0.5

58. A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase its efficiency by 50% of original efficiency ? (a) 280 K

(b) 275 K

(c) 325 K

(d) 250 K

59. An ideal Carnot engine whose efficiency is 40% receives heat at 500 K. If its efficiency were 50%, then in take temperature for same exhaust temperature would be (a) 700 K (c) 800 K

(b) 900 K (d) 600 K

temperature of 27°C and a sink at – 123°C. Its efficiency is (b) 0.25 (d) 0.4

61. Four engines are working between the given temperatures ranges given below. For which temperature range the efficiency is maximum? (a) 100 K, 80 K (c) 60 K, 40 K

(b) 40 K, 20 K (d) 120 K, 100 K

62. An engine has an efficiency of 1/3. The amount of work this engine can perform per kilocalorie of heat input is (a) 1400 cal (c) 700 J

(b) 700 cal (d) 1400 J

63. A Carnot engine works between 600 K and 300 K. In each cycle of operation, the engine draws 1000 J of heat energy from the source. The efficiency of the engine is (a) 50%

(b) 70%

(c) 20%

(c) 270 K

(d) 360 K

ice and room temperatures (17°C). The amount of energy in kWh that must be supplied to freeze 1 kg of water at 0°C is (a) 1.4 (c) 0.058

(b) 1.8 (d) 2.5

68. A Carnot engine has an efficiency of 1/6. When temperature of sink is reduced by 62°C, its efficiency is doubled. Temperature of source and sink are, (a) 99°C, 37°C (c) 37°C, 99°C

(b) 124°C, 62°C (d) 62°C, 124°C

69. A Carnot engine whose low temperature reservoir is at 27°C has an efficiency 37.5%. The high temperature reservoir is at (a) 480°C (c) 307°C

60. A Carnot’s engine works between a source at a (a) 0.5 (c) 0.75

(b) 90 K

67. A refrigerator works between temperature of melting

57. The efficiency of a Carnot engine working between (b) 0.625

(b) 4200 J (d) 500 J

66. A Carnot engine has same efficiency between (i) 100 K (a) 180 K

(a) 0.4

(b) 1344 J (d) 3200 J

If the coefficient of performance is 4, then work done by the motor is

100 kPa

(a) 20 J

591

(d) 80%

(b) 327°C (d) 207°C

70. The coefficient of performance of a refrigerator working between 10°C and 20°C is (a) 28.3 (c) 2

(b) 29.3 (d) Cannot be calculated

1 6 from source into work. When temperature of source is 600 K, temperature at which heat exhausts is

71. A reversible heat engine converts th of heat it absorbs

(a) 500 K

(b) 100 K

(c) 0 K

(d) 600 K

72. A Carnot engine used first ideal monoatomic gas and then an ideal diatomic gas, if the source and sink temperatures are 411°C and 69°C, respectively and the engine extracts 1000 J of heat from the source in each cycle, then (a) area enclosed by the p-V diagram is 10 J (b) heat energy rejected by engine is 1st case is 600 J while that in 2nd case in 113 J (c) area enclosed by the p-V diagram is 500 J (d) efficiencies of the engine in both the cases are in the ratio 21 : 25

592 JEE Main Physics 73. A Carnot engine whose source is at 400 K takes 200 cal of heat and rejects 150 cal to the sink. What is the temperature of the sink? (a) 800 K (c) 300 K

(b) 400 K (d) Cannot say

74. An ideal gas heat engine operates in Carnot cycle between 227°C and 127°C. It absorbs 6 × 104 cal of heat at higher temperature. Amount of heat converted into work is (a) 1.2 × 104 cal (c) 6 × 104 cal

(b) 2.4 × 104 cal (d) 4.8 × 104 cal

75. Two heat engines A and B have their sources at 1000 K and 1100 K and their sinks are at 500 K and 400 K respectively. What is true about their efficiencies? (a) h A = h B (c) h A < h B

(b) h A > h B (d) Cannot say

76. An engine takes compressed steam at 127°C and rejects it at 47°C. Efficiency of the engine is (a) 60% (c) 20%

(b) 35% (d) 40%

77. A Carnot engine has the same efficiency between 800 K to 500 K and x K to 600 K. The value of x is (a) 100 K (c) 846 K

(b) 960 K (d) 754 K

78. What is the temperature of source in Carnot cycle of 10% efficiency when heat exhausts at 270 K? (a) 400 K (c) 300 K

(b) 500 K (d) 600 K

79. Even Carnot engine cannot give 100% efficiency because we cannot (a) prevent radiation (b) final ideal sources (c) reach absolute zero temperature (d) eliminate friction

80. A Carnot engine take 3 ´ 106 cal of heat from a reservoir at 627°C and gives it to a sink at 27°C. The work done by the engine is (a) 4.2 ´ 106 J 6

(c) 16.8 ´ 10 J

(b) 8.4 ´ 106 J (d) zero

81. A Carnot engine has the same efficiency between 800 K to 500 K and x K to 600 K. The value of x is (a) 100 K (c) 846 K

(b) 960 K (d) 754 K

82. What is the value of sink temperature when efficiency of engine is 100%? (a) 0 K (c) 273 K

(b) 300 K (d) 400 K

Specific Heat of Gases and Degrees of Freedom 83. For a gas the difference between the two specific heats is 4150 J/kg-K. What is the specific heats at constant volume of gas it the ratio of the specific heat is 1.4? (a) 8475 J/kg-K (c) 1660 J/kg-K

(b) 5186 J/kg-K (d) 10375 J/kg-K

84. For a gas if the ratio of specific heats at constant pressure and volume is g, then value of degree of freedom is 3g -1 2g -1 9 (c) ( g - 1) 2

2 g -1 25 (d) ( g - 1) 2

(a)

(b)

85. If 70 cal of heat is required to raise the temperature of 2 moles of an ideal gas at constant pressure from 30°C to 35°C, then the amount of heat required to raise the temperature of same gas through same range at constant volume is (a) 50 cal

(b) 70 cal

(c) 60 cal

(d) 65 cal

86. One mole of an ideal gas requires 207 J heat to raise the temperature by 1 K, when heated at constant pressure. If the same gas is heated at constant volume to raise the temperature by the same range, the heat required will be (Take R = 8.3 Jmol–1 K–1) (a) 215.3 J (c) 207 J

(b) 198.7 J (d) None of these

87. One mole of a gas enclosed in a vessel is heated at constant pressure 1 K. Work done by the gas is 1 J R (d) None of these

(a) 1 J

(b)

(c) R J

88. One mole of a monoatomic gas is heated at a constant pressure of 1 atm from 0 K to 100 K. If the gas constant R = 8.32 Jmol–1 K–1, the change in internal energy of the gas is approximately (a) 2.3 J (c) 8.67 × 103 J

(b) 46 J (d) 1.25 × 103 J

89. Which one of the following gases possesses the largest internal energy ? (a) 2 moles of helium occupying 1 m3 at 300 K (b) 56 g of nitrogen at 107 N m –2 at 300 K (c) 8 g of oxygen at 8 atm at 300 K (d) 6 × 1026 molecules of argon occupying 40 m3 at 900 K

90. For the same rise in temperature of one mole of gas at constant volume, heat required for a non linear triatomic gas is K times that required for monoatomic gas. The value of K is (a) 1

(b) 0.5

(c) 2

(d) 2.5

Thermodynamics 91. Value of two principal specific heats of a gas in cal (mol K)–1 determined by different students are given.

Which is most reliable? (a) 5, 2

(b) 6, 5

(c) 7, 5

(d) 7, 4

92. In the above question, if g = 1.5, the gas may (a) monoatomic (b) diatomic (c) a mixture of monoatomic and diatomic gases (d) a mixture of diatomic and triatomic gases

93. Calculate change in internal energy when 5 mole of hydrogen is heated to 20°C from 10°C, specific heat of hydrogen at constant pressure is 8 cal (mol°C)–1. (a) 200 cal

(b) 350 cal

(c) 300 cal

(d) 475 cal

94. A gas expands with temperature according to the 2/ 3

relation V = kT , Calculate work done when the temperature changes by 60 K. (a) 10 R

(b) 30 R

(c) 40 R

(d) 20 R

95. A gaseous mixture contains equal number of hydrogen and nitrogen molecules. Specific heat measurements on this mixture at temperature below 150 K would indicate the value of g = C p/CV for the mixture as (a) 3/2

(b) 4/3

(c) 5/3

(a) 5186 Jkg–1K–1 (c) 1660 Jkg–1K–1

the ratio of two specific heat C p / CV is given by (a)

2 +1 f

Only One Correct Option work done by the system equals the decrease in its internal energy. The system must have undergone an (b) adiabatic change (d) isochoric change

2. The change in internal energy, when a gas is cooled from 927°C to 27°C (a) 300%

(b) 400%

(c) 200%

(d) 100%

enclosed gas, then increase in internal energy and external work done are related as

(a)

3 2

(b)

1 f

(d) 1 -

1 f

4 3

(c) 2

(d)

5 3

99. The adiabatic elasticity of hydrogen gas (g = 1.4) at NTP is

(b) 1 ´ 10 -8 N /m2

(a) 1 ´ 105 N /m2 2

(d) 1.4 ´ 105 N /m2

(c) 1.4 N /m

100. A system is taken through a cyclic process represented by a circle as shown. The heat absorbed by the system is V (in) cc 40 30 20 50

(a) p ´ 103 J

(b)

100 150 200

p J 2

p (in kPa)

(c) 4 p ´ 102 J (d) p J

(Mixed Bag) and 127°C. It absorbs 104 J of heat at the higher temperature. The amount of heat converted into work is (a) 2000 J (c) 8000 J

(b) 4000 J (d) 5600 J

6. In figure two indicator diagrams are shown. If the amounts of work done in the two cases are W1 and W2 respectively, then A

A p

p

(b) DQ = mCV DT + pDV (d) DQ = mC p DT + pDV

helium at NTP is (b) 1.01 × 10–5 Nm–2 (d) 1.69 × 10–5 Nm–2

B

B V

4. Value of adiabatic bulk modulus of elasticity of (a) 1.01 × 105 Nm–2 (c) 1.69 × 105 Nm–2

(c) 1 +

found to be proportional to the cube of its absolute temperature. The ratio C p / CV for the gas is

3. When a small amount of heat DQ is added to an (a) mCV DT = DQ + pDV (c) mCV = DQ + pDV

2 f

5. An ideal gas heat engine is operating between 227°C

1. A given system undergoes a change in which the (a) isothermal change (c) isobaric change

(b) 1 -

98. During an adiabatic process the pressure of a gas is

96. For a gas, the difference between the two principal

Round II

(b) 10375 Jkg–1K–1 (d) 8475 Jkg–1K–1

97. If the degrees of freedom of a gas molecule be f, then

(d) 7/5

specific heats is 4150 Jkg–1K–1. What is the specific heat of the gas at constant volume if, the ratio of specific heat is 1.4?

593

(a) W1 = W2 (c) W1 < W2

V

(b) W1 > W2 (d) Cannot say

594 JEE Main Physics 7. At constant temperature, the volume of a gas is to be decreased by 4%. The pressure must be increased by (a) 4% (c) 8%

(b) 4.16% (d) 3.86%

8. One mole of an ideal monoatomic gas is heated at a constant pressure of 1 atm from 0°C to 100°C. Work done by the gas is 103

10–3

(a) 8.31 × J (c) 8.31 × 10–2 J

(b) 8.31 × J (d) 8.31 × 102 J

9. 500 J of heat energy is removed from 4 moles of a monoatomic ideal gas at constant volume. The temperature drops by (a) 40°C (c) 10°C

(b) 30°C (d) 0°C

10. The ratio of specific heat of a gas at constant pressure to that at constant volume is g. The change in internal energy of one mole of gas when volume change from V to 2 V at constant pressure p is (a) R/( g - 1)

(b) pV g pV (d) g -1

(c) pV /( g - 1)

11. In an

adiabatic change, the pressure and temperature of a monoatomic gas are related as p µ T - c where c equals (a)

2 5

(b)

5 2

(c)

3 5

(d)

5 3

12. If AB is an isothermal, BC is an isochoric and AC is an adiabatic curve; which of the graph correctly represents them in figure p

p A B

(a)

(b)

C V

V p

p A

A

(d) C

C

(b) 66.28 atm (d) 150 atm

15. A gas under constant pressure of 4.5 ´ 105 Pa when subjected to 800 kJ of heat changes the volume from 0.5 m3 to 2.0 m3. The change in the internal energy of the gas is (a) 6.75 × 105 J (c) 3.25 × 105 J

(b) 5.25 × 105 J (d) 1.25 × 105 J

16. A perfect gas goes from state A to state B by

absorbing 8 ´ 105 J of heat and doing 6.5 ´ 105 J of external work. It is now transferred between the same two states in another process in which it absorbs 105 J of heat. In the second process, (a) work done on gas is 105 J (b) work done on gas is –0.5 × 10 5 J (c) work done by gas is 105 J (d) work done by gas is 0.5 × 105 J

17. A Carnot engine is made to work between 200°C and 0°C first and then between 0°C to –200°C. The ratio of efficiencies of the engine in the two cases is (a) 1 : 2 (c) 1.73 : 1

(b) 1 : 1 (d) 1 : 1.73

18. For an engine operating between t1°C and t2 °C, the efficiency will be t1 t2 t1 - t2 (c) t2

(a)

t2 t1 t1 - t2 (d) t1 + 273 (b) 1 -

adiabatic processes are shown in figure. Plots 1 and 2 should p correspond respectively to 2 V

20. An ideal gas is taken through V

13. Ten moles of an ideal gas at constant temperature 600 K is compressed from 100 L to 10 L. The work done in the process is (a) 4.11 × 104 J (b) – 4.11 × 104 J (c) 11.4 × 104 J (d) – 11.4 × 104 J

the cycle A ® B ® C ® A, as shown in figure. If the net heat supplied to the gas in cycle is 5J, work done by the gas in the process C ® A (a) –5 J (b) –10 J (c) 15 J (d) –20 J

2

C

V (m3)

V

1

(a) He and O2 (b) O2 and He (c) He and Ar (d) O2 and N2

B

B

(c)

(a) 20 atm (c) 30 atm

19. p-V plots for two gases during

B

C

1 th of its 20 original volume suddenly. If g = 1.4, the final pressure would be

14. A mass of dry air at NTP is compressed to

B A

1

0

10 5 p(Pa)

Thermodynamics

595

21. 200 cal of heat is given to a heat engine so that it

30. Certain amount of an ideal gas is contained in a

reject 150 cal of heat. If source temperature is 400 K, then the sink temperature is

closed vessel. The vessel is moving with a constant velocity v. The rise in temperature of the gas when the vessel is suddenly stopped is (M is molecular mass, g = C p / CV )

(a) 300 K (c) 100 K

(b) 200 K (d) 50 K

22. Calculate change in internal energy of a system

(a)

(b)

which has absorbed 2 kcal of heat and done 500 J of work

Mv2 ( g - 1) 2R

Mv2 ( g + 1) 2R

(c)

Mv2 2R y

(d)

Mv2 2 R ( g + 1)

(a) 7900 J (c) 6400 J

(b) 8900 J (d) 5400 J

23. During the adiabatic expansion of 2 moles of a gas, change in internal energy was found to be equal to 100 J. Work done in the process will be equal to (a) 100 J (c) 200 J

(b) 50 J (d) 400 J

24. A gas undergoes a process in which its pressure p and volume V are related as Vp = constant. The bulk modulus for the gas in this process is n

(a) np p (c) n

(b) p1/n (d) p n

25. For a monoatomic gas, work done at constant pressure is W. The heat supplied at constant volume for the same rise in temperature of the gas is (a) W /2 (c) 5 W / 2

(b) 3 W / 2 (d) W

26. An ideal gas expands isothermally from a volume V1

31. The specific heat of hydrogen gas at constant

pressure is C p = 3.4 ´ 103 cal / kg° C and at constant volume is CV = 2.4 ´ 103 cal / kg° C. If one kilogram hydrogen gas is heated from 10°C to 20°C at constant pressure, the exterted work done on the gas to maintain it at constant pressure is (a) 105 cal (c) 103 cal

(b) 10 4 cal (d) 5 ´ 103 cal

32. An ideal gas is made to go through a cyclic thermodynamical process in four steps. The amount of heat involved are Q1 = 600 J, Q2 = -400 J, QB = -300 J and Q4 = 200 J respectively. The corresponding work involved are W1 = 300 J,W2 = -200 J,W3 = -150 J and W4 . What is the value of W4 ? (a) –50 J

(b) 100 J

(c) 150 J

(d) 50 J

33. One mole of an ideal monoatomic gas is heated at a constant pressure of one atmosphere from 0°C to 100°C. Then the change in the internal energy is (b) 8.32 ´ 102 J (d) 20.80 J

to V2 and then compressed to original volume V1 adiabatically. Initial pressure is p1 and final pressure is p3. Total work done is W. Then,

34. If the ratio of specific heat of gas at constant pressure

27. Temperature of an ideal gas is 300 K. The change in

to that at constant volume is g. The change in internal energy of a mass of gas when the volume changes from V to 2 V at constant pressure p is

(a) p3 > p1; W > 0 (c) p3 > p1; W < 0

(b) p3 < p1; W < 0 (d) p3 = p1; W = 0

temperature of the gas when its volume changes from V to 2 V in the process p = aV (here a is a positive constant) is (a) 900 K (c) 600 K

(b) 1200 K (d) 300 K

28. The specific heats of an ideal gas at constant pressure and constant volume are 525 J (kg°C)–1 and 315 J (kg°C )–1 respectively. Its density at NTP is (a) 0.64 kgm–3 (c) 1.75 kgm–3

(b) 1.20 kgm–3 (d) 2.62 kgm–3

29. Pressure p, volume V and temperature T of a certain

material are related by p = aT2 / V , where a is constant. Work done by the material when temperature changes from T0 to 2 T0 and pressure remains constant is (a) 3 aT20 3 (c) aT20 2

(b) 5 aT20 (d) 7 aT20

(a) 6.56 J (c) 12.48 ´ 102 J

R ( g - 1) pV (c) ( g - 1) (a)

(b) pV (d)

g pV ( g - 1)

35. When the ideal monoatomic gas is heated at constant pressure fraction of heat energy supplied which increases the internal energy of gas is (a)

2 5

(b)

3 5

(c)

3 7

(d)

3 4

36. A steam engine delivers 5.4 ´ 108 J of work per minute and services 3.6 ´ 109 J of heat per minute from the boiler. What is the efficiency of the engine? How much heat is wasted per minute?

[NCERT Exemplar] 8

8

(a) 1.1 ´ 10 J/min

(b) 2.1 ´ 10 J/min

(c) 1.9 ´ 10 9 J/min

(d) 3.1 ´ 10 9 J/min

596 JEE Main Physics 37. In an isothermal reversible expansion if the volume

45. Two spheres of the same material have radii 1 m and

of 96 g of oxygen at 27°C in increased from 70 L of 140 L, then the work done by the gas will be

4 m and temperatures 4000 K and 2000 K respectively. The ratio of the energy radiated per second by the first sphere to that by the second is

(a) 300 R log10 2 (b) 81 R log e 2 (c) 900 R log10 2 (d) 2.3 ´ 900 R log10 2

(a) 16:1 (c) 1:1

46. Ideal gas undergoes an adiabatic change in its state

38. One mole of O2 gas having a volume equal to 22.4 L at

0°C and 1 atmospheric pressure is compressed isothermally so that its volume reduces to 11.2 L. The work done in this process is (a) 1672.5 J (c) –1728 J

(b) 1728 J (d) –1572.5 J

inside at 90°C. Calculate the coefficient of performance, if the room temperature is 36°C. [NCERT] (b) 11.5 (d) None of these

40. 540 cal of heat converts 1 cubic centimeter of water at 100°C into 1671 cubic centimeter of steam at 100°C at a pressure of one atmosphere. Then, the work done against the atmospheric pressure is nearly (a) 540 cal (c) 200 cal

(b) 40 cal (d) 500 cal

41. The volume of an ideal gas is 1 litre and its pressure is equal to 72 cm of mercury column. The volume of gas is made 900 cm 3 by compressing it isothermally. The stress of the gas will be (a) 8 cm (mercury) (c) 6 cm (mercury)

42. 1 mm 3 of a gas is compressed at 1 atmospheric pressure and from temperature 27°C to 627°C. What is the final pressure under adiabatic condition (g for the gas = 1.5)? (b) 80 ´ 105 N /m2

(c) 36 ´ 105 N /m2

(d) 56 ´ 105 N /m2

p¢ then should be p (a) 1/128 (c) 128

d¢ = 32, d

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

(b) 32 (d) None of these

8 5 of its original volume. If g = , then the rise of 27 3 temperature is (b) 375 K (d) 405 K

starts becoming hotter remains at the same temperature starts becoming cooler may become hotter or cooler depending upon the amount of water vapour present

48. How many times a diatomic gas should be expanded adiabatically so as to reduce the root mean square velocity to half (a) 64 (c) 16

(b) 32 (d) 8

49. At NTP one mole of diatomic gas is compressed adiabatically to half of its volume, g = 1.41. The work done on gas will be (b) 1610 J (d) 2025 J

50. Two moles of an ideal monoatomic gas at 27°C occupies a volume of V. If the gas is expanded adiabatically to the volume 2 V, then the work done 5 æ ö by the gas will be ç g = , R = 8.31 J/ mol K ÷ è ø 3 (b) 2627.23 J (d) –2500 J

51. 5.6 L of helium gas at STP is adiabatically compressed to 0.7 L. Taking the initial temperature to be T1, the work done in the process is (a)

44. An ideal gas at 27°C is compressed adiabatically to

(a) 450 K (c) 225 K

(b) W = m (T1 - T2 ) CV (d) W = m (T1 + T2 ) CV

normal temperature suddenly starts coming out from a puncture. The air inside [NCERT]

(a) + 2767.23 J (c) 2500 J

43. The pressure and density of a diatomic gas ( g = 7 / 5) change adiabatically from ( p, d) to ( p¢, d¢) if

(a) W = m (T1 - T2 ) C p (c) W = m (T1 + T2 ) C p

(a) 1280 J (c) 1815 J

(b) 7 cm (mercury) (d) 4 m (mercury)

(a) 27 ´ 105 N /m2

from ( p1V1T1) to ( p2 V2 T2 ). The work done (W ) in the process is (m = number of moles C p and CV are molar specific heats of gas)

47. Compressed air in the tube of a wheel of a cycle at

39. A refrigerator is to remove heat from the eatable kept

(a) 10.4 (c) 9.8

(b) 4:1 (d) 1:9

9 RT1 8

(b)

3 RT1 2

(c)

15 RT1 8

(d)

9 RT1 2

52. In changing the state of a gas adiabatically from an equillibrium state A to another equillibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal the net work done by the system in latter case will be (a) 5.9 J

(b) 16.9 J

(c) 9.3 J

(d) 4.6 J

Thermodynamics 53. If the temperature of 10 mole of ideal gas is changed from 0°C to 100°C at constant pressure of 2 atm, then the work done in the process is (R = 8.3 J/ mol-K) (a) 8.3 ´ 10 –3 J

(b) 8.3 ´ 10 –2 J

(c) 8.3 ´ 10 4 J

(d) 8.3 ´ 103 J

597

3 to 1 is adiabatic Such a process does not exist because p 1

2

54. 2 kg of water is converted into steam by boiling at atmospheric pressure. The volume changes from 2 ´ 10-3 m 3 to 3.34 m 3. The work done by the system is about (a) –340 kJ (c) 170 kJ

(b) –170 kJ (d) 340 kJ

More Than One Correct Option 55. Figure shows the p-V diagram of an ideal gas undergoing a change of state from A to B. Four different parts I, II, III and IV as shown in the figure may lead to the same change of state. [NCERT Exemplar] A

3 V

(a) heat is completely converted to mechanical energy in such a process, which is not possible. (b) mechanical energy is completely converted to heat in this process, which is not possible. (c) curves representing two adiabatic processes don’t intersect. (d) curves representing an adiabatic process and an isothermal process don’t intersect.

58. An ideal gas is taken from the state A (p, V) to the state B (p/2, 2 V) along a straight line path as shown in figure. Select the correct statement from the following. p

p

p

B

A

C p/2 B

V

(a) DQA ® B = negative (c) DWCAB = negative

(b) DUC ® A = negative (d) DUB ® C = negative

56. Figure shows the p-V diagram of an ideal gas undergoing a change of state from A to B. Four different paths I, II, III and IV as shown in the figure may lead to the same changes of state. [NCERT Exemplar] p

(a) Work done by the gas in going from A to B exceeds the work done in going from A to B under isothermal conditions (b) In the T-V diagram, part AB would become a parabola, (c) In the p-T diagram, path AB would be part of hyperbola (d) In going from A to B, the temperature T of gas first increases to a maximum value 1 and then decreases

are heat added to heat bath T1 and heat T2 taken from in one cycle of engine. W is the mechanical work done on the engine. If W > 0, then possibilities are

IV II B

III

[NCERT Exemplar] V

T1

(a) change in internal energy is same in IV and III cases, but not in I and II (b) Change in internal energy is same in all the four cases (c) Work done is maximum in case I (d) Work done is minimum in case II

57. Consider a cycle followed by an engine (figure) [NCERT Exemplar]

1 to 2 is isothermal 2 to 3 is adiabatic

V0

59. Consider a heat engine as shown in figure. Q1 and Q2

I A

V 2V

V

W

Q1 Q2 T2

(a) Q1 > Q2 > 0 (b) Q2 > Q1 > 0 (c) Q2 < Q1 < 0 (d) Q1 < O, Q2 > 0

598 JEE Main Physics Comprehension Based Questions Passage I The efficiency of a Carnot engine working between source temperature T1 and sink temperature T2 is T h = 1 - 2 . The efficiency cannot be 100% as we T1 cannot maintain T2 = 0. Coefficient of performance of a refrigerator working between the same two temperature is T2 1 -h = T1 - T2 h

27°C and – 73°C is (b) 60%

(c) 33%

temperature of sink is 27°C, what is the source temperature? (b) 400 K

(c) 600 K

(d) 500 K

62. The efficiency of a Carnot engine is 60%. If temperature of source is 127°C. The sink must be maintained at (a) 113 K

(b) – 113°C

(c) 113°C

(d) –113 K

Passage II The changes in pressure and volume of a gas when heat content of the gas remains constant are called adiabatic changes. The equation of such changes is pV g = constant. The changes must be sudden and the container must be perfectly insulting to disallow any exchange of heat with the surroundings. In such changes, dQ = 0, then as per first law of thermodynamics, dQ = dU + W = 0. dU = - dW .

63. A gas in a container is compressed suddenly. Its temperature would (a) increase (b) decrease (c) stay constant (d) change depending upon surrounding temperature.

64. The tyre of a bicycle bursts suddenly. The changes in pressure and volume of air are (a) isothermal (c) isobaric

(b) adiabatic (d) isochoric

Assertion and Reason Direction

is more than the work done by the gas in the same expansion, adiabatically. Reason Temperature remains constant in isothermal expansion, and not in adiabatic expansion. and 227°C cannot have efficiency more than 20%. T Reason Under ideal conditions h = 1 - 2 T1

(d) zero

61. The efficiency of a Carnot engine is 40%. If (a) 300 K

65. Assertion Work done by a gas in isothermal expansion

66. Assertion A reversible engine working between 127°C

60. The efficiency of a Carnot engine working between (a) 100%

(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 correct explanation of the Assertion (c) If Assertion is true but Reason is false (d) If Assertion is false but the Reason is true

Question No. 65 to 74 are Assertion-Reason type. Each of these contains two Statements: Statement I (Assertion), Statement II (Reason). Each of these questions also has four alternative choice, only one of which is correct. You have to select the correct choices from the codes (a), (b), (c) and (d) given below

67. Assertion Reversible systems are difficult to find in real world. ReasonMost processes are dissipative in nature.

68. Assertion A Carnot engine working between 100 K and 400 K has an efficiency of 75%. T Reason It follows from h = 1 - 2 T1

69. Assertion An adiabatic process is an isotropic process. DQ = 0 \ DQ = 0 T which represents an adiabatic process.

Reason DS =

70. Assertion It is not possible for a system unaided by an external agency to transfer heat from a body at a lower temperature to another at a higher temperature. Reason It is not possible to violate the second law of thermodynamics.

71. Assertion Internal energy of an ideal gas depends only on temperature and not on volume. Reason Temperature is more important than volume.

72. Assertion Change of state is an example of isothermal process. Reason Change of state from solid to liquid occurs only at melting point of solid and change of state from liquid to gas occurs only at boiling point of liquid. Thus, there is no change of temperature during change of state.

73. Assertion Efficiency of a Carnot engine decreases with decrease in temperature difference between the source and the sink. Reason h = 1 -

T2 T1 - T2 = T1 T1

74. Assertion First law of thermodynamics is a re-statement of the principle of conservation of energy. Reason Energy is something fundamental.

Thermodynamics

599

Previous Years’ Questions Passage

B (3 p, 4V)

Two moles of helium gas are taken over the cycle ABCDA, as shown in p-T diagram.

p

y 2 × 105 p Pa

A

B

B (p, V)

B (p, 4V) V

1×

105

D 300

(a) 2 pV (c) 6 pV

C 500 T(K)

x

81. The p-V diagram of a gas undergoing a cyclic process

75. Assuming the gas to be ideal, the work done on the gas in taking it from A to B is (a) 300 R (c) 500 R

(b) 3 pV (d) zero

(ABCDA) is shown in the figure, where p is in units of Nm–2 and V in cm3. Identify the incorrect statement. [Kerala CET 2008]

[AIEEE 2009]

(b) 400 R (d) 200 R

y 2 × 105

A

B

p(Nm–2)

76. The work done on the gas in taking it from D to A is [AIEEE 2009]

(a) 415.9 R (c) + 690 R

1 × 105

(b) – 690 R (d) – 414 R

2.0

77. Which of the following p-V diagrams best represents an isothermal process?

[UP SEE 2009]

p

p

(a)

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

(b)

C

D V(cm3)

4.0

0.4 J of work is done by the gas from A to B 0.2 J of work is done on the gas from C to D Now work is done by the gas from B to C Work is done by the gas in going from B to C and on the gas from D to A

82. The work of 146 kJ is performed in order to compress V

V

p

(a) diatomic (b) triatomic (c) a mixture of monoatomic and diatomic (d) monoatomic

p

(c)

(d)

V

V

78. An ideal heat engine exhausting heat at 27°C is to have 25% efficiency. It must take heat at [UP SEE 2008] (a) 127ºC (c) 327ºC

(b) 227ºC (d) None of these

79. A heat engine has an efficiency h. Temperatures of source and sink are each decreased by 100 K. The efficiency of the engine [Karnataka CET 2008] (a) increases (c) remains constant

one kilomole of a gas adiabatically and in this process the temperature of the gas increases by 7°C. The gas [UP SEE 2007] is (R = 8.3 J mol–1 K–1)

(b) decreases (d) becomes 1

80. An ideal monoatomic gas is taken through a cyclic process as shown in p-V diagram, Work done per [Karnataka CET 2008] cycle is

83. A gas is taken through a number of thermodynamic states. What happen to its specific heat? [BVP Engg. 2006]

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

It is always constant It increases It decreases It can have any value depending upon process of heat absorbed or evolved

84. Six moles of O2 gas is heated from 20°C to 35°C at

constant volume. If specific heat capacity at constant pressure is 8 cal (mol-K)–1 and R = 8.31 J(mol-K)–1, what is change in internal energy of gas? [UP SEE 2006] (a) 180 cal (c) 360 cal

(b) 300 cal (d) 540 cal

600 JEE Main Physics 85. In Carnot engine efficiency is 40% at hot reservoir

T

temperature T. For efficiency 50%, what will be temperature of hot reservoir? [UP SEE 2006]

2T0

(a)

2T 5 6T (d) 5

T 5

(b)

(c) 6 T

T0 S

86. During an adiabatic process, the pressure of a gas is found to be proportional to cube of its absolute Cp for the gas is temperature. Then ratio of g = CV [BVP Engg. 2006]

(a) 4/3 (c) 3/2

(b) 5/3 (d) 2

87. In adiabatic expansion of a gas

[BVP Engg. 2005]

2 3 1 (c) 4

1 3 1 (d) 2

(a)

(b)

92. A system goes from A to B via two process I and II as shown in figure. If DU1 and DU2 are the changes in internal energies in the processes I and II respectively, then [AIEEE 2005] p

(a) its pressure increases (b) its temperature falls (c) its density increases (d) its thermal energy increases

II

heating, its pressure is doubled and volume becomes three times. The resulting temperature of the gas [BVP Engg. 2005] will be

89. The value of

(b) 162ºC (d) 600ºC

pV for one mole of an ideal gas is nearly T

equal to mol–1

[Kerala CET 2005]

K–1

(a) 2 J (c) 4.2 J mol–1 K–1

(b) 8.3 cal mol–1 K–1 (d) 2 cal mol–1 K–1

I V

(a) DU2 < DU1 (b) DU2 > DU1 (c) relation between DU1 and DU2 cannot be determined (d) DU1 = DU2

93. The

above p-V diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat, extracted [JEE Main 2013] from the source in a single cycle is

90. A rigid container with thermally insulated walls contains a coil of resistance 100 W carrying current 1 A. Change in internal energy after 5 min will be

2p0

[IIT JEE 2005]

p0

(a) 0 kJ (c) 20 kJ

B

A

88. A gas at 27°C has a volume V and pressure p. On

(a) 1800ºC (c) 1527ºC

2S0

S0

(b) 10 kJ (d) 30 kJ

2T0

2T0 V0

91. The temperature-energy diagram of a reversible engine cycle is given in figure. Its efficiency is [AIEEE 2005]

4T0

2V0

(a) p 0V0

æ13 ö (b) ç ÷ p 0V0 è2ø

æ11ö (c) ç ÷ p 0V0 è2ø

(d) 4 p 0V0

Answers Round I 1. 11. 21. 31. 41. 51. 61. 71. 81. 91.

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

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

2. 12. 22. 32. 42. 52. 62. 72. 82. 92.

3. 13. 23. 33. 43. 53. 63. 73. 83. 93.

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

4. 14. 24. 34. 44. 54. 64. 74. 84. 94.

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

5. 15. 25. 35. 45. 55. 65. 75. 85. 95.

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

6. 16. 26. 36. 46. 56. 66. 76. 86. 96.

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

7. 17. 27. 37. 47. 57. 67. 77. 87. 97.

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

8. 18. 28. 38. 48. 58. 68. 78. 88. 98.

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

9. 19. 29. 39. 49. 59. 69. 79. 89. 99.

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

10. 20. 30. 40. 50. 60. 70. 80. 90. 100.

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

Round II 1. 11. 21. 31. 41. 51. 61. 71. 81. 91.

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

2. 12. 22. 32. 42. 52. 62. 72. 82. 92.

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

3. 13. 23. 33. 43. 53. 63. 73. 83. 93.

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

4. 14. 24. 34. 44. 54. 64. 74. 84.

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

5. 15. 25. 35. 45. 55. 65. 75. 85.

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

6. 16. 26. 36. 46. 56. 66. 76. 86.

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

7. 17. 27. 37. 47. 57. 67. 77. 87.

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

8. 18. 28. 38. 48. 58. 68. 78. 88.

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

9. 19. 29. 39. 49. 59. 69. 79. 89.

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

10. 20. 30. 40. 50. 60. 70. 80. 90.

the Guidance Round I p2 =? p1

1. Slope of p-V graph of adiabatics = g p / V Slope of p-V graph of isothermal = p / V Required ratio = g

2. Here, p1 = 1 atm, T1 = 27° C

As hydrogen is a diatomic gas. 7 \ g = = 14 . 5

= 27 + 273 = 300 K p2 = 8 atm, T2 = ?, g = 3 / 2 As change are adiabatic,

For an adiabatic change,

\ As changes are adiabatic,

\

\

-g

æp ö = ç 1÷ è p2 ø

g

æ V ö =ç ÷ èV / 2ø

g -1

æp ö T2 = T1ç 2 ÷ è p1 ø

1. 4

= (2)1. 4

= 2.64

( g -1 )/ g

= 300 ´ (8)(1.5-1)/1.5 = 300 ´ (8)1/3

T2 = 600 K = (600 - 273)° C = 327° C

3. Let initial volume of the gas in the cylinder be V. \ and

p2 æ V1 ö =ç ÷ p1 è V2 ø

P1g -1T1-g = p2g -1T2- g æ T2 ö ç ÷ è T1 ø

Þ

p1V1g = p2V2g

V1 = V V2 = V / 2

4. As, WAB = - p0V0 , WBC = 0 and \

WCD = 4p0V0 WABCD = WAB + WBC + WCD = - p0V0 + 0 + 4p0V0 = 3p0V0

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

602 JEE Main Physics 5. Here, for hydrogen, C p - CV = m =

R 2

17. Work done by the gas is equal to area of rectangle ABCDA, .

R = 2m R and for nitrogen, C p - CV = n = or R = 28n 28

or

Þ

= AB ´ BC = (2V0) p0 = 2 p0V0 As the trace is anticlockwise, this work is done on the gas. \Work done by the gas = - 2p0V0

18. From symmetry considerations and also from theory,

2m = 28n

Va Vb = Vd Vc

m = 14n

6. Amount of sweat evaporated/minute Sweat produced / Minute Number of cals required for evaporation / kg

= =

4

W1 > W2 > W3 .

20. Work done = Area of DABC

14.5 ´ 10 145 = = 0.25 kg 580 ´ 10 3 580

=

7. As, W = ò p dV = area under the p -V curve

AB ´ BC ( p2 - p1) (V2 - V1) = 2 2

21. Curve IV is parallel to volume axis. It represents isobaric

= minimum along ADB

8. As work done in a process = area under the curve, which increases continuously. Dp 9. As, Ka = gp = DV / V DV Dp =\ V gp

curve. Out of curve II and III, slope of curve III is smaller. Therefore, curve III represents an isothermal curve.

22. AB and CD are isothermal curves therefore Ta = Tb and Tc = Td but all the four temperatures are not equal.

23. Work done during the complete cycle = Area ABCDA = AD ´ AB = p ´ V = pV

10. In curves A and B, pressure and volume both increase. Therefore, temperature must rise and heat must be supplied/work is done. Therefore, A and B cannot be required curves. Out of C and D, slope of D smaller. Therefore, D is isothermal curve and C is adiabatic curve.

11. In a cyclic process, work done is equal to area of the loop ACBDA, representing the cycle of changes.

24. When T is constant, pV = constant. When volume is decreased by 10% i. e. , volume become

90 , the pressure 100

must become100 / 90 . \ % increase in pressure =

(100 - 90) ´ 100 = 11.1% 90

25. As, dW = dU = mCV DT = - CV (T2 - T1) = CV (T1 - T2)

(Q m = 1)

26. According to question, the figure can be drawn as below

12. As, p2V2g = p1V1g g

\

19. As work done by the gas = are under the p-V curve, therefore

g

ær ö æV ö æ2ö p2 = p1ç 1 ÷ = p1ç 2 ÷ = p ç ÷ è 1ø r V è 1ø è 2ø

7 /5

= 2.63p

6 × 105 4 × 105

13. As isothermal curve at T1 is farther from the origin than the isothermal at T2, therefore, T1 > T2.

O

14. Figure shows that loop 1 is anti-clockwise, therefore W1 is negative, loop 2 is clockwise, therefore W2 is positive.

Þ

= 2 ´ 10 ´ (3 - 1) = 4 ´ 10 = 400 J

16. As, it is known, Cp Slope of adiabatic curve =g= CV Slope of isothermal curve

V

= 4 ´ 10 5 ´ 2 +

15. Isobaric expansion is represented by curve AB. 2

D 1m3

E 3m3

= Area under p-V diagram = Area of rectangle BCDE + area of DABC

W2 > W1 Hence, net work done W = - W1 + W2 = W2 - W1 = ( + ) ve

2

B

C

Now, work done by the system

Also, loop 2 is bigger.

Work done = Area under AB

A

2 ´ 10 5 ´ 2 2

W = 10 ´ 10 5 J

27. As dW = pdV (i) dW = p ´ 0 = 0 \ and (ii) dW = p(2V - V ) = pV

(Q change in volume = 0)

Thermodynamics 28. In figure, T is constant and p1 > p2. This situation is represented by curve (iii) in figure, in which p1 > p2 and straight line graph is parallel to pressure axis indicating constant temperature.

29. In adiabatic operation ( e. g . , bursting of tyre)

33. Here, V1 = 1L = 10 -3m3 , V2 = 3, L = 3 ´ 10 -3m3 p1 = 1 atm = 1.013 ´ 10 5 Nm-2, g = 1.40, W = ? As changes are adiabatic. p1V1g = p2V2g

\

g

p1 æ V2 ö = ç ÷ = (3)1.4 = 4.6555 p2 è V1 ø

p(21- g )T2g = p1(1- g )T1g æp ö T2 = T1ç 1 ÷ è p2 ø

or

(1- g )/ g

\ æ 1- 7 /5 ö ç ÷ 7 /5 ø

æ 4öè = 300 ç ÷ è 1ø

p1 1.013 ´ 10 5 = 4.6555 4.6555

= 0.217 ´ 10 5 Nm-2 Now, work done =

30. During adiabatic expansion.

=

TV g -1 = constant Þ T2V2g -1 = T1V1g -1 T1 æ V2 ö =ç ÷ T2 è V1 ø

p2 =

= 300( 4) -2/ 7

(Q atmospheric pressure = 1atm)

or

603

p1V1 - p2V2 g -1

1.013 ´ 10 5 ´ 10 -3 - 0.217 ´ 10 5 ´ 3 ´ 10 -3 1.4 - 1

= 90.5 J

g -1

34. Area under curve III is minimum. Therefore, work done is minimum.

For monoatomic gas, g = 5 / 3 T1 æ AL2 ö =ç ÷ T2 è AL1 ø

5 /3 -1

æV ö è V2 ø

æL ö = ç 2÷ è L1 ø

2/3

g

31. From, p2V2g = p1V1g Þ p2 = p1ç 1 ÷ æ100 ö p2 = ç ÷ è 124 ø

5 /3

p1

p2 = 0.6985p1 \% decrease in pressure p - p2 = 1 ´ 100% p1

35. Heat absorbed by the system at constant pressure, Q = nC pDT Change in internal energy DU = nCV Dt As, W = Q - DU (1st law of thermodynamics) W Q - DU DU \ = =1Q Q Q RC g DT C =1=1- V RC g DT Cp æ = ç1 è

=

p1 - 0.6985p1 ´ 100 p1

=

0.3015p1 ´ 100% p1

= 30.15% » 30%

32. The p -V graphs three given processes are shown in figure.

1ö ÷ gø

36. Change in internal energy does not depend upon path so DU = DQ - DW remains constant.

37. DQ = - 20 J ; DW = -10 J. Now, DQ = (Uf - Ui ) + DW Þ Þ

-20 = (Uf - 40) - 10 Uf = - 10 + 40 = 30 J

38. Given, work done (W ) = - 22 .3 J Work done is taken negative as work is done on the system. In an adiabatic change DQ = 0

2

p0

Isobaric

1

p

Isothermal 3 V1

Adiabatic V

V2

As work done by the gas = area under the p -V graph (between the curve and V axis) Þ \

(Area) 2 > ( Area)1 > ( Area)3 W2 > W1 > W3

Using first law of thermodynamics, DU = DQ - W = 0 - ( -22.3) = 22. 3 J For another process between state A and state B, Heat absorbed ( DQ) = + 9.35 cal = + (9.35 ´ 419 . )J = + 3918 . J Change in internal energy between two states via different paths are equal.

604 JEE Main Physics \ DU = 22.3 J \ From first law of thermodynamics DU = DQ - W W = DQ - DU

or

= 3918 . - 22.3 = 16.88 J » 16.9 J

39. DQ = DU + DW DQ = 0 - 150 J = -150 J So, heat has been given by the system.

40. Here, dQ = 400 cal, dW = -105 J = - 105 / 4.2 cal = - 25 cal; dU = ? Now,

dU = dQ - dW dU = 400 - ( -25) = 425 cal

48. Here dQ = 50 J, dW = -15 J \

49. According first law of thermodynamics, DQ = DU + DW Þ

DQ = 0 + W = W 7 50. Here, n = 5, g = , T1 = 0° C, T2 = 400° C 5 nRdT dU = \ 7 -1 5 5 ´ 8.31 ´ ( 400 - 0) = 41550 J dU = 7 -1 5 dU = 41.55 kJ

51. Internal energy ( DU) does not depend upon path. It depends

Note dW is negative because work is done on the system.

41. Here, dQ = 110 J, dU = 40 J, dW = ? From

52.

dQ = dU + dW dW = dQ - dU = 110 - 40 = 70 J

42. Process 1 is isobaric (p = constant) expansion.

only on initial and final states. dU CVdT CV (3 / 2)R 3 As, = = = = dQ C pdT C p (5 / 2)R 5 (Here, number of moles of gas is constant)

53. When the compression is isothermal for gas in A, p2V2 = p1V1 pV V p2 = 1 1 = p1 1 = 2p1 V2 V1/ 2

Hence, temperature of gas will increase. \ DU1 = negative Process 2 is an isothermal process

For gas in B, when compression is adiabatic,

\ DU2 = 0 Process 3 is an adiabatic expansion. Hence, temperature of gas will fall. \ DU3 = constant, \ DU1 > DU2 > DU3

p2¢ V2g ¢ = p1V1g g

DU = ( -20) - ( -8) = - 12 J DU = Uf - Ui = -12

\

Uf = - 12 + Ui = - 12 + 30 = 18 J

\

then as there is no heat loss to the surroundings therefore heat lost by M1 and M2 = heat gained by M3 M1s (T1 - T2) + M2s(T2 - T) = M3 s(T - T3) M1T1 + M2T2 + M2T3 = (M1 + M2 + M3) T M T + M2T2 + M3T3 Þ T= 11 M1 + M2 + M3

representing cyclic changes, therefore, U = constant and DU = 0 in all the cases.

45. As initial and final states in the two processes are same. As \

55. The internal energy of a system remains constant when the temperature does not change i. e. , when the system is isothermal.

DQ = DU + DW DQ1 > DQ 2

46. As, dU = dQ - dW = mL - p(dV )

56. Work done by the system = Area of shaded portion on the p-V diagram

1.013 ´ 10 5(1671 - 1)10 -6 = 1 ´ 540 4.2 = 540 - 40 = 500 cal

47. In adiabatic expansion, dQ = 0 , \

dW = - dU = - ( -50 J ) = 50 J

p2¢ 2 g p1 = = 2g - 1 p2 2p1

54. If the equilibrium temperature T > T1 and T2 but less than T3,

44. As indicator diagrams in all the three cases are closed curves,

Therefore, DU1 = DU2. As area under curve a > area under curve b, i.e., DW1 > DW2

g

æV ö æ V ö p2¢ = p1 ç 1 ÷ = p1 ç 1 ÷ = 2 g p1 è V2¢ ø è V1/ 2 ø

43. As, DU = DQ - DW \

dU = dQ - dW = 50 - ( -15) = 65 J

= (300 - 100)10 6 ´ (100 - 200)10 3

57.

= - 20 J T2 500 3 As, h = 1 =1= = 0.375 T1 800 8

Thermodynamics 58. As,

T2 40 3 =1- h =1= T1 100 5

66. As, h = 1 -

5 5 T2 = ´ 300 = 500 K 3 3 Increase in efficiency = 50% of 40% = 20%

\

\ New efficiency, h¢ = 40 + 20 = 60% T2 60 2 = 1 - h¢ = 1 = \ T1¢ 100 5

T2 100 T Þ1=1T1 500 900 T 1 = 900 5

\

T1 =

T = 180 K

or

67. Given, T2 = 0°C = 273 K, T1 = 17°C = 17 + 273 = 290 K COP =

5 ´ 300 = 750 K 2 Increase in temperature of source = T1¢ - T1 T1¢ =

Þ

Þ

T2 , T1 T2 40 3 =1- h =1= T1 100 5 3 3 T1 = ´ 500 = 300 K 5 5

\

T2 =

Again

T2 =1- h T1¢

or or

300 50 1 =1= T1¢ 100 2 T1¢ = 600 K

60. Given, T1 = 27° C = (27 + 273) K = 300 K , and \

61. As, h = 1 -

62.

63. 64.

T2 = - 123 + 273 = 150 K T 150 h =1- 2 =1= 0.5 T1 300 T2 T1 - T2 = T1 T1

In all the four cases, T1 - T2 = 20 K. Therefore, h is highest, when T1 is lowest. W We have, h = Q1 1 1000 Þ W = hQ1 = ´ 1000 cal = ´ 4.2 = 1400 J 3 3 300 1 T As, h = 1 - 2 = 1 = = 50% 600 2 T T2 273 - 23 250 COP = = = = 2.5 T1 - T2 (273 + 77) - (273 - 23) 100

Q As, COP = 2 W 1000 ´ 80 ´ 4.2 \ 2.5 = W 1000 ´ 80 ´ 4.2 or W= = 134400 J 2.5 Q 65. Here, Q 2 = 2000 cal. As, COP = 2 W \

4 = 2000 / W W = 500 cal = 500 ´ 4.2 = 2100 J

Q2 T2 = W T1 - T2

80 ´ 1000 ´ 4.2 273 273 = = W 290 - 273 17

= 750 - 500 = 250 K

59. From h = 1 -

605

W=

80 ´ 1000 ´ 4.2 ´ 17 J 273

W=

33.6 ´ 17 ´ 10 4 kWh 273 ´ 3.6 ´ 10 5

= 0.058 kWh

68.

T As, h = 1 - 2 T1 T2 1 5 =1-h =1- = T1 6 6

…(i)

T2 - 62 2 2 = 1 - h¢ = 1 - = T1 6 3

…(ii)

Þ In second case

h¢ = 2 ´

Þ

1 2 = 6 6

From Eqs. (i) and (ii) 2 2 6 Now, T2 - 62 = T1 = ´ T2 3 3 5 Þ

T2 = 310 K = 310 - 273 = 37°C 6 6 T1 = T2 = ´ 310 = 372 K 5 5 = 372 - 273 = 99°C

69. Given, T2 = 27 + 273 = 300 K and h = 37.5% h =1-

As,

T2 T1

\

37.5 300 =1100 T1

or

300 62.5 5 = = T1 100 8 T1 =

2400 = 480 K 5

= 480 - 273 = 207°C

70. Givne, T1 = 273 + 20 = 293 K, T2 = 273 + 10 = 283 K \Coefficient of performance T2 283 = = T1 - T2 293 - 283 =

283 = 28.3 10

606 JEE Main Physics 71. As, h = 1 -

T2 T1

80. As, h = 1 T2 1 5 =1- h =1- = T1 6 6

or

T2 =

5 5 T1 = ´ 600 = 500 K 6 6

72. Here, T1 = 411°C = ( 411 + 273) K = 684 K T2 = 69°C = (69 + 273) K = 342 K Q1 = 1000 J W T 342 1 h= =1- 2 =1= Q1 T1 684 2

and Q Þ

W = hQ1 =

\

\

Þ

T2 =

Q 2 T2 = Q1 T1 Q2 127 + 273 400 = = 4 227 + 273 500 6 ´ 10 Q2 =

4 ´ 6 ´ 10 4 = 4.8 ´ 10 4 cal 5

\ W = Q1 - Q 2 = 6 ´ 10 4 - 4.8 ´ 10 4 = 1.2 ´ 10 4 cal

75. The efficiency of two engines are

600 x 3 600 Since h1 = h 2 therefore = 1 8 x 600 3 =1x 8 600 ´ 8 5 or x = = 960 K 5 8 T Efficiency of an engine is h = 1 - 2 T1

82.

where, T1 is the source temperature (higher) and T2 is the sink temperature (lower) either

h =1 T1 = ¥

or

T2 = 0 K

For

76.

hB = 1 -

Clearly, h A < hB 320 1 T ( 47 + 273) As, h = 1 - 2 = 1 =1= = 20% 400 5 T1 (127 + 273)

77. As, h = 1 \

78.

T2 400 7 =1= T1 1100 11

500 T2 600 =1=1800 T1 x

3 600 =18 x 600 3 5 =1- = x 8 8

5x = 4800 4800 x= = 960 K 5 T T 10 90 As, h = 1 - 2 \ 2 = 1 - h = 1 = T1 T1 100 100 or

100T2 90 100 = ´ 270 = 300 K 90

T1 =

(i. e. , 100%)

83. As, C p - CV = R = 4150 J/kg-K Cp

and

CV

T 500 1 hA = 1 - 2 = 1 = T1 1000 2 and

500 3 = 800 8

and in second case (h 2) = 1 -

1000 = 500 J 2

150 Q2 ´ T1 = ´ 400 = 300 K 200 Q1

æ T ö ì (273 + 27) ü W = ç1 - 2 ÷Q = í1 ý ´Q è T1 ø î (273 + 627) þ æ 300 ö 6 6 6 W = ç1 ÷ ´ 3 ´ 10 = 2 ´ 10 ´ 4.2 J = 8.4 ´ 10 J è 900 ø

81. In first case(h1) = 1 -

Q 2 T2 = Q1 T1

73. As,

74. As,

Þ

T2 W = T1 Q

CV =

= g = 1.4

4150 R = = 10375 J / kg -K g - 1 (1.4 - 1)

2 f

84. As, g = 1 + (where f = degree of freedom) Þ

g -1=

2 f

Þ

f 1 = 2 g -1

Þ

f=

2 g -1

85. From, DQ = m C p( DT) ( m = number of moles) 70 = 2 ´ C p ´ (35 - 30), \

C p = 70 / 10 = 7 cal (mol°C) -1 CV = C p - R = 7 - 2 = 5 cal/mol°C

Now,

DQ ¢ = m CV ( DT) = 2 ´ 5 ´ 5 = 50 cal

86. As, CV = C p - R = 207 - 8.3 = 198.7 J 87. Work done in expansion = C p - CV = R joule

Thermodynamics 88. As, dU = CV dT = æç Rö÷ dT = ´ 8.32 ´ 100 = 1.25 ´ 103 J 3 è2 ø

3 2

Work done,

89. Internal energy U = number of moles ´ number of degrees of 1 RT 2 Out of four cases, product of number of moles (1000) degrees of freedom (3) and T( = 900 K) is maximum for argon gas. freedom ´

90. For a non-linear triatomic gas, CV = 3R

\ For the mixture,g =

Cp CV

\

5 For monoatomic gas, g = = 1.67 3 7 and for diatomic gas g = = 1.40 5

= 1.4, C p = 1.4 CV

1.4 CV - CV = 4150

97. As, h =

Cp CV

=1+

2 2 =1+ n f

98. Given, p µ T3 but we know that, for an adiabatic process the pressure, p µ T g / g -1 g =3 g -1 3 g= 2 Cp 3 = CV 2

So,

As actual g = 1.5. Therefore, gas must be a mixture of monoatomic and diatomic gases.

Þ

-1

93. Given, C p = 8 cal (mol°C) ,

Þ

CV = C p - R = 8 - 2 = 6 cal (mol °C ) -1 dU = mCV (T2 - T1) = 5 ´ 6(20 - 10) = 300 cal

99. We know that, E f = gp = 1.4 ´ (1 ´ 105) = 1.4 ´ 105 N/m 2

94. As V = KT 2/3

\

1 1 æ5 7ö 3 [ g mono + g di ] = ç + ÷ = 2 2 è3 5ø 2

CV = 4150 / 0.4 = 10375 J - kg K -1

must be 2.

\

T1

95. Below 150 K, hydrogen behaves as monoatomic gas

and

91. As, C p - CV = R = 2 cal (mol K) -1. Difference in the two values

\

dV T2 2 dT RT V òT1 3 T 2 2 W = R(T2 - T1) = R ´ 60 = 40R 3 3 T2

W = ò RT

96. Given, C p - CV = 4150

3 and for monoatomic gas, CV ¢ = R 2 Q CV 3R \ = =K = =2 3 Q ¢ CV ¢ R 2

92.

2 (after differentiating) dV = K T -1/3dT 3 2 -1/3 KT dT dV 3 2 dT = = 2/3 V 3 T KT

100. In cyclic process, DQ = Work done = Area inside the closed p ( p2 - p1) (V2 - V1) 4 p p Þ DQ = {(180 - 50) ´ 10 3 ´ ( 40 - 20) ´ 10 -6} = J 4 2 curve treat the circle as an ellipse of area =

Round II 1. According to first law of thermodynamics, dQ = dU + dW As

2. Here, T1 = 927° C = (927 + 273) K = 1200 K As \

T2 (127 + 273) 1 =1= T1 (227 + 273) 5 1 W = hQ1 = ´ 10 4 J = 2000 J 5

5. As, h = 1 -

dW = - dU

Þ dQ = dU - dU = 0 The change must be adiabatic. and

6. As, work done = area under the p-V diagram \

U µT DU U1 - U2 1200 - 300 = = ´ 100 = 300% U2 U2 300

5 4. We know that, ka = gp = æç ö÷ ´ 1.01 ´ 105 Nm-2 è3ø

= 1.69 ´ 10 5 Nm-2

W1 > W2

7. At constant temperature,

T2 = 27° C = (27 + 273) K = 300 K

3. As, DQ = DU + DW = mCV ( DT) + p( DV )

607

or

p1V1 = p2V2 p1 V2 = p2 V1

Fractional change in volume V1 - V2 4 1 = = V1 100 25 1 V 1- 2 = V1 25

608 JEE Main Physics V2 24 = V1 25 p1 V2 24 p 25 or 2 = \ = = p2 V1 25 p1 24 p2 - p1 25 1 = -1= p1 24 24 100 % increase in pressure = = 4.16 24

æV ö è V1 ø

13. As, W = 2.3026 n RT log10 ç 2 ÷ æ 10 ö = 2.3026 ´ 10 ´ 9.3 ´ 600 log10 ç ÷ è100 ø = - 11.4 ´ 10 4 J

14. From

g

Þ

8. As, dW = dQ - dU = C p(T2 - T1) - CV (T2 - T1) = R[T2 - T1]

(Q C p - CV = R)

= 8.31 ´ 100 = 8.31 ´ 10 2 J

9. For monoatomic gas, 3 3 CV = R = ´ 8.31 Jmol-1°C-1 2 2 Q = 500 J, n = 4q = ? Q 500 q= = = 10°C nCV 4 ´ 3 ´ 8.31 2

Given, \

or \

As \

C p - CV CV

1.4

= 66.28 atm

15. Here,

p = 4.5 ´ 10 5 Pa, dV = (2.0 - 0.5) m3 = 1.5 m3

and

dQ = 800 kJ = 8 ´ 10 5 J, dU = ? dW = pdV = 4.5 ´ 10 5 ´ 1.5 = 6.75 ´ 10 5 J dU = dQ - dW = 8 ´ 10 5 - 6.75 ´ 10 5 = 1.25 ´ 10 5 J

16. As, dU = dQ - dW = 8 ´ 105 - 6.5 ´ 105 = 1.5 ´ 105 J \

= g -1 C p - CV

R g -1 RdT npdV = DU = nCVdT = n ( g - 1) g -1 np(2V - V ) npV = = g -1 g -1 CV =

g -1

=

n = 1, DU =

dW = dQ - dU = 10 5 - 1.5 ´ 10 5 = - 0.5 ´ 10 5 J

pV (g - 1)

11. In an adiabatic change, p1- g T g = constant or

æ V1 ö æV ö p2 = p1ç 1 ÷ = 1 ç ÷ è1 / 20V1 ø è V2 ø

In the 2nd process, dU remains the same

10. As, C p / CV = g \

p2V2g = p1V1g

pT g /1- g = constant

p µ T -(1- g )/ g 1- g Thus, c=g 5 For a monoatomic gas, g = 3 1- 5 /3 2 \ -c = =5 /3 3 2 c= Þ 5 or

12. As slope of adiabatic AC is more than the slope of isothermal AB, and BC is isochoric (i. e. , at constant volume), therefore, Fig. (b) represents the curves correctly.

.

17. Given, T1 = 200° C = 200 + 273 = 473 K \

T2 = 0° C = 0 + 273 = 273 K 273 200 T h1 = 1 - 2 = 1 = 473 473 T1

Again, T1¢ = 0° C = (0 + 273) K = 273 K and T2¢ = - 200° C = ( -200 + 273) K = 73 K 73 200 T h2 = 1 - 2 = 1 = \ 273 273 T1¢ Now,

18. As, h =

h1 200 273 273 1 = ´ = = h 2 473 200 473 1.732 T1 - T2 (t1 + 273) - (t 2 + 273) t -t = = 1 2 T1 t1 + 273 t1 + 273

19. As is clear from figure, Slope of curve 2 > Slope of curve 1 ( gp) 2 > ( gp)1 As

g 2 > g1 g He > go 2

\ Adiabatic curve 2 corresponds to helium and adiabatic curve 1 corresponds to oxygen.

20. As, DWAB = pDV = 10(2 - 1) = 10 J DWBC = 0 , because V is constant From first law of thermodynamics,

and

DQ = DW + DU As ABCA is a cyclic process, therefore,

Thermodynamics

\ or

DU = 0 DQ = DWAB + DWBC + DWCA

p3

C

p

= DWAB + DWCA DWCA = DQ - DWAB = 5 - 10 = - 5 J

p1

A

21. Here Q1 = 200 cal, Q 2 = 150 cal, T1 = 400 K As,

B

Q1 T1 = Q 2 T2 T2 =

V

Q2 150 ´ T1 = ´ 400 = 300 K Q1 200

dU = dQ - dW

|WBc| > |WAB| \ W = WAB + WBC = Negative i. e. ,W < 0. From the graph, it is clear that p3 > p1.

= 8400 - 500 = 7900 J

23. Here, dU = -100 J, in adiabatic expansion \

dW = - dU = 100 J

27. The given relation is p = aV

n

24. Given that, Vp = constant \

p µV When V changes from V to 2V , pressure p is also doubled. pV For an ideal gas, = constant T Therefore,

Vp n = (V + DV ) ( p + Dp) n DV ö æ Dp ö æ = Vp n ç1 + ÷ ç1 + n ÷ è V øè p ø

DV Dp DV Dp +n +n V p V p DV Dp (neglecting the product) or = -n , V p p - Dp Þ Bulk modulus, k = = DV / V n 1=1+

25. For the process at constant pressure \

dQ = C pdT + dW dQ - dW dT = Cp (\ dW = 0)

æ dQ - dW ö ÷ = CV ç Cp è ø

or

\

=

dQ - dW C p / CV

=

dQ - dW g

\ T µ pV Hence, T becomes 2 ´ 2 = 4 times i. e. ,

4 ´ 300 K = 1200 K

28. If M is molecular mass of the gas, then from M(C p - CV ) = R 8.31 M= = 0.0392 Þ 210 If r is density of the gas at NTP, then mass of 1 m 2 of gas at NTP = r kg \ Mass of 22.4 L ( = 22.4 ´ 10 -3 m3) of gas at NTP = r ´ 22.4 ´ 10 -3 kg,

For the process at constant volume, dQ = CVdT

In going from A to B, volume is increasing \ WAB = positive In going from B to C volume is decreasing \ WBC = negative As work done is area under p-V graph, therefore,

dW = 500 J

\

V2

V1

22. Given, dQ = 2 kcal = 200 cal = 2000 ´ 4.2 J = 8400 J and

609

( g - 1dQ ) = dW æ5 ö ç - 1÷dQ = W è3 ø 3 dQ = W 2

26. As, slope of adiabatic process at a given state is more than the slope of isothermal process, therefore in figure, AB is isothermal and BC is an adiabatic.

Which is the molecular mass of the gas \ r ´ 22.4 ´ 10 -3 = 0.0392 0.0392 r= = 1.75 kgm-3 22.4 ´ 10 -3

29. The given relation is p = \

V=

aT 2 V aT 2 p

As pressure is kept constant, æ 2aT ö \ dV = ç ÷dT è p ø Now,

W = ò p dV = ò

(after differentiating) 2T 0

T0 2T 0

æ 2at ö pç ÷ dT è p ø

éT2ù = - 2a ê ú = 3aT02 2 ë ûT0

610 JEE Main Physics 1 2 when the vessel is suddenly stopped, the ordered motion of the gas molecules is converted into disordered motion of the molecules increasing thereby the internal energy of the gas. Thus, 1 1 DU = nCV DT = mv 2 = (nM)v 2 2 2 where n is number of moles of the gas in the vessel and M is molecular weight of the gas.

We know that,

30. KE of the vessel = Mv 2

\

Mv 2 DT = 2CV

As

R Mv 2( g - 1) \ DT = CV = 2R g -1

DQ = mC p DT DU = mCV DT DU CV 3 = = DQ C p 5

and Þ

i. e. , Fraction of heat energy to increases the internal energy 3 be × 5

36. Given, work done (W ) = 5.4 ´ 10 8 J/min Total heat energy taken from the boiler, Q = 3.6 ´ 10 9 J/min Efficiency of heat engine (h) =

31. From first law of thermodynamics, DQ = DU + W Work done at constant pressure (DW ) p = ( DQ) p - DU = ( DQ) p - ( DQ)V (As, we know DQV = DU) Also ( DQ) p = MC p DT and

( DQ)V = MCV DT

Þ

( DWp) = M(C p - CV ) DT

=

5.4 ´ 10 8 ´ 100 3.6 ´ 10 9

=

3 ´ 100% 20

= 15% Heat wasted per minute = Q - W = 3.6 ´ 10 9 - 5.4 ´ 10 8

= 1 ´ (3.4 ´ 10 3 - 2.4 ´ 10 3) ´ 10 = 10 4 cal

= (36 - 5.4) ´ 10 8 J/min

32. From the first law of thermodynamics, DQ = DU + DW

= 30.6 ´ 10 8 J/min

..(i)

For a cyclic process, DU = 0 \ DQ = DW Now, DQ = Q1 + Q 2 + Q3 + Q 4 = 600 J - 400 J - 300 J + 200 J = 100 J and DW = W1 + W2 + W3 + W4 Þ

» 31 . ´ 10 9 J/min

Note DW is negative because work is done on the system.

37. As work done in an isothermal process is W = mRT log e

DW = 300 J - 200 J - 150 J + W4

100 J = - 50 J + W4

DU = ( DQ)V = mCV DT (m = number of moles) 3 For monoatomic gas, CV = R 2 3 æ3 ö DU = r ç R ÷ DT = 1 ´ ´ 8.31 ´ (100 - 0) è2 ø 2

96 140 = 2.3 ´ 900R log10 2 R(273 + 27) log10 32 70 V æ 22.4 ö As, W = - mRT log e 2 = - 1 ´ 8.31 ´ (273 + 0) log e ç ÷ è 11.2 ø V1 = 2.3 ´

W4 = 150 J

33. Change in internal energy is always equal to the heat supplied

V2 V1

V m V æmö = ç ÷RT log e 2 = 2.3 ´ RT log10 2 èMø V1 M V1

= - 50 J + W4 Substitute the value of DQ and DW in Eq. (i), we get

at constant volume.

38.

(–ve sign shows compression)

i. e. ,

= 12.48 ´ 10 2 J æ R ö ÷ DT è g - 1ø

34. As, DU = mCV DT = m ç \

DU =

pD V p (2V - v) pV = = ( g - 1) (g - 1) ( g - 1)

35. For monoatomic gas, g =

Cp CV

=

5 3

W ´ 100 Q

= - 8.31 ´ 273 ´ log e 2 = -1572.5 J

[Q log e 2 = 0.693]

39. Given, temperature of source (T1) = (36 + 273) K = 309 K Temperature of sink (T1) = (9 + 273) K = 282 K Coefficient of performance of a refrigerator T2 282 b= = T1 - T2 309 - 282 =

282 27

= 10.4

Thermodynamics 40. Amount of heat given = 540 cal

48. As, v rms =

Change in volume DV = 1670 cc atmospheric pressure p = 1.01 ´ 10 6 dyne/cm 2

Now,

\Work done against atmospheric pressure, W = pD V =

Thus,

41. For isothermal process, p1V1 = p2V2 p1V1 72 ´ 1000 = = 80 cm of mercury V2 900

Þ

p2 =

\Stress,

Dp = p2 - p12 = 80 - 72 = 8 cm of mercury

T 3 RT (v rms)1 = = 1 M (v rms) 2 T2

T1Vgg -1 = T2V2g -1 T1 æ V2 ö =ç ÷ T2 è V1 ø

1.01 ´ 10 6 ´ 1670 = 40 cal 4.2 ´ 10 7

1/ 2

æT ö = ç 2÷ è T1 ø

1/ 2

æ 900 ö =ç ÷ è 300 ø

æ p2 ö ç ÷ è p1 ø æ p2 ö ÷ ç è105 ø

Þ

3/ 2

=

43. As, volume of the gas, V =

æV ö è V2 ø

Þ

45.

M d

50.

|W| = 1815 J

51. Number of moles of He = -1

2

2

æ3ö æ 27 ö 3 = 300 ç ÷ = 300 ç ÷ = 675 K è2ø è8ø

Now,

æ 1ö T1 = T2ç ÷ è8ø

-nR[T2 - T1] = \Work done W = g -1

4

æT ö 1 ç 1÷ = çT ÷ 1 è 2ø

\ \

g= g -1= W=

1 4

Cp

CV

Þ

|W| =

-

1 R[3T1] -9 4 RT1 = 2 8 3

9 RT1 8

52. According to 1st law of thermodynamics, DQ = DU + DW , in adiabatic process DQ = 0 0 = DU - DW (work done on the system negative

CV C p - CV

2/3

4T1 = T2

mR(T1 - T2) Work done during an adiabatic change is, W = ( g - 1) Now,

5 é -1ù ê1 - æç 1 ö÷ 3 ú ê è2ø ú û ë

T1(5.6) g -1 = T2(0.7) g -1

Þ DT = 576 - 300 = 375 K dQ = er sAT 4 = E dt 2

é T2 ù ê1 - T ú ë 1û

g -1 mRT1 é æ V1 ö ù 2 ´ 8.31 ´ 300 = ê1 - ç ÷ ú = ( g - 1) ê è V2 ø ú æ5 ö û ë ç - 1÷ è3 ø

E µ r 2T 4

46.

= 273(2) 0.41 = 273 ´ 1.328 = 363 K

= + 2767.23 J

æ 27 ö 3 T2 = 300 ç ÷ è8ø

ær ö E \ 1 = ç 1÷ E 2 è r2 ø

1

æV ö5 æV ö = ç 1 ÷ = ç 2 ÷ = 25 = 32 è V1 ø è V1 ø

mR(T1 - T2) mRT1 As, W = = ( g - 1) ( g - 1)

g -1

5

Þ

=2

Now, for one mole of the gas, the work done R(T1 - T2) 8.31(273 - 363) |W | = = = - 1815 g -1 1.41 - 1

3/ 2

g

T2 æ V1 ö =ç ÷ T1 è V2 ø

æV V = ç 2÷ V / 2 è V1 ø

g -1

49. As, T2 = T1ç 1 ÷

p¢ æ V ö = ç ÷ = (32) 7 /5 = 128 p è V ¢ø

44. As,

7 -1 5 ö 2

2 1 ´ 2

using pV g = constant, we get

Now,

g -1 2

æV ö5 = ç 2÷ è V1 ø

p2 = 27 ´ 10 5 N / m2

Þ

g -1

(Vrms)1 æ V2 ö =ç ÷ (Vrms) 2 è V2 ø

42. Here, P = 105 N/m 2, T1 = 27 + 273 = 300 K T2 = 627 + 273 = 900 K and g = 1.5 Tg For adiabatic change, g -1 = constant p

611

=

R CV

mR(T1 - T2)CV = m(T1 - T2)CV R

47. Pressure is reduced so the temperature falls.

DU = + DW = + 22.3 [Q work done on the system\internal energy increases] in 2nd process DQ = DU + DW 9.35 ´ 4.18 = 22.3 + DW Work done by system, DW = 16.95 J

612 JEE Main Physics If Q 2 is negative, Q1 is also negative (but less negative as W > 0). \ Q 2 < Q1 < 0 Choices (a) and (c) are correct.

53. From 1st law of thermodynamics, Þ Þ

DQ = DU + DW DW = DQ - DU DW = nC p DT - nCV DT = 10 (C p - CV ) DT = 10 RDT

60. Here, T1 = 27° C = (27 + 273) K = 300 K

= 10 ´ 8.300 J = 8.3 ´ 10 4 J 5

T2 = - 73° C = ( -73 + 273) K = 200 K 200 1 T h =1- 2 =1= = 33% 300 3 T1

\ -3

54. As, W = pDV = 1.01 ´ 10 (3.34 - 2 ´ 10 ) = 337 ´ 10 3 J = 340 kJ

61. From h = 1 -

40 (27 + 273) =1100 T1

55. As shown in figure, during the process A to B, p and V both decrease. As T µ pV , therefore, T must also be decreasing. So internal energy must be decreasing.

300 40 3 =1= T1 100 5

\ DUA ® B is negative. As volume is decreasing, therefore, DWA ® B is also negative. Thus, DQ A ® B = negative. During the process B to C, volume is increasing at constant pressure. Therefore, T( µ V ) must increase and so does the internal energy, DUC ® A = positive. During the process CAB, volume is decreasing. Therefore, DWCAB = negative.

300 ´ 5 = 500 K 3 T 60 T2 From h = 1 - 2 , =1(127 + 273) T1 100 T1 =

62.

56. From the given initial state A to final state B, change in internal energy is same in all the four cases, as it is independent of the path from A to B. As work done = area under p - V curve, therefore, work done is maximum in case I. Choices (b) and (c) are correct.

\

returns to its initial state.

Therefore, work done on the gas increases the temperature.

64. The bursting of tyre is sudden. Therefore, the changes are adiabatic.

65. Adiabatic curve is steeper than isothermal curve. Therefore, area under adiabatic curve is smaller than the area under isothermal curve i. e. , work done by the gas in adiabatic expansion is smaller than the work done by the gas in isothermal expansion.

58. Isothermal curve from A to B will be parabolic with lesser area under the curve than the area under straight line AB. Therefore, work done by the gas in going straight from A to Bis more. Therefore is correct.

66. Here, T1 = 227 + 273 = 500 K T2 = 127 + 273 = 400 K 400 1 T h =1- 2 =1= = 20% 500 5 T1

If p0 , V0 be the intercepts of curve on p and V axes, then its equation is obtained from y = mx + c p p = 0 V + p0 i. e. , V0 or or

RT p0V = + p0 V V0 T=

p0 2 p0V V + , V0R R

T2 60 2 =1= 400 100 5 800 T2 = = 160 K = (160 - 273)°C = - 113°C 5

63. As compression is sudden, changes are adiabatic, dQ = 0.

57. In the given one complete cycle, 1 ® 2 ® 3 ® 1, the system \ dU = 0 and dQ = dW , i. e. , heat is completely converted into mechanical energy, which is not possible in such a process. Further, the two adiabatic curves (2 3) and (3 1) cannot intersect each other. Choices (a) and (c) are correct.

T2 T1

68.

This is the maximum value of efficiency. 100 3 T As, h = 1 - 2 = 1 = = 75% 400 4 T1 DQ × In an adiabatic change, DQ = 0 T DS = 0 S = constant

69. Change in entropy, DS =

Which is the equation of a parabola. Hence T-V curve is parabolic. Therefore (b) is correct.

\ \

Also ( p / 2) ´ (2V ) = pV = constant i. e. , process is isothermal.

i. e. , entropy remains constant, or it is an isotropic process.

59. Figure represents the working of a refrigerator, wherein Q1 = Q 2 + W If W > 0 , Q1 > Q 2 > 0 . Both Q1 and Q 2 are positive.

71. In an ideal gas, we assume that intermolecular force are zero. No work is done in changing the distance between the molecules. Therefore, internal energy is only kinetic and not potential. Therefore, internal energy of an ideal gas depends only on temperature and not on volume.

Thermodynamics 73. As h = 1 -

T2 T1 - T2 = , therefore, h will decrease if (T1 - T2) T1 T1

decreases.

74. First law of thermodynamics is a restatement of the principle of conservation of energy as applied to heat energy.

75. From A to B in figure the process is isobaric. \

W = nRDT = 2 ´ R ´ (500 - 300)

81. In going from B to C and in going from D to A, V = constant. Therefore, dV = 0 , dW = pdV = 0 .

82. For adiabatic process, dQ = 0 dU = - DW

So, Þ

nCVdT = + 146 ´ 10 3 J

Þ

nfR ´ 7 = 146 ´ 10 3 2

Þ

10 3 ´ f ´ 8.3 ´ 7 = 146 ´ 10 3 2

= 400R

76. From D to A, in figure, the process is isothermal \

æp ö W = nRT log e ç 1 ÷ è p2 ø æ 1 ´ 10 5 ö = 2 ´ R ´ 300 log e ç ÷ è 2 ´ 10 5 ø

= 600R ´ 2.303(0 - 0.3010) = - 415.9 R \ Work done on the gas = 415.9 R

77. In this process, p and V changes but T = constant i. e. , change in temperature DT = 0 Boyle’s law is obeyed i. e. , pV = constant Þ

78.

p1V1 = p2V2

According to equation, pV = constant, graph between p and V is a part of rectangular hyperbola. 1 Here, T2 = 27° C = (27 + 273) K = 300 K , h = 25% = 4

(f ® Degree of freedom)

f = 5.02 » 5 Therefore gas is diatomic.

83. The specific heat of a gas depends upon the process and can have any value as specific heat at constant volume is CV and at constant pressure is C p .

84. Consider n moles of a gas which undergo isochoric process, i. e. ,V = constant. From first law of thermodynamics, DQ = DW + DU Here, DW = 0 as V = constant Substituting in Eq. (i), we get

…(i)

DU = nCV DT Mayor’s relation can be written as

…(ii)

C p - CV = R Þ CV = C p - R From Eqs. (ii) and (iii), we have

We know that,

Þ

Given, n = 6, C p = 8 cal (mol-K) -1

1 300 =14 T1

Hence, DU = 6(8 - 2) (35 - 20) = 6 ´ 6 ´ 15 = 540 cal

300 1 =1T1 4

Þ

300 3 = T1 4

or

T1 =

300 ´ 4 3

T1 = 400 K

79. h = 1 -

R = 8.31 Jmol-1 K -1

85. The coefficiency of a heat engine is defined as the ratio of work done to the heat supplied, i. e. , work done W h= = heat input Q T or h =1- 2 T1 where,

T1 = ( 4500 - 273)° C = 127°C T2 T1 - T2 = T1 T1

When T1 and T2 are decreased by 100 K each, (T1 - T2) stays constant and T1 decreases, hence h increases.

80. Work done = area of DABC =

AB ´ BC ( 4V - V ) (3p - p) = = 3pV 2 2

…(iii)

DU = n(C p - R) DT

T h =1- 2 T1

or

or

613

\

T2 = temperature of sink, T1 = temperature of hot reservoir. 40 T =1- 2 100 T1

Þ

T2 = 0.6 T1

Þ

T2 = 0.6T1 50 T =1- 2 100 T1¢

Again, Þ

T2 = 0.5 T1¢

614 JEE Main Physics 0.6T1 = 0.5 T1¢

Þ

T1¢ =

86. From the relation,

( as T1 = T)

Þ h=

Tg = constant p g -1 p µT

But

6 0.6 T1 = T 5 0.5

91. As, h =

g / g -1 3

p µ T (given)

DW Area of DABC = QBC Area under curve BC S 0T0 / 2 1 = 3S 0T0 / 2 3 T 2T0

…(i) …(ii) T0

From Eqs. (i) and (ii), we get 3

T =T

C

A

g /( g -1)

S S0

2S0

or

g =3 g -1

92. As change in internal energy does not depend upon the path

or

3 g= 2

followed between the two given points on p-V diagram, therefore, DU1 = DU2.

87. In an adiabatic expansion of gas energy is consumed in the

88.

B

gas. On account of the consumption of energy the temperature of system falls. p1V1 p2V2 = T1 T2

93. 2p0 p0

(Here, p1 = p, V1 = V , p2 = 2p, V2 = 3V ,T1 = 27°C = 300 K) pV 2p ´ 3V = T T2 Hence,

T2 = 300 ´ 2 ´ 3 = 1800 K = 1800 - 273 = 1527°C

89. For one mole of an ideal gas, pV = R = gas constant for one mole T = 2 cal mol -1 K -1

90. Change in internal energy of gas is equal to the heat produced due to current flowing i. e. , dU = I 2Rt = 12 ´ 100 ´ (5 ´ 60) = 30 , 000 J = 30 kJ

4T0

2T0

2T0 V0

2V0

Heat supplied H = nCV + Dt + nC p DT = nCV (2T0 - T0) + nC p + ( 4T0 - T0) 3R 5R For monoatomic gas CV = and C p = 2 2 3 RT0 æ5Rö H =n +nç \ ÷ 2T è 2 ø 0 2 3R nT0 + 5nRT0 2 13 13 = nRT0 = p0V0 2 2 =

15 Oscillations JEE Main MILESTONE < < <
a times taken by pendulum from B to C

Using,

and C to B is T 1 L L = ´ 2p =p 2 2 g g

t1 =

t2 = 2 t =

and

or

2 æ aö sin -1 ç ÷ èbø w

(a)

I1w12 + I2w22 I1 + I2

(b)

I1 + I2 I12w1 + I22w2

(c)

I1w12 - I2w22 I1 - I2

(d)

I1 - I2 I12w1 - I22w2

I1a = - w12I1q + G

…(i)

- w 22 I2q - G

…(ii)

Interpret (a) The time period of a simple pendulum is given by T = 2p

(I1w12 + I2w22) q I1 + I2

Comparing with a = - w2q, we get Frequency,

w=

I1w12 + I2w22 I1 + I2

Sample Problem 18 Length of a simple pendulum which ticks seconds is (a) 1 m (c) 3 m

(b) 2 m (d) 4 m

L g

Þ L=

gT 2 4 p2

The time period of a simple pendulum which ticks seconds is 2 s. Therefore, for g = 9.8 ms–2 and T = 2 s, L is 9.8 ´ 4 L= » 1m 4 p2

the surface of the moon is 1.7 ms–2. What is the time period of a simple pendulum on the moon, if its time period on the earth is 3.5 s? (g on the earth 9.8 ms–2). (a) 8.4 s (c) 7.4 s

(b) 8.2 s (d) 6.4 s

Interpret (a) Given, g m = 1.7 ms–2, g e = 9.8 ms–2,Tm = ? ; T = 3.5 s

Adding Eqs. (i) and (ii), we get a=-

L é aù p + 2 sin -1 ú g êë bû

Sample Problem 19 The acceleration due to gravity on

Interpret (a) When the pendulums are rigidly joined and set to oscillate, each exerts a torque on the other. These torques are equal and opposite, thus I2a =

q = q0 sin wt a = b sin wt 1 æ aö t = sin -1 ç ÷ èbø w

Time period of motion, T = t1 + t 2 =

Sample Problem 17 Two physical pendulums perform small oscillations about the same horizontal axis with frequencies w1 and w2. Their moments of inertia relative to the given axes are I1 and I2 respectively. In the equilibrium positions, they are joined rigidly. The frequency of small oscillations of the combined pendulum is

625

As,

Te = 2p

l ge

and

Tm = 2p

l gm

\

Tm = Te

or

Tm = Te

ge gm

= 3.5

ge gm 9.8 = 8.4 s 1.7

626 JEE Main Physics Some Important Points (vi) Finally, the graph between T and l/g is also a straight line.

1. The time period of a simple pendulum is T = 2p l/g

Y

Þ

T µ l or T µ

Þ

T µ

1 g

T2

l g

X

O

l/g

Using these relations. We may conclude (i) The graph between T 2 and l is a straight line.

2. In the case of water oscillating in a U-tube

Y T2

h O

l

X

(ii) The graph between T and l is a parabola.

æ hö T = 2p ç ÷ è gø

Y

where, h is the height of liquid column in each limb.

T

3. When a ball of mass m is made to oscillate in the neck of an air chamber having volume V and neck area A, then O

l

mV T = 2 p æç 2 ö÷ è pA ø

X

(iii) The graphs l-T and l-T 2 intersect at T = 1s.

4. When a pendulum is kept in a car which is sliding down, then æ l ö T = 2p ç ÷ è g cos q ø

T = 1s

Y l-T

where, q is the angle of inclination.

5. If a simple pendulum oscillates in a non-viscous liquid of density r, then its time period is l é T = 2p ê æ sö ê ç1 - ÷ g rø êë è

l-T 2 X

O

(iv) The graph between T 2 and 1/g is a straight line. Y

ù ú ú úû

r = density of suspended mass.

6. If the mass m attached to a spring oscillates in a non-viscous liquid of T2

O

1/g

X

density s, then its time period is 1/2 ém æ s ö ù T = 2 p ê ç1 - ÷ ú ë k è r øû where, k = force constant.

7. If a small ball is rolling down in hemispherical bowl. Time period,

2

(v) The graph between T and g is a rectangular hyperbola. Y

T = 2p

R -r g

where, R = radius of the bowl and

T2

r = radius of the ball

8. For a body executing SHM in a tunnel dug along any chord of earth. Time period, T = 2p O

g

X

Re = 84.6 min g

where Re is the radius of earth.

Oscillations

In a freely falling lift, geff = 0 and T = ¥, i .e ., the pendulum will not oscillate.

9. If the time period of simple pendulum is 2s, then it is called as second’s pendulum.

14. If in addition to gravity one additional force F (e .g ., electrostatic force

10. If the simple pendulum is placed in some non-inertial frame of

Fe ) is also acting on the bob, then in that case F geff = g + m Here, m is the mass of the bob.

reference like an accelerated lift, g is replaced by geff whose value can be computed by considering the inertial force. In these cases, the equilibrium position may also change.

11. If the length of simple pendulum is very large, then g can’t be

Torsional pendulum In a torsional pendulum, an object is

taken along vertical direction. In this case, T = 2 p

627

suspended from a wire. If such a wire is twisted due to elasticity, it exerts a restoring torque t = C q

1 1 1 g æç + ö÷ èl Rø

l

where, R = Radius of length of the pendulum.

12. If temperature of system changes, then time period of simple pendulum changes due to change in length of the simple pendulum. θ

13. If a simple pendulum is in a carriage which is accelerating with an acceleration a, then

In this case, time period is given by

g eff = g - a | geff | = g + a and T = 2p

l g+a

where,

phr n 2l

h = modulus of elasticity of wire r = radius of wire l = length of wire

| geff | = g - a T = 2p

I = moment of inertia of the object C = torsional constant of wire =

If the acceleration a is downwards, then (g > a )

and

l C

T = 2p

e .g ., if the acceleration a is upwards, then

l g -a

If the acceleration a is in the horizontal direction, then

|g eff | = a 2 + g 2

Physical Pendulum When a rigid body of any shape is capable of oscillating about an axis (may or may not be passing through it), it constitutes a physical pendulum.

P

θ

[As q and

d

d 2q are oppositely directed] dt 2

C

C Consider an arbitrary shaped body mg whose centre of gravity is at C, being pivoted about a point P at distance d from C. When P and C are in the same vertical line, the body is in its equilibrium position. When the body is displaced (rotated) slightly by an angle q about point P, then gravity force will provide the necessary restoring torque to execute oscillations.

Comparing with the equation

l

t = mgd ´ q

= 2p l

l mgd

The simple pendulum whose time period is same as that of a physical pendulum is termed as an equivalent simple pendulum. T = 2p

Ia = mgdq

where I is moment of inertia of the body about a horizontal axis passing through P.

mgd I

T = 2p

i. e. , body will perform SHM. Þ

d 2q = - w2q, we get dt 2 w=

t = mg ´ d sin q For small, q,

d 2q mgdq = I dt 2

Þ

I mgd l g

The length of an equivalent simple pendulum is given by

l=

I md

628 JEE Main Physics

15.7 Free, Forced, Damped and Resonant Vibrations

In these oscillations, the amplitude of oscillation decreases exponentially and hence, energy also decreases exponentially. If the velocity of an oscillator is v, the damping force

Free Vibrations If a given body is once set into vibrations and then let free to vibrate with its own natural frequency, the vibrations are said to be free vibrations. The natural frequency of free vibrations depends on the nature and structure of the body and in ideal situation, the amplitude, frequency and the energy of the vibrating body remain constant.

Fd = - bv where, b = damping constant. Resultant force on a damped oscillator is given by F = FR + FD = - kx - bv or

md 2x bdx + + kx = 0 dt dt 2

Displacement of a damped oscillator is given by

Forced Vibrations

x = xme- bt / 2m sin (w¢t + f )

The vibrations in which a body oscillates under the effect of an external periodic force, whose frequency is different from the natural frequency of oscillating body are called forced vibrations. In forced vibrations, the oscillating body vibrates with the frequency of external force and amplitude of oscillations is generally small.

where, w¢ = angular frequency of the damped oscillator

If an external driving force is represented by

as

w¢ = w20 - (b /2 m ) 2

For a damped oscillator, if the damping is small then the mechanical energy decreases exponentially with time E=

F ( t ) = F0 cos wd t The motion of particle is under combined action of (i) restoring force (-kx) (ii) damping force (-bv), and (iii) driving force F (t )

Now,

ma = - kx - bv + F0 cos wd t

or

d 2x kx b dx F0 cos wd t =+ 2 m m dt m dt

The solution of this equation gives x = x0 sin (wd t + f ) with F0/ m amplitude x0 = 2 æ bw ö (w20 - wd2 ) + ç ÷ èmø w2 - wd2 tan q = 0 bwd / m and

w0 =

k = natural frequency m

Damped Vibrations When a body is set in free vibrations, generally there is a dissipation of energy due to dissipative causes like viscous drag of a fluid, frictional force, hysteresis, electromagnetic damping force, etc., and as a result amplitude of vibration regularly decreases with time. Such vibrations of continuously falling amplitudes are called damped vibrations.

1 2 - bt / m kxme 2

Resonant Vibrations It is a special case of forced vibrations in which frequency of external force is exactly same as the natural frequency of oscillator. As a result the oscillating body begins to vibrate with a large amplitude leading to the resonance phenomenon to occur. Resonant vibrations play a very important role in music and tuning of station/channel in a radio/TV.

Check Point 2 1. How would the period of spring mass system change, when it is made to oscillate horizontally and then vertically?

2. Glass windows may break due to an explosion far away Explain, why?

3. There are two springs, one delicate and another stout one. For which spring the frequency of oscillation is more?

4. Water in a U-tube executes SHM. Will the time period for mercury filled upto the same height in U-tube be lesser or greater than that in case of water?

5. Why are soldiers asked to break their steps while marching over a bridge?

WORKED OUT Examples Two SHMs are represented by y1 = A sin( wt - f) and y2 = B cos( wt - f). The phase difference between the two, is

Example 1 (a)

p 2

p 4 p (d) 3

(b)

(c) p

Solution

Þ

wµ a 2p 1 T= , hence, T µ w a

As,

A simple comparison of arguments of sine terms shows that phase p p ö æ difference is ç wt + - f ÷ - ( wt - f) = . ø è 2 2

Example 2

A mass of Hg is executing SHM which is given pö æ by x = 6.0 cos ç100t + ÷ cm. What is the maximum kinetic è 4ø

energy? (b) 6 J (d) 18 J

Example 4

Two springs of force constants k1 and k2, have equal highest velocities when executing SHM. Then the ratio of their amplitudes (given their masses are equal) will be (a)

k1 k2

(b)

k1 k2

(c)

k2 k1

(d)

k2 k1

Solution

At highest velocities A1w1 = A2w2 k /m A1 w2 k = = 2 2= 2 A2 w1 k1 k1 /m1

\

Here, m =1kg

The given equation of SHM is pö æ x = 6.0 cos ç100t + ÷ è 4ø Comparing it with equation of SHM x = A cos ( wt + f), we have 6 A = 6.0 cm = m 100 and

w = 100 rad s-1

1 Maximum kinetic energy = m (v max ) 2 2 2

1 1 é 6 ù = m ( Aw) 2 = ´ 1 ´ ê ´ 100 ú =18 J 2 2 100 ë û

Example 3

(d) proportional to a3 / 2

As potential energy U( x ) = k| x|3 , hence maximum value of potential energy 1 Umax = U( a) = mw2a2 = ka2 2

y 2 = B cos ( wt - f) p ö æ = B sin ç wt + - f ÷ ø è 2

Solution

(b) independent of a

Solution

Here, y1 = A sin ( wt - f) and

(a) 3 J (c) 9 J

1 a (c) proportional to a (a) proportional to

A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is U( x) = k| x|3 , where k is a positive constant. If the amplitude of oscillation is a, then its time period T is

(since, m1 = m2)

Example 5 Two point masses of 3.0 kg and 1.0 kg are attached to opposite ends of a horizontal spring whose spring constant is 300 Nm -1 as shown in adjacent figure. The natural frequency of vibration of the system is k = 300 Nm–1 3 kg

1 kg

(a) 4 Hz 1 (c) Hz 3

Solution

(b) 3 Hz 1 (d) Hz 4 Here reduced mass of the system, m =

m1m2 3 ´1 = m1 + m2 3 + 1

= 0.75kg \Vibrational frequency, n=

1 2p

k 1 = m 2p

300 20 10 = = @ 3 Hz 0.75 2p p

630 JEE Main Physics Example 6 A particle is executing SHM of amplitude 25 cm and time period 3 s. What is the minimum time required for the particle to move between two points12.5 cm on either side of the mean position? (a) 0.5 s (c) 1.5 s

Solution

(b) 1s (d) 2 s Here, A = 25 cm; T = 3 s;

Let the particle be at the locating -12.5 cm at time t1 and + 12.5 cm at time t 2. æ 2pt ö Using the relation, x = A cos ç + f÷ è T ø First condition,

æ 2 p t1 ö -12.5 = 25 cos ç + f÷ è 3 ø

æ 2 pt 2 ö Second condition, 12.5 = 25 cos ç + f÷ è 3 ø 1 2p ö -12.5 æ 2 p t1 From Eq. (i), cos ç = - = cos + f÷ = ø è 3 25 2 3 2 p t1 2p +f= 3 3

\

2 p t1 + 3 f = 2 p æ 2 pt 2 ö From Eq. (ii), cos ç + f÷ è 2 ø or

= \

Example 8

The displacement of a particle executing periodic motion is given by æt ö y = 4 cos2 ç ÷ sin(1000t) è2 ø Find independent constituents of SHMs (a) y1 = 2 sin 1000 t , y 2 = sin 1001t , y3 = sin 999 t (b) y1 = 3 sin 1000 t , y 2 = sin 1000 t , y3 = sin 899 t (c) y1 = sin 1001t , y 2 = sin 999 t , y3 = 2 sin 1000 t (d) None of the above Solution y = 4 cos2 æç t ö÷ sin(1000t) è2ø = 2 (1 + cos t ) sin(1000t ) = 2 sin(1000t ) + 2 cos t sin(1000t ) = 2 sin(1000t ) + sin (1001t ) + sin(999t ) Thus, the given periodic motion is a combination of three independent SHMs, which are given by y1 = 2 sin 1000t , y 2 = sin 10001t and y3 = 999t

Example 9

As shown in figure, system consisting of massless pulley, a light spring of force constant k and a block of mass m. If the block is slightly displaced vertically downwards from its equilibrium position and released, find the expressing for frequency of vertical oscillations.

12.5 1 p = = cos 25 2 3 2 pt 2 p + f= 3 3

P

or 2 p t 2 + 3f = p Subtracting Eq. (iv) from Eq. (iii), we get

k

2 p (t1 - t 2) = p (t1 - t 2) = p / 2p =1/2 = 0.5 s

or

Example 7

A man stands on a weighing machine placed on a horizontal platform. The machine reads 50 kg . By means of a suitable mechanism, the platform is made to execute harmonic vibrations up and down with a frequency of 2 vibrations per second. What will be the maximum reading of the weighing machine? The amplitude of vibration of platform is 5 cm. [Take g = 10 ms-2] (a) 64.5 kgf (c) 89.5 kgf

Solution

m

(b) 74.5 kgf (d) 95.5 kgf

1 3p 1 (c) n = 2p

(a) n =

k m k m

1 3p 2 (d) n = 3p (b) n =

Solution

In equilibrium, due to weight mg spring is stretched by y 0 as shown in figure. When further depressed by a small distance y, the restoring force will be

Here, m = 50 kg, n = 2s-1, A = 5 cm = 0.05 m

Maximum, acceleration, amax = w2 A

m

= (2p n) 2 A = 4p 2n 2A 2

æ 22 ö = 4 ´ ç ÷ ´ (2) 2 ´ 0.05 = 7.9 ms-2 è7ø Maximum force on the man = m ( g + amax ) = 50 (10 + 7.9) = 89500 N = 89.5 kgf

m k m k

y0 m y m

Oscillations F = - [k (y + y 0) - mg ] = [k (y + y 0) - ky 0 ] = - ky F k \ Acceleration, a = = y m m As acceleration is proportional to the displacement y and opposite to y, the motion will be SHM. Frequency of oscillations will be 1 n= 2p

k m

Example 10

A uniform rod of mass m and length l0 is pivoted at one end and is hanging in the vertical direction. The period of small angular oscillations of the rod is l0 2

\Time period of oscillation

Þ

T = 2p

I = 2p mgl

T = 2p

2l0 3g

(c) T = 4p

2l0 3g

1 2 ml 0 3 æl ö mg ç 0 ÷ è 2ø

Example 11

The bob of a simple pendulum executes SHM in water with a period t, while the period of oscillation of the bob is t 0 in air. Neglecting frictional force of water and given æ 4ö that the density of the bob is ç ÷ ´ 1000 kgm -3. What è3ø relationship between t and t 0 is true?

l0

(a) t = t 0

O CM

(b) t = 4 t 0 t (d) t = 0 2

(c) t = 2t 0 2l0 (a) T = 3p 3g

631

l (b) T = 4p 0 3g (d) T = 2p

2l0 3g

Solution

Here the rod is oscillating about an end point O. Hence, moment of inertia of rod about the point of oscillating is 1 I = ml02 3

Moreover, length l of the pendulum = distance from the oscillation axis to centre of mass of rod = l0 /2

Solution water,

Here density of bob, r =

4 ´ 1000 kgm-3 and density of 3

s = 1000 kgm-3

\In air t 0 = 2p t = 2p

L and in water g L = 2p æ sö g ç1 - ÷ è rø

= 2 ´ 2p

L = 2t 0 g

L æ 3ö g ç1 - ÷ è 4ø

Start Practice for

JEE Main Round I

(Topically Divided Problems)

Oscillatory Motion and Simple Harmonic Motion with its Characteristics

6. Starting from y = A sin wt and y = A cos wt

1. The displacement of two particles executing SHM are represented by equations y1 = 2 sin (10 t + q ), y2 = 3 cos 10 t. The phase difference between the velocity of these particles is (a) q (c) q + p /2

(b) - q (d) q - p /2

2. Two pendulums of length 121 cm and 100 cm start vibrating. At same instant the two are in the mean position in the same phase. After how many vibrations of the shorter pendulum, the two will be in phase at the mean position? (a) 10

(b) 11

(c) 20

(d) 21

3. The displacement of the particle varies with time according to the relation. [NCERT Exemplar] y = a sin wt + b cos wt, then (a) The motion is oscillating but not SHM (b) The motion is SHM with amplitude a + b (c) The motion is SHM with amplitude a2 + b2 (d) The motion is SHM with amplitude a2 + b2

4. Two pendulums have time period T and 5T/4. They start SHM at the same time from the mean position. What will be the phase difference between them after the bigger pendulum completed one oscillation? (a) 45º

(b) 90º

(c) 60º

(d) 30º

5. The displacement of a particle is represented by

p the equation y = 3 cos æç - 2wt ö÷. The motion of the è4 ø

particle is (a) simple harmonic with period 2p/w (b) simple harmonic with period p /w (c) periodic but not simple harmonic (d) non-periodic

[NCERT Exemplar]

(a) acceleration lags the displacement by a phase p/4 (b) acceleration lags the displacement by a phase p/2 (c) acceleration leads the displacement by a phase p/2 (d) acceleration leads the displacement by a phase p

7. The displacement of a particle is represented by the equation y = sin 3 wt. The motion is (a) (b) (c) (d)

[NCERT Exemplar]

non-periodic periodic but not simple harmonic simple harmonic with period 2p /w simple harmonic with period p /w

8. A simple harmonic oscillator has and amplitude a and time period T. The time required by it to travel a from x = a to x = is 2 T 6 T (c) 3 (a)

T 4 T (d) 2 (b)

9. The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed? [NCERT] (a) 100 m/min (c) 300 m/min

(b) 200 m/min (d) 50 m/min

10. Two points are located at a distance of 10 m and 15 m from the source of oscillation. The period of oscillation is 0.05s and the velocity of the wave is 300 m/s. What is the phase difference between the oscillations of two points? (a) p (c)

p 3

p 6 2p (d) 3 (b)

633

Oscillations 11. A horizontal plank has a rectangular block placed on it. The plank starts oscillating vertically and simple harmonically with an amplitude of 40 cm. The block just loses contact with the plank when the later is momentarily at rest. Then (a) the period of oscillation is 2p/5 s (b) the block weighs double its weight when the plank is at one of the positions of momentary at rest (c) the block weighs 1.5 times its weight on the plank half way down (d) the block weighs its true weight on the plank, when the latter moves fastest

12. A body has a time period T1 under the action of one

force and T2 under the action of another force, the square of the time period when both the forces are acting in the same direction is (a) T12T22

(b) T12T22

(c) T12 + T22

(d) T12T22 /(T12 + T22 )

13. Motion of an oscillating liquid column in a U-tube is [NCERT Exemplar]

(a) periodic but not simple harmonic (b) non-periodic (c) simple harmonic and time period is independent of the density of the liquid (d) simple harmonic and time-period is directly proportional to the density of the liquid

14. A particle is acted simultaneously by mutually perpendicular simple hormonic motions x = a cos wt and y = a sin wt. The trajectory of motion of the particle will be [NCERT Exemplar] (a) an ellipse (c) a circle

(b) a parabola (d) a straight line

15. This time period of a particle undergoing SHM is 16 s. It starts motion from the mean position. After 2 s, its velocity is 0.4 ms–1. The amplitude is (a) 1.44 m (c) 2.88 m

(b) 0.72 m (d) 0.36 m

16. A particle is performing simple harmonic motion along x-axis with amplitude 4 cm and time period 1.2 s. The minimum time taken by the particle to move from x = + 2 to x = 4 cm and back again is given by (a) 0.4 s (c) 0.2 s

(b) 0.3 s (d) 0.6 s

17. The acceleration of a particle performing SHM is 12 cms–2 at a distance of 3 cm from the mean position. Its time period is (a) 2.0 s (c) 0.5 s

(b) 3.14 s (d) 1.0 s

18. The acceleration d2 x / dt2 of a particle varies with displacement x as

d 2x = - kx dt 2

where k is a constant of the motion. The time period T of the motion is equal to (a) 2pk (c) 2 p / k

(b) 2p k (d) 2 p / k

19. Two linear SHMs of equal amplitude A and angular frequencies w and 2w are impressed on a particle along the axes x and y respectively. If the initial phase difference between them is p/2, the resultant path followed by the particle is (a) y2 = x2 (1 - x2 / A2 )

(b) y2 = 2x2 (1 - x2 / A2 )

(c) y2 = 4 x2 (1 - x2 / A2 )

(d) y2 = 8x2 (1 - x2 / A2 )

20. A coin is placed on a horizontal platform, which undergoes horizontal SHM about a mean position O. The coin placed on platform does not slip, coefficient of friction between the coin and the platform is m. The amplitude of oscillation is gradually increased. The coil will begin to slip on the platform for the first time (a) at the mean position (b) at the extreme position of oscillations (c) for an amplitude of mg/ w2 (d) for an amplitude of g/mw2

21. A block is resting on a piston which is moving vertically with SHM of period 1.0 s. At what amplitude of motion will the block and piston separate? (a) 0.2 m (c) 0.3 m

(b) 0.25 m (d) 0.35 m

pt . 2 The distance covered by it in the time interval between t = 0 to t = 3 s is

22. A particle moves according to the law, x = r cos

(a) r

(b) 2r

(c) 3r

(d) 4r

23. A particle is executing SHM of period 24s and of amplitude 41 cm with O as equilibrium position. The minimum time in seconds taken by the particle to go from P to Q, where OP = - 9 cm and OQ = 40 cm is (a) 5 (c) 7

(b) 6 (d) 9

24. Figure shows the circular motion of a particle. The radius of the circle, the period, sense of revolution and the initial position are indicated on the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P is

y P (t = 0) T = 30 s B O

x

[NCERT Exemplar]

634 JEE Main Physics æ 2pt ö (a) x (t ) = B sin ç ÷ è 30 ø

æ pt ö (b) x (t ) = B cos ç ÷ è 15 ø

æ pt p ö (c) x (t ) = B sin ç + ÷ è 15 2 ø

æ pt p ö (d) x (t ) = B ç + ÷ è 15 2 ø

25. A 1.00 ´ 10-20 kg particle is vibrating with simple

harmonic motion with a period of 1.00 ´ 10-5 s and a maximum speed of 1.00 ´ 103 m/s. The maximum displacement of the particle is (a) 1.59 mm (c) 10 m

(b) 1.00 cm (d) None of these

26. Which one of the following equations does not represent SHM, x = displacement and t = time. Parameters a, b and c are the constants of motion? (a) x = a sin bt (b) x = a cos bt + c (c) x = a sin bt + c cos bt (d) x = a sec bt + c cosec bt

27. A particle executing SHM has a maximum speed of

30 cm/s and a maximum acceleration of 60 cm/s2 . The period of oscillation is [NCERT Exemplar] p s 2 p (d) s t

(a) p s

(b)

(c) 2p s

at time t = 0 from a position of small angular displacement. Its linear displacement at time t is given by

(c) X = a sin

L ´t g g ´t L

(b) X = a cos 2p (d) X = a cos

particle which represents SHM (1) y = sin wt - cos wt (2) y = sin 3 wt æ3p ö (3) y = 5 cos ç - 3 wt ÷ è 4 ø (4) y = 1 + wt + w2t 2 (a) Only (1) and (2) (b) Only (1) (c) Only (4) does not represent SHM (d) Only (1) and (2)

33. A large horizontal surface moves up and down in SHM with an amplitude of 1 cm. If a mass of 10 kg (which is placed on the surface) is to remain continuously is in contact with it, the maximum frequency of SHM will be (a) 5 Hz (c) 1.5 Hz

g ´t L

g ´t L

(b) 0.5 Hz (d) 10 Hz

34. The composition of two simple harmonic motions of equal periods at right angle to each other and with a phase difference of p results in the displacement of the particle along (a) circle (c) straight line

28. The bob of a simple pendulum of length L is released

(a) X = a sin 2p

32. Out the following functions representing motion of a

(b) figure of eight (d) ellipse

35. A horizontal platform vibrates with simple harmonic motion in the horizontal direction with a period 2 s. A body of mass 0.5 kg is placed on the platform. The coefficient of static friction between the body and platform is 0.3. What is the maximum frictional force on the body when the platform is oscillating with an amplitude 0.2 m? Assume p2 = 10 = g. (a) 0.5 N (c) 1.5 N

(b) 1 N (d) 2 N

29. Displacement-time equation of a particle executing SHM is , x = 4 sin w t + 3 sin (wt + p/ 3 ). Here x is in centimetre and t in second. The amplitude of oscillation of the particle is approximately (a) 5 cm (c) 7 cm

(b) 6 cm (d) 9 cm

30. A particle in SHM is described by the displacement function x( t) = A cos( wt + f), w = 2p/ T. If the initial ( t = 0) position of the particle is 1 cm, its initial velocity is p cm s–1 and its angular frequency is ps–1, then the amplitude of its motion is (a) p cm

(b) 2 cm

(c) 2 cm

(d) 1 cm

31. A particle moves in xy-plane according to the rule x = a sin wt and y = a cos wt. The particle follows (a) an elliptical path (b) a circular path (c) a parabolic path (d) a straight line path inclined equally to x and y-axis

Energy in SHM 36. The angular velocity and the amplitude of a simple pendulum is w and a respectively. At a displacement x from the mean position if its kinetic energy is T and potential energy is V, then the ratio of T to V is (a) ( a2 - x2 w2 ) / x2 w2

(b) x2 w2 /( a2 - x2 w2 )

(c) ( a2 - x2 ) / x

(d) x2 / ( a2 - x2 )

37. A particle of mass m is executing oscillations about the origin on the x-axis with amplitude A. Its potential energy is given as U( x) = a x4 , where a is positive constant. The x-coordinate of mass where potential energy is one-third of the kinetic energy of particle, is A 3 A (c) ± 3

(a) ±

A 2 A (d) ± 2 (b) ±

Oscillations 38. A particle starts SHM from the mean position. Its amplitude is a and total energy E. At one instant its kinetic energy is 3E/4 its displacement at this instant is (a) y = a / 2

a (b) y = 2

a 3 /2

(d) y = a

(c) y =

39. A point particle of mass 0.1 kg is executing SHM of amplitude 0.1 m. When the particle passes through the mean position, its kinetic energy is 8 ´ 10 -3 J. The equation of motion of this particle, if its initial phase of oscillation is 45°, is pö ær (a) y = 0.1sin ç + ÷ è4 4ø pö æ (c) y = 0.1sin ç 4 t - ÷ è 4ø

æt pö (b) y = 0.1sin ç + ÷ è2 4 ø pö æ (d) y = 0.1sin ç 4t + ÷ è 4ø

40. A particle is vibrating in a simple harmonic motion with and amplitude of 4cm. At what displacement from the equilibrium position is its energy half potential and half kinetic? (a) 1cm

(b) 2 cm

(c) 3 cm

Springs and their Oscillations 46. A body of mass 500 g is attached to a horizontal spring of spring constant 8 p2 Nm -1. If the body is pulled to a distance of 10 cm from its mean position then its frequency of oscillation is (a) 2 Hz (b) 4 Hz (e) 4 p Hz

a 4

(b)

a 3

(c)

a 2

(d)

2a 3

42. The potential energy of a particle (U X ) executing SHM is given by

k ( x - a )2 2 (c) Ux = Ae - bx (a) Ux =

(b) Ux = k1x + k2 x2 + k3x3 (d) Ux = constant

43. If a simple pendulum of length l has maximum angular displacement q, then the maximum kinetic energy of bob of mass m is (a)

1 ælö ´ç ÷ 2 è gø

1 mg ´ 2 l 1 (d) ´ mgl sin q 2

(b)

(c) mgl ´ (1 - cos q)

44. When the displacement is half of the amplitude, then

(d) 0.5 Hz

cut into two springs of length l1 and l2 such that l1 = nl2 (n = an integer). The force constant of the spring of length l2 is ( n + 1) k n (d) k / ( n + 1)

(a) k (1 + n )

(b)

(c) k

48. Two springs of force constants k and 2 k are connected to a mass as shown below. The frequency of oscillation of the mass is 2k

k m

41. When the potential energy of a particle executing

(a)

(c) 8 Hz

47. A simple spring has length l and force constant k. It is

(d) 2 2 cm

simple harmonic motion is one-fourth of its maximum value during the oscillation, the displacement of the particle from the equilibrium position in terms of its amplitude a is

1 k/ m 2p 1 3k (c) 2p m

1 2k/ m 2p 1 m (d) 2p k

(b)

(a)

49. A weightless spring which has a force constant k oscillates with frequency n when a mass m is suspended from it. The spring is cut into two equal halves and a mass 2 m is suspended from one part of spring. The frequency of oscillation will now become (a) n

(b) 2n

(c)

n 2

(d) n (2)1 /2

50. An object suspended from a spring exhibits oscillations of period T. Now, the spring is cut in two halves and the same object is suspended with two halves as shown in figure. The new time period of oscillation will become

what fraction of the total energy of a simple harmonic oscillator is kinetic? (a) 2/7th

(b) 3/4th

(c) 2/9th

(d) 5/7th

45. For a particle executing SHM, the kinetic energy K is

given by K = K 0 cos2 wt. The equation of its displacement can be æ K ö (a) ç 02 ÷ è mw ø

1 /2

æ 2w2 ö (c) ç ÷ è mK0 ø

1 /2

sin wt

æ 2K ö (b) ç 02 ÷ è mw ø

sin wt

æ 2K ö (d) ç 0 ÷ è mw ø

1 /2

m

sin wt

1 /2

sin wt

635

(a)

T 2 2

(b)

T 2

m

(c)

T 2

(d) 2T

636 JEE Main Physics 51. On a smooth inclined plane, a body of mass M is

57. A uniform spring of force constant k is cut into two

attached between two springs. The other ends of the springs are fixed to firm support. If each spring has force constant k, the period of oscillation of the body (assuming the springs as massless) is

pieces, the lengths of which are in the ratio 1 : 2. The ratio of the force constants of the shorter and longer piece is (a) 1 : 2 (c) 1 : 3

(b) 2 : 1 (d) 2 : 3

58. What will be the force constant of the spring system shown in figure?

M

θ

k1

k1

(a) 2p[ M/ 2k ]1 /2

(b) 2p[2M/ k ]1 /2

(c) 2p [ Mg sin q/ 2k ]1 /2

(d) 2p[2Mg/ k ]1 /2

k2

52. A mass M, attached to a spring, oscillates with a period of 2 s. If the mass is increased by 4 kg, the time period increases by 1 s. Assuming that Hooke’s law is obeyed, the initial mass M was (a) 3.2 kg (c) 2 kg

(b) 1 kg (d) 8 kg

53. A mass M is suspended from a light spring. An additional mass m added displaces the spring further by a distance X. Now the combined mass will oscillate on the spring with period (a) T = 2p (c) T = p / 2

mg X (M + m) mg X( M + m )

(b) T = 2p

(M + m)X mg

(d) T = 2p

(M + m) mg

(a)

k1 + k2 2

é 1 1ù (b) ê + ú ë 2k1 k2 û

(c)

1 1 + 2k1 k2

é2 1ù (d) ê + ú k k ë 1 2û

vertical wall and the other to a block of mass m resting on a smooth horizontal surface. There is another wall at a distance x0 from the block. The spring is then compressed by 2 x0 and then released. The time taken to strike the wall is M

the spring is cut to one-half and made to oscillate by suspending double mass, the time period of the mass will be (b) 4T (d) T

55. Two blocks with masses m1 = 1 kg and m2 = 2 kg are connected by a spring of spring constant k = 24 Nm–1 and placed on a frictionless horizontal surface. The block m1 is imparted an initial velocity v0 = 12 cms–1 to the right, the amplitude of oscillation is (a) 1 cm (c) 3 cm

(b) 2 cm (d) 4 cm

56. A mass 1 kg suspended from a spring whose force

constant is 400 Nm–1, executes simple harmonic oscillation. When the total energy of the oscillator is 2 J, the maximum acceleration experienced by the mass will be ms–2

(a) 2 (c) 40 ms–2

ms–2

(b) 4 (d) 400 ms–2

-1

59. One end of a spring of force constant k is fixed to a

54. Time period of mass m suspended by a spring is T. If

(a) 8T T (c) 2

-1

B

A 2x0

(a)

1 k p 6 m

(b)

k m

(c)

2p 3

C x0

m k

(d)

p 4

k m

Simple Pendulum and other Systems 60. A simple pendulum of length l and mass (bob) m is suspended vertically. The string makes an angle q with the vertical. The restoring force acting on the pendulum is (a) mg tan q (c) mg sin q

(b) - mg sin q (d) - mg cos q

61. A simple pendulum has a length l. The inertial and gravitational masses of the bob are m1 and m g respectively. Then the time period T is given by (a) T = 2p (c) T = 2p

mg l mig mi ´ mg ´ l g

(b) T = 2p

mil mg g

(d) T = 2p

l mi ´ mg ´ g

Oscillations

637

62. A man measures the period of a simple pendulum

68. Two simple pendulums of length 0.5 m and 20 m

inside a stationary lift and finds it to be T second. If the lift accelerates upwards with an acceleration g/4, then the period of pendulum will be

respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed ....oscillations.

(a) 2T 5 2T (c) 5

(b) T T (d) 4

(a) 5

the same elastic support as shown in figure. A and C are of the same length, while B is smaller than A and [NCERT Exemplar] D is larger than A. If A is

A C D

(a) D will vibrate with maximum amplitude (b) C will vibrate with maximum amplitude (c) B will vibrate with maximum amplitude (d) All the four will oscillate with equal amplitude

64. If the length of second’s pendulum is increased by 2%. How many seconds it will lose per day? (b) 3427 s (d) 864 s

65. A pendulum bob of mass m is hanging from a fixed point by a light thread of length l. A horizontal speed v0 is imparted to the bob so that it takes up horizontal position. If g is the acceleration due to gravity, then v0 is (b) 2gl

(c) gl

(d) gl

66. A tunnel is made across the earth of radius R, passing through its centre. A ball is dropped from a height h in the tunnel. The motion will be periodic with time period. (a) 2p

R h +4 g g

(b) 2p

R 2h +4 g g

(c) 2p

R + g

R (d) 2p + g

(d) 3

reduced to one-third of its initial value a0 at the end of 100 oscillation. When the oscillation completes 200 oscillations, its amplitude must be a0 2 a0 (c) 6

a0 4 a0 (d) 9

(a)

(b)

70. A and B are fixed points and

B

(a) mgl

(c) 2

69. In damped oscillation the amplitude of oscillations is

63. Four pendulums A, B, C and D are suspended from

(a) 3927 s (c) 3737 s

(b) 1

B

A

the mass M is tied by strings at A and B. If the mass M is displaced slightly out of this plane and released, it will execute oscillations with period. (Given, AM = BM = L, AB = 2 d)

M

(a) 2p

L g

(b) 2p

( L2 - d 2 )1 /2 g

(c) 2p

( L2 + d 2 )1 /2 g

(d) 2p

(2d 2 ) 3 /2 g

71. A simple pendulum of length l has been set up inside a railway wagon sliding down a frictionless inclined plane having an angle of inclination q = 30° with the horizontal. What will be its period of oscillation as recorded by an observer inside the wagon? (a) 2p

2l 3g

(b) 2p 2l / g (d) 2p

(c) 2p l / g

3l 2g

72. If a simple pendulum is taken to a place where g decreases by 2%, then the time period (a) increases by 0.5% (c) increases by 2.0%

h g

(b) increases by 1% (d) decreases by 0.5%

73. A heavy sphere of mass m is suspended by string of

2h g

67. The bob of a pendulum of length l is pulled a side from its equilibrium position through an angle q and then released. The bob will then pass through its equilibrium position with a speed v, where v equals (a) 2gl(1 - cos q)

(b) 2gl(1 + sin q)

(c) 2gl(1 - sin q)

(d) 2gl(1 + cos q)

length l. The sphere is made to revolve about a vertical line passing through the point of suspension in a horizontal circle such that the string always remains inclined to the vertical at an angle q . What is its period of revolution? (a) T = 2p

l g

(b) T = 2p

l cos q g

(c) T = 2p

l sin q g

(d) T = 2p

l tan q g

638 JEE Main Physics 74. A piece of wood has dimensions a, b and c. Its relative

æ ba ö (b) T = 2p ç ÷ è dg ø

æ gö (c) T = 2p ç ÷ è dc ø

æ dc ö (d) T = 2p ç ÷ è gø

(2)

75. A pendulum clock is placed on the moon, where object weighs only one-sixth as much as on the earth, how many seconds the clock tick out in an actual time of 1 min the clock keeps good time on the earth?

(3)

(b) 24.5 (d) 0.245

76. A uniform cylinder of length L and mass M having cross-sectional area A is suspended with its vertical length, from a fixed point by a massless spring, such that it is half submerged in a liquid of density d at equilibrium position. When released, it starts oscillating vertically with a small amplitude. If the force constant of the spring is k, the frequency of oscillation of the cylinder is 1 æ k - Adg ö (a) ç ÷ 2p è M ø (c)

1 /2

1 æ k - dgL ö ç ÷ 2p è M ø

(b)

1 æ k + Adg ö ç ÷ 2p è M ø

(d)

1 æ k + AgL ö ç ÷ 2p è Adg ø

1 /2

1 /2

1 /2

Time

Displacement

(a) 12.25 (c) 2.45

Time

Displacement

æ abc ö (a) T = 2p ç ÷ è g ø

Displacement

density is d. It is floating in water such that the side c is vertical. It is now pushed down gently and released. The time period is

Time

(4)

(a) Fig. 1 alone (c) Fig. 4 alone

(b) Fig. 2 alone (d) Fig. 3 and 4

79. A particle of mass m is released from rest and follows

a parabolic path as shown. Assuming that the displacement of the mass from the origin is small. Which graph correctly depicts the position of the particle as a function of time v(x)

77. A particle, with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force, F = F0 sin wt If, the amplitude of the particle is maximum for w= w1 and the energy of the particle is maximum for w = w2 , then (a) w1 = w0 and w2 (b) w1 = w0 and w2 (c) w1 ¹ w0 and w2 (d) w1 ¹ w0 and w2

¹ w0 = w0 = w0 ¹ w0

O x(t)

(a)

Displacement

(1)

t

(b)

t

O

x(t)

(c)

Time

x(t)

O

78. Which of the following figure represent(s) damped simple harmonic motions?

(x)

x(t)

t

O

(d)

t

O

80. The amplitude of damped oscillator becomes in 2 s. Its amplitude after 6 s is 1/n times the original. Then n is equal to (a) 23

(b) 32

(c) 31/3

(d) 33

Oscillations

Round II

639

(Mixed Bag)

Only One Correct Option 1. A simple pendulum of length l has a bob of mass m, with a charge q on it. A vertical sheet of charge, with surface charge density s passes through the point of suspension. At equilibrium, the spring makes an angle q with the vertical. Its time period of oscillations is T in this position. Then sq (a) tan q = 2e 0 mg

sq (b) tan q = e 0 mg

1 (c) T > 2p g

1 (d) T = 2p g

2. An instantaneous displacement of a simple harmonic oscillator is x = A cos ( wt + p / 4). Its speed will be maximum at time (a) p / 4 w (c) p / w

(b) p /2w (d) 2p / w

3. The period of oscillation of a mass m suspended from a spring is 2 s. If along with it another mass 2 kg is also suspended, the period of oscillation increases by 1 s. The mass m will be (a) 2 kg (c) 1.6 kg

(b) 1 kg (d) 2.6 kg

4. Two pendulums begin to swing simultaneously. The first pendulum makes 9 full oscillations when the other makes 7. The ratio of lengths of the two pendulums is (a) 9/7 (c) 49/81

(b) 7/9 (d) 81/49

5. A bottle weighing 220 g and area of cross-section

50 cm2 and height 4 cm oscillates on the surface of water in vertical position. Its frequency of oscillation is (a) 1.5 Hz (c) 3.5 Hz

(b) 2.5 Hz (d) 4.5 Hz

6. A body of mass 4.9 kg hangs from a spring and oscillates with a period 0.5 s on the removal of the body, the spring is shortened by (Take g =10 ms -2 , p 2 = 10) (a) 6.3 m (c) 6.25 cm (e) 0.625 cm

(b) 0.63 m (d) 63.5 cm

7. A ball of mass (m) 0.5 kg is attached to the end of a string having length (L) 0.5m. The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that string can bear is 324 N.The maximum possible value of angular velocity of ball (in rad/s) is

L

m

(a) 9

(b) 18

(c) 27

(d) 36

8. The period of particle in SHM is 8 s. At t = 0, it is at the mean position. The ratio of the distances travelled by it in Ist second and 2nd second is (a) 1.6 : 1

(b) 2.4 : 1

(c) 3.2 : 1

(d) 4.2 : 1

9. A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period? (a) 2p

3l g+

(c) 2p

v2 r

2l æ 2 v2 ö çg + 2 ÷ r ø è

(b) 2p

æ v4 ö l ç g2 + 2 ÷ r ø è

(d) 2p

2l ( g 2 + v 2 /r )

10. A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body? [NCERT] (a) 222.13 N (c) 193.13 N

(b) 200.13 N (d) 219.13 N

11. One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. The suction pump is removed, the column of mercury in the U-tube will show [NCERT]

(a) periodic motion (b) oscillation (c) simple harmonic motion (d) None of the above

640 JEE Main Physics 12. The time period of a particle in simple harmonic

18. A highly rigid cubical block A of small mass M and

motion is 8 s. At t = 0, it is at the mean position. The ratio of the distances travelled by it in the first and second, seconds is

side L is fixed rigidly on the another cubical block of same dimensions and low modulus of rigidity h such that the lower face of A completely covers the upper face of B. The lower face of B is rigidly held on a horizontal surface. A small force F is applied perpendicular to one of the side faces of Z. After the force is withdrawn, block A executes small oscillations, the time period of which is given by

(a) 1/2

(c) 1 / ( 2 - 1) (d) 1 / 3

(b) 1 / 2

13. The equation of SHM is given by x = 3 sin 20pt + 4 cos 20t where x is in cm and t is in second. The amplitude is (a) 7 cm

(b) 4 cm

(c) 5 cm

(d) 3 cm

14. When a body of mass 1.0 kg is suspended from a certain light spring hanging vertically, its length increases by 5 cm. By suspending 2.0 kg block to the spring and if the block is pulled through 10 cm and released, the maximum velocity of it, in ms–1 is (g = 10 ms–2) (a) 0.5

(b) 1

(c) 2

(d) 4

15. A pendulum is made to hang from a ceilling of an elevator.It has period of Tsec . (for small angles). The elevator is made to accelerate upwards with 10 m/s2.The period of the pendulum now will be (assume g =10 m/s2) (a) T 2

(b) infinite

(c) T/ 2

(d) zero

16. A mass M is attached to a horizontal spring of force constant k fixed on one side to a rigid support as shown in figure. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, another mass m is gently placed on it. What will be the new amplitude of oscillations? m

k

M

æ M ö (a) A ç ÷ è M - mø

æ M - mö (b) A ç ÷ è M ø

æ M ö (c) A ç ÷ è M + mø

æ M + mö (d) A ç ÷ è M ø

17. Lissajous figure shown in figure corresponds to which one of the following?

(a) 2p MLh

(b) 2p Mh /L

(c) 2p ML / h

(d) 2p M / h L

19. The bob of a simple pendulum is of mass 10 g. It is suspended with a thread of 1 m. If we hold the bob so as to stretch the string horizontally and release it, what will be the tension at the lowest position? (g = 10 ms–2) (a) Zero (c) 0.3 N

(b) 0.1 N (d) 1.0 N

20. A block of mass M is suspended from a light spring of force constant k. Another mass m moving upwards with velocity v hits the mass M and gets embedded in it. What will be the amplitude of the combined mass ? (a)

mv k (M - m)

(b)

mv (M - m)k

(c)

mv k (M + m)

(d)

mv (M + m)k

21. Two pendulums of length 1 m and 16 m start vibrating one behind the other from the same stand. At some instant, the two are in the mean position in the same phase. The time period of shorter pendulum is T. The minimum time after which the two threads of the pendulum will be one behind the other is (a) T/4

(b) T/3

(c) 4 T/3

(d) 4 T

22. Two pendulums of lengths 1m and 1.21m respectively start swinging together with same amplitude. The number of vibrations that will be executed by the longer pendulum before the two will swing together again are (a) 9

(b) 10

(c) 11

(d) 12

23. An elastic string has a length l when tension in it is 5 N. Its length is h when tension is of 4 N. On subjecting the string to a tension of 9 N, its length will be (a) Phase difference p/2 and period 1 : 2 (b) Phase difference 3p/4 and period 1 : 2 (c) Phase difference p/4 and period 2 : 1 (d) Phase difference 2p/3 and period 2 : 1

(a) l + h (b) l - h (c) (5 l - 4 h) (d) (l + h)/(h – l)

Oscillations 24. A point mass is subjected to two simultaneous sinusoidal

displacement

in X-direction 2 p ö÷ . Adding X1( t) = A sin wt and X2 ( t) = A sin æç wt + è 3 ø a third sinusoidal displacement X 3( t) = B sin( wt + f) brings the mass to a complete rest. The value of B and f 3p 4 5p (c) 3 A, 6

(a) 2 A,

4p 3 p (d) A, 3

when mass M is suspended from one end of each spring. If both springs are taken in series and the same mass M is suspended from the series combination, the time period is T, then

(c) T 2 = T12 + T22

1 1 1 = + T T1 T2 1 1 1 (d) 2 = 2 + 2 T T1 T2 (b)

maximum value at a distance of 4 cm from the mean position. Find the amplitude of motion. (b) 2 / 6 cm

(c) 2 cm

(d) 6 / 2

27. If a spring extends by x on loading, then the energy stored in the spring is (if T is the tension and k is the force constant of the spring) 2

T 2x 2k (c) 2 T (a)

2

T 2k 2T 2 (d) k

(b)

(b) t 0-2 = t1-2 + t2-2 (c) t 0-1 = t1-1 + t2-1 (d) t 0 = t1 + t2

(a) 6.4 s (c) 9.4 s

(b) 7.4 s (d) 8.4 s

32. The total energy of a particle executing SHM is 80 J. What is the potential energy when the particle is at a distance of 3/4 of amplitude from the mean position? (b) 10 J (d) 45 J

33. Two simple harmonic motions act on a particle. These harmonic motions are x = A cos( wt + a ); y = A cos( wt + a ), when (a) (b) (c) (d)

an ellipse and the actual motion is counter clockwise an ellipse and the actual motion is clockwise a circle and the actual motion is counter clockwise a circle and the actual motion is clockwise

34. The time period of a mass suspended from a spring is 5 s. The spring is cut into four equal parts and the same mass is now suspended from one of its parts. The period is now

35. A block whose mass is 650 g is fastened to a spring

(a) 1.1 ms–1 (c) 1.30 ms–1 k1

(b) 2.5 s 1 (d) s 16

whose spring constantly is 65 Nm–1. The block is pulled a distance x = 11 cm from its equilibrium position at x = 0. On a frictionless surface and released from rest at t = 0. The maximum velocity of the vibrating block is

1 mw2 A2 ,0 2 1 1 (d) mw2 A2 , mw2 A2 4 4 (b)

29. A mass m is suspended separately

(a) t 02 = t12 + t22

(b) 2 cm (d) 2.5 cm

moon is 1.7 m/s2 . What is the time period of a simple pendulum on the surface of moon, if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 m/s2 .) [NCERT]

(c) 1.25 s

over entire time period is

by two different springs in successive order, then time periods is t1 and t2 respectively. If m is connected by both springs as shown in figure, then time period is t0 , the correct relation is

(a) 1 cm (c) 2 cm

(a) 5 s

28. Average value of kinetic energy and potential energy 1 (a) 0, mw2 A2 2 1 1 (c) mw2 A2 , mw2 A2 2 2

function x( t) = A cos ( wt + q ). If the initial ( t = 0) position of the particle is 1 cm and its initial velocity is p cms–1, what is its amplitude? The angular frequency of the particle is p s–1.

(a) 60 J (c) 40 J

26. A particle is having kinetic energy 1/3 of the

(a) 2 6 cm

30. A particle in SHM is described by the displacement

31. The acceleration due to gravity on the surface of

(b) A,

25. Let T1 and T2 be the time period of spring A and B

(a) T = T1 + T2

641

k2

(b) 0.65 ms–1 (d) 2.6 ms–1

More Than One Correct Option 36. Motion of a ball bearing inside a smooth curved bowl,

m

when released from a point slightly above the lower [NCERT Exemplar] point is (a) simple harmonic motion (b) non-periodic motion (c) periodic motion (d) periodic but not SHM

642 JEE Main Physics 40. A metal rod length L and mass m is pivoted at one

SHM is shown in figure. Choose the correct statements. [NCERT Exemplar]

end. A thin disc of mass M and radius R ( < L) is attached at its centre to the free end of the rod. Consider two ways the disc is attached ( case A). The disc is not free to rotate about its centre. The rod-disc system perform SHM in vertical plane after being released from the same displaced position which of the following statements is (are) true?

Displacement

37. Displacement vs. time curve for a particle executing

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

0

1

2

3

4 5

7

6

Time (s)

Phase of the oscillator is same at t = 0 s and t = 2 s Phase of the oscillator is same at t = 2 s and t = 6 s Phase of the oscillator is same at t =1 s and t =7 s Phase of the oscillator is same at t =1 s and t = 5 s

38. The displacement time graph of a particle executing

Displacement

SHM is shown in figure. Which of the following statement is/are true ? [NCERT Exemplar]

0

2T/4 3T/4 T

T/4

5T/4

Time (s)

(a) Restoring torque in case A = Restoring torque in case B (b) Restoring torque in case A < Restoring torque in case B (c) Angular frequency for case A > Angular frequency for case B (d) Angular frequency for case A < angular frequency for case B

Comprehension Based Questions Passage I

(a) The force is zero at t =

3T 4

(b) The acceleration is maximum at t = (c) The velocity is maximum at t =

4T 4

T 4

(d) The PE is equal to KE of oscillation at t =

T 2

A uniform cylindrical metal rod A of length L and radius R is suspended at its mid-point from a rigid support through a strong metal wire of length l. The rod is given a small angular twist and released so that it oscillates to and fro about its mean position with a time period T1. Another body B of an irregular shape is suspended from the same rigid support using the same length of given suspension wire and its time period is found to be T2 .

39. A particle is in linear simple harmonic motion between two points A and B, 10 cm apart (Fig). Take the direction from A to B as the +ve direction and choose the correct statements. [NCERT Exemplar] B

O

C

A

B

A

AO = OB = 5 cm BC = 8 cm (a) The sign of velocity, acceleration and force on the particle when it is 3 cm away from A going towards B are positive (b) The sign of velocity of the particle at C going towards O is negative (c) The sign of velocity, acceleration and force on the particle when it is 4 cm away from B going towards A are negative (d) The sign of acceleration and force on the particle when it is at point B is negative

41. The rotational inertia of metal rod about the wire as an axis is ML2 12 é L2 R2 ù (c) M ê + 2 úû ë 12

(a)

MR2 2 é L2 R2 ù (d) M ê + 4 úû ë 12 (b)

42. The motion of rod is (a) periodic but non-oscillatory (b) oscillatory but non-simple harmonic (c) linear harmonic motion (d) angular harmonic oscillation

Oscillations 43. Time period of oscillations of rod is given by m k I (c) T = 2p k

k m k (d) T = 2p I

(a) T = 2p

(b) T = 2p

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

46. Assertion In SHM, the motion is to end fro and periodic.

Passage II A particle performs harmonic oscillation along the x-axis about the equilibrium position x = 0. The oscillation frequency is w= 4.00 s–1. At a certain moment of time the particle has a coordinate x0 = 25.0 cm and its velocity is equal to n x = 100 cms–2. 0

44. Find the amplitude of oscillation. (a) 13 3 cm (c) 27 5 cm

643

(b) 25 2 cm (d) 2 3 cm

Reason Velocity of the particle (v) = w K 2 - x2 ,where x is the displacement and k is amplitude.

47. Assertion Soldiers are asked to break steps while crossing the bridge. Reason The frequency of marching may be equal to the natural frequency of bridge and may lead to resonance which can break the bridge.

48. Assertion The percentage change in time period is

45. Find the equation of motion of the particle. pö pö æ æ (a) y = 13 3 sin ç 4t + ÷ (b) y = 25 2 sin ç 4t + ÷ è è 4ø 4ø pö pö æ æ (c) y = 27 2 sin ç 4t + ÷ (d) y = 27 5 sin çt + ÷ è è 4ø 2ø

Assertion and Reason Direction

Question No. 46 and 50 are Assertion-Reason type. Each of these contains two Statements : Statement I (Assertion), Statement II (Reason). Each of these questions also has four alternative choice, only one of which is correct. You have to select the correct choices from the code (a), (b), (c) and (d) given below (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion

1.5%, if the length of simple pendulum increases by 3%. Reason Time period is directly proportional to length of pendulum.

49. Assertion If the length of a spring is made n times, the spring factor of the spring becomes 1/nth of its original value. Reason Time of oscillation of a spring pendulum is m T = 2p . k

50. Assertion The amplitude of a particle executing SHM with a frequency of 60 Hz is 0.01 m. The maximum value of acceleration of the particle is ±144 p2 ms -2 .

Reason Acceleration amplitude = w2 A,where A is displacement amplitude.

Previous Years’ Questions 51. If a simple pendulum has significant amplitude (up

1 of original) only in the period between e t = 0 s to t = ts. Then t may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportionality the average life time of the pendulum is (assuming damping is small ) in seconds

to a factor of

[AIEEE 2012]

0.693 b 1 (c) b (a)

(b) b (d) 2/b

52. Two particles are executing simple harmonic motion of the same amplitude A and frequency w along the x-axis. Their mean position is separated by distance X 0 ( X 0 > A). If the maximum separation between

them is ( X 0 + A) the phase difference during their motion is [AIEEE 2011] p 2 p (c) 4 (a)

p 3 p (d) 6 (b)

53. A mass M attached to a horizontal spring executes SHM with a amplitude A when the mass M passes through its mean position then a smaller mass m is placed over it and both of them move together with æA ö amplitude A2 .The ratio of ç 1 ÷ is è A2 ø [AIEEE 2011] M M+m 1 /2 æ M ö (c) ç ÷ è M + mø

(a)

M+m M 1 /2 æ M + mö (d) ç ÷ è M ø (b)

644 JEE Main Physics 54. If x, v and a denotes the displacement, the velocity

62. The particle execute simple harmonic motion with a

and the acceleration of a particle executing simple harmonic motion of time period T. Then which of the following does not change with time? [AIEEE 2009]

time period of 16 s. At time t = 2 s, the particle crosses the mean position while at t = 4 s, its velocity is 4 ms–1. The amplitude of motion in metre is [Kerala CET 2007]

(a) a2T2 + 4 p2 v2

(b) aT /x

(c) aT + 2pv

(d) aT / v

55. If the maximum velocity and acceleration of a particle executing SHM are equal in magnitude, then the time period will be [BVP Engg. 2008] (a) 1.57 s (c) 6.28 s

an amplitude A and time period T. The displacement of the particle after 2 T period from its initial position is [EAMCET 2008] (b) 4 A

(c) 8 A

(d) zero

57. Two springs are joined and attached to a mass of 16 kg. The system is then suspended vertically from a rigid support. The spring constant of the two springs are k1 and k2 respectively. The period of vertical oscillations of the system will be [WB JEE 2008] (a)

1 8p

(c)

p 2

k1 + k2

(b) 8p

k1 - k2

(d)

p 2

k1 + k2 k1k2 k1 k2

[Karnataka CET 2008]

(c) 2 g

(d) 3 g

59. The total energy of a simple harmonic oscillator is proportional to

[Kerala CET 2008]

(a) square root of displacement (b) velocity (c) frequency (d) amplitude (e) square of the amplitude

60. The amplitude of SHM y = 2 (sin 5 pt + 3 cos 5 pt) is (a) 2 (c) 4

[Kerala CET 2008]

(b) 2 2 (d) 2 3

lift is T. The lift accelerates upwards with an acceleration of g/3. The time period of pendulum will be [Kerala CET 2008] (b)

T 2

(c)

3 T 2

63. A mass oscillates along the x-axis according to the law, (a) A = x0 w2 , d = 3p / 4 2

(c) A = x0 w , d = p / 4

(b) A = x0 , d = - p / 4 (d) A = x0 w2 , d = - p / 4

64. A particle of mass m executes simple harmonic motion with amplitude a and frequency n. The average kinetic energy during its motion from the position of equilibrium to the end is [AIEEE 2007] (a) 2 p2 ma2 v2 1 (c) p2 ma2 v2 4

(b) p2 ma2 v2 (d) 4 p2 ma2 v2

65. A particle executes simple harmonic oscillation with

T (a) 4 T (c) 12

T (b) 8 T (d) 2

66. A mass of 2.0 kg is put on a flat pan

m

attached to a vertical spring fixed on the ground as shown in the figure. The mass of the spring and the pan is negligible. When pressed slightly and released the mass executes a simple harmonic motion. The spring constant is 200 Nm–1. What should be the minimum amplitude of the motion, so that the mass gets detached from the pan? (Take g = 10 ms–2) [UP SEE 2007] (a) 8.0 cm (b) 10.0 cm (c) Any value less than 12.0 cm (d) 4.0 cm

67. The kinetic energy and potential energy of a particle

61. The period of a simple pendulum inside a stationary

(a) 2 T

(e) 24 2p

[UP SEE 2007]

lift. When the lift is at rest its time period is T. With what acceleration should the lift be accelerated upwards in order to reduce its period to T/2? (g is the acceleration due to gravity) (b) g

(d) 4 / p

an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

58. A simple pendulum is suspended from the ceiling of a

(a) 4 g

(b) 16 2p

(c) 32 2 /p

x = x0 cos ( wt - p/ 4). If the acceleration of the particle is written as a = A cos( wt + d), then [AIEEE 2007]

(b) 3.14 s (d) 12.56 s

56. A particle is executing simple harmonic motion with

(a) A

(a) 2 p

(d)

T 3

executing SHM of amplitude a will be equal when displacement is [BVP Engg. 2007] (a)

a 2

(c) 2a

(b) a 2 (d)

a 2

Oscillations

645

68. One end of a long metallic wire of length L is tied to

73. A particle starts SHM from the mean position. Its

the ceiling. The other end is tied to massless spring of spring constant k. A mass m hangs freely from the free end of the spring. The area of cross-section and Young’s modulus of the wire are A and Y respectively. If the mass is slightly pulled down and released, it will oscillate with a time period T equal to [DCE 2006]

amplitude is a and total energy E. At one instant its kinetic energy is 3E/4. Its displacement at that instant is [Kerala CET 2005]

(a) 2p m / k (c) 2p

mYA KL

m(YA + KL) YAK ML (d) 2p YA

(b) 2p

[Kerala CET 2006]

(b) 1 : 1 (d) 4 : 1

(a) at the highest position of the platform (b) at the mean position of the platform (c) for an amplitude of g/w2 (d) for an amplitude of g2/w2

harmonically with a period of 2 s. After what time will its kinetic energy be 75% of the total energy? [AIEEE 2006]

1 s 6 1 (d) s 3 (b)

72. A particle executes SHM with an amplitude of 2 cm. When the particle is at 1 cm from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in second is [Kerala CET 2005]

(b) 2p 3

(c)

2p 3

(d)

(e)

2 p

3 2p

(c) 4 T

(d) 2 T

75. A simple pendulum has time period T1. The point of

suspension is now moved upward according to the relation) y = kt2 . ( k = 1 ms -2 ) where y is the vertical displacement. The time period now becomes T2 . The T2 ratio of 12 is (g = 10 ms–2) T2 (b) 5/6

(c) 1

(d) 4/5

ball filled with water. A plugged hole near the bottom of the oscillation bob gets suddenly unplugged. During observation, till water is coming out, the time [AIEEE 2005] period of oscillation would (a) (b) (c) (d)

increase towards a saturation value remain unchanged first decrease and then increase to the original value first increase and then decrease to the original value

77. If a simple harmonic motion is represented by

71. Starting from the origin, a body oscillates simple

1 2p 3

(b) 3 T/2

76. The bob of a simple pendulum is a spherical hollow

undergoes vertical simple harmonic motion of angular frequency w . The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time [AIEEE 2006]

(a)

74. If the length of the pendulum is made 9 times and

(a) 6/5

70. A coin is placed on a horizontal platform which

1 s 12 1 (c) s 4

(d) a / 3

(a) 3 T

connected first in series and then in parallel. A mass M is suspended from them. The ratio of their frequencies of vertical oscillations will be

(a)

(b) a/2

mass of the bob is made 4 times, then the value of time period becomes [BHU 2005]

69. Two identical springs, each of spring constant k are

(a) 2 : 1 (c) 1 : 4 (e) 1 : 2

(a) a / 2 (c) a / ( 3 / 2)

d2 x + ax = 0, its time period is dt2 (a) 2p a

(b) 2pa

2p (c) a

78. The function sin2 (wt) represents

[AIEEE 2005]

2p (d) a [AIEEE 2005]

(a) a simple harmonic motion with a period p / w (b) a simple harmonic motion with a period 2p / w p (c) a periodic with a period , but not simple harmonic 3 motion 2p (d) a periodic with a period , but not simple harmonic 3 motion

79. Two simple harmonic motions are represented by the

p equations y1 = 0.1sin æç100 pt + ö÷ and y2 = 0.1cosp t. è 3ø The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is [AIEEE 2005] p 6 p (c) 3

(a)

-p 3 -p (d) 6 (b)

646 JEE Main Physics 80. A particle is executing simple harmonic motion with

that the system is completely isolated from its surrounding the piston executes a simple haromonic motion with frequency. [JEE Main 2013]

amplitude of 0.1 m. At a certain instant when its displacement is 0.02, its acceleration is 0.5 ms–2. The maximum velocity of the particle is (in ms–1) [BVP Engg. 2005]

(a) 0.01 (c) 0.5

(b) 0.05 (d) 0.25

81. 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 V0 and its pressure is p0 . The piston is slightly displaced from the equilibrium position and released. Assuming

(a)

1 Ag p 0 2p V0 M

(c)

1 2p

A2 g p 0 MV0

(b)

1 V0 Mp 0 2p A2 g

(d)

1 2p

MV0 Ag p 0

82. The amplitude of a damped oscillator decreases to 0.9 times its original magnitude is 5 s. In another 10 s it will decrease to a time its original magnitude, [JEE Main 2013] where a equals (a) 0.7 (c) 0.729

(b) 0.81 (d) 0.6

Answers Round I 1. 11. 21. 31. 41. 51. 61. 71.

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

2. 12. 22. 32. 42. 52. 62. 72.

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

3. 13. 23. 33. 43. 53. 63. 73.

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

4. 14. 24. 34. 44. 54. 64. 74.

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

5. 15. 25. 35. 45. 55. 65. 75.

(b) (a) (a) (b) (b) (b) (b) (b)

3. 13. 23. 33. 43. 53. 63. 73.

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

4. 14. 24. 34. 44. 54. 64. 74.

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

5. 15. 25. 35. 45. 55. 65. 75.

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

6. 16. 26. 36. 46. 56. 66. 76.

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

7. 17. 27. 37. 47. 57. 67. 77.

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

8. 18. 28. 38. 48. 58. 68. 78.

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

9. 19. 29. 39. 49. 59. 69. 79.

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

10. 20. 30. 40. 50. 60. 70. 80.

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

10. 20. 30. 40. 50. 60. 70. 80.

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

Round II 1. 11. 21. 31. 41. 51. 61. 71. 81.

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

2. 12. 22. 32. 42. 52. 62. 72. 82.

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

6. 16. 26. 36. 46. 56. 66. 76.

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

7. 17. 27. 37. 47. 57. 67. 77.

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

8. 18. 28. 38. 48. 58. 68. 78.

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

9. 19. 29. 39. 49. 59. 69. 79.

(b) (c) (b) (a,c,d) (a) (e) (e) (d)

the Guidance Round I 1. We can find the velocities, v1 =

dy1 = 2 ´ 10 cos (10 t + q) dt

6. Given, y = A sin wt

2.

121 100 The time periods, T1 = 2p and T2 = 2p g g

Acceleration

So, T1 > T2. , Let the shorter pendulum makes n vibrations, then the longer pendulum will make less than n vibrations to come in phase again. nT2 = (n - 1)T1

So, or

10n = (n - 111 )

or

3. Given,

a = A cos q and b = A sin q

then

y = A cos q sin wt + A sin q cos wt

…(i)

y = A sin ( wt + q) which is in the form of SHM From Eq. (i) a2 + b 2 = A2 cos2 q + A2 sin 2 q Þ

v=

dy ö æp = 3 ´ 2 w sin ç - 2 wt ÷ ø è4 dt

Acceleration,

A=

dv æp ö = - 4 w2 ´ 3 cos ç - 2wt ÷ = - 4 w2y è4 ø dt

As A µ y and - ve sign shows that it is directed towards equilibrium (or mean position), hence particle will execute SHM Comparing Eq. (i) with equation

w¢ = 2 w or

or

T¢ =

p w

2p =2 w T¢

(as per question)

t=

T 6

Stroke length = 1m \Amplitude of SHM, A =

Now,

Stroke length 1 = 2 2

= 0.5 m v max = wA = 200 ´ 0.5 = 100 m/min

... (i)

Velocity,

we have,

a = a cos wt 2 p wt = 3 2p p ×t = Þ T 3

9. Given, angular frequency of the piston, w = 200 rad/min

vibrations, the smaller pendulum will complete (5 / 4) vibrations. It means the smaller pendulum will be leading the p bigger pendulum by phase T /4 sec = rad = 90° 2

y = r cos ( f - w¢ t )

Þ

Þ

A = a2 + b 2

ö æp y = 3 cos ç - 2wt ÷ ø è4

1 4

7. Given, y = sin3 wt = [3 sin wt - sin 3 wt ]

Þ

4. When bigger pendulum of time period (5T / 4) completes one

5. Given,

= w2A sin( wt + p )

Hence in this case, equation of displacement of particle can pö æ be written as x = a sin ç wt + ÷ = a cos wt è 2ø

y = a sin wt + b cos wt

Let

= A w sin ( wt + p / 2) dv a= = - w2A sin wt dt

8. It is required to calculate the time for extreme position.

n = 11

or

dy = Aw cos wt dt

As this motion is not represented by single harmonic function, hence it is not SHM. As this motion involves sine and cosine functions, hence it is periodic motion.

100 121 = (n - 1) ´ 2p g g

n ´ 2p

v=

\

and v 2 = -3 ´ 10 sin 10t = 30 cos(10t + p / 2) \ Phase difference = (10t + q) - (10t + p / 2) = q - p / 2

10. As, i. e. , or

Wavelength = Velocity of wave ´ time period l = 300 ´ 0.05 l = 15 m

According to the problem path difference between two points = 15 - 10 = 5 m \

Phase difference = Df = =

2p ´ path difference l 2p ´ Dx l 2p 2p ´5 = 15 3

648 JEE Main Physics 11. At one of the extreme position, weight of block = restoring force. At the other extreme position, weight of block and restoring force both act downward direction. So the force on block at is doubled than its weight.

12. As, F1 =

F = F1 + F2 =

4p 2ma 4p 2ma + T12 T22

æ 1 1ö = 4p 2maç 2 + 2 ÷ è T1 T2 ø or

or

m 4 p 2a m 4p 2a and F2 = 2 T22 p

Net force,

æ 1 4p 2ma 1ö = 4p 2ma ç 2 + 2 ÷ 2 T è T1 T2 ø

or

1 1 1 = + T 2 T12 T22

or

1 = T2

T12 + T22 T12T22

or

T2 =

It is an equation of a circle. Thus trajectory of motion of the particle will be a circle.

15. Velocity, v = rw cos wt Þ or

2p 2p 2p 1 cos ´2 = r ´ ´ 0.4 = r ´ 16 16 16 2 r=

0.4 ´ 16 ´ 2 3.2 2 = = 1.44 m 2p p

16. When particle is at x = 2, the displacement is y = 4 - 2 = 2 cm. If t is the time taken by the particle to go from x = 4 cm to 2 pt 2 pt = a cos x = 2 cm, then y = a cos wt = a cos T 1.2 2pt y 2 or cos = = 1.2 a 4 1 p = = cos 2 3 2t 1 or = 1.2 3 1.2 or t= = 0.26 6 Time taken to move from x = +2 cm to x = +4 cm and back again = 2t = 2 ´ 0.2 s = 0.4 s

y a

22 3 ´ = 3.14 s 7 12

T = 2p

displacemant acceleration

So,

T = 2p

x 1 = 2p kx k

19. As,

x = Asin( w+ p / 2) = Acoswt

\

cos wt = x / A

and

sin wt = 1 - ( x2 / A2) y = A sin 2wt = 1 - ( x2 / DA2) y = A sin 2 wt = 2 A sin wt cos wt

or

y 2 = 4 A2 sin 2w t cos2 wt = 4 A2 ´

l , where l is the height of liquid column g in one arm of U tube in equilibrium position of liquid. Therefore, T is independent of density of liquid. x2 + y 2 = a2(cos2 wt + sin 2 wt ) = a2

= 2p

and

with period, T = 2p

\

1/ 2

18. As, d 2x / dt 2 = - kx

13. Motion of an oscillating liquid column in a U tube is SHM

14. x = a cos wt and y = a sin wt

æ 4p 2y ö T=ç ÷ è a ø =2´

T12T22 T12 + T22

- 4p 2 y T2

17. Acceleration, a = - w2y =

x2 æ A2 - x2 ö ´ç ÷ A2 è A2 ø

æ x2 ö = 4x2ç1 - 2 ÷ è A ø

20. Let, O be the position and x be the distance of coin from O. The coin will slip if centrifugal force on coin just becomes equal to force of friction i. e., mxw2 = mmg The coin will slip if, x = maximum = amplitude A m Aw2 = m mg or

A = mg / w2

21. Weight kept on the system will separate from the piston when the maximum force just exceeds the weight of the body. Hence, mw2y = mg or

y = g / w2 = 9.8 /(2p ) 2 = 0.25 m

p ´0 =r 2 p ´3 when t = 3s, x = r cos =0 2 p Here w= 2 2p p or = 2 T

22. When t = 0 , x = r cos

or

T = 4s

\ In 3 s, the particle goes from one extreme to other extreme and then back to mean positing. So, the distance travelled = 2r + r = 3r

Oscillations 23. For displacemant OQ = 40 cm; let t1 be the time taken, then

29. Given, x = 3 sin wt + 4 sin ( wt + p / 3)

2p 40 = 41sin t1 , 24

Comparing it with the equation

t1 = 516 . s

On solving,

For displacement OQ = -9 cm, let t 2 be the time taken, then 2p 9 = 41 sin t1 12 On solving t 2 = 0.84s

x = r1 sin wt + r2 sin( wt + f) We have, r1 = 3 cm, r2 = 4 cm and f = p /3 The amplitude of combination is r = r12 + r22 + 2rr 1 2 cos f = 3 2 + 4 2 + 2 ´ 3 ´ 4 ´ cos p / 3 = 37 » 6 cm

Total time = 5.16 + 0.84 = 6.00 s

24. Given, T = 30 s, OQ = B. The projection of the radius vector on the diameter of the circle when a particle is moving with uniform angular velocity ( w) on a circle of reference is SHM. Let the particle go from P to Q in time t. Then ÐPOQ = wt = ÐOQP. The projection of radius OQ on x-axis will be OR = x (t ) say. P (t = 0)

\

p = - A ´ p sin (0 + f) = - pA sin f - 1 = A sin f Squaring and adding Eqs. (ii) and (iii), we have

25. v max Þ

x

x (t ) In DOQR , sin w t = B 2p 2p x (t ) = B sin wt = B sin t = B sin t T 30 2p = 1 ´ 10 3 = a ´ T v max ´ T a= 2p a=

1 ´ 10 3 ´ 1 ´ 10 -5 = 1.59 mm 2p

26. As, sec bt is not define for bt = p / 2 and

x = a sec bt + c cosec bt =

x a

31. Given, = sin wt and

v max = wr = 30 cm/s Amax = w r = 60 cm/s

\

28.

This is equation of a circle having radius a.

32. For SHM,

d 2y µ -y dx2

33. Here, a = 1cm = 0.01m . The mass will remain in contact with surface, if mg = mw2a or

w= g /a

or

2pn = g / a n=

1 2p

7 980 g = 2 ´ 22 1 a

= 4.9 Hz » 5 Hz

34. As, x = a sin wt and or

y = b sin( wt + p ) = - b sin wt x y b =- Þy=- x a b a

It is an equation of a straight line. 2

35. Maximum force on body while in SHM

w2r 60 2p = 2 or T = ps. = = 2 or w = 2 or wr 30 T

L 2p and w = Time period, T = 2p = g T \Displacement, x = a cos wt = a cos

y = cos wt a

y 2 + x2 = a2

Þ

a sin bt + c cos bt sin bt cos bt

2

…(ii)

y 2 x2 + =1 a2 a2

or

This equation cannot be modified in the form of simple equation of SHM i. e., x = a sin( wt + f) So, it cannot represent SHM

27. Here,

…(ii)

A = 2 cm

Now

or

…(i)

1 = A cos( p ´ 0 + f) = A cos f d [ x (t )] velocity = = - Aw sin ( wt + f) dt

or

ωt x(t) R

O

x (t ) = A cos ( wt + f)

30. As,

1 + 1 = A2(cos2 w + sin 2 w) = A2

Q ωt

649

g t L

g L

= mw2a = 0.5 (2p /2) 2 ´ 0.2 = 1 N Maximum force of friction = m mg = 0.3 ´ 0.5 ´ 10 = 1.5 N Since the maximum force on the body due to SHM of the platform is less than the maximum possible frictional force, so the maximum force of friction will be equal to the maximum force acting on body due to SHM of platform i. e., 1 N.

650 JEE Main Physics 1 2

43. From the figure, AC = l cos q

36. Potential energy, V = mw2x2

O

1 mw2( a2 - x2) 2

and kinetic energy E, T =

θ

T a2 - x2 = V x2

\

T

l

37. Energy of oscillation,E = aA4

(at maximum displacement)

C

Kinetic energy of mass at x = x is K = E - U = a ( A2 - x4)

A

K = 3U

As

a ( A4 - x4) = 3 ax4

OC = OA - AC = l - l cos q = l (1 - cos q) Maximum KE of bob at O = Maximum PE of bob at B = ma ´ OC = mgl (1 - cos q)

1 mw2a2 2 3E 1 KE = = mw2( a2 - y 2) 4 2 E=

38. Total energy,

2

39.

\

A 2

x=±

or

So,

3 a -y = 4 a2

or

y 2 = a2 / 4

or

y = a/2

1 2

44. Kinetic energy, E k = m ( a2 - y 2) =

2

or

é 2 ´ 8 ´ 10 -3 ù =ê 2ú ë 0.1 ´ (0.1) û

Equation of SHM is,

1/ 2

=4

maximum kinetic energy, 1 K0 = mw2r 2 2 æ 2K ö r = ç 02 ÷ è mw ø

Þ

2

2x = a

2

Þ

x=

a y = = 2 2 cm 2 2

41. As,

Þ

U Umax

1 mw2y 2 1 2 = = 1 mw2a2 4 2

y2 1 = a2 4 a y= 2

42. Potential energy of body in SHM at an instant, U=

1 2 ky 2

if the displacement, y = ( a - x), then 1 1 U = k ( a - x) 2 = k ( x - a) 2 2 2

1/ 2

The displacement equation can be æ 2K ö y = r sin wt = ç 0 ÷ è mw ø

40. Let x be point, where KE = PE 1 1 Hence, mw2( a2 - x2) = mw2x2 2 2

1 mw2a2 2

45. If m is the mass and r is the amplitude of oscillation, then

or

pö æ y = a sin( wt + q) = 0.1sin ç 4 t + ÷ è 4ø

1 æ 2 a2 ö 1 3 m ç a - ÷ = mw2a2 2 è 4ø 2 4

Ek / E = 3 / 4

So,

1/ 2

E=

Total energy,

1 Kinetic energy at mean position = mw2a2 = 8 ´ 10 -3 2 æ 2 ´ 8 ´ 10 -3 ö w=ç ÷ ma2 ø è

mg mg cos θ mg sin θ

1/ 2

sin w t

46. Here, mass of body, m = 500 g = 500 ´10 -3 kg Spring constant k = 8 p 2 Nm–1 The frequency of oscillation is v=

1 2p

k 1 = m 2p

8 p 2 Nm–1 = 2 Hz 500 ´ 10 -3 kg

47. Let k be the force constant of spring of length l2. Since, l2 = nl2, where n is an integer, so the spring is made of (n + 1) equal parts in length, each of length l2. 1 (n + 1) = \ k k k = (n + 1)k

or

The spring of length l2( = n l2) will be equivalent to n spring k (n + 1) k = n n

connected in series where spring constant k¢ =

48. As, n =

1 2p

keffective 1 = m 2p

k + 2k 1 = m 2p

3k m

Oscillations 49. We have,

n=

1 2p

k ; m

n¢ =

1 2p

k' 1 = 2m 2p

2k =n 2m

a complete spring, the spring constant k¢ = k / 2 (spring in series). When two splitted parts of a spring are connected to the body, then spring are in parallel. Their effective spring constant, k¢ = k + k = 2 k. T =2p

As

Tµ

or

Putting the value of v = 4 cms–1 and solving, we get x = 2 cm.

56. As, energy stored = work done

m k

1 k

1 2 kr 2

Þ

E=

or

r=

Now,

æ kö 1 a = w2r = ç ÷ ´ è m ø 10

(where, r = displacement)

2E 2 ´2 1 = = m k 400 10 2

(for a fixed value of m)

T¢ k/2 1 = = T 2k 2 T T¢ = 2

\

Using law of conservation of energy, we have 1 1 1 m1v 02 = kx2 + (m1 + m2)v 2 2 2 2

(Qk' = 2k )

50. Let, k be the spring constant of each half part of the spring. For

651

æ 400 ö 1 = 40 ms-2 =ç ÷´ è 1 ø 10

57. Let k be the force constant of the shorter part of the spring of

Therefore slope is irrelevent. Here, the effective spring constant = k + k = 2 k

length l / 3. In a complete spring, three springs are in series each of force constant k. 3k k1 = 2 k 3k \ = =2 k1 3 k / 2

Thus time period, T = 2p M / 2k

or

or

51. It is a system of two springs in parallel. The restoring force on the body is due to springs and not due to gravity.

52. As, T = 2 = 2p

58. Two spring each of spring constant k1 in parallel, given

M k M+ 4 2 + 1 = 2p k

and

3 = 2p

So,

4 M = 9 M+ 4

equivalent spring constant of 2k1 and this is in series with spring of constant k2, so equivalent spring constant

(from questions)

k+ 4 k

or

æ1 1 ö k=ç + ÷ è k2 2k1 ø

or

t AC = t AB + t AC = (T /4) + t AC where, T = time period of oscillation of spring mass system Now, t AB can be obtained from, BC = AB sin(2p /T) tBC BC 1 Putting = AB 2 T We obtain tBC = 12 T T 2p m Þ t AC = + = 4 12 3 k Period of oscillation, m 3m / 4 3m T = 2p = 2p =p k k k

53. A total restoring force, F = kX = mg or

k = mg / X

\

T = 2p

54. As, T = 2p and

(M + m) (M + m) = 2p mg / X mg

m k T ¢ = 2p

m 2m = 2p =T k 2k

60.

S θ

55. The amplitude of oscillations will be the maximum when compression in the spring is maximum. At the time of maximum compression, velocities of both the blocks are equal say v, then using law of conservation of momentum, m1v 0 = (m1 + m2) v or or

1 ´ 12 = (1 + 2) v v = 4 cms-1

-1

59. The total time to go from A to C

4 M + 16 = 9 M 16 M= = 3.2 kg 5

or

k : k1 = 2 : 1

l

T

C O

P θ mg mg cos θ B mg sin θ

652 JEE Main Physics When the bob is displaced to position P, through a small angle q from the vertical, the various forces acting on the bob at P are (i) the weight mg of the bob acting vertically downwards (ii) the tension T in the string acting along PS Resolving mg into two rectangular components, we get (a) mg cos q acts along PA, opposite to tensions, we get (b) mg sin q acts along PB, tangent to the arc OP and directed towards O. If the string neither slackens nor breaks but remains taut, then T = mg cos q The force mg sin q tends to bring the bob back to its mean position O. \Restoring force acting on the bob is F = - mg sin q

where,

mg g mi l

\T =

4 pR 2r 3 4 M pr = 3 3 R GMmx F= R3

Mass of the earth, M = or

F µx

As, this force, F is directed towards the centre of earth i. e. , the mean position. So, the ball will execute periodic motion about the centre of earth.

æ mg g ö ÷ q = - w2q a = -ç è mi l ø w2 =

into the tunnel and reaches on the other side of earth and goes again upto a height h from that side of earth. The ball again returns back and thus executes periodic motion. Outside the earth ball crosses distance h four times. When the ball is in the tunnel at distance x from the centre of the earth, then gravitational force acting on ball is Gm æ 4 ö æ4 ö F = 2 ´ ç px2r ÷ = G ´ ç pr ÷mx è3 ø è3 ø x

i. e. ,

(mi l 2) a = - (mg g ) l q

or

earth in time, t = 2h/g . Its velocity is v = 2gh. It then moves

\

61. Torque acting on the bob = Ia = - (mg ) l sin q or

66. When the ball m falls from a height h, it reaches the surface of

inertia factor = mass of ball = m GMm gm Spring factor = = R R3

Here,

mi l 2p = 2p mg g w

l . When lift is accelerated upwards with g acceleration a ( = g /4), then effective acceleration due to gravity inside the lift g 5g g1 = g + a = g + = 4 4 l l 2 2T \ T1 = 2p = 2p ´ = g 5g /4 5 5

\Time period of oscillation of ball in the tunnel is

63. Since length of pendulums A and C is same and T = 2p l/ g ,

Therefore, total time period of oscillation of ball is

62. As, T = 2p

hence their time period is same and they will have same frequency of vibration. Due to it, a resonance will take place and the pendulum C will vibrate with maximum amplitude.

= 2p

l + 2 l / 100 l

be = 4 2h/g

= 2p

v 0 = 2 gl

R 2h +4 g g

67. When the bob of pendulum is brought is brought to a position

u = 0 , a = g , g = h = l (1 - cos q), v = ? Now, or

v 2 = u 2 + 2 gh = 0 + 2 gl (1 - cos q) f v = 2 gl (1 - cos q)

68. Let T1 and T2 be the time period of shorter length and longer length pendulums nT1 = (n - 1)T2 So

65. According to the law of conservation of

Þ

m R =p gm / R g

Time spent by ball outside the tunnel on both the sides will

1/ 2

2 ö 1 ö æ æ = T ç1 + ÷ ÷ = 2ç1 + è 100 ø è 100 ø 2 1 T¢ -T = = s \ 100 50 Therefore, loss in seconds per day 1 / 50 = ´ 24 ´ 60 ´ 60 = 864 s 2 mechanical energy, we get 1 mv 02 = mgl 2

inertia factor spring factor

making and angle q with the equilibrium position, then height of bob of pendulum will be, h = l - l cos q = l (1 - cos q). Taking free fall of the

64. According to question, T =T

T ¢ = 2p

or

l v0

As

per

question,

0.5 20 = (n - 1) 2p g g n = (n - 1) 40 » (n - 16 )

5n = 6 Hence, after 5 oscillations they will be in same phase.

Hence,

m

n 2p

reapectively.

Oscillations 69. In damped oscillation amplitude goes on decaying exponentially where, b = damping coefficient initially a0 = a0 e- b ´ 100T , 3 T = time of one oscillation 1 or = e-100 bT 3

…(i)

- b ´ 200T

Finally

a = a0 e

or

a = a0[ e-100bt ]2

or

æ 1ö a = a0 ´ ç ÷ è3ø a=

T = 2p

r g tan q

= 2p

l sin q g tan q

= 2p

l cos q g

74. Let the distance of vertical disc c of block be pushed in liquid, when black is floating, then buoyancy force = abxdw g = abxg The mass of piece of wood = abcd

2

[ \from Eq. (i) ]

a0 9

So,

(L2 - d 2)1/ 2 T = 2p g

(Q dw = 1)

æ g ö acceleration = - abxg / abcd = - ç ÷ x è cd ø

Hence, time period,

70. The motion of M is SHM, with length cm = L2 - d 2

T = 2p

dc g

75. We have, T ¢ = 2p l/( g /6) = 6T Hence, the clock will tick in one minute.

C

A

g tan q = rw2 = r ´ 4p 2/ T 2

or or

a = a0 e- bt

653

= 60 / 6 =24.5 times

B

76. When the cylinder is given a small downward displacement,

2d L

say y, the additional restoring force is due to (i) additional extension y, which is, F1 = ky (ii) Additional buoyancy which is F2 = AYd g

L

Total restoring force, - F = F1 + F2 = (k + Adg ) y = new force constant

M

71. On the inclined plane, the effective acceleration due to gravity

\

g ¢ = g cos 30° = g ´ 3 /2 T = 2p

l = 2p g¢

=

1 2p

k + Adg M

force which is less than the frequency of undamped maximum vibration, velocity-resonance takes place ( i. e. , maximum energy) when frequency of external periodic force is equal to natural frequency of undamped vibrations.

1 1 log T = log 2 + log p + log l - log g 2 2 ( \l is constant )

1 æ -2 ö =- ç ÷ ´ 100 = 1% 2 è100 ø

80. For damped motion a = a0e-bt For first case,

T cos q = mg

or For second case,

a0 = a0 eb ´ 6 n

T sin q = mr w2 T sin q rw2 = tan q = T cos q g

a0 = a0 e- b ´ 2 3 1 = e-2b 3

(increase)

73. Resolving tension T in string into two rectangular components, we get

78. Fig. (1) alone represents damped SHM. 79. Motion given here is SHM starting from rest .

% change in time period dT 1 dg = ´ 100 = ´ 100 T 2 g

So,

k¢ k

77. Amplitude resonance takes place at a frequency of external

Differentiating it, we get dT 1 dl 1 dg 1 dg = =T 2 l 2 g 2 g

and

1 2p

2l 3g

72. As, T = 2p l /g Þ

n=

8

or

1 æ 1ö = e-6b = ( e-2b)3 = ç ÷ n = 33 è3ø n

654 JEE Main Physics

Round II Substituting the given values, we get 4 ´ 10 ´ 4.9 Nm–1 K= (0.5) 2

1. Electric intensity at B due to sheet of charge, 1 sq E= 2 e0 Force on the bob due to sheet of charge 1 sq F = qE = 2 e0

O

On the removal of the body the spring is shortened by x

θ T B

As, the bob is in equilibrium, so C F mg F T = = mg OC CB BO Sheet of charge 1 sq /e 0 CB F sq tan q = = =2 = \ 2e 0mg OC mg mg dx = - Aw sin ( wt + p / 4) dt Velocity will be maximum, when

\

mg = kx

Þ

x=

mg 4.9 ´ 10 ´ (0.5) 2 [ from Eq. (i) ] = k 4 ´ 10 ´ 4.9 0.25 = = 0.0625 m = 6.25 cm 4

7. From the figure, T sin q = mL sin qw2 324 = 0.5 ´ 0.5 ´ w2 324 w2 = 0.5 ´ 0.5

2. Velocity v =

or

wt + p / 4 = p / 2 wt = p / 2 - p / 4 = p / 4

or

t = p /4w 2 = 2p

m k

and

3 = 2p

m+2 k

So,

3 m+2 = 2 m

3. Here,

or or

4.

t0 = 2p 9

9m = 4m + 8 m = 1.6 kg t l1 and 0 = 2p g 7

l2 g

l1 æ 7 ö æ 49 ö =ç ÷ =ç ÷ è 81 ø l2 è 9 ø

5. Let h be the depth of in water, then or

Þ

w=

8. As, x1 = a sin( w ´1) = a sin w and Now

x2 = a sin( w ´ 2) - a sin w x2 sin(2 w) - sin w = x1 sin w = sin 2 ´ (2p / 8) - sin 2p / 8

or

x1 1 2 +1 = = x2 2 - 1 ( 2 - 1)( 2 + 1) =

2 +1 = 2.414 = 2.4 2 -1

perpendicular to each other viz acceleration due to gravity g v2 and radial acceleration aR = towards the centre of the R circular path.

Ah r g = mg m 200 h= = = 4 cm Ar 50 ´ 1 2 ar = v R

1 1 n= = T 2p

g

g h

7 980 = = 2.5 Hz 2 ´ 22 4

6.

(from question)

1 - (1 / 2) 2 -1 = = 1 (1 / 2)

h T = 2p g Now,

324 18 = = 36 rad/s 0.5 ´ 0.5 0.5

9. The bob is subjected to two simultaneo us, accelerations

2

\

…(i)

m Time period of oscillation, T = 2p k where m is the mass of body suspended from a spring and K is 4 p 2m spring constant of the spring and K = T2

\Effective acceleration aeff

æv2ö = g +ç ÷ èRø

2

2

\Time period of the simple pendulum T = 2p = 2p

l 2 2

æv ö g2 + ç ÷ èRø

= 2p

l aeff

l g2 +

v4 R2

Oscillations 10. As the length of the scale is 20 cm and it can read upto 50 kg. The maximum extension of 20 cm will correspond to maximum weight of 50 kg ´ 9.8 m/s 2.

655

where, m = mass of the mercury column of length l If r Hg is density of mercury, m = Alr

then \

T = 2p

Alr l = 2p 2g 2Arg

12. When t = 1s, y1 = r sin w ´1 = r sin w when \ F = - kx

Using,

t = 2 s, y 2 = r sin w ´ 2 = r sin 2w y1 r sin w = y 2 r sin 2w 1 1 1 = = 2 cos w 2 cos 2p / T 2 cos 2p / 8 1 1 1 = = = 2 cos p / 4 2(1 / 2) 2

=

|F| = F = kx x = 20 ´ 10 -2 m

Here,

k=

50 ´ 9.8 20 ´ 10 -2

Distance covered in 2nd second = y 2 - y1 = ( 2 - 1)y1

m=

13. As, x = 3 sin 20 p t + 4 cos 20 p t =5

T 2k (0.60) 2 ´ 2450 = 4p 2 4 ´ (3.14) 2

4 é3 ù sin 20 pt + cos 20 pt êë 5 úû 5

= 5 (cos q sin 20 pt + sin q cos 20 pt ) = 5 sin(20 pt + q) It is a SHM of amplitude 5 cm

= 22.36 kg Weight = mg = 22.36 ´ 9.8 = 219.13 N

\

Ratio = 1 : ( 2 - 1)

\

m k 2 2m T = 4p k T = 2p

or

y 2 = 2y1

\

= 2450 N/m We have for loaded oscillation

14. As, F = mg = kx For first case,

11. Density of mercury column = r

k=

Acceleration due to gravity = g

mg 1 ´ 10 N = = 200 Nm-1 x 0.05

For second case, Area of cross-section = A

w=

2h

r= \ and

Clearly,

m¢g 2 ´ 10 = = 0.1 m k 200

v max = rw = 0.1 ´ 10 = 1 ms-1 T ¢ = 2p

l g net

g net = g + a = 10 + 10

Restoring force, one arm

k 200 = = 100 = 10Hz m 2.0

F = - Weight of mercury column in excess of = - (Volume ´ density ´ g) = - ( A ´ 2h ´ r ´ g ) = - 2Argh = - k ´ Displacement in one arm (h) 2Arg = constant = k

(say) (As F = - kx)

F µ-h Hence, motion in SHM, k = 2Arg \Time period, T = 2p

m m = 2p k 2Arg

g net = 20 m /s2 T¢ =

T 2

16. When a mass m is placed on mass M, the new system is of mass = (M + m), attached to the spring. New time period of oscillation, T ¢ = 2p T = 2p

M+m k M k

656 JEE Main Physics Let v = velocity of mass M while passing through the mean position.

mv 2 0.01 ´ 20 = = 0.20 N r 1 Net tension = weight + centrifugal force

Centrifugal force =

v ¢ = velocity of mass (M + m), while passing through the mean position. According to law of conservation of linear momentum Mv = (M + m)v' v = A w = and v ¢ = A¢w¢

At mean position, \

MA w = (M + m) A' w' M T¢ æ M ö w A¢ = ç ´ ´A ÷ A= è M + m ø w' M+m T

or

M+m æ M ö ´A =ç ÷´ è M + mø M =A

M M+m

= (0.01 ´ 10 + 0.20) = 0.30 N

20. If v and v' are the velocities of the block of mass M and (M + m) while passing from the mean position when executing SHM. Using law of conservation of linear momentum, we have mv = (M + m) v ¢ v ¢ = mv / (M + m)

or

Also, maximum PE = maximum KE 1 1 \ k A¢2 = (M + m) v ¢2 2 2 æ M + mö A¢ = ç ÷ è k ø

or

17. The Lissajous figure will be parabola if period ratio is 1 : 2 and

=

phase difference is p / 2. Let x = a sin(2wt + p / 2) and y = b sin wt sin wt = y / b x Now, = sin(2wt + p / 2) = cos 2w / t a

\

= 1 - 2 sin 2 wt = 1 or or

2y b2

or

b2 ( x - a) 2a

It is an equation of a parabola as given in figure.

18. When the force F is applied to one side of block A, let the upper face of A be displaced through distance DL. Then ∆L

θ

A

or

mv (M + m)

mv (M + m)k

16 l1 = =4 1 l T1 = 4 T

t t -T = 4T T t = 4t - 4T

or or or

3 t = 4T t = 4T / 3

22. Let T1 and T2 be the time periods of the pendulum with lengths 1.0 m and 1.21 m respectively T2 = T1

L

θ

l2 1.21 = = 1.1 l1 1

… (i)

Let v1 and v 2 be the vibrations made by two pendulum to swing together.

B

\ F / L2 h= or F = hLDL DL / L So, F µ DL and this force is restoring one. So, if the force is removed, the block will execute SHM. From Eq. (i)

spring factor = hL

Here,

inertia factor = M T = 2p

\Time period,

T1 = T

´

Let after time t , the pendulum be in the same phase. It will be so, then t t t -T = -1 = T1 T T

2

x 2y 2 æ x - aö = 1- = -ç ÷ è a ø a b2 y2 = -

21. As,

1/ 2

M hL

19. When the bob falls through a vertical height of h, the velocity

…(ii)

For the two pendulum to swing together, required condition is v1 - v 2 = 1 or

v1 = v 2 + 1

\

(v 2 + 1)T1 = v 2T2

or

(v 2 + 1)v 2 = T2 / T1 = 1.1 1 1+ = 1.1 v2

or or

1 1 = 1.1 - 0.1 = v2 10

or

v 2 = 10

acquired at the lowest point, v = 2gh = 2 ´ 10 ´ 1 = 20 ms-1

v1T1 = v 2T2

Oscillations 23. If L is the original length of spring, and k is a spring constant of the spring, then L + (5 / k) = l L + ( 4 / k) = h

and \ or

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

l - h = 1/ k k = 1/ ( l - h)

L = (5 h - l) \Length of spring when subjected to tension 9 N is and

= L + 9 / k = (5h - 4l ) + 9 ( l - h) = (5l - 4h)

24.

4.9

240°

120° A

1 2

28. Maximum KE = mw2A2; minimum KE = 0 1 mw2A2 1 2 Averege KE = = mw2A2 2 4 1 æ 2 2ö ç 0 + mw A ÷ 1 2 2 2 Similarly average PE = ç ÷ / 2 = mw A 2 4 ÷ ç ø è 0+

29. As, t1 = 2p

m k1

or

t12 =

2p 2m k1

or

k1 =

4p 2m t12

Similarly,

k2 =

4p 2m t 22

B=A

A

(k1 + k2) =

and It is clear from figure, that B = A, f = 240 =

25. As, T1 = 2p

M k1

4p 2M

or k1 =

4p 2M T22

Velocity, k1k2 4p 2M = 2 k1 + k2 T1 + T22

T = 2p 1 2

M = T12 + T22 keff

1 1 3 2

1 r - y = r2 3 2

or

2

or

3r 2 - 3y 2 = r 2

or

2r 2 - 3y 2 = 0 r=

3 3 ´y = ´4 2 2 = 2 6 cm

27. In equilibrium, T = mg Work done = mg = mgx = or

x=

1 2 kx 2

2mg 2T = k k

Energy stored = mg x = T x =T ´

2T 2T 2 = k k

v=

dx = - Aw sin( wt + q) dt

= - Aw 1 - cos2( wt + q) = - A w 1 - x2/A2 = - w A2 - x2 Here, v = p cms-1, x = 1 cm, w = p s-1 So,

26. As, mw2(r 2 - y 2) = ´ mw2r 2

or

1 1 1 = + t 02 t12 t 22

30. Given, (t) = A cos ( wt + q)

In series combination, keff =

4p 2m t 02

4p 2m 4p 2m 4p 2m = 2 + 2 t 02 t1 t2

\ or

T12

k2 =

and

2p 3

657

or or or

p = - p A2 - 12 ( -1) 2 = A2 - 1 A2 = 2 A = 2 cm

31. Given, acceleration due to gravity on moon ( g m) = 17 . m/s 2 Acceleration due to gravity on earth ( g e) = 9.8 m/s 2 Time period on earth Te = 3.5 s Time period on moon Tm = ? On the surface of the earth, time period = Te \

Te = 2p

l ge

…(i)

On the surface of the moon, time period = Tm \

Tm = 2p

l gm

g e , g m are acceleration due to gravity on the earth and moon surface respectively.

…(ii)

658 JEE Main Physics On dividing Eq. (i) by Eq. (ii), Te 2p = Tm 2p

Force is zero, when cos wt = 0 or wt = l gm ´ l ge

Te = Tm

gm ge

If

Þ

æ ge ö Tm = ç ÷ Te è gm ø

If

Acceleration is maximum if cos wt =1or 2 p or

Putting the values, we get Tm =

9.8 ´ 3.5 1.7

PE = =

1 1 æ3 ö mw2y 2 = mw2 ´ ç r ÷ è4 ø 2 2

2

9 æ1 ö 9 ´ 80 = 45 J ç mw2r 2÷ = ø 16 16 è 2

33. Given, x = A cos ( wt + a) = - A sin (w + a) y = A cos( wt + a)

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

Squaring and adding Eqs. (i) and (ii), we get x2 + y 2 = A2[sin 2( wt + a) + cos2( wt + a)] = A2 It is an equation of a circle. The given motion is anti-clockwise.

34. As,

T = 2p

m k

and

T' = 2p

m T 5 = = s = 2.5 s 4k 2 2

35. Maximum KE = Maximum PE 1 1 1 mv 2 = kx2 = ´ 65 ´ (0.11) 2 2 2 2 or

v2 =

2p t = 2p or T

4T (given) s 4 Velocity is maximum if sin ( wt + p ) =1or wt + p = p /2 p T 2p p or wt = - p = - p / 2 or t = - or t = - s 2 4 T 2 1 2 2 1 2 2 2 PE = m w y = mw a cos wt 2 2 1 KE = m w2a2 sin 2 wt 2

1 Given, mw2r 2 = 80 J; 2 \

(given)

t =T =

= 8.4 s

32.

2p p 3p t = or T 2 2 2p p T t = , then, t = T 2 4 2p 3p 3T t= , then t = s T 2 4

i. e. ,

Þ

p 3p or , 2 2

65 ´ (0.11) 2 or v = 11 . ms-1 650 ´ 10 -3

36. A ball bearing when released a little above the lower limit

If PE = KE, then cos2 wt = sin 2 wt or cos wt = sin wt or tan wt =1 2p p p T or t = or t = s wt = or T 4 4 8

39. As per question, the direction from A to B, i. e. , from A towards mean position O is positive, therefore if a particle starting from A reaches at C, where AC = 3 cm, then its direction of motion is towards the mean position O. Hence, its velocity is positive, acceleration is positive and force is positive. 3 cm B

O

C

4 cm

5 cm

When a particle from B reaches point D, where BD = 4 cm, then its direction if motion is towards BA i. e. , along BO, then velocity, acceleration and force are negative. When particle reaches at B, its velocity becomes zero but its acceleration and force are towards BA, i. e. , negative.

40. Torque is same for both the cases

inside a smooth curved ball, will execute SHM with a definite period.

T = 2p

37. Phase is the state of a particle as regards its position and direction is motion w.r.t. mean position. In the given curve phase is same when t = 1s and t = 5 s. Also phase is same when t = 2 s and t = 6 s.

D

\

I mgd

IA > IB WA < WB

L

mg

38. For the given SHM, the displacement is given by y = a cos wt dy = - w sin wt = aw sin( wt + p ) dt dV Acceleration, A = = aw2 cos wt dt

Velocity,

v=

Force = mass ´ acceleration = - m a w2 cos wt

41. The moment of inertia of a cylindrical rod about axis of wire (i. e. , an axis passing through the centre of rod and perpendicular to its length) is é L2 R 2 ù I =Mê + ú ë12 4 û

Oscillations 42. As in torsional vibrations of this type restoring torque is directly proportional to angular displacement (e.g., t = - kq, where, k = restoring torque per unit twist = spring factor), hence the vibrational are angular harmonic oscillations.

43. As torsional vibrations are angular SHMs, hence m will be replaced by I, the moment of inertia of oscillating body. l T = 2p k

\

44. We can write,

v 2x0

2

= w (A

2

- x02)

or

v 2x A = 20 + x20 w

or

A = 25 2 cm

(\w = 4)

We have, \

\

(\ A = 25 2 )

v(0) = Aw cos f 0 p 100 1 cos f 0 = = \ f0 = 4 Aw 2

where, I = ml 2 d 2q g bn = a = - sin q + 2 l l dt

q = q0 e

\

l g

Tµ l DT 1 Dl = T 2 l DT 1 = ´ 3 = 15 . % T 2

\ Þ

Þ

bt 2

q = q0 e 2 e bt =1 2 2 t= b

f + f2 ö é æ æ f - f2 ö ù X1 - X 2 = A ê2 sin ç wt + 1 ÷ sin ç 1 ÷ è è 2 ø úû 2 ø ë æ f - f2 ö A = 2 A sin ç 1 ÷ è 2 ø f1 - f 2 p = 2 6 p f1 - f 2 = 3

53. Law of conservation of momentum gives, \ Now,

Mv max = (m + M) v new and v max = A1w1 Mv max v new = (m + M) v new = A2 × w2 MA1 k k = A2 (m + M) M (m + M)

given material and thickness so, inversely proportional to its 1 length, i. e., k µ l = w2A = 4p 2v 2A = 4p 2 ´ (60) 2 ´ 0.01 = 144p 2 ms-2

-

52. X1 = A sin ( wt + f1) = X 2 = A sin ( wt + f 2)

49. Correct explanation of the assertion is that spring constant of a

50. Maximum acceleration

sin ( wt + f)

bt

pö æ y = 25 2 sin ç 4 t + ÷ è 4ø

48. Time period of simple pendulum of length l is

bt 2

According to question, in t time (average life time) 1 angular amplitude drops to value of its original value ( q) e

Þ

steps the frequency of marching steps of soldiers may match with the natural frequency of oscillations of the suspended bridge. In that situated resonance will take place then the amplitude of oscillation of the suspended bridge will increase enormously which may cause the collapsing of the bridge. To avoid such situation the soldiers are advised to break steps on suspended bridge.

-

\ Angular amplitude will be q0 × e

pö æ x (t ) = 25 2 sin ç 4 t + ÷ è 4ø

T = 2p

= - mgl sin q + mbvl bn Ia = - mgl sin q + l

\

47. If the soldiers while crossing a suspended bridge march in

\

\ Net restoring torque when angular displacement q, is given by

\

The equation of motion

or

\ Retardation force = mbv

For small damping the solution of the above differential equation will be

v(t ) = Aw cos( wt + f 0) x (0) = A sin f 0 = 25 25 1 sin f 0 = = A 2

51. As, retardation= bv

\

45. Let x (t) = A sin( wt + f 0) and

As during SHM the direction of deflection is opposite of displacement. It may be +ve or -ve. Hence, maximum acceleration = ±144p 2 ms-2

\

2

659

A2 = A1 Þ

M (m + M)

A1 æ m + M ö =ç ÷ A2 è M ø

1/ 2

660 JEE Main Physics 54. Acceleration a = w2x

61. When lift acceleration upwards with acceleration g /3, then effective acceleration due to gravity is g ¢ = g + g /3 = 4g /3.

2

\

4 p2 aT w2xT æ2 p ö = = w2T = ç ÷ T= è T ø x x T

Now,

T = 2p

l g

and

T ¢ = 2p

3 3 l l = 2p = T 2 g 2 ( 4 g /3)

It is a constant term for SHM i. e. ,it does not change with time.

55. According to question, |v max| = | amax| wa = w2a 2p w =1= T T = 2p T = 2 ´ 3.14 T = 6.28 s

i. e. , Þ \ Þ Þ

62. As, y = a sin wt = a sin

when t = 2 s, let y = y1. Then a é 2p ù æpö y1 = a sin ´ 2 = a sin ç ÷ = è 4ø 2 ëê 16 ûú

56. The particle completes one oscillation in time T. Therefore, in time 2T, it will complete two oscillations and will reach to its starting point, i. e. , initial position. Therefore, the displecement is zero.

57. Here the two springs are in series. Their effective spring constant, k =

k1 k2 k1 + k2 M T = 2p k

T = 2p

l g

Since, velocity = w a2 - y12 \

a2 p a æ 2p ö = ´ 4 = ç ÷ a2 è 16 ø 2 8 2

or

a=

velocity, …(i)

…(ii)

g+a æ aö = ç1 + ÷ è gø g

\

a 4 = 1+ g

or

a=3g

1/ 2

60. Equation, is y = 2 (sin 5pt + 3 cos 5pt) é1 ù 3 = 2 ´ 2ê sin 5pt + cos 5pt ú 2 ë2 û p p é ù = 4 cos sin 5pt + sin cos 5pt 3 3 ëê ûú

It represents a SHM. Its amplitude is 4.

pö dx æ = - x0 w sin ç w t - ÷ è 4ø dt pö dv æ = - x0 w2 cos ç wt - ÷ è 4ø dt

= x0 w2 cos [ w t + 3p / 4)] Comparing it with acceleration, a = A cos( w t + d)

59. As, total energy, E = 2p 2m2a2, i. e. , E µ a2

pö æ = 4 sin ç5 p t + ÷ è 3ø

v=

= x0 w2 cos [ p + ( w t - p / 4)]

Dividing Eq. (i) by Eq. (ii), we have 2=

32 2 m p

x = x0 cos( wt - p / 4)

Acceleration, a =

When lift moves upward with acceleration a, let time period becomes T / 2. Then T l = 2p 2 g+a

velocity = 4 ms-1

and

63. Given,

(k + k ) 16 = 2p = 8p 1 2 k1k2 / (k1 + k2) k1 k2

…(i)

After 4 second from mean position, a y1 = 2

\The period of oscillation,

58. Here,

2p t T

we have,

A = x0 w2, d = (3p / 4)

64. Kinetic energy of the particle executing SHM at an instant when its displacement from the mean position is y, is 1 K = mw2( a2 - y 2) 2 1 At the mean position, y = 0; K1 = mw2a2 2 At the extreme position, y = a; 1 K2 = mw2( a2 - a2) = 0 2 K + K2 æ 1 ö1 \ Average kinetic energy = 1 = ç mw2a2 + 0 ÷ ø2 è2 2 1 mw2a2 4 1 = m( 4p 2v 2) a2 = mp 2v 2a2 4 =

Oscillations 65. Let displacemant equation of particle executing SHM is y = a sin wt As particle travels half of the amplitude from the equilibrium position, so a y= 2 a Therefore, = sin wt 2 1 p or sin wt = sin 2 6 p p or or t = wt = 6 6w 2p ö p æ ç as w = ÷ è 2 p T ø æ ö 6ç ÷ èT ø T or t= 12 Hence, the particle travels half of the amplitude from the T equilibrium in seconds. 12 t=

or

66. Let the minimum amplitude of SHM is a. Restoring force on

Restoring force is balanced by weight mg of block. For mass to execute simple harmonic motion of amplitude a. \

ka = mg mg a= k

or Here,

m = 2 kg , k = 200 Nm-1, g = 10 ms-2 a=

2 ´ 10 10 10 = m= cm = 10 cm 200 100 100

Hence, minimum amplitude of the motion should be 10 cm, so the mass gets detached from the pan.

67. Kinetic energy = Potential energy Þ Þ

1 1 mw2( a2 - y 2) = mw2y 2 2 2 a y= 2

68. Let x be the extension in the wire, when mass m hangs from the end of spring. Then F F F YA y = ´ or ´ A x x L The wire and the spring are in series, so total spring constant, YA k´ kk¢ L = kY A K= = k + k¢ k + YA / L kL + Y A As time period, T = 2p

m k

\

m m(kL + YA) = 2p kYA / (kL + YA) kYA

T = 2p

k ´k k = k+k 2

kp = k + k = 2k In series, frequency of vertical oscillation,

In parallel,

vs =

1 2p

ks M

In parallel, frequency of vertical oscillation, vp =

1 2p

kp M

vs k k/2 1 = s = = vp kp 2k 2

\

70. The coin will be leaving its constant at the lowest point when restoring force is equal to or greater than the weight of the coin, i. e. , mw2 ³ mg or a ³ g / w2

71. As per question, 75 75 1 E= ´ m a2w2 100 100 2 1 75 1 m a2w2 cos2 w t = ´ m a2w2 2 100 2 3 cos2 wt = 4 3 p p cos w t = = cos or w t = 2 6 6 p p 1 t= = = s 6p 6(2p / 2) 6 KE =

\

spring F = ka

ks =

69. In series,

661

or or or

72. Given, w a2 - u2 = w2y 2p y T 2py 2p ´1 2p = = s 2 2 2 2 3 a -y 2 -1

or

a2 - y 2 = w y =

or

T=

73. Kinetic energy of a particle while displacement is y, will be = or or

74. As, T = 2p

1 3 3 1 mw2( a2 - y 2) = E = ´ mw2a2 2 4 4 2 3 a2 2 2 a -y = 4 2 a a or y = y2 = 2 4 l i. e. , T µ l g

or

T¢ 9l = =3 T l

or T' = 3T

dy = 2kt dt d 2y and = 2k = 2 ´ 1 = 2 ms-2 dt 2 So point of suspension of pendulum is moving upwards with acceleration a = 2 ms-2.

75. Given, y = kt 2,

662 JEE Main Physics Then, effective acceleration due to gravity on pendulum, g ¢ = ( g + a) = 10 + 2 = 12 ms-2

and

l g

T2 = 2p

l g'

T12

Þ

76. T = 2p

T1 = 2p

T22

=

g ' 12 6 = = g 10 5

l ; so T µ l g

80. Accleration, a = 0.5 ms-2 amplitude, a = 0.1 a displacement, y = 0.1 m Using the formula of maximum acceleration a = w2y 0.5 or = 25 0.5 = w2 ´ 0.02 or w2 = 0.02 So, w=5 Now, maximum velocity is v = aw = 0.1x5 = 0.5 ms-1

81. Pressure applied by piston Mg = p0 A

When plugged hole near the bottom of the oscilliating bob gets suddenly unplugged, water flows out, the value of l increase because when level of the position of centre of gravity of bob and water falls down. It is so water falls upto the centre of the bob. After that as water comes out, the position of the centre gravity of bob and water rises and finally it reaches at the centre of bob when whole water leakes out of bob. Therefore, T first increases and then decreases to the original value.

Mg = p0 A

…(i) x

x0

77. The equation of SHM, d 2x d 2x + ax = 0 or = - ax 2 dt dt 2 Comparing it with the equation of SHM d 2x = - w2x = ax dt 2 w2 = a or w = a 2p 2p or T= = w a 1 - cos 2wt 1 1 = - cos 2 w t 2 2 2 dy 1 v= = ´ 2w sin 2wt = w sin wt dt 2 dv Acceleration = = 2w2 cos 2w t dt

78. Given, y = sin2 wt =

As accleration is directly proportional to displacement and directed towards the mean position, hence motion is not in SHM but a periodic motion. Its period, 2p p T' = = 2w w

79. As, y1 = 0.1sin(100 pt + p / 3) dy1 = 0.1 ´ 100 p cos (100 pt + p /3) dt and y 2 = 0.1cos pt = 0.1sin ( pt + p /2) dy \Velocity, v 2 = 2 = 0.1 ´ p cos ( pt + p /2) dt

\Velocity, v1 =

\ Phase difference of the velocity of particle 1 w.r.t. particle 2. = (100 pt + p / 3) - ( pt + p / 2) = 99pt + p / 3 = p / 2 p p p At t = 0 , Phase difference, = - = 3 2 6

Here, the system is completely isolated, so the process will be adiabatic. p0V0g = pV g p0 Ax0g = pA ( x0 - x ) g p=

p0 x0g ( x0 - x) g

Let piston is displaced by x æ p xg ö Mg - ç 0 0 g ÷ A = frestoring è ( x0 - x) ø æ x0g ö p0 A ç1÷ = Frestoring è ( x0 - x) g ø g p Ax F=- 0 x0 \

f=

1 g p0 A = 2p x0M

1 2p

82. Amplitude of damped oscillator A = A0e After 5 s, 0.9 A0 = A0 e

-

g p0 A2 MV0

bt 2m

b (5 ) 2m

0.9 = e Þ After 10 more second

-

b (5 ) 2m

A = A0 e

-b

…(i) (15 ) 2m

æ - 5b ö A = A0 çç e 2m ÷÷ è ø From Eqs. (i) and (ii) A = 0.729 A0 Hence,

-

( x0 - x » x0)

a = 0.729

3

…(ii)

16 Waves JEE Main MILESTONE < < < <
= rw2 A2 < cos2 (kx - wt ) >

transverse waves by means of an oscillator coupled to one end. The string has a diameter of 4 mm. The amplitude of the oscillation is10 -4 m and the frequency is 10 Hz. Tension in the

1 = rw2 A2 2

Power (P) If we consider a transverse wave on a string, then the instantaneous rate at which energy is transferred along the string is called power. Thus, P = Energy density ´ Volume 1 = rw2 A2 ´ Dv 2 1 = rw2 A2Dv 2

Intensity (I) Flow of energy per unit area of cross-section of the string in unit time is known as intensity of the wave. Thus, P Power I = = Area of cross - section S 1 or I = rw2 A2v 2

Note The intensity of sound waves is given by I=

2 Pmax

2rv

where, Pmax is the maximum change of pressure in the medium. The intensity of waves emitting in all directions due to a point source varies inversely as the square of the distance (r). 1 i.e., Iµ 2 r The intensity of waves from a linear source varies inversely as the 1 distance (r) perpendicular to the source, i.e., I µ r

with standard equation

y ( x, t ) = a sin (kx - wt )

Sample Problem 7 A stretched string is forced to transmit

string is 100 N and mass density of wire is 42 . ´ 103 kgm -3. Find (a) the equation of the waves along the string (b) the energy per unit volume of the wave (c) the average energy flow per unit time across any section of the string

Interpret (a) Speed of transverse wave on the string is, v=

T rS

(as m = rS)

Substituting the values, we have v=

100 pö æ ( 4.2 ´ 10 3) ç ÷ ( 4.0 ´ 10 –3) 2 è 4ø

= 43.53 ms–1 w = 2 pf = 20 p rad/s = 62.83 rad /s w k = = 1.44 m–1 v \ Equation of the waves along the string y ( x, t ) = A sin (kx - wt ) = (10 -4 m) sin [(1.44 m–1) x - (62.83 rad s–1) t ] (b) Energy per unit volume of the string, 1 u = energy density = rw2A2 2 Substituting the values, we have æ 1ö u = ç ÷ ( 4.2 ´ 10 3) (62.83) 2 (10 -4) 2 è2ø = 8.29 ´ 10 –2 Jm–3

668 JEE Main Physics (c) Average energy flow per unit time

16.4 Sound Waves

P = power æ1 ö = ç rw2A2÷ ( Sv) = (u) ( Sv) è2 ø

Sound wave is defined roughly as any longitudinal wave. These waves travel with a speed of 332 ms–1 (approx). They are classified into following categories

Substituting the values, we have æpö P = (8.29 ´ 10 –2) ç ÷ ( 4.0 ´ 10 –3) 2 ( 43.53) è 4ø

(i) Infrasonics The longitudinal waves having frequencies below 20 Hz are called infrasonics. These waves cannot be heard. These waves can be heard by snakes.

= 4.53 ´ 10 –5 Js–1

Sample Problem 8 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 the amplitude, frequency and velocity of the wave. s–1,

cms–1

(a) 8 cm, 0.8673 cycle 200 (b) 5 cm, 0.673 cycle s–1, 200 cms–1 (c) 5.8 cm, 0.673 cycle s–1, 250 cms–1 (d) None of the above

(ii) Audible waves The longitudinal waves having the frequency between 20 Hz and 20000 Hz are called audible waves. Human can hear these waves. (iii) Ultrasonics The longitudinal waves having the frequencies above 20000 Hz are called ultrasonics. These waves are also called supersonic waves or supersonics.

Check Point 1

Interpret (b) Given, y = 5 sin ( 4.0 t - 0.02 x ). Comparing this with the standard equation of wave motion, y = A sin æç2 pft è

2p ö x÷ l ø

where, A, f and l are amplitude, frequency and wavelength respectively. Thus, amplitude A = 5 cm, 2 pf = 4 4 Frequency, f= = 0.673 cycle s–1 2p

2p = 0.02 l 2p or wavelength, l = = 100 p cm 0.02 4 2p Velocity of the wave v = f l = ´ 2 p 0.02 Again,

= 200 cms–1

1. What type of mechanical waves do you expect to exist in (a) vacuum, (b) air, (c) inside the water, (d) rock and (e) on the surface of water?

2. Is it possible to have longitudinal waves on a string? 3. What is the phase difference between the waves y = A cos ( wt + kx ) and y = A sin ( wt + kx ) ?

4. Why is sound heard in CO 2 more intense in comparison to sound heard in air?

5. What will be the speed of sound in a perfectly rigid rod? 6. What do mechanical waves transfer energy, matter, both or neither?

7. Which characteristic of the medium determines the velocity of sound waves?

8. The speed of sound waves depends on temperature but speed

Sample Problem 9 In above example, find the maximum transverse speed and acceleration of a particle in the rope. (a) 40 cms–2 (b) 50 cms–2 (c) 80 cms–2 (d) 75 cms–2

16.5 Reflection and Transmission of Waves

Interpret (c) Transverse velocity of the particle, ¶y u= = 5 ´ 4 cos ( 4.0 t - 0.02 x ) ¶t = 20 cos ( 4.0 t - 0.02 x ) Maximum velocity of the particle = 20 cms–1 Particle acceleration, a =

of light waves does not. Why?

¶y = 20 ´ 4 cos ( 4.0 t - 0.02 x) ¶t

Maximum particle acceleration = 80 cms–2

When sound waves are incident on a boundary separating two media, a part of it is reflected back into the initial medium while the remaining is partly absorbed and partly transmitted into the second medium.

Characteristics (i) In case of reflection and transmission of sound, the frequency of the wave remains unchanged, i.e., wi = wr = wt = w.

Waves (ii) The incident ray, the reflected ray, normal and the refracted ray are always in the same plane. (iii) In case of reflection of sound, angle of incidence = angle of reflection (iv) In case of refraction of sound,

sin i vi = sin r ¢ vt (v) In case of reflection from a denser medium or rigid support or fixed end there is inversion of the reflected displacement wave, i.e., if the incident wave is y = At sin (wt - kx) the reflected wave will be y = - Ar sin (wt + kx) = Ar sin (wt + kx + p ) i.e., in case of reflection from a denser medium, displacement wave changes in phase by p while in case of reflection from a rarer medium, no inversion of wave or phase change occurs. The transmitted wave is never inverted. (vi) On reflection, the amplitude and intensity of wave may decrease. (vii) When a transverse wave is reflected from a denser medium, the trough is reflected as crest and vice-versa. (viii) When a transverse wave is reflected from a rarer medium, crest and trough do not invert after reflection.

16.6 Principle of Superposition of Waves Two or more waves can traverse the same space independently of one another. Thus, the resultant displacement of each particle of the medium at any instant is equal to the vector sum of displacements produced by the two waves separately. This principle is called principle of superposition of waves.

Interference When two waves of the same frequency, superimpose each other, there occurs redistribution of energy in the medium which causes either a minimum intensity or maximum intensity. This phenomenon is called interference of waves. Let at a given point two waves arrive with a phase difference f and the equation of these waves is given by y1 = A1 sin wt, y2 = A2 sin (wt + f ) Then by principle of superposition y = y1 + y2 y1 = A1 sin wt + A2 sin (wt + f )

Þ

= A sin (wt + f )

(ix) When a longitudinal wave is reflected from a denser medium, the compression and rarefaction do not invert after reflection.

A2

A

(x) When a longitudinal wave is reflected from a rarer medium, compression is reflected as rarefaction and vice-versa. (xi) Waves on reflection from a fixed end undergo a phase change of 180°.

θ

φ A1

Incident wave

where,

A=

Reflected wave

tan q = (xii) While a wave reflected from a free end is reflected without a change in phase.

669

Incident wave

Reflected wave

A12 + A22 + 2 A1 A2 cos f A2 sin f A1 + A2 cos f

If I1 and I 2 are intensities of the interfering waves and f is the phase difference, then the resultant intensity is given by I = I1 + I 2 + 2 I1I 2 cos f Now,

Imax = ( I1 + I 2 ) 2

(xiii) In case of pressure wave, there is no phase change when reflected from a denser medium or fixed end.

for

Note The concept of rarer or denser medium for a wave is through

for

speed (and not density of medium). For example, water is rarer for sound and denser for light than air, as for soundvw > v a , while for lightvw < v a .

Moreover, Intensity (I ) µ Amplitude ( A) 2

f = 2 pn Imin = ( I1 - I 2 ) 2 f = (2 n + 1) p

670 JEE Main Physics Sample Problem 10 Two coherent sound sources are at distances x1 = 0.2 m and x 2 = 0.48 m from a point. The intensity of the resultant wave at that point if the frequency of each wave is f = 400 Hz and velocity of wave in the medium is v = 448 ms -1 is (The intensity of each wave is I 0 = 60 Wm -2 .) (a) 120 Wm–2 (c) 130 Wm–2

(b) 125 Wm–2 (d) 135 Wm–2

Interpret (a) Path difference, Dx = x2 - x1 = 0.48 – 0.28 m Phase difference, 2p æ 2 pf ö Dx = ç ÷ Dx è v ø l 2 p ( 400) (0.28) p = = 448 2

f=

I = I1 + I2 + 2 I1I2 cos f or

I = I0 + I0 + 2 I0 cos ( p /2) = 2 I0 = 2 (60) = 120 Wm–2

Important Points 1. In this equation, it is seen that a particle at any particular point x executes simple harmonic motion and all particles vibrate with the same frequency.

2. The amplitude is not the same for different particles but varies with the location x of the particle.

3. The points having maximum amplitudes are those for which 2a sinkx has a maximum value of 2A, these are at the positions, kx = p /2 , 3p /2 , 5 p /2 , ¼ or

x = l /4 , 3l /4 , 5 l /4 , ¼

These points are called antinodes.

4. The amplitude has a minimum value zero at positions where, kx = p , 2 p , 3 p , ¼ or

x = l / 2 , l , 3l / 2 , 2 l , ¼

These points are called nodes.

5. Energy is not transported along the string to the right or to the left, because energy cannot flow past the nodes points in the string which are permanently at rest.

16.7 Standing or Stationary Waves

6. Standing wave is an example of interference. Node means

A standing wave is formed when two identical waves travelling in the opposite directions along the same line, interfere.

7. Due to persistance of vision these waves appear in the form of loops.

On the path of the stationary wave, there are points where the amplitude is zero, these points are known as nodes. On the other hand, there are points where the amplitude is maximum, these points are known as antinodes.

8. Stationary waves may be transverse or longitudinal. 9. As in stationary waves nodes are permanently at rest, so energy

l

l

The distance between two consecutive nodes or two consecutive antinodes is l /2. The distance between a node and the next antinode is l /4.

Consider two waves of the same frequency, speed and amplitude, which are travelling in opposite direction along a string. Two such waves may be represented by the equations

destructive interference and antinode means constructive interference. All the particles in a loop are in the same phase. But the particles in adjacent loops differ in phase by p.

cannot be transmitted across them.

10. Two identical waves moving in opposite directions along the string will still produce standing waves even if their amplitudes are unequal (as shown in figure).

Antinode Node

y1 = A sin (kx - wt ) and

y2 = A sin (kx + wt )

Hence, the resultant may be written as

The standing wave ratio (SWR) is defined as A = max Amin =

y = y1 + y2 = A sin (kx - wt ) + A sin (kx + wt ) y = 2 A sin kx cos wt This is the equation of a standing wave.

A1 + Ay A1 - Ay

For 100% reflection SWR = ¥ and for no reflection SWR = 1

Waves i. e.,

16.8 Fundamental Mode and Harmonics A string of length L is stretched between two points. When the string is set into vibrations, a transverse progressive wave begins to travel along the string. It is reflected at the other fixed end. The incident and the reflected waves interfere to produce a stationary transverse wave in which the ends are always nodes. Various modes of vibrations of a stretched string are shown below

(iii) The same string under the same conditions may also vibrate in three segments. N

\

second overtone. Thus, a stretched string in addition to the fundamental mode, also vibrates with frequencies which are integral multiples of the fundamental frequencies. These frequencies are known as harmonics.

N

A

N

A

N

L (b)

If f2 is the frequency of vibrations, then the velocity of transverse waves is given as, v = l 2 f2 v = Lf2

or

f2 = v/ L 1 T f2 = L m

v = pf1 2L

It is called the (p – 1)th overtone or the pth harmonic.

\ l2 = L

\

2L p

and frequency is given by

1 T 2L m

is minimum and is called fundamental frequency. The sound or note so produced is called fundamental note or first hormonic.

l2 2

l3 2 \ l3 = L 2 3

fp = p

string

L =2

N

The frequency f3 is known as the third harmonic or

…(i)

This (first) normal made of vibration is called fundamental mode. The frequency of vibration ( f1 ) of

\

L=3

lp =

v = 2 Lf1 T v= m

(ii) The same string under the same conditions may also vibrate in two loops, such that the centre is also the node.

A

(iv) In general, when the string vibrates in p loops, wavelength of the pth mode of vibration is given by

\ Frequency of vibration v f1 = 2L

f1 =

N

A

If f3 is the frequency in this mode of vibration, then 2 v = l3 f3 \ v = Lf3 3 3v or …(iii) f3 = 2L

\ l1 = 2 L

v = l1 f1

As we know,

N

L

The velocity of transverse waves is given as,

Þ

A

(c)

(i) In the simplest form, when the N A N string is plucked in the middle, it vibrates in one loop in which L the ends are the nodes and the (a) centre is the antinode. This mode of vibration is known as the fundamental node and the frequency of vibration is known as the fundamental frequency of first harmonic. Since, the distance between consecutive nodes is l/2. l1 2

f2 = 2 f1

It is called the first overtone or second harmonic. It is of twice the frequency of fundamental and is called an octave higher than the fundamental frequency.

Stationary Waves in Strings

L=

671

…(ii)

Note 1. Harmonics on a string When you need to obtain information on a stretched string of given length l, we draw harmonics. If you are asked about, say, the Vth harmonic, you need to draw five loops between the fixed support points. That would mean that five loops, each of length l / 2, occupy the length l of the string. Thus, 5 ( l / 2) = l and l = 2l / 5. You can then use n = v / l to find the frequency of the harmonic. Keep in mind that the wavelength of a harmonic is set only by the length l of the string, but the frequency depends also on the wave speed v, which is set T . by the tension and the linear density of the string via, v = m 2. If a string is vibrating in nth mode of vibration, then (a) the number of harmonics = n (b) fn = nf , where f is frequency of first or fundamental mode of vibration. (c) the number of loops = n (d) the number of antinodes = n (e) the number of nodes = n + 1 (f) the number of overtones = n - 1 But at n = 1, overtone is fundamental.

672 JEE Main Physics Laws of Vibration of Stretched String (i) Law of length For a given wire under a given tension, the frequency of wire varies inversely as its vibrating length i.e., 1 v µ or v1l1 = v2l2 l

(c) The particle velocity is equal to æ ¶y ö = 4 sin æ px ö (96 p) ( - sin96 pt ) ç ÷ ç ÷ è15 ø è ¶t ø px = -384 sin æç ö÷ sin (96 pt ) è15 ø x = 7.5 and t = 0.25, we get æ ¶y ö = -384 p sin æ px ö sin (96 pt ) ç ÷ ç ÷ è15 ø è ¶t ø

at,

where, T and m are constants. (ii) Law of tension For a uniform wire of given length and material, the frequency of the wire varies directly as the square root of tension vµ T or

v1 = v2

æ T1 ö ç ÷ è T2 ø

p = -384 p sin æç ö÷ sin (24 p) = 0 è2 ø (d) The equations of the component waves are px y1 = 2 sin æç + 96 pt ö÷ è15 ø px y 2 = 2 sin æç - 96 pt ö÷ è15 ø

and

where l and m are constants. (iii) Law of mass When l and T are constants, the frequency of vibration of the wire varies inversely as the square root of mass per unit length of the wire i.e.,

1 vµ m where l and T are constant. 1 1 So, vµ µ l m

Stationary Waves in Air Column Open pipe If both ends of a pipe are open and a system of air is directed against an edge, standing longitudinal waves can be set up in the tube. The open end has a displacement antinode First harmonic A

Hence, a graph between l and m is a straight line.

Sample Problem 11 The vibrations of a string of length 60 cm fixed at both ends are represented by the equation æ px ö y = 4 sin ç ÷ cos (96 pt), where x and y are in cm and t in sec. è15 ø (a) What is the maximum displacement of a point at x = 5 cm? (b) Where are the nodes located along the string? (c) What is the velocity of the particle at x = 7.5 cm and at t = 0.25 s ? (d) Write down the equations of the component waves whose superposition gives the above wave.

px px é ù y = 2 ê sin æç + 96pt ö÷ + sin æç - 96 pt ö÷ ú è15 øû ø ë è15

Thus, the waves are of the same amplitude and frequency but travelling in opposite directions which thus, superimpose to give a standing wave, (a) At x = 5 cm the standing wave equation gives y = 4 sin ( 5p /15) cos (96 pt ) = 4 ´ 3 / 2 cos (96 pt ) \ Maximum displacement = 2 3 cm (b) The nodes are points permanently at rest. Thus, they are those points for which i. e.,

sin ( px /15) = 0 px /15 = np, n = 0, 1, 2, 3, 4, …

x = 15n, i. e., at x = 0, 15, 30, 45 and 60 cm.

λ —3 2

N

λ N —1 2

L

A

λ —2 2

A N

A

λ —3 2

λ —2 4

N

(a)

A N

λ —3 A 4

A

(b)

(c)

(i) For fundamental mode or first harmonic,

l1 \ l1 = 2 L 2 v = l1 f1 \ v = 2 Lf1 v f1 = 2L L=

Interpret Given, y = 4 sin ( px /15) cos (96 pt ). It can be broken up into

Second Third harmonic harmonic A A λ3 λ — —2 N 4 4

velocity, Þ

(ii) For the second harmonic or first overtone, L = l2 velocity, v = l 2 f2 \ v = Lf2 v f2 = Þ 2L

...(i)

…(ii)

(iii) For the third harmonic or second overtone,

L = 3´

l3 2 \ l3 = L 2 3

velocity,

v = l3 f3 \ v =

Þ

f3 =

3v 2L

2 Lf3 3

…(iii)

Waves From Eqs. (i), (ii) and (iii), we get f1 : f2 : f3 : ¼ = 1 : 2 : 3 : ¼ i.e., for a cylindrical tube, open at both ends, the harmonics excitable in the tube are all integral multiples of its fundamental. 2L In the general case, l = , where n = 1, 2, ¼ n v nv Frequency, f = = , where n = 1, 2, ¼ l 2L

In open organ pipe, all (even and odd) harmonics are present. Ratio of harmonics is f1 : f 2 : f3 ¼ = 1 : 2 : 3 ¼ and ratio of overtones = 2 : 3 : 4 : 5 l 3l 5l Position of nodes from one end x = , , ,¼ 4 4 4 3l l Position of antinodes from one end x = 0, , l, ,¼ 2 2

If one end of a pipe is closed, the reflected wave is 180° out of phase with the incoming wave. The displacement of the small volume elements at the closed end must always be zero. Hence, the closed end must be a displacement node. First harmonic

A λ —2 4

λ —1 4

L

N (a)

Fifth harmonic

Trird harmonic A

N A

λ —2 2

λ —3 4

velocity, \

v = l 2 f2 v f2 = l2

Þ

f2 =

3v 4L

(iii) It is the fifth harmonic or second overtone, l L = 5´ 3 4 4 l3 = L \ 5 Velocity, v = l3 f3 4 v = Lf3 \ 5 5v f3 = Þ 4L

...(ii)

...(iii)

N

(b)

(c)

ends, harmonic mode of the pipe which resonates at 1.1 kHz source is [Given, speed of sound in air is 330 ms -1] [NCERT] (a) first (c) third

v = l1 f1

(ii) It is the third harmonic or first overtone, 3 l2 L= 4

(b) second (d) four

Interpret (b) The first harmonic frequency is

l1 = 4 L

v = 4 Lf1 v f1 = 4L

3l l , l, ,¼ 2 2

Sample Problem 12 A pipe, 30 cm long is open at both A

N

(2n - 1)v , where n = 0, 1, 2, ¼ 4L

A N

λ —3 2

Frequency =

Position of antinodes from closed end l 3l 5l , ,¼ x= , 4 4 4

N

λ —3 2

f1 : f2 : f3 ¼ = 1 : 3 : 5 ¼ 4l In the general case, l = , where n = 0, 1, 2, ¼ (2n - 1)

Position of nodes from closed end x = 0,

If f1 is the fundamental frequency, then the velocity of waves is given as,

\

l2 =

A

(i) It represents the fundamental mode or first harmonic. l L= 1 4

\

4L 3

\

From Eqs. (i), (ii) and (iii), we get

Close pipe

\

673

f1 =

(open pipe)

whereL is the length of the pipe. The frequency of itsnth harmonic is nv for n = 1, 2, 3, ¼ (open pipe) fn = 2L Given, L = 30 cm,

v = 330 ms–1 fn =

…(i)

v n = li 2 L

n (330 ms–1) = 550 ms–1 0.6 m

The frequencies of 2nd harmonic, 3rd hormonic, 4th hormonic, … are 2 ´ 550 = 1100 Hz, 3 ´ 550 = 1650 Hz, 4 ´ 550 = 2200 Hz. Clearly, a source of frequency 1.1 kHz will resonate at f2, i. e. , the second harmonic.

674 JEE Main Physics End Correction It was found that the antinode is not formed exactly at the open end of the organ pipe but actually due to finite momentum of the particles the reflection takes place a little above the open end; that is why the antinode is formed a little above the open end. For this, a correction is applied being known as end correction. This is denoted by e. If length of organ pipe is l and end correction is e, then length of air-column in closed pipe will be (l + e) and in open pipe, ( l + 2 e). Thus, for a closed organ pipe. e

A l1 = λ/4

A

N l2 = 3 λ/4

N

A

e N

λ — 4

l

λ — 2

l

(a)

and

Þ

e (a)

(b)

v f1 = 4 ( l + e) and for an open organ pipe, f2 =

v 2 ( l + 2 e)

Note The value of end correction e is 0.6r for closed organ pipe and 1.2r for an open organ pipe, where r is the radius of the pipe.

Resonance Tube It is a closed organ pipe in which length of air-column can be increased or decreased. When a vibrating tuning fork is brought at its mouth as shown in figure, then forced vibrations are set up in its air-column. If we adjust the length of air-column as such its any natural frequency equals to the frequency of tuning fork, then the amplitude of forced vibrations of air-column increases very much. This is the state of resonance. When length of air-column is l = l/4, then the first resonance occurs. As shown in Fig. (a), antinode is formed at an open end and a node is formed at the water surface. Now, when length of air-column is l2 = 3l/ 4, then second resonance occurs. In this condition, two antinodes and two nodes are formed as shown in Fig. (b).

End correction In resonance tube, antinode is not formed exactly at open end but it is formed a little above the open end known as end correction (e). So, in first and second state of air-column, the lengths are l1 + e and l2 + e.

Hence, end correction,

(b)

l1 + e = l/ 4 l2 + e = 3l/ 4 l2 + e =3 l1 + e l2 = 3 l1 + 2 e l - 3 l1 e= 2 2

16.9 Musical Sound and Noise A musical sound consists of a quick succession of regular and periodic compressions and rarefactions without any sudden change in amplitude. While a noise consists of slow succession of irregular and periodic rarefactions and compressions accompanied by a sudden change in amplitude.

Characteristics of Musical Sound Musical sounds differ from each other due to the following three characteristics

Pitch Pitch is the characteristic of sound that depends on frequency. It determines the shrillness or graveness of sound. Smaller the frequency smaller is the pitch, greater the frequency greater is the pitch. Frequency of ladies voice is usually higher than that of gents’. Therefore, ladies voice has higher pitch (sharper) than gents’.

Loudness The loudness being a sensation, depends upon the sensitivity of the listener’s ears. Therefore, loudness of a sound of given intensity may be different for different listeners. Hence, it depends on intensity of sound.

Waves æ I ö L = 10 log10 ç ÷ è I0 ø

decibel (dB)

Here, I 0 is the intensity of minimum audible sound which is 10-12 Wm-2. Humming of mosquito has a high pitch (high frequency) but low intensity (low loudness) while the roar of a lion has high intensity (loudness) but low pitch.

Quality Quality is that characteristic of sound by which we can differentiate between the sounds coming from different sources. Quality of sound depends on the number of overtones and their relative intensities. If same note is played on different instruments say sitar and veena, at same loudness and same frequency, they produce different sensation on our ears due to their different quality.

Musical Interval The ratio between the frequencies of two notes is called the musical interval. The combined effect of two tones is musical, if the interval can be expressed as a ratio of two small numbers æ 2 3 ö ç as , , ¼÷ and the combined effect of two tones is a è ø 1 2

noise, if the interval is given by the ratio of two large æ 21 22 ö numbers ç as , , ¼÷ . è 20 20 ø Following are the names of some musical intervals (a) (b) (c) (d) (e) (f)

n Unison 2 = 1 n1 n Octave 2 = 2 n1 n 9 Major tone 2 = n1 8 n 10 Minor tone 2 = 9 n1 n2 16 Semi tone = n1 15 n 3 Fifth tone 2 = n1 2

Musical Scale The arrangement of notes having a definite ratio with respect to fundamental frequencies is called a musical scale.

Musical scales are of two types Diatonic scale It is known as ‘Sargam’ in Indian system. It contains eight notes with definite ratios in their frequencies. The note of lowest frequency is called key

675

note and the highest (which is double of first) is called an

octave. Harmonium, piano, etc., are based on this scale. Tempered scale It contains 13 notes. The ratio of frequencies of successive notes is 21/12. This scale is used in harmonium. The frequencies of successive tones of an equal tone temperature scale, form a geometric series.

Sample Problem 13 An open pipe is suddenly closed at one end with the result that the frequency of third harmonic of the closed pipe is found to be higher by 100 Hz, than the fundamental frequency of the open pipe. The fundamental frequency of the open pipe is (a) 100 Hz (c) 200 Hz

(b) 150 Hz (d) 250 Hz

Interpret (c) For fundamental mode in open pipe, Þ and

L = l /2 l = 2L v v fF = = l 2L

…(i)

For third harmonic in closed pipe 3l L= 4 4L l= Þ 3 3v v v fH 2 = = = l 4 L 4L 3 From Eqs. (i) and (ii), we have fH 3 = fF 2 3 Þ fH = fF ´ 2 but fH - fF = 100 3 fF - fF = 100 2 Þ fF = 200 Hz

…(ii)

…(iii)

Sample Problem 14 A window whose area is 2 m2 opens on a street where the street noise results in an intensity level at the window of 60 dB. Now, if a sound absorber is fitted at the window, how much energy from the street will it collect in a day? (a) 0.73 J (c) 2.73 J

(b) 0.173 J (d) 1.73 J

Interpret (b) By definition sound level = 10 log or

I = 60 I0

I = 10 6 I0 I = 10 -12 ´ 10 6 = 1 m Wm–2 [I0 = 10 -12 Wm–2]

Power entering the room = 1 ´ 10 -6 ´ 2 = 2 mW Energy collected in a day = 2 ´ 10 -6 ´ 86400 = 0.173 J

676 JEE Main Physics

16.10 Beats Beats is an interesting phenomenon arising from interference of waves. When two sound waves of nearly same frequency are produced simultaneously, then the intensity of resultant sound wave increases and decreases with time. This change in the intensity of sound is called as the phenomenon of ‘beats.’ The time interval between two successive beats is called beat period and the number of beats per second is called the beat frequency. If f1 and f 2 are the frequencies ( f1 > f 2 ) of the two waves, then the beat frequency b = f1 - f 2

Important Points 1. At frequency difference greater than about 6 or 7 Hz, we no longer hear individual beats. For example, if you listen to a whistle that produces sounds at 2000 Hz and 2100 Hz, you will hear not only these tones but also a much lower 100 Hz tone.

2. If the frequency of a tuning fork is f and it produces Df beats per second with a standard fork of frequency f 0 , then f = f 0 ± Df If on filing the arms of an unknown fork, the beat frequency decreases, then f = f 0 - Df This is because filing an arm of a tuning fork increases its frequency. Similarly, if on loading/waxing of the unknown fork, the beat frequency decreases, then the frequency of the unknown fork is f = f 0 + Df . This is because loading/waxing decreases the frequency of tuning fork. Similarly, f = f 0 + Df , if on filing beat frequency decreases and f = f 0 - Df if on loading/waxing beat frequency increases.

Sample Problem 15 The first overtone of an open pipe and the fundamental note of a pipe closed at one end, gives -1 5 beats s , when sounded together. If the length of the pipe, closed at one end is 25 cm, what are the possible lengths of the open pipe? (Neglect end corrections and take the velocity of sound in air to be 340 ms-1). (a) 90.5 and 120.5 cm (c) 95.5 and 102.5 cm

(b) 98.5 and 101.5 cm (d) 95.5 and 200 cm

Interpret (b) Let the fundamental frequency of the closed end pipe of length 25 cm be f0 . Then 340 ´ 100 v = = 340 Hz f0 = 4l 4 ´ 25 Possible frequencies of first overtone of the required open pipe are 340 ± 5, i.e., 345 or 335 Hz.

For the first overtone of an open pipe, the length of the pipe l equals the wavelength of the vibration. v Hence, 345 = l 34000 or l= = 101.5 cm 335 Other possible length l ¢ is given by v 335 = l¢ 34000 l¢ = = 101.5 cm 335 Hence, possible lengths of the open pipe are 98.5 and 101.5 cm.

Sample Problem 16 Two tuning forks A and B sounded together give 8 beat s-1. With an air resonance tube closed at one end, the two forks give resonance when the two air columns are 32 cm and 33 cm respectively. The frequencies of forks are (a) 260 Hz, 250 Hz (b) 264 Hz, 256 Hz (c) 274 Hz, 256 Hz (d) 2709 Hz, 250 Hz

Interpret (b) Let the frequency of the first fork be f1 and that of second be f2. Then, we have v v and f2 = f1 = 4 ´ 32 4 ´ 33 We also see that f1 > f2 \ and

f1 - f2 = 8 f1 33 = f2 32

…(i) …(ii)

Solving Eqs. (i) and (ii), we get f1 = 264 Hz and

f2 = 256 Hz

Sample Problem 17 Two sitar strings A and B playing the note ‘Dha’ are slightly out of tune and produce beats of frequency 5 Hz. The tension of the string B is slightly increased and the beat frequency is found to decrease to 3 Hz. The original frequency of B, if the frequency of A is 427 Hz is [NCERT]

(a) 427 Hz (c) 5 Hz

(b) 422 Hz (d) 10 Hz

Interpret (b) Increase in the tension of a string increases its frequency. If the original frequency of B ( nB) were greater than that of A ( n A ), further increase in nB should have resulted in an increase in the beat frequency. But, the beat frequency is found to decrease. This shows that nB < n A . Since, We get,

n A - nB = 5 Hz and n A = 427 Hz nB = 427 - 5 = 422 Hz

Waves

Hot Spot

677

Doppler’s Effect In SOUND

When there is a relative motion between the source and the observer the apparent frequency changes. This change in apparent frequency because of relative motion is called Doppler’s effect. Doppler’s effect is a wave phenomena, it holds not only for sound waves but also for electromagnetic waves. Change in frequency can be analyzed under three different situations. 1. Observer is stationary, but the source is moving. 2. Observer is moving but the source is stationary. 3. Both the observer and source are moving.

In our focus, we take the first case in detail 1. Observer is stationary but source is moving Taking the direction from the observer to the source as the positive direction of velocity. Consider a source S moving with velocity vS and an observer who is stationary in a frame in which the medium is also at rest. vS O

S1

S2

v T = T 0 æç1 + S ö÷ è v ø

...(i)

1 and apparent T frequency (fo ) that would be measured, if the source and observer were stationary and the frequency f observed, when the source is moving as The above equation is terms of actual frequency f =

v f = fo æç1 + S ö÷ è v ø

-1

If vS is small compared with the wave speed v, taking Binomial v expansion to terms in first order is S and neglecting higher powers v the above equation becomes v f = f 0 æç1 - S ö÷ è v ø

For a source approaching the observer, then L + vST0

v f = fo æç1 + S ö÷ è v ø

vST0

Let the speed of a wave of angular frequency w and period T0 both measured by an observer at rest w.r.t. the medium be v. At time t = 0, the source is at point S1, located at a distance L from the observer and emits a crest. This reaches the observer at time t1 =

L v

At time, t = T0 , the source has moved a distance vST0 and is at point S2 , located at distance ( L + vST0 ) from the observer. At S2 , the source emits a second crest. This reaches the observer at t2 = T 0 +

(L + v ST 0 ) v

At time, t = nT0 , the source emits its ( n + 1)th crest and this reaches the observer at time tn + 1 = nT 0 +

(L + nv ST 0 ) v

Hence, in time interval, énT + (L + nv ST 0 ) - L ù êë 0 v v úû

The observer’s detector counts n crests and the observer records the period of the wave as T given (L + nv ST 0 ) L ù - ú n T = éênT 0 + v vû ë v ST 0 T = T0 + v

Note The observer thus measures a lower frequency, when the source recedes from him, then he does when it is at rest. He measures a higher frequency when the source approaches him.

Sample Problem 19 A rocket is moving at a speed of 200 ms -1 towards a stationary target. While moving it emits a wave of frequency 1000 Hz. Some of the sound reaching the target gets reflected back to the rocket as an echo. The frequency of sound as detected by the target is (a) 2540 Hz (c) 4240 Hz

(b) 1500 Hz (d) 6200 Hz

Interpret (a) The observer is at rest and the source is moving with a speed of 200 ms -1. Since, this is comparable with the velocity of sound 330 ms -1, using equation æ v ö f = f0 ç1 + S ÷ è vø

-1

Since, the source is approaching a stationary target v 0 = 0 and v S must be replaced by -v S. We have -1 æ v ö f = f0 ç1 - S ÷ è vø é 200 ù Hz f = 1000 ê1 ë 330 úû f = 2540 Hz

678 JEE Main Physics Sample Problem 20 A siren emitting a sound of

Interpret (b) From Doppler’s effect frequency fapp

frequency 1000 Hz moves away from you towards a cliff with a speed of 10 ms-1. What is the frequency of the sound you hear coming directly from the siren?

800 m

v

vS cos θ

(a) 33/34 ×1000 Hz (b) 34/33 × 1000 Hz (c) 35/34 × 1000 Hz (d) 34/35 × 1000 Hz

P

v s = 10 ms–1 æ 330 ö f1 = ç ÷ ´ 1000 è 330 + 10 ø 33 ´ 1000 Hz 34

Sample Problem 21 A bullet passes past a person at a speed 220 ms -1. The fractional change in the frequency of the whistling sound heard by the person as the bullet crosses the person is (Speed of sound = 330 ms-1) (a) 0.67 (b) 0.8 (c) 1.2 (d) 3.0

Interpret (b) Limiting cases when it is just at the verge of crossing and when it has just crossed are taken. From Doppler’s effect, we have æ v ö f1 = ç ÷ f = 0.6 f è v + vS ø f2 = fnet

or

v f =3f v - vS

f +f = 1 2 2 3.6 f = = 1.8 f 2

Df = 0.8 f Df = 0.8 f

Sample Problem 22 A person P is 600 m away from the station, when train is approaching station with 72 km/hr it flows a whistle of frequency 800 Hz when 800 m away from the station. Frequency heard by the person is [Given, speed of sound = 340 ms–1] (a) 800 Hz (b) 839.5 Hz (c) 829.5 Hz (d) 834.5 Hz

v

fapp =

æ v ö f1 = f0 ç ÷ èv + v s ø

=

600 m

1000

Interpret (a) Sound heard directly,

\

S

=

v - v S cos q

f

340 ´ 800 = 839.5 Hz 340 - 16

Sample Problem 23 A source S of acoustic wave of the frequency, n 0 = 1700 Hz and a receiver R are located at the same point. At the instant, t = 0, the source starts from rest to move away from the receiver with a constant acceleration a. The velocity of sound in air is 340 m/s. If a = 10 m / s2, the apparent frequency that will be recorded by the stationary receiver at t = 10 s will be (a) 1700 Hz (c) 2.89 Hz

(b) 1.35 Hz (d) 1300 Hz

Interpret (b) Source frequency, n 0 = 1700 Hz. Source (coinciding with observer at t = 0) moves away with uniform acceleration a. Consider the wave which is received by the observer at instant t = t. It will have left the source at an earlier instant of time, say t ( < t), when the distance of source was r (say), if u be æ 1ö velocity of source at instant t, then r = ç ÷ at 2 and u = at . The è2ø relation between t and t, is t =t +

r at 2 =t + v 2v

This is a quadratic equation in t, giving the solution - 2v + 4 v 2 + 8 vat 2 æ 2 at ö u = at = v ç 1 + - 1÷ v è ø

at =

æ 2 ´ 10 ´ 10 ö u = 340 ç 1 + - 1÷ 340 è ø æ 27 ö = 340 ç - 1÷ è 17 ø Then apparent frequency is given by æ v ö na = ç ÷n èv + uø 0 Putting the values v = 340 m/s, t = 10 s, a = 10 m/s2, we have æ 340 ö na = ç ÷ 1700 è 340 + u ø n a = 1700

17 = 1.35 Hz 27

Waves After Cases of Doppler’s Effects 2. Observer is moving and source at rest (a) When observer is coming towards the source, then apparent frequency is v + v0 ö f0 = f æç ÷ è v ø (b) When observer is going away from the source, then apparent frequency is v - v0 ö f0 = f æç ÷ è v ø

3. Both observer and source are moving (a) If observer and source are moving in same direction and observer is ahead of the source, then apparent frequency is æ v - vo ö f0 = f ç ÷ è v - vs ø

Transverse Doppler’s Effect (i) The Doppler’s effect in sound does not take place in the transverse direction. (ii) From the figure, the position of a source is S and of observer is O. The component of velocity of source towards the observer is v cosq. For this situation, the approach frequency is

(b) If both are moving in the same direction and source is ahead of the observer, then apparent frequency is æ v + vo ö f0 = f ç ÷ è v + vs ø (c) If both are moving towards each other, then apparent frequency is æ v + vo ö f0 = f ç ÷ è v - vs ø (d) If both are moving away from each other, then apparent frequency is æ v - vo ö f0 = f ç ÷ è v + vs ø

Note Frequency is abbreviated as either n or f.

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 speed u and the speed of sound is v, then ö æ -1 ç 1 ÷ v ö æ uö f ¢ = æç ÷ f = ç u ÷ f = ç1 - ÷ f è vø èv - u ø ç1 - ÷ è vø

θ v

P

v

S

θ

θ

S

v

Using the Binomial expansion, we have

v

æ1 - u ö ç ÷ è vø

co

T

s

θ

v ´f v - v s cosq

f ¢ which will now be a function of q so, it will no more be constant. Similarly, if the source is moving away from the observer as shown above, with velocity component v s cos q, then v f¢ = ´f v + v s cosq (iii) If q = 90°, the v s cos q = 0 and there is no shift in the frequency. Thus, at point P, Doppler’s effect does not occur.

Note 1. If wind blows at a speed vw from the source to the observer, take v ® v + vw (both in numerator and denominator) and if in opposite direction (i.e., from observer to source), take v ® v - vw . Thus, the modified formula is æ v ± vw ± v o ö f¢ = ç ÷f è v ± vw ± v s ø 2. Change in frequency depends on the fact that whether the source is moved towards the observer or the observer is moved towards the source. But when the speed of source and observer

-1

» 1+

u u2 + v v2

æ u u2 ö f ¢ = çç1 + + 2 ÷÷ f è v v ø

O

f¢ =

679

u f ¢ » æç1 + ö÷ f , ifu n1 and given that (n2 - n1) = 3

n2 - n1 = 4 On solving, we get n2 = 24 Hz

and

n1 = 20 Hz and maximum loudness = ( 2a + 2a) 2 = 4a2 = 4 I

…(i)

nµ T

78.

n2 T T = = n1 16 4

If n1 corresponds to 4; then n2 corresponds to 3 + 4 = 7, which is T . Therefore, T = 49 N

72. Here, n1 = 200 Hz Number of beats s-1; m = 4 \

when n = 200 + 5 = 205 n1 l2 25 5 = = = n2 l1 30 6

77. Beat frequency = number of beats per second = n1 - n2

As per the choice given, T2 > T1

\

n = 36 and 2 n = 72

75. Here, n = 200 ± 5 and 2 n = 420 ± 10. This is possible only

Frequency of overtones are 2n,3n, 4n,…= 1320,1980 2640 Hz.

Now

79.

n2 = 200 ± 4 = 204 or 196 Hz

n2 101 æ 1 ö = = ç1 + ÷ n1 100 è 100 ø

1/ 2

= 1+

n2 = 200 - 4 = 196 Hz

=

81.

N

82. Here, n1 = 480 ,m = 10

In fundamental mode of vibration, l = l or l = 2l 2 v = nl n = 45 Hz m m = = 4 ´ 10 -2 kg/m l m = 3.5 ´ 10 -2 kg \

l=

frequency decreases on loading, therefore, original frequency of unknown fork = 260 Hz. Number of beats per second = n1 - n2 \Time interval between two successive beats/successive 1 maxima = n1 - n2

λ/2 e

As

v é l + Dl = l ù v Dl = 1 êë l( l + Dl) úû 4 l 2

80. Frequency of unknown fork = 256 ± 4 = 260 or 252. As

A N

1 200

(From Binomial Expansion) n n2 = n1 + 1 200 200 n Number of beats s-1 = n2 - n1 = 1 = =1 200 200 v v Number of beats s-1 = n1 - n2 = 3 l 4( l + Dl)

On loading fork 2, its frequency decreases. And number of beats per second increases to 6. Therefore, m is negative.

73.

[Q l = 2l]

T m

68. Octave stands for an interval 2 :1. Three octaves will have a

n=

703

m 3.5 ´ 10 -2 = 0 .875 m = m 4 ´ 10 -2

\

n2 = n1 ± m = 480 ± 10 = 490 or 470.

when tension is increased, n2 will increase …(i)

As number of beats s

83. As,

-1

decrease, n2 = 470 Hz

n1 l2 51 = = n2 l1 50

n1 - n2 = 5 On solving, we get and

n2 = 250 ,n1 = 255

(\ n2 µ T ).

704 JEE Main Physics 84. Two possible frequencies of source are = 100 ± 5 = 105 or 95 Hz Frequencies of 2nd harmonic = 210 or 190 Hz 5 beats with source of frequency 205 Hz are possible only when 2nd harmonic has frequency = 210 Hz \Frequency of source = 105 Hz T 8 l 36 D 4 85. Here, 1 = , 1 = , 1 = T2 1 l2 35 D2 1

\ frequency of sound reaching the wall is f ¢ =

where v is the velocity of sound in the air and VS is the velocity of source. On reflection the wall is the source of sound of frequency f ¢ at rest and bat is an observer approaching the wall

Now,

n2 l1D1 T2 r1 = n1 l2D2 r 2T1

90 ´ 10 3 ´ 334 326 = 92.1 ´ 10 3 Hz

=

n2 36 4 1 1 36 = ´ ´ = n1 35 1 8 2 35 Clearly

92. As source and observer both are moving in the same direction

n2 > n1.

n2 = 360 Hz; n1 = 350 Hz Number of beats per second = n2 - n1 = 360 - 350 = 10 u 2u Number of extra waves received per second= - ( -u / l) = l l When

with the same velocity, their relative velocity is zero. Therefore, n ¢ = n = 200 Hz.

93. From Doppler’s effect, æ v - v0 ö æ 340 - 10 ö n¢ = nç ÷ = nç ÷ = 1950 è 340 + 10 ø v v + è sø

87. n = 165 Hz, and n¢ =

94. Large vertical plane acts as listener moving per second.

\Number of beats per second = n - n = 170 - 165 = 5

88. Whistling train is the source of sound, v s = V . Before crossing a stationary observer on station, frequency heard is vn vn n¢ = = = constant and n¢ > n. (v - v s ) v - V Here, v is velocity of sound in air and n is actual frequency of whistle. After crossing the stationary observer, frequency heard is vn vn n¢ = = = constant and n¢ < n (v + v s ) v + V Therefore, the expected curve is (c). 1 Dn 89. As n= c 2 n 1 Dn 0.2c = c \ 2 ( 4 ´ 10 7)

n¢ =

95. As source is moving towards observer, \

v¢ =

uv 333 ´ 450 = = 400.5 » 500 u - vs 333 - 30

96. Here, v s1 = 34 ms -1, v = 340 ms -1 340 ´ n 340 v ´n = = f1 = n v - v s1 340 - 34 306 v ´n 340 ´ n 340n = = v - v s 2 (340 - 17) 323

f1 323 19 = = f2 306 18

97. Let n be the actual frequency of sound of horn.

As the rocket is receding away n ¢ = n - Dn = 4 ´ 10 - 1.6 ´ 10

( c + v)n c This is the number of waves striking the surface per second.

\

f2 =

Dn = 1.6 ´ 10 7 Hz 7

(from question)

n = 2068 Hz

335 + 5 335 ´ ´ 165 = 170 Hz 335 330 ,

\

[using Eq. (i)]

æ 330 + 4 ö = 90 ´ 10 3 ç ÷ è 330 - 4 ø

n1 = 360 Hz, n2 = ?

and

vf v - vS

\ frequency heard by the bat is f ¢(v + v 0) (v + v 0) =f f ¢¢ = v (v - v s )

r1 1 = r2 2

86.

91. Here bat is a source of sound and the wall is observer at rest

7

= 2.4 ´ 10 7 Hz

90. No beat is heard, because frequency received by listener directly from the source and that received on reflection from 256 ´ 330 Hz the wall is same i. e. , = 330 - 5

If v s is velocity of car, then frequency of sound striking the cliff (source moving towards listener) (v + v s )n ¢ (v + v s ) v ´ n = ´ n¢ = v v (v - v s ) n ¢¢ v + v s or = =2 n v - vs v + vs = 2 v - 2 vs v 3v s = v ,v s = 3

Waves 98. Here, us = 50 ms-1,vL = 0 ,v = 350 ms-1

When source is moving away from observer, the apparent frequency 350 6000 u ´v = ´ v¢ = 7 u + v s (350 + 50)

When source is moving towards observer, frequency

\

705

n¢ = 1000 u ´v n¢ = u - us

= 750 Hz u ´v 99.From Doppler’s effect, v ¢ = u - us

(u - us )v ¢ u (350 - 50)1000 6000 Hz = = 350 7

n¢ =

=

330 ´ 500 = 550 H 330 - 30

Round II 1. As, energy E µ (amplitude) 2(frequency) 2 Amplitude is same in both the cases; but frequency 2w in the second case is two times the frequency ( w) in the first case. Hence

E 2 = 4E1

2. Frequency of 1st overtone of A 2 T 2 = n1 = 2l1 m l1 D1

T pr

Frequency of 2nd overtone of B; n2 = As \

3 T 2 = 2 l2 m l2 D2

n1 = n2 2 T 2 = = l1D1 m l1D2

T pr (resonance condition)

T pr

l1D1 2 = l1D2 3

v 4l Now,

4. nc =

v 2l n0 - nc = 2 v v =2 2l 4l v =8 l v v n0¢ = = 2l / 2 l v v nc¢ = = 4 (2l) 8 l n0 =

and

\ or Also and

Number of beats per second = n0¢ - nc¢ v v 7v = = l 8l 8l =

5. Let speed of observer be vL = v along Y-axis and speed of

l1 2D2 2 l = = or 1 = 1: 3 l2 D2 3 l2

source the v s = 2vL = 2v along X-axis Y vL = v

3. As, it is clear from figure, at t = 0, x = 0 , displacement y = 0. Therefore, options (a) or (d) may be correct. In case of (d);

P

y = A sin (kx - wt ) dy = A cos(kx - wt ) [ - w] dt

O

(after differentiating w.r.t. t ) and

\

dy dt = - wA cos (kx - wt ) = - w = -v dy / dx kA cos (kx - wt ) k dy æ dy ö = -v ç ÷ è dx ø dt

i. e. , particle velocity = - (wave speed) ´ slope And slope at x = 0 and t = 0 is positive, in figure. Therefore, particle velocity is in negative y- direction.

β θ

S

vs = 2 vL

\

X

PS = 2 (OL) 2 5 2 cos b = 5 cos a =

dy = A cos(kx - wt ) [k] dx (after differentiating w.r.t. x )

\

7 ´8 = 7 8

and

Now, apparent frequency n ¢ is given by (v - vL cos b)n n¢ = (v + vL cos a) where v is velocity of sound. n¢ =

(v - v 5)n (v + 4v 5)

Clearly, n ¢ is constant but n ¢ < n. This is shown in curve (b).

706 JEE Main Physics 6. When O is a fixed end, the formation of reflected pulse is equivalent to overlapping of two inverted pulses travelling in opposite direction as shown in figure. 3 cm

\

t=0

v1 Y ra Ya . = = v2 r T T t = 3s

Hence, at t = 3s, net displacement of all particles of the string will be zero i. e. , the string will as shown in figure. 3l 2 2l 2(0.6) l= = = 0.4 m 3 3 l=

v=

T m

Y=

Þ

v1 = v2

v=

= (0.5 ´ 10 -2) ´ 2p ´ 50 = 1.57 ms-1

8. As the string vibrates in n loops therefore, nl 2 1 therefore, v would become time. 2 l=

vµ T

As

We are given,

1 Therefore, to make v half -time , T must be made time i. e. , 4 M / 4. T = tension in the rope at a distance xfrom the lower end \ T = (mg ) x = weight of x metre of rope

i. e. ,

T m

\

v=

-1/ 2

Dl 1 = l n v1 æ 1 ö =ç ÷ v2 è n ø

\

-1/ 2

= n

v1 = f1 l

mgx = gx m

vµ x

\

11. Let m be the total mass of the rope of length l. Tension in the rope at a height h from lower end = weight of rope of length h mg T= (h) l

i. e. , As

v=

T (m / l)

\

v=

mg (h) = gh l(m / l)

v 2 = gh which is a parabola. Therefore, h versus v graph is a parabola option (a) is correct.

12. As is known, frequency of vibration of a stretched string n µ T µ mg µ g 80 na = 0.8 na 100

As

nw =

\

g ¢ æ nw ö = ç ÷ = (0.8) 2 = 0.64 g è na ø

2

If, r w = relative density of water (= 1)

10. Velocity of longitudinal waves, Y v1 = r and velocity of transverse waves v2 =

v 2 = f2 l v1 f1 = = n v 2 f2

and

9. Let m = mass per unit length of rope

As v =

T a æ Dl ö =ç ÷ æ Dl ö T è l ø aç ÷ è l ø

If f1, f2 are the corresponding fundamental frequencies of longitudinal and transverse vibrations, then

80 = 20 ms-1 2.0 v 20 n= = = 50 Hz l 0.4 Amplitude of particle velocity æ dy ö = ( amax ) w = amax (2p n) =ç ÷ è dt ø max

\

F T = aDl / l a( Dl / l)

As

7. As the string is vibrating in three segments, therefore,

Now,

T ra

v2 =

3 cm

or

mass mass = ´ area = ra length volume

m=

T m

If a is area of cross-section of string, then

r m = relative density of mass r t = relative density of liquid, then g ¢ æ rw ö = ç1 - ÷ = 0.64 g çè rm ÷ø rw = 1 - 0.64 = 0.36 rm

…(i)

Waves Similarly, in the liquid

17. Proceeding as in above question, 2

g ¢ æ nL ö = ç ÷ = (0.6) 2 = 0.36 g è na ø

tan q = tan 60° = \

g ¢ æ rL ö = ç1 ÷ = 0.36 g è rm ø …(ii)

\

Hence, specific gravity of liquid = 1.77 VO r O + VH2r H2 Density of mixture = r mix = 2 2 VO2 + VH2 V (r O2 + r H2 ) r H2

2v + 16 r H2 ) 2

Þ As

=

r O2 + r H2 2

[ since VO2 = VH2 = V ]

= 8.5 r H2

Vµ

1 r

r H2 r H2 Vmix 2 = = = VH2 17 r mix 8.5 r H2

p æ100 p y = 5 sin ç tè 2 2

ö x÷ ø

=

T = 0.04 s 70 22 As, v s = rw = r ´ 2pn = ´2 ´ ´ 5 = 22 ms-1 100 7

n=

20. As,

Frequency is minimum when source is moving away from listenery. Therefore from Doppler’s effect, u ´ v 352 ´ 1000 n' = = = 941Hz u + us 352 + 22

Number of beats per second = nA - nB = 6 102 97 xx=6 100 100 6 ´ 100 x= = 120 Hz 5

1 2l

æ Dl ö gç ÷A è l ø Ar

gDl lp

1 9 ´ 10 10 ´ 4.9 ´ 10 -4 = 35 Hz 2 ´1 1 ´ 9 ´ 10 3

y = 4 cos2(t / 2) sin (1000t ) = 2 [2 cos2(t / 2) sin (1000t )] = 2 [2(1 cos t ) sin (1000t )] = 2 sin 1000t + 2 sin 1000t cos t y = 2 sin 1000t + sin (1001)t + sin (999t )

or

102 97 x and nB = x 100 100

1 ´ 10 11 Y = = 10 7Nm-2 (vL / vT ) 2 (100) 2

1 T 1 n= = 2l M 2l

The general equation y = a sin ( wt - kx) On comparing w = 50 p 2p 2p 1 T= = = w 50 p 25

16. Let the frequency of standard fork = x

Stress =

Y stress

M = Mass, r = density, A = Area of cross-section

p ö æ y = 5 sin ç50 pt - x÷ è 2 ø

nA =

T pr 2r

V = volume, l = length, Dl = change in length M Alr The mass per unit length m = = = Ar l l T/A And Young's modulus of elasticity = Y = Dl / l YDlA T= l Hence, lowest frequency of vibration

p 2

\

Y T and vT = = m r

19. For wire, if

14. As, y = 5 sin (100t - x )

15.

v

vL Y pr 2r Y = ´ = = r vT T T / pr 2

Dividing Eq. (i) by Eq. (ii), we get rL 0.64 = = 1.77 r w 0.34

=

vp

v p = v ´ tan 60 º = v 3

18. As, vL =

rL = 1 - 0.36 = 0.64 rm

13.

707

\ The given wave equation represents the super position of three waves.

21. In a wave equation, x and t must be related in the form ( x - vt). Therefore, we rewrite the given equation as 1 y= 1 + ( x - vt ) 2 1 For t = 0, it becomes Y = 1 + x2 and for t = 2, it becomes y= \ or

1 1 = 2 [1 + ( x - 2v) ] 1 + ( x - 1) 2

2v = 1 v = 0.5 ms-1

708 JEE Main Physics 22. When source is moving towards observer, then apparent frequency n¢ =

Þ

vn v + vs

When source is moving away from observer, then apparent frequency vn n ¢¢ = v - vs Now, change in frequency, é v + v s - v + v s ù (2v s ) vm n ¢ - n ¢¢ = vn ê ú= 2 2 v 2 - v s2 û v - vs ë 2 v sn When v S,n ¢ - n ¢¢ = v n ¢ - n ¢¢ 2 2v 2 vs Now, = = s = n 100 v 300 \ v s = 3 ms-1

23. The motorist receives two sound waves, one direct from the band and second reflected from the wall which is shown is figure. For direct sound waves, apparent frequency (v + v m) f f¢ = v + vb Motorist (Listener)

vm

Band Master (Source)

For rotational equilibrium, net torque should be equal to zero

Wall vb

25. SupposeIi andIr are intensities of incident and reflected waves. Reflection coefficient = where ,

Frequency of reflected waves as received by the moving motorist, (v + v m) f ¢¢ (v + v m) f = f ¢¢ = v v - vb \ Beat frequency = f ¢¢¢ - f ¢ (v + v m) f (v + v m) f 2 v b(v + v m) f = = v - vb v + vb v 2 - v b2 1 T1 1 T2 = 2l m l m

26. Here,A1 = A2; n1 = w, n2 = w2 \ and Now,

where,

D

T2 O C

M

y1 = A sin 2 pw1t , y 2 = A sin 2 pw2 t (in case of superposition) y = y1 + y 2 cos 2p ( w2 - w1) t 2p ( w2 + w1) t = 2A sin 2 2 = A¢ sin p ( w2 + w1) t A¢ = 2A cos p ( w2 - w1)t cos p ( w2 - w1) t = max = ±1 p ( w2 - w1)t = 0 , p, 2p 1 2 t = 0, ; ;… w2 - w1 w2 - w1

Time interval between two successive maxima 1 2 = = = 10 -3 s w2 - w1 10 3

27. Resultant displacement along X-axis is x = y1 - y3 = 8 - 2 = 6 Resultant displacement along Y-axis is y = y 2 - y 4 = 4 - 1 = 3 Net displacement, r = x2 + y 2 = 6 2 + 3 2 = 45 y 3 1 = = x 6 2 q = tan -1(1 / 2)

tan q =

vp vn and n ¢¢ = v - vs v + vs n v \ = 1- s , n¢ v n vs = 1+ n ¢¢ v Adding the two, we get n n + =2 n ¢ n ¢¢ 2n ¢n ¢¢ \ n= n ¢ + n ¢¢

28. As, n¢ =

T2 = T1 / 4

x

25 5 = 9 3 2

Also,

24. According to the question

L

v1 T / m1 m2 = = v2 T / m2 m1

Sound heard will be of maximum intensity ( > 2A2)

Frequency of sound wave, reflected from the wall v´f f ¢¢ = v - vb

B

2

1 æ 5 / 3 - 1ö \ Reflection coefficient = ç ÷ = è 5 / 3 + 1ø 16

For reflected sound waves,

T1

m=

Ir æ m - 1ö =ç ÷ Ii è m + 1ø

=

when

A

T1 x = T2 (L - x ) L x= 5

Waves 29. When the stone is suspended in air

32. Equation of stationary wave is y1 = a sin kx cos wt , and equation of progressive wave is

1 Wa n= 2L m When the stone is suspended in water, 1 Ww n= 2L ¢ m Wa Ww = L L¢ 2 Wa L = Ww L ¢2

\ or

Specific gravity of stone Wa 1 1 = = = Wa - Ww 1 - Ww L ¢2 1- 2 Wa L ( 40) 2 L2 = 2 2= L - L ¢ ( 40) 2 - (22) 2

30. When aeroplane is at P2 vertically above the observer O, sound comes along P1 O at 60° with the vertical. (vs) t

P1

P2

y 2 = a sin ( wt - kx) = a (sin wt cos kx - cos wt sin kx) p 3p At and x2 = x1 = 3k 2k sin kx1 or sin kx2 is zero. so, neither x1 nor x2 is node. 3p p 7p Dx = x2 - x1 = = = 2k 3k 6k 7p As Dx = , 6x 2p p Therefore, > Dx > k k 2p But =l k l So, l > Dx > 2 In case of a stationary wave, phase difference between any two points is either zero or p. \ and

vt 60°

O

\

P1O = v ´ t , P1 P2 = v pt P1 P2 v pt vp = = sin 60° = P1O v ´ t v

\

709

v p = v sin 60° = v 3 /2

31. Here, y1 = 0.05sin(3pt - 2x ) y 2 = 0.05sin(3pt + 2x ) According to superposition principle, the resultant displacements is y = y1 + y 2 = 0.05[sin (3pt - 2x)+ sin (3pt +2x)] y = 0.05 ´ 2 sin 3pt cos 2x y = (0.1cos 2x) sin 3pt = R sin 3pt where, R = 0.1 cos 2x = amplitude of the resultant standing wave. At x = 0.5 m R = 0.1cos 2x = 0.1 cos 2 ´ 0.5 180° = 0.1cos 1(radian) = 0.1 cos p = 0.1 cos 57.3° or R = 0.1 ´ 0.54 m = 0.054 m = 5.4 cm

\

f1 = p

7p 7 = p 6k 6 f1 p 6 = = f2 7 p 7 6 f 2 = kD x = k

33. As intensity µ a2w2 here,

aA 2 w 1 = and A = aB 7 wB 2

Þ

1 iA æ 2 ö æ 1 ö = ç ÷ ´ç ÷ = ø è ø è 1 2 1 iB

2

2

34. If a is amplitude of each wave, I0 = k( 2a + 2a) 2 = 4ka2 Let f be the phase difference to obtain intensity I0 / 2 I0 = kar2 = k( a2 + a2 + 2aa cos f) \ 2 f = k 2a2(1 + cos f) = k4a2 cos2 2 = I0 cos2 f/2 f 1 \ = = cos 45° cos 2 2 \ f = 90° If Dx is path difference between the two waves, then l l æpö l f= Dx = ç ÷= 2p 2p è 2 ø 4 1 Therefore, displacement of sliding tube ( Dx) = l / 8 2

710 JEE Main Physics 35. Let, v be the speed of sound in air, vL velocity of observer at time t. As, the observer approaches the source, therefore, apparent frequency (v + vL) é v + (0 + at ) ù æ f0 at ö f= f0 = ê úû f0 = f0 + çè v ÷ø v v ë This is the equation of a straight line with a positive intercept æ f aö ( f0) and positive slope ç 0 ÷ . Therefore, option (d) is correct. è v ø vT µ strain

36.

1 ´ 15 v0 = 30 ´ 1 vT vT = 2 v 0

37. Sound waves in air are longitudinal while light transverse. 38. The time taken by the plate falling through a distance y is given by æ 2 ´ 10 ö æ 1 ö t = (2yg ) = ç ÷ =ç ÷s è 980 ø è 7 ø 1 s is 8. 7 \Frequency = number of oscillations completed in one sec 8 = = 56 Hz 1/ 7 The number of oscillations completed

39. Displacement and amplitude both, are added vectorially in superposition principle.

40. Compare the given equation with the standard equation y = a sin( wt - x ), we get

i. e. ,

Therefore, velocity at x = will become

-1

41. In the given equation as x is positive, therefore, the wave is travelling along negative direction of x-axis in which, 2p 2p = 10 p , l = = 0.2m l 10 p 2p 2p 2 and = 15p , T = = s T 15p 15 l 0.2 v= = = 1.5 ms-1 T 2 /15 A will be –2F. Potential energy U µ x2 2 A i. e. , potential energy at x = will become 4U. 2 2

2

v = w A -x 2

(b) Compare the given equation with the standard from é 2 pt 2 px ù y = r sin ê + + fú l ë T û 2p = 36, T 2p = 0.018 l l 36 Speed of wave, v = = = 2000 cm/s = 20 m/s T 0.018 2p p (c) Again, T = = 36 18 1 18 Frequency v = = Hz = 5.7Hz T p

44. As the string is rigidly clamped at its two ends, therefore, y = 0 at x = 0. This can be satisfied only by the term np x sin = , L Therefore, options (a) and (b) correct.

45. A travelling wave is of the form F ( ax ± bt). Therefore, choice (a), (b), (c) are correct. 2 px ö ÷ cos (120 pt ) è 3 ø

46. (a) The given equation is y ( x, t) = 0.06 sin æç

(b) As terms involving x and t are independent of each other, the given equation represents a stationary wave. (c) Compare the given equation with the standard form of equation of stationary wave y ( x, t ) = 2r sin kx cos wt 2p 2p k= = l 3 \

\

2

v µ A -x

x=

4û

(a) As positive direction is from left to right and x is positive, therefore, wave is travelling from right to left.

Speed of particle is given by

and at

4 times or 0.8 times. 5 ë

at x =

x=-

A 4 may be ± v or kinetic energy 2 5

p 43. The given equation is y ( x, t) = 3.0 sin éê36 t + 0.018x + ùú

42. Force increases linearly therefore, force acting on the particle

at

4 times 5

where m is an integer.

v = 200 ms

i. e. ,

A2 - x2 has become

15 A A , A2 - x2 = 4 16

3 A A , A2 - x2 = 2 4

and

l = 3m w = 120 p w 120 p v= = = 60 Hz 2p 2p v = vl = 60 ´ 3 = 180 m/s

Hence the given stationary wave is the result of superposition of two waves of wavelength 3 m and frequency 60 Hz each, travelling with a velocity of 180 m/s in opposite directions.

Waves 47. According to Newton’s formula for velocity of sound in a fluid, v=

vµ

(b) Speed of sound for observer standing on platform, v ¢ + w = 340 + 10 = 350 m/s.

1 r

54. Here, w1 = 100 p and w2 = 92p ,

Choices (c) and (d) are correct.

Hence n1 =

48. Compare the given equation

with the standard form 2 pt ö æ 2p y = a sinç x+ ÷ , we get è l T ø a = 20 2p p = ,l =8 l 4 2p p = ,T = 4 T 2 1 1 n = = = 0.25 T 4 v v 330 ´ 100 , l= = = 31.25 cm 4l 4n 4 ´ 264

sin i v cooler T 273 + 27 3 50. m = = = 1 = = sin r v hotter T2 273 + 127 4

\

sin r =

4 4 4 1 1 ´ sin i = ´ = sin 30° = 3 3 3 2 3 -1

r = sin (1 / 3) 2 px ö ÷ cos (120 pt ) è 3 ø

51. The given equation is y ( x, t) = 0.06 sin æç

It represents a stationary wave. Therefore, all the points between two consecutive nodes. (a) vibrate with same frequency (b) in same phase, but (d) different amplitudes. The amplitude is zero at nodes and maximum at antinodes (between the nodes).

52. When the sources are coherent, 2

2

2

R = a + b + 2ab cos f For constructive interference, f = 0 \

I = I0 + I0 + 2 I0I0 cos 0° = 4I0

When the sources are incoherent, intensities just add I = I0 + I0 = 2I0

56. At x = 0 , y = y1 + y 2 = A cos100 pt + A cos 92 pt = 2 A cos 96 pt cos 4 pt y = 0 when either cos 96 pt = 0 or cos 4 pt = 0 p p Þ 96pt = (2n + 1) and 4pt = (2m + 1) , 2 2 where n and m are integers.

In second normal mode of vibration, 3v 3v n= = 3 ´ 31.25 = 93.75 cm , l= 4l 4n

\

92p = 46 Hz 2p \ Number of beats per second = n1 - n 2 = 50 - 46 = 4 w 100 p 55. Wave velocity n = 1 = = 200 ms-1 k2 0.5p = A cos (0.5 px - 100 pt ) + A cos (0.46 px - 2 pt )

49. In first normal mode of vibration n=

100 p 2p

= 50 Hz and n 2 =

pt ö æp y = 20 sinç x + ÷ è4 2ø

\

53. Here, n = 400Hz, v = 340 m/s, w = 10 m/s (a) The frequency of sound as heard by an observer standing on platform = 400 Hz only, as it is not affected by motion of wind alone.

Ba r

v µ Ba and

711

For 0 < t < 1, n can have 96 integer values and m can have 4 integer values. Hence net amplitude becomes zero 96 + 4 = 100 times

58. When the source moves at 90°to the line joining the source and the listener, apparent frequency remains unaffected.

60. The correct formula for velocity of sound in a gas is v =

gp r

For monoatomic gas, g = 1.67 For diatomic gas, g = 1.40 \ v is larger in case of monoatomic gas compared to its value in diatomic gas. v 350 61. As, l = = = 0.7m n 500 p p f = 60° = 60 ´ = rad 180 3 l As, x= f 2p 0.7 é 60 p ù x= ´ = 0.12 m = 12 cm \ 2p êë 180 úû

62. As lv < lr \ Violet shift means apparent wavelength of light from a star decreases. Obviously, apparent frequency increases. This would happen when the star is approaching the earth.

64. Let n be the frequency of fork. and \

n1 - n = 4 n - n2 = 4 n1 - n2 = 8

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

712 JEE Main Physics Also, \ From Eq. (ii),

From Eq. (i),

n1 l2 50 = = n2 l1 49 50 n1` = n2 49 50 1 n2 - n2 = 8, n2 = 8 49 49 n2 = 49 ´ 8 = 392. n = 4 + n2 = 4 + 392 = 396 Hz

65. The equation of stationary waves is px y = 20 sin cos wt 4 Compare with y = 2a sin kx cos wt p k= 4 2p As l= k 2p l= =8m \ p/4 Distance between two consecutive antinodes l 8 = = =4m 2 2

66. Equations show that phase difference between two waves f = p /2 \ Resultant amplitudes R = a2 + b 2 + 2ab cos p / 2 = a2 + a2 + 2a2 cos 90° = 2 a2 = a 2

67. When

b = a, then from 2

Þ

é v + v0 ù é v - ( -v 0) ù n¢ = nê ú ú = nê v v 0 û ë v - v0 û ë é 340 + 20 ù = 480 ê = 540 Hz ë 340 - 20 úû

71. fclosed = fclosed = =

v 2lopen v

68. Number of beats per second n = n1 - n2

n¢ =

v ´ n 330 ´ 600 = = 660 Hz v - vs 330 - 30

On reflection frequency of sound heard by driver, (v + vL)n ¢ (330 + 30)600 n ¢¢ = = = 720 Hz v 330

73. As the listener on motor cycle is moving away from the source (siren), therefore Þ

1-

Þ

vL2 19.8 ´ 19.8 = 2a 2 ´2

= 98 m

74. If we assume that all the three waves are in same phase at t = 0, we shall hear only one beat per second.

75. As,

v He = v oxygen

g HeMoxygen g oxygenMHe v He 460

69. If the lenght of the wire between the two bridges is l, then the frequency of vibration is

=

(5 / 3) ´ 32 (7 / 5) ´ 4

=

200 21

v He = 460

T pr 2d

æ radius ö In the length and diameter ç = ÷ of the wire are doubled è 2 ø keep in the tension same, then new fundamental frequency n will be . 4

n ¢ v - vL 94 = = n v 100

94 vL = v 100 vL 94 6 = 1= v 100 100 6 ´ v 6 ´ 330 = = 19.8 ms-1 vL = 100 100

Distance covered =

é1 1ù v v = vê - ú l1 l 2 ë l1 l 2 û é1 1 ù = 336.6 ê ú =3 1 1.01 û ë

1 1 T = 2l m 2 l

4lopen / 2

72. Frequency of sound reaching the hill

=

n=

lopen ù é ú ê As lclosed = 2 û ë

v

v = fopen 2fopen

Þ

a2 = a2 + a2 + 2a a cos f = 2a2(1 + cos f) 1 1 + cos f = 2 1 1 cos f - 1 = - , f = 120° 2 2

=

4lclosed

2

R = a + b + 2ab cos f, we get

Þ

70. The frequency of reflected sound heard by the girl,

200 21

= 1420 ms–1

76. Here, l1 = 18 cm During summer, v increases, g increases. \ l1 increases. As \

l2 > 3 l1 l2 > 54 cm

Waves 77. Given, y ( x, t) = 0.005 cos ( ax - bt) Compare it with standard equation 2p ù é 2p y ( x, t ) = r cos ê xt , we get T úû ë l 2p 2p a= = = 25.00 p l 0.08 2p 2p b= = =p T 2.0

Þ or

= 260 ms-1

…(i)

v 330 = 0.66 m = 66 cm n 500 The successive resonance lengths are at l 3l 5l 7l and so on. , , , 4 4 4 4

82. As, l = = …(ii)

Within one metre, length of the tube, total number to 7l resonances is 3(as is more than1.0 m). 4

83. The apparent frequency of sound striking the wall, n¢ =

cos ( wt - kx ) = 1 v p = Aw ´ 1Þ v p = Aw

p . 2

So, the resultant amplitude

\ Number of beats s-1 = n ¢¢ - n = 170 - 165 = 5

84. Here, n = 200 Hz, v = 360 ms-1

A = a12 + a22 + 2a1a2 cos f

or

\

pù é êëHere, a1 = a, a2 = a and f = 2 úû p A = a2 + a2 + 2as cos 2

85.

A = a2 + a2 + 0 Þ A = 2 a

80. The speed of sound in air is defined as, gRT M vµ T v1 T = 1 v2 T2 v=

Þ \

But according to question, v1 = v ,v 2 = 2v , T1 = 27° C = (27 + 273) K = 300 K, On pulling these value in Eq. (i), we have v 300 = 2v T2

335 ´ 165 335 v ´n Hz = = 2 (v - v s ) (335 - 5)

For reflected sound, wall is the source and passenger in the bus is as listener. (v + nL)n ¢ \ n ¢¢ = v (335 + 5) 335 = ´ = 170 Hz 335 2

y = a sin ( wt - kx ) y = a cos( wt - kx )

Here, the phase difference between the two waves is

\

T2 = 1200 K T2 = (1200 - 273)° C = 927° C

\ Frequency of sonometer wire,n = 260 Hz v = nl = 260 (2 l) = 260 (2 ´ 0.5)

79. Given that, two waves and

1 300 = T2 4

Possible frequencies of sonometer wire = (250 ± 10) Hz On filling the fork, number of beats per second decreases

Þ v p = Aw cos( wt - kx ) For maximum particle velocity, So,

or

81. Frequency of fork = 250 Hz.

78. Given that, the displacement of a particle is y = A sin ( wt = kx ) The particle velocity dy vp = dt Now, on differentiating Eq. (i) with respect to t, dy = A cos( wt - kx ) ´ w dt dy Þ = Aw cos ( wt - kx ) dt From Eq. (ii)

1 300 = T2 2

or

713

l=

v 360 = = 1.8 m n 200

Distance between two consecutive antinodes l 1.8 = = = 0.9 m 2 2 v dl From, = (where, dl = change in wavelength) ; c l dl v= ´c Þ l 0.32 = ´ 2 ´ 10 8 ms-1 100 = 9.6 ´ 10 5 ms-1

…(i)

As wavelength increases, apparent frequency decreases. The star must be moving away from rarth.

86. Here, l1 = 80 cm, l2 = 70 cm \

n2 l1 80 8 = = = n1 l2 70 7

If n is frequency of tuning fork, then n2 - n = 8

714 JEE Main Physics and \

\

n - n1 = 8 n2 - n1 = 16 8n1 - n1 = 16,n1 = 112 7 n = 120 Hz

92. Musical interval produced between two notes of frequencies is given as

93. Suppose, m1, v1 and m2, v 2 are the masses and velocities of the balls. Since, kinetic energy is same

87. The frequency of fork 2 = 200 ± 4 = 196 or 204 Hz. Since, on attaching the tape on the prong of fork 2, its frequency decreases, but now the number of beats per second is 6 i.e., the frequency difference now increases. It is possible only when before attaching the tape, the frequency of fork 2 is less than the frequency of tunning fork 1. Hence the frequency of fork 2 is 196 Hz.

88. As,

æIö B1 = 10 log ç ÷ è I0 ø

and

æ I¢ ö B2 = 10 log ç ÷ è I0 ø

Given,

i. e. , So,

(m1v12) (m2v 2) 2 = m1 m2

or

p12 p22 = m1 m2

Therefore, the heavier ball will have greater momentum.

94. The fundamental frequency of an open prgan pipe is given by n= =

I ¢ = 100I

f1 =

v 2l

v v = 4l / 4 l f1 1 = f2 2 f2 =

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

54 90. Frequency, n= 60 9 n= Þ 10 9 \ Velocity, v = nl = ´ 10 = 9 ms-1 10

91.

v The frequency of an open organ pipe is given by, n = 2l v …(i) n1 = 2 ´ 0.5 v and …(ii) n2 = 2 ´ 0.505 (Given l2 = 50 cm = 0.5 m, l2 = 50.5 cm = 0.505 cm) From Eqs. (i) and (ii), we get \

or

v 2l 350 = 350 Hz 2 ´ 0.5

95. Between two fixed points, resonance is obtained, when one

89. According to problem

We, get

pµ m

i. e. ,

æ I¢ ö 20 = 10 log ç ÷ è I0 ø

and

KE1 = KE2 1 1 m1v12 = m2v 22 2 2

or

B2 - B1 = 20

Þ

320 = 1.33 240

n1 - n2 = 3 v ö 0.01v æ v 3=ç ÷= è1.0 1.01ø 1.01 0.01v = 3.03 3.03 v= = 303 ms-1 0.01

loop, two loops, three loops etc., are formed. The resonant frequencies are in the ratio 1 :2 :3 :4. As the two resonance frequencies are 315 Hz and 420 Hz, with highest commom factor = 105.Therefore, the lowest resonant frequency for this string = 105 Hz.

96. Here, n = 9500 Hz, v S = ? v = 300 ms–1, n = 1000 Hz As source is moving towards the listener, v´n (from Doppler effect) n¢ = \ v - vS Þ or or

1000 =

300 ´ 9500 300 - v S

300 - v S = 285 v S = 300 - 285 = 15 ms–1

97. 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 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 than the beats are reduced to 2, therefore, 254 is not possible. Therefore, the required frequency must be 246 Hz. (This is true, because on filling the frequency may increase to 248, giving 2 beats with Q of frequency 250 Hz).

Waves 98. When the source and observer are approaching towards each

102. Time taken by stone to drop into lake is obtained from

other, then apparent frequency heard by the observer is æ v + v0 ö n¢ = ç ÷n è v - v0 ø

or

vö æ çv + ÷ è 2ø = n v æ ö çv - ÷ è 2ø æ3 2ö n ¢ = ç ´ ÷ n = 3n è 2 1ø

s=

Þ \

1æ v ö ç ÷ = 100 2 è2 lø n1 = 200 Hz

100.The node and antinodes are formed in a standing wave pattern as a result of the interference of two waves. Distance between two nodes is half of wavelength (l). Standard equation of standing wave is 2 px 2pvt …(i) y = 2a sin cos l l where a is amplitude, l the wavelength, v the velocity and t the time. Given equation is 2 px …(ii) y = 5 sin cos 20 pt 3 Comparing Eqs. (ii) with (i), we have 2 px 2 px = l 3 Þ

l = 3 cm

101. From Doppler’s effect, the perceived frequency ( f ¢ ) is given by f ¢= f

v v - vs

where v S is velocity of source, v is the speed of sound and f the original frequency. Given, f = 500 Hz, v s = 30 ms-1, v = 330 ms-1 330 f ¢ = 500 ´ , 330 - 30 330 f ¢ = 500 ´ = 550 Hz 300

1 2 gt 2

2s 2 ´ 500 = = 10 s g 10

t=

Sound so produced travels to the top of the tower. s 500 Time taken t¢ = = = 1.5 s v 330 Total time = t + t ¢ = 10 + 1.5 = 11.5 s

103. For first resonance,

99. An open pipe forms antinode at both ends. If length of pipe is l and v the velocity, then the fundamental frequency is given by v n1 = 2l 3v Frequency of third harmonic of closed pipe is n ¢ = 4l Given, n ¢ - n = 100 3v v \ - = 100 3l 2l v Þ = 100 4l

715

30.7 =

l +x 4

…(i)

For second resonance, 3l +x 4 Solving Eqs., (i) and (ii), l = 65.0 cm 63.2 =

…(ii)

The effort in measuring the length using metric scale would be 0.1 cm, which is the least count of metric scale. Therefore , l = (65.0 ± 0.1) cm. v = nl = 512 (65.2 ± 0.1)

As,

Therefore, error in velocity = 51.2 cms-1

104. In case of open pipe, the frequency of second harmonic is, f1 = 2v / 2L = v / L In case of closed pipe, the frequency of nth harmonic is f2 = nv / 4L = nf1 / 4 where, n = 1, 3, 5, ¼ , i. e. ,n is odd and f1 > f2 It will be so if n = 5 5 \ f2 = f1 4 1 v+ v v ¢ (v + vL) 5 5 105. As, = = = v v v 6 Percentage increase in frequency (v ¢ - v) (6 - 5) = ´ 100 = = 20% v 100 1 - cos 2wt 1 1 106. As, y = sin2 wt = = = wt 2 2 2 dy 1 velocity, = ´ 2w sin 2wt dt 2 Acceleration,

d 2y æ1 ö = 2w2 cos 2wt = 4 w2 ç - y ÷ 2 ø è2 dt

As acceleration µ displacement and negative sign shows that it is direction towards mean position. \ Motion is simple harmonic and its period = p / w

107. Given, y1 = 0.1sin(100 pt+ p / 3) \velocity,

v1 =

dy1 = 0.1 ´ 100 p cos (100 pt - p / 3) dt

y 2 = 0.1cos pt = 0.1sin( pt + p / 2)

716 JEE Main Physics dy 2 = 0.1 ´ p cos ( pt+ p / 2) dt \ Phase difference of the velocity of particle 1 w.r.t. particle 2 = ( pt + p / 3) - ( pt + p / 2) = - p / 6 v 108. As, f = 2l Velocity,

v2 =

Now, it will act like one end opened and other closed v v v So, f0 = 0 = = =f 4 l ¢ 4l2 2l ax + bt )2

109. Given, y = ( xt) = e-(

2

It is transverse type y( x, t ) = e-( ax+ bt ) Speed, v =

111. Maximum number of beats = ( n + 1) - ( n -1) = 2 112. As,

v =f 4 ( l + e)

Þ

v 4f v l= =e 4f

( l + e) =

or

e = (0.6) lr = (0.6) (2) = 1.2 cm

Here,

113. Here, fincident = freflected Þ

=

320 ´ 8 Hz 320 - 10

b a 36 km/h = 10 m/s

The wave is moving along x-direction. é

t x öù ÷ è 0.04 0.50 ø úû

110. Given, y = 0.02 sin ê2p æç ë

v=

T w T 1 / 0.04 = = = M k 0.04 1 / 0.50 2

æ 0.05 ö 2 T=ç ÷ ´ 0.04 = (12.5) ´ 0.04 = 6.25 N è 0.04 ø

Now,

320 + 10 freflected 320 330 =8´ = 8.51kHz = 8.5 kHz 310

fobserved =

Part - II Chapters from Class 12

th

Syllabus

17 Electrostatics JEE Main MILESTONE <
EB = EB = 2EB

A

+2 q B

O C +q

conducting sphere (figure). The electric field is best [NCERT Exemplar] given by

43. A long charged cylinder of linear charged density l is

(a)

+q A

48. A point positive charge brought near an isolated

42. Three concentric spherical shells have radii a, b and

(a) VC ¹ VB ¹ VA (c) VC = VB = VA

(b) 6 cm from the charge 5 ´ 10 -8 C

r

B

(d) +q

758 JEE Main Physics 49. A charge 5 mC is placed at a point. What is the work

56. An electric dipole is placed at the centre of a hollow

required to carry 1C of charge once round it in circle of 12 cm radius?

conducting sphere. Which of the following option is correct?

(a) 100

(b) 0

(d) ¥

(c) 1

50. Two metallic spheres A and B of same radii one solid and other hollow are charged to the same potential. Which of the two will hold more charge? (a) Sphere A (c) Both spheres

57. A glass rod rubbed with silk is used to charged a gold

(b) Sphere B (d) None of these

51. The tangential component of electrostatic field is continuous from one side of a charged surface to another is [NCERT] (a)

1 æ1 1 1ö - ÷ ç + rB rC ø 4 pe 0 è rA

(b) Zero 1 æ1 1 1ö (c) + ÷ ç rB rC ø 4 pe 0 è rA (d)

(a) Electric field is non-zero anywhere on the sphere (b) The flux of electric field is zero through the sphere (c) Options (a) and (b) both are true (d) No option is correct

leaf electroscope and the leaves are observed to diverse. The electroscope thin, charged is exposed to X-rays for short period. Then, (a) the leaves will diverge further (b) the leaves will melt (c) the leaves will not be affected (d) None of the above

58. A uniform electric field of 100 N/C exist in the vertically downward direction. The increase in the electric potential as one goes up the through a height of 50 cm is

1 æ1 1 1ö + ÷ ç + rB rC ø 4 pe 0 è rA

52. Which of the following lines of force is uniform field? (a)

(b)

(c)

(d)

(a) 10 V (c) 0 V

(b) 5 V (d) 0.5 V

59. A test charge q0 is moved without acceleration from

A to C and covers the path ABC as shown in figure. The potential difference between A and C is E C

45°

A

d

53. Two plates are 1 cm apart, and potential difference between them is 10 volt. The electric field between the plates is (a) 10 N/C (c) 103 N/C

54. A positively charged ball hangs from a silk thread. We put a positive test charge q0 at a point and F then it can be predicted that the electric measure q0 field strength E is (b) =

(c)
F/q0

(a) Ed (c) 2 Ed

(d)

l 2 p e0a

(a) - 2 2

(b) -1

(c) 1

(d) r = C1 e C2x

61. A charge +q is fixed at each of the points x = x0 ,

x = 3 x0 , x = 5x0 ...¥, on the x-axis and a charge -q is fixed at each of the points the x = 2x0 , x = 4 x0 x = 6x0 K ¥. Here, x0 is constant. Take the electric potential at a point due to a charge Q at a distance r from it to be Q / 4 pe 0 r. Then, the potential at the origin due to the above system of charges is q log e 2 4 pe 0 x0 q (b) log e 2 8pe 0 x0 (a)

(c) 0 (d) ¥

Electrostatics 10 ´ 10-9 C are placed at each of the 3 four corners of a square of side 8 cm. The potential at the point of intersection of the diagonals, is

62. The charge of +

(a) 1500 2 V (b) 1800 2 V (c) 600 2 V

(d) 900 2 V

63. Below figures. (i) and (ii) represent

field lines. Which of the following correct statement?

(i)

(ii)

(a) Fig. (i) represents magnetic lines of force (b) Fig. (ii) represents magnetic lines of force (c) Fig. (i) represents electric lines of force (d) Fig. (ii) represents electric lines of force

(d) 10 -15 cm

Electric Flux and Gauss’s Theorem 65. A cylinder of radius, R and length, L is placed in a uniform electric field, E parallel to the cylinder axis. The total flux for the surface of the cylinder is given by (a) zero

(b) pR2 / E

(c) 2 pR2 E

(d) None of these

66. The Electric flux through the surface [NCERT Exemplar] S S +q

(i)

S

S

(ii)

(iii)

(c) ò (E1 + E2 + E3 ) × dA = q 1 + q2 + q3 / 2 e 0 (d) None of the above

68. Five charges q1, q2 , q3, q4

and q5 are fixed at their positions as shown in figure S is a Gaussian surface. The Gauss’s law is given by q ò E × ds = e 0

(iv)

(a) in figure (iv) is the largest (b) in figure (iii) is the least (c) in figure (ii) is same as figure (iii) but is smaller than figure (iv) (d) is the same for all the figures

as shown in figure and S is the spherical gaussian surface of radius, R. Which of the following is true according to the Gauss’s law?

q1 q4 q3

q4 q5

q3

69. A square surface of side lm in the plane of the paper. A uniform electric field E (V /m) also in the plane of the paper, is limited only to the lower half of the square surface, the electric flux (in SI units) associated with the surface is E

(b) El 2 (d) EL/2

(a) zero (c) El 2 /2

70. A hollow cylinder has a charge q coulomb within it. If f is the electric flux in unit of V-m, associated with the curved surface B, the electric flux linked with the plane surface A in unit of V-m, will be B C

A

æ q ö (a) ç - f ÷ è e0 ø

(b)

(c) q / e f

(d) q - e f / f

ö 1æ q ç - f÷ è ø 2 e0

71. The adjacent diagram shows a charge + Q held on an

R q2

q2

(a) E on the LHS of the above equation will have a contribution from q1, q5 and q3 on the RHS will have a contribution from q2 and q4 only (b) E on the LHS of the above equation will have a contribution from q2 and q3 only (c) E on the LHS of the above equation will have a contribution from all charges while q on the RHS will have a contribution from q1, q2 and q5 only (d) Both E on the LHS and q on the RHS will have contributions from q2 and q4 only.

67. q1 , q2 , q3 and q4 are point charges located at points

S

S

q1

+q

+q

+q

(b) ò (E1 + E2 + E3 ) × dA = ( q 1 + q2 + q3 + q 4 ) / e 0

[NCERT Exemplar]

through 180° by a fixed uranium nucleus. The distance of closest approach is of the order of (b) 10 -10 cm (c) 10 -12 cm

(a) ò (E1 + E2 + E3 ) × dA = ( q1 + q2 + q3 ) / e 0

Which of the following statements is correct?

64. An alpha particle of energy 5 MeV is scattered (a) 1Å

759

insulating support S and enclosed by a hollow spherical conductor, O represents the centre of the spherical conductor and P is a point such that OP = x and SP = r. The electric field at point, P will be

760 JEE Main Physics Charge + Q

76. Two insulated metal spheres of radii 10 cm and

r

S

15 cm charged to a potential of 150 V and 100 V respectively, are connected by means of a metallic wire. What is the charge on the first sphere?

P

x O

(a) 2 esu

(b) 4 esu

(c) 6 esu

(d) 8 esu

77. The variation of potential V with distance x from a (a) zero (c)

fixed point charge is shown in figure. The electric field strength between x = 0.1 m and 0.3 m is

Q (b) 4 pe 0 x2

Q e 0 x2

y

(d) None of these 3

72. The electrostatic potential inside a charged

spherical ball is given by f = ar2 + b, where, r is the distance from the centre, a and b are constants. Then the charge density inside the ball is (a) - 24 p a e 0 r (c) -24 p e 0

(b) - 6 ae 0 (d) - 6 ae 0 r

placed at the vertices of an isoscales right angled triangle. Which of the numbered vectors coincides in direction with the electric field at the mid-point M of the hypotaneous? 3

EB

4

EC

1

(b) 3

B

(c) 2

(d) 1

74. Two electric dipoles of moment P and 64 P are placed in opposite direction on a line at a distance of 25 cm. The electric field will be zero at point between the dipoles whose distance from dipole of moment P is (a) 10 cm

(b) 5 cm

(c) + 10 Vm

0.3

x

(b) - 0.4 Vm -1

-1

(d) - 10 Vm -1

78. The electric potential V at any point (x, y, z) in space

is given by V = 4 x2 . The electric field at (1, 0, 2) m in Vm -1 is

79. A hollow conducting sphere of radius, R has a charge

A

(a) 4

(a) + 0.4 Vm -1

0.2

(a) 8, along negative X-axis (b) 8, along positive X-axis (c) 16, along negative X-axis (d) 16, along positive Z-axis

2

M

1

0.1

73. Three identical point charges as shown in figure, are

C

2

(c) 8 cm

(d) 20 cm

(+Q) on its surface. What is the electric potential within the sphere at a distance, r = R/3 from its centre? 1 4 pe 0 1 (c) 4 pe 0

(a)

Q r Q × R ×

(b)

1 Q × 4 pe 0 r2

(d) Zero

80. Two point charges -q and + q/2 are situated at the

Electric Potential, Electric Flux and Capacitor 75. The electrostatic potential on the surface of a charged conducting sphere is 100 V. Two statements are made in this regard. S1 : At any point inside the sphere, electric intensity is zero. S2 : At any point inside the sphere, the electrostatic potential is 100 V. Which of the following is a correct statements. [NCERT Exemplar]

(a) S1 is true but S2 is false. (b) Both S1 and S2 are false (c) S1 is true, S2 is also true and S1 is the cause of S2 (d) S1 is true, S2 is also true but the statements are independent.

origin and at the point ( a, 0, 0), respectively. The point along the X-axis, whereas the electric field vanished, is (a) x =

2a 2 -1

(c) x = ( 2 - 1) 2a

(b) X = 2a - 2 - 1 (d) None of these

81. A positively charged particle is released from rest in an uniform electric field. The electric potential energy of the charge [NCERT Exemplar] (a) (b) (c) (d)

remains a constant because the electric field is uniform increases because the charge moves along the electric field decreases because the charge moves along the electric field decreases because the charge moves opposite to the electric field

Electrostatics 82. A ball of mass 1 kg carrying a charge 10-8 C moves from a point A at potential 600 V to a point B at zero potential. The change in its kinctic energy is (a) -6 ´ 10 -6 erg (b) -6 ´ 10 -6 J

761

89. An electric field is given by E = ( yi$ + x$j ) NC -1. The work done in moving a 1 C charge from rA = (2$i + 2$j)

m to rB = (4$i + 2$j) m is (a) 2 y (c) zero

(b) 3 y (d) infinity

90. The work done by electric field done during the

(c) 6 ´ 10 -6 J (d) 6 ´ 10 -6 erg

83. A charge (– q) and another charge (+ Q) are kept at two points A and B, respectively. Keeping the charge ( + Q) fixed at B, the charge (– q) at A is moved to another point C such that ABC forms an equilateral triangle of side l. The net work done in moving the charge (– q) is 1 Qq 4 p e0 l 1 (c) Qql 4 p e0

(b)

(a)

1 Qq 4 p e 0 l2

(d) zero

displacement of a negatively charged particle towards a fixed positively charged particle is 9 J. As a result the distance between the charges has been decreased by half. What work is done by the electric field over the first half of this distance? (a) 3 J (c) 1.5 J

(b) 6 J (d) 9 J

91. A parallel plate capacitor is connected to a battery. A metal sheet of negligible thickness is placed between the plates at their centre. Which of the following is correct?

(a) - 886 . ´ 10 -7 C

(b) 7 . 86 ´ 10 -7 C

(a) Equal and opposite charges will appear in the low these faces of metal sheet (b) Capacity remain same (c) Potential difference between the plates increases (d) Battery supplies more charge

(c) 6 . 85 ´ 107 C

(d) 6.85 ´ 10 -7 C

92. A sphere of radius r is charged to a potential V. The

84. The flux entering and leaving a closed surface are

5 ´ 105 and 4 ´ 105 in MKS unit respectively, then the charge inside the surface will be

85. The potential difference applied to an X-ray tube is 5 kV and current through it is 3.2 mA. Then, the number of electrons striking the target per second is (a) 2 ´ 1010

(b) 3 ´ 1018

(c) 2 ´ 1016

(d) 5 ´ 1015

86. Equipotentials at a great distance from a collection of charges whose total sum is not zero are [NCERT Exemplar] approximately. (a) spheres (c) paraboloids

(b) planes (d) ellipsoids

outward pull per unit area of its surface is given by 4 pe 0V 2 r2 2pe 0V 2 (c) r2

e 0V 2 2r 2 e V2 (d) 0 2 4r (b)

(a)

93. Figures shown below regular hexagons, with charges at the vertices. In which of the following cases the electric field at the centre is not zero? q

q

q

q

q

–q

q

q

87. The work done in carrying an electron from point A to a point B in an electric is 10 MJ. The potential difference ( VB - V A ) is then (a) + 2 kV (c) + 200 V

(b) - 2 kV (d) None of these

q

q

q

(a) 2q

2q

2q

q

q

88. Shown below is a distribution of charges. The flux of electric field due to these charges through the surface is

2q

2q

–q (b)

q

2q

–q

(d)

94. n small drops of same size are charged to V volt each. If they coalesce to form a single large drop, then its potential will be

+

(a) 3q / e 0 (c) 2q / e 0

q

2q

(c) +q

q

(b) zero (d) q / e 0

(a) Vn (c) Vn1/3

(b) Vn -1 (d) Vn 2 /3

762 JEE Main Physics 95. A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube. [NCERT] 4q 3 pe 0 b 3q (c) 2 pe 0 b2

3q 2 p e0b 2q (d) 3 pe 0 b (b)

(a)

simultaneously to the same potential of 10 V each. Assuming drops to be spherical, if all the charged drops are made to combine to form one larger drop, then the potential of larger drop would be (a) 45 V

(b) 135 V

(c) 270 V

(d) 90 V

103. The electric field intensity at a point P due to long

96. Potential energy of two equal negative point charges 2 mC each held 1 m apart in air is (a) 2 J (c) 4 J

102. 27 identical drops of mercury are charged

uniformly charged wire as shown in figure (charge per unit length is l)

(b) 2 eV (d) 0.036 J X

97. In a region of space, the electric field is given by $ . The electric flux through a surface E = 8 $i + `4 $j + 3 k of area of 100 units xy-plane is (a) 800 units (c) 400 units

R

P

Y

l l 2l (b) (c) (a) 2 2p e 0 R 2p e 0 R 2p e 0 R

(b) 300 units (d) 1500 units

98. Figures shows some equipotential lined distributed in space. A charged object is moved from point A to [NCERT Exemplar] point B. 30V40V

B

A

l 4 2p e 0 R

104. Which one of the following graphs figure shows the variation of electric potential V with distance r from the centre of a hollow charged sphere of radius R? V

V

30V

(a) A

(d)

B

A

(b)

B

R

r

r

R V

V 10V 20V 30V 40V 50V 10V 20V 50V 10V 20V 40V 50V Fig.(i) Fig.(ii) Fig.(iii)

(c) (a) The work done in Fig. (i) is the greatest (b) The work done in Fig. (ii) is least (c) The work done in Fig. (i), Fig. (ii) and Fig. (iii) (d) The work done in Fig. (iii) is greater than Fig. (ii) but equal to that in Fig. (i)

99. In a region of space having a unifrom electric field E, a hemispherical bowl of radius r is placed. The electric flux f through the bowl is (a) 2p rE

(b) 4 p r2 E

(c) 2p r2 E

(d) p r2 E

If one proton is kept at least distance and the other is released, the kinetic energy of second proton when it is at infinite separation is (a) 23.0 ´ 10 (c) 2.3 ´ 10

-19

J

J

(b) 115 . ´ 10

-19

J

(d) zero

101. A regular hexagon of side 10 cm has a charge 5 mC at each of its vartices. The potential at the centre of the hexagon is? [NCERT] (a) 3.7 ´ 106 V 6

(c) 4 ´ 10 V

R

r

R

(b) 2.7 ´ 106 V (d) 5 ´ 107 V

r

105. A capacitor connected to a 10 V battery collects a charge of 40 mC with air as dielectric and 100 mC with a given oil as dielectric. The dielectric constant of the oil is (a) 1.5 (b) 2.0 (c) 2.5 (d) 3.0

106. A capacitor of 4 mF is

100. Two free protons are separated by a distance of 1 Å.

-19

(d)

connected as shown in the circuit. The internal resistance of the battery is 0.5 W. The amount of charge on the capacitor plates will [NCERT Exemplar] be (a) 0

(b) 4 mC

(c) 16 mC

4µ F

10 Ω

2.5 V

2Ω

(d) 8 mC

107. The capacitance of a spherical condensers is 1 mF. If the spacing between two spheres is 1 mm, the radius of the outer sphere is (a) 3 m (b) 7 m (c) 8 m (d) 9 m

Electrostatics 108. A parallel plate capacitor has a capacitance of 50 mF in air and 100 mF when immersed in an oil. The dielectric constant K of the oil is (a) 2.2

(b) 1.1

(c) 0.45

(d) 5.0

109. A

parallel plate d K1 capacitor is made of two 1 dielectric blocks in d K2 2 series. One of the blocks has thickness d1 and dielectric constant K 1 and the other has thickness d2 and dielectric constant K 2 as shown in figure. This arrangement can be thought as a dielectric slab of thickness d( = d1 + d2 ) and effective dielectric constant K . The K is [NCERT Exemplar] K1d1 + K2 d2 d1 + d2 K1K2 ( d1 + d2 ) (c) ( K1d1 + K2 d2 )

(a)

K1d1 + K2 d2 K1 + K2 2K1K2 (d) K1 + K2

(b)

(a) 8 V (c) 12 V

2F

12V

(b) 4 V (d) 6 V

1F

111. The equivalent capacity between points A and B in figure will be, while capacitance of each capacitor is 3 mF. d A

(b) 4 mF

(c) 7 mF

(b) 2 CV 2

(c) 3 / 4 CV 2

(d) 1 / 2 CV 2

113. The force on each plate of parallel plate capacitor has

1 a magnitude equal to QE, where Q is the charge on 2 the capacitor and E is the magnitude of electric field [NCERT] between the plates. Then E contributes to the force against which the plates are 2 moved E (b) contributes to the force against which the plates are 3 moved (c) E contributes force against which the plates are moved (d) None of the above

(a)

2 mF in a circuit across a potential difference of 1 kV. A large number of 1 mF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of [NCERT] capacitors. (a) six rows having 3 capacitors in each row (b) three rows having 6 capacitors in each row (c) nine rows having 2 capacitors in each row (d) Two rows having 9 capacitors in each row

Then, it is disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process ?

(d) 9 mF

112. Two condenser one of capacity C and other of V+ capacity C/2 are connected to 9V – battery, as shown in

(a) 1 / 4 CV 2

115. A 600 pF capacitor is charged by a 200 V supply.

B

(a) 2 mF

figure. The work done in charging fully both condensers is

114. An electrical technician requires a capacitance of

110. In a circuit shown in figure, the potential difference across the capacitor of 2 F is

[NCERT] C

C/2

(a) 4 ´ 10 -6 J

(b) 6 ´ 10 -6 J

(c) 5 ´ 10 -6 J

(d) 8 ´ 10 -6 J

Round II Only One Correct Option 1. Two equally charged small balls placed at a fixed

distance experiences a force F. A similar uncharged ball after touching one of them is placed at the middle point between the two balls. The force experienced by this ball is (a)

F 2

(b) F

(c) 2 F

763

(d) 4 F

2. The maximum field intensity on the axis of a uniformly charged ring of charge q and radius R will be

1 4 pe 0 1 (c) 4 pe 0

(a)

q 3 3R2 2q × 3 3R2 ×

(Mixed Bag) 1 4 pe 0 1 (d) 4 pe 0

(b)

2q 3R2 3q × 2 3R2 ×

3. Charges 2q, - q and – q lie at the vertices of an

equilateral triangle. The value of E and V at the centroid of the triangle will be (a) E ¹ 0 and V ¹ 0 (c) E ¹ 0 and V = 0

(b) E = 0 and V = 0 (d) E = 0 and V ¹ 0

764 JEE Main Physics 4. If a positively charged

9. Work done in carrying a charge Q1 once round a circle

pendulum is oscillating in a uniform electric field as shown in figure. Its time period as compared to that when it was uncharged will

of radius R with a charge Q2 at the centre is (a)

B of an equilateral triangle ABC of side a. The magnitude of electric field intensity at the point C is 2q q q 3 2q (c) (a) (b) (d) 4 pe 0 a2 4 pe 0 a2 4 pe 0 a2 4 pe 0 a2

6. Six charges, three positive and three negative of equal magnitude are to be placed at the vertices of a regular hexagon such that the electric field at O is double the electric field when only one positive charge of same magnitude is placed at R. Which of the following arrangements of charges is possible for P, Q, R, S, T and U, respectively? Q

U

R O

(d) infinite

(a)

V 2r

(b)

v 3r

(c)

V 6r

(d)

V 4r

11. The electric potential at a point (x, y) in the xy-plane is given by V = – Kxy

The electric field intensity at a distance r from the origin varies as (a) r2 (c) 2r

(b) r (d) 2r2

12. If the electric flux entering and leaving an enclosed surface are f1 and f 2 respectively, then charge enclosed in closed surface is f - f1 f +f (b) 1 2 (a) 2 e0 e0 f1 - f2 (c) (d) e 0 (f2 - f1 ) e0

(– d, 0) and (d, 0) respectively of a (x, y) coordinate system, then

(b) +, –, + –,+, – (d) –, +, +, –, +, –

7. Three charges - q1, + q2 and -q3 are placed as shown in figure. The x component of the force on -q1 proportional to

is

y

–q 3

Q1Q2 4pe 0 R

13. Two point charges + q and - q are held fixed at

S

(a) +, –, +, –, –, + (c) +, +, –, +, –, –

(c)

potential difference between its surface and a point at a distance 3r from the centre is V, then electric field intensity at a distance 3r is

5. Equal charges q each are placed at the vertices A and

T

(b) zero

10. A hollow charged metal sphere has radius r. If the

(a) increase (b) decrease (c) not change (d) first increase and then decrease

P

Q1Q2 4pe 0 R2

(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) work has to be done in bringing a test charge from infinity to the origin (d) the dipole moment is 2 qd directed along $i

14. In which of the states shown in figure, is the a

potential energy of a electric dipole maximum?

θ

+q

b –q1

(a) (c)

q2 b2 q2 b2

+

q3 a2 q3 a2

x

+q2

sin q

(b)

sin q

(d)

q2 b2 q2 b2

+

–q

q3 a2 q3 a2

cos q

+q

(a)

E

(b) E

cos q

–q q

8. Electric field on the axis of a small electric dipole at a distance r is E1 and E2 at a distance of 2r on a line of perpendicular bisector. Then E1 8 E1 (c) E2 = 4

(a) E2 = -

(b) E2 = (d) E2 =

(c)

E

(d) q

E1 16

E1 8

q E

–q

Electrostatics 15. The magnitude of electric field E in the annual region of a charged cylindrical capacitor l

765

20. 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 the condition

b a

(a) VA < VB (c) VA < VC

(b) VA > VB (d) VA > VC

21. A charged body has an electric flux f associated with (a) is same throughout (b) is higher near the outer cylinder than near the inner cylinder 1 (c) varies as , where r is the distance from the axis r 1 (d) varies as 2 , where r is the distance from the axis r

16. A positive point charge q is carried from a point B to a point A in the electric field of a point charge + Q at O. If the permittivity of free space is e 0 , the work done in the process is given by (where, a = OA and b = OR) qQ 4 pe 0 qQ (c) 4 pe 0 (a)

æ 1 1ö ç + ÷ è a bø 1ö æ 1 ç 2 - 2÷ èa b ø

qQ 4 pe 0 qQ (d) 4 pe 0 (b)

æ 1 1ö ç - ÷ è a bø 1ö æ 1 ç 2 + 2÷ èa b ø

17. Three large parallel plates have uniform surface charge densities as shown in the figure. Find the electric field at point P. ^

k

σ P –2 σ –σ

- 4s $ k e0 - 2s $ (c) k e0

(a)

z=a z = –a

it. The body is now placed inside a metallic container. The electric flux, f1 associated with the container will be (a) f1 = 0 (c) f1 = f

22. 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 be due to (a) q2 (c) all the charges

+q

corners of a square and a charge q is at its centre. If the system is in equilibrium, the value of q is Q (1 + 2 2) 4 Q (c) - (1 + 2 2) 2

Q (1 + 2 2) 2 Q (d) (1 + 2 2) 4

(a) -

(b)

24. In the electric field of a point charge q, a certain point charge is carried from point A to B, C, D and E as shown in figure. The work done is A

4s $ k e0 2s $ (d) k e0

(b)

(a) 0.22 mm (b) 0.44 mm (c) 0.66 mm (d) 0.88 mm

+q1

(b) only the positive charges (d) + q1 and - q1

z = – 2a

metal plate having surface charge density2 ´ 10-6 cm -2 . The distance from where the electrons be projected so that it just fails to strike the plate is

q2

23. Four charges equal to - Q are placed at the four

E q B D

18. A 100 eV electron is fired directly towards a large

C

(a) least along the path AE (b) least along the path AC (c) zero along any of the paths (d) least along AB

25. Two spheres of radii R1 and R2 joined by a fine wire

19. Three concentric conducting spherical shells carry charges as follows : + Q on the inner shell, - 2Q on the middle shell and - 5Q on the outer shell. The charge on the inner surface of the outer shell is (a) zero (c) - 2Q

(b) 0 < f1 < f (d) f1 > f

(b) + Q (d) - 3Q

are raised to a potential V. Let the surface charge densities at these two spheres be s1 and s2 , s respectively. Then the ratio 2 has a value s1 (a)

R1 R2

(b)

R2 R1 2

(c) 1

æ R2 ö (d) ç ÷ è R1 ø

766 JEE Main Physics 26. A non-conducting ring of radius 0.5 m carries total charge of 111 . ´ 10-10 C distributed non-uniformly on its circumference producting an electric field everywhere in space.

The value of the line integral

l=0

òl = ¥ - E × dl

O

(l = 0,

being centre of ring) in volt is (a) + 2 (b) –1 (c) – 2 (d) zero

1 C (V12 - V22 ) 4 1 (c) C (V1 - V2 )2 4

(a)

1 C (V12 + V22 ) 4 1 (d) C (V1 + V2 )2 4 (b)

32. Seven capacitors each of the capacitance 2 mF are be connected in a configuration to obtain an effective 10 capacitance of mF. Which of the combination(s) 11 shown in figure will achieve the desired result?

27. A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge - 3Q, the new potential difference between the same two surfaces is (a) V (c) 4V

(a)

(b) 2V (d) –2V

28. A slab of copper of thickness, b is inserted in between the plates of parallel plate capacitor as shown in figure. The separation between the plates is d if b = d/2, then the ratio of capacities of capacitors after and before inserting the slab will be (a) 2 : 1

(b) 2 : 1

(c) 1 : 1

d

(b)

b

(c)

(d) 1 : 2

29. For the circuit shown figure, which of the following statements is true? S1

(d)

V1 = 30V S3

V2 = 20V S 2

C1 = 2pF

C2 = 3pF

(a) With S1 closed, V1 = 15 V , V2 = 20 V (b) With S3 closed, V1 = V2 , V2 = 20 V (c) With S1 and S3 closed, V1 = V2 = 0 (d) With S1 and S3 closed, V1 = 30 V , V2 = 20 V

33. Consider the arrangement of three metal plates A, B and C of equal surface area and separation d as shown in figure. The energy stored in the arrangement, when the plates are fully charged, is A d

30. Two identical metal plates are given positive charges

B

Q1 and Q2 (< Q1), respectively. If they are now brought

C

close together to form a parallel plate capacitor with capacitance C, the potential difference between them is (a) (c)

Q1 + Q2 2C Q1 - Q2 C

(b) (d)

Q1 + Q2 C Q1 - Q2 2C

31. Two identical capacitors have the same capacitance C. One of them is charged to potential V1 and the other to V2. The negative ends of the capacitors are

connected together. When the positive ends are also connected, the decrease in energy of the system is

– + d

e 0 AV 2 2d 2 e 0 AV 2 (c) d

(a)

V

e 0 AV 2 d 3 e 0 AV 2 (d) 2d (b)

34. A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants K1 , K 2 and K 3 as shown. If a single dielectric material is to be used to have the same capacitance C is this capacitors, then its dielectric constant K is given by

Electrostatics A/2

A/2

39. 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) in Vm -1 is

d/2

K2

K1 d K3

(a)

(c) K =

K1 K2 K1 + K2

(a) - 8 $i (c) - 16 i$

d/2

1 1 1 1 = + + K K1 K2 2K3

(b)

+ 2K3

767

1 1 1 = + K K1 + K2 2K3

(b) + 8 $i (d) 16 k$

40. In the given circuit of figure with steady current, the potential drop across the capacitor must be R

–

(d) K = K1 + K2 + 2K3

V

35. In the circuit arrangement shown in figure, the value of C1 = C2 = C3 = 30 pF and C4 = 120 pF. If the combination of capacitors is charged with 140V DC supply, the potential differences across the four capacitors will be respectively C3 C1

C4

V 2R 2V

V 2 2V (d) 3

(a) V (c)

C2

C

–

(b)

V 3

41. Two particles A and B having charges 8 ´ 10-6 C and +

- 2 ´ 10-6 C respectively, are held fixed with a separation 20 cm. Where should a third charged particle be placed so that it does not experience a net electric force?

–

140 V

(a) 80 V, 40V, 40V and 20 V (b) 20V, 40V, 40V and 80 V (c) 35V, 35V, 35V and 35 V (d) 80V, 20V, 20V and 20 V

J A

36. In the arrangement of capacitors

20cm

B

shown in figure, each capacitor is A of 9 mF, then the equivalent

C1

capacitance between the points A C3

C2

and B is (a) 9 mF (c) 4.5 mF

(b) 18 mF (d) 15 mF

C4

37. Two spherical conductors A and B of radii 1 mm and 2  mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire, then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is (a) 4 : 1 (c) 2 : 1

(b) 1 : 2 (d) 1 : 4

(a) 0.2 m (c) 0.6 m

B FCF C FCA x

(b) 0.5 m (d) 0.1 m

42. An electric dipole with electric dipole moment 4 ´ 10-9 cm is aligned at 30° with the direction of a uniform electric field of magnitude 5 ´ 104 NC -1. The magnitude of the torque acting on the dipole is (a) 10 -6 Nm (c) 10 -10 Nm

(b) 10 -4 Nm (d) 10 -2 Nm

43. In the electric field shown in figure, the electric lines in the left have twice the separation as that between those on right. If the magnitude of the field at point A is 40 NC -1. The force experienced by a proton placed at point A is

38. The equivalent capacitance of the combination of three capacitors, each of capacitance C shown in figure between points A and B is C1 A

C3

(a) 6.4 ´ 10 -18 N

B

(b) 3.2 ´ 10 -15 N

C2

C (a) 2

3C (b) 2

A

1 (c) 3C

(c) 5.0 ´ 10 -12 N (d) 2C

(d) 1.2 ´ 10 -18 N

768 JEE Main Physics 44. Two insulated metallic sphere of 3 mF and 5mF capacitances are charged to 300 V and 500 V, respectively. The energy loss, when they are connected by a wire, is (a) 0.0375 J (c) 0.375 J

(b) 0.235 J (d) 375 J

percentage of the stored energy dissipated after the switch S is turned to position 2 is 2 S 8 µF

V 2 µF

(a) 20%

(b) 80%

(c) 10%

(d) 100%

46. Two parallel plane sheets 1 and 2 carry uniform charge densities s1 and s2 , as shown in figure. The magnitude of the resultant electric field in the region marked III is (s1 > s2 ) σ1 + + + + I + + + +

(a)

s 1 + s2 2 e0

(b)

σ2

II

+ + + + + + + +

Sheet 2

s 1 s2 e0

(c)

s1 2 s2 e 0

(d)

s2 2 e0

47. A 4 mF capacitor and a resistance of 2.5 W are in series with 12 V battery. Find the time after which potential difference across the capacitor in 3 times the potential difference across the resher. [Given, log (2) = 0693 ] . (a) 14 s (c) 13.86 s

æ qq 1 a ö (c) ç × 2× ÷ 2 2ø è 4p e0a a

(d) zero

expression for the equivalent capacitance of the system shown in figure is ( A is the cross-sectional area of one of the plates) e A (a) 0 3d 3e 0 A (b) d e0 A (c) 6d (d) None of the above

(b) 16. 93 s (d) 8 s

48. As per figure a point charged + q is placed at the origin O work done in taking another point charge - Q from the point M coordinates (0, a) to another point N coordinate(a, 0) along the strength path MN is Y

X

A

d

d

A 3A

and the charge per unit length is l . The electric field at the centre is l2 2pe 0 a l (c) 2pe 0 a2

l 2pe 0 a l (d) 4e0a

(b)

(a)

q is hanging in between two parallel plates by a string of lenght l. Time period of pendulum is T0 . When parallel plates are charged, the time period changes to T. The ratio of 1/2

æ ö ç g ÷ (a) ç ÷ ç g + qE ÷ è mø

+++++++++++++

L m –––––––––––––

T is equal to T0 3/4

æ ö ç g ÷ (b) ç ÷ ç g + qE ÷ è mø

3/2

qE ö æ ÷ çg+ m÷ (d) ç ç g ÷ ø è

(d) None of these

52. Two positive ions, each carrying a charge q, are separated by a distance. If F is the force of repulsion between the ions, the number of electrons missing from each ion will be (e being the charge on an electron). (a)

N

d

50. A semi circular arc of radius a in charged uniformly

M

O

A

51. A small sphere carrying a charge

III

Sheet 1

æ - qq 1ö (b) ç × 2 ÷ × 2a 2 è 4p e0a a ø

49. The

45. A 2mF capacitor is charged as shown in figure. The

1

æ qq ö (a) ç ÷ × 2a è 4 p e 0 a2 ø

4 pe 0 Fd 2

e 4 pe 0 Fd 2 (c) e2

(b)

4 pe 0 Fd 2

e2 4 pe 0 Fd 2 (d) e

Electrostatics

769

53. An infinite number of charges, each 1 mC are placed

60. If on the concentric hollow sphere of radii r and

on the x-axis with coordinates x = 1, 2, 4, 8 K ¥. If a charge of 1 C is kept at the origin, then what is the net force acting on 1 C charge

R( > r) the charge Q is distributed such that their surface densities are same, then the potential at their common centre is

(a) 10000 N (c) 12000 N

(b) 32000 N (d) 18000 N

(a)

54. Two particles of equal mass m and charge q are placed at a distance of 16 cm. They do not experience q any force. The value of is m (a)

pe 0 G

(b)

G pe 0

(c) 4 pe 0 G

(d) l

55. Three charges each of magnitude q are placed at the corners of an equilateral triangle. The electrostatic force on the charge placed at the centre is 1 3q2 4 pe 0 L2

(a)

1 q2 4 pe 0 L2

(b)

(c)

1 q2 12pe 0 L2

(d) zero

Q( R2 + r 2 ) 4 pe 0 ( R + r )

Q( R + r ) 4 pe 0 ( R2 + r 2 ) QR (d) R+r

(b)

(c) zero

61. Two equal charges q of opposite sign separated by a distance 2a constitute an electric dipole of dipole moment p. If P is a point at a distance r from the centre of the dipole and the line joining the centre of the dipole to this point makes and angle q with the axis of the dipole, then the potential at P is given by ( r > > 2a) (where, p = 2qa) p cos q 2pe 0 r2 p cos q (c) V = 4 pe 0 r

p sin q 4 pe 0 r p cos q (d) V = 4 pe 0 r2

(a) V =

(b) V =

56. Two point charges placed at a certain distance r in

62. Two infinitely long parallel wires having linear

air exert a force F on each other. Then the distance r at which these charges will exert the same force in a medium of dielectric constant K is given by

charge densities l1 and l2 respectively are placed at a distance of R metres. The force per unit length on æ 1 ö either wire will be ç K = ÷ 4 pe 0 ø è

(a) r/K

(b) r/ K

(c) r K

(d) r

57. Five balls numbered 1 to 5 are suspended using separate threads. Pairs (1, 2), (2, 4) and (4, 1) show electrostatic attraction,while pair (2, 3) and (4, 5) show repulsion, therefore ball 1 must be (a) neutral (c) negatively charged

(b) positively charged (d) None of these

58. Three plates of common surface area A are connected as shown in figure. The effective capacitance will be d A

B d

(a) 3 e 0 A /d (c) 2 e 0 A /d

(b) e 0 A /d 3 (d) e 0 A /d 2

59. A solid metallic sphere has a charge + 3Q. Concentric with this sphere is a conducting spherical shell having charge - Q. The radius of the sphere is a and that of the spherical shell is b ( b > a). What is the electric field at a distance R( a < R < b) from the centre? 3Q 4pe 0 R2 Q (c) pe 0 R

(a)

Q 2 pe 0 R Q (d) 4pe 0 R2

(b)

(a) K

2 l1l2 R2

(b) K

2 l1l2 R

(c) K

l1l2 R2

(d) K

l1l2 R

63. Suppose, an imaginary cube is with a charge situated at the centre of it. The total electric flux passing through each of the faces of the cube will be q 6e0 q (c) 12e 0

(a)

(b)

q 2e 0

(c) None of these

64. Two equal negative charges - q are fixed at the points (0, a) and (0, - a) on the Y-axis. A positive charge Q is released from rest at the point (2a, 0) on the X-axis. The charge Q will (a) execute simple harmonic motion about the origin (b) move to the origin and remain at rest (c) move to infinity (d) execute oscillatory but not simple harmonic motion

65. An electric line of force in the xy-plane is given by

equation x2 + y2 = 1. A particle with unit positive charge, initially at rest at the point x = 0, y = 0 in the xy-plane (a) not move at all (b) will move along straight line (c) will move along the circular line of force (d) information is insufficient to draw any conclusion

770 JEE Main Physics 3µF

66. In the figure below, what is the potential difference between the points A and B and between B and C respectively, in steady state? (a) VAB (b) VAB (c) VAB (d) VBC

= VBC = 50V = 25 V, VBC = 75 V = 75 V, VBC = 25 V = VAB = 100V

1µF

B

1µF

3µF

(a) -8 ´ 103 C

(b) 854 . ´ 10 -4 C

(c) 8 ´ 10 -3 C

(d) 0.85 ´ 10 -6 C

71. Three infinitely charged sheets are kept parallel to 1µF

xy-plane having charge densities as shown in figure. Then the value of electric field at point P is

10Ω

100V

20Ω

C

A

(c) 3 : 5

(d) 5 : 2

68. A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is connected to another battery and is charged to potential difference 2V. The charging batteries are now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is (a) infinite

(b) zero

(c)

3 C V2 2

(d)

6 C V2 2

69. A finite ladder is constructed by connecting several sections of 2mF, 4mF capacitor combinations as shown in the figure. 4µF

4µF

4µF

A

2µF

2µF

–2

Z=0

identical parallel plate B A capacitors connected to a battery with switch S closed. The switch is now opened and the free space between the plate of capacitors is filled with a dielectric constant 3. What will be the ratio of total electrostatic energy stored in both capacitors before and after the introduction of the dielectric? (b) 1 : 5

P

Z = 2a

67. Figure given below shows two

(a) 1 : 2

Z

Z = 3a

2µF

(a) -

2s $ k e0

(b)

–

2s $ k e0

(c) -

4s $ k e0

4s $ k e0

72. An electric dipole consists of two opposite charges of

magnitude q = 1 ´ 10-6 C separated by 2.0 cm. The dipole is placed in an external field of 1 ´ 105 NC -1. What maximum torque does the field exert on the dipole? How much work must an external agent do to turn the dipole end, starting from position of alignment (q = 0° ) ? (a) 4.4 ´ 106 N-m, 32 . ´ 10 -4 J (b) 2 ´ 103 N-m, -4 ´ 10 -3 J (c) 4 ´ 103 N-m, 2 ´ 10 -3 J (d) 2 ´ 10 -3 N-m, 4 ´ 10 -26 J

73. Two insulated charged conducting spheres of radii 20 cm and 15 cm, respectively and having an equal charge of 10 mC are connected by a copper wire and then they are separated. Then (a) both spheres will have equal charges (b) surface charge density on the 20 cm sphere will be greater than that on the 15 cm sphere (c) surface charge density on the 15 cm sphere will be greater than that on the 20 cm sphere (d) surface charge density on the two spheres will be equal

74. A point charge q moves from point P to point S along the path PQRS in a uniform electric field E pointing parallel to the positive direction of the x-axis as shown in figure. y

C

E

P(a, b, 0) B

x

It is terminated by a capacitor of capacitance C. What value should be chosen for C such that equivalent capacitance of the ladder between the points A and B becomes independent of the number of sections in between. (a) 4 mF

(d)

(b) 8 mF

(c) 12 mF

(d) 16 mF

70. An electric dipole is placed at an angle of 60° with an

electric field of intensity 105 NC -1. It experiences a torque equal to 8 3 N-m. Calculate the charge on the dipole, if the dipole length is 2 cm.

(0, 0, 0)S

Q(2a, 0, 0) R(a, –b, 0)

The coordinates of the points P, Q, R and S are ( a, b, 0) (2a, 0, 0), (a, – b, 0) and (0, 0, 0) respectively. The work done by the field in the above process is given by the expression (b) - qaE

(a) qE 2

2

(c) q ( a + b ) + E

(d) 3qE ( a2 + b2 )

Electroststics 75. The electron is projected from a

P

2de 0 m u 2 (a) el 2 de m u 2 (c) 0 el

separation of the capacitor. The capacity of the capacitor is

u

distance d and with initial velocity u parallel to a uniformly charged flat conducting plate as l shown in figure. It strikes the plate after travelling a distance l Y along the direction. The surface charge density of conducting plate is equal to

X

2de 0 m u (b) el de m u (d) 0 el

(a)

2 e 0 d æ K1 + K2 ö ç ÷ A è K1K2 ø

(b)

2 e 0 A æ K1K2 ö ç ÷ d è K1 + K2 ø

(c)

2 e0d ( K1 + K2 ) A

(d)

2 e 0 A æ K1 + K2 ö ç ÷ d è K1K2 ø

80. The effective capacitance between points X and Y shown in figure. Assuming C2 = 10 mF and that outer capacitors are all 4 mF is C4

76. A large insulated sphere of radius r charged with

X

Y C1

Q units of electricity is placed in contact with a small insulated uncharged sphere of radius r¢ and in then

separated. The charge on smaller sphere will now be Qr ¢ r¢ + r Q (d) r¢ + r

(a) Q ( r + r ¢ )

(b)

(c) Q ( r + r ¢ )

–q

(a) 1 mF (c) 4 mF

(b) 3 mF (d) 5 mF

shown in figure. The DC voltmeter reads 200 V. The charge on each plate of capacitor is

q

4

3

1 q2 × 4 pe 0 a

(c) +

V – +

– +

– +

– +

2

+q

a

(a)

C3 C5

corners of a square of side a, as shown in figure, is 1

C2

81. The four capacitors, each of 25 mF are connected as

77. The work required to put the four charges at the q

771

2.6 q2 × 4 pe 0 a

(b) -

2.6 q2 × 4 pe 0 a

(d) None of these

78. If a point charge q is placed at a point inside a hollow conducting sphere, then which of the following electric lines of force pattern is correct?

(a) ± 2 ´ 10 -3 C

(b) ± 5 ´ 10 -3 C

(c) ± 2 ´ 10 -2 C

(d) ± 5 ´ 10 -2 C

82. A 10 mF capacitor and a 20 mF capacitor are connected in series across 200 V supply line. The charged capacitors are then disconnected from the line and reconnected with their positive plates together and negative plates together and no external voltage is applied. What is the potential difference across each capacitor? 800 800 V (b) V (c) 400 V (d) 200 V (a) 9 3

83. A parallel plate capacitor is made by stocking n (a)

(b)

+q

equally spaced plates connected alternately. If, the capacitance between any two plates is x, then the total capacitance is, (a) nx

(b) n/x

(c) nx2

(d) (n–1)x

84. Six identical capacitors are joined in parallel, (c)

+q

(d) None of these

79. A parallel plate capacitor has the space between its d each and 2 and K 2. d is the plate

charged to a potential difference of 10 V, separated and then connected in series, i.e., the positive plate of one is connected to negative plate of other. Then potential difference between free plates is

plates filled by two slabs of thickness

(a) 10 V

dielectric constants K1

(c) 60 V

(b) 30 V 10 (d) V 6

772 JEE Main Physics More Than One Correct Option 85. ò E × dS = 0 over a surface, then [NCERT Exemplar] (a) the electric field inside the surface and on it is zero (b) the electric field inside the surface is necessarily uniform (c) the number of flux lines entering the surface must be equal to the number of lux lines leaving it (d) all charges must necessarily be outside the surface

86. A positive charge Q is located at the centre of a thin metallic spherical shell. Select statement(s) from the following.

the

correct

90. If there were only one type of charge in the universe, then

[NCERT Exemplar]

(a) ò E × dS ¹ 0 on any surface (b) ò E × dS = 0 if the charge is outside the surface (c) ò E × dS could not be defined (d)

q

ò E × dS = e 0 if charges of magnitude q where inside the surface

91. A parallel plate capacitor is connected to a battery. A

(a) The electric field at any point outside the shell is zero (b) The electrostatic potential at any point outside the shell Q is , where, r is the distance of the point from centre 4 pe 0 r

metal sheet of negligible thickness is placed between two plates at their centre. Which of the following is correct?

(c) The outer surface of the spherical shell is an equipotential surface (d) The electric field at any point inside the shell, other than centre point is, zero

(a) Equal and opposite charges will appear on the two faces of metal sheet (b) Capacity remains same (c) Potential difference between the plates increases (d) Battering supplies more charge

87. A circular ring of radius R with uniformly distributed

92. Each plate of a parallel plate capacitor has a charge q

charge q is placed in their yz-plane with its centre at the origin. Select the correct statement(s) out of the following.

on it. The capacitor is now connected to a battery. Which of the following statements are true?

(a) The electric intensity is maximum at x = ± 2R 2 R 2 q (c) The maximum intensity has a magnitude 6 3pe 0 R2

(b) The electric intensity is maximum at x = ±

q (d) The maximum intensity is 6 6 pe 0 R2

88. The electric field at a point is

[NCERT Exemplar]

(a) always continuous (b) continuous if there is no charge at that point (c) discontinuous only if there is a negative charge at that point (d) discontinuous if there is a charge at that point

89. A parallel plate capacitor of plate area A and plate separation d is charged to potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q, E and W denotes respectively the magnitude of charges on each plate, the electric field between the plates (after the slab is inserted) and work done on the system, in the process of inserting the slab, then e 0 AV d V (c) E = Kd

(a) Q =

e 0 KAV d e 0 AV 2 (d) W = 2d

(b) Q =

(a) The surfaces of capacitor plates facing each other have equal and opposite charges (b) Capacity remains same (c) Battery supplies more charge (d) Potential difference between two plates increases

93. Consider a region inside which there are various types of charges but the total charge is zero. At points outside the region [NCERT Exemplar] (a) the electric field is necessarily zero (b) the electric field is due to the dipole moment of the charge distribution only 1 (c) the dominant electric field is µ 3 , for large r, where r is r the distance from a origin in this region (d) the work done to move a charged particle along a closed path, away from the region, will be zero

94. A parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved farther apart by means of insulating handles, then which of the following is correct? (a) The charge on the capacitor increases (b) The voltage across the plates increases (c) The capacitance increases (d) The electrostatic energy stored in the capacitor increases

95. Refer to the arrangement of charges in figure and a 1ù é êë1 - K úû

Gaussian surface of radius R with Q at the centre. [NCERT Exemplar] Then

Electroststics Gaussian surface

+

96. A large non-conducting sheet S

+

A

B

+

– +

– +

+ +

97. A closed surface S is constructed around a metal wire connected to a battery and a key, K . On pressing the key, number of free electrons entering per second is S equal to number of free electrons leaving per second. The electric flux through the closed surface is (b) increased (d) remaining zero

98. The electrostatic potential ( fr ) of a spherical

symmetric system kept at origin, is shown in the q adjacent figure and given as f r = ( r ³ R0 ) and 4 pe 0 r fr =

(a) If q > 0 and is displaced away from the centre in the plane of the ring, it will be pushed back towards the centre (b) If q < 0 and is displaced away from the centre in the plane of the ring, it will never return to the centre and will continue moving till it hits the ring (c) q < 0, it will perform SHM for small displacement along the axis (d) q at the centre of the ring is in an unstable equilibrium within the plane of the ring for q > 0

100. The work done to move a charge along an equipotential from A to B (a) cannot be defined as (b) must be defined as -

(b) S attracts B (d) A repels B

(a) decreased (c) remains constant

R

+ ++ + +

(c) flux through the surface of sphere due to 5Q is zero (d) field on the surface of sphere due to - 2 Q is same everywhere

(a) S attracts A (c) A attracts B

q

q ( r £ R0 ). Which of the following option(s) 4 pe 0r

+

-Q (a) total flux through the surface of the sphere is × e0 -Q (b) field on the surface of the sphere is × 4 pe 0 R2

S is given a uniform charge density. Two uncharged small metal rods A and B are held near the sheet as shown in figure. Which of the following statement is true?

+ + + + + + Q

+

– 2R

+ + + + +

R

R/ 2

+

+ + + +

distributed along a circular ring of radius R. A small test charge q is placed at the centre of the ring [NCERT Exemplar] figure. Then

+

Q

R

99. A positive charge Q is uniformly

773

[NCERT Exemplar] B

òA E × d l B

òA E × d l

(c) is zero (d) can have a non-zero value

101. A positive charged thin metal ring of radius R is fixed in the xy-plane with the centre at the origin O. A negatively charged particle P is released from rest at the point (0, 0, z0 ) where, z0 > 0. Then, the motion of P is (a) periodic, for all values of z0 satisfying 0 £ z0 < ¥ (b) simple harmonic, for all values of z0 satisfying 0 £ z0 £ R (c) approximately simple harmonic, provided z0 R, the electric field is given by

1 -Q × 4 pe 0 r 2 1 Q2 (d) E = × 2 4 pe 0 r

1 Q × 4 pe 0 r 2 1 Q (c) E = × 4 pe 0 r 3

(a) E =

(b) E =

104. For a point situated inside the sphere at a distance r from its centre i.e., r < R, the electric field is given by 1 4 pe 0 1 (c) E = 4 pe 0

(a) E =

Q r R3 Q × ( R - r )2 ×

1 4 pe 0 1 (d) E = 4 pe 0

(b) E =

×

Q r2

×

Q (R - r 3) 3

105. If radius of sphere be 0.1 m and the sphere contains 1 mC charge, then electric field at its surface has a magnitude (a) 9 ´ 1011 NC-1

(b) 9 ´ 105 NC-1 1 (d) ´ 10 -5 NC-1 9

(c) 3 ´ 105 NC-1

106. Given four possible E-r curves for the charged sphere. The correct curve is

Passage II It is defined as the electrostatic potential at a point in an electrostatic field as the amount of work done in moving a unit positive test charge from infinity to that point against the electrostatic forces, along any path. Due to single charge q, potential at a point distant r q . The potential can be from the charge is V = 4 pe 0 r positive or negative. However, it is a scalar quantity. The total amount of work done is bringing various changes to their respective positions from infinitely large mutual separation gives us the electric potential energy of the system of charges. Whereas electric potential is measured in volt, electric potential energy measured in J. We have given a square of each side 1 m with four charge + 1 ´ 10-8 C, -2 ´ 10-8 C, +3 ´ 10-8 C and + 2 ´ 10-8 C placed at 4 corners of square with the help of the passage given above, choose the correct answer of the following questions.

107. Electric potential and electric potential energy both are (a) scalars (b) vectors (c) both (a) and (b) (d) neither (a) nor (b)

108. Potential at the centre of square is (a) 5.09 ´ 105 V

(b) 5.09 ´ 10 2 V

(c) 5.09 V

(d) 5.09 ´ 10 –2 V

109. Electric potential energy of the system of four charges is

E

(a)

(a) -6.4 ´ 10 –7 J

(b) 6.4 ´ 10 –5 J

(c) 8.5 ´ 107 J

(d) 0

110. The electric potential at a point due to a given charge –

–R

+

R

O

r

E

varies inversely as the square of the distance of the point from the charge. The statement is (a) true (c) neither true nor false

(b) –

Passage III –R

+

R

O

r

E

(c) –

–R

O

The given figure shown an arrangement of four parallel, conducting plate of area A each. All the plates are equally separated by d. The plates A and D are joined together and a battery of emf E volt, is attached between the plates B and C.

+

R

A

r

d C

(d)

D

–R

O

+

R r

d

B + E –

E

–

(b) false (d) None of these

d

Electroststics 111. The arrangement of plates is equivalent to capacitors as shown in figure. B A + – C (a)

D C + – C

A D + – C

Column II

+ –

(A) |E| =

II. Electric field

(B) t = pEsin q

C

– +

+

+ –

III. Torque

E

E A D

IV. Gauss’s theorem C – C

1 q 4pe0 r 2

I. Dipole moment

(b) + –

(c) B

116. Match the following Column I with Column II. Column I

B A + – C

D C + – C

775

(C) |p| = q ´ 2 a q òS E × dS = e0

(D)

Code

(d) None of these

(a) I-C, II-A, III-B, IV-D (c) I-D, II-A, III-C, IV-B

C + –

(b) I-A, II-B, III-C, IV-D (d) I-D, II-B, III-C, IV-A

E

112. The equivalent capacitance of the system is given by 2 e0 A 3 d e A (c) 0 d

3 e0 A 2 d 3e A (d) 0 d

Direction (Question No. 117 to 126) Assertion-Reason type. Each of these contains two Statements : Statement I (Assertion), Statement II (Reason). Each of these questions also has four alternative choice, only one of which is correct. You have to select the correct choices from the codes (a), (b), (c) and (d) given below (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 the Assertion (c) If Assertion is true but Reason is false (d) If Assertion is false but the Reason is true

(b)

(a)

113. Total charge on the arrangement is 3 e0 A E 2 d 3 e0 A 2 (c) E 4 d (a)

e0 A E d e AE (d) 0 d 2 (b)

Matching Type 114. Match the following Column I with Column II. Column I

Column II

I.

Electrical capacity

(A)

II.

Permittivity of free space

(B)

III.

Electrical potential

(C)

IV.

–1 –3

4

2

[M L T A ] [M1L2 T –3 A –1 ] [M1L2 T –2 ]

when air separating the charges is replaced by water. Reason Medium intervening between the charges has no effect on force.

[M L T A ]

(b) I-D, II-A, III-C, IV-C (d) I-A, II-D, III-C, IV-B

119. Assertion A point charge is brought an electric field.

(D)

–1 –3 4

2

115. Match the following Column I with Column II. Column I

Column II

I. Coulomb force and gravitational force, both fallow II. In series combination of capacitors

(A) charge on each capacitor is same

III. In parallel combination of capacitons IV. Electric dipole

(B) potential difference across each capacitor is same (C) stable equilibrium (D) inverse square law of distance

Code (a) I-D, II-A, III-B, IV-C (c) I-B, II-C, III-D, IV-A

117. Assertion Force between two charges decreases,

charging by rubbing, the insulating material with lower work function becomes positively charged. Reason Electrons are negatively charged.

Electrical energy

Code (a) I-A, II-D, III-B, IV-C (c) I-D, II-A, III-C, IV-B

Assertion and Reason

(b) I-A, II-B, III-C, IV-D (d) I-D, II-A, III-C, IV-A

118. Assertion During

The field at a nearby point will be increase. Whatever the nature of charge. Reason The electric field is independent of the nature of charge.

120. Assertion The displacement current goes through the gap between the plates of the capacitor when the charge of the capacitor does not change. Reason The displacement current arises in region in which the electric field and hence the electric flux does not change with time.

121. Assertion No two electric field lines can intersect one another at any point in space. Reason Electric field lines always start from a positive charge and end on a negative charge.

776 JEE Main Physics 122. Assertion Net electric flux over a cube enclosing an electric dipole is zero. Reason Total net charge on an electric dipole is zero.

Reason Capacity increases on introducing dielectric slab and hence more charge is drawn from the battery.

123. Assertion The electric field and hence electric field

125. Assertion When charges are shared between two

lines are everywhere at right angle to an equipotential surface.

bodies, there occurs no loss of charge. However, there is a loss in electrical energy.

Reason Equipotential surfaces are closer together where the electric field is stronger and farther apart where the field is weaker.

Reason Electrostatic potential energy does not come under the preview of the conservation law of energy.

124. Assertion The plates of a parallel plate capacitor are connected to a battery. Charge on the plates increases on introducing a dielectric slab between the plates.

126. Assertion An electric dipole is placed in a uniform electric field. Its equilibrium will be stable when dipole is set along the direction of electric field. Reason In stable equilibrium energy of dipole should be least possible.

Previous Years’ Questions 127. In a uniformly charged sphere of total charge Q and radius R, the electric field E is plotted as function of distance from the centre. The graph which would [AIEEE 2012] correspond to the above will be E

129. Combination of two identical capacitors, a resistor R

E

(a)

(b)

R

r

R

E

r

E

(c)

(d)

R

r

(c) Statement 1 is true, Statement II is true Statement II is the correct explanation for statement I (d) Statement 1 is true, Statement II is true, Statement II is not the correct explanation of Statement I

and a DC voltage source of voltage 6 V is used in an experiment on C-R circuit. It is found that for a parallel combination of the capacitor the time in which the voltage of the fully charged combination reduces to half its original voltage is 10 s. For series combination, the time needed for reducing the voltage of the fully charged series combination by [AIEEE 2011] half is (a) 20 s

R

r

128. This question has Statement I and statement II. Of the four choices given after the statements, choose the one that best describes the two statements. An insulating solid sphere of radius R has a uniform positive charge density r. As a result of this uniform charge distribution, there is a finite value of electric potential at the centre of the sphere, at the surface of the sphere and also at a point outside the sphere. The electric potential at infinity is zero. [AIEEE 2012] Statement I When a charge q is taken from the centre of the surface of the sphere its potential energy qe . changes by 3 e0 Statement II The electric field at a distance r ( r < R) rr from the centre of the sphere is . 3 e0 (a) Statement I is false, Statement II is true (b) Statement 1 is true, Statement II is false

(b) 10 s

(c) 5 s

(d) 2.5 s

130. An electric charge + q moves with velocity v = 3 i$ + E = 3 $i +

$ , in an electromagnetic field given by 4 $j + k $ , B = i$ + $j - 3 k $ . The y-component of $j + 2 k

the force experienced by + q is (a) 2 q (c) 5 q

[AIEEE 2011]

(b) 11q (d) 3 q

131. Two positive charges of magnitude q are placed at the end of a side 1 of a square of side 2 a. Two negative charges of the same magnitude are kept at the other corners. Starting from rest, if a charge Q moves from the middle of side 1 to the centre of square, its kinetic energy at the centre of square is [AIEEE 2011] (a)

1 2 qQ æ 1 ö ç1 ÷ è 4 pe 0 a 5ø

(b) zero 1 2 qQ æ 1 ö (c) ÷ ç1 + 4 pe 0 a è 5ø 1 2 qQ æ 2 ö (d) ç1 ÷ 4 pe 0 a è 5ø

Electroststics

777

132. Two identical charged spheres suspended from a

138. Let there be a spherically symmetric charge

common point by two massless strings of length l are initially a distance d ( d R, è 4 Rø where, r is the distance from the origin. The electric field at a distance r ( r < R) from the origin is given by

-1

1/2

(b) v µ x (d) v µ x-1/2

(a) v µ x (c) v µ x

133. A fully charged capacitor C with initial charge q0 is

connected to a coil of self-inductance L at t = 0. The time at which the energy is stored equally between the electric and the magnetic fields is [AIEEE 2011] (a)

p 4

(b) 2p LC

LC

(d) p LC

(c) LC

134. A resistor R and 2mF capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s after the switch has been closed (log10 2.5 = 0.4)

[AIEEE 2010]

4 pr 0 r æ 5 r ö (a) ç - ÷ 3 e0 è 3 R ø 4 r0 r æ 5 r ö (c) ç - ÷ 3 e0 è 4 R ø

r r æ5 r ö (b) 0 ç - ÷ 4 e0 è 3 R ø r r æ5 r ö (d) 0 ç - ÷ 3 e0 è 4 R ø

139. The question contains Statement I and Statement II. Of the four choices given after the statements, choose the one that best describes the two statements. Statement I For a charged particle moving from point P to point Q, the net work done by an electrostatic field on the particle is independent of the path connecting point P to point Q. Statement II The net work done by a conservative force on an object moving along a closed loop is zero. [AIEEE 2009]

[AIEEE 2011] 5

6

(a) 1.7 ´ 10 W

(b) 2.7 ´ 10 W

7

4

(c) 3.3 ´ 10 W

(d) 1.3 ´ 10 W

135. The electrostatic potential inside a charged spherical

ball is given by f = ar2 + b, where, r is the distance from the centre a, b are constants. Then the charge density inside the ball is [AIEEE 2011] (a) - 6 a e 0 r (c) - 6 ae 0

(b) -24 pae 0 (d) -24 pae 0 r

136. Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t1, is the time taken for the energy stored in the capacitor to reduce to half its initial value and t2 is the time taken for the charge to reduce to one-fourth its initial value. Then, the ratio t1 will be [AIEEE 2010] t2 (a) 1

(b)

1 2

(c)

1 4

(d) 2

(a) Statement I is true, Statement II is false (b) Statement I is true, Statement II is true, Statement II is the correct explanation of Statement I (c) Statement I is true, Statement II is true, Statement II is not the correct explanation of Statement I (d) Statement I is false, Statement II is true

140. A spherical metal shell A of radius R A and a solid

metal sphere B of radius RB ( < R A ) are kept far apart and each is given charge ‘ + Q’. Now they are connected by a thin metal wire. Then, [IIT JEE 2011] (a) EAinside = 0

(b) QA > QB

s R (c) A = H sB RA

(d) EAon surface < EBon surface

141. A few electric field lines for a system of two charges Q1

and Q2 fixed at two different points on the x-axis are shown in the figure. These lines suggest that [IIT JEE 2010]

137. A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net field E at the centre O, is [AIEEE 2010]

Q1 Q2

^

j

^

i

O

q (a) 2 2 $j 4p e0r q (c) - 2 2 $j 2p e 0 r

q $j 4 p e 0 r2 q (d) 2 2 $j 2p e 0 r

(b) -

2

(a) |Q1 | > |Q2 | (b) |Q1 | < |Q2 | (c) at a finite distance to the left of Q1 the electric field is zero (d) at a finite distance to the right of Q2 the electric field is zero

778 JEE Main Physics 142. Which of the field patterns given below, is valid for electric field as well as for magnetic field? [IIT JEE 2010]

147. Two charges are at distance d apart. If a copper plate (connecting medium) of thickness d/ 2 is placed between them, the effective force will be [J & K CET 2005]

(a) 2 F (a)

(b) F/2

(c) 0

(d) 2 F

148. A comb run through one’s dry hair attracts small bits

(b)

of paper. This is due to

[Karnataka CET 2006]

(a) comb is a good conductor (b) paper is a good conductor (c) the comb possess magnetic properties (d) the atoms in the paper set polarised by the charged comb (c)

(d)

149. Four metal conductors having different shapes I. A sphere

II. Cylidner

III. Rear

IV. Lightning conductor

Qr 143. Let, r( r) = 4 be the charge density distribution for pR a solid sphere of radius R and total charge Q. For a point P inside the sphere at a distance r1 from the centre of the sphere, the magnitude of electric field is

are mounted on insulating stands and charged. The one which is best suited to retain the charges for a [K CET 2005] longer time as

[AIEEE 2009]

150. Two small spheres of masses, M1 and M2 are

Q (a) 4 pe 0 r12 (c)

Qr12 3pe 0 R4

Qr12 (b) 4pe 0 R4 (d) zero

144. A solid spherical conductor of radius R has a spherical cavity of radius a ( a < R) as its centre. A charge + Q is kept at the centre. The charge at the inner surface, enter and at a position r ( a < r < R) are respectively, [VITEEE 2008] (a) + Q1 - Q, 0 (c) 0, - Q, 0

(b) - Q, + Q, 0 (d) + Q, 0, 0

145. Consider a neutral conducting sphere. A positive point charge is placed outside the sphere. The net charge on the sphere is then [IIT JEE 2007] (a) negative and distributed uniformly over the surface of sphere (b) negative and appears only at the point on the sphere closed to the point charge (c) negative and distributed non-uniformly over the entire surface of the sphere (d) zero

146. Two identical charges repel each other with a force equal to 10 mgwt when they are 0.6 m apart in air ( g = 10 ms -2 ). The value of each charge is [Karnataka CET 2007]

(a) 2mC (b) 2 ´ 10 -7 C (c) 2 nC (d) 2mC

(a) I (c) III

(b) II (d) IV

suspended by weightless insulating threads of lengths L1 and L2 . The sphere carry charges Q1 and Q2 , respectively. The spheres are suspended such that they are in level with another and the threads are inclined to the vertical at angles of q 1 and q 2 as shown below, which one of the following conditions is essential, if Q1 = Q2 ?

L1 θ1 M1

θ2 L2

M2

+Q1

(a) M1 ¹ M2 but Q1 = Q2 (c) Q1 = Q2

+Q2

(b) M1 = M2 (d) L1 = L2

151. A conductor has been given a charge -3 ´ 10-7 C by transferring electron. Mass increase (in kg) of the conductor and the number of electrons added to the [AMU Engg. 2010] conductor are respectively, (a) 2 ´ 10 -16 and 2 ´ 1031 (c) 3 ´ 10 -19 and 9 ´ 1016

(b) 5 ´ 10 -31 and 5 ´ 1019 (d) 2 ´ 10 -18 and 2 ´ 1012

152. The electric potential V at any point O (x, y, z all in

metres) in space is given by V = 4 x2 volt. The electric field at the point (1m, 0, 2m) in V/m is (a) 8 along negative x-axis (b) 8 along positive x-axis (c) 16 along negative x-axis (d) 16 along positive x-axis

Electroststics

779

153. Consider a system of three charges q/ 3, q/ 3 and -2q/ 3 placed at points A, B and C, respectively as shown in figure. Take O to be the centre of the circle of radius, R and Angle, ÐCAB = 90°. Then B

(a)

C

A

q directed along the 4 pe 0 R2

negative x-axis (b) Potential energy of system will be zero q (c) The potential at point O is 12pe 0 R (d) The magnitude of force between the charges and B is q2 54pe 0 R2

154. A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric constant K = 2, the level C d of liquid is , initially. 3 R d d/3 Suppose, the liquid level decreases at a constant speed v, the time constant as a function of time t is

(c)

1 s2 e0 R

(d)

1 s2 e 0 R2

158. Consider a thin spherical shell of radius R with its centre at the origin, carrying uniform positive surface charge density. The variation of the magnitude of the electric field|E ( r)|and the electric potential V ( r) with the distance r from the centre, is [IIT JEE 2012] best represented by which graph |E(r)|

(15 d + 9 vt ) e 0 R 2d 2 - 3dvt - 9 v 2t 2 (15 d - 9 vt ) e 0 R (d) 2 2d + 3dvt - 9 v 2t 2

V(r)

(a)

|E(r)|

V(r)

(b) O

O

r

R

|E(r)|

V(r)

r

R

|E(r)|

V(r)

(b)

(c)

parallel and charged upto potential V. The battery is removed and the condenser of capacity C is filled completely with a medium of dielectric constant K . The potential difference across the capacitors will [IIT 1988; Similar AMU (Engg. 2009] now be 3V (b) K V (d) K

156. A uniformly charged thin spherical shell of radius R carries uniform surface charge density of s per unit area. It is made of two hemispherical shells, held together by pressing them with force F (see figure) [IIT JEE 2010] then F is proportional to

(d) O

155. Two condensers of capacities 2C and C are joined in

3V (a) K+2 V (c) K+2

1 2 sR e0

(a) If the electric field due to a point charge varies as r -2 s instead of r -2 , then the Gauss' law will still be valid (b) The Gauss' law can be used to calculate the field distribution around an electric dipole (c) If the electric field between two point charges is zero somewhere, then the sign of the two charges is the same (d) The work done by the external force in moving a unit positive charge from point A at potential VA to point B at potential VB is (VB - VA )

[IIT JEE 2008]

6 e0R 5 d + 3 vt 6e0R (c) 5 d - 3vt

(b)

[IIT JEE 2011]

90°

(a)

1 2 2 sR e0

157. Which of the following statement(s) is/are correct?

X

(a) The electric field at point O is

F

F

O

r

R

r

R

159. To form a composite 16 mF, 1000 V capacitor from a supply of identical capacitors marked 8 mF, 250 V, we required a minimum number of capacitors. [Karnataka CET 2008]

(a) 40

(b) 32

(c) 18

(d) 22

160. Given, R1 = 1 W, R2 = 2 W, C1 = 2 mF, C2 = 4 mF C1

C 1 C2 R1

V

R1 R1

C2

R2 C 1 C2

R2 (I)

R2 (II) V

V (III)

780 JEE Main Physics 3µF

The time constants (in ms) for the circuits I, II, III are respectively, [IIT JEE 2006] (a) 18, 18/9, 4 (c) 4, 8/9, 18

S

(b) 18, 4, 8/9 (d) 8/9, 18, 4

3Ω

6Ω Y

161. A spherical portion has been removed

9V

from a solid sphere having a charge distributed uniformly in its volume in the figure. The electric field inside the emptied space is [IIT JEE 2007] (a) zero everywhere (c) non-uniform

6µF

X

(b) 54 mC

(a) 0

(c) 27 mC

(d) 81 mC

163. A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. The electric potential at the point O lying at distance L from the end A is [JEE Main 2013]

(b) non-zero and uniform (d) zero only at its centre

162. A circuit is connected as shown in the figure with the

A

O

switch S open. When the switch is closed, the total amount of charge that flows from Y to X is

(a)

[IIT JEE 2007]

Q 8p e 0 L

B

L

(b)

L

3Q 4p e0L

(c)

Q Q ln 2 (d) 4 pe 0 L ln 2 4p e0L

Answers Round I 1. 11. 21. 31. 41. 51. 61. 71. 81. 91. 101. 111.

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

2. 12. 22. 32. 42. 52. 62. 72. 82. 92. 102. 112.

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

3. 13. 23. 33. 43. 53. 63. 73. 83. 93. 103. 113.

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

4. 14. 24. 34. 44. 54. 64. 74. 84. 94. 104. 114.

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

5. 15. 25. 35. 45. 55. 65. 75. 85. 95. 105. 115.

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

6. 16. 26. 36. 46. 56. 66. 76. 86. 96. 106.

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

7. 17. 27. 37. 47. 57. 67. 77. 87. 97. 107.

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

8. 18. 28. 38. 48. 58. 68. 78. 88. 98. 108.

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

9. 19. 29. 39. 49. 59. 69. 79. 89. 99. 109.

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

10. 20. 30. 40. 50. 60. 70. 80. 90. 100. 110.

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

Round II 1. (b) 11. (b) 21. (c) 31. (c) 41. (a) 51. (a) 61. (d) 71. (a) 81. (b) 91. (a,b) 101. (a,c) 111. (c) 121. (c) 131. (a) 141. (a,d) 151. (d) 161. (b)

2. (c) 12. (d) 22. (c) 32. (a) 42. (b) 52. (b) 62. (b) 72. (b) 82. (a) 92. (a,b,c) 102. (a,d) 112. (b) 122. (a) 132. (d) 142. (c) 152. (a) 162. (a)

3. (c) 13. (b) 23. (d) 33. (b) 43. (a) 53. (c) 63. (a) 73. (c) 83. (d) 93. (c,d) 103. (a) 113. (a) 123. (b) 133. (a) 143. (c) 153. (d) 163. (d)

4. (a) 14. (a) 24. (c) 34. (b) 44. (a) 54. (c) 64. (d) 74. (b) 84. (c) 94. (b,d) 104. (a) 114. (a) 124. (a) 134. (b) 144. (b) 154. (a)

5. (c) 15. (c) 25. (a) 35. (a) 45. (b) 55. (d) 65. (c) 75. (a) 85. (c,d) 95. (a,c) 105. (b) 115. (a) 125. (c) 135. (c) 145. (d) 155. (a)

6. (d) 16. (b) 26. (a) 36. (d) 46. (a) 56. (b) 66. (b) 76. (b) 86. (b,c) 96. (a,b,c) 106. (b) 116. (a) 126. (a) 136. (c) 146. (d) 156. (a)

7. (b) 17. (c) 27. (a) 37. (c) 47. (c) 57. (a) 67. (c) 77. (b) 87. (a,c) 97. (c,d) 107. (a) 117. (d) 127. (c) 137. (b) 147. (c) 157. (c)

8. (b) 18. (b) 28. (b) 38. (a) 48. (d) 58. (c) 68. (c) 78. (a) 88. (b,d) 98. (a-d) 108. (a) 118. (b) 128. (c) 138. (b) 148. (d) 158. (d)

9. (b) 19. (a) 29. (d) 39. (a) 49. (d) 59. (a) 69. (c) 79. (b) 89. (a,c,d) 99. (a,b,c) 109. (a) 119. (d) 129. (d) 139. (d) 149. (a) 159. (b)

10. (c) 20. (b) 30. (d) 40. (c) 50. (b) 60. (b) 70. (c) 80. (c) 90. (b,d) 100. (b,c) 110. (b) 120. (b) 130. (b) 140. (a-d) 150. (b) 160. (d)

the Guidance Round I 1. Charge (C) has the largest magnitude since maximum

10. If same charges on spheres A and B are q,

number of field lines are associated with it.

2. For equilibrium, net force on q = 0 kQQ kqQ + 2 =0 (2x) 2 x

\ \

1 q1q 2 99 = (110/100) (90/100) times i. e. , 4pe 0 r 2 100

times.

4.

q = 4 pe 0r 2E

Þ

q=

=

(0.25) 2 ´ 2 = 1.39 ´ 10 –11 C 9 ´ 10 9

y +q2

y +q2 Q (–) q1

x

O

(x, 0) +q3

+q2 (a)

(b)

When a positive charge Q is added at ( x, 0), it will attract ( - q1) along + x direction, Fig. (b). Therefore, force on q1 will increase along the positive x-axis.

As, \

F qE = m m

v = u + at = 0 + KE =

11. Here, F1 =

k ( +10) ( -20) -k ´ 200 = R2 R2

As spheres are of equal radii, their capacities are same. On touching, the net charge = ( +10 - 20) mC = -10 mC, is shared equally between them, i. e. , each sphere carries -5 mC charge. K ( -5) ( -5) F2 = R2 K ´ 25 = R2 F1 - 8 \ = F2 1 Þ

F1 : F2 = - 8 : 1 R = 3 mm = 3 ´ 10 -3 m

12. Here,

E=

As,

6. Inside the sphere at any point, E = 0. 8. Here, u = 0 and a =

æ 1 qö 1 q2 1 q2 -2 = -ç × ÷ 2 4pe 0r r 4pe 0 r 2 è 4pe 0 r 2 ø

= -3 ´ 10 -5 N = 3 ´ 10 -5 N

direction, therefore q1 must be negative as shown in figure.

O

B

C

F = FA + FB 1 (q /2) (q /2) 1 (q /2) ´ q F= 4pe 0 (r /2) 2 4 pe 0 (r /2) 2

\

5. As, q 2, q3 are positive charges and net force on q1 is along + x

(–) q1

FA

FB A

\Net force on C,

99 ´ 100 N = 99 N \ Net force = 100 q As, E= 4 pe 0r 2 or

1 q2 = 3 ´ 10 -5 N 4pe 0 r 2

Charge on A and C after touching q q A¢ = qC = 2

q = - Q /4

3. As, F =

F=

Force,

Q 4 pe 0R 2

or

Q = 4 pe 0R 2E

For the maximum value of E = 10% dielectric strength \

qE t m 2 2 2

2 2 2

1 mq E t Eqt = mv 2 = 2 2m 2 m2

9. In a uniform electric field, field line should be straight but line of force cannot pass through the body of metal sphere and must end/start from the sphere normally. All these conditions are fulfilled only in (d).

E = 10% of 2 ´ 10 7 = 2 ´ 10 6 NC–1 Q=

1 (3 ´ 10 -3) 2 ´ 2 ´ 10 6 = 2 ´ 10 -9 C = 2 nC 9 ´ 10 9

13. In figure, spacing between electric lines of force increases from left to right. Therefore, E on left is greater that E on right. Force on + q charge of dipole is smaller and to the right. Force on - q charge of dipole is bigger and to the left. Hence, the dipole will experience a net force towards the left.

782 JEE Main Physics 14. When a point charge + q is placed at a distance (d) from an

20. As,

t max = pE sin 90° = q (2a) E = 10 -6 ´ 2 ´ 20 -2 ´ 10 5

isolated conducting plane, some negative charge developes on the surface of the plane towards the charge and an equal positive charge develops on opposite side of the plane.

+q

–

+

–

+

–

+

= 2 ´ 10 -3 N-m

21. From Coulomb’s law F= P

–

+

–

+

15. We have, E = 16. As, E = å

3.7 ´ 10 -9 = 9 ´ 10 9 ´

Þ

Hence, the field at a point P on the other side of the plane is directed perpendicular to the plane and away from the plane as shown in figure. ö æ e = K÷ çQ ø è e0

s s Gauss’s law = 2 e 2 e 0K

l dl l ( pa) l dq =å = = 4 pe 0 a2 4 pe 0 a2 4 pe 0 a2 4 e 0 a

q = 3.2 ´ 10 -19 C

As,

q = ne

\

n=

18. On equatorial line of electric dipole, E µ

= 5.20 ´ 10 –15 N

23. As, E =

F mg 3.2 ´ 10 –27 ´ 9.8 = = q q 1.6 ´ 10 –19 = 19.6 ´ 10 –8 NC–1

24. As, F =

1 r3

1 q1 q 2 4pe 0 r 2 e0 =

Þ

19. Here, q = ± 6.0 nC = ± 6.0 ´10 –9 C 2 a = 6 cm = 6 ´ 10

-2

m

F ¢ - 6k r 2 1 = 2 ´ =F 24 k 4 r F

Þ

F F F

q2

= 4 ´ 10 -2 m q 0 = 2 nC = 2 ´ 10

-9

C,F = ?

F = F1 cos q + F2 cos q 1 qq 0 cos q =2´ 4pe 0 r 2 = 2 ´ 9 ´ 10 9 ´

6 ´ 10 -9 ´ 2 ´ 10 -9 3 ´ 5 (5 ´ 10 -2) 2

F = 5.18 ´ 10 –5 N Clearly, this force is along $j. So, F = -51.8 $j mN or

F¢ = -

F 40 == -10 N 4 4

q 4 ´ 10 -7 = 2.5 ´ 10 12 = e 1.6 ´ 10 –19 1 1 From s = ut + at 2 = at 2 2 2

26. As, n =

r = 4 cm (on equatorial line) and

k (3) (8) 24 k = 2 r2 r k (3 - 5) (8 - 5) k ( -2) (3) - 6k F¢ = = = 2 r2 r2 r

25. Here, F = 40 = and

q

1 q1 q 2 [ A 2T 2] = 4 pF r 2 [MLT –2][L2]

= [M –1L–3 T 4 A 2]

q1

F

q 3.2 ´ 10 -10 =2 = e 1.6 ´ 10 -10

22. As, F = qE = 1.6 ´10 –19 ´ 3.25 ´10 4

to 60°. R = F 2 + F 2 + 2 F F cos 60° = F 3

q2 (5 ´ 10 -10) 2

\

17. Angle between two forces due to individual charges is equal \

1 q1q r 4pe 0 r 2

27.

t=

2s a

\

tµ

1 a

\

t2 = t1

or

(Q u = 0)

As s is same,

Mp a1 F /M = e e = a2 Fp /Mp Me

28. In case of spherical conductor, the whole charge is concentrated at the centre. Now, the distance between two charges will be (R + x). Thus, the force will be proportional to 1 / (R + x) 2.

Electrostatics q 2¢ = +6 - 4 = 2 mC, F ¢ = ? F ¢ q1¢ q 2¢ ( -2) ´ (2) 1 = = =F q1 q 2 2 ´6 3

29. Charge enclosed in the sphere of radius r q=

4 3 pr r 3

As

4 3 pr r q rr 3 E= = = 2 2 3 e0 4pe 0r 4pe 0r

30. As, E =

2

1 Q 1 4 pR s s = ´ = 4pe 0 R 2 4pe 0 e0 R2

\

32.

the centre of hemisphere charged uniformly positively, the electric field is perpendicular to the diameter. The components of electric intensity parallel to the diameter cancel out. 1 In case of an electric dipole, F µ 3 r

\

39.

9 ´ 10 9 (10 -6 ´ 10 -8) p = 3 4pe 0r (10 -1)3

= 9 ´ 10 -2 NC–1 = 0.09 NC–1

34. The force of interaction, F µ

1 r2

When length is increased by 10% then, 1 100 F= times = times = 0.03 times 121 (110 /100) 2

t max = pE sin 90°

= 6 ´ 10 -30 ´ 1.5 ´ 10 4 ´ 1 = 9 ´ 10 -26 N-m 1 As, E e = E a \ E a = 2 E e 2

40. As, s 1 = s 2 \

Q1 Q2 = 4pr12 4p 0r22

or

Q1 Q2 = 4pe 0r12 4pe 0r22

\ or

33. The point lies on equatorial line of a short dipole. E=

41.

E1 = E 2 E1 /E 2 = 1

1 q1 q 2 From, F = 4pe 0R r 2 For a brass sheet, K = ¥ q1 q 2 1 F= \ 4pe 0( ¥) r 2 Þ

F =0

35. Inside hollow sphere, E = 0. On the surface of hollow sphere,

VA =

s ( a - b + c) e0

VB =

ö s æ a2 ç - b + c÷ e0 è b ø

VC =

ö s æ a2 b 2 + c÷ ç c e0 è c ø

E = maximum and outside the sphere, E µ1/r 2.

36. As, F =

and

4pe 0r 2

e In a medium of dielectric constant, K = e0 F¢ = As, Þ or, \

q1 q 2 4peR 2

F¢ = F q1 q 2 q1 q 2 = 2 4peR 4pe 0r 2 e0 2 r 2 r = K e r R= K

R2 =

37. Here, q = +2 mC, q 2 = + 6 mC, F = 12 N q1¢ = + 2 - 4 = -2 mC,

σ –σ

42. We have,

\ Decrease in force = (1 - 0.83)100% = 17%

q1 q 2

-F 12 == - 4 N (approx) 3 3

E = 1.5 ´ 10 4 N/C

\ New force = F /23 = F /8

\

F¢ =

38. Here, p = 6 ´10 -30 C-m and

This independent of radius and depends on s. Hence, the electric field on the surface of new sphere will be E.

31. When the point is situated at a point on diameter away from

783

Putting these, c = a+ b VA = VC ¹ VB

Þ or

VC = VA ¹ VB

43. The charge on cylinder A, q = ll Long charged cylinder

– – – – – B – –

+ + A + + + +

r + + a + b+ +

c

C B A

– – – l – – – –

Hollow co-axial, conducting cylinder

a

b

784 JEE Main Physics Total charge = Linear charge density ´ Length

46. Each sphere having -2 mC each

This charge spreads uniformly on A and a charge -q is induced on B. Let E be the electric field produced in the space between the two cylinders. Consider a Gaussian cylindrical surface of radius r between the two given cylinders.

\

n=

2 ´ 10 -6 q = 1.25 ´ 10 13 =e -1.6 ´ 10 –19

Thus, 1.25 ´ 10 13 electrons in excess.

47. The unit positive charge at O. The resultant force due to

Electric flux linked with the Gaussian surface fE = ò E × dS = ò E × dS cos 0° = E òdS = E ´ 2prl

charge placed at A and C is zero. Force due to charge ( +2q) on B will be direction of OD. Force due to charge ( -2q) on D will be in direction OD.

[As angle between the direction of electric field and area vector is zero]

Therefore, the resultant force will be along the direction OD.

48. Fig. (i) represent field lines between two equal positive

According to Gauss’s theorem,

charges.

fE = E ´ 2prl =

q e0

49. Because, all the points on the circular path are at same

E ´ 2 prl =

ll e0

50. It is because the two spheres A and B possess equal

potential. capacitance. The capacitance of the sphere depends only on its radius. It does not matter, whether the sphere is hollow or solid.

l 2pe 0r

E=

44. The density of lines of force µ E.

51. The tangential component of electrostatic field is continuous

Here, the density of lines of force at A is greater than at B. E A > EB

Thus,

45. Let the potential be zero at point C at a distance x from point B. Let us consider two charges q1 and q 2. According to the question, 16 cm A q1

(16 – x)

C

x

from one side of a charged surface to another, we use that the work done by electrostatic field on a closed-loop is zero. Let ABA be a charged surface in the field of a point charge q lying at origin. Let rA and rB be its positive vectors at points A and B respectively. Y

B q2

A

q1 = 5 ´ 10 -8 C, q 2 = - 3 ´ 10 -8 C, AB = 16 cm = 16 ´ 10 -2 m

rA rB

q

…(i)

E

B

The potential at point C due to charge q1 1 9 ´ 10 -9 ´ 5 ´ 10 -8 q VA = × 1 = 4pe 0 AC (16 - x ) ´ 10 -2

P Q dl

X

O Z

The potential at point C due to charge q 2 VB =

1 q - 9 ´ 10 9 ´ 3 ´ 10 -8 × 2 = 4pe 0 RC (16 - x ) ´ 10 -2

…(ii)

Now the net potential at point C is zero i. e. ,VA + VB = 0 Putting the values from Eqs. (i) and (ii), we get 9 ´ 10 9 ´ 5 ´ 10 -8 æ - 9 ´ 10 9 ´ 3 ´ 10 -8 ö +ç ÷ =0 (16 - x ) ´ 10 -2 x ´ 10 -2 è ø or

5 3 - =0 16 - x x

or

5x - 3(16 - x) = 0

or

5x - 48 + 3x = 0

or

8x = 48 x = 6 cm

Thus, the electric potential is zero at the distance of 6 cm from q 2 (- 3 ´ 10 -8 C)

Let E be the electric field at point P, thus E cos q is the tangential component of electric field E. To prove that E cos q is continuous from one to another side of the charge surface, we have to find the value of ò E × dl. If it comes to be zero then we can say that AaBba

tangential component of E is continuous. B

òA and \

E × dl =

æ 1 1ö 1 q×ç - ÷ 4pe 0 è rA rB ø 1

A

1ö

æ1

òB E × dl = 4pe 0 q × çè rB - rA ÷ø B

òAaBbA E × dl = òA =

A

E × dl + ò E × dl B

æ 1 1 1 1ö 1 ×q × ç - + - ÷ 4pe 0 è rA rB rB rA ø

=0

Electrostatics 52. When the lines of force are equidistant straight lines, the field 53.

61. As we have,

is uniform. V 10 E = = -2 = 10 3 Vm-1 or NC-1 d 10

V= =

54. Here, the electric field due to test charge + q 0 opposes the

55.

field of positive charge being measured. Therefore, the measured value is less than the actual value or actual value of field E is greater than the measured vlaue (F / q 0). dq l dl l( p a) As, E = S =S = 4 p e 0 a2 4 p e 0 a2 4 p e 0 a2 l E= 4 e 0a

q q æ 1 1 ö æ1 1 1 ö ç1 + + + ¼÷ ç + + ¼÷ ø 4pe 0 x0 è 2 4 6 ø 4pe 0 x0 è 3 5 q q æ 1 1 1 ö log e 2 ç1 - + - + ¼÷ = ø 4pe 0 x0 4pe 0 x0 è 2 3 4

62. Potential at the centre O, V = 4´ where, and

q=

58. As, E = - dV /dr \ Þ

1 dV = -EdR = ( -10) ´ = - 5 V 2 | dV | = 5V

59. As it is clear from figure As \

AB + BC = AC dV (V - VA ) VA - VC E==- C = dr d d

–q

q

Q F43

Q

q

Magnitude of resultant of F41 + F43 = F42 1 Q ´q Here, F41 + F43 = 2 × 4pe 0 d 2 (where, d = side of square) Resultant on Q becomes zero only when q charges are of negative value 1 QQ \ F42 = 4pe 0 ( 2 d) 2 dQ Q ´ Q Q ´Q 2 2 = x 2 = 2q = Þ 2 d 2d Q Q or \ q== -2 2 q 2 2

+q

10 ´ 10 -9 9 3 Then, V = 4 ´ 9 ´ 10 ´ = 1500 2 V 8 ´ 10 -2 2

63. Magnetic lines of force always makes a closed loop. 64. From conservation of mechanical energy decrease in kinetic energy = increase in potential energy 1 ( Ze) (2e) or = 5 MeV 4pe 0 rmin = 5 ´ 1.6 ´ 10 -13 J \

rmin = =

60. Three forces F41, F42 and F43 acting on Q are shown below. F42

–q

u/√2

VA - VC = Ed

F41

10 ´ 10 -9 C, in magnitude 3

a = 28 cm = 28 ´ 10 -2 m

not zero anywhere on the sphere. However, net electric flux through the sphere is zero. When a capacitor is charged using a battery, the battery supplies equal and opposite charges to two plates. Outer surfaces of plates have equal charges. But the inner surfaces to plates have equal and opposite charges. leaves will also be positive. Due to X-rays, more electrons from leaf will be emitted, so leaf becomes more positive and diverse further.

1 q × 4pe 0 a 2

+q

56. When electric dipole is held in the sphere, electric field is

57. Charge on the glass rod is positive, so charge on the gold

785

1 2Ze2 4pe 0 5 ´ 1.6 ´ 10 -13 (9 ´ 10 9) (2) (92) (1.6 ´ 10 -19) 2 5 ´ 1.6 ´ 10 -13 +Ze

+Ze

( z = 92)

rmin

= 5.3 ´ 10 -14m = 5.3 ´ 10 -12 cm i. e. , rmin is of the order of10 -12 cm.

65. Flux through surface A, ds

A

f A = E ´ pR 2

B

and fB = -E ´ pR 2

Electric flux through the curved surface C = ò E × ds = ò E dS cos 90° = 0 \ Total flux = EpR 2 - EpR 2 + 0 = 0

786 JEE Main Physics 66. The electric flux through a surface depends only on amount

74. Let us consider a neutral point O lies at a distance x from the

of charge enclosed by the surface. It does not depend on size and shape of the surface, as per Gauss’s theorem in electrostatic. Therefore, electric flux through the surface is the same for all the figures. 1 By using, ò E × dA = (Q in) e0

dipole of moment p or at a distance (25 - x) from dipole of 64 p.

67.

q × e0 Here, E is due to all the charges q1, q 2, q3 , q 4 and q5 . As q is charge enclosed by the Gaussian surface, therefore, q = q 2 + q 4.

25

Þ

conducting sphere, E = 0 and V =

V1 = 150 V,V2 = 100 V Common potential, C V + C 2V2 4pe 0(rV 1 1 + r2V2) V= 11 = C1 + C 2 4pe 0(r1 + r2)

ö 1æ q ç - f÷ 2 è e0 ø

= 120 V \

ò E × ds =

Q in e0

Þ

E × 4px2 =

Q e0

77. As, E = -

or

E=

Q 4pe 0 x2

78.

Þ

10 -1 ´ 120 C 9 ´ 10 9

12 ´ 3 ´ 10 9 esu = 4 esu 9 ´ 10 9 (1 - 3) =Vm–1 = 10 Vm–1 0.3 – 0.1 d = - ( 4x2) = - 8x = - 8 (1) = -8 Vm–1 dx

dV dr dV As, E = dr

Negative sign indicates E is along negative direction of X-axis.

79. Electric potential inside the hollow conducting sphere is

df = - 2ar dr q ò E × ds = e 0 E=-

constant and equal to potential at the surface of the sphere Q . i. e. , 4pe 0R

80. Let us consider the field varies at distance x, as we have

q q = - 8 e 0 a p r3 e0

y

…(i)

q 4 3 pr 3

r = - 6 ae 0

q1 = C1V = 4pe 0rV 1 = =

72. Here, f = ar 2 + b

r=

dV d (constant) = Zero =dr dr

76. Here, r1 = 10 cm, r2 = 15 cm,

71. According to Gauss’s theorem,

- 2ar × 4pr 2 =

E=-

Infact,

Q fB = f and f A = fC = f ¢(say) q 2f ¢ + f = \ e0

\

kq = potential on the R

surface = 100 V

Thus, the lines are parallel to the surface. q We have, f total = f A + fB + fC = e0

As

x = 5 cm

75. As we known from theory, at any point inside a charged

= ò E ds cos q = ò E ds cos 90° = 0

f¢ =

1 64 = x3 (25 - x)3

Þ

69. As, electric flux, fE = ò E × ds

Þ

2

At O, E.F due to dipole (1) = 1E.F due to dipole (2) 1 2p 1 2 (64 p) Þ = 3 4pe 0 x 4pe 0 (25 - x)3

68. According to Gauss’s theorem in electrostatics, ò E × ds =

70.

t 64 p

x

q1 + q 2 + q3 e0

Þ oò (E1 + E 2 + E3) d A =

O

p

1

q/2

–q

N (0, 0, 0)

[from Eq. (i)]

kq kq /2 = x2 ( x - a) 2

73. Let, E A = Electric field at M due to charge placed at A. EB = Electric field at M due to charge placed at B EC = Electric field at M due to charge placed at C As seen from the figure|EB| = |EC|, so that net electric field at M, E net = E A in the direction of vector 2.

(a, 0, 0) x

or

2 ( x - a) 2 = x2 2 ( x - a) = x or ( 2 - 1) x = 2 a

Þ

x=

2a 2 -1

Electrostatics \

81. In a uniform electric field, when a positively charged particle is released from rest, it moves along the electric field (i. e. , from higher potential to lower potential). Therefore, electric potential energy of charge decreases.

Work done = U1 - U3 Q ( -q) Qr 4 = + ´ 4pe 0r 4pe 0r 3 =

82. As work is done by the field, KE of the body increased \

DKE = W = q (VA - VB)

Qq Qq = zero 4pe 0l 4pe 0l

and \

84. Net flux leaving the surface, f = 4 ´ 10 5 - 5 ´ 10 5 = -10 5 MKS units q = f e 0 = -10 5 ´ 8.86 ´ 10 –12 = -8.86 ´ 10 –7 C

94.

86. From a collection of charges, whose total sum is not zero, equipotentials at large distances must be spheres only.

87. Work done by the field W = q ( -dV ) = e (VA - VB) = e (VB - VA ) = e (VC - VA )

p=

1 V2 e0 2 2 r

93. Electric field at a point due to positive charges acts away

Negative sign indicates charges must be negative

Þ

1 e 0E 2 2 V E= r

p=

Þ

83. As, net work done = final PE - initial PE

Now,

Qq 1 9 ´ = =3 J 4pe 0r 3 3

92. Electrical pressure (force/area)

= 10 -8 (600 - 0) = 6 ´ 10 -6 J

=

(Q VB = VC )

from the charge and due to negative charge it acts towards the charge. 4 3 4 As p R = n ´ p r3 3 3 R = n1/3r nq New potential V ¢ = = n 2/3V 4pe 0r

\

95. Let there is a cube of side b and its centre is O. The charge q is placed at each of the corners. Side of the cube = b

W (10 ´ 10 6 J) (VC - VA ) = = e 1.6 ´ 10 –19 C

q

q q

= 6.62 ´ 10 25 V

q O

88. Charges present outside the surface made no contribution to

q q

electric flux \Wet flux,

f=

q q =0 e0 e0

q

= b2 + b2 + b2 = 3b

Dr = ( 4 - 2) $i + (2 - 2) $j = 2$i

Distance of centre O from each of the vertices is r.

DV = E × Dr = (y$i + x$j) × 2$i = 2 y

r=

90.

\

Q ( -q) é 1 2 ù 4pe 0 êë r r ûú Qq = =9J 4pe 0

b 3 2

Potential at point O due to one charge is V =

\Work done, W = q( DV ) = q ×2 y = 1 - 2y = 2y J Q ( -q) Q ( -q) Here, U1 = ; U2 = 4pe 0 4pe 0(r /2)

q

Length of the main diagonal of the cube

89. Here, displacement would be,

Now,

787

(Q charge = 1C)

U1 - U2 =

When negative charge travels first half of distance, i. e. ,r /4, potential energy of the system. Q ( -q) Qq 4 U3 = =´ 4pe 0(3r / 4) 4pe 0 r 3

1 q × 4pe 0 r

Potential at point O due to all charges placed at the vertices of the cube 8q ´ 2 8 ´1´ q [From Eq. (i)] = V ¢ = 8V = 4pe 0 × r 4pe 0 × b 3 =

…(i)

…(i)

4q 3pe 0 b

The electric field due to one vertex is balanced by the electric field due to the opposite vertex because all charges are positive in nature. Thus, the resultant electric field at the centre O of the cube is zero.

96. As, PE =

1 q1 q 2 9 ´ 10 9 (2 ´ 10 -6) 2 × = = 0.036 J r 4pe 0 1

788 JEE Main Physics 97. Here,

E = 8$i + 4$j + 3k$ S = 100 k$

102. Radius of bigger drop, 4 ù é 4 3 Q pR = 27 ´ pr3 úû êë 3 3 27 q 27 q V= = 4pe 0R 4pe 0 (3r) R = 3r

(direction of area is perpendicular to xy-plane) f = E × s = (8$i + 4$j + 3k$ ) ×100 k$

\

= 300 unit

æ q ö =9 ç ÷ è 4pe 0R ø

98. We observe that in all the three parts,VA = 20 V and VB = 40 V. Work done in carrying a charge q from A to B is W = q(VB - VA ) = same in all the three figures.

99. As flux, f = E (ds) cos q = E (2pr 2) cos 0° = 2pr 2E 100. As, energy of second proton = PE of the system =

1 q1q 2 × 4pe 0 r

1.6 ´ 10 –19 ´ 1.6 ´ 10 –19 1 ´ 10 -10 –19 = 23.0 ´ 10 J

= 9 ´ 10 9 ´

= 9 ´ 10 = 90 V

103. Elemental charge, dq = l dl dq (r 2 + l 2) dq Then, dE x = 2 2 cos q (R + l ) On integration, we get l Ex = Þ 4pe 0R Field at point p is, dE = K ×

x

101. ABCDEF is a regular hexagon of side 10 cm each. At each

a

corner the charge q = 5 mC is placed. O is the centre of the hexagon.

x y

D

E q

l

q

θ O

F q q

dEy

10 cm B

Given, AB = BC = CD = DE = EF = FA = 10 cm As the hexagon has six equilateral triangles, so the distance of centre O from every vertex is 10 cm. i.e., OA = OB = OC = OD = OE = OF = 10 cm Potential at point O = Sum of potential at centre O due to individual point charges. \

dq sin q (R + l 2) l On integration we get, E y = 4pe 0R Also,

q A

VO = VA + VB + VC + VD + VE + VF 1 é q q q q q q ù VO = × + + + + + 4pe 0 êë OA OB OC OD OE OF úû æ 1 qö × ÷ çQ V = 4pe 0 r ø è

Similarly, \Resultant, and Þ

dE y =

é 5 ´ 10 - 6 5 ´ 10 - 6 5 ´ 10 - 6 5 ´ 10 - 6 + + + VO = 9 ´ 10 9 ê -2 -2 -2 10 ´ 10 10 ´ 10 10 ´ 10 - 2 ë10 ´ 10 5 ´ 10 - 6 5 ´ 10 - 6 ù + + -2 10 ´ 10 10 ´ 10 - 2 úû

or

VO = 9 ´ 10 9 ´

l

E = E x2 + E y2 = tan a =

VO = 27 ´ 10 4 VO = 2.7 ´ 10 6 V

l 2 2 pe 0R

Ex Ey

a = 45°

104. Potential difference across both the lines, i. e. , 2 V. Then, charge in line 2 is Q =

(2 ´ 2) mC = 2 mC (2 + 2)

2µF

2µF Line 2 1µF Line 1

-6

6 ´ 10 ´ 5 10 ´ 10 - 2

2

ò dEy = Ey = 4pe 0R

Putting the values,

or

dEx

R

C q

+

– 2V

Thus, charge on each capacitor in line 2 is 2 mC. Charge line 1, Q = 2 ´ 1 = 2 mC.

Electrostatics 105. As, charge on capacitor, Q = CV Since, V is constant, therefore Q µ C 100 Hence, C becomes = 2.5 times 40 \ K = 2.5

106. No current is allowed to flow through 10W resistance because of capacitor. The current drawn by 2 W resistance. E 2.5 I= = = 1A R + r 2 + 0.5

789

112. The two condensers are connected in parallel, we have C 3C = 2 2 \Total work done in charging both the condensers 1 1 3C 2 3 W = C pV 2 = ´ V = CV 2 2 2 2 4 Cp = C +

113. Let the distance between the plates be increased by a very small distance Dx. The force on each plate is F. ∆x

A

Terminal potential difference of battery, V = E - Ir = 2.5 - 1 ´ 0.5 = 2 volt d

\Charge on capacitor plates, Q = CV = 4 ´ 2 = 8 mC

107. As, rb - ra = 1mm = 10 -3 m From

C= 10 -6 =

Þ

4pe 0rarb rb - ra 1(rb - 10 -3) rb 9 ´ 10 9 (10 -3)

rb2 = 9

The amount of work done in increasing the separation by Dx = Force × Increased distance = F. Dx Increase in volume of capacitor = Area of plates × Increased distance = A. Dx Energy u = Energy density = Volume Energy = u ´ volume = u × A × Dx

or rb = 3 m

108. We know that, C µ K \

C1 120 =K = = 2.2 C2 50

109. The capacities of two individual condensers are k1e 0 A ke A and C 2 = 2 0 d1 d2 The arrangement is equivalent to two capacitors joined in series. Therefore, the combined capacity (C s ) is given by 1 1 1 d d2 = + = 1 + C s C1 C 2 k1e 0 A k2e 0 A C1 =

or \

1 1 é d1 d 2 ù d1k2 + d 2k1 = + = C s e 0 A êë k1 k2 úû e 0 Ak1k2 e Ak k ke 0 A Cs = 0 1 2 = d1k2 + k2k1 (d1 + d 2) k=

k1k2(d1 + d 2) (d1k2 + d 2k2)

110. Here two capacitor are in series, and their equivalent is is green by, Þ Now, \

1 1 1 3 = + = Cs 2 1 2 2 Cs = F 3 2 ´ 12 = 8 C 3 Q 8 V1 = = = 4V C1 2

Q = C sV =

111. Positive plate of all the three condensers is connected to one point A and negative plate of all the three condensers is connected to point B, i. e. , they are joined in parallel. \

C p = 3 + 3 + 3 = 9 mF

…(i)

As

…(ii)

Energy = Work done F × Dx = u × A × Dx = u× A 1 = e 0E 2 × A 2

[From Eqs. (i) and (ii)] æçQ u = 1 e E 2 and E = V ö÷ 0 è dø 2

1 V2 e0 × 2 × A 2 d e A e A æ ö V 1 æç = ç 0 ×V ÷ ´ Q C = 0 , CV = q ö÷ ø è d ø d 2 è d 1 1 = × E × C × V = QE 2 2 =

1 in the force can be explained by the fact that 2 the field is zero inside the conductor and outside the conductor, field is E. So, the average value of the field i. e. , E contributes to the force against which the plates are moved. 2 The factor of

114. The required capacitance C = 2 mF Potential difference V = 1kV = 1000 V Capacitance of each capacitor C1 = 1mF and it can withstand a potential difference of V1 = 400 V Let the n capacitors are connected in series and there are m rows of such capacitors. As the potential difference across each row is 1000 V. 1000 So, the potential difference across each capacitor = n

790 JEE Main Physics Minimum number of capacitors that must be connected in series in a row are 1000 = 400 Þ n = 2.5 n n C1

C1

C1

C1

Þ

C¢ =

1 3

1 m = 3 3 According to question, the total capacitance required is 2 mF. So, m =2 3 The total capacitance of m rows is m ´

m=6 C1

C1

C1

m rows

C1

Thus, the total number of capacitor = m ´ n = 3 ´ 6 = 18 So, 1 mF capacitors are connected that of 6 rows having 3 capacitors in each row.

115. Given, capacitance of capacitor C1 = 600 pF = 600 ´ 10 -12 F C1

C1

C1

and supply voltage V1 = 200 V

C1

C 2 = 600 pF = 600 ´ 10 - 12 F and V2 = 0

1000 V

Here n is the number of capacitors, so it should be a whole number. If we take n = 2, then potential difference across each capacitor is 500 V.

Loss in energy (E) =

Here according to question a capacitor can bear only 400 V, so they burst. We take the value of n = 3. So, the capacitance of each row (in series) 1 1 1 1 3 = + + = C¢ 1 1 1 1

E=

C1C 2(V1 - V2) 2 2 (C1 + C 2) 600 ´ 10 - 12 ´ 600 ´ 10 - 12 (200 - 0) 2 2 (600 + 600) ´ 10 - 12

= 6 ´ 10 - 6 J Thus, the 6 ´ 10 - 6 J amount of electrostatic energy is lost in the sharing of charges.

Round II 1. Initially, FAB =

1 q. q 1 q2 . 2 = . 4pe 0 r 4pe 0 r 2 A

A B

q

2q

r

q r

A q r

But

FC = FBC - FCA æq ö æq öæq ö ç ÷(q) ç ÷ç ÷ 1 è2ø 1 è 2 øè 2 ø 1 q2 = . . = . 4pe 0 æ r ö 2 4pe 0 æ r ö 2 4pe 0 r 2 ç ÷ ç ÷ è2ø è2ø Þ FC = FAB 1 qx . 4pe 0 ( x2 + R 2)

and E is maximum when x =

R 1 2q Þ E max = . 4pe 0 3 3R 2 2

3. In an equilateral triangle distance of centroid from all the vertices is same (say r). \ V = V1 + V2 + V3 1 é q1 q 2 q3 ù 1 = = + + r r úû 4pe 0 4pe 0 êë r

é 2q q q ù =0 - êë r r r úû

O

r

B –q

r

Finally, force on, C

2. For a ring E =

r

B q

C q

–q

C

EA =

1 2q along AO, . 4pe 0 r 2

EB =

1 q × along OB and 4pe 0 r 2

EC =

1 q . along OC. 4pe 0 r 2

Obviously, EB + EC will also be in the direction of AO (extended) and hence, EA and (EB + EC ) being in same direction will not give zero resultant.

4. Bob will experience an additional force, F = q E in vertically upward direction and hence, effective acceleration due to gravity is reduced from g to ( g - a). Consequently, time l period of oscillation will become, T = 2p ( g - a) i. e. , time period will increase.

Electrostatics 5. As, E1 = E 2 =

q ´1 , acting at 60°. 4pe 0 a2

\Resultant intensity, E = E12 + E 22 + 2E1E 2 cos q =

E12 + E12 + 2E12 cos 60°

= E1 3 =

q 3 4pe 0 a2

Fx = F12 + F12 sin q q ù éq = Kq1 22 + 32 sin q úû êë b a

8. As,

Þ

E1 =

ù é q 2 q3 sin q + úû êë b 2 a2

E2 = -

1 1 p p . =. 4pe 0 (2r3) 4pe 0 8r3

E2 = -

E1 16

9. Since, electrical potential at any point of circle of radius (R) Q2 × Hence 4pe 0R work done in carrying a charge Q1 round the circle is zero. q 1 2q q and Here, V = = . 4pe 0. r 4pe 0(3r) 4pe 0 3r due to charge, Q 2 at its centre is same, V =

1 q 1 q E= . . 2 = 2 4pe 0 (3r) 4pe 0 9r E 1 V On simplification, we get, = or E = V 6r 6r

11. Coordinates of the point are ( x, y)

1 (q) e0

q = e 0( f 2 - f1)

14. Potential energy of electric dipole, U = - p × E = -pE cos q. 15. 16.

In Fig. (a), q = p rad, hence,U = - pE cos p = + pE = maximum. 1 As, E µ , where r is the distance from the axis. r é Q Q ù As, work done WBA = q (VA - VB) = q ê ú ë 4pe 0 a 4pe 0 b û qQ é 1 1 ù = 4pe 0 ëê a b ûú

17. Here, E = E1 + E2 + E3 æ s ö $ æ 2s ö $ æ s ö $ = ç+ ÷ ( - k) ÷ ( - k) + ç ÷ ( - k) + ç è 2e 0 ø è 2e 0 ø è 2e 0 ø

1 2p and . 4pe 0 r3

(Here negative sign means direction)

10.

E µr

any point on y-axis is at the equatorial line. Hence, E at all points on y-axis will be in a direction opposite to p and p is along negative x-axis. So, E is along positive x-axis, i. e. , along $i.

along x-axis

\x component of force on - q1 is

Fx µ

i. e. ,

13. The two charges form an electric dipole and for this dipole

at Ð q with negative direction of y-axis

i. e. ,

E = E x2 + E y2 = k y 2 + x2 = kr

Þ

7. Force on - q1 due to q 2 is, b2 Force on - q1 due to - q3 is, kq1 q3 F13 = a2

\

( f 2 - f1) =

1ö æ çQ cos 60° = ÷ è 2ø

ER = EU along ROU direction. kq1 q 2

dV ( -kxy) = kx dy

12. Net electric flux of surface,

6. The net field at OE = 2 ER, when EP + ES = 0, EQ + ET = 0 and

F12 =

Ey = -

and

791

æ 2s ö = - ç ÷ k$ è e0 ø

18. Here, KE = 100 eV = 100 ´1.6 ´10 -19 J = 1.6 ´10 -17 J. This is lost when electron moves through a distance (d) towards the negative plate. æs ö (KE) e 0 KE = work done = F ´ s = qE ´ s = e ç ÷ d = \ es è e0 ø Þ d=

1.6 ´ 10 -17 ´ 8.86 ´ 10 -12 = 4.43 ´ 10 -4m 1.6 ´ 10 -19 ´ 2 ´ 10 -6 = 0.44 mm

19. Electric charge on the inner surface of a hollow conducting spherical shell is always zero. dV × Hence dx potential at A must be greater than that at B i. e. ,VA > VB.

20. As, electric field is along positive x-axis and E = C(0, 1)

\ Distance of point from origin, r = x2 + y 2 and given V = - kxy Þ

dV d Ex = = - ( -kxy) = ky dx dx

A(0, 0)

B (1, 0)

21. Total electric flux, f1 = f = of given body is constant.

E

1 (charge enclosed) i.e., charge e0

792 JEE Main Physics 22. Electric flux may be due to the charges present inside the

27. Due to additional charge of -3Q, given to external spherical

gaussian surface, but for the purpose of calculation of electric field E at any point, we shall have to consider contribution of all the charges.

shell, the potential difference between conducting sphere and the outer shell will not change because by presence of charge on outer shell, potential everywhere inside and on the surface of the shell, will change by same amount. Therefore, the potential difference between sphere and shell will remain unchanged. Ae 0 …(i) As, C = d After inserting the slab, Ae 0 Ae 0 C¢ = = (d - b) d - d 2 2Ae 0 …(ii) or C¢ = d From Eqs. (i) and (ii), we get C¢ 2 = = 2 :1 C 1

23. For equilibrium net electric force on any charge (say charge - Q at A) should be zero. Hence, FA = FAB + FAD + FAC + FAO = 0 1 Q2 along BA, FAB = 4pe 0 a2 FAD =

1 Q2 along DA, × 4pe 0 a2

1 Q2 × 2 along CA 4pe 0 2 a 1 2 Qq along AO =× 4pe 0 a2

FAC = and

FOA

a

–Q A

–Q B

29. With S1 and S3 closed, the capacitors C1 and C 2 are in series

O q a D –Q

–Q

Resultant of FAB and FAD =

C

1 Q2 × 2 along COA, 4pe 0 a2

1 Q2 1 Q2 1 2 Qq × 2 2+ × × =0 4pe 0 a 4pe 0 2a2 4pe 0 a2 Q Þ q = (1 + 2 2) 4

\ FA =

24. For charge q placed at the centre of circle, the circular path is an equipotential surface and hence work done along all paths AB or AC or AD or AE is zero.

25. Since the two spheres are joined by a wire, their potential are equal i. e. ,

Þ Now, and

q1 q2 = 4pe 0R1 4pe 0R2 q1 R1 = q 2 R2 q1 s1 = 4pe 0R12 q2 s2 = 4pe 0R22 s 2 q 2 R12 æ R2 ö æ R1 ö = ´ = ç ÷ç ÷ s 1 q1 R22 è R1 ø è R2 ø s 2 R1 = s 1 R2

hence Þ

26. As, ò

l =0

l =¥

and

arrangement. In series arrangement, potential difference developed across capacitors are in the inverse ratio of their capacities. Hence, V1¢ C 2 3 pF 3 = = = and V2¢ C1 2 pF 2 V1¢+ V2¢ = V1 + V2 = 30 + 20 = 50 V On simplification, we get V1¢ = V1 = 30 V and V2¢ = V2 = 20 V

30. On bringing the charged metal plates

Q2

Q1

closer, electric field E in the intervening space is \

E=

Q1 Q2 s1 s 2 = 2e 0 2e 0 2Ae 0 2Ae 0

or

E=

(Q1 - Q 2) V = d 2Ae 0

or

V=

(Q1 - Q 2)d 2Ae 0

Þ

V=

Q1 - Q 2 2C

E

d

e Aö æ çQ C = 0 ÷ è d ø

31. On sharing of charges loss in electrical energy, DU = 2

-E. dl = V , the potential at the centre of ring, V=

28.

1 Q 9 ´ 10 9 ´ 1.11 ´ 10 -10 . = = 2V 0.5 4pe 0 R

C1C 2 (V1 - V2) 2(C1 + C 2)

In present case, C1 = C 2 = C \

DU =

1 C2 (V1 - V2) 2 = C(V1 - V2) 2 2(2C) 4

32. Net capacity of 5 capacitors joined in parallel = 5 ´ 2 = 10 mF Now it is connected with two capacitors of 2 mF each in 10 series, thus, equivalent capacitance is mF. 11

Electrostatics 33. The arrangement behaves as a combination of 2 capacitors

34.

e A each of capacitance, C = 0 . d Thus, equivalent capacity = 2C 1 e A \ Total energy stored U = ´ (2C)V 2 = CV 2 = 0 V 2 d 2 A Capacitance of two capacitors each of area , plate 2 separation d but dielectric constants K1 and K2, respectively joined in parallel æ Aö æ Aö K1e 0 ç ÷ K2e 0 ç ÷ è2ø è 2 ø (K1 + K2) e 0 A + C1 = = d /2 d d /2

It is in series with a capacitor of plate area A, plate separation Ke A d / 2 and dielectric constant K3 , i. e. ,C 2 = 3 0 . d /2 Ke 0 A If resultant capacitance be taken as C = , d 1 1 1 then, = + C C1 C 2 d d d /2 = + Ke 0 A (K1 + K2) e 0 A K2e 0 A

\

1 1 1 = + K (K1 + K2) 2K3

Þ

35. Here, C 23 = 30 + 30 = 60 pF. Total equivalent capacitance is given by 1 1 1 1 1 1 1 7 = + + = + + = C C1 C 23 C 4 30 60 120 120 120 pF Þ C= 7 120 ´ 140 pC = 2400 pC \ Total charge, Q = CV = 7 Q 2400 pC V1 = \ = = 80 V C1 30 pF Q 2400 pC V2 = V3 = V23 = = = 40 V C 23 60 pF Q 2400 pC and V4 = = = 20 V C4 120 pF

r1 1mm 1 = = . When the spheres are connected by a r2 2 mm 2 conducting wire then, V1 = V2 q1 q2 or = 4pe 0r1 4pe 0r2

37. Here,

q1 r1 1 = = q 2 r2 2

Þ

2

38. In the arrangement shown both plates of capacitor C3 are joined to point B. Hence, it does not act as a capacitor and is superfluous. Now, C1 and C 2 are in parallel, hence C AB = C1 + C 2 = C + C = 2C

39. QV = 4x2 dV = -8x$i dr Thus, value of E at (1 m, 0, 2 m) will be E = - 8 ´ 1 $i = - 8 $i Vm-1

Hence,

C1

A C4

and

\

E=-

40. In steady state, no current flows through the capacitor segment. The steady current in remaining loop V 2V - V (anti-clockwise). Now, applying Kirchhoff’s I= = 2R + R 3R second law to loop containing 2V , 2R, C and V , we have V V VC = 2V - 1.2V .2R - V = . 3 3R

41. The net electric force on C should be equal to zero, the force due to A and B must be opposite in direction. Hence, the particle should be placed on the line AB. As, A and B have charges of opposite nature also A has longer magnitude of charge than B. Hence, C should be placed closed in B than A. From figure BC = x (say) and charge on C is Q. 1 8.0 ´ 10 –6 Q $ Then, FCA = × i 4pe 0 (0.2 + x) 2 and \

FCB = -

1 (2.0 ´ 10 –6) $ × i 4pe 0 x2

FC = FCA + FCB 1 é (8.0 ´ 10 –6) Q (2.0 ´ 10 –6) Q ù $ = úi 4pe 0 êë (0.2+ x) 2 x2 û

|FC| = 0 1 é (8.0 ´ 10 –6) Q (2.0 ´ 10 –6) Q ù Then × ú =0 4pe 0 êë (0.2 + x) 2 x2 û But

C2

\

2

E1 q1 æ r2 ö 1 æ2ö 4 = . ç ÷ = ´ ç ÷ = = 2 :1 2 è 1ø 2 E 2 q 2 è r1 ø

Now,

36. The arrangement can be redrawn as shown in the adjoining figure. B

C3

C13 = C1 + C3 = 9 + 9 = 18 mF C ´C C 2-13 = 2 13 C 2 + C13 9 mF ´ 18 mF = = 6 mF (9 + 18) mF C = C 2-13 + C 4 = 6 mF+9 mF = 15 mF

793

which gives, x = 0.2 m

42. The magnitude of torque acting on the dipole, t = pE sin q = 4 ´ 10 -9 ´ 5 ´ 10 4 sin 30° 1 = 2 ´ 10 -4 ´ = 10 -4 N-m 2

43. Force on proton at point A, FA = qE A = 1.6 ´ 10 –19 ´ 40 = 6.4 ´ 10 –18 N

794 JEE Main Physics DU =

44. We have,

45. We have, Þ

1 C1C 2 (V2 - V1) 2 2 (C1 + C 2)

shown in figure.

-12

=

1 (3 ´ 5) ´ 10 ´ (500 - 300) 2 (3 + 5) ´ 10 -6

=

1 15 ´ 10 -12 ´ 4 ´ 10 4 ´ 2 8 ´ 10 -6

2

Uf =

2

1C O

F=

…(i) 2

=

2

…(ii)

+ + ++ E

–

+ + +

– –

–

–

–

qE mg

Þ

mg ¢ = mg + qE Effective acceleration, qE ö æ g ¢ = çg + ÷ è mø

2p T = T0

l 1/ 2 qE ö æ æ ö çg + ÷ ç ÷ è g mø =ç qE ÷ l çg + ÷ 2p è mø g

d

q

Þ

or

l l = 2p qE ö g¢ æ çg + ÷ è mø

52. We have, F= n=

q 2

(ne) 4 pe 0d 2 4 pe 0Fd 2 e2

G

m 1 q2 = × -2 2 4pe 0 (16 ´ 10 -2) 2 (16 ´ 10 ) q = 4 pe 0G m

55. In the figure, since |FA| =|FB| =|FC|

56. We have,

new time period, T = 2p

2

A

and they, are equally inclined with each other. Then, there resultant will be zero.

Net downward force

Force,

10 -6 æ 1 1 1 1 ö + + + +¼¥ ÷ ç ø 4 pe 0 è 1 4 16 64

54. They will not experience any force if

Þ

\

ù 1 ´ 1 ´ 10 -6 + ¼¥ ú (8) 2 û

|FG | = |Fe|

l As, T0 = 2p g

Thus,

1 é1 ´ 1 ´ 10 -6 1 ´ 1 ´ 10 -6 1 ´ 1 ´ 10 6 + + 4pe 0 êë (1) 2 (2) 2 ( 4) 2

(Q series is infinite geometric series) 4 = 9 ´ 10 9 ´ 10 -6 ´ 3 4 3 = 9 ´ 10 N = 12000 N 3

46. The resultant electric field is given by

51.

n=8

n=4

æ 1 ö = 9 ´ 10 9 ´ 10 -6 ç ÷ è1 - 1/4 ø

V2 V 2 U - Uf 5 ´ 100 = 4 V / 5 ´ 100 = 80% Now, i ´ 100 = 2 Ui V V2 2

or

1µC

n=1 n=2

+

V 1 5V æV ö æV ö (2 + 8) ç ÷ = 5 ç ÷ = = è5ø è5ø 2 25 5

E = E1 + E2 s s s + s2 E= 1 + 2 = 1 2 e0 2 e0 2 e0

1µC 1µC 1µC

That force acting on1C charge is given by

30 ´ 10 -2 = J = 0.0375 J 8 1 1 Ui = CV 2 = ´ 2 ´ V 2 2 2 Ui = V 2 2

and

53. The schematic diagram of distribution of charges on x-axis as

Þ

F = F¢ Q1Q 2 Q1Q 2 = 4 pe 0r 2 4pe 0r ¢2K r r¢ = K

q

FB Q

B

q

FC

q

C

57. Suppose, 1 ball has any type of charges, 1 and 2 must have different charges, 2 and 4 must have different charges, i. e. , 1 and 4 must have same charges.

58. The circuit is equivalent to parallel combination of two identical capacitors. Each having capacitance e A C= 0 d 2 e 0A Thus, C eq = 2 C = d

59. The electric field at a distance R is only due to sphere because electric field due to shell inside it is always zero. 1 3Q 3Q Thus, electric field = × 2 = 4pe 0 R 4pe 0R 2

Electrostatics 60. As, q1 + q 2 = Q

A Q

q1 q2 = 4 pr 2 4 pR 2 Qr 2 q1 = 2 2 \ R +r QR 2 and q2 = 2 2 R +r \Potential at component centre 1 æ Qr 2 QR 2 ö = ç 2 2 + 2 2 ÷ 4pe 0 è (R + r ) r (R + r ) R ø 1 Q (R + r) = 4pe 0 (R 2 + r 2) Here,

θ

B –q

C

i. e. ,

P

x

Fµ

Þ

Fnet = -

x ( a2 + x2)1/ 2

1 2qQ x 4pe 0 ( a2 + x2)3 / 2

As the resulting force Fnet is not linear the motion will be oscillating, but not simple harmonic.

65. Charge will be move along the circular line of force become

p'

x2 + y 2 = 1 is equation of circle in xy-plane.

q +q 2a p

66. We have, (3 + 3) ´ (1 + 1) +1 (3 + 3) ´ (1 + 1) 6 ´2 12 = +1= +1 6+2 8 12 + 8 20 5 = = = mF 8 8 2

C eq =

p ¢ = p cos q Then, the electric potential at point P will be p cos q V= 4pe 0r 2

62. The force on l length of the wire 2 is

Also

Q

a

line OP will be –q

F

a

61. From the figure, the component of dipole moment along the

Þ

795

2 kl1 f2 = QE1 = ( l 2l) R f2 2 kl1l 2 = l R f1 f2 f1 2 Kl1l 2 = = = l l l R

λ1

λ2

Q = CV 5 Q = ´ 100 = 250 mC 2

Q Q

\

B R

6 µF

1 µF

2 µF

63. An imaginary cube can be made by considering charge q at the centre and given square is one of its face. Then, flux from the given square i. e. , one face of the cube. q f= 6 e0

q

100 V

Change in 6 mF branch æ6 ´2ö = CV = ç ÷ ´ 100 = 150 mC è6 + 2ø

a/2

VAB = R

64. By symmetry of problem, the components of force on Q due to charges at A and Balong y and will cancel each other while long x-axis will added up and will be along co. under the action of this force is charge Q will move towards O. If at any time charge Q at a distancex from O. Net force on charge Q. - qQ 1 Fnet = 2F cos q = 2 ´ 4pe 0 ( a2 + n 2)

and

150 = 25 V 6

VBC = 100 - VAB = 75 V

…(i)

67. The energy of the system when the both capacitors are same. U1 =

1 1 CV 2 + CV 2 = CV 2 2 2

In 2nd case, when K is opened and dielectric medium is filled between the plates, capacitance of the both the capacitance of both the capacitors becomes 3 C, while potential difference across A is V and potential difference across B is V/3.

796 JEE Main Physics V2 1 1 (3 C) V 2 + (3 C) 2 3 3 10 5 = CV 2 = CV 2 6 3

74. As the electrostatic forces are conservative so, work done is

U2 =

Then,

independent of path \

…(ii)

Dividing Eq. (i) by Eq. (ii) we get, U1 3 = = 3 :5 \ U2 5

75.

W = F × ds = qE$i × [(0 - a) $i + (0 - b) $j ]

= - qEa s l and t = Here, E = u e0 Along Y-axis, u = 0 , a =

69. If the value of C is chosen 4 mF, the equivalent capacity across every part of the section will be 4 mF.

\

s =d =

68. The total charge = (2 C) ´ (2 V ) + (C) ( -V ) s=

or,

= 4 CV - CV = 3 CV 3 CV = 3V C 1 3 \Energy = (3 C) V 2 = CV 2 2 2 \Common potential =

V=

or

=

t 8 3 = 2 aE sin q 20 -2 ´ 10 5 ´ sin 60° 8 3 2 ´ 10 3 ´ 3 /2

=

q = 8 ´ 10 -3 C

of charge is independent of distance of point from the sheet. Applying the principle of superposition, we get s 2s $ s Ep = ( - k$ ) + ( - k) + ( - k$ ) 2 e0 2e 0 2e 0 Ep = -

2s $ k e0

and

79.

Now,

W = ?, q1 = 0° , q2 = 180° t max = pE = q (2a) E ´ 2.0 ´ 10

–2

Þ 5

´ 1 ´ 10 = 2 ´ 10

-3

N-m

= - (10

-2

5

´ 2 ´ 10 ) (10 ) (cos 0° - cos180° )

= - 4 ´ 10 -3 J

73. On connecting, potential becomes equal q µ C µ r and s=

2 K2e 0 A d 1 1 1 d d = + = + C s C1 C 2 2K1e 0 A 2K2e 0 A =

d æ K1 + K2 ö ÷ ç 2e 0 A è K1K2 ø

Cs =

2e 0 A æ K1K2 ö ÷ ç d è K1 + K2 ø

80. The arrangement shows a Wheatstone bridge.

W = - pE (cos q1 - cos q2) -6

é 4q 2 2q 2 ù 2.6 q 2 = + ê a 4pe 0 a a 2 úû ë

1 surface of the sphere. Outside the sphere, V µ . Hence, r Fig. (a) represents the correct graph. K e A 2K e A As, C1 = 1 0 = 1 0 d /2 d

E = 1 ´ 10 5 NC–1, t max = ?

= 1 ´ 10

1 4pe 0

( -q) ( -q) + q 2 4pe 0 a 2

78. Inside the hollow sphere, V = constant = potential on the

2a = 2.0 cm = 2.0 ´ 10 –2 m

-6

1 [q ( -q) + (9 - q) q + q ( -q) + ( -q) (q)] 4pe 0 a

C2 =

72. Given, q = ±1´10 -6 C

\

Qr ¢ r + r¢

-

71. It is know that the electric field due to an infinite plane sheet

or

Q+0 total charge =total capacity 4pe 0 (r + r ¢ )

77. As, work done = Potential energy of configuration of charge

From, t = pE sin q = q (2a) E sin q

=

2 de 0mu 2 el 2

\ Charge on smaller sphere = 4pe 0r ¢ ´ V =

t = 8 3 Nm,q = ?,2 a = 2 cm = 2 ´ 10 -2 m

q=

1 2 1 eE 2 1 es l 2 at = t = 2 2 m 2 me 0 u 2

76. Common potential,

70. Here, q = 60° , E = 105 NC–1

or

F eE (from potential of motion) = m m

q r 1 µ = A r2 r

\ Surface charge density on 15 cm sphere will be greater than that on 20 cm sphere.

As

C1 C 4 = = 1, therefore the bridge is balanced. C3 c5 1 1 1 2 = + = , C s1 = 2 mF C s1 4 4 4

Similarly, C s 2 = 2 mF \Effective capacitance = C p = C s1 + C s 2 = 2 + 2 = 4 mF

Electrostatics 81. Charge on each plate of each capacitor will be Q = ± CV = ± 25 ´ 10 -6 ´ 200 = ± 5 ´ 10 -3 C

82. For equivalent capacitance, 1 1 1 1 1 3 = + = + = ; C s C1 C 2 10 20 20 20 Cs = mF 3 \Charge on each capacitor 20 4000 = C sV = ´ 200 = mC 3 3 total charge Common potential = total capacity 2 ´ 4000 / 3 800 V = = 10 + 20 9

83. (n - 1) capacitors are made by n plates and all are convex in parallel because plates are connected alternately. \Total capacitance = (n - 1)x

84. If C is capacity of each condenser, then charge on each capacitor = 10 C When connected in series, potential difference between free total charge plates = total capacity 10 C = = 60 V C /6

85. ò E × ds represents electric flux over the closed surface. When

ò E × ds = 0, it means the number of flux lines entering the surface must be equal to number of flux lines leaving it. q Further, as ò E × ds = where q is charge enclosed by the e0 surface. When ò E × ds = 0; q = 0, i. e. , net charge enclosed by

797

89. On inserting the dielectric slab electric charge on plates

e 0 AV in accordance with d 1 conservation of charge principle. But, new electric field is K V . As now potential times of original value. Hence, E = Kd V difference V ¢ = , hence work done K 1 1 QV é 1 ù 1W = QV - QV ¢ = 2 2 2 êë K úû remains unchanged at Q =

or

W=

e 0 AV V é 1 ù e 0 AV 2 é 1 ù 1= × 12 êë K ûú 2 d ëê K ûú d

90. As per Gauss’s theorem in electrostatics, ò E × ds =

q , where e0

q is charge enclosed by the surface. If the charge is outside the surface, q inside = 0. Therefore, ò E × ds = 0

91. On introducing the metal sheet, equal and opposite charges will appear on the two faces of metal sheet. Therefore, capacity of each condenser becomes twice and the two are connected in series. 1 1 1 1 = + = \ Cs 2 C 2 C C Þ

C s = C,

i. e. , the capacity remains unchanged. Potential difference V remains constant and hence, battery does not supply more charge.

92. When a capacitor is charged using a battery, the battery supplies equal and opposite charges to the two plates. Outer surfaces of plates have equal charges. But, the inner surfaces of plates have equal and opposite charges.

93. When there are various types of charges in a region, but the

the surface must be zero. Therefore, all other charges must necessarily be outside the surface. This is because charges outside the surface do not contribute to the electric flux. Q At any point outside the shell, V = , where r is the 4pe 0r radius or spherical shell. The outer surface of the spherical shell is an equipotential surface.

total charge is zero, the region can be supposed to contain a number of electric dipoles. Therefore, at points outside the region (which may be anywhere w.r.t. electric dipoles), 1 the dominant electric field µ 3 for large r. Choice (c) is r correct. Further, as electric field is conservative, work done to move a charged particle along a closed path, away from the region will be zero.

87. For a uniformly charged circular ring the electric field

94. Battery is disconnected, hence charge remains conserved.

86.

R intensity is maximum at its axial line at a distance, x = ± 2 from the centre of ring and q 2q 1 = |E max| = . 4pe 0 3 3R 2 6 3pe 0R 2

88. From the general knowledge of theory, electric field at a point is continuous if there is no charge at that point. And the field is discontinuous if there is charge at that point.

The plates are parallel apart, so, capacity will decrease on account of increase of voltage. Now capacity decreases, hence, energy will increase as charge is constant.

95. Charge enclosed by the Gaussian surface Q - 2 = - Q. Therefore, as per Gauss’s theorem, total flux through the Q surface of the sphere = × Further, as charge 5 Q lies e0 outside the surface, it makes no contribution to electric flux through the given surface.

798 JEE Main Physics 96. If the sheet S is given some positive charge density, by

103. Consider a gaussian surface of radius, r with centre of sphere

induction, negative charge developes on ends of A and B, closer to S and an equal positive charges developes on farther ends of A and B as shown in question figure. Therefore, S attracts A and B, also A attracts B.

as the centre. From symmetry E is constant everywhere on this surface and directed normally on the surface. Q R

97. Electric flux through S is zero and continuous to remain zero only i. e. , unchanged.

E

E

98. As, for r £ R0 electric potential is constant, hence, E inside the sphere is zero and consequently electrostatic energy is not stored at all inside the sphere. This indicates that the sphere is a thin spherical shell and charge lies only on its surface i.e., at r = R0 . No charge is outside. Moreover, as for r < R0 , E = 0 and at r = R0 electric field suddenly appears, it shows that E is discontinuous at r = R0 .

99. As is known from theory, at the centre of the ring, E = 0 when a positive charge (q > 0) is displaced away from the centre in the plane of the ring, say to the right, force of repulsion on q, due to charge on right half increases and due to charge on left half decreases. Therefore, charge q is pushed back towards the centre. Along the axis of the ring, at a distance r from the centre, Qr 4pe 0 (r 2 + a2)3 / 2

E=

If charge q is negative (q < 0), it will perform SHM for small displacement along the axis.

r

Now, from Gauss’s law. fE = E × ds = E × 4pr 2 = \

E=

B

W = - ò E × dl A

On equipotential surface, E ^ dl B

W = - ò E(dl) cos 90° = Zero

surface. Let electric field at any point P on this surface be E. Q Q Then fE = E × ds = E × 4pr 2 = e0 Þ

E=

towards the centre of ring and is variable as 1 qx . Consequently, motion of charged E= 4pe 0 ( x2 + R 2)3 / 2 particle will definitely be periodic. Moreover, if x 0 and for semiconductors a < 0.

Low Temperature Resistivity The temperature dependence of resistivity at temperatures around room temperature is characterized by a linear increase with temperature. Microscopic examination of the conductivity shows it to be proportional to the mean free path between collisions (d), and for temperatures above about 15 K, d is limited by thermal vibrations of the atoms. The general dependence is summarized in the proportionalities 1 p µ µ A2 µ T d

Impurity dominated region ρ = constant

Linear region Transition to Superconductor

r = resistivity, d = mean free path of the electrons, A = amplitude of atomic vibrations, T = temperature (K ). At extremely low temperatures, the mean free path is dominated by impurities or defects in the material and becomes almost constant with temperature. With sufficient purity, some metals exhibit a transition to a superconducting state.

18.6 V-I Characteristics of Ohmic and Non-ohmic Conductor V-I Characteristics of Ohmic Conductors R (the slope of the V -I graph) is a constant for a given conductor under these conditions. The circuit element is said to be ohmic, commonly known as an ohmic conductor. Materials and components that obey Ohm’s law are described as ohmic which means they produce

Current Electricity Vö æ the same value for resistance ç R = ÷ regardless of the è Iø value of V or I which is applied and whether the applied voltage or current is DC (direct current) of either positive or negative polarity or AC (alternating current). In true ohmic V circuit, the ratio of is constant, and when current is I plotted as a function of voltage, the curve is linear (straight line). I

the first three bands indicate the first three digits of the resistance. Value and the fourth band indicates the number of zeros. In the five band code, the fifth band is gold for 1% resistors and silver for 2%. Resistor

The current is directly proportional to the potential difference

Code

Colour

Resistance value

0

Black (B)

First three bands

1

Brown(B)

1st band-1st digit

2

Red (R)

2nd band-2nd digit

3

Orange (O)

3rd band-number of zeros

4

Yellow (Y)

5

Green (G)

6

Blue (B)

7

Violet (V)

8

Grey (G)

9

White (W)

The resistance is R= 1 Slope V

V-I Characteristics of Non-ohmic Conductor Components of electrical resistance which do not obey Ohm’s law, i. e. , their relationship between current and voltage (their I-V curve) is non-ohmic (or non linear) are known as or non-ohmic conductor. In use, non-ohmic circuit. The curves are not linear and don’t have a well defined slope

I

809

Shortcut to learn the series B B R O Y Great Britain Very Good Wife.

Example

A resistor with bands of yellow, violet, red and gold will have first digit 4 (for yellow) second digit 7 (violet) followed by 2 (red) zeros. Gold signifies that the tolerance is ± 5% so the real resistance could lie anywhere between 4,465 and 4,935 ohms.

Diode Light bulb filament V

When current is plotted as a function of voltage, the curve is non-linear

Example p-n junction diode, light bulbs such as car headlamps.

Note All coded components will have at least two value bands and a multiplier, other bands are optional.

The standard colour code per EN 60062 : 2005 is as follows Colour Black Brown

18.7 Colour Code for Resistors The electronic colour code is used to indicate the values or ratings of electronic components. The resistance value and tolerance can be determined from the standard resistor colour code. The following diagram shows a carbon resistor.

Red Orange Yellow Green

A variation on the colour code is used for precision resistors which may have five colour bands. In that case,

1 2 3 4 5

0

´ 10

Tolerance –

1

= 1%

2

= 2%

´ 10 ´ 10

3

´ 10

–

4

± 5%

5

± 0.5%

6

´ 10

´ 10

6

´ 10

+ 0.25%

Violet

7

´ 107

± 0.1%

8

8

± 0.05% (± 10%)

9

–

-1

± 5%

-2

± 10%

White R

0

Multiplier

Blue

Grey A B C

Significant figure

Gold Silver

9 – –

´ 10

´ 10 ´ 10 ´ 10

810 JEE Main Physics

18.8 Electric Cell

SO42– ions move towards Pb electrode give negative charge

An electric cell is a device which maintains a continuous flow of charge (or electric current) in a circuit by a chemical reaction.

charge.

and H+ ions move to the PbO 2 electrode, given up positive

In an electric cell, there are two rods of different metals called electrodes.

There are two basic types of electric cell

(i) Primary Cell A cell is called primary, if it used only for discharge. The current leaves the cell at the positive (+) terminal goes through the external circuit and enters the cell at the negative (–) terminal, Examples of primary cells are daniell cell, laclanche cell and dry cell.

Dry cell The most popular cell is dry cell. It is a special type of laclanche cell in which both NH4Cl are MnO 2 are prepared in the form of a paste. The paste is contained in a zinc container which is negative electrode. The internal resistance of a dry cell is very small generally 0.1 W. Its emf is generally 1.5 V. – C

NH4OH

(ii) Secondary Cell In a secondary cell, the current pass in both direcions When current leaves the cell at the positive (+ve) terminal and enters the cell at the negative (–ve) terminal, the cell is discharge. This is called normal working of the cell. In this case, the chemical energy is converted into electrical energy. The most commonly used secondary cell is a lead accumulator. – + – PbO2

Pb

H+ – SO4

(a) Discharging

i +

– PbO2

Pb

PbO + H2SO4 ¾® PbSO4 + H2O

Therefore, PbSO4 is formed at the both electrodes. In charging process, a current forced from the positive to the negative inside the cell. The H+ ions move towards the negative electrode and react with the PbSO4. PbSO4 +2H ¾® Pb + H2SO4 At the positive electrode, the reactions PbSO4 + SO4 +2H2O ¾® PbO 2 +2H2SO4

Electromotive Force (EMF) of Cell The potential difference between the two poles of the cell in an open circuit is called the electromotive force (emf) of the cell. It is denoted by E. Its S.I. unit is volt (V) or joule coloumb -1 (JC -1)

Internal Resistance 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. As the current moves through r, it experiences associated drop in potential equal to ir. Thus, when a current is drawn through a source, the potential difference between the terminal of the source is V = E - ir

Zn

MnO2

i +

dil

i. e. ,

Types of Electric Cell

+

PbO 2 +2H ¾® PbO + H2O The PbO so formed reacts with dil. H2SO4 to get PbSO4 and H2O

H+ –

SO4

(b) Charging

A lead commulator consists of electrodes made of PbO 2 and Pb immered is dil. H2SO4. Discharging process, the

This can also be shown as below E

r

A

i

B

V A - E + ir = VB or

V A - VB = E - ir

Following three special cases are possible (i) If the current flows in opposite direction (as in case of charging of 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 or E = ir i= r \ or

E - ir = 0 V =0

Current Electricity

In order to have maximum current, the cell should be connected in series, when the total internal resistance of the cell is negligible as compared to the external resistance of the circuit.

Short circuited r

E

Thus, we can summarise it was follows E

In parallel

r

Consider the following three cases i

Case I Let n cells each of emf E and internal resistance r are connected in parallel as shown in figure. Then,

V = E - ir or V < E E

r i

V = E + ir or V > E E

Net emf = E

r

E

r

E

r

E

r

i

i

r Total resistance = + R n

V = E, if i = 0

R

\ Current in the circuit, i

Net emf Total resistance E i= R + r/n

E r

i=

or

r

E

811

V = 0 is short circuited

Case II Let n cells have different E and r. A

18.9 Combination of Cells in Series and Parallel

E1

r1

E2

r2

E3

r3

F

i1

i3

Cells are usually grouped in following ways

E

i2

B

i

i

In series

R

Suppose n cells each of emf E and internal resistance r are connected in series as shown in figure. Then E

r

E r

E

C

D

Net emf = Eeq =

r

i

Total resistance = Req = R + R

Net emf = nE Total resistance = nr + R Net emf \Current in the circuit = Total resistance nE or i= nr + R

Hence,

i=

1 S(1/r )

Eeq Req

or

i=

S (E/r ) R + S (1/r )

Case III This is 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 other as shown. A

E1

r1

E2

r2

E3

r3

i1

Note If polarity of m cells is reversed, then equivalent emf E = (n - 2m)E (n - 2m)E while total resistance is still nr + R i = nr + R In series grouping of cells their emf’s are additive or subtractive while their internal resistances are always additive.

S (E/r ) S (1/r )

i2 i3

i

i R

812 JEE Main Physics Net emf = Eeq

Total resistance = Req = R +

Hence,

i=

Eeq Req

1 æ 1 1 1ö + ÷ ç + è r1 r2 r3 ø

(E1/r1 ) - (E2/r2 ) + (E3/r3 ) æ 1 1 1ö + ÷ 1+ Rç + è r1 r2 r3 ø

=

Note In order to have maximum current, the cells should be connected in parallel, when the external resistance of the circuit is negligible as compared to the total internal resistance of the cells.

In mixed grouping There are n identical cells in a row and number of rows are m. E

E = 2 V , r = 1.5W, and R = 10 W

Here,

(E /r ) - (E2/r2 ) + (E3/r3 ) = 1 1 æ 1 1 1ö + ÷ ç + è r1 r2 r3 ø

r

The emf of the two cells in a row = 2E = 2 ´ 2 = 4 V The total emf E¢ of the 4 cells is equal to that of a row, i. e. , E¢ = 4 V The internal resistance of the two cells in a row = 1.5 + 1.5 = 3W If r¢ is internal resistance of the four cells, then 1 1 1 2 = + = r¢ 3 3 3 3 or r¢ = = 1.5W 2 Total resistance of the circuit, R ¢ = R + r ¢ = 10 + 1.5 = 11.5W Therefore, current in the circuit, e¢ 4 i= = = 0.348 A R ¢ 11.5 As the two branches are identical, i 0.348 Current in each branch = = = 0.174 A 2 2

i

i

R

Potential difference across external resistance = iR = 0.348 ´ 10 = 3.48 V

Net emf = nE nr Total resistance, Req = R + m Hence,

i=

nE nr r+ m

Note In order to have the maximum current, the cells should be mixed group in such a way that the external resistance of the circuit is equal to the two resistance of the cells in mixed grouping.

Sample Problem 6 A set of 4 cells, each of emf 2 V and internal resistance 1.5 W, are connected across an external load of 10 W with 2 rows, 2 cells in each branch. The current in each branch and potential difference across 10 W is (a) 0.174 A, 3.48 V (c) 2.174 A, 3.48 V

(b) 1.174 A, 2.48 V (d) 3.174 A, 4.48 V

Interpret (a) The four cells, each of emf E and internal resistance r are connected to the external resistance R as shown in figure. r

r

E

E r

i

r E

E R

18.10 Series and Parallel Combination of Resistors Series Combination A series circuit is a circuit in which resistors are arranged in a chain, so the current has only one path to take. The current is the same through each resistor. The total resistance of the circuit is found by simply adding up the resistance values of the individual resistors. Equivalent resistance of resistors in series R = R1 + R2 + R3 + K

V

R1

R2

R3

I

The current flows through each resistor in turn. If the values of the three resistors are R1 = 8 W, R2 = 8 W and R3 = 4W, then total resistance is R = 8 + 8 + 4 = 20 W. With a 10 V battery, by V = IR, the total current in the circuit is V 10 I = = = 0.5 A. The current through each resistor R 20 would be 0.5 A.

Current Electricity

Parallel Combination A parallel circuit is a circuit in which the resistors are arranged with their heads connected together and their tails connected together. The current in a parallel circuit breaks up, with some flowing along each parallel branch and recombining, when the branches meet again. The voltage across each resistor is parallel is the same. The total resistance of a set of resistors in parallel is found by adding up the reciprocals of the resistance values, and then, taking the reciprocal of the total. The equivalent resistance of resistors in parallel, 1 1 1 1 = + + +K R R1 R2 R3 I V

I1 I2

R1 I R2

R1 I3

R3

A parallel circuit is shown in the diagram above. In this case, the current supplied by the battery splits up, and the amount going through each resistor depends on the resistance. If the values of the three resistors are R1 = 8 W, R2 = 8 W and R3 = 4W, the total resistance is found by 1 1 1 1 1 = + + = R 8 8 4 2 This gives,

R = 2W

With a 10 V battery by V = IR, the total current in the circuit V 10 is I = = = 5A R 2 The individual currents can be found using I =

V × The R

voltage across each resistor is 10 V, 10 10 = 1.25 A, I 2 = = 1.25 A 8 8 10 I3 = = 2.5 A 4 I1 =

and

Short-cut for Calculating Equivalent Resistance in Parallel Combination If the resistor in parallel are identical, it can be very easy to work out the equivalent resistance. In this case, the equivalent resistance of N identical resistors is the resistance of one resistor divided by N, the number of resistors. So, two 40 W resistors in parallel are equivalent to one 20 W resistor, five 50 W resistors in parallel are equivalent to one 10 W resistor etc. When calculating the equivalent resistance of a set of parallel resistors, one 1 1 often forget to flip the upside down, putting of an ohm instead of 5 W. R 5 Here’s a way to check your answer. If you have two or more resistors in parallel, look for the one with the smallest resistance. The equivalent resistance will always be between the smallest resistance divided by the number of resistors and the smallest resistance. Here’s an example you have three resistors in parallel, with values 6 W, 9 W and 18 W. The smallest resistance is6 W, so the equivalent resistance must be between2 W and6 W 6 ( 2 = , where 3 is the number of resistors). 3 1 1 1 6 Doing the calculation gives + + = 6 9 18 18 18 Flipping this upside down gives = 3 W which is certainly between 2 and 6. 6

General Rules for Reduction Process 1. Two (or more) resistors with their heads directly connected together and their tails directly connected together are in parallel, and they can be reduced to one resistor using the equivalent resistance equation for resistors in parallel. 2. Two resistors connected together so that the tail of one is connected to the head of the next, with no other path for the current to take along the line connecting them, are in series and can be reduced to one equivalent resistor. Finally, remember that for resistors in series, the current is the same for each resistor and for resistors in parallel, the voltage is the same for each one.

Sample Problem 7 In the figure, if a battery is connected between points A and B, emf E = 18 V, then the current following through the battery is

Note That the currents add together to 5 A, the total current.

D 18Ω

Circuits with Series and Parallel Components Many circuits have a combination of series and parallel resistors. Generally, the total resistance in a circuit like this is found by reducing the different series and parallel combinations step-up-step to end up with a single equivalent resistance for the circuit. This allows the current to be determined easily. The current flowing through each resistor can then be found by undoing the reduction process.

813

A

B 9Ω

4/5Ω 3Ω

3Ω E

1/5Ω

6Ω C

814 JEE Main Physics (a) 10 A

(b) 20 A

Interpret

(c) 5 A

(d) 15 A

(a) As observed from the circuit the resistances 9W

and 3W are in parallel, hence equivalent resistance is 9 1 1 1 1 1 12 4 = + = + = = ÞR = W 4 R R1 R2 9 3 27 9 The complicated circuit can now to be reduced as follows Hence, resistors 18W , 3W , 6W and 9/4W are in parallel. 18Ω 4/5Ω

B

9/4Ω

Taking their equivalent, we have 1 1 1 1 4 1 + 6 + 3 + 8 18 = + + + = = = 1W R 18 3 6 9 18 18 \

V = V1 + V2 = I(R1 + R2) V 12 From Ohm’s law, current I = = = 1.2 R1 + R2 10 Hence,

V1 = (1.2) (6) = 7.2 V

and

V2 = (1.2) ( 4) = 4 . 8 V

Conductance and Conductivity Conductance

3Ω 6Ω

A

Potential drop across branch

The reciprocal of resistance of a conductor is called its conductance. It is denoted by G. Thus, conductance of a conductor having resistance R is given by G=

R = 1W

This is connected in series with Req. \

4 W, hence we have 5

4 9 = R1 + R2 = + 1 = W 5 5

E = IReq. E 18 I= = = 10 A Req 9/5

Þ

Sample Problem 8 In the given circuit, the potential drop

1 R

…(i)

SI unit of conductance is ohm -1 (W -1 ), which is also called

mho. The unit of conductance in SI is also called simen and is denoted by the symbol S.

Conductivity The reciprocal of resistivity of the material of a conductor is called its conductivity. It is denoted by s Thus, 1 …(ii) s= r From Eq. (ii), it follows that

across R1, R2 is R = 6Ω

R = 4Ω e

f

SI unit of conductivity =

1 ohm metre

= ohm -1 metre -1 = mho -1 metre -1 = simen metre -1. g

d 6V

17Ω

2Ω

2Ω

12V

h

c

mho metre -1 or simen metre -1 (Sm-1 ).

12Ω 6V

1Ω a

b 12V

(a) 3.2 V, 5 V (c) 1 V, 10 V

Thus, SI unit of conductivity is ohm -1 metre -1 (W -1m-1 ) or

(b) 7.2 V, 4.8 V (d) 5 V, 5 V

Sample Problem 9 A battery of 10 V and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance 1W. The equivalent resistance of the network is [NCERT]

Interpret

(b) A close look at the circuit, makes it clear that points a, h, g , and f have same potential. They are connected by conducting wires without any circuit elements between them. Similarly, points b, c, d and e have the same potential. Hence, the potential drop across branch e and f , and a and b is same. The two resistors 6W and 4W in series are directly connected across the

D′

C′ I/2 A′

B′

I

I/2 I/2 I

C

D

R = R1 + R2 = 6 + 4 = 10 W

I/2

I

The complex circuit in the middle has no effect on the potential drop across the upper 10 W branch. If the current through it is I.

B

A 3I

Potential drop across (R1 = 6W) is V1 = IR1

Potential drop across (R2 = 4W) is V2 = IR2

I I/2

terminals of 12 V battery. Hence, resistance is arm fe is

\

I

I

I/2 10V

E

(a)

6 R 5

(b)

2 R 3

(c)

3 R 2

(d)

5 R 6

Current Electricity Interpret (d) From the figure, it is clear that the network is not reducible to a simple series and parallel combination of resistors. There is, however, a clear symmetry in the problem which we can exploit to obtain the equivalent resistance of the network. The path AA¢ , AD and AB are obviously symmetrically placed in the network. Thus, the current in each must be the same, say, I. Further at the corners A¢ , B and D, the incoming current must split equally into the two outgoing branches. In this manner, the current in all the 12 edges of the cube are easily written down in terms of I, using Kirchhoff’s first rule and the symmetry in the problem. Next take a closed loop, say ABCC ¢ EA and apply Kirchhoff’s second rule æ 1ö - IR - ç ÷ - IR + e = 0 è2ø where R is the resistance of each edge and e the emf of the 5 IR. 2 e 5 The equivalent resistance Req of the network is Req = = R. 3I 6 battery. Thus, e =

Sample Problem 10 A resistor of 5W resistance is connected in series with a parallel combination of a number of resistors each of 6W. If the total resistance of the combination is 7W, how many resistors are in parallel? (a) 2

(b) 3

(c) 4

(d) 5

Interpret (b) Let n resistors each of 6W be connected in parallel and then the combination be connected in series with resistor of 6W. The resistance of the parallel combination of n resistors each of 6W is given by n 1 1 1 1 = + + K + n times = 6 Rp 6 6 6 6 Rp = W n As this parallel combination is connected in series with the resistor of 5W, the total resistance of the combination is given by 6 R = Rp + 5 = + 5 n Since, resistance of the combination is 7W, it follows that 6 6 + 5 = 7 or = 2 or n = 3 n n

Interpret (b) Here, load resistance, RL = 500 W EMF of source, E = 50 V Total resistance of the rheostat between points A and C, RAC = 2 kW = 2000 W Resistance of the rheostat between points A and B, RAB = 500 W Therefore, resistance between points B and C, RBC = RAC - RAB = 2000 - 500 = 1500 W Now, RBC and RL are connected in parallel, If R¢ is resistance of their combination, then 1 1 1 = + R ¢ RBC RL =

1 1 4 + = 1500 500 1500

R¢ = 375W Now, RAB and R¢ are in series. If i is current in the circuit, then E 50 2 i= = = A RAB + R ¢ 500 + 375 35

or

Potential drop across RL is same as the potential drop across the parallel combination of RBC and RL. Therefore, potential drop across RL = applied potential difference - potential drop across RAB 2 = 50 - iRAB = 50 ´ 500 = 21.43 V 35

Sample Problem 12 Two conductors have the same resistance at 0°C but their temperature coefficients of resistance are a 1 and a 2. The respective temperature coefficients of their series and parallel combinations are nearly a1 + a 2 2 a1 + a 2 a1 + a 2 (d) , 2 2

a1 + a 2 , a1 + a 2 2 a1a 2 (c) a1 + a 2, a1 + a 2

(b) a1 + a 2,

(a)

Interpret (d) Let R0 be the initial resistance of both the conductors.

Sample Problem 11 As shown in figure, a variable rheostat of 2 kW is used to control the potential difference across a 500W load. If the resistance AB is 500W, what is the potential difference across the load?

At temperature q their resistances will be R1 = R0(1 + a1 q) R2 = R0(1 + a 2 q)

i

For series,

50V

B 2kΩ

R2

Rs = R1 + R2 R L = 500Ω

C

where, \

(a) 15 V (c) 36 V

R1

Rs = R1 + R2 \ Combined resistance is

A

(b) 21 V (d) 45 V

815

Þ

Rs (1 + a s q) = R0(1 + a1q) + R0(1 + a 2q) Rs0 = R0 + R0 = 2R0 2R0(1 + a s q) = 2R0 + R0 q( a1 + a 2) a + a2 as = 1 2

816 JEE Main Physics For parallel combination, RR Rp = 1 2 R1 + R2 Rpo(1 + a p q) = where, \

Rpo = R0 (1 + a p q) = 2

R0(1 + a1 q) R0(1 + a 2 q)

R1

R2

R0(1 + a1 q) + r0(1 + a 2 q) R0R0 R = 0 2 R0 + R0

A B C

=

a1 + a 2 é æ a1 + a 2 ö ù ê1 - çè 2 ÷ø qú 2 ë û

as ( a1 + a 2) 2 is negligible, hence ap =

a1 + a 2 2

Check Point 1 1. A bulb is connected, by a pair of long straight conductors of

effective resistance R , to a source of emf E. What time after turning on the switch the bulb glows and after what time the electrons from the source reach the bulb? Answer it qualitatively. R

E

2. A rectangular metallic plate has its dimensions as shown in the figure. The resistance of the plate across the length, breadth and thickness are R 1, R 2 and R 3 , respectively. A range of these resistance in increasing order.

D

R1 R2

R0(2 + a1 q + a 2 q)

a1a 2 is negligible a1 + a 2 ap = 2 + ( a1 + a 2) q

or

Many electric circuits cannot be reduced to simple series parallel combinations. For example, two circuits that cannot be so broken down are shown in figure

R02(1 + a1q + a 2 q + a1 a 2 q2)

As a1 and a 2 are small quantities. \

18.11 Kirchhoff’s Laws and Their Applications

A

R3

R4

B R1

E1

R2

E2

D

E1

C

E2

E3 F H

E

F

E

G

I

R3 (a)

R5 (b)

However, it is always possible to analyze such circuits by applying two rules, devised by Kirchhoff. First here are two terms that we will use often (i) 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. (ii) Loop A loop of any closed conducting path. For example, in Fig. (a) ABCDA, DCEFD and ABEFA are loops. Similarly, in Fig. (b), CBFEC, BDGFB are loops. Kirchhoff’s rules statements

consist

of

the

following

two

Junction Rule The algebraic sum of the currents at any i1 junction is zero. i. e. , S i=0

i2

junction

i4

i3

This law can also be written as, ‘‘the sum of all the currents directed towards a point in 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 bases on conservation of electric charge.

3. Two wires A and B are of the same metal and of the same

length have their areas of cross-section in the ratio of 2 : 1.. If the same potential difference is applied across each wire in turn, what will be the ratio of the currents flowing in A and B? 4. Why is manganin used for making standard resistors? 5. It is easier to start a car engine on a warm day than on a chilly day?

Loop Rule The algebraic sum of the potential difference in any loop including those associated emf’s and those of resistive elements, must be equal to zero. That is,

S

closed loop

DV = 0

This law represent conservation of energy. Applying Kirchhoff’s law for the following circuit, we have

Current Electricity Resulting equation is Vr1 + Vr2 + Vr3 - 10 = 0. Vr The equation is obtained by – + traversing a circuit loop in R1 either direction and writing + assumed current R2 Vr down unchanged the 10V +– direction –2 R3 voltage of each element whose ‘+’ terminal is –Vr + entered first and writing down the negative of every elements voltage where the minus sign in first met. The loop must start and end at the same point. It does not matter where you start on the loop.

Summation of voltage terms may be done in either direction. +

Vr

–

10V + –

R2

of internal resistances 1W and 2W respectively have their positive terminals connected by a wire of 10W resistance and their negative terminals by wire of 4W resistance. Another coil of 10W is connected between the middle points of these wires. The potential difference across the 10W coil is (a) 1.07 V (c) 3.45 V

The distribution of currents in various branches is shown in the figure

+ V – r2

E1 = 2V R1 = 5 Ω

Resulting equation,

F

-Vr3 - Vr2 - Vr1 + 10 = 0

B R3 = 2 Ω i1 C R4 = 2 Ω

R = 10 Ω

i1 + i2

R2 = 5 Ω E

Note For both summations, the assumed current direction was the

D E2 = 1V r2 = 2 Ω

same.

In closes part ABCFA of the circuit,

Sign Convention in Kirchhoff’s Laws

i1 ´ r1 + i1 ´ R1 + (i1 + i2)R + i1 ´ R3 = E1

In applying the loop rule, we need sign convention as discussed below (a) When we travel through a source in the direction from –ve to +ve, the emf is considered to be positive.

i1 ´ 1 + i1 ´ 5 + (i1 + i2) ´ 10 + i1 ´ 2 = 2 or 9i1 + 5i2 = i In the closed part CDEFC of the circuit,

…(i)

i2 ´ r2 + i2 ´ R2 + (i1 + i2) ´ R + i2 ´ R4 = E 2

E B V

r1 = 1 Ω

A

– Vr +

Path (a)

(b) 2.03 V (d) 4.25 V

Interpret (a) The positive terminals of the cells E1 and E 2 are connected to the wire AE of resistance10 W and negative terminals to the wire BD of resistance 4W. The resistance of10 W is connected between the middle points F and C of the wires AE and BD respectively. Therefore, 10 R1 = R2 = = 5W; 2 4 R3 = R4 = = 2W 2

R3

A

817

or

E

(b) When we travel from +ve to -ve, the emf is considered to be negative.

or 10i1 + 19i2 = 1 Solving Eqs. (i) and (ii), we have

E A

B Path (b)

R (c) When we travel through a resistor A in the same direction as the Path assumed current, the iR term is (c) negative because the current goes in the direction of decreasing potential.

V

E

i

i1 = and

(d)

emf’s 2V and 1V and

14 A 121

i2 = -

B

(d) When we travel through a resistor in the direction opposite to the assumed current, R i the iR term is positive because A B this represents a rise of potential. Path

Sample Problem 13 Two cells of

i2 ´ 2 + i2 ´ 5 + (i1 + i2) ´ 10 + i2 ´ 2 = 1

1 A 121

Therefore, current through resistance R, 14 æ 1 ö + ç÷ 121 è 121ø 13 = A 121

i1 + i2 =

Potential difference across the resistance, 13 R = (i1 + i2)R = ´ 10 = 1.0744 V 121

…(ii)

818 JEE Main Physics Sample Problem 14 The number of electrons moving per

18.12 Joule’s Law Whenever the electric current is passes through a conductor, it becomes hot after some time. This indicates that the electric energy is being converted into heat energy. This effect is known as heating effect of current or Joule heating effect. This effect forms the basis of various electric appliances such as electric bulb, electric furnace, electric press, immersion rod etc. Joule in 1941, found experimentally the heating effect of current and stated that the amount of heat produced (H ) when a current i flows through a conductor of resistance R for a time t is given by H µ i2Rt

…(i)

This Eq. (i) is known as Joule’s law of heating. Consider a conductor AB of resistance R. Let V = potential difference (in volt) applied across the ends of AB, i = current (in ampere) flowing through AB, t = time (in second) for which the current is flowing. Total charge flowing from A to B in time t is V B

A R

i

i E

By definition of potential difference, work done in carrying unit charge from A to B = V Total work done in carrying a charge q from A to B is W = V ´ q = Vit J = i2Rt J

(Q V = iR)

If this entire work is dissipated as heat, then amount of heat produced (H ) is given by

(b) 3.1 ´ 10 19

(a) 3.1 ´ 10 18 (c) 4.8 ´ 10

(d) 4.8 ´ 10 19

18

Interpret (a) Here, charge on electron, e = 1.6 ´ 10 -19 C Power of the lamp, P = 100 W; operating voltage, V = 200 volt Now, P = Vi P 100 i= = = 0.5 A \ V 200 Charge passing through the lamp in 1s, q = i ´ t = 0.5 ´ 1 = 0.5 C Therefore, number of electrons moving through the filament per second, N=

q 0.5 = = 3.125 ´ 10 18 e 1.6 ´ 10 -19

18.13 Electrical Power The electrical power in watts associated with a complete electric circuit or a circuit component represents the rate at which energy is converted from the electrical energy of the moving charges to some other from e. g, heat, mechanical energy or energy stored in electric fields. For a resistor in a DC circuit the power is given by the product of applied voltage and the electric current. P = VI Power = Voltage ´ current The detailed units of power are Power = Volts ´ Amperes joule coulomb = ´ coulomb second joule = = watt second Power dissipated in a resistor can be obtained by the use of Ohm’s law.

H = W = i2Rt J =

second through the filament of a lamp of 100 watt, operating at 200 volt is (Given, charge on electron, e = 1.6 ´ 10 -19 C )

i2RT cal 4.18

…(iii) (Q 1 cal = 4.18 J )

Eq. (iii) is a statement of Joule’s law of heating.

P = VI =

V2 = I 2R R

Note These relationships are valid for AC applications also if the voltages and currents are rms or effective values.

Current Electricity Power Relationship The power relationship is one of the main tools for the analysis of electric circuits, along with Ohm’s law, the voltage law and the current law. The determining of the voltages and currents associated with a particular circuit along with the power allows you to completely describe the electrical state of a direct circuit.

For parallel circuit The powers for individual parallel elements can be calculated from the voltage across the element times the current through it. The sum of those powers will equal the power supplied by the battery, as is clear from the following circuit. PT = 12 V ´ 8 A = 96 W

VB

= 1000 W ´ 1h 1 kWh = 1000 Wh = (1000 W ) ´ (60 ´ 60) = 3.6 ´ 106 J Thus, electric energy = Vit = i2Rt =

V 2t R

Note Number of units of electricity consumed = Number of kWh =

watt ´ hour 1000

230 volt. If the supply voltage drops to 115 V, what is the heat and light energy produced by the bulb in 20 min?

I1 12V 12V I2 2A 6A R1 R2 6Ω

The commercial unit of electric energy is called a kilowatt hour (kWh) or Board of Trade Unit (BTU) or unit of electricity, in brief, where 1 kWh = 1 kW ´ 1 h

Sample Problem 15 An electric bulb is marked 100 W,

IB = 8A

12V

819

(a) 20000 J (c) 30000 J

(b) 25000 J (d) 35000 J

Interpret (c) Here, P = 100 W; V = 230 volt

2Ω

Let R be resistance of the filament of the bulb. Now, electric power, P = Vi = P1 = 12V × 2A = 24W

P2 = 12V × 6A = 72W

For series circuits The current is the same at any point in the circuit. Multiplying that current times the voltage drop across the resistor gives the power dissipated.

Therefore,

12V

I1 I 12V R1 9V drop 6Ω

VB

3V

R2 1.5A I

2Ω

3V drop OV

P1 = 9V ´ 1.5 A P1 = 13.5 W P2 = 3V ´ 1.5 A P2 = 4.5 W

Electric Energy The total work done or energy supplied by the source in maintaining the current in electric circuit for a given time is called electric energy consumed in the circuit. Electric energy, W = Vit = Pt Thus, electric energy = electric power ´ time Sl unit of electric energy is joule, where

1J = 1V ´ 1A ´ 1s = 1W ´ 1s

V 2 (230) 2 = = 529 W P 100

When the voltage drops to 115 volt, heat and light energy produced by the bulb in 20 min is given by

IB = 1.5A PT = 12V × 1.5A = 18W

R=

V2 R

W = Vit =

V2 (115) 2 t= ´ 20 ´ 60 = 30000 J R 529

Power Transformation Rule When bulbs are connected in parallel Let R1, R2, R3, K be resistances of given bulbs meant to operate at same voltage V to consume powers P1, P2, P3, K Then

R1 =

V2 V2 V2 … , R2 = , R3 = P1 P2 P3

or

P1 =

V2 V2 V2 … , P2 = , P3 = R1 R2 P3

When connected in parallel, their combined resistance R is 1 1 1 1 = + + +K R R1 R2 R3 Power consumed P=

æ 1 ö V2 1 1 = V 2ç + + + K÷ è R1 R2 R3 ø R

P=

V2 V2 V2 + + +K R1 R2 R3

820 JEE Main Physics (i) P = P1 + P2 + P3 + K (ii) If n identical bulbs are in parallel, then Ptotal = nP 1 (iii) Pconsumed (Brightness) µ RR µ i µ i. e., in parallel R combination bulb of greater wattage will give more bright light and more current will pass through it.

Important Points 1. A heater coil has a resistance R. It can boil certain amount of water in t. Its power is P. (i) When the coil is cut in two halves, power of each half = 2P (ii) When the coil is stretched to double its length, power = P/4

When bulbs are connected in series

(iii) When the coil is twisted to double its area power = 4P

Net resistance in series R = R1 + R2 + R3 + K

(iv) When the coil is twisted to double its thickness, power = 16P 1 (v) Time taken to boil t µ × P

(i) Net power consumed is given by

1 1 1 = + +K Ptotal P1 P2

(ii) If n bulbs are identical, then Ptotal = (iii) Pconsumed (Brightness) µ V µ R µ

P n 1

Prated

(i) When they are used together in series, time ts = t1 + t2 tt (ii) When they are used together in parallel time tP = 12 t1 + t2

i. e., in series

combination, bulb of lesser wattage will give more bright light.

Sample Problem 16 An electric heater and an electric bulb are rated 500 W-220 V and 100 W-220 V respectively. Both are connected in series to a 220 AC mains. The power consumed by (i) the heater and (ii) electric bulb are (a) 69.89 W, 13.98 W (c) 29.95 W, 89.58 W

2. Two coils boil separately a certain amount of water in timest1 andt2.

(b) 13.98 W, 69.89 W (d) 89.5 W, 29.95 W

Interpret (b) Power of electric heater, P1 = 500 W; operating

Maximum Power Transfer Theorem It states that the power output across load due to a cell or or battery is maximum if the load (external) resistance is equal to the effective internal resistance of cell of battery. It means, when the effective internal resistance of cell or a battery is equal to external load resistance in a circuit, the efficiency of battery or cell is maximum. The circuit is shown in figure.

voltage V1 = 200 V, power of electric bulb, P2 = 100 W; operating voltage, V2 = 220 V.

R r

Let R1 and R2 be the resistance of electric heater and bulb V2 respectively. Since, P = , we have R R1 = and

V12 (220) 2 = = 96.8 W P1 500

V 2 (220) 2 R2 = 2 = = 484W P2 100

When the series combination of electric heater and bulb are connected to 220 V mains, current in the circuit is given by V 220 i= = = 0.38 A R1 + R2 96.8 + 484 Now, power consumed by heater, P1¢ = i 2R1 = (0.38) 2 ´ 96.8 = 13.98 W

E i

The power consumed across R is P = i2R where

i=

E R+ r

\

P=

E 2R (R + r ) 2

or

P=

E 2R (R - r ) 2 + 4Rr

For P to be maximum.

and power consumed by bulb, P2¢ = i 2R2 = (0.38) 2 ´ 484 = 69.89 W

Note If in the glowing bulbs connected in series, one bulb gets fuses, or one bulb is switched off, the other bulbs also do not glow. But in the glowing bulbs connected in parallel, if one bulb get fused or one bulb is switched off the other bulbs remains, glowing.

R-r =0 Þ \

R=r Pmax =

E2 E2 = 4r 4R

Current Electricity Sample Problem 17 A house is served by a 220 V supply line in a circuit protected by fuse marked 9 A. The maximum number of 60 W lamps in parallel that can be turned on is (a) 44 (c) 22

(b) 33 (d) 20

Interpret (b) In parallel, power of n bulbs = n ´ 60 W Current,

9 = n ´ 60 /220

or

n = 220 ´ 9/60 = 33

821

Voltmeter It is an instrument used to find the potential difference across two points in a circuit. It is essential that the resistance RV of a voltmeter by very large compared to the resistance of any circuit element with which the voltmeter is connected otherwise, the meter itself becomes an important circuit element and alters the potential difference that is measured. RV >> R For an ideal voltmeter, RV = ¥

18.11 Different Electrical Instruments

Since, the resistance of coil of galvanometer is low, hence to convert galvanometer to voltmeter high resistance R is connected with the galvanometer.

Galvanometer It is used to detect very small current. It has negligible resistance. With suitable modifications, it can be used to measure current and potential difference.

ig

V RV

V -G ig

ig (G + R) = V Þ R =

Ammeter It is an instrument used to measure current. It is put in series with the branch in which current is to be measured. An ideal ammeter has zero resistance. A galvanometer with resistance G and current rating ig can be converted into an ammeter of rating i by connecting a suitable resistance S in parallel to it. G

i

ig i

ig

Sample Problem 19 A galvanometer having a coil resistance 100W gives a full scale deflection when a current of 1 mA is passes through it. What is the value of the resistance which can convert this galvanometer into a meter giving full scale deflection for a potential difference of 10 V? (a) 8.9 kW

(b) 9.9 kW

i

S (i - ig ) = igG igG S= i - ig

or

Sample Problem 18 A galvanometer having a coil resistance of 100 W gives a full scale deflection when a current of 1mA is passes through it. What is the value of the resistance which can convert this galvanometer into an ammeter giving full scale deflection for a current of 10 A? (a) 10 -2W (c) 2 ´ 10 -2W

æ 10 ö RV = ç -3 ÷ - 100 = 9,900 W = 9.9 kW è10 ø

Wheatstone Bridge For measuring accurately any resistance Wheatstone bridge is widely used. There are two known resistors, are variable resistor and one unknown resistors, one variable resistor and one unknown resistor connected in bridge form as shown. T1

(10 -3 A) (100 W) 0.1 Interpret (a) S = = = -3 9. 99 i - ig (10 - 10 ) A S = 1/99.99W » 10 -2W

B

P

Q P

(b) 10 -3 W (d) 3 ´ 10 -2W ig × G

(d) 10 kW

10 = (10 -3) (100 + RV )

A

\

Thus,

(c) 7.9 kW

Interpret (b) As, V = ig (G + RV )

i

S

Þ

G

A I2

S2 R

I

I1 C I2

G

D

E S1

S

I

822 JEE Main Physics By adjusting the variable resistor the current through the galvanometer is made zero. When the current through the galvanometer becomes zero, the ratio of the two known resistors is exactly equal to the ratio of adjusted value of variable resistance and the value of unknown resistance. In this way the value of unknown resistance can easily be measured by using a Wheatstone bridge. It is a four arms bridge circuit where arm AB, BC, CD and ADare consisting of resistances P, Q, S and R respectively. Among these resistances P and Q are known fixed resistances and these two arms are referred as ratio arms. An accurate and sensitive galvanometer is connected between the terminals B and D through switch S2. The voltage source of this Wheatstone bridge is connected to the terminals AC via switch S1. A variable resistor S is connected between points C and D. The potential at D can be varied by adjusting the value of variable resistor. Suppose currents I1 and I 2 are flowing through paths ABC and ADC respectively. If we vary the electrical resistance value of arm CD the value of current I will also be varied as the voltage across AC is fixed. If we continue to adjust the variable resistance one situation may come when voltage drop across the resistor S that is I 2 becomes exactly equal to voltage drop across resistor Q that is I1 Thus, the potential at point B becomes equal to the potential at point D hence potential difference between these two points in zero hence current through galvanometer is nil. Then, the deflection in the galvanometer is nil when the switch S is closed. Now, from Wheatstone bridge circuit V and Current, I1 = P+Q Current,

V I2 = R+ S

Now, potential of point B in respect of point C is nothing but the voltage drop across the resistor Q and this is V ×Q …(i) I1 × Q = P+Q Again potential of point D in respect of point C is nothing but the voltage drop across the resistor S and this is V ×S …(ii) I2 × S = R+ S Equating Eqs. (i) and (ii), we get V ×Q V ×S = P +Q R+ S Þ

Q S = P +Q R+ S

Þ

P +Q R+ S = Q S

Þ

P R + 1= + 1 Q S

Þ

P R = Q S

Þ

R=S ´

P × Q

In the above equation, the value of S and

P are known, so Q

value of R can be easily determined. The resistances P and Q of the Wheatstone bridge are made of definite ratio such as 1 : 1, 10 :1 or 100 : 1 known as ratio arms and S the rheostat arm is made continuously variable from 1 to 1000 W or from 1 to 10000 W.

Sample Problem 20 In the circuit in the figure, if no current flows through the galvanometer when the key k is closed, the bridge is balanced. The balancing condition for bridge is (a)

C1 R1 = C 2 R2

(b)

C1 R2 = C 2 R1

(c)

C12 R12 = C 22 R22

(d)

C12 R2 = C 22 R1

Interpret (b) In the steady state, no current is passing through capacitor. Let the change on each capacitor be q. Since, the current through galvanometer is zero \

I1 = I2

The potential difference between ends of galvanometer will be zero. \ \ Similarly,

VA - VB = VA - VD q I1R1 = C1 VB - VC = VD - VC q I2R2 = C2

…(i)

…(ii)

Dividing Eq. (i) by Eq. (ii), we get I1 R1 q / C1 C 2 = = I2 R2 q / C 2 C1 Þ

C1 R2 = C 2 R1

Sample Problem 21 Four resistances of 15 W, 12 W, 4 W and 10W respectively are connected in cyclic order to form. Wheatstone network. Is the network balanced? If not calculate the resistance to be connected in parallel with the resistance of 10 W to balance the network. (a) 4 W (c) 12 W

(b) 10 W (d) 15 W

Current Electricity Interpret (b) Let P = 15, W , Q = 12 W , S = 4W , R = 10 W The resistances of 15 W , 12W , 4W and 10 W are connected in cyclic order as shown in figure.

15

Ω

B

P

=

Q = 12 Ω

823

There are two gaps; in one of whose value is to be determined is connected. The galvanometer is connected with the help of jockey across BD and the cell is connected across AC. After making connections, the jockey is moved along the wire and the null point is found. Wheatstone bridge, wire used is of uniform material and cross-section. the resistance can be found with the help of the following relation S

R C

A

B R = 10 Ω S=4Ω

G A

C

D

D

100

I1

Now,

P 15 = = 1.25 Q 12

Metro Scale

R 10 = = 2.5 S 4 P R Since, is not equal to , the bridge is not balanced. Q S

I1

V

K1

R l1 l1 or R = S = S (100 - l1 ) 100 - l1

To balance the network, suppose a resistance X is connected in parallel to R = 10 W, so that the bridge is balanced. If R¢ is effective resistance between points A and D, then 1 1 1 …(i) = + R ¢ 10 X

Sample Problem 22 Fig. (a) shows a meter bridge (which

Since, the bridge is now balanced P R¢ = Q S

is nothing but a paractical Wheatstone bridge) consisting of two resistors X and Y together in parallel with one metre long constant wire of uniform cross-section.

or

P Q

R¢ = S ´ =4´

X

Y B

15 = 5W 12

Substituting for R¢ in Eq. (i), we have 1 1 1 = + 5 10 X 1 1 1 1 or = = X 5 10 10 or

where l1 is the length of the wire from one end where null point is obtained. The bridge is most sensitive, when null point is somewhere near the middle point of the wire.

X = 10 W

Meter Bridge (Special Case of Wheatstone Bridge) This is the simplest form of Wheastone bridge and is specially useful for comparing resistances more accurately. The construction of the meter bridge is shown in the figure. It consists of one metre resistance wire clamped between two metallic strips bent at right angles and it has two points for connection.

G A

D

C

(a)

With the help of a movable contact D, one can change the ratio of the resistances of the two segments of the wire, until a sensitive galvanometer G X Y connected across B and D 12 Ω B shows to deflection. The null point is found to be at a G distance of 33.7 cm from the end A. The resistance Y D C is shunted by a resistance Y ¢ A of 12.0 W [Fig. (b)] and the null point is found to shift (b) by a distance of 18.2 cm. Determine the resistances of X and Y.

824 JEE Main Physics Applications of Potentiometer

(a) Y = 13.5 W and X = 6.86 W (b) Y = 13.5 W and X = 5.86 W (c) Y = 11.5 W and X = 6.86 W (d) Y = 12.5 W and X = 6.86 W

(i) To find emf of an unknown battery E1

E1

Interpret (a) Since, the wire is of uniform cross-section, the resistances of the two segments of the wire AD and DC are in the ratio of the lengths of AD and DC. Using the null-point conditions of a Wheatsone bridge, we have

i

…(i)

l1

l1

B i

B

A i

C1

C2 i2

i2

When Y is shunted by a resistance of12.0 W net resistnace changes

G

G EK

Y ¢ = 12Y(Y + 12) Since, Y ¢ is less than Y , the ratio X / Y ¢ is greater than

X . Thus, the Y

null point must shift towards the end C i. e. , æ X ö æ 51. 9 ö ç ÷=ç ÷ è Y ¢ ø è 48.1ø

EU

We calibrate the device by replacing E2 by a source of known emf Ek and then by unknown emf Eu . Let the null points are obtained at lengths l1 and l2. Then, E K = i( rl1 ) and EU = i ( rl2 ) Here, r = resistance of wire AB per unit length

X(Y + 12) / 12 Y = (519 . / 481 .) Y + 12 æ 51.9 ö 66. 3 =ç ÷´ è 48.1ø 33. 7 12

i. e. ,

i

l A

æ X ö æ 33.7 ö ç ÷=ç ÷ è Y ø è 66.3 ø

or

i

i

which give Y = 13.5 W and X = 6. 86 W using Eq. (i)

æl ö E K l1 = or EU = ç 2 ÷ E K è l1 ø EU l2

\

So, by measuring the lengths l1 and l2, we can find the emf of an unknown battery.

(ii) To find the internal resistance of a cell

Potentiometer

Firstly, the emf E of the cell is balanced against a length AD = l1. For this, the switch S¢ is left opened and S is closed. A known resistance R is then connected to the cell as shown. The terminal voltage V is now balanced against a smaller length AD¢ = l2. Here, now switch is opened and S¢ is closed. Then,

Potentiometer is an ideal device to measure the potential difference between two points. It consists of a long resistance wire AB of uniform cross-section in which a steady direct current is set up by means of a battery.

Principle

E1

l1 i

i

l2 S

L

E

B

A i

C i

D′ D

A

G E

r

Potential gradient, Potential difference across AB k= Total length V AB iRAB = = = il L L R where, l = AB = resistance per unit length of L potentiometer wire. The emf of source balanced between points B and C R E2 = kl = i CB ´ l = iRCB l

B

E, r G S′ R

E l1 = V l2 Since, or Þ

E R+ r = V R R + r l1 = R l2 æl ö r = ç 1 - 1÷ R è l2 ø

{Q E = i(R + r ) and V = iR}

Current Electricity Sample Problem 23 The potentiometer wire AB is 100 cm long. When AC = 40 cm, no deflection occurs in the galvanometer. Find R. 10Ω

R

(b) r1 - r2

B

C

(a) 20 (c) 18

i=

(c) r2 - r1

(d) rr 12

E+E 2E = r1 + r2 + R r1 + r2 + E

At terminal potential drop across 1st cell is zero, hence 2E V1 = E - ir1 = E r1 = 0 (r1 + r2 + R)

(b) 15 (d) 25 (b)

different internal resistances r1 and r2 are connected in series with an external resistance cell. The potential drop across the first cell is found to be zero. The external resistance R is

Interpret (b)As both cells are in series, the circuit current,

A

Interpret

Sample Problem 25 Two cells of same eff E but of

(a) r1 + r2 G

825

10 AC = R CB

E + –

E + –

æ CB ö R = 10 ç ÷ è AC ø

\

æ100 – 40 ö = (10) ç ÷ è ø 40

R

æ 60 ö = 10 ç ÷ è 40 ø

Þ

= 15 W

or

E=

2Er1 (r1 + r2 + r3)

r1 + r2 + R = 2r1 Þ R = (r1 - r2)

Sample Problem 24 A potentiometer wire of length 100 cm has a resistance of 10 W. It is connected in series with a resistance and a cell of emf 2V 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 external resistance? (a) 620 W

Interpret

(b) 690 W

(c) 720 W

Check Point 2 1. What’s the effective resistance of the circuits shown in the adjacent figures?

(d) 790 W

R R

(d) From the theory of potentiometer, E1

R A

i

D

R

R

R

R B

C R

R

R

R D

C

R

2. An ammeter is always connected in series and voltmeter B

A C E G

VCB = E, if no current is drawn from the battery or Here,

and

æ E1 ö ç ÷ RCB = E è R + RAB ø E1 = 2V , RAB = 10 W æ 40 ö RCB = ç ÷ ´ 10 = 4 W è100 ø E = E = 10 ´ 10 -3 V

Substituting in above, we get R = 790 W

connected in parallel to any circuit element, why?

3. How will you compare the sensitivity of two potentiometers? 4. The emf of the cell used in the main circuit of the potentiometer should be more than the potential difference to be measured. Why?

5. The variation of potential difference V with length l in case of two potentiometers X and Y is as shown in figure. Which one of these two will you prefer for comparing emf’s of two cells and why? X Y V

I

WORKED OUT Examples Example 1 In the circuit shown in figure, a 12V power supply with unknown internal resistance r is connected to a battery with unknown emf E and internal resistance1W and to a resistance of 3 W carrying a current of 2A. The current through the rechargeable battery is 1A in the direction shown. The internal resistance r and the emf E is 12V

r

i

For aluminium wire Density of the wire, r1 = 2.7 ´ 103 kg m-3,

and resistivity of the wire, r Al = 2.63 ´ 10 -8 W m

For copper wire Density of the wire, r 2 = 8.9 ´ 103 kg m-3 2A

(a) 1 W, - 2V (c) 2 W, - 5V

Resistivity of the wire, r Cu = 1.72 ´ 10 -8 W m

(b) 1 W, 2V (d) 2 W, 5V

Then,

Applying Kirchhoff's junction law at C, we have 12V

1Ω

a

i

m2 A2l ´ 8.9 ´ 10 3 8.9 A2 = ´ = m1 A1l ´ 2.7 ´ 10 3 2.7 A1

\

c

Since, two wires are of equal resistances, we have

d

2.63 ´ 10 -8 ´ l 1.72 ´ 10 -8 ´ l = A1 A2

3Ω 2A

i = (1 + 2) A = 3 A From Kirchhoff’s loop law in eabde, we have + 12 - ir - 3 ´ 2 = 0 12 - 6 6 r= = = 2W i 3

Further applying second law in loop efcde, we have -E + 1 ´ 1 - 3 ´ 2 = 0 E = -5 V

Example 2

Two wires of equal lengths, one of aluminum and the other of copper have the same resistance. Which of the two wires is lighter? (r Al = 2.63 ´ 10 -8 Wm, r Cu = 1.72 ´ 10 -8 Wm, relative density of Al = 2.7, and Cu = 8.9) (a) Aluminium wire (b) Copper wire (c) Both have equal weights (d) None of the above

l 1.72 ´ 10 -8 ´ l = A2 A2

b

1A e

R2 = r Cu

Mass of the copper wire, m2 = A2l r 2 = A2l ´ 8.9 ´ 10 3

1Ω

f

\

l 2.63 ´ 10 -8 ´ l = A1 A1

Mass of the aluminium wire, m1 = A1 lr1 = A1l ´ 2. 7 ´ 10 3

3Ω

\

R1 = r Al

1Ω 1A

E

Let A 1 and A2 be areas of cross-section of the wires

made from aluminium and copper respectively. Let R1 and R2 be resistances of the same length l of the wires made of aluminium and copper respectively.

Then,

E

Solution

Solution

A2 1. 72 = A1 2. 63

or

From Eqs. (i) and (ii), we have m2 8.9 1.72 = ´ = 2.16 m1 2.7 2.63 m1 =

or

m2 2.16

It follows that aluminium wire is lighter. Since, for the same value of resistance and length, aluminium wire has lesser mass, aluminium wire are preferred for overhead power cable.

Example 3

A piece of copper and another of germanium are cooled from room temperature to 40 K. The resistance of (a) each of them decreases (b) each of them increases (c) copper increases and germanium decreases (d) copper decreases and germanium increases

Current Electricity Solution

Since, on decreasing temperature resistance of conductors also decreases while for semiconductors (Ge) increases because for semiconductors coefficient of thermal resistance (a) is negative.

Example 4

An ideal battery of emf 2 V and a series resistance R are connected in the primary circuit of a potentiometer of length 1 m and resistance 5 W . The value of R, to give a potential difference of 5 mV across 10 cm of potentiometer wire is (a) 180 W (b) 190 W (c) 195 W (d) 200 W Current

in the circuit is i =

Therefore, potential difference across l = 10 cm of the wire is V ´l 10 ´ 10 1 volt V¢ = = = L (R + 5) ´ 100 (R + 5) Given, V ¢ = 5 mV = 5 ´ 10 -3 V. Hence, we have 5 ´ 10 -3 =

1 Þ R = 195 W (R + 5)

Example 5 The range of voltmeter of resistance 300 W is 5 V. The resistance to be connected to convert it into an ammeter of range 5 A is (a) 1 W in series (b) 1 W in parallel (c) 0.1 W in series (d) 0.1 W in parallel

Solution

A voltmeter is a galvanometer having a high resistance connected in series with it. The current through the galvanometer is 5V 1 A ig = = 300 W 60 An ammeter is a galvanometer having a low resistance connected in parallel with it. The shunt resistance S is determined from ig S = i - ig G where G = 300 W (given). For i = 5 A, we have 1 / 60 S = 5 - 1 / 60 300 Þ

Example 6 The balancing length for a cell is 560 cm in a potentiometer experiment. When an external resistance of 10 W is connected in parallel to the cell, the balancing length changes by 60 cm. The internal resistance of the cell in ohm, is (a) 1.6 (c) 1.2

Solution

(b) 1.4 (d) 0.12 If the cell E is in open circuit and balancing length is l1,

then E = kl1. The potential difference V is balanced by length l2, then V = kl2. æl - l ö æE - V ö Internal resistance of cell r = ç ÷R = ç 1 2 ÷ R è V ø è l2 ø

E 2 = A (R + r) (R + 5) Therefore, the potential difference across the potentiometer wire of length L = 100 cm is 2 10 volt V = ir = ´5 = (r + 5) (R + 5)

Solution

827

S = 1W

r l1 - l2 = R l2 Here, R = 10 W, l1 = 560 cm and l2 = 560 - 60 = 500 cm æl ´ l ö Therefore, r = R ´ ç 1 2÷ è l2 ø æ 560 - 500 ö = 10 ´ ç ÷ = 1.2 W è ø 500

Example 7 A conductor of resistance 3 W is stretched uniformly till its length is doubled. The wire is now bent in the form of an equilateral triangle. The effective resistance between the ends of any side of the triangle in ohm is 9 2 (c) 2

8 3 (d) 1

(a)

(b)

Solution

The resistance of a conductor of length l, cross-sectional area A and made of a material of resistivity r is given by rl æ r ö 2 R= = ç ÷l A èV ø

4Ω

4Ω

A

B 4Ω

where V = Al is the volume of the conductor. Since, r is a constant and volume V cannot change if the conductor is stretched, if follows that R is proportional to l 2. Thus, if l is doubled, R becomes four times. Hence, the new resistance is 3 ´ 4 = 12 W. Hence, each side of the equilateral triangle has a resistance of 4 W. Therefore, the effective resistance between the ends of any side of the triangle (such as side AB) is equal to the resistance to a parallel combination of R1 = 4 W and R2 = 4 + 4 = 8 W which is given by R ´ R2 8 ´ 4 8 Re = 1 = = W R1 + R2 8 + 4 3

Start Practice for

JEE Main Round I

(Topically Divided Problems)

Electric Current and Drift Velocity, Electromotive Force 1. The current flowing through a wire depends on time as I = 3 t2 + 2 t + 5. The charge flowing the cross section of the wire in time from t = 0 to t = 2 s in (a) 21 C

(b) 10 C

(c) 22 C

19

(d) 1 C 19

2. In a region, 10 a-particles and 10 protons move to the left, while 1019 electrons move to the right per second. The current is (a) 3.2 A towards left (c) 6.4 A towards left

(b) 3.2 A towards right (d) 6.4 A towards right

3. Every atom makes one free electron in copper. If 1.1 A

current is flowing in the wire of copper having 1 mm diameter, then the drift velocity (approx.) will be (density of copper = 9 ´ 103 kg m -3 and atomic weight of copper = 63 ) -1

(a) 0.1 mms (c) 0.3 mms -1

-1

(b) 0.2 mms (d) 0.2 cms -1

4. Consider a current carrying wire (current I) in the

shape of a circle. Note that as the current progresses along the wire, the direction of j (current density) changes in an exact manner, while the current I remain unaffected. The agent that is essentially responsible for is [NCERT Exemplar] (a) source of emf (b) electric field produced by charges accumulated on the surface of wire (c) the charges just behind a given segment of wire which push them just the right way by repulsion (d) the charges ahead

5. The electron of hydrogen atom is considered to be revolving round a proton in circular orbit of radius h2/me2 with velocity e 2 / h where h = h/2p. The current i is 2

5

4 p me h2 2 2 2 4p m e (c) h3

(a)

2

2

4 p me h3 2 4 p me 5 (d) h3

(b)

6. Two batteries of emf e1 and e2 ( e2 > e1) and internal

resistances r1 and r2 respectively are connected in parallel as shown in figure. [NCERT Exemplar] e1

r1 (a) The equivalent emf e eq of the two cells is between e1 and e2 , A i. e., e1 < e eq < e2 r2 e2 (b) The equivalent emf e eq is smaller than e1 (c) The eeq is given by e eq = e1 + e2 always (d) e eq is independent of internal resistances r1 and r2

B

7. Two wires of the same material but of different diameters carry the same current i. If the ratio of their diameters is 2 : 1, then the corresponding ratio of their mean drift velocities will be (a) 4 : 1 (c) 1 : 2

(b) 1 : 1 (d) 1 : 4

8. A straight conductor of uniform cross-section carries a current i. If s is the specific charge of an electron, the momentum of all the free electrons per unit length of the conductor, due to their drift velocity only is (a) is

(b) i / s

(c) i /s

(d) ( i /s ) 2

9. In a neon gas discharge tube Ne + ions moving through a cross-section of the tube each second to the right is 2.9 ´ 1018 , while 1.2 ´ 1018 electrons move towards left in the same time. The electronic charge being 1.6 ´ 10-19 C, the net electric current is (a) 0.27 A to the right (c) 0.66 A to the left

(b) 0.66 A to the right (d) zero

10. A metallic resistor is connected across a battery. If the number of collisions of the free electrons with the lattice is some how decreased in the resistor (for example by cooling it), the current will (a) remains constant (c) decrease

(b) increase (d) become zero

Current Electricity 11. A capacitor of 10 mF has a potential difference of 40 V across it. If it is discharged in 0.2 s, the average current during discharge is (a) 2 mA

(b) 4 mA

(c) 1 mA

(d) 0.5 mA

area of cross-section is 10-6 m 2 . If the number of free electrons per m 3 is 8.4 ´ 1028 , then find the drift velocity, (e = 1.6 ´ 10–19 C) (a) 2 × 10 -5 ms -1 (b) 1.56 × 10 -5 ms -1 (c) 1 × 10 -5 ms -1 (d) 0.64 × 10 -5 ms -1

(a) 0.049/°C (c) 0.0039/°C

[NCERT Exemplar]

(a) high resistivity (b) low temperature coefficient of resistance (c) low resistivity (d) both (a) and (b)

14. All the edges of a block with parallel faces are unequal. Its tangent edge is twice its shortest edge. The ratio of the maximum to minimum resistance between parallel faces is (b) 4 (d) none of these

15. The resistance of a 10 m long wire is 10W. Its length is increased by 25% by stretching the wire uniformly. The resistance of wire will change to (approximately) (b) 14.5 W (d) 16.6 W

16. Two plates R and S are in the form of a square and have the same thickness. A side of S is twice the side of R. Compare their resistances. The direction of current is shown by an arrow head in figure. S

(a) r1 r2 r + r2 (c) 1 2

(a) The resistance of R is twice that of S (b) Both have the same resistance (c) The resistance of S is four times that of R (d) The resistance of R is half that of S

17. Masses of the three wires of same material are in the ratio of 1 : 2 : 3 and their lengths in the ratio of 3 : 2 : 1. Electrical resistance of these wires will be in the ratio of (c) 9 : 4 : 1

(b) (r1 + r2 ) (d) None of these

20. In cosmic rays 0.15 protons cm -2 sec-1 are entering the earth’s atmosphere. If the radius of the earth is 6400 km, the current received by the earth in the form of cosmic rays is nearly. (a) 0.12 A

(b) 1.2 A

(c) 12 A

(d) 120 A

21. The temperature coefficient of resistance for a wire is 0.00125°C-1. At 300 K its resistance is 1W. The temperature at which the resistance becomes 1.5W is? (a) 450 K

(b) 727 K

(c) 454 K

(d) 900 K

22. The current i and voltage V

I T2

graphs for a given metallic wire at two different temperatures T1 and T2 are shown in the figure. It is concluded that (a) T1 > T2 (c) T1 = T2

T1

(b) T1 < T2 (d) T1 = 2T1

V

23. There are two concentric spheres of radius a and b respectively. If the space between them is filled with medium of resistivity r, then the resistance of the inter gap between the two spheres will be r 4 p( b + a) r æ1 1ö (c) ç - ÷ 4 p è a2 b2 ø (a)

(a) 1 : 1 : 1

(b) 0.0049/°C (d) 0.039/°C

r2 are connected in series. The equivalent resistivity of the combination is

make standard resistance became they have

R

resistance of 2.7 W at 100 °C. Determine the temperature coefficient of resistivity of silver.

19. Two wires of same dimensions but resistivities r1 and

13. The alloys constantan and manganin are used to

(a) 12.5 W (c) 15.6 W

Specific Resistance, Current Density, Electrical Conductivity and Effect of Temperature 18. A silver wire has a resistance of 2.1 W at 27.5 °C and a

12. There is a current of 0.21 A in a copper wire whose

(a) 8 (c) 2

829

r 4p r (d) 4p

(b)

æ 1 1ö ç - ÷ è b aø æ 1 1ö ç - ÷ è a bø

24. A copper wire of length 1 m and radius 1 mm is joined in series with an iron wire of length 2 m and radius 3 mm and a current is passed through the wires. The ratio of the current density in the wires. The ratio of the current density in the copper and iron wires is

(b) 1 : 2 : 3

(a) 2 : 3

(b) 6 : 1

(d) 27 : 6 : 1

(c) 9 : 1

(d) 18 : 1

830 JEE Main Physics 25. A conductor with rectangular cross-section has dimensions (a ´ 2a ´ 4 a) as shown in figure. Resistance across AB is R1, across CD is R2 and across EF is R3. Then

31. The variation between V and i has been shown by V-i graph for heater filament is represented by I

I

C F 2a

A

(a)

(b)

B

4a

V

V E

(a) R1 = R2 = R3 (c) R2 > R3 > R1

D

I

(b) R1 > R2 > R3 (d) R1 > R3 > R2

26. A resistance of 2W is to be made from a copper wire

I

(c)

(d)

-8

(specific resistance = 1.7 ´ 10 W m) using a wire of length 50 cm. The radius of the wire is (a) 0.0116 mm (c) 0.116 mm

(b) 0.367 mm (d) 0.267 mm

27. A metal rod of length 10 cm and a rectangular

1 cm is connected to a battery 2 across opposite faces. The resistance will be

cross-section of 1 cm ´

[NCERT Exemplar]

(a) maximum when the battery is connected across 1 1 cm ´ cm 2 (b) maximum when the battery is connected across 10 cm ´ 1cm faces (c) maximum when the battery is connected across 1 10 cm ´ cm faces 2 (d) same irrespective of the three faces

V

Grouping of Resistors 32. An electric cable of copper has just one wire of radius 9 mm. Its resistance is 5 W. This single copper wire of cable is replaced by 6 different well insulated copper wires each of radius 3 mm. The total resistance of the cable will now be equal to (a) 7.5W

Rt = R0 (1 + at + bt )

6Ω

6Ω

P

(a) 3 V

2

(d) 270W

6Ω

6Ω

have the same length. Conductor A is a solid wire of diameter 1 mm. Conductor B is a hollow tube of outer diameter 2 mm and inner diameter 1mm. What is the ratio of resistances R A to RB ? [NCERT Exemplar]

29. Resistance of a resistor at temperature t° C is

(c) 90W

shown in figure with the current 0.5A as shown in figure, the potential difference, VP - V Q is

28. Two conductors are made of the same material and

(b) 3 : 1 (d) 3 : 2

(b) 45W

33. Resistances of 6W each are connected in the manner

0.5 A

(a) 1 : 3 (c) 2 : 3

V

Q 6Ω

6Ω

(b) 5 V

(c) 4 V

(d) 3.9 V

34. The equivalent resistance of n resistors each of same resistance when connected in series is R. If the same resistances are connected in parallel, the equivalent resistance will be (a) R /n2

(c) n2 R

(b) R /n

(d) nR

35. The resistance of the following circuit figure between

Here R0 is the resistance at 0°C. The temperature coefficient of resistance at temperature t° C is

A and B is E

2

(1 + at + bt ) a + 2bt a +2 bt (c) (1 + at + bt2 )

(a)

(b) ( a + 2 bt ) (d)

2Ω C

( a + 2 bt ) 21 ( + at + bt )

30. A given resistor has the following colour scheme of the various strips on it, brown, black, green and silver. Its value in ohm is (a) 1.0 ´ 10 4 ± 10%

(b) 1.0 ´ 105 ± 10%

(c) 1.0 ´ 10 6 ± 10%

(d) 1.0 ´ 107 ± 10%

2Ω

2Ω

D

2Ω 2Ω

2Ω

2Ω

F A

B 2Ω

(a) ( 3/2) W

(b) 2 W

(c) 4 W

(d) 8 W

Current Electricity 36. What is the equivalent resistance across the points A

831

40. The resistance across R and Q in the figure.

and B in the circuit given below ?

A

C 10Ω

r

B

r r

r r

10Ω

16Ω P

12Ω 2.5Ω

10Ω

(b) r /2 (d) 6r

form a figure. The resistance between two corners A

(b) 12 W (d) 32 W

and B is

37. Three resistances each of 4W are connected in the form

4Ω

A

of an equilateral triangle. The effective resistance between any two corners is

B

4Ω 4Ω

(a) (3/8) W (b) (8/3) W (c) 8 W (d) 12 W

O 4Ω 4Ω C

D

38. In the circuit figure, the voltmeter reads 30 V. What is the resistance of the voltmeter? 30 V V

4Ω

(a) 4 W (c) 12 W

(b) 4/3 W (d) 2 W

42. If each of the resistances in the network in figure. R, the equivalent resistance between terminals A and B is

400 Ω

300 Ω

Q

RΩ

60 V RΩ

RΩ

O

(a) 1200 W (b) 700 W (c) 400 W (d) 300 W

A

RΩ P

39. The effective resistance between points A and B is R A

2R

C

B

B S

RΩ

(a) 5R (c) 4R

(b) 2R (d) R

43. The resistance between the points A and C in the figure below is RΩ B

A

2R

RΩ

R

RΩ RΩ

2R (c) 3

E RΩ

D

(a) R

Q

41. Six equal resistances each of 4W are connected to

E

(a) 8 W (c) 16 W

C

r

(a) r /3 (c) 2 r

A

D

B

R 3 3R (d) 5

RΩ

RΩ C

D RΩ

(b)

(a) R W (c)

2 RW 3

4 RW 3 8R (d) 3

(b)

832 JEE Main Physics 44. In the circuit shown in figure, the point F is grounded. Which statement?

of

the

following

1Ω

A

wrong

1Ω 1Ω

5Ω

B

is

(a) infinite (c) 2 W

2Ω

3Ω

A 10 V

D 3V

1Ω

(b) zero (d) (1 + 5 )/2 W

49. The effective resistance between points A and B in figure. 4Ω

E

F

1Ω

B

C

4Ω

1Ω

6Ω

(a) D is at 5 V (b) E is at zero potential (c) The current in the circuit will be 0.5 A (d) The potential at E is same whether or not F is rounded

45. In the circuit shown, the cell is ideal, with emf = 10 V. Each resistance is of 2W. The potential difference across the capacitor is R

C = 3 µF

G

B

H R

R R

R A

3Ω

(a) 12 V (c) 8 V

(a) 10 W (c) 9.85 W

their cross-sections are in the ratio 3 : 1. They are joined in series. The resistance of the thicker wire is 10W. The total resistance of the combination is (a) 5/2 W (c) 40 W

(b) 40/3 W (d) 100 W

47. What is the equivalent resistance between points A and B in the circuit of figure, if R = 3 W? A

and a current I2 through a resistance R2 . The internal resistance of a cell is (a) R2 - R1 ( i + i2 ) (b) 1 R1R2 i1 –i2 i1R2 - i2 R1 i1 - i2 i2 R2 - i1R1 i1 - i2

Grouping of Cells 51. To get a maximum current through a resistance of 2.5 W, one can use m rows of cells each row having n cells. The internal resistance of each cell is 0.5W. What are the values of m and n, if the total number of cells are 20? (a) m = 2, n = 10 (b) m = 4, n = 5 (c) m = 5, n = 4 (d) n = 2, m = 10

52. Two identical cells connected in series send 1.0 A

B

R

R R D

C R

R

current through a 5 W resistor. When they are connected in parallel, they send 0.8 A current through the same resistor. What is the internal resistance of the cell? (a) 0.5 W (c) 1.5 W

(a) 8 W (c)12 W

(b) 12 W (d) 10.85 W

50. A cell supplies a current i1 through a resistance R1

(d)

46. Two wires of same metal have the same length but

B 24Ω

(c)

(b) 10 V (d) zero

5Ω

A

D 10 V

12Ω

(b) 9 W (d)15 W

48. The equivalent resistance between points A and B of an infinite network of resistances each of 1W connected as shown in figure, is

(b) 1.0 W (d) 2.5 W

53. Figure shows a circuit with known resistances R1 and

R2 . Neglect the internal resistance of the sources of current and resistance of the connecting wire. The magnitude of electromotive force E1 such that the current through the resistance R is zero will be

Current Electricity

833

58. The emf of the battery shown in figure, is

R

2Ω

R1

2Ω

1Ω

R1

E 2Ω

4Ω

E

E1

1Ω

1A

(a) ER1 /R2 (c) E ( R1 + R2 )/R2

(b) ER2 /R1 (d) ER1 /( R1 + R2 )

54. Under what condition will the strength of current in a wire of resistance R be the same for connection is n series or in parallel of n identical cells each of the internal resistance r, when (a) R = nr (c) R = r

(b) R = r /n (d) R ® ¥, r ® 0

(a) 12 V

(b) 13 V

arms of a quadrilateral ABCD. Across AC is the

battery circuit, the emf of the battery being 4V and internal resistance negligible. The potential difference across BD is B 40Ω

resistance r, are connected in series a cell A is joined with reverse polarity. The potential difference across each cell, except A is

network shown in figure is given by

C

90Ω

(b)

56. The equivalent resistance between points a and b of a

60Ω

A

( n - 2) E n 2E (d) n

2nE n -2 ( n - 1) E (c) n

(d) 18 V

59. Four resistances 40 W, 60 W, 90 W and 110 W make the

55. n identical cells, each of emf E and internal

(a)

(c) 16 V

110Ω D

4V

(a) 1 V (c) –0.2 V

(b) –1 V (d) 0.2 V

60. A storage battery of emf 8.0 V and internal resistance R R O R

0 . 5 W is being charged by a 120 V DC supply using a series resistor of 155 . W. What is the terminal voltage of the battery during charging?

b a

R

(a) 11.5 V 3 R 4 5 (c) R 6

57. For what value of R in the circuit as shown in figure, current passing through 4W resistance will be zero. C

B

D

R

4Ω

2Ω

(c) 12.5 V

(d) 10.5

61. A, B, C and D are four resistances of 2 W, 2 W, 2 W and

4 R 3 5 (d) R 4 (b)

(a)

(b) 13.5

3 W respectively. They are used to form a Wheatstone bridge. The resistance D is short circuited with a resistance R in order to get the bridge balanced. The value of R will be (a) 4 W

(b) 6 W

(c) 8 W

(d) 3 W

62. A battery of internal resistance 4 W is connected to the network of resistances as shown. In order to given the maximum power to the network, the value of R (in W) should be R

R 6V A 9V

(a) 1 W (b) 2 W (c) 3 W (d) 4 W

F

R

6R

R

E

4Ω

3V

4R

R

(a) 4/9

(b) 8/9

(c) 2

(d) 18

834 JEE Main Physics 67. Two cells of emf ’s approximately 5 V and 10 V are to

Effect of Current 63. In the given figure when galvanometer shows no deflection current flowing through 5 W resistance will be B 8Ω

2Ω

2.1A

G

A

C

20Ω

5Ω D

(a) 0.5 A (c) 1.5 A

64. In a Wheatstone bridge, P = 90 W, Q =110 W, R = 40 W and S = 60 W and a cell of emf 4 V. Then the potential difference between the diagonal along which a galvanometer is connected, is (b) + 0.2 V (d) + 1 V

Potentiometer 65. The circuit shown here is used to compare the emf of two cells E1 and E2 ( E1 > E2 ). The null point is at C when the galvanometer is connected to E1. When the galvanometer is connected to E2 , the null point will be

C

A

B

(a) He should measure l1 more accurately (b) He should change S to1000 W and repeat the experiment (c) He should change S to 3 W and repeat the experiment (d) He should give up hope of a more accurate measurement with a meter bridge

Thermal Effect of Current 69. A resistor R1 dissipates power P, when connected to a

certain generators. If the resistor R2 is put in series with R1, the power dissipated by R1 (b) decreases (d) None of these

70. An electric kettle boils some water in 16 min. Due to some defect, it becomes necessary to remove 10% turns of heating coil of the kettle. Now, how much time will it take to boil the same of water?

G

E2

(a) to the left of C (c) at C itself

bridge. Student chooses the standard resistance S to be 100 W. He finds the null point at l1 = 2.9 cm. He is told to attempt to improve the accuracy. Which of the [NCERT Exemplar] following is a useful way?

(a) increases (c) remains constant

V

E1

(a) The battery that runs the potentiometer should have voltage of 8V (b) The battery of potentiometer can have a voltage of 15V and R adjusted so that the potential drop across the wire slightly exceeds 10V (c) The first portion of 50 cm of wire itself should have a potential drop of 10v (d) Potentiometer is usually used for comparing resistances and not voltages

68. A resistance R is to be measured using a meter

(b) 0.6 A (d) 2.0 A

(a) – 0.2 V (c) –1 V

be accurately compared using a potentiometer of length 400cm. [NCERT Exemplar]

(b) to the right of C (d) none where on AB

66. AB is a potentiometer wire (figure), if the value of R is increased, in which direction will the balance point J shift? E

(a) 17.7 min (c) 20.9 min

(b) 14.4 min (d) 13.7 min

71. If two identical heaters each rated as (1000 W-220 V) are connected in parallel to 220 V, then the total power consumed is (a) 200 W (c) 250 W

(b) 2500 W (d) 2000 W

72. The resistance of hot tungsten filament is about 10 R J

A G

(a) Towards B (b) Towards A (c) No current flowing in circuit (d) None of the above

times the cold resistance. What will be the resistance of 100 W and 200 V lamp, when not in use? B

(a) 40 W (c) 400 W

(b) 20 W (d) 200 W

73. A 4 mF conductor is charged to 400 V and then its plates are joined through a resistance of 1kW. The heat produced in the resistance is (a) 0.18 J (b) 0.21 J (c) 0.25 J (d) 0.32 J

Current Electricity

Round II Only One Correct Option

835

(Mixed Bag) 5. In the arrangement shown in figure, the current

1. Which of the following circuits is correct for

through 5 W resistor is

verification of Ohm’s law?

2Ω

A

2Ω

5Ω

12V

12V

A

(a)

(b)

V

V

(a) 2A

(b) zero

(c)

12 A 7

(d) 1A

6. A heating element using nichrome connected to a A (c)

(d) None of these

V

2. A capacitor of capacitance 2 mF is connected as shown in figure. The internal resistance of the cell is 0.5 W. The amount of charge on the capacitor plates is

(a) 967°C

(b) 867°C

(c) 853°C

(d) 937°C

7. The effective resistance between points P and Q of

2Ω 2µF

230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element, if the room temperature is 27.0°C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 ´ 10- 4 °C - 1.

the electrical circuit shown in the figure.

10Ω

2R

2R 2R

2.5 V

Q

P r

(b) 2 mC (d) 6 mC

(a) zero (c) 4 mC

2R 2R

3. A copper wire of length L and radius r is nickel plated till its final radius becomes R but length remains L. If the resistivity of nickel and copper be r n and r c respectively, the conductance of the nickel wire is 2

2

r

2

(a)

pr L × rc

(b)

p( R - r ) Lr n

(c)

p é r2 ( R2 - r 2 ) ù + ê ú L ë rc rn û

(d)

Lr c Lr n + 2 pr p ( R2 - r 2 )

(a)

2Rr R+ r

(b)

2R

8R( R + r ) (c) 2R + 4 r (3R + r )

(d)

5R + 2R 2

8. The total current supplied to the circuit by the battery as shown in figure is

2Ω 6Ω 3Ω

4. The current through the circuit shown in figure is 1A. 1.5Ω

If each of 4 W the resistors is replaced by 2 W resistor, the current in circuit will become nearly 15Ω

(a) 1A

4Ω 2Ω

15Ω

(b) 6A

(c) 4A

(d) 2A

9. The equivalent resistance of the figure i. e., infinite network of resistors between the terminals A and B is

4Ω 15Ω

R1

10V

R1

R1

R1

R2

A R3

(a) 1.11 A (c) 1.34 A

(b) 1.25 A (d) 1.67 A

R3=∞

R3

R3

B R2

R2

R2

R2

836 JEE Main Physics (a) zero (b) infinite R1 + R2 + R3 (c) 3 1 (d) [( R1 + R2 ) + 2

Q

P S

( R1 + R2 ) ( R1 + R2 + 4 R3 ]

G

R

10. Two bars of radius r and 2r are kept in contact as shown. An electric current i is passed through the bars. Which one of the following is correct? l2 I

C A

B

(a) Heat produced in bar BC is 4 times the heat produced in bar AB (b) Electric field in both halves is equal (c) Current density across AB is double that of across BC (d) Potential difference across AB is 4 times that of across BC

11. A wire of length l is drawn such that its diameter is reduced to half of its original diameter. If initial resistance of the wire were 10 W, its resistance would be (a) 160W (c) 140W

(a) IR = RG

(b) IP = IG

(a) P and Q (b) Q and R (c) P and R (d) any two points

(b) 0.2v/m (d) 0.3v/m

13. The potential difference across 8 W resistance is 48V as shown in figure. The value of potential difference across points A and B will be 3Ω 60Ω

30Ω

16. In the given circuit shown in figure, it is observed that the current i is independent of the value of resistance R6 . Then, the resistance values must satisfy R5

I

R1

B

(a) 62 V (c) 128 V

R3

8Ω

R4

R2

(a) R1 R2 R5 = R3 R4 R6 1 1 1 1 (b) + = + R5 R6 R1 + R2 R3 + R4 (c) R1 R 4 = R2 R3 (d) R1 R3 = R2 R4 = R5 R6

17. Resistors of resistance 20W and 30W are joined in

series with a battery of emf 3V. It is desired to measure current and voltage across the 20W resistor with the help of an ammeter and voltmeter. Identify the correct arrangement of ammeter (A) and voltmeter (V) out of four possible arrangement shown in figure. Given below 3V

A 24Ω

R

R6

series with a battery of emf 3V (negligible internal resistance) and a resistance of 10W find the potential gradient along the wire

20Ω

P

Q

(b) 120W (d) 100W

(a) 3v/m (c) 01 . v/m

(d) IQ = IR

connected between points P, Q and R as shown in the figure. Then the net resistance will be maximum between

12. A 10 m long wire of resistance 20W is connected in

A

(c) IQ = IG

15. Six equal resistances are

l2 r

2r

V

V

V

48V

(a)

1Ω

20Ω

30Ω

(b)

3V

galvanometer G is same with switch open or closed. Then

A

30Ω

3V

3V V

(b) 80 V (d) 160 V

14. In the circuit shown, as P ¹ R and the reading of the

20Ω

(c)

A

A 20Ω 3V

30Ω

(d)

20Ω 3V

V

30Ω

Current Electricity 18. A 100 W bulb B1 and two 60 W bulbs B2 and B3 are connected to a 250 V source as shown in figure. Now W1, W2 and W3 are the output powers of the bulbs B1, B2 and B3 respectively, then B1

837

22. Four identical resistors of 4 W each are joined in circuit as shown in figure. The cell B has emf 2V and its internal resistance is negligible. The ammeter reading is 4Ω

B2

B A 4Ω

B3

4Ω

250V

(a) W1 > W2 = W3 (c) W1 < W2 = W3

4Ω

(b) W1 > W2 > W3 (d) W1 < W2 < W3

19. In a Wheatstone’s bridge, three resistances P, Q and R connected in the three arms and the fourth arm is formed by two resistances S1 and S2 connected in parallel. The condition for the bridge to be balanced will be (a)

P 2R = Q S1 + S2

(b)

P R ( S1 + S2 ) (c) = Q 2S1S2

3 A 8 1 (c) A 2

(b) 2A

(a)

(d)

1 A 8

23. In the circuit shown in figure, the potential difference between the points A and B will be 5Ω

5Ω

B

P R ( S1 + S2 ) = Q S1S2

2V

5Ω

P R (d) = Q S1 + S2

5Ω

A

20. In the circuit shown in figure, the heat produced in the 5 W resistor due to the current flowing through it is 100 Js -1. The heat generated in the 4 W resistor is 4Ω

6Ω

5Ω

2 (a) V 3 4 (c) V 3

5Ω

(b)

8 V 9

(d) 2V

24. A house wife uses a 100 W bulb 8 hour a day, and an i

i

electric heater of 300 W for 4 hour a day. The total cost for the month of June at the rate of 0.05 rupee per unit will be

5Ω

(a) 10 Js -1

(b) 20 Js -1

-1

(d) 40 Js -1

(c) 30 Js

(a) ` 20 (c) ` 30

21. Arrange the order of power dissipated in the given circuits, if the same current is passing through all the circuits. The resistance of each resistor is r. I P1

P2 P3

B

(b) ` 25 (d) ` 30 paise 50

25. The temperature of the cold junction of a thermocouple is 0 °C and the temperature of hot junction is T °C. The emf is E = 16T - 004 . T2 mV. The inversion temperature Ti is (a) 200 °C

(b) 400 °C

(c) 100 °C

(d) 300 °C

26. A lamp having tungsten filament consumes 50 W. I B I B

I P4

(a) P1 > P2 > P3 > P4 (c) P4 > P3 > P2 > P1

Assume the temperature coefficient of resistance for tungsten is 4.5 ´ 10 -3 °C -1 and temperature of the surrounding is 20 °C. When the lamp burns, the temperature of its filament becomes 2500 °C, then the power consumed at the moment switch is on, is (a) 608 W

B

(b) P2 > P3 > P4 > P1 (d) P1 = P2 = P3 = P4

(b) 710 W

(c) 215 W

(d) 580 W

27. What must be the efficiency of an electric kettle marked 500 W, 230 V, if it was found to bring 1 kg of water at 15 °C to boiling point in 15 min? (Given specific heat capacity of water = 4200 J/kg°C) (a) 79%

(b) 81%

(c) 72%

(d) 69%

838 JEE Main Physics 28. A fuse wire of circuit cross-section and having

33. Two identical batteries each of emf 2 V and internal

diameter of 0.4 mm, allows 3 A of current to pass through it. But if another fuse wire of same material and circular cross-section and having diameter of 0.6 mm is taken, then the amount of current passed through the fuse is

resistance 1 W are available to produce heat in an external resistance by passing current through it. The maximum Joulean power that can be developed across the resistance using these batteries it.

3 (b) 3 ´ A 2

(a) 3 A

32

æ 3ö (c) 3 ´ ç ÷ è 2ø

æ 3ö (d) 3 ´ ç ÷ A è 2ø

A

29. The charge supplied by source varies with time t as Q = at - bt2 . The total heat produced in resistor 2R is R

Source

R

a3R 6b a3R (c) 3b

(b)

(a)

(a) 2W (b) 3.2 W (c) 1.28 W (d) 8/9 W

34. The resistors P, Q and R in the circuit have equal resistance. The battery, of negligible internal resistance, supplies a total power of 12 W. What is the power dissipated by heating in resistor R?

2R

P

a3R 27b

Q

R

(d) None of these

30. One junction of a thermocouple is at a particular temperature Tr and another is at T. Its thermo emf is expressed as

35. A torch bulb rated at 4.5 W, 1.5 V is connected as shown in figure. The emf of the cell needed to make the bulb glow at full intensity if

1 ü ì E = K (T - Tr )íT0 - (T + Tr )ý 2 þ î

At a temperature T =

4.5 W 1.5 V

T0 , the value of thermoelectric 2

1Ω

power will be

E

1 KT0 2 1 (c) KT20 2

(a)

(b) KT0 (d)

1 K (T0 - Tr )2 2

(a) 4.5 V (c) 2.67 V

(b) 1.5 V (d) 13.5 V

31. The ratio of the amounts of heat developed in the four

36. A cell of internal resistance r is connected to a load of

arms of a balanced Wheatstone bridge, when the arms have resistance P = 100 W; Q = 10 W; R = 300 W and S = 30 W respectively is

resistance R. Energy is dissipated in the load, but some thermal energy is also wasted in the cell. The efficiency of such an arrangement is found from the expression

(a) 3 : 30 : 1 : 10 (c) 30 : 10 : 1 : 3

(b) 30 : 3 : 10 : 1 (d) 30 : 1 : 3 : 10

r

32. Four resistances carrying a current shown in the circuit diagram are immersed in a box containing ice at 0 °C. How much ice must be put in the box every 10 min to keep the average quantity of in the box constant? 10 Ω

5Ω

10A

P 5Ω

10 Ω

(Latent heat of ice is 80 cal g -1) (a) 5 kg (c) 3 kg

h= 10A

Q

(b) 1.19 kg (d) 2.29 kg

R

energy dissipated in the load energy dissipated in the complete circuit

Which of the following gives the efficiency in this case ? r R (b) (a) R r r R (d) (c) R+ r R+ r

Current Electricity 37. A dry cell of emf 1.5 V and internal resistance 0.10 W is connected across a resistor in series with a very low resistance ammeter. When the circuit is switched on, the ammeter reading settles to a steady rate of 2A. Find (i) chemical energy consumption of the cell (ii) energy dissipation inside the cell (iii) energy dissipation inside the resistor (iv) power output of source is (a) (i) 3 W (ii) 0.4 W (b) (i) 0.4 W (ii) 3 W (c) (i) 2.6 W (ii) 0.4 W (d) None of the above

(iii) 2.6 W (iii) 2.6 W (iii) 9 W

(iv) 2.6 W (iv) 2.6 W (iv) 1 W

38. The number density of free electrons in a copper

conductor estimated at 8.5 ´ 1028 m - 3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 20 . ´ 10- 6 m2 and it is carrying a current of 3.0 A. (a) 6 h 23 min (c) 7 h 43 min

density of 10 - 9 C/m 2 . The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe. Radius of earth = 6.37 ´ 106 m) (b) 263 s (d) 205 s

40. A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire due to the current, the temperature of the wire is raised by DT is a time t. A number N of similar cells is now connected in series with a wire of the same material and cross-section but of length 2L. The temperature of the wire is raised by the same amount DT in the same time t. The value of N is (a) 4 (c) 8

(b) W1 > W2 > W3 (d) W1 < W2 < W3

42. The thermo emf of a thermo-couple is found to depend on temperature T (in degree celsius) as T2 , where T is the temperature of the hot E = 4T 200 junction. The neutral and inversion temperatures of the thermocouple are (in degree celsius) (a) 100, 200 (c) 300, 600

(b) 200, 400 (d) 400, 800

43. The wiring of a house has resistance 6W. A 100 W bulb is glowing as shown in figure. If a geyser of 1000 W is switched on, the change in potential drop across the bulb is nearly 6Ω

Bulb

Geyser

(b) 7 h 33 min (d) 6 h 53 min

39. The earth’s surface has a negative surface charge

(a) 273 s (c) 283 s

(a) W1 > W2 = W3 (c) W1 < W2 = W3

(b) 6 (d) 9

41. A 100 W bulb B1 and two 60 W

bulb 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

839

44. In a thermocouple, one junction which is at 0 °C and the other at t°C, the emf is given by E = at2 - bt 3. The neutral temperature is given by (a) a b (c) 3a 2b

(b) 2a 3b (d) b 2a

45. A 10 mF capacitor is charged to 500 V and then its plates are joined together through a resistance of 10 W. The heat produced in the resistance is (a) 500 J (c) 125 J

(b) 250 J (d) 1.25 J

46. In India electricity is supplied for domestic uses at 220 V. It is supplied at 110 V in USA. If the resistance of a 60 W bulb used in India is R, the resistance of a 60 W bulb used in USA will be (a) R (c) R/4

(b) 2 R (d) R/2

47. For ensuring dissipation of same energy in all three resistors (R1, R2 , R3) connected as shown in figure, their values must be related as R1

(a) R1 = R2 - R3 (b) R2 = R3 and R1 = 4 R2 (c) R2 = R3 and R1 = R2 4

V

R2

R3

(d) R1 = R2 + R3 B1

B2 B3 250 V

48. An electric lamp is marked 60 W, 230 V. The cost of kilowatt hour of power is s ` 1.25. The cost of using this lamp 8 hour a day for 30 days is (a) ` 10

(b) ` 16

(c) ` 18

(d) ` 20

840 JEE Main Physics 49. Consider four circuits shown in figure. In which circuit power dissipated is greatest. (Neglect the internal resistance of the power supply).

(a) 420 s (c) 105 s

(b) 210 s (d) 50 s

52. In copper voltameter, mass deposited in 30 s is m gram. If the time-current graph is as shown in figure, ECE of copper is

E

R

R

(Current in mA)

(a)

R

(b)

200 mA A

B

10

20

100 mA

E

0

(a) m (c) 0.6 m

R

C 30

(Time in sec)

(b) m/2 (d) 0.1 m

53. The temperature of hot junction of a thermocouple changes from 80 °C to 100 °C, the percentage change in thermo electric power is R

(c)

(a) 25

R

(b) 20

(c) 10

(d) 8

54. A capacitor of capacitance 3 mF is first charged by

E

connecting across 10 V battery, then it is allowed to get discharged through 2W and 4W resistor by closing the key K as shown in figure. The total energy dissipated in 2W resistor is equal to

R

C = 3 µF

R

(d)

E

R 2Ω

R

K 4Ω

50. When an electric heater is switched on, the current flowing through it is plotted against time (t). Taking into account the variation of resistance with temperature, which of the following best represents the variations I

(a) 0.15 mJ (c) 0.05 mJ

(b) 0.5 mJ (d) 1.0 mJ

55. The resistor in which maximum heat will be produced is

I

2Ω

4Ω

3Ω

(a)

5Ω

(b)

6Ω

t

t

I

I

(c)

(b) 3 W

(c) 4 W

(d) 6 W

56. Two bulbs consume same power when operated at 200 V and 300 V respectively. When these bulbs are connected in series across a DC source of 400 V, then the ratio of power consumed across them is

(d)

t

(a) 2 W

t

51. An electric heater of 1.08 kW is immersed in water. After the water has reached a temperature of 100 °C, how much time will be required to produce 100 g of steam? (Latent heat of steam = 540 calg -1)

(a) 2/3 (c) 4/9

(b) 3/2 (d) 9/4

57. One junction of a certain thermocouple is at a fixed temperature Tr and the other junction is at temperature T. The thermoelectric force for this is expressed by

Current Electricity 1 E = K ( T - Tr ) éê T0 + ( T2 + Tr2 ) ùú. 2 ë û

62. Figure shows a network of resistances and batteries. Select the correct statements out of the following.

At temperature T = T0 /2, the thermoelectric power is 1 (a) KT0 2

3 (b) KT0 2

1 (c) KT20 2

4Ω

6V

r

V

semiconductors, insulators and metals is significantly based on the following factors [NCERT Exemplar] (a) number of charge carriers can change with temperature T (b) time interval between two successive collisions can depend on T (c) length of material can be a function of T (d) mass of carriers is a function of T

60. The potential difference between the points A and B in the circuit shown in figure is 16 V. Which is/are the correct statements out of the following? 1Ω

3Ω B

1Ω

2Ω

(a) The current through the 2W resistor is 3.5 A (b) The current through the 4W resistor is 2.5 A (c) The current through the 3W resistor is 1.5 A (d) The potential difference between the terminals of the 9 V battery is 7 V

61. In the figure, galvanometer reads zero. The resistance x is 10Ω 40Ω

G

7Ω

XΩ

(a) 7 W W

(b) 21 W (d) 28 W

(c) 14

D

B R

59. Temperature dependence of resistivity r( T) of

A

2Ω

2Ω

(a) Potential drop across AB is nearly constant as R¢ is varied (b) Current through R¢ is nearly a constant as R¢ is varied (c) Current I depends sensitively on R¢ V (d) I ³ always r+R

3V 1Ω

C

2V

R′

[NCERT Exemplar]

9V 1Ω

4Ω

A

58. Consider a simple circuit

4Ω

B

1 (d) K (T0 - Tr )2 2

More Than One Correct Option shown in figure stands for a variable resistance R ¢ . R ¢ can A vary from R0 to infinity. r is internal resistance of the battery ( r R2 2

26. As, A = pr = rl / R or r = (rl / pR)1/ 2 æ1.7 ´ 10 –5 ´ 0.5 ö r=ç ÷ 3.14 ´ 2 ø è

1/ 2

= 0.367 mm

848 JEE Main Physics rl A (a) When the battery is connected across 1 cm ´ 1/2 cm faces, then r ´ 10 l = 10 cm; A = 1 ´ 1/2 cm2, R1 = = 20 rW 1 ´ 1/2

27. We know that R =

34. Effective resistance of n resistance each of the resistance r in series Rs = r ´ n = R (as per question); so r = R / n. When these resistances are connected in parallel, the effective resistance r R /n R = 2 Rp = = n n n

35. Since resistance connected in arms CE, ED , CF and FD will 1 cm 2

form a balanced Wheatstone bridge, therefore, the resistance of arm EF becomes ineffective. Now resistance of arm CED or CFD = 2 + 2 = 4 W. Effective resistance of these two parallel 4´4 arm = =2 W 4+ 4

1cm 10cm

(b) When the battery is connected across 10 cm ´ 1cm 1 r ´ 1/2 r faces, then l = cm; A = 10 ´ 1 cm2,R2 = = W 2 10 ´ 1 20 1 (c) When the battery is connected across 10 cm ´ cm faces, 2 r ´1 r then l =1cm A = 10 ´ 1/2 cm2, R3 = = W (10 ´ 1/2) 5 RA =

28. We have RB =

Now 2 W and 10 W are in series. R2 = 10 + 2 = 12 W 1 1 1 = + Þ R3 = 6 W R2 and 12 W are in parallel R3 12 12

rl p [(10 –3) 2 - (0.5 ´ 10 –3) 2] RA (10 -3) 2 - (0.5 ´ 10 –3) 2 =3 = RB (0.5 ´ 10 –3) 2

Þ \

rl ¢ p (10 -3 ´ 0.5) 2

RA : RB = 3 : 1

29. Temperature coefficient of resistance =

1 dR Rt dt

1 R0(1 + at + bt 2) = R0(1 + at + bt 2) dt =

a + 2 bt 1 + at + bt 2

Now R3 and 6 W are in series R4 = 10 + 6 = 16 W Now R4 and 16 W are in parallel 1 1 1 \ = + Þ R = 32 W R 16 16

37. The equivalent resistance between two corners of equilateral triangle having resistance R in each arm = 2R /3 = 2 ´ 4 /3 = 8 /3 W

38. The potential difference across 300 W = 60 - 30 = 30 V

30. Number attached for brown, black, green and silver are 1, 0, 5, ± 10%. Therefore, the resistance of given resistor = 10 ´ 10 5 W ± 10% = 1.0 ´ 10 6 W ± 10%

31. As the current in heater filament increases, it gets more heated, hence its temperature increases and thereby its resistance increases. Due to which the current will decrease. Hence, the variation of V and i for heater filament will as shown in Fig.(a).

32. For one wire cable, resistance,

36.

Now resistance of arm ACDB = 2 + 2 + 2 = 6 W, is in parallel with resistance arm AB = 2 W. Thus, effective resistance 6 ´2 3 between A and B = = W 6+2 2 1 1 1 5 1 As, = + = = Þ R1 = 2 W R1 10 2.5 10 2

-3 2

R ¢ = rl / p (9 ´ 10 ) = 5 W

Therefore the effective resistance of voltmeter resistance R and 400 W in parallel will be equal to 300 W, as 60 V is R ´ 400 equally divided between two parts. So, 300 = R + 400 or

300 R + 120000 = 400 R or R = 1200 W

39. Here points Band D are common. So, 2 R in arm DC and 2 R in arm CB are in parallel between C and B. Their effective 2R ´2R resistance = =R 2R + 2R The modified and simpler circuit will be shown in figure. The effective resistance between A and B is R

A

C

For other wire of cable, resistance R ¢ = rl / p (3 ´ 10 -3) 2 = 9 2 ´ 5 / 3 2 = 45 W When six wires each of resistance R ¢ are connected in parallel, their effective resistance will be R ¢ 45 = = 7.5 W Rp = 6 6

33.

æ 6 12 ´ 6 ö As Vp - Vq = ç + ÷ (0.5) è 3 12 + 6 ø = (2 + 4) ´ 0.5 V = 6 ´ 0.5 V = 3 V

R

R

B

Reff =

R ´ (R + R) 2 = R R + (R + R) 3

Current Electricity 40. Two resistances of each side of triangle are connected in parallel. Therefore, the effective resistance of each arm of the r ´r r triangle would be = = . The two arms AB and AC are in r+r 2 series and they together are in parallel with third one. \ R ¢ = (r / 2) + (r / 2) = r Total resistance 1 1 2 3 = + = R r r r r R= 3 shown in figure. The effective resistance of arm 4´4 EG = =2 W 4+ 4 4Ω B 4Ω E

4Ω

4Ω

G

4Ω 4Ω

D

C

Total resistance between A and B will be 1 1 1 1 3 4 = + + = or R = W R 4 4 4 4 3

42. The equivalent circuit of this network is as shown in figure, which is a balanced Wheatstone bridge. Therefore no current will flow in the resistance of arm PQ. When cell is connected to points A and B. Therefore effective resistance of arm APS = (R + R = 2 R) will be in parallel to the total resistance of arm AQS ( = R + R = 2 R) P RΩ

RΩ

44. Effective emf of circuit = 10 - 3 = 7 V Total resistance of circuit = 2 + 5 + 3 + 4 = 14 W i = 7 / 14 = 0.5 A Potential difference between A and D = 0.5 ´ 10 = 5 V Potential at D = 10 - 5 = 5 V Hence, E cannot be at zero potential, as there is potential drop at E.

Current,

45. A fully charged capacitor draws no current. Therefore, no

41. Equivalent circuit of this combination of resistances is as

A

849

current flows in arm GHF. So, the R of arm HF is ineffective. The total resistance of the resistors in circuit is (R + R) ´ R R¢ = +R (R + R) ´ R (2 + 2) ´ 2 10 = +2 = W (2 + 2) + 2 3 E 10 Total current, i = = =3A R ¢ (10 /3) In parallel circuit, the current divides in the inverse ratio of resistance, so current in arm ABGD = 1A and current in arm AD = 2 A. Potential difference between G and D = VG - VD = 1 ´ 2 = 2 V Potential difference between D and F = VD - VF = 3 ´ 2 = 6 V \ VG - VF = (VG - VD ) + (VD - VF ) = 2 + 6 + 8 V Potential difference across capacitor = VG - VF = 8 V

46. For the same length and same material, R2 A1 3 = = or R2 = 3 R1 R1 A2 1 The resistance of thick wire, R1 = 10 W The resistance of thin wire = 3 R1 = 3 ´ 10 = 30 W Total resistance = 10 + 30 = 40 W

47. Effective resistance of three resistances between C and D RΩ

A

S RΩ

RΩ

=

B

R ´2R 2 = R R + 2R 3

Total resistance between A and B = R +

Q

2R ´2R =R W \ Equivalent resistance = 2R + 2R

43. If a cell is connected between points A and C, no current will flow in arms BE and ED. Therefore, the resistance of arms BE and ED can be removed. Now resistance between points A and C will be the resistance of three parallel arms, each of resistance = R + R = 2 R \Total resistance Rp will be 1 1 1 1 3 = + + = Rp 2 R 2 R 2 R 2 R or

Rp =

2R 3

2 8 R + R = ´3 = 8 W 3 3

48. The x be the total resistance of infinite network of resistance connected to points A and B. Therefore, the addition of one step of resistances in the infinite network of resistances will not change the total resistance x of the network. Therefore equivalent circuit will be as shown in fiugre. Then total resistance between A¢ and B¢ is x given by A'

1Ω A

1Ω

B'

x

B

850 JEE Main Physics x = 1+

54. In series combination of cells current, i =

x + x2 = 1 + 2 x

or or

1+ x 1+ 2 x = 1+ x 1+ x

If i = i ¢ then

1± 5 1+ 5 = 2 2 Since negative value of R is not possible. x=

It will be so if r = R.

49. Here resistances 4 W, 6 W, 12 W and 24 W are in parallel.

55. When one cell is wrongly connected in series, the emf of

Their effective resistance, Rp will be 1 1 1 1 1 = + + + Rp 4 6 12 24 6 + 4 + 2 + 1 13 = 24 24 24 or Rp = 13 Total resistance between A and B 24 128 =3+ +5 = = 9.85 W 13 13 =

56.

cells decreases by 2 E, but internal resistances of cells remains the same for all the cells. (n - 2) E Current in the circuit is i = ´r nr Potential difference across each cell is (n - 2) E 2E V = E - lr = E ´r = nr n 1 1 1 1 3 As = + + = R¢ R R R R R R

50. E = i1 (R1 + r) = i2 (R2 + r) On solving, r =

A

(i1 - i2)

From Eq. (i) or or Therefore,

52. Case (i) or

Case (ii)

resistance 4 W becomes ineffective in current. B

m2 = 4

R 6V A

E + E = (r + r + 5) ´ 1.0

or

E = 0.4 r + 4.0

…(i)

1.2 r = 3 3 r= = 2.5 W 1.2

53. Current through resistance R will be zero if

F

9V

…(ii)

Current through resistance 2 W is 9 -6 3 i= = A 2 2

E1 =

E (R1 + R2) R2

3 2

12 = 6 + 3 R R =2 W

or or

58. The distribution of current is as shown in figure. As per question, A

2Ω

2Ω

B

D i1/2

i1/2

i–i1 i1 E

1Ω

C

2Ω

4Ω

E E1 = R2 R1 + R2 or

E

3V

In circuit ABCDEFA,9 - 3 = (2 + R) ´

Multiplying Eq. (ii) by 2 and equating with Eq. (i), we get 2 r + 5 = 0.8 r + 8

D

2Ω

2Ω

m=2 n = 5 ´ 2 = 10

ær ö E = ç + 5÷ 0.8 è2 ø

or

C

m ´ 5 m = 20

or

or

R 3R + R 4 = = R 3 3 3

RN = R +

57. Since, no current is to flow in the 4 W resistance, hence

2.5 = n ´ 0.5 /m n =5m

2E = 2r + 5 ö ær ´r E=ç + 5÷ ´ 0.8 ø èr + r

B

R

\

For maximum current R = nr/m or

R

i2R 2 - i1R1

51. Here, mn = 20 or

E (r /n) + R nE E nE = = nr + R (r /n) + R r + nR

In parallel combination of cells, i ¢ =

x2 - x - 1 = 0

or

nE nr + R

1Ω

A E

i

G

i1

F

i1/2

E1

Current Electricity i1 = 1 or i1 = 2 A 2

R R

4R

E, 4 Ω

R ¢ = 2R = 4 or

arm ABC = (2.1 - i). As there is no deflection in the galvanometer, hence (20 + 5) i = (8 + 2) (2.1 - i)

Current through resistance R and S, 4 1 i2 = = A 40 + 60 25 1 VA - VD = Ri2 = 40 ´ = 1.6 V 25 VB - VD = (VA - VD ) - (VA - VB)

=

Since, the network of resistances is a balanced Wheatstone bridge, so resistance between points A and B of network figure (b) is given by 1 1 1 2 +1 1 = + = = R¢ 3 R 6 R 6 R 2 R

35 i = 21 21 3 i= = = 0.6 A 35 5 4 1 = A 90 + 110 50 1 VA - VB = Pi1 = 90 ´ = 1.8 V 50

E = V - e = 120 - 8 = 112 V Current in circuit, Effective emf E I= = Total resistance r + R

62. The equivalent circuit is as shown in figures (a) and (b).

or

i1 =

Since, the battery is bring changed, so effective emf in the circuit

3´S . On solving, S = 6 W 3+ S

25 i = 21 - 10 i

64. Current through resistance P and Q.

60. Emf of the battery e = 8 V, emf of DC supply V = 120 V

value 2 W, i. e. ,2 =

or

or

= 1.8 - 1.6 = 0.2 V

61. The bridge will be balanced when the shunted resistance of

R = 4 /2 = 2 W

63. Let i be the current through arm ADC. Then current through

Potential difference between A and D, 4 VA - VD = ´ 90 = 1.8 V 200

Terminal potential difference V = E + Ir = 8 + 7 (0.5) = 115 . V

(b)

R¢ = 2 R For maximum power to the network. R ¢ should be equal to internal resistance of the battery. So,

= 4 /(90 + 110) = 4 /200 A

The battery of 8 V is being charged by 120 V, so the terminal potential across battery of 8 V will be greater than its emf

E, 4 Ω

(a)

or

VA - VB = 0.04 ´ 40 = 1.6 V Current through arm ADC,

112 0.5 + 15.5 112 = =7A 16

4R

2R

R

59. Current through arm ABC,

VB - VD = (VA - VD ) - (VA - VB)

B

R

Total resistance of the circuit between A and H is 4 ´ 3 26 =2+ = 4+3 7 7 26 EMF of cell is E = ´ = 13 V 2 7

\

A

6R

7 i1 = 4 i 7 7 7 i = i1 = ´ 2 = A 4 4 2

= 4 /( 40 + 60) = 0.04 A Potential difference across A and B

2R

R R

In a closed circuit ACFG. i 2 i + 2 ´ 1 - 4 (i - i1) = 0 2

or

851

65.

= 1.6 –1.8 = – 0.2 V E1 l1 E1 = . As , therefore l1 > l2. Therefore, the null point for E 2 l2 E2 the cell of emf E 2 must be at shorter length than that of cell E1. Thus, the null point on potentiometer wire should shift towards left of C.

66. If R is increased, the current through the wire will decrease and hence the potential gradient will atom decrease, which will result in increase in balance length. Therefore, J will shift towards B.

67. In a potentiometer experiment, the emf of a cell can be measured if the potential drop along the potentiometer wire is more than the emf of the cell to be determined. As values of emfs of two cells are approximately 5 V and 10 V, therefore, the potential drop along the potentiometer wire must be more than10 V.

852 JEE Main Physics 68. As meter bridge is balanced, so

100 ´ 2.9 ~ - 3W 97.1 The meter bridge (i. e. , Wheatstone bridge) is most sensitive and accurate when the resistance of all the four arms of the bridge is of the same order. Therefore, a student must use S = 3 W to improve the accuracy of the measurement of resistance R.

\

R=

or

Þ

71. When a single heater (resistance R1 = R2 ) is connected to 220 V, then it will consume a power P1 = 1000 W. If two such identical heaters are connected in parallel (total resistance R2 = R 2) to same source, then it will consume power P2. P2 R1 = Þ P2 = 2P1 P1 R2

69. The equivalent resistance in the second case = (R1 + R2) W Pµ

We know, that

1 R

In second case, the resistance (R1 + R2) is higher then thent in first case R1. Thus, the power dissipation in the second case decreases. V 2t = Q = ms dq 4.2R R = resistance of the coil

70. We know, Let,

t = constant N t1 t = 2 N1 N2 9 N ´ 16 = 14.4 min t 2 = 2 ´ t1 = 10 N1

2.9 R R = = 100 - 2.9 S 100

P2 = 2000 W

72. As, P =

... (i)

N = initial number of turns , L r ´ N ´ 2 pr As, R =r = A A where r = radius of coil. V 2t A = Q = ms dq 4.2 ´ r ´ N ´ 2pr

... (ii)

[from Eqs. (i) and (ii)]

2

V R

V 2 200 ´ 200 = = 400 W P 100 400 Rcold = = 40 W 10 1 The energy stored in the capacitor = CV 2 ; 2 This energy will be converted into heat in the resistor. 1 H = ´ 4 ´ 10 -6 ´ 400 ´ 400 2 Rhot =

\

73.

= 32 ´ 10 -2 = 0.32 J

Round II 1. The circuit arrangement shown in figure (b) is the correct arrangement for verification of Ohm’s law. For convenience the same figure has been redrawn here. In the figure, R is the resistance, for which Ohm’s law is to be verified. Voltmeter V is connected to its parallel and ammeter, cell and rheostate arrangement in the series. +

A

–

B 2.5V I

0.5Ω

I

3. Resistance of coper part of wire, Rc =

V –

Rh

2. According to loop rule, 2.5 – 0.5 l - 2 I = 0 Þ

10Ω

A

rc × L rc × L and = Ac pr 2 r ×L rn × L resistance of nickel portion of wire, Rn = n = An p (R 2 - r 2)

+ R

2Ω

I

I = 1A q VA - VB = 0 = 2 I = 2 V C q 0 = C ´ 2 = 2 ´ 10

-6

´ 2 = 4 mC

As these two resistances are conductance of the nickelled wire 1 1 1 C= = + R Rc Rn =

pr 2 p (R 2 - r 2) + rc × L rn × L

=

p æ r 2 R2 - r 2 ö ç + ÷ L è rc rn ø

in

parallel,

hence

Current Electricity

853

4. In the circuit shown total external resistance R = 2 W +

7. In the circuit arrangement PSTQ is a balanced Wheatstone

parallel combination of two 4 W resistors + parallel combination of three15 W resistors. 4 15 =2+ + =2+2+5 =9 W 2 3 As E = 10 V and i = 1A, hence internal resistance r of the cell should have a value given by

bridge, hence resistance 2 R joined in arm AB be omitted. Similarly, resistance 2 R joined in arm BC may also be omitted. 1 1 1 1 m + 2 R + m (R + r) \ = + + = = Req 4 R 2 r 4 R 4 mR 2 Rr

E = i (R + r) E 10 or r = -R = -9 =1W i 1 If 4 W resistors are replaced by 2 W resistors, then as before 2 15 R¢ = 2 + + = 2 + 1+ 5 = 8 W 2 3 E 10 \New circuit current i ¢ = = = 1.11 A R¢ + r 8 + 1

S

A

T

2R

2R 2R

P

Q B

r

2R 2R

2R

K

C

Req =

Þ

5. The circuit may be redrawn as shown in the adjacent figure. Here

2 ´2 =1W 2+2 12 12 = = =2 W 5 +1 6

E eq = 12 V, req =

\

i=

5Ω

E eq R + req

+ + 2Ω – 12 V 2Ω –12 V

r

L

2 Rr R+r

8. The given circuit may be redrawn as shown in adjacent figure. Resistance of parallel combination of 2 W and 6 W. 2 ´6 R1 = = 1.5 W 2+6 Now resistance of ABC arm = 1.5+1.5 = 3 W 3 ´3 and total network resistance R = = 1.5 W 3+3 +

– 6V

2Ω

A 1.5Ω

6. Given, potential difference = 230 V

B 6Ω

Initially current at 27°C = I27°C = 3.2 A Finally current at t°C = It °C = 2.8 A Room temperature = 27°C

3Ω

Temperature coefficient of resistance a = 1.70 ´ 10 - 4 / °C Resistance at 27 °C, R27 °C = Resistance at t°C,

Rt °C =

V I27°C

=

230 2300 = W 3.2 32

\Total current supplied by the battery i =

9. Let R be the equivalent resistance. Then addition/subtraction of one more set of resistors R1, R2 and R3 will not affect the total resistance. Thus,

V 230 2300 = = W It °C 2.8 28

R2

Temperature of coefficient of resistance R - R27 a= t R27 (t - 27) Þ

or

2300 2300 32 1.7 ´ 10 = 28 2300 (t - 32) 32 82.143 – 71.875 t - 27 = = 840.347 71.875 ´ 1.7 ´ 10 - 4

R

R3

R

R1

R = R1 + (parallel combination of R and R2) + R3 æ RR3 ö R = R1 + ç ÷ + R2 è R + R3 ø

-4

t = 840.3 + 27 = 867.3 ° C Thus, the steady temperature of heating element is 867.3 °C. or

6V =4A 1.5 W

Þ Þ Þ

R 2 + RR3 = RR1 + R1R3 + RR3 + RR2 + R2R3 R 2 - R (R1 + R2) = (R2 + R1) R3 = 0 R=

(R1 + R2) ± (R1 + R2) 2 + 4 (R1 + R2) R3 2

854 JEE Main Physics R = 3 + 10 + 6 + 1 = 20 W. As potential difference across R2 ( = 6 W) is 48 V, hence R 48 ´ 20 VAB = 48 ´ = = 160 V R2 6

As R cannot be negative, hence 1 R = [(R1 + R2) + (R1 + R2) 2 + 4 (R1 + R2) R3 ] 2 1 = [(R1 + R2) + (R1 + R2) (R1 + R2 + 4 R3) ] 2

10. Current flowing through both the bars is equal. Now, the heat produced is given by H = I 2Rt H µR

or

=

open or closed, hence the bridge circuit is balanced. Hence, IP = IQ and IR = IG . However, as P ¹ R , hence IP ¹ IR.

15. The circuit given in figure can be redrawn as shown here.

HAB RAB (1 / 2r) 2 = = HBC RBC (1 / r) 2

or

14. As galvanometer deflection remains unaffected with switch S

1 1ö æ çQ R µ µ 2 ÷ è A r ø

1 4

here two resistances are joined in series and the combination is joined in parallel with the third resistance. Since in parallel grouping effective resistance is even less than the smallest individual resistance, hence net resistance will be maximum between the points P and Q. P

HBC = 4 HAB

or

11. Here, original volume of a wire pD 2 l 4

V = pr 2l =

New volume of a wire, V ¢ = pr 2l ¢ =

pD 2l ¢ 16

r/2 Q

V¢ = V

Since \

Original resistance, R =

the circuit is of a balanced Wheatstone bridge. As per condition of balance, we have R1 R2 = Þ R1R4 = R2R3 R3 R4

4 rl 4 D2

æ 4 rl ö and new resistance, R ¢ = 16 ç ÷ = 16 R è pD 2 ø

17. As ammeter must be connected in series of 20 W resistor and

= 16 ´ 10 W = 160 W

the voltmeter in parallel of 20 W resistor, the correct arrangement is as shown in figure (c).

12. As potential gradient Potential difference across wire Length of wire Rwire ´ I = l E 3 Current, I= = = 0.1 Rwire + Rext (20 + 10) =

…(i)

30 W, 60 W is R1, where 1 1 1 1 3 + 2 +1 6 1 = + + = = = R1 20 30 60 60 60 10 R1 = 10 W Similarly effective value of parallel combination of 24 W and 8 W resistances is given by 24 ´ 8 R2 = =6 W 24 + 8

A

6Ω

Hence,

Now the combined potential difference across B1 and B2 is same as the potential difference across B3 . Hence, W3 is more than W1 and W2. Now B1 and B2, being in series, carry same current and R1 < R2, therefore W1 < W2, \

W1 < W2 < W3

19. Here S consists of S1 and S 2 arranged in parallel, hence S=

S1S 2 S1 + S 2

So, the balance condition will be

P R R ( S1 + S 2) = = Q S S1S 2

20. As 5 W resistor is joined in parallel to series combination of

48 V 10 Ω

V2 , P 1 1 1 R1 : R2 : R3 = : : 100 60 60

18. As resistance of a bulb, R =

13. Effective value of resistance of parallel combination of 20 W,

3Ω

R

16. Since current i is independent of the value of R6, it is clear that

l¢ = 4 l

\

r/3

r

1Ω B

Hence, the circuit may be redrawn as shown in the adjacent figure, where total resistance across A and B,

4 W and 6 W (i. e. , total resistance10 W), V = constant. i1 R2 10 and = = =2 i2 R1 5 or

i2 =

i1 2

Current Electricity Now heat produced per second in 5 W resistor H1 = i12R1 = i12 ´ 5 = 100 Js–1

25. As, E = 16T - 0.04T 2 …(i)

and for 4 W resistor

At temperature of inversion, E = 0 16Ti - 0.04Ti 2 = 0

\ 2

æi ö H2 = i22R2 = ç 1 ÷ ´ 4 = i12 è2ø

21.

…(ii)

Ti =

Similarly powers, P0 and Pt , Here, voltage remains the same

é 2 r ´ 2r ù 2 =i r P4 = i 2 ê ë 2r + 2r úû

22. Here three resistances of 4 W each are connected in parallel 4 W. It is in series with 3 ammeter, battery and last 4 W resistance. 4 16 Net resistance R = + 4 = \ W 3 3 so that their combined resistance =

Current in main circuit = ammeter reading, E 2V 3 i= = = A R 16 W 8 3

\

P0 =

V2 R0

or

R0 =

V2 V2 and Rt = P0 Pt

Also

Rt = R0(1 + a[2500 - 20)]

and

P0 = Pt [1 + a(2500 - 20)] = 50[1 + 4.5 ´ 10 -3(2500 - 20)] = 608 W

27. Heat produced H = Vit = Pt J where, P = Vi watt \ H = 500 ´ 15 ´ 60 = 45 ´ 10 4 J Heat absorbed by water = mass ´ specific heat capacity ´ rise in temperature. Þ

1 ´ 4200 ´ (100 - 15) = 4.2 ´ 85 ´ 10 -3 Efficiency =

23. The circuit diagram may be redrawn as shown here. D

10 Ω

=

B

A

Obviosuly,

5Ω

ICAD = ICBD =

3 2 æ 0.02 ö =ç ÷ I22 è 0.03 ø

C

2 A 15

2 2 VC - VA = A ´5 W = V \ 15 3 2 4 and VC - VB = A ´ 10 W = V 15 3 4 2 2 \ VA - VB = (VC - VB) - (VC - VA ) = V - V = V 3 3 3

32

A

29. Q = at - bt 2 A

R

B

C

(I – I1) R

2R I1

I F

E

(kWh). Therefore the total number of units consumed is

Total cost = ` 60 ´ 0.5 = `30

3

æ3ö I2 = 3 ´ ç ÷ è2ø

24. We know that, the unit of electrical energy is kilowatt hour N = (0.1 ´ 8 + 0.3 ´ 4) ´ 30 (Q June has 30 days) N = 60 units

85 ´ 4.2 ´ 10 3 ´ 100 = 79% 45 ´ 10 4

I12 r13 = I22 r22

2V

10 Ω

heat absorbed ´ 100 heat produced

I 2 µ r3

28. For a fuse, 5Ω

16 = 400 °C 0.04

Rt = resistance of filament at 2500 °C.

So, it is obvious that P2 > P3 > P4 > P1.

\

Þ

26. Let R0 = resistance of filament at room temperature

Simplifying Eqs. (i) and (ii), we get H2 1 = 100 5 1 or H2 = ´ 100 = 20 Js–1 5 r Here, P1 = i 2 ´ ,P2 = i 2 ´ 3 r 3 ær ö 3 P3 = i 2 ç + r ÷ = i 2r è2 ø 2 and

855

\ when,

I=

D

dQ = a - 2bt dt

t = t 0 , I = 0, i. e. , a - 2 bt = 0

…(i)

856 JEE Main Physics In loop BCDEB I1(2R) - (I - I1)R = 0 or 3I1 = I i a - 2bt I1 = = 3 3

33. When two batteries are in series to the external resistance R, total resistance of the circuit Þ R + 2r = R + 2 ´ 1 = R + 2W Total emf of batteries = 2 + 2 = 4 V

t0

H = ò (I12 (2h) 0

\Joulean power across R

2R t 0 ( a - 2bt ) 2dt = 9 ò0 2R é t 0 2 = ( a - 4b 2t 2 - 4dt )dt ù ûú 9 ëê ò0

2

æ 4 ö 4 2R =ç ÷ ´R = èR + 2ø (R - 2) 2 + 8R

t0 2R éì 2 4b 2t 2 4bat 2ü ù êí a t + = ý ú 9 êî 3 2 þ0 ú û ë 2 3 ù 2R é 2 4b t 0 = - 2bat 02 ú êa t0 + 9 ë 3 û a [from Eq. (i)] t0 = 2b 2R é 2 a 4b 2 a3 a2 ù + - 2ab 2 ú H= êa ´ 3 9 ë 2b 3 8b 4b û

=

The total resistance across the battery is r 3 rtotal = r + = r 2 2 Current through P,

I=

12 3 r = 8 r 2

1 2 Ip = r 2

Power dissipated in R is thus æ2ö PR = I 2r = ç ÷r = 2 W èrø

T = T0 2 1 S = kT0 2

35. Resistance of the bulb = æç15 . ´ è

31. Let I be the total current passing through balanced Wheat stone bridge. Current through arms of resistances P and Q in series is I ´ 330 3 I1 = = I and current through arms of resistances 330 + 110 4 I ´ 110 1 = I R and S in series is I2 = 330 + 110 4 \ Ratio of heat developed per sec HP : HQ : HG : HS 2

.

Current through R,

1 dE ö æ = k çT - 0 - ´ 2T + 0 ÷ = k (T0 - T) ø è 2 dT

2

42 ´ 2 =2W (2 - 2) 2 + 8 ´ 2

34. Let the resistance of P, Q and R be r.

Power Ip = = rtotal

1 1 ö æ E = k çTT0 - T0Tr - T02 + Tr2÷ è 2 2 ø

at

\ Maximum Joulean power =

2R é a3 a3 a3 ù a3R + - ú= ê 9 ë 2b 6b 2b û 27b

30. Thermoelectric power

S=

\ Joulean power will be maximum if R - 2 = 0 or R = 2W

2

2

æ1 ö æ1 ö æ3 ö æ3 ö = ç I ÷ ´ 100 : ç I ÷ ´ 10 : ç I ÷ ´ 300 : ç I ÷ ´ 30 è4 ø è4 ø è4 ø è4 ø

15 1 . ö ÷ = 0.5 = W 4.5 ø 2

1 1 2 Resistance of the circuit R = = W 1 3 1+ 2 E -V Now, r= R V 8 E - 15 1 = ´ 3 15 3 . 1´

E = 13.5 volt

or

36. Assuming current I flows through the circuit.

= 30 : 3 : 10 : 1

r

32. Total resistance between points P and Q, R=

10 ´ 5 5 ´ 10 20 + = W 10 + 5 5 + 10 3

If m gram of the ice melts in given time t, then as per question 20 (10) 2 ´ ´ (10 ´ 60) = m ´ 80 ´ 4.2 3 100 ´ 20 ´ 10 ´ 60 or m= . ´ 10 3 g = 119 3 ´ 80 ´ 4.2 = 119 . kg

R

Energy dissipated in load = I 2R Energy dissipated in the complete circuit = I 2(r + R) \

The efficiency =

I 2R R = 2 I (R + r) R + r

Current Electricity 37. (i) Rate of chemical energy consumption

Now,

= 15 . ´ 2 = 3W (ii) Rate of energy dissipation inside the cell

and

= 2 ´ 2 ´ 0.1 = 0.4 W

(iv) Power output of source = (3 - 0.4) W = 2.6 W

38. Given, number density of electrons n = 8.5 ´ 10 28/m3 Length of wire l = 3 m

42.

Area of cross-section of wire A = 2 ´ 10 - 6 m 2 Current I = 3 A and charge on electron e = 1.6 ´ 10 -19 C Time taken by electron to drift from one end to another of the wire, Length of the wire l …(i) t= = vd Drift velocity

or

or

…(ii)

When geyser is also switched on, effective resistance of bulb and geyser 484 ´ 48.4 = = 44W 484 + 48.4 220 ´ 44 Vbulb = = 193.6 V ( 44 + 6)

Negative surface charge density s = 10 - 9 C/m 2 Potential difference V = 400 kV = 400 ´ 10 3 V Current on the globe I = 1800 A

Hence, the potential drop = 217.4 ´ 193.6

Surface area of earth A = 4pR 2 = 4 ´ 3.14 ´ (6.37 ´ 10 6) 2

× Surface charge density Q = A s = 509.64 ´ 10 12 ´ 10 - 9 = 509.64 ´ 10 3 C We know that Q = I t \Time required to neutralize earth’s surface Q 509.64 ´ 10 3 = I 1800

t = 283.1 s or t = 4 min 43 s Thus, the time required to neutralize the earth’s surface is 283.1 s. rL r2L As, R = ; R1 = = 2R A A m = ALd ; m1 = A2Ld = 2m where d is density.

(220) = 484 W 100

(220) 2 = 48.4 W 1000 When only bulb is on, 220 ´ 484 Vbulb = = 217.4 V ( 484 + 6)

39. Given, radius of earth R = 6.37 ´ 10 6 m

Charge on earth surface Q = Area of earth surface

2

Rgeyser =

t = 2.72 ´ 10 4 s = 7 h 33 min

= 509.64 ´ 10 12 m 2

brightness. In series combination of bulbs, the bulb of lesser wattage will glow more bright. Hence W2 > W1. So, W1 < W2 < W3 . dE 2T T ; = 4= 4dT 200 100 At neutral point,T = Tn dE = 0 = 4 - Tn 100 dT or Tn = 400° C,

43. Rbulb =

Thus, the time taken by an electron to drift from one end to another end is 7 h 33 min.

40.

…(ii)

= 2 ´ 400 - 0 = 800°C

l ne A 3 ´ 8.5 ´ 10 28 ´ 1.6 ´ 10 - 19 ´ 2 ´ 10 - 6 = I 3

t=

(NV ) 2t = 2mcDT 2R

Ti = 2Tn - T0

Putting the value in Eq. (ii) from Eq. (i), t=

...(i)

41. Voltage across B3 is greatest, hence B3 will show maximum

= (3 - 0.4)W = 2.6 W

I = ne A v d I vd = ne A

(3V ) 2t = mcDT R

Solving Eqs. (i) and (ii), we get N = 6

(iii) Rate of energy dissipation inside the resistor

Using the relation,

857

44.

= 23.8 V = 24 V dE d As, = ( at 2 - bt 3) = 2at - 3bt 3 dt dt dE When t = t n (i.e., neutral temperature), =0 dt \

0 = 2at n - 3bt n2

or

tn =

2a 3b

.

45. Heat produced = energy stored in capacitor. =

1 1 CV 2 = ´ (10 ´ 10 -6) ´ (500) 2 = 1.25 J 2 2

46. In India, power P1 =

V 2 (220) 2 = R R

In USA, power P2 =

V 2 (110) 2 = R R

As

P1 = P2

858 JEE Main Physics (220) 2 (110) 2 = R1 R2

So,

53. Thermo electric power, S µ q(q = temperature)

or

(as R1 = R)

show same dissipation of energy, so using the relation for V2 energy, H = t , we have R2 = R3 . Thus, the current in each R resistor R2 and R3 will be I 2 i.e., I1 = I 2 and I2 = I 2 I

R1

F

V

= 25

I1

I2

R2

R3

E

D

Since, the energy dissipation is same in all the three resistors, So,

I 2R1t = I12R2 t

or

I 2R1t = (I 2) 2R2 t or R1 = R2 4

48. Cost = æç è

60 ´ 8 ´ 30 ö . = ` 18 ÷ ´ 125 1000 ø

49. Power dissipated P =

ö æ5 ç S 80 - S 80 ÷ 4 =ç ÷ ´ 100 S 80 ÷ ç ø è

B

I C

100 5 = S 80 80 4 Therefore % change in thermo electric power æ S - S 80 ö = ç 100 ÷ ´ 100 è ø S 80 S100 = S 80 ´

or

47. As voltage across the resistors R2 and R3 is same and they

A

S100 100 = S 80 80

\

(110) 2 R2 = R (220) 2 R R = 1= 4 4

V2 Reff

Reff is least in case of figure (a). Hence, power dissipated in circuit (a) is maximum.

1 2

54. Total energy stored in capacitor, E total = CV 2 1 ´ 3 ´ 10 -6 ´ 10 2 = 15 . ´ 10 -4 J 2 2 Energy dissipated in 2W = ´ E total (2 + 4) 2 = ´ 15 . ´ 10 -4 = 0.5 ´ 10 -4 J = 0.05 mJ 6 =

55. Here 2W, 3W and 6W are in parallel. So, potential drop across them will be the same. As heat produced, H =

1 H µ , so maximum heat will be generated across 2 W R resistance. Similarly 4W and 5W are also in parallel, so more heat will be generated across 4W. Now the effective circuit will become 1Ω

50. The filament of the heater reaches its steady resistance when

200 Ω

Vi

Vs

the heater reaches the steady temperature, which is much higher than room temperature. The resistance at room temperature is then much lower than the resistance of its steady state. When the heater is switched on, it draws a larger current than its steady state current as the filament heats up, its resistance increases and current falls to steady state value.

V

20 29 = W 9 9 V 9V A I= = 29 29 9 9V 9 V1 = V ´1 = 29 29 9V 20 20V V2 = ´ = 29 9 29

Total resistance = I +

51. Heat produced by heater per second = 108 . ´ 10 3 J Heat taken by water to form steam = mL.

Current

= 100 ´ 540 cal = 100 ´ 540 ´ 4.2 J \ or

108 . ´ 10 3 ´ t = 100 ´ 540 ´ 4.2 100 ´ 540 ´ 4.2 = 210 s t= 1.08 ´ 10 3

52. z = m It , From graph,

\

and

Power spent across 2W It = Area OABC 1 100 = (10 + 30) ´ =2 2 1000 z = m 2 = 0.5 m

V2 t i.e., R

P1 =

V12 2

=

æ 9V ö ç ÷ è 29 ø 2

2

=

40.5V 2 (29) 2

Current Electricity Power spend across 4W P2 =

V22 = 4

æ 20V ö ÷ ç è 29 ø 4

56. R =

æ P Rö = ÷ of a èQ S ø Wheatstone bridge. There exists a potential difference between points Band D due to cell of emf 2 V joined between these points. Hence VB - VD = 2 V.

62. The circuit is satisfying the condition of balance ç =

50V 2 (29) 2

\ P2 > P1. Hence maximum heat is produced in 4W resistance. 2

859

2

4 (200) V R or R µ V 2 \ 1 = = P R2 (300) 2 9

63. For charging a battery the external voltage generator must

When bulbs are connected in series, the current I is same through each. As P = I 2R or P µ R (as I is same in series). P1 R1 4 So, = = P2 R2 9 1 57. As, E = k (T - Tr )T0 + k (T 2 - Tr2) 2 dE 1 \ = kT0 + k ´ 2T = kT0 + kT dT 2 At temperature T = T0 2, dE 3 T Thermoelectric power is = kT0 + k 0 = kT0 dT 2 2 V 58. Current, I = As R ¢ >> R , so R + R ¢ ³ R ¢ RR ¢ r+ R + R¢ V V Hence, I = ³ r + RR ¢/R ¢ r + R RR ¢ Potential difference across A and B, VAB = I ´ R + R¢ V RR ¢ V ´ RR ¢ \ VAB = ´ = r + (RR ¢/R + R ¢ ) (R ´ R ¢ ) r(R + R ¢ ) + RR ¢ V ´ RR ¢ VR [QR + R ¢ » R ¢] = = rR ¢ + RR ¢ r + R \VAB is constant if R¢ is varied. V VR VR Current through R ¢ , I ¢ = AB = = R ¢ (r + R) R ¢ rR ¢ + RR ¢ Thus I¢ is not constant as R¢ is varied. The current I does not depend on the sensitivity of R¢. m , where the ne2t various terms have their usual meanings. As temperature changes, n changes and also t changes. Hence, r changes with temperature.

59. Resistivity of a conductor is given by r =

60. Potential difference between the terminals of 9V battery = 16 V - 9 V = 7 V Consequently, current in 2 W resistor =

7V = 3.5 A 2W

61. In the balanced condition of Wheatstone bridge, 10 X = 7 ´ 40 7 ´ 40 X= = 28 W 10

A current also flows in branch BD due to this cell of emf 2 V in the direction from D to B. provide an output voltage V greater than emf of battery E. In charging state æV -E ö or I=ç ÷ èR+rø

64. As, H =

V2 V2 t1 = t2 R1 R2

or and

H V 2t1 H R2 = 2 V t2 R1 =

When heaters are in series, V 2t H= (R1 + R2)

...(i)

When heaters are in parallel, æV 2 V 2ö H=ç + ÷t è R1 R2 ø

...(ii)

Putting the values of R1 and R2 in Eq. (i), we have t = t1 + t 2 When the values of R1 and R2 is put in Eq. (ii), we get tt t = 12 t1 + t 2

65. From figure, For first student R2 = 10 W , R1 = 5 W , R3 = 5 W

For other student R2 = 1000 W , R1 = 500 W , R3 = 5 W According to Wheatstone bridge principle R1 R1 = R R3 R or R = R3 ´ 2 R1

R

R2 A

C

G R1

R3 D

Putting values in above relation, we note that the value of R for both the students is10 W. The Wheatstone bridge is most sensitive and accurate if the resistance of all the four arms of bridge is of the same value. Therefore, the errors of measurement of the two students depend on the accuracy and sensitivity of the bridge,which inturn depends on the accuracy with which R2 and R1 can be measured. When R2 and R1 are large, the currents through the arms of bridge will be feeble. This will make the determination of null point accurately more difficult.

860 JEE Main Physics 66. Let V be the potential at S. VPS = 70 - V = 10I1 VSQ = V - 0 = 20I2 VSR = V - 10 = 30(I1 - I2) On solving these equations, we get, V = 40V, I1 = 3 A, I2 = 2 A then

Total power = I12 ´ 10 + I22 ´ 20 + (I1 - I2) 2 ´ 30 = 200 W

67. When there is a neutral point at D in meter-bridge, then R l1 = S (100 - l1) For the given values of R and S, there will be only one value of l1 for which we shall get the neutral point on bridge wire. In this case VA - VB = VA - VD or VB = VD . Therefore the galvanometer shows no deflection when jockey contacts a point at D. There is no current in galvanometer arm. When jockey contacts a point D1 on meter-bridge wire left of D, the resistance of arm AD1 becomes smaller than previous value. Due to it, VA - VB > VA - VD1 or VD1 > VB. Therefore current flows to B from the wire through galvanometer. When jockey contacts a point D2 on meter bridge wire right of D, the resistance of arm AD2 becomes more than first value. Due to it, VA - VD 2 > VA - VB or VB > VD 2 Therefore the current will flow from B to the wire through galvanometer. When R is increased, the neutral point will shift to the right instead of left on bridge wire.

68. The power supplied by cell = EI When this power is supplied to R and r, it is divided in the r æ R ö ratioç ÷ and èR + r ø (R + r)

69. E =

V 3.0 V = = 5 Vm–1 L 0.6 m

\ vd =

et 1.6 ´ 10 –19 ´ 2.5 ´ 10 –14 ´ 5 = 2.2 ´ 10 –2 ms–1 E= m 9.11 ´ 10 31

1 2

74. As, E = aq + bq2 dE = a + bq dq dE At neutral temperature, q = qn , =0 dq

\

\

a + bqn = 0

or

qn = -

75. If T is the temperature in kelvin corrsponding to q°C then

E = a (T - 273) +

E 5 Vm–1 = = 2.37 ´ 10 –8 W-m J 2.11 ´ 10 8 Am–2

71. With only 30 V battery, as shown in figure, total current will be =

30 V = 4 A (Since resistance of network = 7.5 W) 7.5 W

1 b (T - 273) 2 2

dE = a + b (T - 273) dT dE \ Peltier coefficient, p = = T[ a + b (T - 273)] dT Þ

= (273 + 27)[14 + ( -0.04) ´ 27] = 300(14 - 1.08) = 4176 mV = 4.2 mV

77. The equivalent circuit is represented

2R

as Now, the resistances 2 R , 2 R and R are connected in parallel combination. Hence, equivalent resistance is given by 1 1 1 1 3 2R = + + = Þ Rp = Rp 2 R 2 R 2 R 2R 3

2R 2R

78. The metallic body of the electrical applicances is connected to the third pin which is connected to the earth. This is actually a safety precaution and avoids eventual electric shock. In this process, the extra charge flowing through the body is passed to earth. The three pin connections do not affect on heading of connecting wires.

79. Drift velocity of free electrons is given by eE t m V E= l

vd =

70. Resistivity r=

a -14 = = 350°C b -0.04

Here, vd =

\

eV 1 æ eV ö t or v d µ = t is constant ÷ ç ø l èm ml

80. The equivalent circuit is represented as, M

72. With only 15 V battery, as shown in figure (b), total current will be =

15 V = 3 A (Since now resistance of network = 5 W) 5W 1 2

73. As, 1250 = 14 q + ( -0.04) q2or 0.02q2 -14q + 1250 = 0 On solving, we get q = 105°C or 595°C

R

R

R

A

R

B

R N

Current Electricity This is balanced Wheatstone bridge hence, resistance in branch MN is not taken into consideration. Hence, the equivalent resistance between points A and B is given by 1 1 1 = + RAB (R + R) (R + R) 1 2 1 = = Þ RAB = R RAB 2 R R

or

81. From the relation for V-i graph

or

R1 = tan q = R0 (1 + aT1)

…(i)

R2 = tan (90° - q) = cot q = R0 (1 + aT2)

…(ii)

2

Req =

Þ

IL = 2 A VP - VQ

\

VQ > VS

i. e. ,

86.

rl As R = A Þ

cot q - tan q = R0 a (T2 - T1) cos q sin q = R0 a (T2 - T1) sin q cos q

So,

6 ´ 12 =4W 18 12 IL = =3A 4 æ 12 ö IL = ç ÷ ´3 è 6 + 12 ø

\

Current through PQ = 6 VS - VQ = -4

R = tan q = R0 (1 + aT) where, q is angle made by V-i graph with i axis. So,

R=

rL r = independent of L tL t

87. The equivalent circuit is a shown in Figs. (a) and (b).

2

cos q - sin q 2 cos 2q = = R0 a (T2 - T1) sin q cos q sin 2q

A

5Ω

10Ω

83.

DRe DR = DRe R

Þ

DRe = Re

5Ω

DR 5 ´ 100 = = 5% R 100

Thus statement I is true. Hence R227 - R27 = 150 - 100 = 50 W

same potential. S 2Ω

2Ω 1Ω

1Ω

I1

4Ω 4Ω

It is not very much less than100 W, hence R - R0 > R, x2 + R2 = x2

m m i (dl ´ r) dB = r 0 4p r3

\

where m r is the relative permeability of the medium and is a dimensionless quantity.

B=

m 0 NiR2 3

2x

m = æç 0 ö÷ è 4p ø

æ 2 NipR2 ö æ m 0 ö ÷÷ = ç ÷ çç 3 ø è 4p ø è x

Let us consider few applications of Biot-Savart law

Here, M = magnetic moment of the loop = NiA = NipR2

dipole is

m 0 2M × 4p x 3

(c) Magnetic field due to an arc of a circle at the centre is

i P

i

q m i B = æç ö÷ 0 è 2p ø 2r or

1. Magnetic field due to a straight thin conductor is

m i B = æç 0 ö÷ æç ö÷ q è 4p ø è Rø

θ O

m i B = 0 (sin q1 + sin q2 ) 4pd (a) For an infinitely long straight wire,

(iii) Magnetic Field along the axis of a solenoid is

q1 = q2 = 90° m i B= 0 2pd

x A B R

α

(b) When wire is semi-infinite,

\

æ2 M ö ç 3 ÷ è x ø

Note This result was expected since, the magnetic field on the axis of

Applications of Biot-Savart Law

θ2 θ1

φ

β P

p 2

i

m 0i æ pö ç sin 0° + sin ÷ 4pd è 2ø m i = 0 4pd

B=

dx

d

1 , i. e., B-d graph for an infinitely long straight d wire is a rectangular hyperbola as shown in figure.

(c) B µ

B

d

P

x = the distance of point P from centre

where m 0 is a constant and is called, permeability of free space.

q1 = 0° and q2 =

x

R

Here, R = radius of the coil

m 0 idl sin q Wb/m2 or tesla 4p r2

d

i

m 0 NiR2

B=

m 0 Ni (sin a + sin b ) 2

(a) For a long solenoid (L > > R). p a=b = i. e., 2 B = m 0 Ni (b) At the ends of solenoid. p a = 0, b = 2 1 we get, B = m 0 Ni (for L > > R). 2

R Inwards

Magnetic Effect of Current

865

Sample Problem 1 Find the magnitude and direction of

Sample Problem 3 A current path shaped as shown in

magnetic field at point P due to the current carrying wire as shown. a

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 is the magnitude of the field at P, if the current is 3.0 A?

m 0i [ -1 / 2 - 3 /2 ] 4 pa m i (b) B = 0 [1 /2 + 3 /2] 4 pa m i (c) B = 0 [ -1 /2 + 3 /2] 4 pa m i (d) B = 0 [ -1 /2 + 5 /2] 4 pa (a) B =

60

P

30° °

i

P

i

30° D E

m i Interpret (c) As, B = 0 [sin q1 + sin q2] 4 pa q1 = - 30° , q2 = 60°

Here,

A

C

(a) 2.62 ´ 10

-6

T

(b) 2.62 ´ 10 -7 T

(c) 3.62 ´ 10

-7

T

(d) 2.62 ´ 10 -8 T

Interpret (b) Magnetic field at centre of the circular loop,

1 3 and sin q2 = 2 2 Putting these values, we get m i B = 0 [ -1 /2 + 3 /2 ] 4 pa sin q1 = -

B=

^

Sample Problem 2 An element DL = Dx × i is placed at the origin and carries a large current I = 10 A. The magnetic field on the y-axis at a distance of 0.5 m is (Dx = 1cm) [NCERT]

m 0Ni 2R

Magnetic field due to an arc of a circle at the centre is æ q ö m i æm ö æ i ö B=ç ÷ 0 =ç 0÷ç ÷q è 2p ø 2R è 4p ø è R ø Here, q = 30° and i = 3 A , R = 0.6 m æm ö æ 3 ö æpö B=ç 0÷ç ÷ç ÷ è 4p ø è 0.6 ø è 6 ø =

P

10 -7 ´ 3 ´ p = 2.6 ´ 10 -7 T 0.6 ´ 6

0.5m x ∆x

(a) 2 ´ 10 -8 T (c) 8 ´ 10

-8

(b) 6 ´ 10 -8 T (d) 4 ´ 10 -8 T

T

Interpret (d) According to Biot-Savart law, | dB| =

m 0 l dl sin q 4p r2

Given, Dx = dl = 10 -2 m, l = 10 A, r = 0. 5 m = y q = 90°, sin q = 1 \

|dB| =

10

-7

´ 10 ´ 10 25 ´ 10 -2

-2

^ ^ ^ dl ´ r = Dx i ´ y$j = yDx ( i ´ j) = y × Dx k

[Using cyclic property of cross products) ^ ^

^

^ ^

^

^

i ´ j = k , j ´ k = i, k ´ i = j ]

Note Field is small in magnitude.

If we spread our right hand in such a way that thumb is towards the direction of current and fingers are towards that point where we have to find the direction of field, then the direction of field will be perpendicular to the palm. i B

Current carrying conductor

= 4 ´ 10 -8 T

^

^

(i) Right hand rule

P

The direction of the field is in the + z direction. Since,

^

19.3 Rules to Find the Direction of Magnetic Field

(ii) Maxwell’s right handed screw rule If a right handed cork screw is rotated so that its tip moves in the direction of flow of current through the conductor, then the rotation P2 of the head of the screw gives the direction of magnetic lines of force.

Current carrying i conductor Magnetic line P1 of force

866 JEE Main Physics Note By convention the direction of magnetic field B perpendicular to paper going inwards is shown by Ä and the direction perpendicular to the paper coming out is shown by.

Sample Problem 4 Shown in the figure is a conductor carrying a current i. Find the magnetic field intensity at the point O.

0.8 m from the centre of the path. Now, half of the charge is removed from one end and placed on the other end. The rod is rotated in a vertical plane about a horizontal axis passing through the mid-point of the rod with the same angular frequency. Calculate the magnetic field at a point on the axis at a distance of 0.4 m from the centre of the rod. (a) 2.13 ´ 10 -3 , 2.26 ´ 10 -3

(b) 1.13 ´ 10 -3 , 2.26 ´ 10 -3

I

(c) 1.13 ´ 10 -3 , 1.26 ´ 10 -3 (d) 1.13 ´ 10 -3 , 3.26 ´ 10 -3

Interpret (b) A revolving charge is equivalent to a current, i = q ×r = q ×

O

5m 0iq $ k 26pr 5m iq (d) - 0 k$ 24pr

5m 0iq $ k 24pr 7m iq (c) - 0 k$ 24pr

The field at a distance z from the centre of the axis of a current carrying coil is given by

(b) -

(a)

B=

Interpret (d) Since, magnetic field at the centre of an arc is m iq equal to, B = 0 4 pr i

4p ´ 10 -7 ´ 5 ´ 10 3(0.6) 2 = 1.13 ´ 10 -3 T 2 [(0.6) 2 + (0.8) 2]3 / 2 q = 1C

B4

r

B5

0.6 m

y

B3

r

B7

m i B= 0 4p =

B

0.8 m

x z

B1 = B3 = B5 = 0 m iq Magnetic field due to arc1, B2 = 0 Ä 4p 3r m iq Magnetic field due to arc 2, B4 = 0 . 4p 2r m iq Magnetic field due to arc 3, B6 = 0 Ä 4 pr Net magnetic field,

P

O

B6

r

or

=

m 0 a2i 2( a + z 2)3 / 2 2

B2

B1

w 10 4 =1´ = 5 ´ 10 3 A 2p 2

B = B2 - B4 + B6 1 1 1ù é êë - r + 2r - 3r úû q

5m 0iq 5m iq ^ Ä=- 0 k 24pr 24pr

Sample Problem 5 A charge of 1 C is placed at one end of a non-conducting rod of length 0.6 m. The rod is rotated in a vertical plane about a horizontal axis passing through the other end of the rod with angular frequency 10 4 p rads -1. Find the magnetic field at a point on the axis of rotation at a distance of

q/2

P

O

0.6 m

0.4 m

q/2

Equivalent current, q q i ¢ = f + f = qf = 5 ´ 10 3 A 2 2 Magnetic field at a point P in this case B¢ =

m 0ia2 2( a + z 2)3 / 2 2

Here, a = 0.3 m, z = 0.4 m which gives, B¢ = 2.26 ´ 10 -3 T

B

Magnetic Effect of Current

1

Hot Spot

867

Magnetic Fieldon an Axial Point of a Circular Current Loop

The figure shows a circular loop carrying a steady current . The loop is placed in the y-z plane with its centre at the originO and has a radius R. The x-axis is the axis of the loop. To calculate the magnetic field at the point P on this axis, let x be the distance of P from the centre O of the loop. Considering an element dl of the loop, magnitude of dB due to dl from Biot-Savart law is y

dB⊥

dl

dB

r I O

x

P

θ x dBx

dl Z

dB = 2

m 0 l |d l ´ r | 4p r3 2

2

r = x + R , |dl ´ r | = rdl

Also,

dB =

l dl m0 2 4 p x + R2

The direction of dB is perpendicular to the plane formed by dl and r. Components perpendicular to x-axis are summed they cancel out,only x-axis component survives. The net contribution along x-direction can be obtained by integrating dBx = dB × cos q over the loop. R ( x + R2 )1/2

\

cos q =

\

dBx =

2

• Magnetic field due to a circular arc of radius r at the centre P is m I( a ) (take ain radian) B= 0 4 pr p For a semicircular loop a = 2 m 0I B= \ 4r

Sample Problem 6 Two concentric coils, each of radius equal to 2p cm, are placed at right angles to each other. Currents of 3 and 4 ampere respectively, are flowing through the two coils. The magnetic induction in Wbm -2 at the centre of the coils will be [m = 4p ´ 10 -7 Wb ( Am) -1] (a) 5 ´ 10 -5

(b) 7 ´ 10 -5

(c) 12 ´ 10 -5

(d) 10 -5

Interpret (a) Field at the centre of the loop is given by B=

y

m 0 l dl R × 2 4p ( x + R 2 )32/

2π I = 3A O

2π

x

I′ = 4A

m 0 IR2 i$ 2 ( x2 + R2 )32/ Bx =

Field at centre of loop ( x = 0 ) is m I B = 0 $i 2R The magnetic field lines due to circular wire form closed loops and is shown as follows

m0 I 2R

where, R is radius given, R = 2 p cm = 2p ´ 10 -2 m

The summation of elements dl over the loop yields 2 pR, the circumference of the loop. Thus magnetic field at P due to entire loop is B = Bx$i =

r α

m0 I1 2 2p ´ 10 -2

I1 = 3 A \

Bx =

m 0 3 ´ 10 2 × = 3 ´ 10 -5 T 2 2p

By =

m0 I2 × , I2 = 4 A 2 2p ´ 10 -2

By = 4 ´ 10 -5 \

Bnet = Bx2 + By2 = (3 2 + 4 2) ´ 10 -10 Bnet = 5 ´ 10 -5 T

868 JEE Main Physics Sample Problem 7 The magnetic field at the centre of the circular loop as shown in figure, when a single wire is bent to form a circular loop and also extends to form straight section is

B1 =

m 0I 4 pR

p p ù m 0I é êë sin 2 - sin 4 úû = 4pR

Magnetic field due to circular loop m I B2 = 0 2R Magnetic field due to straight wire BC m I é p pù B3 = 0 ê sin + sin ú 4 pR ë 2 4û

R

B C

90°

=

I

m 0I 2R

(c)

m 0I 2R

1 ö m 0I æ ÷ ç1 + è 2R p 2ø m Iæ 1 ö (d) 0 ç1 ÷ R è p 2ø

(b) 1 ö æ ÷ ç1 è p 2ø

m 0I 4 pR

1 ù é êë1 + 2 úû

\Resultant magnetic field, B = B1 + B2 + B3

A

(a)

1 ù é êë1 - 2 úû

æ m I 2m 0I 1 ö m 0I =ç 0 + ÷= è 2 R 4pR 2 ø 2R

é ê1 + ë

1 ù ú 2 pû

Interpret (b) From the figure, magnetic field due to AB,

19.4 Ampere’s Law It states that the line integral of B around any closed path or circuit is equal to m 0 times the total current crossing the area bounded by the closed path provided the electric field inside the loop remains constant. Thus,

ò B × dl = m 0 (inet ) Its simplified form is Bl = m 0inet This equation can be used only under following conditions

æ\ n = N ö ç ÷ è Lø

B = m 0ni

19.5 Solenoid A long straight coil of wire can be used to generate a nearly uniform magnetic field similar to that of a bar magnet. Such coils called solenoids have an enormous number of practical applications. The field can be strengthened by the addition of an iron core. Such cores are typical electromagnets.

(a) at every point of the closed path B || dl (b) magnetic field has the same magnitude B at all places on the closed path.

Applications of Ampere’s Law (i) Magnetic field due to a long metallic circular wire rod of radius R carrying a current i m i (a) If r < R, B = æç 0 2 ö÷ r, i. e, B µ r è 2pR ø (b) If r = R (i. e, at the surface) m i B = æç 0 ö÷ è 2pRø Magnetic field of a solenoid wounded in the form of a helix is

B = µnl

I

I

The magnetic field is concentrated into a nearly uniform field in the centre of a long solenoid. The field outside is weak and divergent. In the expression B = mnI , B is the magnetic field, n is the number of turns per unit length. The expression is an idealization to an infinite length solenoid, but provides a good approximation to the field of a long solenoid.

Magnetic Effect of Current Field due to Solenoid

B=

Taking a rectangular path about which to evaluate Ampere’s law such that the length of the side parallel to the solenoid field is L gives a contribution BL inside the coil. The field is essentially perpendicular to the sides of the path, giving negligible contribution. If the end is taken so far from the coil that the field is negligible, then the length inside the coil is the dominant contribution. Contribution Ampere’s law path

Dominant

869

m NI 2 pr

Toroid is a useful device used in everything from tape heads to marks.

Note Ampere’s law is valid only for steady current. Further more, it is a useful only for calculating the magnetic fields of current configurations with high degrees of symmetry, just as Gauss’s law is useful only for calculating the electric fields of highly symmetric charge distributions.

Sample Problem 8 A solenoid of length 0.4 m and diameter 0.6 m consists of a single layer of 1000 turns of fine wire carrying a current of 5 ´ 10 -3 A. Calculate the magnetic field on the axis at the middle and at the end of the solenoid. (a) 8.7 ´ 10 -6T , 6.28 ´ 10 -6T

B = µnl

(b) 6.28 ´ 10 -6T , 8.7 ´ 10 -6T

I

I

(c) 5.7 ´ 10 -6 T , 6.28 ´ 10 -6T

From Ampere’s law gives

(d) 8.7 ´ 10 -6T , 8.28 ´ 10 -6T

BL = m NI N B=m I L

Interpret (a) In case of solenoid the field at a point on the axis as shown in figure is given by B =

B = m nI This turns out to be a good approximation for the solenoid field, particularly in the case of an iron core solenoid.

α

At the centre of a long solenoid

m0 2pni (sin a + sin b) 4p

α

β

β

h

B

B

L/2

B = mnI α=β (b)

(a)

µ = kµ0

r

I

I

m 0 = 4 p ´ 10-7 T/A-m

β

C

B L

k = relative permeability α=0 ©

19.6 Magnetic Field of Toroid Finding the magnetic field inside a toroid is a good example of Ampere’s law. The current enclosed by the dashed line is just the number of loops times the current in each loop. Ampere’s law, then gives the magnetic field by I I

n=

So,

B = 10 -7 ´ 2p ´ 2.5 ´ 10 3 ´ 5 ´ 10 -3(sin a + sin b)

i. e,

B = 2.5p ´ 10 -6(sin a + sin b)

So, (a) when the point is at the middle on the axis a = b with L 4 sin a = = 2 2 7.2 L + 4r Then,

b B

a r

B × 2pr = m NI

B

N 1000 = = 2.5 ´ 10 3 turns m-1 L 0.4

Here,

B = 2.5p ´ 10 -6 ´ 2 ´

4 7.2

= 8.7 ´ 10 -6 T and (b) when the point is at the end on the axis a = 0 with L sin b = 2 L + r2 0.4 4 = = 2 2 5 (0.4) + (0.3)

870 JEE Main Physics Then,

B = 2.5p ´ 10 -6 ´ = 6.28 ´ 10

-6

4 5

T

19.8 Motion of Charged Particle in Magnetic Field The path of charged particle in uniform magnetic field depends on angle between v and B. Therefore, following cases are possible

Check Point 1 1. Looking at a circular coil, the current is found to be flowing in anticlockwise direction. Predict the direction of magnetic field produced at a point on the axis of the coil on the same side as the observer.

2. Consider the circuit as shown in figure, where APB and AQB are semicircles. What will be the magnetic field at the centre C of the circular loop?

The magnetic force is, F = Bqv sin 0° or

B C Q

3. What kind of magnetic field is produced due to straight solenoid?

19.7 Lorentz Forces 1. If a charge q is moving with velocity v enters in a region in which electric field E and magnetic field B both are present, it experiences force due to both fields simultaneously. The force experienced by the charged particle is given by the expression F = q(v ´ B) + qE Here, magnetic force Fm = q(v ´ B) = Bqv sin q and electric force Fe = qE.

The magnetic force is F = Bqv sin 90° = bqv. This magnetic force is perpendicular to the velocity at every instant. Hence, path is circle. The necessary centripetal force is provided by the magnetic force hence, if r be the radius of the circle, then mv2 = Bqv r mv or r= Bq This expression of r can be written in following different ways 2 qVm mv p 2 Km r= = = = Bq Bq Bq Bq Here, p = momentum of particle p2 or p = 2 Km K = Kinetic energy of particle = 2m Further, time period of the circular path will be æ mv ö 2p ç ÷ è Bq ø 2 pm 2 pr T = = = v v bq

2. The direction of magnetic force is same as v ´ B if charge is positive and opposite to v ´ B, if charge q is negative.

Cases (i) If v = 0, then F = 0 i. e, no force is exerted on a stationary charge, in a magnetic field. (ii) If q = 0, then F = 0 i. e, when the charge is moving parallel to the field then no force will be exerted by the field. (iii) If q = 90° , then sin q = sin 90° = 1 or

sin180° = 0

Hence, path of the charged particle is a straight line (undeviated) when it enters parallel or antiparallel to magnetic field.

Case II When q = 90°

P I A

Case I When q is 0° or 180°

F = qvB ´ 1 = qvB

i. e, when the charged particle is moving perpendicular to the field, the force exerted by the field will be maximum.

or

T =

2pm Bq

or the angular speed (w) of the particle is w = Bq m 1 Bq Frequency of rotation is, f = or f = T 2pm

\

w=

2 p Bq = T m

Magnetic Effect of Current v sin θ

Special Case of Motion of Charged Particle in Magnetic Field

r

θ v cos θ

If angle q is other than 0°, 180° or 90°, then velocity of charged particle can be resolved in two components one along B and another perpendicular to B. Let the two components be v || and v ^ . Then

p

T=

And as B

871

v sin θ

2pr 2p ´ 1.2 ´ 10 -2 = v sin q 4 ´ 10 5 ´ ( 3/2)

= 2.175 ´ 10 -7 s

v

v cos θ

p = v cos q ´ t

So, pitch

= 4 ´ 10 5 ´

q, m

v || = v cos q

1 ´ 2.175 ´ 10 -7 2

p = 4.35 ´ 10 -2 m = 4.35 cm

i. e,

v ^ = v sin q

and

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, mv ^ mv sin q = r= Bq Bq Time period and frequency do not depend on velocity and so they are given by 2pm Bq and f = T = Bq 2pm 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 2 pm Bq 2pmv cos q P = Bq

P = (v cos q)

or

19.9 Cyclotron Cyclotron is a device used for accelerating positively charged particle (like a-particles, deutrons etc.) by the help of uniform magnetic field upto energy of the order of MeV. It consists of two hollow metallic dees D1 and D2. These are placed in a uniform magnetic field perpendicular to the plane of dees. An alternating voltage is applied between the dees. The charged particle to be accelerated is produced at centre point S between the dees. The particle moves along circular path. The frequency of revolution of particle is made equal to the frequency of the alternating voltage source. This is called the condition of resonance in a cyclotron. The particle is accelerated twice in a revolution. If n is the number of revolutions, the energy gained by the particle, E K = n ×2 qV N Deflector

Sample Problem 9 A beam of protons with a velocity

4 ´ 10 5ms-1 enters a uniform magnetic field of 0.3 T at an angle of 60° to the magnetic field. Find the pitch of the helix (which is the distance travelled by a proton in the beam parallel to the magnetic field during one period of the rotation). Mass of the proton = 1.67 ´ 10 -27 kg (a) 2.35 cm (c) 4.35 cm

S Dess

D1 B

D2

(b) 5.35 cm (d) 6.35 cm

Interpret (c) When a charged particle is projected at an angle q to a magnetic field, the component of velocity parallel to the field is v cos q while perpendicular to the field is v sin q; so the particle will move in a circle of radius r=

Target

m(v sin q) 1.67 ´ 10 -27 ´ 4 ´ 10 5 ´ ( 3 / 2) = qB 1.6 ´ 10 -19 ´ 0.3 2 = ´ 10 -2 m = 1.2 cm 3

E High frequency oscillator

S

The accelerated beam finally comes out of a hole at the periphery of the dee to hit the target.

Cyclotron Frequency Time taken by ion to describe a semicircular path is given by t=

pr pm = v qB

é mv ù êQ r = ú qB û ë

872 JEE Main Physics If T = Time period of oscillating electric field, then 2pm T = 2t = qB \ Cyclotron frequency, n =

1 dB = T 2 pm

Maximum Energy of Particle Maximum energy gained by the charged particle æ q2B2 ö 2 Emax = ç ÷ r0 è 2m ø

19.10 Force on a Current Carrying Conductor in a Magnetic Field When a current carrying conductor is placed in a magnetic field, the conductor experience a force in a direction perpendicular to both the direction of magnetic field and the direction of current flowing in the conductor. i

where, r0 is maximum radius of the circular path followed by positive ion.

F B

Note 1. Cyclotron is suitable only for accelerating heavy particles like proton, deutron, a-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. As a result of it, the electrons go quickly out of step with oscillating electric field. 2. When a positive ion is accelerated by the cyclotron, it moves with greater and greater speed. As the speed of ion becomes comparable with that of light, the mass of the ion increase according to the relation m0 m = æ v2 ö çç1 - 2 ÷÷ è c ø where m0 = the rest mass of the ion m = the mass of the ion while moving velocity v and c = velocity of light Now, the time taken by the ion to describe semicircular path is m0 pm p t = = × qB qB æ v 2 ö çç1 - 2 ÷÷ è c ø It shows, that as v increases, t increases. It means, the positive ion will take longer time to describe semicircular path than the time for half-cycle of oscillating electric field. As a result of it, the ion will not arrive in the gap between the two dees exactly at the instant, the polarity of the two dees is reversed and hence, will not be accelerated further. Therefore, the ion cannot move with a speed beyond a certain limit in a cyclotron.

i

The direction of this force can be found out either by Fleming’s left hand rule or by right hand palm rule. The magnetic force is F = ilB sin q In vector form, where,

F = i (l ´ B)

B = intensity of magnetic field i = current in the conductor l = length of the conductor

and q = angle between the length of conductor and direction of magnetic field.

Cases (i) If q = 90° of sin q = 1, then F = ilB (maximum). Therefore, force will be maximum when the conductor carrying current is perpendicular to magnetic field. (ii) If q = 0° or sin q = 0 then, F = ilB ´ 0 = 0 Thus, the force will be zero, when the current carrying conductor is parallel to the field.

Magnetic Effect of Current

2

Hot Spot

873

Force on a Current Carrying Conductor in a Uniform Magnetic Field

The diagram shows two wires 1 and 2 kept parallel to each other at a distance r and carrying currents i 1 and i 2 respectively in the same direction. Magnetic field at wire 2 from current in wire 1 is B=

m 0 I1 2 pr

Interpret (c) From the figure, due to FABC, the magnetic field at O is along y-axis and due to CDEF, the magnetic field is along axis. Hence, the field will be of the form A( $i + $j) C

I2 I1 Electric current

B

r B

I

π/4 π/4

O

F F

X Z

B

Magnetic field

Force on length DL of wire 2 is

A

F

F = I2 DLB

Calculating the field due to FABC

Force per unit length in terms of the currents F m II = 0 12 DL 2 pr

Due to AB, BAB =

The direction is obtained from the right hand rule.

other, and they repel, if the currents are is opposite direction.

Due to BC,

BAB

loop of side l, but is folded in two equal parts so that half of it lies in the xz-plane. The origin O is centre of the frame also. The loop carries current i, the magnetic field at the centre is y

=

D

BAB =

C B

x

O

Þ F

m 0i $ i 2 2 pl

BFABC =

m 0i é 1 1 2ù $ + + ê ú i pl ë 2 2 2 2 2 û 2 m 0i $ i pl

Similarly, due to CDEP, we have

A

Z

m 0i $ $ (a) (i - j ) 2 2 pl

m i (b) 0 ( - $i + $j ) 4 pl

2 m 0i $ $ (i + j ) pl

m 0i $ $ (d) (i + j ) 2 pl

(c)

m 0i 2 2 pl

Hence, the field due to FABC, BFABC =

E

m 0i 2 pl m 0i (sin 0° + sin 45° ) = ælö 4 pl ç ÷ è2ø

= 2

Note Two wires carrying current in the same direction attract each

Sample Problem 10 In the figure, ABCDEFA was a square

m 0i (sin 45° + sin 45° ) $i ælö 4p ç ÷ è2ø

BCDEF = \

Bnet =

2 m 0i $ j pl 2 m 0i $ $ (i + j ) pl

874 JEE Main Physics Sample Problem 11 The length of a conductor ab

Interpret (a) Let a small element dy at a distance y from the

carrying current I2 is l. The force acting on it due to a long current carrying conductor is

long conductor carrying current be taken force on this element is I1

I1 a

I2

x

I2 dy

a

l b

b

y x x+l

m 0l x+ l log e 2p x m 0I x-2 (c) log e 4p x

m 0l x log e p x+ l m I x-2 (d) 0 log e 4p x

(a)

Sample Problem 12 A straight wire of length 30 cm and mass 60 mg lies in a direction 30° east of north. The earth’s magnetic field at this site is horizontal and has a magnitude of 0.8 G. What current must be passed through the wire so that it may float in air? (a) 10 A

dF =

(b)

(b) 20 A

(c) 40 A

(d) 50 A

Interpret (d) As shown in figure, if a current i is passed through the wire from end A towards B it will experience a force BiL sin q vertically up and hence, will float if B F

i B

A

\

F=

Bil sin q = mg i. e,

Sample Problem 13 The horizontal component of the

earth’s magnetic field at a certain place is 3 ´ 10 -5 T and the direction of the field is from the geographic south to the geographic north. A very long straight conductor is carrying a steady current of 1 A. The force per unit length on it when it is placed on a horizontal table and the direction of the current is east to west is (a) 3 ´ 10 -5 Nm-1

(b) 6 ´ 10 -3 Nm-1

(c) 9 ´ 10 -2 Nm-1

(d) 12 ´ 10 -6 Nm-1

m 0I 2p

x+l

òx

Rules to Find the Direction of Force (i) Right hand palm rule If we stretch the right hand palm such that the fingers and the thumb are mutually perpendicular to each other and the fingers point in the direction of magnetic field and thumb points in the direction of motion of positive charge, the direction of force will be along the outward normal on the palm. Force F

i

Field B

When the current is flowing from east to west, q = 90°, hence, fI × B = 1 ´ 3 ´ 10 -5 = 3 ´ 10 -5 Nm-1

Current or motion of positive charge

(ii) Fleming’s left hand rule If we spread the forefinger, central finger and thumb of our left hand in such a way that these three are perpendicular to each other then, if first forefinger is in the direction of magnetic field, second central finger is in the direction of current, then thumb will represent the direction of force. Force F

Interpret (a) As, F = Il ´ B or, F = IlB sin q The force per unit length is, F f = = IB sin q l

dy m I x+l = 0 log e y 2p x

This is larger than the value 2 ´ 10 -7 Nm-1 quoted in the definition of ampere. Hence, it is important to eliminate the effect of the earth’s magnetic field and other stray fields while standardising the ampere. The direction of the force is downwards. This direction may be obtained by the directional property of cross product of vectors.

mg

60 ´ 10 -6 ´ 10 mg i= = 50 A = BL sin q 0.8 ´ 10 -4 ´ 30 ´ 10 -2 ´ 1 / 2)

m 0 Idy 2 py

Field B Current or motion of positive charge v

Note To learn this rule, remember the sequence of Father, Mother, Child. Thumb ® Father ® Force Forefinger ® Mother ® Magnetic field Central finger ® Child ® Current of direction of positive charge

Magnetic Effect of Current

F Mg = L L m 0 2i1i2 Mg é dF m 0 2i1i2 ù as = = 4p d L êë dL 4p d úû 2 ´ 50 ´ 25 1 d = 10 -7 ´ = ´ 10 -2 m 3 0.075 é Mg -1 ù êë as, L = 0.075 Nm úû

Sample Problem 14 A straight wire of mass 200g and length 1.5 m carries a current of 2 A. It is suspended in mid air by a uniform horizontal magnetic field B. The magnitude of the magnetic field is [NCERT]

mg

F = Mg , i. e. ,

or or

I ×B l

(a) 0.35 T

(b) 0.65 T

(c) 0.25 T

(d) 0.88 T

Interpret (b) Applying Flaming’s rule, we find that upward force F of magnitude IlB acts. For mid-air suspension this must be balanced by the force due to gravity. mg \ mg = I lB Þ B = Il Given,

we have,

m = 200 g = 0.2 kg, g = 9.8 m/s 2 I = 2A, l = 1.5 m 0.2 ´ 9.8 B= = 0.65 T 2 ´ 1.5

Note The earth’s magnetic field is approximately 4 ´ 10-5 T and it has been ignored.

Sample Problem 15 A long horizontal wire P carries a current of 50 A. It is rigidly fixed. Another fine wire Q is placed directly above and parallel to P. The weight of wire Q is 0.075 Nm -1 and carries a current of 25 A. Find the positive of wire Q from P so that wire Q remains suspended due to the magnetic repulsion. Also indicate the direction of current in Q with respect to P. 1 ´ 10 -2 m 2 1 (c) ´ 10 -2 m 4 (a)

(b) (d)

1 ´ 10 -2 m 3

1 ´ 10 -2 m 5

Interpret (b) As force per unit length between two parallel current carrying wires separated by a distance d is given by dF m 0 2i1i2 = dL 4p d i2

875

F Q Mg

d P i2

and is repulsive if the current in the wires is in opposite direction (otherwise attractive). So, in order that wire Q may remain suspended, the force F on it must be repulsive and equal to its weight, i. e, the current in the two wires must be opposite directions and

19.11 The Moving Coil Galvanometer (MCG) Current Sensitivity It is a device whose principle is based on the torque on a current carrying loop. The MCG consists of a multi-turn coil free to rotate about a vertical axis, in a uniform radial magnetic field. There is a cylindrical soft iron core to increase the sensitivity of the MCG. When a current flows through the coil, a torque acts on it. This torque is given by t = NiAB. Torsion head Phosphor Bronze strip Concave mirror Coil

Soft Iron core A

T2

N

D S

B

C

T1

Hair spring Levelling Screw

where, the symbols have their usual meaning, since the field is radial by design, we have taken sin q » 1 in the above expression for the torque. A spring S0 provides a counter torque resulting in a steady angular deflections f. In equilibrium, kf = NiAB where, k is the torsional constant of the spring. The deflection f is indicated on the scale by a pointer attached to the spring. æ NAB ö f=ç ÷i è k ø f NAB The current sensitivity, = i k If a pulse of current carrying charge Dq is passed through the galvanometer, then angular impulse applied by the NABDq magnetic field on the current loop = ò t dt = , which k would provide an instantaneous angular velocity about the pivot to the coil.

876 JEE Main Physics Conversion of Galvanometer to 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. From the figure, we note that at G full scale deflection the total ig current through the parallel S i – +a combination is i, the current b i – ig through the galvanometer is ig and current through shut S i - ig . The potential difference Vab = (Va - Vb ) is the same for both the paths, hence igG = (i - ig ) × S

1. Percentage error in measuring the potential difference by a voltmeter is ö ö æ æ æV - V ¢ ö ´ 100 = ç 1 ÷ ´ 100 , % error = ç 1 ÷ ´ 100 ÷ ÷ ç ç ÷ ç è V ø ç1 + r ÷ ç1 + r ÷ è è RV ø RV ø 2. Resistance of voltmeter RV = R + G.

Sample Problem 16 A moving coil galvanometer has 100 turns and each turns has an area of 2 cm 2. The magnetic field produced by the magnet is 0.01 T. The deflection in the galvanometer coil is 0.05 rad when a current of 10 mA is passed through it. Find the torsional constant of the spiral spring. (a) 3.0 ´ 10 -4 Nm rad -1 (c) 5 ´ 10

-6

(i) Percentage error in measuring a current through an ammeter is æ1 - 1 ö ÷ ç i i ¢ æ ö ´ 100 = ç R R + A ÷ ´ 100 ç ÷ 1 è i ø ÷ ç ÷ ç R ø è æ A ö % error = ç ÷ ´ 100 è R + Aø GS G+S

A voltage measuring device is called a voltmeter. It measures the potential difference between two points. A galvanometer can be converted into 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 is to be measured.

(d) 7 ´ 10 -7 Nm rad -1 k q NAB

k=

NABi q

or

k=

100 ´ 2 ´ 10 -4 ´ 0.01 ´ 10 ´ 10 -3 0.05

\

k = 4.0 ´ 10 -5 Nmrad -1

Sample Problem 17 A current of 5.0 A is passed through the coil of a galvanometer having 500 turns and each turn has an average area of 3 ´ 10 -4 m 2. If a torque of 1.5 Nm is required for this coil carrying same current to set it parallel to a magnetic field, calculate the strength of the magnetic field. (a) 20 T

Conversion of Galvanometer to Voltmeter

(b) 4 ´ 10 -5 Nm rad -1

Þ

Note

(ii) Resistance of ammeter A =

Nm rad

-1

Interpret (b) We have, i =

æ ig ö ÷÷G S = çç è i - ig ø

Þ

Note

(b) 25 T

(c) 23 T

(d) 21 T

Interpret (a) The magnetic moment of a current loop M = NiA = 500 ´ 0.5 ´ 3 ´ 10 -4 = 0.075 Am2. Also, t = M ´ B of | t| = MB sin q where, q = angle between B and A. Here, q = 90° \ t = MB sin 90° ; t 1.5 B= = = 20 T M 0.075

G

Sample Problem 18 In the given

R ig

ig + i a

– Circuit element

i b

V

Fog a voltmeter with full scale reading V , we need a series resistor R such that or

V = ig (G + R) V R= -G ig

circuit, the current is to be measured. The value of the current if the ammeter shown is a galvanometer with a resistance Rg = 60 W is [NCERT] (a) 0.99 A (c) 0.02 A

A 3Ω 3V

(b) 0.048 A (d) 0.06 A

Interpret (b) Total resistance in the circuit is R = RG + 3 Given, RG = 60 W, therefore R = 60 + 3 = 63 W V 3 From Ohm’s is law I = = = 0.048 A R 63

Magnetic Effect of Current So,

Check Point 2 1. If a charged particle is deflected either by an electric or a magnetic field, how can we as certain the nature of the field?

2. If an electron is not deflected in passing through a certain region of space, can we be sure that thee is no magnetic field in that space. What possible conclusions could be drawn regarding the existence of electric and magnetic fields?

3. Why should a solenoid tends to contract, when a current

F=

ò IdL ´ B = I[ò dL ´ B]

Here it must be kept in mind that in this situation different parts of the loop may experience elemental force due to which the loop may be under tension or may experience a torque as shown in figure.

×

×

×

passes through it? I

5. A rectangular current loop is in an arbitrary orientation in an

Applications of a Current Carrying Conductor 1. As the force BI dL sinq is not a function of position r, the magnetic force on a current element is non-central [a central force is of the form F = Kf (r )n r ] 2. The force dF is always perpendicular to B and idL to each other. 3. In case of current-carrying conductor in a magnetic field if the field is uniform i.e., B = constant.,

ò IdL ´ B = IL ´ B and as for a conductor ò dL represents the vector sum F=

of all the length elements from initial to final point which in accordance with the law of vector addition is equal to the length vector L¢ joinitial to final point, so a current-carrying conductor of any arbitrary shape in a uniform field experiences a force F = i [ ò dL ´ B = IL ´ B

× F=0 t=0

…(i)

where L¢ is the length vector joining initial and final points of the conductor as shown in figure.

Fin B

Fout

×

dF = 0

×

×

external magnetic field. Is any work is required to rotate the loop about an axis perpendicular to its plane?

…(ii)

i.e., the net magnetic force* on a current loop in a uniform magnetic field is always zero as shown in figure.

4. Why earth’s magnetic field does not affect the working of a moving coil galvanometer?

877

dF = 0

× Current loop in a uniform feild

F=0 t=0

5. If a current- carrying conductor is situated in a non-uniform field, its different elements will experience different forces; so in this situation, FR ¹ 0 but t may or may not be zero

…(iii)

6. The net force on a current-carrying conductor due to its own field is zero; so if there are two long parallel current-carrying wires 1 and 2 as shown in figure. Wire-1will be in the field of wire-2 and vice-versa. So, force on length of wire-2 due to field of wire-1, dF2 = I2dL2B1 = or

m 0 2 I1 I2 dL2 4p d

é asB = m 0 2 I1 ù 1 êë 4p d úû

dF2 m 0 2 I1 I2 = dL2 4p d

Same will be true for wire-1 in the field of wier-2. The direction of force in accordance with Fleming' s left hand rule will be as shown in figure. So, force per unit length in case of two parallel current-carryingwires separated by a distance d is given by

×

×

×

B

×

×

×

and the force between the wires is attractive if the current in them is in the same direction, otherwise repulsive [ this is opposite to that of what happens between two charges].

×

×

×

×

×

×

Note Through this concept the SI unit current, ampere, is defined as the

L′ A

×

×

×

×

×

×

4. If the current-carrying conductor in the form or a loop of any arbitrary shape is placed in a uniform field, F=

ò IdL ´ B = I[ò dL ´ B]

and as for a closed loop, the vector sum of dL is always zero.

current which when passed through each other, produces between them a force of 2 ´ 10 -7 newton for one metre of their length. 7. In case of a current-carrying conductor in a magnetic field if the conductor experiences a force and is free to move, work will be done and hence its kinetic energy or speed will change, i.e,. W = DKE with W = ò F . ds

WORKED OUT Examples Example 1

A flat dielectric disc of radius R carries an excess charge on its surface. The surface charge density is s. The disc rotates about an axis prependicular to its plane passing through the centre with angular velocity w. Find the torque on the disc if it is placed in a uniform magnetic field B directed perpendicular to the rotation axis. swpBR3 3 swp (c) 4BR 4

swpBR 4 4 swp (d) 5BR5

(b)

(a)

Solution

Consider an angular ring of radius r and of thickness dr on this disc. Charge within this ring. dq = (s ) (2prdr) i = swrdr

or

Solution

So,

As the field due to an arc at the centre is given by m iq B= 4p r m i q m i (2p - q) B0 = 0 1 + 0 2 Ä r 4p r 4p

But as, (VA - VB) = i1R1 = i2 R2 iR i l i2 = 1 1 = 1 1 i. e. , l2 R2 or

i2 = i1

So,

B0 =

q (2p - q )

(as l = rq)

m 0i1q m 0i1pq + Ä=0 4 pr 4 pr

Example 3

In the figure, a charged sphere of mass m and charge q starts sliding from rest on a vertical fixed circular track of radius R from the position shown. There exists a uniform and constant horizontal magentic field of induction B. The maximum force exerted by the track on the sphere is

ω

B dr

m q

r

θ ×

B

Magnetic moment of this annual ring, M = iA = (swrdr) ( pr 2) (along the axis of rotation). Torque on this ring, dt = MB sin 90° = (swpr3B)dr \Total torque on the disc is,

(a) mg

(b) 3mg - qB 2gR

(c) 3mg + qB 2gR

(d) mg - qB 2gR

Solution

R

swpBR 4 t = ò d t = (swpB) ò r dr = 0 4 0

Magnetic force, Fm = qvB and directed radially inward.

3

N- mg sin q + qvB =

Example 2

Magnetic field intensity B at the cetnre of the circular loop is i (a) zero m (2p - q )i (b) 0 4 pR m 0iq (c) 4 pR m 0i 2( p - q ) (d) 4 pR

Þ

2

At O R i

i i1

mv 2 R

mv 2 + mg sin q - qvB R p q= 2 2mgR Mmax = + mg - qB 2gR R N=

= 3mg - qB 2gR

Magnetic Effect of Current = 100 ´ 0.5(0.08) ´ 0.04i = 16 ´ 10 -2 ( $i) Am2

Example 4

A straight conductor of mass m and carrying a current i is hinged at one end and placed in a plane perpendicular to the magentic field of intensity B as shown in the figure. At any moment if the conductor is left free, then the angular acceleration of the conductor will be (Assume gravity free region)

879

The torque acting on the coil is t = M ´ B = MB( $i ´ $j) 0.05 $ k 2 = 5.66 ´ 10 -5 (Nm) k$

= 1.6 ´ 10 -2 ´

× × × ×B × × × ×i

Example 6

The magnetic moment of an electron orbiting in a circular orbit of radius r with a speed v is equal to

× × × ×

(a) evr/2 (c) er/2v

× × × × l

2iB 3m 3B (c) 2m

3iB 2m 3i (d) 3mB

(b)

(a)

Solution

(b) evr (d) None of these

Magnetic moment M = NiA

where, N = number of turns of the current loop and i = current

Solution

The force acting on the elementary portion of the current carrying conductor is given as, dF = i (dr) B sin 90° Þ dF = iB dr

Since, the orbiting electron behaves as a current loop of current i, we can write

dF

i=

O r

The torque applied by dr about O = dt = rdF. Þ The total torque about O = t = ò dF = ò r (iBdr) l

Þ

t = iBò rdr = 0

iBl 2 2

In the figure shown, the magnetic field at the

point P is y

The angular acceleration, a = Þ Þ

2 4 3

3iB a= 2m z

Example 5

The rectangular coil having 100 turns is turned in 0.05 $ a uniform magnetic field of j as shown in the figure. The 2 torque acting on the loop is

i

P

x

a/2 3a/2

5

m 0i 4 + p2 3 pa 2m 0i (d) ( 4 - p 2) 3 pa

2m 0i 4 - p2 3 pa 2m 0i (c) ( 4 + p 2) 3 pa (a)

Solution

Z

e e ev = = T 2 pr 2 pr v

A = area of the loop = pr 2 evr æ ev ö 2 M = (1) ç Þ ÷ ( pr ) = è 2 pr ø 2

Example 7

t (where, MI = moment of inertia) l æ ibl 2 ö æ ml 2 ö a=ç ÷ ÷/ç è 2 ø è 3 ø

v

(b)

BP = (B1)P + (B2)P + (B3)P + (B4)P + (B5)P y

0.8 m i = 0.5 A

(a) 11.32 ´ 10 -4 (Nm) k$ (c) 5.66 ´ 10 -5 (Nm) k$

Solution

2 4

Y

–0.04 m

3

(b) 22.64 ´ 10 -4 (Nm) k$ (d) zero

The magnetic dipole moment of the current carrying coil is given by M = NiAn$

z

where,

(B1) =

1

P

a/2 3a/2

m 0i ( - $j) æ 3a ö 4p ç ÷ è2ø

x 5

(semi-finite wire)

880 JEE Main Physics m 0i ( + k$ ), (B3)P = 0 4(3a/ 2) m 0i (B4)P = ( - k$ ) 4(3a/ 2) m 0i (B5)P = ( - $j) æ aö 4p ç ÷ è2ø

Solution

(B2)P =

Þ

The gravitational torque must be counter balanced by the magnetic about O, for equilibrium of the sphere. The gravitational torque = tgr = mg ´ r sin q Þ

tgr = mgr sin q

The magnetic torque, t m = M ´ B where the magnetic moment of the coil M = (ipr 2)

Bnet = (B4 - B2) 2 + (B5 - B1) 2 =

m 0i 3 pa

p2 + 4

θ r O mg

Example 8

An electron moves straight inside a charged parallel plate capacitor of uniform surface charge density s. The space between the plates is filled with constant magenetic field of induction B, as shown in the figure. Neglecting gravity, the time of straight line motion of the electron in the capacitor is +σ ×

×

×

em ×

×

×

×

×

×

s e o lB

Solution

(b)

e 0 lB s

(c)

s e oB

t m = pir 2B sin q

Þ

pir 2B sin q = mgr sin q

Example 10

(d)

e0 B s

A long straight wire along the z-axis carries a current i in the negative z-direction. The magnetic vector field B at a point having coordinates ( x, y) in the z = 0 plane is (a)

m 0 (y$i - x$j) 2p ( x2 + y 2)

(b)

m 0i ( x$i - y$j) 2p ( x2 + y 2)

(c)

m 0i ( x$j + y$i) 2p ( x2 + y 2)

(d)

m 0i ( x$i - y$j) 2p ( x2 + y 2)

The net electric field s s s + = E = E1 + E2 Þ E = 2e 0 2e 0 e 0

The net force acting on the electron is zero because it moves with cosntant velocity Fnet = Fe + Fm = 0 | Fe | = | Fm | or eE = evB E s v= = B e 0B

Þ or

B = mg/pir

or

l

(a)

Þ

\The time of motion inside the capacitor = t =

Solution l e 0 lB . = v s

Magnetic field| B| =

m 0i x2 + y 2

2p

Unit vector perpendicular to the position vector is

Example 9 In the figure shown, a coil of single turn is wound on a sphere of radius r and mass m. The plane of the coil is parallel to the inclined plane and lies in the equatorial plane of the sphere. If the sphere is in rotational equilibrium, the value of B is (current in the coil is i)

P

E B

a

O

2a

Q

(yi - x$j) x2 + y 2

O B

(a)

mg pir

(b)

θ

mg sin q mgr sin q (c) (d) None of these pi pi

;

\

m 0i (y$i - x$j) 2p ( x2 + y 2)

Start Practice for

JEE Main Round I

(Topically Divided Problems)

Biot-Savart Law and Magnetic Field

4. Biot-Savart law indicates that the moving electrons

1. Two charged particles traverse identical helical paths in a completely opposite sense in a uniform $. magnetic field B = B0 k [NCERT Exemplar] (a) They have equal z-components of momenta (b) They must have equal charges (c) They necessarily represent a particle-antiparticle pair  e  e (d) The charge to mass ratio satisfy :   +   = 0  m  1  m 2

2. A horizontal overhead power line carries a current of 90 A in east to west direction. What are the magnitude and direction of the magnetic field due to the current 1.5 m below the line? [NCERT]

(velocity v) produce a magnetic field B such that [NCERT Exemplar]

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

B⊥v B || v it obeys inverse cube law it is along the line joining the electron and point of observation

5. A pair of stationary and infinite long bent wires are placed in the xy-plane. The wires carrying currents of 10 A each as shown in figure. The segments L and M are parallel to x-axis. The segments P and Q are parallel to y-axis, such that OS = OR = 0.02 m. The magnetic field induction at the origin O is y

(a) 1.2 × 10 −5 T, perpendicularly outward to the plane of paper (b) 1.9 × 10 −5 T, perpendicularly outward to the plane of paper (c) 2.6 × 10 −5 T, perpendicularly inward to the plane of paper (d) 2.6 × 10 −5 T, perpendicularly inward to the plane of paper

3. Which of the following graph represents the variation of magnetic flux density B with distance r for a straight long wire carrying an electric current ? B

(b)

r

r

B

i Q L i

R P

M∞ O S

x

i ∞

(a) 10 −3 T (c) 2 × 10

−4

(b) 4 × 10 −3 T T

(d) 10 −4 T

6. A uniform electric and magnetic fields are produced pointing in the same direction. If an electron is projected with its velocity pointing in the same direction. [NCERT Exemplar]

B

(a)

∞

B

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

The electron velocity will decrease in magnitude The electron velocity will increased in magnitude neither (a) nor (b) None of the above

7. Two long and parallel straight wires A and B (c)

carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A?

(d)

r

r

(a) 1.5 × 10 −5 N

(b) 2 × 10 −5 N

(c) 4 × 10 −5 N

(d) 3.2 × 10 −5 N

882 JEE Main Physics 8. A length l of wire carries a steady current i. It is bent

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

one-third of its value unaltered three times of its initial value nine times of its initial value

0 1 2 3 4 5 6 7 8 9101112 1314 15

(a) zero mark (c) 3 cm mark

a current through a loop as shown in figure (θ < 180° )

turns and radius r which carries a current i is measured on the axis of the coil at a small distance h from the centre of the coil. This is smaller than the magnetic field at the centre by the fraction 2

2

A θ

O

2

(a) (2 / 3) r / h

(b) (3 / 2) r / h

(c) (2 / 3) h2 / r 2

(d) (3 / 2) h2 / r 2

10. The magnetic field of the earth can be modelled by that of a point dipole placed at the centre of the earth. The dipole axis makes an angle of 11.3° with the axis of the earth. At mumbai declination is nearly zero. Then [NCERT Exemplar] (a) the declination varies between 11.3° W to 11.3° E (b) the least declination is 0° (c) the plane defined by dipole and the earth axis posses through greenwich (d) declination average over the earth must be always negative

11. An element, dl = dx ^i (where dx = 1 cm) is placed at the origin and carries a large current i = 10 A. What is the magnetic field on the y-axis at a distance of 0.5 m? (a) 2 × 10 −8 k$ T (b) 4 × 10 −8 k$ T

(c) −2 × 10 −8 k$ T (d) −4 × 10 −8 k$ T

B

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

zero perpendicular to paper inwards perpendicular to paper outwards perpendicular to paper outwards if 90 ° ≤ θ < 180 °

15. A current carrying circular loop of radius R is placed in the x− y plane with centre at the origin. Half of the loop with x > 0 is now bent so that it now lies in the [NCERT Exemplar] y− z plane. (a) (b) (c) (d)

The magnitude of magnetic moment now diminishes The magnetic moment does not change The magnitude of B at (0.0 z), z>>R increases The magnitude of B at (0.0.z), z>>R is unchanged

16. Two parallel long wires A and B carry currents i1 and

i2 (< i1). When i1 and i2 are in the same direction, the magnetic field at a point mid way between the wires is 10µT. If i2 is reversed, the field becomes 30 µT. The ratio, i1/ i2 is (a) 1

12. A circular coil A of radius r carries current i. Another circular coil B of radius 2r carries current of i. The magnetic fields at the centres of the circular coils are in the ratio of (a) 3 : 1 (c) 1 : 1

(b) 9 cm mark (d) 7 cm mark

14. Net magnetic field at the centre of the circle O due to

9. The magnetic field normal to the plane of a wire of n

2

3i

i

first to form a circular plane coil of one turn. The same length is now bent more sharply to give three loops of smaller radius. The magnetic field at the centre caused by the same current is

(b) 4 : 1 (d) 2 : 1

Magnetic Field due to Various Current Carrying Conductors 13. Two parallel long straight conductors are placed at right angle to the meter scale at the 2 cm and 6 cm marks as shown in the figure. If they carry currents i and 3i respectively in the same direction, then they will produce zero magnetic field at

(b) 2

(c) 3

(d) 4

17. Two straight long conductors AOB and COD are perpendicular to each other and carry currents i1 and i2 . The magnitude of the magnetic induction at a point P at a distance a from the point O in a direction perpendicular to the plane ABCD, is µ0 ( i1 + i2 ) 2πa µ (c) 0 ( i21 + i22 )1/2 2πa (a)

µ0 ( i1 − i2 ) 2πa i1i2 µ (d) 0 2πa ( i1 + i2 )

(b)

18. Two wires PQ and QR, carry equal currents i as shown in figure. One end of both the wires extends to infinity ∠PQR = θ. The magnitude of the magnetic field at O on the bisector angle of these two wires at a distance r from point Q, is

P O

i

Q

θ i

R

Magnetic Effects of Current µ0 4π µ0 (c) 4π

(a)

i θ sin r 2 i θ tan r 2

µ0 i cot θ 4π r µ i (1 + cos θ/2) (d) 0 2π r (sin θ/2)

segments of a circular loop in the A direction shown in figure. Radius of the loop is r. The magnitude of magnetic field induction at the centre of the loop is

windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near [NCERT] its centre.

i C

(a) 1.5 × 10 −2 T, opposite to the axis of solenoid B

θ

r

(d) 1.5 × 10 −2 T, opposite to the axis of solenoid

24. Let the magnetic field on the earth be modelled by

D

(b)

µ0 i (π − θ) 2π r

20. In the given diagram two

i

Y

that of a point magnetic dipole at the centre of the earth. The angle of dip at a point the geographical [NCERT Exemplar] equator i

long parallel wires carry equal currents in O X x = +d opposite direction. Point x = –d O is situated midway Z 2d between the wires and the XY -plane contains the two wires and the positive Z-axis comes normally out of the plane of paper. The magnetic field, B at O is non-zero along (a) X, Y and Z-axes (c) Y-axis

(b) X-axis (d) Z-axis

the axis of a current carrying long solenoid. Which of the following is true? [NCERT Exemplar] (a) The electron will be accelerated along the axis (b) The electron path will be circular about the axis (c) The electron will experience a force at 45° to the axis and hence execute a helical path (d) The electron will continue to move with uniform velocity along the axis of the solenoid

Ampere’s Circuital Law and Solenoid 22. The magnetic flux density B at a distance r from a long straight rod carrying a steady current varies with r as shown in figure.

r

B

B

(c)

(d) r

25. A long straight, solid metal wire of radius 2 mm carries a current uniformly distributed over its circular cross-section. The magnetic field induction at a distance 2 mm from its axis is B. Then, the magnetic field induction at distance 1 mm from axis will be

r

(b) B/ 2 (d) B

26. A current of i ampere flows along an infinitely long straight thin walled tube, then the magnetic induction at any point inside the tube is (a) infinite µ 2i (c) 0 T 4 πr

(b) zero µ i (d) 0 T 2r

Forces on Charged Particle in Electric and Magnetic Fields 27. A circular current loop of magnetic moment M is in an arbitrary orientation in an external magnetic field B. The work done to rotate the loop by 30° about an axis perpendicular to its plane is [NCERT Exemplar]

(c)

(b) r

is always zero can be zero at specific points can be positive or negative is bounded

(a) MB

B

(a)

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

(a) B (c) 2 B

21. An electron is projected with uniform velocity along

B

(b) 2.5 × 10 −2 T, along the axis of solenoid (c) 3.5 × 10 −2 T, along the axis of solenoid

O

µ 0 iθ 3πr µ0 i (d) (2π − θ ) 2π r

(a) zero (c)

23. A closely wound solenoid 80 cm long has 5 layers of

(b)

19. Equal current i flows in two

883

MB 2

(b)

3

MB 2

(d) zero

28. A deutron of kinetic energy 50 keV is describing a circular orbit of radius 0.5 m, is plane perpendicular to magnetic field B. The kinetic energy of proton that describes a circular orbit of radius 0.5 m in the same plane with the same magnetic field B, is (a) 200 keV (c) 100 keV

(b) 50 keV (d) 25 keV

884 JEE Main Physics 29. A proton, a deutron and an α-particle enter a

35. Two charged particles M and N enter a space of

magnetic field perpendicular to field with same velocity. What is the ratio of the radii of circular paths?

uniform magnetic field, with velocities perpendicular to the magnetic field. The paths are as shown in figure. The possible reason (s) is/are

(a) 1 : 2 : 2 (c) 1 : 1 : 2

(b) 2 : 1 : 1 (d) 1 : 2: 1

30. A particle of charge q and mass m starts moving from the origin under the action of an electric field, E = E0 ^i and B = B0 ^i with a velocity, v = v0 ^j. The speed of the 5 particle will becomes v0 after a time 2 (a) (c)

mv 0 qE 3mv 0 2qE

(b) (d)

mv 0 2qE 5 mv 0 2qE

31. A proton, a deutron and an α-particle with the same kinetic energy enter a region of uniform magnetic field, moving at right angle to B. What is the ratio of the radius of their circular paths ? (a) 1 : 2 : 1 2 : 1: 1

(d)

2 : 2 :1

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

the charge of M is greater than that of N the momentum of M is greater than that of N specific charge of M is greater than that of N the speed of M is greater than that of N

36. A particle of mass, m and charge, q is placed at a rest in a uniform electric field, E and then released. The kinetic energy attained by the particle after moving a distance, y is (a) q Ey2

(b) q E2 y

(d) q2 Ey

(c) q Ey

37. A beam of protons is moving parallel to a beam of electrons. Both the beams will tend to (a) repel each other (c) move more apart

(b) 1 : 2 : 2 (c)

N

M

32. A proton of mass 1.67 × 10−27 kg and charge

1.6 × 10−19 C is projected with a speed of 2 × 106 ms −1 at an angle of 60° to the x-axis. If a uniform magnetic field of 0.104 T is applied along y-axis, the path of proton is (a) a circle of radius = 0.2 mand time period = 2π × 10 −7 s

(b) a circle of radius = 0.1 mand time period = 2π × 10 −7 s

(b) come closer (d) either (b) or (c)

Force and Torque on a Current Carrying Conductor/Coil in a Magnetic Field 38. A conducting rod of length, l and mass, m is moving down a smooth inclined plane of inclination, θ with constant speed, v. A vertically upward magnetic field, B exists in space there. The magnitude of magnetic field, B is

(c) a helix of radius 0.1 m and time period = 2π × 10 −7 s

P B

(d) a helix of radius 0.2 m and time period = 2π × 10 −7 s

33. An electron and a proton enter a magnetic field

v

perpendicularly. Both have same kinetic energy. Which of the following is true ? (a) (b) (c) (d)

Trajectory of electron is less curved Trajectory of proton is less curved Both trajectories are equally curved Both more on straight line path

34. A uniform magnetic field, B = B0 ^j exists in space. A particle of mass m and charge, q is projected towards x-axis with speed, v from a point ( a, 0, 0). The maximum value of v for which the particle does not hit the yz-plane is Bqa m Bq (c) am

(a)

Bqa 2m Bq (d) 2am (b)

Q

(a)

θ

mg mg mg sin θ (b) cos θ (c) tan θ il il il

39. A current i1 carrying wire AB is

placed near an another long wire CD carrying current i2 as shown i in figure. If free to move, wire AB 2 will have (a) (b) (c) (d)

(d)

mg il sin θ

D

A

C rotational motion only translational motion only rotational as well as translational motion neither rotational nor translational motion

i1

B

Magnetic Effects of Current

885

40. A wire of length, l is bent in the form of circular coil of

45. A candidate connects a moving coil ammeter A and a

some turns. A current, i flows through the coil. The coil is placed in a uniform magnetic field, B. The maximum torque on the coil can be

moving coil voltmeter, V and a resistance, R as shown in figure. If, the voltmeter reads 20 V and the ammeter reads 4 A, then R is

(a)

iBl2 2π

(b)

iBl2 4π

(c)

iBl2 π

(d)

2iBl2 π

41. A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil? [NCERT Exemplar] (a) (b) (c) (d)

0.96 N-m 2.06 N-m 0.23 N-m 1.36 N-m

µ 0 iqv π r

3 µ 0 iqv 2π r µ 0 iqv (d) 2π r (b)

43. A metal wire of mass m slides without friction on two rails placed at a distance l apart. The track lies in a uniform vertical magnetic field B. A constant current i flows along the rails across the wire and brack down the other rail. The acceleration of the wire is B mi l Bil (c) m (a)

(b) mBi l

equal to 5 Ω greater than 5 Ω less than 5 Ω greater or less than 5 Ω depending upon its material

46. A voltmeter has resistance of 2000 Ω and it can measure upto 2 V. If we want to increase its range by 8 V, then required resistance in series will be (b) 6000 Ω (d) 8000 Ω

deflection for a current of 10−5 A. To convert it into a ammeter capable of measuring upto 1 A, we should connect a resistance of

deflection, when a current of 0.005 A is passed through its coil. It is converted into a voltmeter reading upto 5 V by using an external resistance of 975 Ω. What is the resistance of the galvanometer coil? (b) 25 Ω (d) 40 Ω

(b) 10 −3 Ω in parallel (d) 100 Ω in series

(a) 1 Ω in parallel (c) 105 Ω in series

48. A microammeter has a resistance of 100 Ω and full scale range of 50µA. 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 combinations (a) (b) (c) (d)

50 V range with 10 kΩ resistance in series 10 V range with 200 kΩ resistance in series 10 mA range with 1 Ω resistance in parallel 10 mA range with 0.1 Ω resistance in parallel

49. A candidate connects a moving coil voltmeter V and a moving coil ammeter A and resistor R as shown in figure ? If the voltmeter reads 10 V and the ammeter reads 2 A, then R is +

mil (d) B

44. A moving coil galvanometer gives full scale

(a) 30 Ω (c) 50 Ω

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

RΩ

47. A galvanometer of resistance 100 Ω gives a full scale

and 2i in opposite directions. The distance between the wires is r. At a certain instant of time a point charge, q is at a point equidistant from the two wires in the plane of the wires. Its instantaneous velocity v is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is

(c)

A

(a) 4000 Ω (c) 7000 Ω

42. Two very long straight parallel wires carry currents i

(a) zero

20V V

+ C

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

A

–

V

–

R D

equal to 5 Ω greater than 5 Ω less than 5 Ω greater or less than 5 Ω depending upon its material

50. An ammeter has resistance R0 and range I. What

resistance should be connected in parallel with it to increase its range by nI ? (a) R0 /( n − 1) (c) R0 /n

(b) R0 /( n + 1) (d) None of these

886 JEE Main Physics

Round II Only One Correct Option 1. A current i flows along the length of an infinitely long, straight, thin-walled pipe. Then (a) the magnetic field at all points inside the pipe is the same, but not zero (b) the magnetic field at any point inside the pipe is zero (c) the magnetic field as zero only on the axis of the pipe (d) the magnetic field at different at different points inside the pipe

(Mixed Bag) µ 0 iqv 2πd 2µ 0 iqv (c) πd (a)

E

(d) zero

5. A current ( i) carrying circular wire of radius R is placed in a magnetic field B perpendicular to its plane. The tension T along the circumference of wire is i

2. A straight rod of mass m and length L is suspended from the identical springs as shown in figure. The spring is stretched a distance x0 due to the weight of the wire. The circuit has total resistance R. When the magnetic field perpendicular to the plane of paper is switched on, springs are observed to extend further by the same distance. The magnetic field strength is

µ 0 iqv πd

(b)

R B

i

(a) BiR (c) πBiR

(b) 2πBiR (d) 2BiR

6. A particle of mass m and charge q released from the origin in a region occupied by electric field E and magnetic field B, B = − B0 ^j, E = E0 ^i The velocity of the particle will be

L

2mgR LE mgR (c) 2LE

(a)

(a) mgR LE mgR (d) E

(b)

3. In a chamber, a uniform magnetic field of 6.5 G −4

(1G = 10 T) is maintained. An electron is shot into the field with a speed of 4.8 × 106 m/s normal to the field explain why the path of the electron is a circle. If ( e = 16 . × 10−19 C, me = 9.1 × 10−31 kg), then obtain the frequency of revolution of the electron in its circular orbit. [NCERT] (a) 6 × 106 Hz (b) 18.18 × 106 Hz (c) 10.10 × 106 Hz

(c)

2qE0 m qE0 2m

(b)

qE0 m

(d) None of these

7. Two long parallel wires carry equal current i flowing in the same direction are at a distance 2d apart. The magnetic field B at a point lying on the perpendicular line joining the wires and at a distance x from the mid point is µ 0 id π ( d2 + x2 ) µ ix (c) 2 0 2 (d + x )

µ 0 ix π ( d2 − x2 ) µ id (d) 2 0 2 (d − x )

(b)

(a) EQ

8. For the arrangement as shown in the figure, the magnetic induction at the centre is

(d) 12.10 × 106 Hz a

4. Two very long, straight, parallel wires carry steady currents i and −i respectively. The distance between the wires is d. At a certain instant of time, a point charge q is at a point equidistant from the two wires, in the plane of the wires. Its instantaneous magnitude of the force due to the magnetic field acting on the charge at this instant is

O 90° i

3µ 0 iπ 4a µ 0i (c) 4 πa (a)

µ 0i (1 + π ) 4 πa 3µ 0 i (d) 8al

(b)

887

Magnetic Effects of Current 9. An equilateral triangle of side l is formed from a piece

14. An electron is shot in steady electric and magnetic

of wire of uniform resistance. The current i is fed as shown in the figure. The magnitude of the magnetic field at its centre O is Q

fields such that its velocity v, electric field E and magnetic field B are mutually perpendicular. The magnitude of E is 1 Vcm −1 and that of B is 2 T. Now if it so happens that the Lorentz (magnetic) force cancels the electrostatic force on the electron, then the velocity of the electron is

3 µ 0i 2π l 3 3 µ 0i (b) 2πl µ 0i (c) 2πl (d) zero (a)

O P

R

i

1012 G at its surface. The maximum magnetic force experienced by an electron moving with velocity 0.9 c is (b) 4.32 × 10 −3 N

(c) 4.32 × 103 N

(d) zero

11. Four wires each of length 2.0 m are bent into four P, Q, R and S and then suspended into a uniform magnetic field. Same current is passed in each loop Q

S

P

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

R

as shown in figure. The magnetic interactions (a) (b) (c) (d)

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

experiences no force experiences a force towards wire experiences a force away from wire experiences a torque but no force

16. Current i0 is passes through a solenoid of length l having number of turns N when it is connected to a DC source. A charged particle with charge q is projected along the axis of the solenoid with a speed v0 . The velocity of the particle in the solenoid (b) decreases (d) becomes zero

17. An infinitely long wire carrying current i is along Y -axis such that its one end is at point (0, b) while the wire extends upto ∞. The magnitude of magnetic field strength at point P( a, 0) is x

Y B i2 A i1

push i2 away from i1 pull i2 closer to i1 turn i2 clockwise turn i2 counter-clockwise

placed near a long straight fixed wire carrying strong current such that long sides are parallel to wire. If the current in the nearer long side of loop is parallel to current in the wire. Then the loop

(a) increases (c) remain same

couple on loop P will be maximum couple on loop Q will be maximum couple on loop R will be maximum couple on loop S will be maximum

12. Two wires A and B carry currents

(b) 2 cms −1 (d) 200 cms −1

15. A rectangular loop carrying current is

i

10. A pulsar is a neutron star having magnetic field at

(a) 43.2 N

(a) 50 ms −1 (c) 0.5 cms −1

i

X

A (0, b) P(a, 0)

13. Three infinite straight wires A, B and C carry

(a)

  a2 + b2 

(b)

currents as shown. The net force on the wire B is directed

µ 0i  1 + 4 πa 

µ 0i  1 − 4 πa 

  a2 + b2 

(c)

µ 0i  1 − 4 πa 

  a +b

(d)

µ 0i  1 + 4 πa 

  a +b

1A

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

2A

towards A towards C normal to plane of paper zero

3A

b

a

2

2

b

a

2

2

18. An electron having kinetic energy E is moving in a circular orbit of radius R perpendicular to a uniform magnetic field induction B. If kinetic energy is doubled and magnetic field induction is tripled, the radius will become (a) R 9 / 4

(b) R 3 / 2

(c) R 2 / 9

(d) R 4 / 3

888 JEE Main Physics 19. Consider the following statements regarding a

25. In a square loop PQRS made with a wire of

charged particle in a magnetic field (i) straight with zero velocity, it accelerates in a direction perpendicular to the magnetic field. (ii) while deflecting in the magnetic field, its energy gradually increases. (iii) only the component of magnetic field perpendicular to the direction of motion of the charged particle is effective in deflecting it. (iv) direction of deflecting force on the moving charged particle is perpendicular to its velocity. Of these statements.

cross-section current i enters from point P and leaves from point S. The magnitude of magnetic field induction at the centre O of the square is

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

(ii) and (iii) are correct (iii) and (iv) are correct (i), (iii) and (iv) are correct (i), (ii) and (iii) are correct

path of diameter 0.1 nm. It produces a magnetic field 14 T at a proton. Then the angular speed of the electron is (b) 4.4 × 1016 rad s −1 (d) 1.1 × 1016 rad s −1

21. A thin disc having radius r and charge q distributed uniformly over the disc is rotated n rotations per second about its axis. The magnetic field at the centre of the disc is µ qn (a) 0 2r

µ qn (b) 0 r

µ qn (c) 0 4r

3 µ 0 qn (d) 4r

22. The torque required to hold a small circular coil of 10

turns, 2 × 10−4 m 2 area and carrying 0.5 A current in the middle of a long solenoid of 103 turns m −1 carrying 3 A current, with its axis perpendicular to the axis of the solenoid, is −7

−7

(a) 12π × 10 Nm

(b) 6 π × 10 Nm

(c) 4 π × 10 −7 Nm

(d) 2π × 10 −7 Nm

23. A steady current i flows in a small square loop of wire of side l in a horizontal plane. The loop is now folded about its middle such that half of it lies in a vertical plane. Let M1 and M2 respectively denote the magnetic moments due to current loop before and after folding. Then (a) M2 = 0 (b) M1 and M2 are in the same direction (c) M1/ M2 = 2 (d) M1/ M2 = 1 / 2

24. A square frame of side 1 m carries a current i, produces a magnetic field B at its centre. The same current is passed through a circular coil having the same perimeter as the square. The magnetic field at the centre of the circular coil is B′. The ratio B / B′ is (a)

8 π2

(b)

8 2 π2

(c)

16 π2

(d)

16 2π2

R

S

P i

i

µ 0 2 2i 4π a µ 0 2 2i (c) 4 a

(a)

20. An electron is revolving around a proton in a circular

(a) 8.8 × 106 rad s −1 (c) 2.2 × 1016 rad s −1

a

Q

(b)

µ 0 4 2i 4π a

(d) zero

More Than One Correct Option 26. A particle of charge + q and mass m moving under the influence of a uniform electric field

^

E i and

a ^

uniform magnetic field B k follows a trajectory from P to Q as shown in figure. The

y E P

v

B

a Q

x

2a 2v

^

velocities at P and Q are v i

and −2v ^i, respectively. Which of the following statement(s) is/are correct? (a) E =

3 4

 µv2     qa 

(b) Rate of work done by the electric field at P is =

3  mv2    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

27. Consider a wire carrying a steady current, I placed in a uniform magnetic field B perpendicular to its length. Consider the charges inside the wire. It is known that magnetic forces do no work. This implies [NCERT Exemplar] that, (a) motion of charges inside the conductor is unaffected by B since they do not absorb energy (b) some charges inside the wire move to the surface as a result of B (c) if the wire moves under the influence of B, no work is done by the force (d) if the wire moves under the influence of B, no work is done by the magnetic force on the ions, assumed fixed within the wire

Magnetic Effects of Current 28. Two identical current carrying coaxial loops, carry

current I in an opposite sense. A simple amperian loop passes through both of them once. Calling the loop as C, [NCERT Exemplar] (a)

∫ B ⋅ dI = m2 µ 0 I C

32. Refer to above question, the magnetic field at a distance x from the axis where b < x < c is µ 0 i( c2 − x2 ) 2πx ( c2 − a2 )

(b)

(c)

µ 0 i( c2 − x2 ) 2πx ( c2 − b2 )

(d) zero

Passage II

C

29. A cubical region of space is filled with some uniform

electric and magnetic fields. An electron enters the cube across one of its faces with velocity v and a positron enters via opposite face with velocity − v. At this instant, [NCERT Exemplar] (a) the electric forces on both the particles cause identical accelerations (b) the magnetic forces on both the particles cause equal accelerations (c) both particles gain or loose energy at the same rate (d) the motion of the centre of mass (CM) is determined by B alone

Comprehension Based Questions Passage I Ampere’s gives a method to calculate the magnetic field due to given current distribution. The circulation

i1

µ 0i 2πx

i

i

33. What

is the angular acceleration of the wire just after it is released from the position shown in figure ? (a) (b) (c) (d)

−8

6.2 × 10 2.1 × 10 −4 4.2 × 10 −5 9.3 × 10 −6

(2l – x) dx

−1

rad s rad s −1 rad s −1 rad s −1

x i

34. We want to keep the suspended wire stationary by

(c)

µ 0 ix2 2πa3

(b) 1.9 m

(b)

µ 0i 2π ( b − a )

(d) zero

(c) 1.3 m

(d) 2.4 m

(d)

µ 0 ix 2πa2

(having the same current) should be placed to keep it stationary ? (a) 2.9 m

(b) 1.9 m

(c) 1.3 m

(d) 2.4 m

Matching Type 36. Match the following of Column I with Column II. Column I I. II. III. IV.

Column II

Lorentz force

A.

Gauss’s law

B.

Biot-Savart law Coulomb’s Law

C. D.

distance x from the axis where a < x < b is µ 0i 2πx µ 0 ix (c) 2π ( b2 − a2 )

o

35. At what distance from suspended wire, the new wire

31. Refer to above question, the magnetic field at a (a)

l

l

(a) 2.9 m

i2

∫ B ⋅ dl = µ 0 ( i1 − i2 ) The contributions of current i3 to magnetic field cancel out because the integration is made around the full loop. 30. Consider a coaxial cable which consists of a wire of radius a and outer cylindrical shell of inner and outer radii b and c respectively. The inner wire carries a current i and outershell carries an equal and opposite current. The magnetic field at a distance x from axis from x < a is (b)

A wire carrying a current i of length l, mass m is suspended from point O as shown. An another infinitely long wire carrying the same current i is at a distance l below the lower end of the wire. Given, i = 2 A , l = 1m, and m = 0.1 kg, ln 2 = 0.693

placing a third long wire carrying an upward current. Then the wire should be placed

i3

(a) zero

µ 0 i( c2 − x2 ) πx ( c2 − a2 )

(a)

(b) the value of ∫ B ⋅ dI is independent of sense of C (c) there may be a point on C where B and dI are perpendicular. (d) B vanishes everywhere on C

889

q ε0 µ idl × r dB = 0 4π r 3 F = q (E + ( v × B )) 1 q1q2 F= 4πε 0 r 2

O ∫ E. dA =

Code (a) 1-C, II-A, III-B, IV-D (c) 1-D, II-C, III-B, IV-A

(b) 1-A, II-B, III-C, IV-D (d) 1-B, II-A, III-D, IV-C

890 JEE Main Physics 39. Assertion A current carrying conductor produces only

Assertion and Reason Direction

Question No. 37 to 43 are Assertion-Reason type. Each of these contains two Statements : Statement I (Assertion), Statement II (Reason). Each of these questions also has four alternative choice, only one of which is correct. You have to select the correct choice from the codes (a),(b), (c) and (d) given below (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 the Assertion (c) If Assertion is true but Reason is false (d) If Assertion is false but the Reason is true

37. Assertion A force of 1 kg-wt acts on 1 m long wire carrying 10 A current held at 90° to a magnetic field of 0.98 T. Reason F = Bil sin θ

38. Assertion The resistance of an ammeter is R. The shunt required to increase its range four fold is R/3. Reason Shunt has to be used to increase the range.

an electric field. Reason Electrons in motion give rise to an electric field.

40. Assertion When two long parallel wires, hanging freely are connected in series to a battery, they come closer to each other. Reason Wires carrying current in opposite direction repel each other.

41. Assertion A charged particle moves perpendicular to a magnetic field. Its kinetic energy remains constant, but momentum changes. Reason Force acts on the moving charged particles in the magnetic field.

42. Assertion We cannot accelerate neutrons by a cyclotron. Reason Neutrons are too heavy.

43. Assertion Out of galvanometer, ammeter and voltmeter, resistance of ammeter is lowest and resistance of voltmeter is highest. Reason An ammeter is connected in series and a voltmeter is connected in parallel, in a circuit.

Previous Years’ Questions 44. 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

of the centre of the disc will be represented by the [AIEEE 2012] figure.

(a)

(b) B

[IIT JEE 2012]

B R

|B|

R

|B|

(a)

(b)

(c)

(d) B

r R/2

R

R/2

|B|

R

R

R

46. A loop carrying current I lies in the xy-plane in the

|B|

(c)

B

r

^

figure. The unit vector k is coming out of the plane of the paper. The magnetic moment of the current loop [IIT JEE 2012] is

(d)

y

r

r R/2

R

R/2

R

45. A charge Q is uniformly distributed over the surface of non-conducting disc of radius R. The disc rotates about an axis perpendicular to its plane and passing through its centre with an angular velocity ω. As a result of this rotation a magnetic field of induction B is obtained at the centre of the disc. If we deep both the amount of charge placed on the disc and its angular velocity to be constant and vary that radius

I a

x a

(a) a2 l k$

^ π  (b)  + 1 a2 l k$ 2 

π  (c) −  + 1 a2 l k$ 2 

(d) (2π + 1) a2 l k$

Magnetic Effects of Current

891

47. A current I flows in an infinitely long wire with

51. A charge particle is moving along a magnetic field

cross-section in the form of a semicircular ring of radius R. The magnitude of the magnetic induction along its axis is [AIEEE 2011]

line. The magnetic force on the particle is [DCE 2009]

(a)

µ 0I π2 R

(b)

µ 0I 2π2 R

(c)

µ 0I 2πR

(d)

µ 0I 4 πR

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

along its velocity opposite to its velocity perpendicular to its velocity zero

48. Two long parallel wires are at a distance 2d apart.

52. Magnetic field intensity H at the centre of circular

They carry steady equal currents flowing out of the plane of the paper, as shown. The variation of the magnetic field B along the line XX ′ is given by

loop of radius r carrying current i emu is [WB JEE 2009]

[IIT JEE (Screening) 2000; AIEEE 2010] B X′ X

(b)

c d

53. Which

X c d

d

X

X′

(d)

c d

X′ d

B

B

(c)

X′

X c d

d

of the following Biot-Savart’s law ? µ 0 id l $r 4π r µ idl × r (c) dB = 0 4 π r3

a

D b

54. A galvanometer having a coil resistance of 60 Ω

(a) zero µ I( b − a ) (c) 0 4 πab

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

figure. The magnetic field at the centre O is [Orissa JEE 2008]

(a)

I C

(c)

[AIEEE 2009]

µ 0 I( b − a ) 24 ab µ 0I (d) [2( b − a ) + π /3 ( a + b)] 4π (b)

putting in series a resistance of 15 Ω putting in series a resistance of 240 Ω putting in parallel a resistance of 15 Ω putting in a parallel a resistance of 240 Ω

55. A current i flowing through the loop as shown in

(b)

49. The magnitude of the magnetic field (B) due to the loop ABCD at the origin (O) is

idl r r2 idl $r r4

[BVP Engg. 2009]

B

30°

(b) dB =

represent

shows full scale deflection. When a current of 1.0 A passes through it. It can be converted into an ammeter to read currents upto 5.0 A by

d

A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius = b) and DA (radius = a) of the loop are joined by two straight wires AB and CD. A steady current I is flowing in the loop. Angle made by AB and CD at the origin O is 30°. Another straight thin wire with steady current I1 flowing out of the plane of the paper is kept at the origin. I1 O

relations

µ0 4π µ0 (d) dB = 4π

(a) dB =

Passage

A

(b)

(d)

7µ 0i acting downwards 12r 5 µ 0i acting upwards 12r 7µ 0i acting upwards 12r 5 µ 0i acting downwards 12r

i O r

56. The resultant force on the current loop PQRS due to a long current carrying conductor will be [Karnataka CET 2008] Q

P

50. Due to the presence of the current I1 at the origin [AIEEE 2009]

(a) the force on AB and DC are zero (b) the force on AD and BC are zero (c) the magnitude of the net force on the loop is given by I1I µ 0 [2( b − a ) + π / 3( a + b)] 4π (d) the magnitude of the net force on the loop is given by µ 0 II1 ( b − a) 24 ab

2r

i

i2

20 A S 2 cm C

(a) 1.8 × 10 −4 N (c) 10 −4 N

15 cm

(a)

B

2πi oersted r 2πr (d) oersted i

r oersted i i oersted (c) 2πr (a)

20 A 20 A

R

100 cm

(b) 5 × 10 −4 N (d) 3.6 × 10 −4 N

892 JEE Main Physics 57. The resistance of the shunt required to allow 2% of

62. A circular coil of 5 turns and of 10 cm mean diameter

the main current through the galvanometer of resistance 49 Ω is [Kerala CET 2008]

is connected to a voltage source. If the resistance of the coil is 10 W, the voltage of the source so as to nullify the horizontal component of earth’s magnetic field of 30 A turn m −1 at the centre of the coil should be [Kerala CET 2007]

(a) 1 Ω (c) 0.2 Ω (e) 0.01 Ω

(b) 2 Ω (d) 0.1 Ω

58. Oscillating frequency of a cyclotron is 120 MHz. If the radius of its dees is 0.5 m, the kinetic energy of a proton, which is accelerated by the cyclotron is [Kerala CET 2008]

(a) 10.2 MeV (c) 20.4 MeV (e) 21.6 MeV

(b) 2.55 MeV (d) 5.1 MeV

59. Two particles of equal charge after being accelerated through the same potential difference enter a uniform transverse magnetic field and describe circular paths of radii R1 and R2 respectively. Then the ratio of their masses ( M1/ M2 ) is [Kerala CET 2008] (a) R1/ R2

(b) ( R1/ R2 )2

(c) ( R2 / R1 )

(d) ( R2 / R1 )2

60. A galvanometer of resistance 50 Ω is connected to a

battery of 3 V along with a 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 resistance in series should be [Kerala CET 2008] (b) 4450 Ω (d) 5550 Ω

61. A closed loopPQRS carrying a current is placed in a uniform magnetic field. If the magnetic force on segments PS, SR and RQ are F1, F2 and F3 respectively and are in the plane of the paper and along the direction shown, the force on the segment [Kerala CET 2008] QP is Q

F3

S

R F2

( F3 − F1 )2 − F22

(b) F3 − F1 + F2 (c) F3 − F1 + F2 (d)

( F3 − F1 )2 + F22

63. A conducting rod of 1 m length and 1 kg mass is suspended by two vertical wires through the ends. An external magnetic fields of 2 T is applied normal to the rod. Now the current to be passed through the rod so as to make the tension in the wires zero is [Kerala CET 2007] (Take g = 10 ms −2 ) (a) 0.5 A (c) 5 A (e) 15 A

(b) 15 A (d) 1.5 A

of 10 divisions. When a current of 1 mA is passed through it. If a shunt of 4 Ω is connected and there are 50 divisions on the scale, the range of the [Kerala CET 2007] galvanometer is (a) (b) (c) (d) (e)

1A 3A 10 mA 30 mA 11 mA

65. The strength of the magnetic field around a long straight wire, carrying current is

[Kerala CET 2007]

(a) same everywhere around the wire at any distance (b) inversely proportional to the distance from the wire (c) inversely proportional to the square of the distance from the wire (d) directly proportional to the square of the distance from the wire (e) None of the above

66. A current i flows along the length of an infinitely

P F1

(a)

6 V, plane of the coil normal to the magnetic meridian 2 V, plane of coil normal to the magnetic meridian 2 V, plane of the coil along the magnetic meridian 4 V, plane of the coil normal to magnetic meridian

64. A galvanometer of resistance 20 Ω shown deflection

(e) None of these

(a) 6050 Ω (c) 5050 Ω (e) 5578 Ω

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

long, straight, thin walled pipe. Then

[AIEEE 2007]

(a) the magnetic field at all points inside the pipe is the same, but not zero (b) the magnetic field is zero only on the axis of pipe (c) the magnetic field is different at different points inside the pipe (d) the magnetic field at any point inside the pipe is zero

67. Under the influence of a uniform magnetic field a charged particle is moving in a circle of radius R. with constant speed v. The time period of the motion [Kerala CET 2007]

Magnetic Effects of Current (a) depends on both R and v (b) is independent of both R and v (c) depends on R and not on v (d) depends on v and not on R (e) None of these

72. A long solenoid has 200 turns per cm and carries a current i. The magnetic field at its centre is 6.28 × 10−2 Wbm −2 . Another along solenoid has 100 turns per cm and it carries a current i/3. The value of the magnetic field at its centre [AIEEE 2006]

68. In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an electric potential V and then made to describe semicircular paths of radius R using a magnetic field B. If V and B are kept constant, the ratio  charge on the ion    will be proportional to  mass of the ion  [Kerala CET 2007] (a) 1 / R2 (c) R (e) R−3

(b) R2 (d) 1 / R

69. A solenoid of 0.4 m length with 500 turns carries a current of 3 A. A coil of 10 turns and of radius 0.01 m carries a current of 0.4 A. The torque required to hold the coil with its axis at right angles to that of solenoid in the middle point of it is [Kerala CET 2006] (a) 6 π2 × 10 −7 Nm (b) 3π2 × 10 −7 Nm (c) 9 π2 × 10 −7 Nm

(a) 1.05 × 10 −4 Wb m−2 (b) 1.05 × 10 −2 Wb m−2 (c) 1.05 × 10 −5 Wb m−2 (d) 1.05 × 10 −3 Wb m−2

73. Two insulating plates are both uniformly charged in such a way that the potential difference between them is V2 − V1 = 20 V ( i. e, plate 2 is at a higher potential). The plates are separated by d = 0.1 m and can be treated as infinitely large. An electron is released from rest on the inner surface of plate 1. [AIEEE 2006] What is its speed when it hits plate 2 ? (a) 32 × 10 −19 ms −1 (b) 2.65 × 106 ms −1 (c) 7.02 × 1012 ms −1 (d) 1.87 × 106 ms −1

74. Two long parallel wires P and Q are both perpendicular to the plane of the paper with distance 5 m between them. If P and Q carry current of 2.5 A and 5 A respectively in the same direction, then the magnetic field at a point half way between the wires [Kerala CET 2005] is

(d) 12 π2 × 10 −7 Nm (e) 15 π2 × 10 −7 Nm

70. When deuterium and helium are subjected to an accelerating field simultaneously then, [Karnataka CET 2006]

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

893

both acquire same energy deuterium accelerates faster helium accelerates faster neither of them is accelerated

3 µ0 2π 3µ 0 (c) 2π (a)

µ0 π µ0 (d) 2π

(b)

75. A straight conductor of length l carrying a current i, is bent in the form of a semi-circle. The magnetic field in tesla at the centre of the semi-circle is

71. A galvanometer coil has a resistance of 15 Ω and gives full scale deflection for a current of 5 mA. To convert it to an ammeter of range 0 to 6 A [Karnataka CET 2006]

(a) 10 mΩ resistance is to be connected in parallel to galvanometer (b) 10 mΩ resistance is to be connected in series with galvanometer (c) 0.1 Ω resistance is to be connected in parallel to galvanometer (d) 0.1 Ω resistance is to be connected in series with galvanometer

the the the the

[Kerala CET 2005]

πi × 10 −7 l iπ (b) × 10 −7 l πi (c) 2 × 10 −7 l π2 (d) × 10 −7 l (e) None of the above 2

(a)

894 JEE Main Physics Answers Round I 1. 11. 21. 31. 41.

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

2. 12. 22. 32. 42.

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

3. 13. 23. 33. 43.

(c) (c) (b) (b) (c)

4. 14. 24. 34. 44.

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

5. 15. 25. 35. 45.

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

6. 16. 26. 36. 46.

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

7. 17. 27. 37. 47.

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

8. 18. 28. 38. 48.

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

9. 19. 29. 39. 49.

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

10. 20. 30. 40. 50.

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

10. 20. 30. 40. 50. 60. 70.

(b) (b) (d) (d) (b) (b) (d)

Round II 1. 11. 21. 31. 41. 51. 61. 71.

(b)

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

2. 12. 22. 32. 42. 52. 62. 72.

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

3. 13. 23. 33. 43. 53. 63. 73.

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

4. 14. 24. 34. 44. 54. 64. 74.

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

5. 15. 25. 35. 45. 55. 65. 75.

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

6. 16. 26. 36. 46. 56. 66.

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

7. 17. 27. 37. 47. 57. 67.

(b) (b) (b,d) (a) (a) (a) (b)

8. 18. 28. 38. 48. 58. 68.

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

9. 19. 29. 39. 49. 59. 69.

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

the Guidance Round I 1. In a uniform magnetic field, the two charged particles will traverse identical helical paths in a completely opposite sense if the charge/mass ratio of these two particles is same and charges on them are of opposite character. In this situation ( e/m)1 + ( e/m) 2 = 0 , holds good.

2. Given, I = 90 A and r = 1.5 m Here, point P is below the power line, where we have to find the magnetic field and its direction. The magnitude of magnetic field µ 2I B= 0 ⋅ 4π r

East 1.5 m P

=

10

−7

distance r from a charge q moving with a velocity v is given by, µ q ( v × r) B = mb 0 4π r3 µ v sin θ or B= 0 4π r 2 The direction of B is along ( v × r), i. e. , perpendicular to the plane containing v and r ⋅ B at a point obeys inverse square law and not inverse cube law.

5. Total magnetic field induction at O is

Overhead line 90 A West

4. According to Biot Savart’s law, the magnetic field B at a point

× 2 × 90 = 1.2 × 10− 5 T 1.5

The direction of magnetic field is given by Maxwell’s right hand rule. So, the direction of magnetic field at point P due to the flowing current is perpendicularly outwards to the plane of paper.

3. Magnetic field induction at a point due to a long current 1 carrying wire is related with distance r by relation B ∝ . r

B = BLR + BRP + BMS + BSQ µ i µ i µ 2i =0 + 0 + 0 + 0 = 0 2π r 2π r 2π 2 =

2 × 10 −7 × 2 × 10 = 2 × 10 −6 T 0.20

6. Since election is moving parallel to the magnetic field, hence, magnetic force on it. Fin = 0 –e

E v

F = eE

Then is only force aching on the direction is electric force which reduces its speed.

7. Given, I1 = 8 A, I2 = 5 A and r = 4 cm = 0.04 m Force per unit length on two parallel wire carrying current µ 2I ⋅ I F= 0⋅ 1 2 r 4π

Magnetic Effects of Current 10 − 7 × 2 × 8 × 5 = 2 × 10 − 4 N 0.04

=

The force on A of length 10 cm is F′ = F × 0.1 (Q 1m = 100 cm) F′ = 2 × 10 − 4 × 0.1

⇒

= 2 × 10

−5

N

12. B1 =

B 5A 4 cm F

8. Magnetic field induction at the centre of circular coil carrying µ 0 2πni i. e. ,B ∝ n / r 4π r

r 3 3 ×r B1 n1 r = × = =9 B r1 n  r    ×1  3

2πr = 3 × 2πr1 or r1 =

So,

9. B1 =

3/ 2

loop along z-direction. When half of the current loop is bent in y-z plane, then magnetic moment due to half current loop in x-y plane, M1 = I ( πR 2/ 2) acting along z-direction. Magnetic

 3 h2 3 h2  = 1 − 1 − 2  = 2   2r  2r

moment due to half current loop in y-z plane, M2 = I ( πR 2/ 2) along x-direction. Effective magnetic moment due to entire bent current loop,

N

N

S (a)

E N

i = 10 Am, r = 0.5 m µ i(dl × r) dB = 0 4π r3

The magnitude of B at a point on the axis of loop, distance z from the centre of current loop in x-y plane is B=

S (b)

Since the axis of the magnetic field pleura at the center of earth maker an angle 11.3° with the axis of the earth. The two situations are given in figure (a) and figure (b) respectively clearly chelation on Vanier from 11.3° W to 11.3° E.

11. Here, dl = dx = 1cm = 10 −2 m;

IπR 2 2 a and r > > l. then denominates is approximated by 2

2 32 /

[( r - x) + a ] 2

B=

and

m 0 nIa 2r3

l

m 0 nI 2la2 × 3 2 r

Also magnetic moment, m = n(2l ) I( p a2 ) B=

Thus,

Sample Problem 4 A long solenoid is formed by winding

20 turns cm -1. What current is necessary to produced a magnetic field of 20 mT inside the solenoid? (b) 6 A

(c) 4 A

(d) 10 A

-1

Interpret (a) Given, n = 20 turns cm = 20 ´ 10 B = 20 mT = 20 ´ 10

-3

-2

-1

m ,

T

The magnetic field inside the solenoid is B = m 0 ni or or

20 ´ 10 -3 T = 4p ´ 10 -2 Tmn -1 ´ (20 ´ 10 3 m-1) i = 8A

Sample Problem 6 A closely wound solenoid of

(a) 0.048 J (c) 48 J

m 0 2m 4 p r3

This is also the far axial magnetic field of a bar magnet. Thus, a bar magnet and a solenoid produce similar magnetic fields.

(a) 8 A

ö æm N I t = (I2 AN2) ´ ç 0 1 1 = 0.4 ´ p ´ (0.01) 2 ´ 10 ÷ ø è l 500 ö æ -7 ´ ç 4p ´ 10 ´ ´ 3÷ ø è 0.4 2 -7 = 6p ´ 10 Nm

2000 turns and area of cross-section 1.6 ´ 10 -4 m 2, carrying a current of 4 A is suspended through its centre allowing it to turn in a horizontal plane. If a uniform horizontal magnetic field of 7.5 ´ 10 -2 T is set up at angle of 30° with the axis of the solenoid, then the torque acting on the solenoid is

3

»r

= ò-dx l

N1 I1 l

(b) 0.012 J (d) 12 J

Interpret (a) Let M = magnetic moment of the solenoid i.e., M = NIA where ,

N = 2000 , I = 4A , A = 1.6 ´ 10 -4 m2

\

M = 2000 ´ 4 ´ 1.6 ´ 10 -4 M = 1.28 JT -1

Torque, where ,

t = MB sin q q = 30° , B = 7.5 ´ 10 -2 T , M = 1.28 JT -1 t = 1.28 ´ 7.5 ´ 10 -2 ´ sin 30° 1 = 1.28 ´ 7.5 ´ 10 -2 ´ 2 t = 0.048 J

910 JEE Main Physics

20.5 Magnetic Field Lines The magnetic field lines is defined as the path along which the compass needles are aligned. l

They are used to represent magnetic field in a region.

l

They are closed continuous curves.

l

l

l

Tangent drawn at any point gives the direction of magnetic field. They cannot intersect. Outside a magnet, they are directed from north to south pole and inside a magnet they are directed from south to north. B

So

F=

m 0 m0m n 4p r 2

B=

m m F = 0 n m0 4p r 2

…(ii)

Regarding Magnetic Field Vector B, it is Worth Noting That (i) It is a vector quantity having dimensions [B] =

F [MLT -2] = = [MT -2A -1] m0 [AL]

(ii) Its Si unit is N Wb = = tesla (T) A ´ m m2 while the CGS unit is gauss (G).

S

(iii) If in the field B = constant, i. e., same at every point, the field is said to be uniform, e. g ., the magnetic field of the earth on its surface at a given place is assumed to be uniform with value of the order of 10-4 T.

N

B

Magnetic Pole Strength When a magnetic pole is kept in magnetic field B, it experience a force mB. It is equivalent to charge in electrostatics. The preferred regions of attraction near the two ends of a magnet, where the magnetic force due to a bar magnet is maximum, are called poles of the magnet. The strength of a magnetic pole is represented by the symbol m.

(iv) If in a field at a point B = 0, the point is called neutral point. Neutral point or point in a field exist only if there is superposition of field and they cancel each other’s effect at certain points.

20.6 Magnetic Dipole Two equal and opposite poles separated by a distance 2l are said to constitute a dipole. Here, it is worthy to note that (i) As magnetic monopoles do not exist, sources of magnetic field are usually dipoles.

Magnetic Field Vector (B) In a magnetic field, the strength of the field at a point is normally represented by vector B which is known as field vector, flux density or magnetic induction and in terms of pole concept can be defined as the force experienced by a unit test north pole supposed to be placed at that point. So if a test pole of strength m0 at a point P in a magnetic field experiences a force F, the field vector at that point will be F …(i) B= , i. e. , F = m0 B m0

(ii) The dipole moment of a dipole is a vector given by M = m(2l) n where n is a unit vector along the axis of the magnet directed from its south to north pole. (iii) The earth for its magnetic effects behaves as a dipole with its magnetic south pole near the earth’s geographical north and magnetic north pole near the earth’s geographical south.

Now, if the field is produced by a pole of strength m from which the field point is at a distance r, by inverse square law. F P r n m S (a)

m0

N S

(b)

N′ S

E M

S′

C

N M B

N

Magnetic-diploe M=m(2l) n

Current loop M=I S

Earth

S

|B| = [C/M ] (c)

(iv) For a distant point a current carrying coil behaves as a magnetic dipole of moment M = IS.

Magnetostatics

M

M1

1

2

2

N 2l

m0 m M M × = 0 4p (r 2 )3/ 2 4p r3

B=

The magnetic field at point P is directed parallel to the length of the magnet from its N to S pole.

2

+M

θ S

r 2 >> l 2

= M

2

N m

–m

r

r

+l

If

2

N

+m

l

r

–m S

Magnetic field strength at point P due to bar magnet at a distance r from its centre is m M B= 0× 2 4p (r + l 2 )3/ 2

+

(a) On bending a magnet, its pole strength remains unchanged while its magnetic moment changes.

Equitorial or Broad Side-on Position

2

Regarding magnetic moment of magnet following points are worth noting

911

2

M2

(b) On cutting a magnet, its magnetic moment decreases. (c) When two bar magnets are placed across each other as shown in figure, then

At Any Point (P) At an angle q with the axis of magnet, the magnetic field at point P at a distance r from centre of magnet in vacuum (or air) is B

M12 + M22

M=

= 2 ml (if, M1 = M2 ).

P

r

(d) When two coils, each of radius r and carrying current i are placed coaxially but their planes are perpendicular as shown in figure, then M2

θ

+M

2

S

2

2 1

M

m0 M 1 + 3 cos2 q × 4p r3 1 tan a = tan q 2 B=

M

O

M N

O

=

r

Br

α

Bθ

and

M1

r

For axial position of point P, q = 0° and for equitorial position, q = 90° M12 + M22 = 2ipr 2

M=

(if M1 = M2 )

Magnetic Field Strength at a Point due to Magnetic Dipole (or Bar Magnet) Axial or End-on Position

Torque on Bar Magnet in Magnetic Field In figure, a uniform magnetic field B is represented by equidistant parallel lines. NS is a bar magnet of length 2l and strength of each pole is m. The magnet is held at angle q with the direction of B.

Magnetic field strength at point P due to bar magnet at a distance r from its centre is

N

N

S

P

2l

2l

N

θ

mB B

r

B=

2 Mr m0 × 2 2 2 4p (r - l )

If magnet is short, l 2 < < r 2 B=

m 0 2 Mr m 0 2 M = 4p r 2 4p r3

The direction of magnetic field at point P is along NP.

mB

S

θ

A

Bar Magnet

Force on N-pole = mB, along B Force on S-pole = mB, opposite to B

912 JEE Main Physics These forces being equal, unlike and parallel form a couple, which tends to rotate the magnet clockwise so as to align it along B. Draw NA perpendicular to B and SA | | B \ Torque acting on the bar magnet t = mB ´ NA NA NA sin q = = NS 2l

In DNAS,

NA = 2l sin q t = mB ´ 2l sin q M = m ´ 2l t = MB sin q

\

magnetic dipole is perpendicular to the field. Above equation shows that at q = 0°, potential energy is minimum ( = - MB ), which is the most stable position. Further, at q = 180°, potential energy is maximum (= + MB), which is most unstable position.

…(i)

\ Eq. (i) becomes As,

Note When q = 90° , U = 0 i .e ., potential energy is zero, when the

In vector form, we can rewrite this equation as t = M ´B The direction of t is perpendicular to the plane containing M and B are is given by right handed screw rule.

Important Points 1.

Magnet attracts the magnetic substances towards them. Magnetic moment is a vector quantity. It is directed from south pole to north pole. If a bar magnet of magnetic moment M is divided into two equal halves transversely, then magnetic moment of each part M becomes , but pole strength remains the same. 2

2. According to molecular theory of magnetism, every atom of

magnetic substance behaves as a total magnet. When they are in the unmagnetised state, then they form closed chains. So, they do not exhibit magnetic property. But when the magnetic substances are magnetised, then the closed chains get broken and get arranged in a definite pattern. As a result, all the molecules or atoms of a magnet get arranged in a definite order.

Potential Energy of a Magnetic Dipole in a Magnetic Field

3. On hammering, on heating or cooling, magnet loses its magnetic

When a magnetic dipole of moment M is held at an angle q with the direction of a uniform magnetic field B, the magnitude of the torque acting on the dipole is

4. There can be magnets with no poles. for example, a magnetised ring

t = MB sin q This torque tends to align the dipole in the direction of the field. Work has to be done in rotating the dipole against the action of the torque. This work done is stored as potential energy of the dipole.

property. If a magnet is kept open for a long time, then also it loses its magnetic property. (called toroid) or a solenoid of infinite length has properties of a magnet, but no poles, figure (a). Figure (b) represents a magnet with two similar poles or with three poles. This may be due to faulty magnetisation of a bar. We find identical poles at the two ends with an opposite pole of double strength at the centre of the bar.

Now, small amount of work done in rotating the dipole through a small angle dq is

N

dW = t dq = MB sin q × dq

S

S

N

Total work done in rotating the dipole from q = q0 to q = q is W =ò

W

0

dW = ò

q

q0

MB sin q dq = MB [- cos q]qq 0

W = - MB [cos q - cos q0] \ Potential energy of the dipole is

Toroid Magnet with no poles (a)

5. The magnetic potential due to a magnetic dipole at distance r is

V =

U = W = - MB (cos q - cos q0 ) When q0 = 90° , then U = W = - MB (cos q - cos 90° ) W = - MB cos q In vector notation, we may rewrite this equation as U = - M ×B

Magnet with similar poles or with three poles (b)

m 0 M cos q × 4p r2

Now, (a) on the axis of magnet q = 0° m M V = 0 2 \ 4p d (b) On the neutral axis, q = 90° \

V =0

Magnetostatics Sample Problem 7 The length of a magnetized steel wire is l and its magnetic moment is M. It is bent into the shape of L with two sides equal. The magnetic moment now will be l √2

l 2

Sample Problem 10 A bar magnet when placed at an angle of 30° to the direction of magnetic field induction of 5 ´ 10 -2 T, experiences a moment of couple 2.5 ´ 10 -6 N-m. If the length of the magnet is 5 cm, its pole strength is (a) 2 ´ 10 2 A-m

(b) 2 ´ 10 -3 A-m

(c) 5 A-m

(d) 5 ´ 10 -2 A-m

Interpret

(b) Here, q = 30°, B = 5 ´ 10 -2 T, t = 2.5 ´ 10 -6 N-m

l 2

M (a) 2

(b) 2M

(c) 2 M

2l = 5 cm = 0.05 m, m = ? t = MB sin q = m(2l) B sin q

M (d) 2

m=

Interpret (d) If m is strength of each pole, then M = m ´ l When the wire is bent into L shape, effective distance between the poles 2

2

l M = 2 2

M¢ = m ´

( m will remain unchanged)

Sample Problem 8 A straight wire carrying current i is turned into a circular loop. If the magnitude of magnetic moment associated with it MKS unit is M, the length of wire will be (a) (c)

4p M

(b)

4pM i

(d)

Mp 4i 4pi M

Interpret (c) Magnetic moment,M = iA = i( pr 2), where l = 2 pr r= \

2.5 ´ 10 -6 t = B(2l) sin q 5 ´ 10 -2(0.05) sin 30°

m = 2 ´ 10 -3 A-m

\

l ælö ælö = ç ÷ +ç ÷ = è2ø è2ø 2 \

913

1. A magnetised needle in a uniform magnetic field experiences a torque but no net force. An iron nail near a bar magnet, however, experiences a force of attraction in addition to a torque why?

2. Two identical looking iron bars A and B are given, one of which

is definitely known to be magnetised. How would one ascertain, whether or not both are magnetised? If only one is magnetised, how does one ascertain which one? Using nothing else but the two bars A and B.

3. What is the basic difference between magnetic lines of force and electric lines of force?

4. Compare the magnetic fields due to a straight solenoid and a bar magnet.

5. A bar magnet is stationary in magnetic meridian. Another

M pi

l = 2p

Check Point 1

M = pi

4p M i

similar magnet is kept parallel to it, such that the centres lie on their perpendicular bisectors. If the second magnet is free to move, then what type of motion it will have translatory, rotatory or both?

Sample Problem 9 The pole strength of 12 cm long bar magnet is 20 A-m. The magnetic induction at a point 10 cm away from the centre of the magnet on its axial line is é m0 -7 -1ù ê 4p = 10 Hm ú û ë (a) 1.17 ´ 10 -3 T

(b) 2.20 ´ 10 -2 T

(c) 1.17 ´ 10 -2 T

(d) 2.21 ´ 10 -2 T

Interpret (a) Here, 2l = 12 cm = 0.12 m On axial line,

m = 20 A-m, d = 10 cm = 0.1 m m 2Md m 2Mld B= 0 = 0 2 22 4p (d - l ) 4p (d 2 - l 2) 2 B = 10 -7 ´

2(20) (0.12) ´ 0.1 [(0.1) 2 - (0.06) 2]2

= 1.17 ´ 10 -3 T

20.7 Earth’s Magnetism The earth is a natural source of magnetic field. One magnetic field present everywhere near the surface of the earth. A freely suspended magnet always points in the north-south direction even in the absence of any other magnet. This suggests that the earth itself behaves as a magnet which causes a freely suspended magnet (or magnetic needle) to point always in a particular direction : north and south. The shape of earth’s magnetic field resembles that of a bar magnet of length one-fifth of earth’s diameter buried at its centre.

914 JEE Main Physics Geographic axis

Geographic north

Geographic N-pole Magnetic S-pole

Geographical meridian

B'

φ

Magnetic north

Equator

B

θ

H

A

S

Magnetic N-pole

N Magnetic equator

Magnetic meridian

C'

D

C

Magnetic axis GeographicS-pole

In figure, ABCD is the magnetic meridian and AB ¢ C ¢ D is the geographical meridian. The angle B¢ AB = f is the angle of declination.

The south pole of a earth’s magnet is towards the earth’s north pole (geographical north), while the north pole of earth’s magnet is towards earth’s south pole (geographical south). Thus, there is a magnetic S-pole near the geographical north and a magnetic N-pole near the geographical south. The positions of the earth’s magnetic poles are not well defined on the globe, they are spread over an area.

2. Angle of Dip or Inclination

Magnetic equator The great circle whose plane is

θ

The angle which the axis of needle makes with the horizontal, is called angle of dip (q). In other words, the angle of dip at a place is the angle which the resultant magnetic field of the earth at that place makes with the horizontal. B

perpendicular to the earth’s magnetic axis is called earth’s magnetic equator.

H C

V

Geographical equator The great circle whose plane is perpendicular to geographical axis is called geographical equator.

Magnetic meridian The line joining the earth’s magnetic poles is called the magnetic axis and a vertical plane passing through it is called the magnetic meridian. Geographical meridian The line joining the geographical north and south poles is called the geographic axis and a vertical plane passing through it is called the geographical meridian.

Magnetic Elements To have a complete knowledge of the earth’s magnetism at a place, the following three elements must be known 1. Angle of declination 2. Angle of dip or inclination 3. Horizontal component of the earth’s field

1. Angle of Declination The angle between the magnetic meridian and geographical meridian at a place is called the angle of declination (or simply the declination) at that place.

A

S

D

N

In figure, AC shows the direction of resultant magnetic field of the earth and the angle BAC = (q) between it and the horizontal AB is the angle of dip.

3. Horizontal Component of the Earth’s Magnetic Field The direction of the earth’s field at the magnetic poles is normal to the earth’s surface (i. e. , in vertical direction) and at magnetic equator it is parallel to the earth’s surface, (i. e. , in horizontal direction). Thus, the resultant earth’s field can be resolved in two components as shown in above figure. (a) The horizontal component BH along AB and (b) The vertical component BV along AD. From figure, Horizontal component

BH = Be cos q

…(i)

and vertical component

BV = Be sin q

…(ii)

Magnetostatics From Eqs. (i) and (ii), we get BV B sin q = e = tan q \ BH Be cos q or

BV = BH tan q

Again Eqs. (i) and (ii) give

BH2 + BV2 = Be2 (cos2 q + sin2 q) or

Be = BH2 + BV2

More about Angle of Dip q 1. At a place on the poles, the earth’s magnetic field is perpendicular to the surface of the earth i. e., vertical. \ As \ As

Be = BV BV = Be sin q sin q = 1 Þ q = 90° BH = Be cos q = Be cos 90° = 0

Therefore, at poles q = 90° and BH = 0. i. e., earth always has a horizontal component except at poles. A freely suspended magnet at poles will stand vertical with its north pole pointing towards earth’s north pole (which is magnetic south), and vice-versa. 2. At a place on the equator, earth’s magnetic field is parallel to the surface of earth i. e., horizontal. \ As, \

Be = BH BH = Be cos q cos q = 1 Þ q = 0°

As BV = Be sin q = Be sin 0° = 0 Therefore, at the equator q = 0 and BV = 0 i. e., earth always has a vertical component except at equator. A freely suspended magnet at equator will stand horizontal. 3. In a vertical plane, at an angle a to magnetic meridian. B¢H = BH cos q and B¢V = - BV Therefore, angle of dip q¢ in a vertical plane making angle a with magnetic meridian is given by BV B¢ tan q = tan q¢ = V = B¢H BH cos a cos a For a vertical plane, other than magnetic meridian, a > 0°, cos a < 1 \ q¢ > q i. e., angle of dip increases. tan q when a = 90°, tan q¢ = =¥ 90° \ q¢ = 90° i. e., in a plane perpendicular to magnetic meridian, dip needle will stand vertical. 4. If q1 and q2 are observed angles of dip in two arbitrary vertical planes, which are perpendicular to each other, the true angle of dip q is given by the relation. cot 2 q = cot 2 q1 + cot 2 q2 5. If l is magnetic latitude at a place, then the angle of dip q at the place is given by tan q = 2 tan l

915

Note 1. At magnetic equator BH = Be cos 0° = Be and at poles BH = Be cos 90° = 0. Similarly, it magnetic equator BV = Be sin0° = 0 and at poles BV = Be sin 90° = Be . 2. Magnetic maps show variation of magnetic elements from place to place. Some important lines drawn on magnetic maps are (i) Isoclinic lines These are the lines joining points of equal dip or inclination. A line joining places of zero dip is called aclinic line or magnetic equator. (ii) Isogonic i.e., lines These are the lines joining places of equal declination. The line joining places of zero declination is called agonic line. (iii) Isodynamic lines These are the lines joining places having the same value of horizontal component of earth’s magnetic field.

Neutral Points A neutral point is a point at which the resultant magnetic field is zero. Following two cases are of special interest.

1. When a bar magnet is placed along the magnetic meridian with its north pole pointing towards geographic north, the horizontal component of earth’s magnetic field and the magnetic field due to the bar magnet cancel on the equatorial line of magnet and these points are called neutral points. If rbe the distance of neutral point from centre of the magnet. Then, the magnetic field due to the magnet at the neutral point is given by m M B eq = 0 × 2 4 p ( r + l 2 )3/2 where, M is magnetic dipole moment of the magnet, Since, at the neutral point, magnetic field due to the magnet is equal to BH , we have M m0 = BH 4 p ( r2 + l 2 )3/2 In case, if the magnet is so small thatl2 can be neglected as compared to r 2, then m0 M × = BH 4 p r3

2. When a bar magnet is placed along the magnetic meridian with its north pole pointing towards geographic south, two neutral points are obtained on either side of the magnet along its axial line. Hence, we have 2Mr m B axial = 0 2 4 p ( r - l 2 )2 (neutral point is at distancerfrom the centre of magnet). Since at the neutral point, the magnetic field due to the magnet is equal toBH , we have 2Mr m0 × = BH 4 p ( r2 - l 2 )2 In case, if the magnet is so small thatl2 can be neglected as compared to r2, then m 0 2M × = BH 4 p r3

916 JEE Main Physics Sample Problem 11 If a magnet is suspended at an angle 30° to the magnetic meridian, the dip needle makes an angle of 45° with the horizontal. The real dip is (a) tan -1 ( 3 / 2)

(b) tan -1( 3)

(c) tan -1( 3 / 2)

(d) tan -1(2 / 3)

Interpret (d) As, tan q¢ =

tan q tan 45° 1 2 = = = cos q cos 30° 3 /2 3

Sample Problem 15 A bar magnet 30 cm long is placed in the magnetic meridian with its north pole pointing south. The neutral point is observed at a distance of 30 cm from its one end. Calculate the pole strength of the magnet. Given, horizontal component of earth’s field = 0. 34 G. (a) 4.3 Am

(b) 0.42 G (d) 0.80 G True north

From the figure,

P

cos q = Given,

\

He θ

He Be

Be

Sample Problem 13 The earth’s magnetic field at the equator is approximately 0.4 G. The earth’s dipole moment is (Re = 6.4 ´ 106 m) (a) 4 ´ 10 2 Am2

(b) 1.05 ´ 10 23 Am2

(c) 4 ´ 10 -5 Am2

(d) 1.05 ´ 10 -20 Am2

Interpret (b) The equitorial magnetic field is given by m 0m 4 p r3

Given, BE ~0.4 G = 4 ´ 10 -5 T , re = 6.4 ´ 10 6 m \

4 ´ 10 -5 ´ (6.4 ´ 10 6)3 mE = m0 4p mE = 4 ´ 10 2 ´ (6.4 ´ 10 6)3 = 1.05 ´ 10 23 Am2

Sample Problem 14 At a place of latitude 5°, the angle of dip is nearly (a) 5° (c) 2.5°

(b) 10° (d) 7.5°

Interpret (b) Angle of dip, q = 2l = 2 ´ 5° = 10°

or

Baxial = BH 2Mr m0 ´ = BH 4p (r 2 - l 2) 2 M= =

Ve He = 0.26 G, q = 60° , 1 cos 60° = 2 He 0.26 Be = = = 0.52 G cos 60° æ 1 ö ç ÷ è2ø

BE =

l = 15 cm = 0.15 m, r = 30 cm = 0. 30 cm,

When magnet is placed with its north pole pointing south, neutral point is obtained on its axial line. Therefore, at the neutral point. or

Interpret (c) The earth’s magnetic field is Be and its horizontal and vertical components are He and Ve.

(d) 8.6 Am

BH = 0.34 G = 0.34 ´ 10 -4 T

Sample Problem 12 In the magnetic meridian of a certain place, the horizontal component of the earth’s magnetic field is 0.26 G and the angle of dip is 60°. The magnetic field of the earth at this location is [NCERT]

(c) 6.9 Am

Interpret (d) Here, 2l = 30 cm or

q¢ = tan -1 (2 / 3)

(a) 0.26 G (c) 0.52 G

(b) 5.2 Am

4p BH (r 2 - l 2) 2 ´ 2r m0

1 0.34 ´ 10 -4 ´ (0.30 2 - 0.15 2) 2 ´ -7 2 ´ 0.30 10

=

0.34 ´ 10 -4 ´ (0.0675) 2 10 -7 ´ 2 ´ 0.30

= 2.582 Am2 The pole strength of the magnet, M 2.582 m= = = 8.606 Am 2l 0.30

20.8 Vibration Magnetometer It is an instrument used to compare magnetic moments of two bar magnets or to determine the horizontal component of earth’s magnetic field. It is based on the principle that when a bar magnet suspended freely in a uniform magnetic field is displaced from its equilibrium position, it starts executing simple harmonic motion about the equilibrium position. The time period of vibration of the magnet of moment of inertia I and magnetic moment M vibrating in uniform I . magnetic field of strength BH is given by T = 2p MBH

Uses of Vibration Magnetometer 1. To compare magnetic moments of two bar magnets Consider two bar magnets A and B of same size and mass. Let M1 and M2 be the magnetic moments of the magnets A and B respectively and I be the moment of inertia of each magnets.

Magnetostatics Then,

T1 = 2 p

I M1BH

and

T2 = 2 p

I M2BH

Sample Problem 16 The time period of vibration of two magnets in sum position is 3s. When polarity of weaker magnet is reversed, the combination makes 12 oscillations per minute. The ratio of magnetic moments of two magnets is

M1 T22 = M2 T12

\

(a)

Net moment of inertia, IS = I1 + I2 Time period of oscillation of this pair in earth’s magnetic field (BH ) IS = 2p MS BH

I1 + I2 ( M1 + M2 )BH

…(i)

(b) When the two bar magnets are placed with their unlike poles in the same direction. M2

N

Td = 2p

Md BH

Magnetic meridian φ

= 2p

I1 + I2 ( M1 - M2 )BH

2 2 M1 Td + TS = 2 2 M2 Td - TS

3. To compare horizontal components of earth’s magnetic field at two places Consider a magnet having moment of inertia I and magnetic moment M. Let BH and B¢H be the value of horizontal components of earth’s magnetic field at two places A andB respectively, then T = 2p

4 5

1 60 s min = = 5s 12 12

I MBH

B T ¢2 I and T ¢ = 2p \ H = 2 B¢H T MB¢H BH can be found. Knowing T and T ¢ the ratio B¢H To determine M and BH from the expression I T = 2p MBH

M BC

O Axis of coil

…(ii)

From Eqs. (i) and (ii), we get TS M1 - M2 = Td M1 + M2 Þ

(d)

M1 5 2 + 3 2 34 17 = = = M2 5 2 - 3 2 16 8

Net magnetic moment, Md = M1 - M2 Net moment of inertia, Id = I1 + I2 and

3 5

In case of tangent galvanometer, a magnetic compass needle is placed horizontally at the centre of a vertical fixed current-carrying coil whose plane is in the magnetic meridian. So, if the needle in equilibrium subtends an angle f with the earth’s magnetic field,

M1

N

Id

(c)

20.9 Tangent Galvanometer

S

S

17 8

M2

N

TS = 2p

\

M1

N

(b)

M1 T22 + T12 = M2 T22 - T12

(a) If the two magnets are placed with their poles in the same direction. Net magnetic moment MS = M1 + M2 S

16 17

Interpret (b) Here, T1 = 3s, T2 =

2. To compare the magnetic moments by sum and difference method

S

917

or, i. e. , or, i. e. ,

| M ´ BH | = | M ´ BC | MBH sin f = MBC sin (90 - f) BC = BH tan f é m 0 2 pNI m 2 pnl ù = BH tan f as BC = 0 ê 4p R 4p R úû ë I = K tan f with K =

4p RBH × ¾® m 0 2 pN

Reduction factor of the tantent galvanometer, i. e. , in case of a tangent galvanometer when the plane of coil is in magnetic meridian, current in the coil is directly proportional to the tangent of deflection of magnetic needle. Current carrying coil (fixed) Compass needle

Tangent Galvanometer

918 JEE Main Physics Table 20.1 Comparative Study of Magnetic Materials S.No.

Diamagnetic Substances

Paramagnetic Substances

Ferromagnetic Substances

1.

These substances when placed in a magnetic field, acquire feeble magnetism opposite to the direction of the magnetic field.

These substances when placed in a magnetic field, acquire feeble magnetism in the direction of the magnetic field.

These substances when placed in a magnetic field are strongly magnetised in the direction of the field.

H M

2. 3.

These substances are repelled by a magnet. When a diamagnetic solution is poured into a U-tube and one arm is placed between the poles of strong magnet, the level of solution in that arm is lowered.

H

H

M

M

These substances are feebly attracted by a magnet. The level of the paramagnetic solution in that arm rises.

These substances are strongly attracted by a magnet. No liquid is ferromagnetic.

S

N

S

N

Paramagnetic solution

Diamagnetic solution

4.

If a rod of diamagnetic material is suspended freely between two magnetic poles, its axis becomes perpendicular to the magnetic field. n

Paramagnetic rod becomes parallel to the magnetic field. N

s B

N

S s

5.

6.

S

n

Axis

n B

S

Axis

In non-uniform magnetic field, they move from weaker to stronger part of the magnetic field slowly.

In non-uniform magnetic field, they move from weaker to stronger magnetic field rapidly. Their permeability is much greater than one (m >> 1.) Their susceptibility is large and positive. They also follow Curie’s law. 1 i .e., c µ . T At Curie temperature, ferromagnetic substances change into paramagnetic substances. No liquid is ferromagnetic.

7.

Their susceptibility is small and negative. Their susceptibility is independent of temperature.

8.

Shape of diamagnetic liquid in a glass crucible and kept over two magnetic poles.

Shape of paramagnetic liquid in a glass crucible and kept over two magnetic poles. Paramagnetic liquid

Diamagnetic liquid

10.

s

Axis

Their permeability is slightly greater than one (m > 1.) Their susceptibility is small and positive. Their susceptibility is inversely proportional to absolute temperature which is Curie’s 1 law, i .e., c µ . T

9.

N

B

In non-uniform magnetic field, the diamagnetic substances are attracted towards the weaker fields, i .e., they move from stronger to weaker magnetic field. Their permeability is less than one (m < 1.)

N

Ferromagnetic rod also becomes parallel to the magnetic field.

S

In these substances, the magnetic lines of force are farther than in air. The resultant magnetic moment of these substances is zero.

N

S

In these substances, the magnetic lines of force are closer than in air. These substances have a permanent magnetic moment.

In these substances, magnetic lines of force are much closer than in air. These substances also have a permanent magnetic moment.

Magnetostatics

20.10 Magnetisation of Materials To describe the magnetic properties of materials, we have to understand the following terms

Relation between m r and c m we have,

or

It is the degree or extent to which a magnetic field can magnetise a substance. Its SI unit is Am -1.

or But

The magnetic moment induced in unit volume of a magnetic substance placed in a magnetic field is called the intensity of magnetisation. It is denoted by I. Thus,

M m ´ 2l m Pole strength I = = = = V A ´ 2l A Area

i. e. , the intensity of magnetisation may also be defined as the pole strength per unit cross-sectional area. The unit of I is Am -1.

Magnetic Susceptibility The magnetic susceptibility is defined as the intensity of magnetisation per unit magnetising field i. e. , I cm = H Since, it is the ratio of two quantities having same units of Am-1, hence, it has no unit.

B = m 0 (I + H ) ö æ I B = m 0H ç + 1÷ ø èH

or

Magnetic Intensity

Intensity of Magnetisation

919

\

B = B0 (c m + 1) B = cm + 1 B0 B m = = m r = relative permeability B0 m 0 m r = cm + 1

20.11 Magnetic Materials According to behaviour of magnetic substances, they are classified into three classes

Diamagnetic Substances Those substances when placed in an external magnetic field acquire a very low magnetism in a direction opposite to the field, are called diamagnetic substances. These substances when brought near the end of a strong magnet, they get repelled.

Examples Copper (Cu), silver (Ag), bismuth (Bi), zinc (Zn), diamond (C), salt (NaCl), water (H2O ), mercury (Hg) , nitrogen (N2 ), hydrogen (H2 ), magnesium (Mg), gold (Au) etc.

Paramagnetic Substances

The magnetic flux density (B) inside a magnetised substance is given by the sum of magnetic field (B0 ) and magnetic field m 0I produced due to magnetisation.

Those substances when placed in a external field, acquire a feeble magnetism in the direction of field, are called paramagnetic substances. These substances when brought near the end of a strong magnet, get attracted towards it.

Thus,

Examples Aluminium (Al), sodium (Na), potassium (K),

Magnetic Flux Density

B = B0 + m 0I = m 0H + m 0I = m 0 (H + I )

Magnetic Permeability The magnetic permeability of a material is the measure of degree to which the materials can be permeated by a magnetic field and is defined as the rate of magnetic induction (B) in the material of the magnetising field. B m= i. e. , H Its SI units is Wb/Am.

Note Relative permeability is represented as mr =

m , where, m 0 = 4p ´ 10 -7 Wb/Am m0

platinum (Pt), manganese (Mn), copper sulphate (CuSO4), oxygen (O 2), etc.

Ferromagnetic Substances Those substances, when placed in a magnetic field, acquire a strong magnetism in the direction of field, are called ferromagnetic substances. These substances when brought near the end of a strong magnet get radially attracted towards it.

Examples Iron (Fe), nickel (Ni), cobalt (Co), magnetite (Fe3O4 ) or natural magnet etc.

920 JEE Main Physics N=

Important Points

7.9 ´ 10 -12 ´ 6.023 ´ 10 23 55

N = 8.65 ´ 10 10 atoms

1. I-M curve L

Paramagnetic

(a) For paramagnetic materials

The maximum possible dipole moment mmax is achieved for the case when all the atomic moments are perfectly aligned. Thus, mmax = (8.65 ´ 10 10) (9.27 ´ 10 -24) = 8 ´ 10 -13 Am2

M L

20.12 Curie’s Law and Curie Temperature

(b) For diamagnetic materials, Diamagnetic M

Curie’s Law

L

According to Curie’s law, the magnetic susceptibility of paramagnetic substances is inversely proportional to absolute temperature, i. e. , 1 cµ T

Ferromagnetic

(c) For ferromagnetic materials,

M

2. I-H curve I

Here, T = absolute temperature.

Ferromagnetic Paramagnetic

On increasing temperature, magnetic susceptibility of paramagnetic substances decreases or vice-versa.

H

Curie Temperature or Curie Point The magnetic susceptibility of these substances decreases on increasing the temperature and above a particular temperature, a ferromagnetic substance behaves like a paramagnetic substance.

Diamagnetic

3. c-T curve

χ

χ

χ

Diamagnetic Paramagnetic

T

(a)

(b)

Ferromagnetic

T

(c)

T

Sample Problem 17 A domain in ferromagnetic iron is in the form of cube of side length 1 mm. The molecular mass of iron is 55 g/mole and its density is 7.9 g/cm3. Assume that each iron atom has a dipole moment of 9.27 ´ 10 -24 Am 2. The maximum [NCERT] possible dipole moment is 5

(a) 8 ´ 10 Am (c) 8 ´ 10

-13

-1

Am

5

-1

13

2

(b) 4 ´ 10 Am 2

(d) 8 ´ 10

Am

Interpret (c) The volume of the cubic domain is V = (10 -6m)3 = 10 -18 m3 = 10 -12 cm3 Its mass is volume ´ density = 7.9 g cm-3 ´ 10 -12 cm3 = 7.9 ´ 10 -12 g It is given that Avogadro’s number (6.023 ´ 10 23) of iron atom have a mass of 55 g. Hence, the number of atoms in the domain is

This particular temperature temperature of the substance.

is

called

the

Curie

For example, the Curie temperature of iron is 770°C. It follows that at a temperature below 770°C, the iron is ferromagnetic and at a temperature above 770°C, the iron is paramagnetic. Similarly, the Curie temperature of nickel is 369°C and that of cobalt is 1150°C.

Curie-Wiess Law At temperature above, Curie point, the magnetic susceptibility of ferromagnetic substances is inversely proportional to (T - TC ) , i. e. , 1 cµ T - TC or

c=

C T - TC

Here, TC = Curie temperature.

χ

Tc

T

Magnetostatics

20.13 Hysteresis The lag of intensity of magnetisation behind the magnetising field during the process of magnetisation and demagnetisation of a ferromagnetic material is called hysteresis. Figure below show the magnetisation curve of a ferromagnetic material, when it is taken over a complete cycle of magnetisation. B

921

If the magnetising field is repeatedly changed between H 0 and - H 0, the curve ABCDEFA is retracted. This curve is called the hysteresis loop. The energy loss in magnetising and demagnetising a specimen is proportional to the area of hysteresis loop.

Note For steel coercivity is large. However, retentivity is comparatively smaller in case of steel. Due to high value of coercivity and fairly large value of retentivity steel is used to make permanent magnets. For soft iron, coercivity is very small and area of hysteresis loop is small. Because of these characteristics, soft iron is an ideal material for making electromagnets.

A

Check Point 2 Retentivity H0 C O H

H0 F

H

E

1. Suppose a man proposes a theory that the earth’s magnetic field is due to permanent magnetism of the molten iron core of the earth. Will you accept this theory? Give reasons in support of your answer.

2. A magnetic needle is placed on a cork floating on a still lake in the northern hemisphere. Does this needle together with the cork move towards the north of the lake.

D Coercivity

The graph follows that (i) Corresponding to point O, the magnetisation ( H ) is zero and likewise intensity of magnetisation ( I ) is also zero. (ii) As magnetising field is increased, intensity of magnetisation I also increases along OA and becomes maximum at A. This maximum value is called saturation value. (iii) If magnetic field is now decreased slowly, intensity of magnetisation decreases along the path AB. Corresponding to point B, magnetising field becomes zero, but some magnetisation equal to OB is still left in the specimen. Here, OB gives the measure of retentivity of the material of the specimen. (iv) If magnetic field H is reversed, the magnetisation decreases along BC, till it becomes zero corresponding to point C. Thus, to make I to be zero, magnetising field equal to OC has to be applied in reverse direction. Here, OC gives the measure of coercivity of the material of the specimen. (v) When field H is further increased in reverse direction i. e., along CD, the intensity of magnetisation attains saturation value corresponding to point D. (vi) Finally, when magnetising field is increased in original direction, the point A is reached via EFA.

3. Is the permeability of a ferromagnetic material independent of magnetic field? If not, is it more for lower or higher field?

4. An iron bar magnet is heated to 1000°C and then cooled in a magnetic field free space. Will it retain magnetism?

20.14 Demagnetisation It is clear from the hysteresis loop that the intensity of magnetisation I does not reduce to zero on removing the magnetising field H, it is zero when the magnetising field H is equal to the coerceive field. To demagnetise a substance, it is subjected to several cycles of 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. I

H

922 JEE Main Physics

20.15 Electromagnets Electromagnets are usually in the form of iron core solenoids. The ferromagnetic property of the iron core causes the internal magnetic domains of the iron to line up with the smaller driving magnetic field produced by the current in the solenoid. The effect is the multiplication of the magnetic field by factors of ten to eleven thousands. The solenoid field relationship is B = k m 0nI , where m = k m 0 and k is the relative permeability of the iron, the figure shows the magnetic effect of the iron core.

N = North pole N Iron core S S = South pole

Electromagnets are widely used in electric and electromechanical devices, including motors and generators. The main advantage of an electromagnet over a permanent magnet is that the magnetic field can be rapidly manipulated over a wide range by controlling the amount of electric current.

20.16 Permanent Magnets Substances which at room temperature retain their ferromagnetic property for a long period of time are called permanent magnets. Permanent magnets can be made in a variety of ways. An efficient way to make a permanent magnet is to place a ferromagnetic rod in a solenoid and pass a current. The magnetic field of the solenoid magnetises the rod. The hysteresis curve allows us to select the suitable materials for permanent magnets. The material should have high retentivity so that the magnet is strong and high coercivity so that the magnetisation is not erased by stray magnetic fields, temperature fluctuations or minor mechanical damage. Further, the material should have high retentivity. Other suitable materials for permanent magnets are alnico, cobalt steel and ticonal.

WORKED OUT Examples Example 1

The intensity of magnetic field at a point X on the axis of a small magnet is equal to the field intensity at another point Y on its equatorial axis. The ratio of distances of X and Y from the centre of the magnet will be (a) (2) -2

(b) (2) -1/3

(c) 23

(d) 21/3

Solution

If d1 is distance of point X on axial line and d 2 is distance of point Y on equatorial line m 2M m M then B1 = 0 3 , B2 = 0 3 4p d1 4p d 2 B1 = B2 m 0 2M m 0 M = 4p d13 4p d 32

As \

Example 2

The magnetic moment of a bar magnet of semilength 20 cm is 4 ´ 10 -6 Am 2. Its pole strength is (c) 80 ´ 10

Solution

Am

(b) 10 ´ 10 -6 Am2

2

(d) 40 ´ 10

-6

Am

Here, M = 4 ´ 10 -6 Am2, l = 20 cm =

\

2

1 m 5

m=? M = m ´ 2l

As m=

M 4 ´ 10 -6 = = 10 ´ 10 -6 Am 2l 2 ´ 1/ 5

A magnetic needle lying parallel to a magnetic field required W units of work to turn it through 60°. The torque required to maintain the needle in this position will be

Solution

Þ

t = 3W

Example 4

Units of pole strength of a magnet is

(a) Am-1 (c) Am-2

Solution

(b) Am2 (d) Am

Pole strength m =

M Am2 or = Am m 2l

(a) 2 ´ 10 -3 N

(b) 2 ´ 10 -4 N

(c) 2 ´ 10 5 N

(d) 2 ´ 10 -5 N

Solution

Force, F =

30 ´ 60 m 0 m1 m2 = 10 -7 ´ = 2 ´ 10 -3 N 4p r 2 (0.3)

Example 6

The total intensity of the earth’s magnetic field at equator is 5 units. What is its value at the poles? (a) 5

Solution

(b) 4

(c) 3

(d) 2

Equator, q = 0° , BH = Be cos q = 5 cos 0° = 5

At poles, q = 90° , V = Be sin q = 5 sin 90°= 5 Thus, the total intensity of earth's magnetic field at poles is also 5 units, but in the vertical direction. At equator, value is the same, but the direction is horizontal.

Example 7

Example 3 (a) 3W 3 (c) W 2

æ MB ö t=ç ÷ 3 è 2 ø

The isolated point poles of strength 30 Am and 60 Am are placed at a distance of 0.3 m. The force of repulsion is

d1 = 21/3 d2

-6

\

MB 3 2

Example 5

d13 = d 32

(a) 20 ´ 10 -6 Am2

t = M. B sin q = MB sin 60° =

and

(b) W (d) 2W Work done W = MB (cos q1 - cos q2) = MB(cos 0° - cos 60° ) æ 1 ö MB = MB ç1 - ÷ = è 2ø 2

A dip needle lies initially in the magnetic meridian when it shows an angle of dip q at a place. The dip circle is rotated through an angle x in the horizontal plane and tan q ¢ is then it shows an angle of dip q ¢. Then tan q (a)

1 cos x

Solution

(b)

1 sin x

Here, tan q =

(c)

1 tan x

(d) cos x

BV BV , or tan q¢ = BH BH cos x tan q¢ 1 = tan q cos x

924 JEE Main Physics Example 8

The earth's magnetic field may be considered to be due to a short magnet placed at the centre of earth and oriented along magnetic south-north direction. The ratio of magnitude of magnetic field on earth’s surface at magnetic equator and that at magnetic poles is (a) 1 : 2 (c) 1 : 4

(b) 2 : 1 (d) 4 : 1

Example 11

A dip needle vibrates in the vertical plane perpendicular to the magnetic meridian. The time period of vibration is found to be 2s. The same needle is then allowed to vibrate in the horizontal plane, and the time period is again found to be 2s. Then the angle of dip is (a) 0° (c) 50°

Solution

Point on magnetic equator is on equatorial line of magnet and at poles, it is on axial line Bequatorial 1 / 2 1 \ = = Baxial 1 2

Solution

(b) 30° (d) 90° Time period t1 = 2 = 2p

I MH V V = H tan d = = 1, H t1 = 2 = 2 p

Similarly,

Example 9

At a given place on the earth's surface, the horizontal component of earth’s magnetic field is 3 ´ 10 -5 T and resultant magnetic field is 6 ´ 10 place is (a) 30° (c) 50°

Solution

-6

T. Angle of dip at this

BH = Be cos q B 3 ´ 10 -6 1 = Þ q = 60° cos q = H = Be 6 ´ 10 -5 2

moment

of

(15 cm ´ 2 cm ´ 1 cm ) is 1.2 Am . What is its magnetisation? (a) 4 ´ 10 Am (c) 10 4 Am-1

-1

4

(b) 2 ´ 10 Am

0.0045 0.0030 0.015 0.0075

Solution

a

2

4

Example 12 (a) (b) (c) (d)

Horizontal component of earth's magnetic field

The

d = 45°

The magnetic susceptibility of a paramagnetic material at -73 ° C is 0.0075, Its value at -173° C will be

(b) 40° (d) 60°

Example 10

I MV

magnet intensity

c m2

of As

-1

(d) None of these

Magnetic susceptibility

Þ

c m1 = 0.0075, T1 = - 73° C = ( -73 + 273) K = 200 K = ?, T2 = - 173° C = ( -173 + 273) K = 100 K 1 cm µ T cm 200 T 2 = 1 = =2 cm T2 100 1

Solution

Intensity of magnetisation M 1.2 I= = = 4 ´ 10 4 Am-1 V (15 ´ 2 ´ 1) 10 -6

\

c m = 2 c m = 2 ´ 0.0075 = 0.015 2

1

Start Practice for

JEE Main Round I

(Topically Divided Problems)

Fields due to Magnetic Dipole, Torque on Dipole and its Potential Energy 1. A magnet of magnetic moment m and pole strength m is divided in two equal parts, the magnetic moment of each part will be (a) M (c) M / 4

(b) M /2 (d) 2 M

400 ab-amp cm2 and 800 ab-amp cm2 are placed with their axis in the same straight line with similar poles facing each other and with their centres at 20 cm from each other. Then the force of repulsion is (a) 12 dyne (c) 800 dyne

2. The magnetic potential due to a magnetic dipole at a point on its axis distant 40 cm from its centre is found to be 2.4 ´ 10–5 JA–1m–1. The magnetic moment of the dipole will be (a) 28.6 Am2 (c) 38.4 Am2

6. Two short bar magnets with magnetic moments

(b) 32.2 Am2 (d) None of these

3. A magnetic needle lying parallel to a magnetic field required W units of work to turn it through 60°. The torque required to maintain the needle in this position will be

(b) 6 dyne (d)150 dyne

7. Two magnets have the same length and the same pole strength. But one of the magnets has a small hole at its centre. Then (a) both the equal magnetic moment (b) one with hole has smaller magnetic moment (c) one with hole has large magnetic moment (d) one with hole loses magnetism through the hole

8. Two magnets of equal magnetic moments M each are placed as shown in figure. The resultant magnetic moment is

(b) W

(a) 3 W W (c) 3 2

S

(d) 2 W

4. A bar magnet of length 3 cm has a point A and B along axis at a distance of 24 cm and 48 cm on the opposite ends. Ratio of magnetic fields at these points will be A 24 cm

(a) 8 (c) 4

48 cm

(b) 3 (d) 1/2 2

5. Rate of change of torque t with deflection q is maximum for a magnet suspended freely in a uniform magnetic field of induction B, when (a) q = 0º (c) q = 60º

S

B O

(b) q = 45º (d) q = 90º

60°

N

(a) M (c) 2 M

N

(b) 3 M (d) M/2

9. A short bar magnet placed with its axis at 30º with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10 – 2 J, then the magnitude of magnetic moment of the magnet is? [NCERT]

(a) 0.36 J/T (c) 8.6 J/T

(b) 3.6 J/T (d) 0.86 J/T

926 JEE Main Physics 10. In which orientation the resultant magnetic moment

16. A short bar magnet with the north pole facing north

of two magnets, will be zero, if magnetic moment of each magnets is M in the following figures?

forms a neutral point a P in the horizontal plane. If the magnet is rotated by 90° in the horizontal plane, the net magnetic induction at P is (Horizontal component of earth’s magnetic field= BH )

N

S

(a)

N

(a) zero 5 (c) BH 2

(b) S

N S

N

S

S N

S N

(d) 60° N

S

11. A closely wound solenoid of 800 turns and area of

cross-section 2.5 × 10 - 4 m2 carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment? [NCERT] (a) 6 J/T

(b) 0.9 J/T

(c) 9 J/T

(d) 0.6 J/T

12. A bar magnet is cut into two equal halves by a plane parallel to the magnetic axis. Of the following physical quantities, the one which remains unchanged is (a) pole strength (b) magnetic moment (c) intensity of magnetisation (d) moment of inertia

13. A toroid of n turns, mean radius R and cross-sectional radius a carries current I. It is placed on a horizontal table taken as x-y plane. Its magnetic [NCERT Exemplar] moment m (a) is non-zero and points in the z-direction by symmetry (b) points along the axis of the toroid (m = m) 1 (c) is zero, otherwise there would be a field falling as 3 at r large distances outside the toroid (d) is pointing radially outwards

Magnetism of Earth and Neutral Points 14. The vertical component of earth’s magnetic field is zero at or the earth’s magnetic field always has a vertical component except at the [NCERT Exemplar] (a) Magnetic poles (c) Every place

(d) 5 B H

17. The earth’s magnetic induction at a certain point is

N N (c) S

(b) 2 B H

(b) Geographic poles (d) Magnetic equator

7 ´ 10–5 Wbm–2. This is to be annulled by the magnetic induction at the centre of a circular conducting loop of radius 15 cm. The required current in the loop is (a) 0.56 A (c) 0.28 A

(b) 5.6 A (d) 2.8 A

18. A bar magnet is placed north-south with its north pole due north. The points of zero magnetic field will be in which direction from centre of magnet (a) north and south (b) east and west (c) north-east and south-west (d) north-east and south-east

19. The magnetic field of the earth can be modelled by that of a point dipole placed at the centre of the earth. The dipole axis makes an angle of 11.3° with the axis of the earth. At Mumbai, declination is nearly zero.Then. [NCERT Exemplar] (a) the declination varies between11.3o W to11.3o E (b) the least declination is 0 o (c) the plane defined by dipole axis and Earth axis passes through Greenwich (d) declination averaged over Earth must be always negative

20. A magnet is placed on a paper in a horizontal plane for locating neutral points. A dip needle placed at the neutral point will be horizontal at the (a) magnetic poles (c) latitude angle 45º

(b) magnetic equator (d) latitude angle of 60º

21. A bar magnet 20 cm in length is placed with its south pole towards geographic north. The neutral points are situated at a distance of 40 cm from centre of the magnet. If horizontal component of earth’s field 3.2 ´ 10–5 T , then pole strength of magnet is (a) 5 Am

(b) 10 Am

(c) 45 Am

(d) 20 Am

22. In a permanent magnet at room temperature

15. The earth’s magnetic field at a certain place has a

[NCERT Exemplar]

horizontal component of 0.3G and total strength 0.5 G. Find angle of dip in tan -1.

(a) magnetic moment of each molecule is zero (b) the individual molecules have non-zero magnetic moment which are all perfectly aligned (c) domains are partially aligned (d) domains are all perfectly aligned

4 3 -1 5 (c) d = tan 3

(a) d = tan -1

3 4 -1 3 (d) d = tan 5

(b) d = tan -1

Magnetostatics

927

23. At a certain place, the horizontal component of the

30. The time period of a thin bar magnet in earth’s

earth’s magnetic field is B0 and the angle of dip is 45°. The total intensity of the field at that place will be

magnetic field is T. If the magnet is cut into four equal parts perpendicular to its length, the time period of each part in the same field will be

(a) B0

(b) 2 B0

(d) B02

(c) 2 B0

24. Consider the two idealized systems: (i) a parallel

(a) T/2

(b) T/4

(c) 2T

(d) 2 T

plate capacitor with large plates and small separation and (ii) a long solenoid of length L >> R, radius of cross-section. In (i) E is ideally treated as a constant between plates and zero outside. In (ii) magnetic field is constant inside the solenoid and zero outside. These idealised assumptions, however, contradict fundamental laws as below

31. A magnet freely suspended in a vibration

[NCERT Exemplar]

32. Two magnets held together in earth’s magnetic field

(a) 9 : 1

(d) case (ii) contradicts ò H . dl = Ien

of magnetisation ( I ) with respect to the magnetising field (H) in a diamagnetic substance is described by the graph in figure. (a) OD (c) OB

A H

O

C D

26. A

rod of ferromagnetic material with dimensions 10 cm × 0.5 cm × 0.2 cm is placed in a magnetic field of strength 0.5 × 104 A-m–1 as a result of which a magnetic moment of 5 A-m–2 is produced in the rod. The value of magnetic induction will be (a) 0.54 T (c) 0.358 T

(b) 6.28 T (d) 2.591 T

27. The space inside a toroid is filled with tungusten

shoes susceptibility is 6.8 × 10–5. The percentage increase in the magnetic field will be (a) 0.0068% (c) 0.68%

28. The horizontal component of flux density of earth’s magnetic field is 1.7 × 10–5 T. The value of horizontal component of intensity of earth’s magnetic field will be (b) 13.5 Am–1 (d) 0.35 Am–1

29. A loop of area 0.5m2 is placed in a magnetic field of strength 2 T in direction making an angle of 30° with the field. The magnetic flux linked with the loop will be 1 (a) Wb 2

(b)

3 Wb 2

(c) 2Wb

(b) 1 : 3

(c) 1 : 9

(d) 5 : 4

(d)

horizontal plane at a place where the angle of dip is 45° and the total intensity is 0.707 CGS units. The number of oscillations per minute at a place where dip angle is 60° and total intensity is 0.5 CGS units will be (a) 5

(b) 7

(c) 9

(d) 11

34. A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 5.0 × 10 - 2 T. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0/s. What is the moment of inertia of the coil about its axis of rotation? (a) 1.2 ´ 10 -4 g-cm2

(b) 3 ´ 10 -4 kg-m2

(c) 0.3 ´ 10 -4 kg-m2

(d) 1.2 ´ 10 -4 kg-m2

35. At a certain place a magnet makes 30 oscillations per minute. At another place where the magnetic field is double, its time period will be

(b) 0.068% (d) None of these

(a) 24.5 Am–1 (c) 1.53 Am–1

(b) 9 × 10–6 T (d) 228 × 10–6 T

33. A magnet performs 10 oscillations per minute in a B

+1

–1

(b) OC (d) OA

(a) 36 × 10–6 T (c) 144 × 10–6 T

with same polarity together make 12 vib-min–1 and when opposite poles together make 4 vib-min–1. The ratio of magnetic moments is

(a) case (i) contradicts Gauss's law for electrostatic fields. (b) case (i) contradicts Gauss's law for magnetic fields. (c) case (i) agrees with ò E. dl = 0.

25. The variation of the intensity

magnetometer makes 40 oscillations per minute at a place A and 20 oscillations per minute at a place B. If the horizontal component of earth’s magnetic field at A is 36 × 10–6 T, then its value at B is

3 Wb 2

(a) 2 s

(b) 2 s

(c) 4 s

(d)

1 s 2

36. Two bar magnets having same geometry with magnetic moments M and 2 M are firstly placed in such a way that their similar poles are same side. Time period of oscillations is T1. Now the polarity of one of the magnets is reversed, and time period of oscillations is T1. Now the polarity of one of the magnets is reversed and time period of oscillation is T2 (a) T1 < T2 (c) T1 > T2

(b) T1 = T2 (d) T2 = ¥

928 JEE Main Physics 37. The magnetic needle of a tangent galvanometer is

44. The time of vibration of a dip needle vibrating in the

deflected at an angle 30° due to a magnet. The horizontal component of earth’s magnetic field 0.34 ´ 10–4 T is along the plane of the coil. The magnetic intensity is

vertical plane is 3s. When magnetic needle is made to vibrate in the horizontal plane, the time of vibration is 3 2 s. Then the angle of dip is

(a) 1.96 × 10–4 T (c) 1.96 × 10–5 T

(b) 1.96 × 104 T (d) 1.96 × 105 T

38. The period of oscillation of a freely suspended bar magnet is 4 s. If it is cut into two equal parts in length, then the time period of each part will be (a) 4 s

(b) 2 s

(c) 0.5 s

(d) 0.25 s

39. Two tangent galvanometers having coils of the same radius are connected in series. A current flowing in them produces deflections of 60° and 45° respectively. The ratio of the number of turns in the coils is (a) 4/3

(b) ( 3 + 1) /1

3 +1 (c) 3 -1

3 (d) 1

(a) 30º (b) 45º (c) 60º (d) 90º

45. A compass needle placed at a distance r from a short magnet in tan A position shows a deflection of 60°. If the distance is increased to r (2)1/ 3, the deflection of compass needle is (a) 30º (c) 45º

(b) 60º (d) 0º

46. The correct I-H curve for a paramagnetic material is represented by, figure. Y

Y

I

I

(a)

(b)

40. A paramagnetic sample shows a net magnetisation of

8 Am -1 when placed in an external magnetic field of 0.6 T at a temperature of 4K. When the same sample is placed in an external magnetic field of 0.2 T at a temperature of 16 K, the magnetisation will be

O

[NCERT Exemplar]

32 (a) Am-1 3 (c) 6 Am-1

2 (b) Am-1 3 (d) 2.4 Am-1

(b) 0.14 A (d) 3.6 × 10–5 A

field with a period T. What happens to its period of motion if its mass is quadrupled? Motion remains SHM with time period = T/2 Motion remains SHM and period remains nearly constant Motion remains SHM with time period = 2T Motion remains SHM with time period = 4 T

43. A vibration magnetometer consists of two identical bar magnets placed one over the other such that they are perpendicular and bisect each other. The time period of oscillation in a horizontal magnetic field is 2 5 / 4 s. One of the magnets is removed and if the other magnet oscillates in the same field, then the time period in second is (b) 21/2

(c) 2

Y

I

I

H

X

(d)

H

X

O

H

X

47. A circular current loop of magnetic moment is in an arbitrary orientation in an external magnetic field B. The work done to rotate the loop by 30° about an axis [NCERT Exemplar] perpendicular to is planers (a) MB

42. A bar magnet is oscillating in the Earth’s magnetic

(a) 21/4

O

Y

O

radius of 15 cm. The horizontal component of the earth’s magnetic field is 3 ´ 10–5 T. The current required to produce a deflection of 45° in it is

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

X

(c)

41. A tangent galvanometer has a coil of 25 turns and a

(a) 0.29 A (c) 1.2 A

H

(d) 4

(b) 3

MB 2

MB 2 (d) 0 (c)

48. Two bar magnets of the same mass, same length and breadth but having magnetic moments M and 3M are joined together pole and suspended by a string. The time period of assembly in a magnetic field of strength H is 3 s. If now the polarity of one of the magnets is reversed and the combination is again made to oscillate in the same field, the time of oscillation is (a) 3 s (c) 3 / 3 s

(b) 3 3 s (d) 6 s

Magnetostatics 49. The variation of magnetic susceptibility ( c) with

+ve

temperature for a diamagnetic substance is best represented by figure

+ve

c

(c) c

(d)

(b) (0, 0)

(0, 0)

T

O

T

T

50. A uniform magnetic field parallel to the plane of paper, exsisted in space initially directed from left to right. When a bar of soft iron is placed in the field parallel to it, the lines of force passing through it will be represented by figure.

(a)

–ve

magnetic field region. The rod when in equilibrium will align itself

c c

(d)

T

52. A copper rod is suspended a non-homogeneous

T

T

O

c

c

(a)

(c)

929

(a) in the region where magnetic field is strongest (b) in the region where magnetic field is weakest and parallel to direction of magnetic field there (c) in the direction in which it was originally suspended (d) in the region where magnetic field is weakest and perpendicular to the direction of magnetic field there

53. The relative permeability of a substance X is slightly less than unity and that of substance Y is slightly more than unity, then (a) X is paramagnetic and Y is ferromagnetic (b) X is diamagnetic and Y is ferromagnetic (c) X and Y both are paramagnetic (d) X is diamagnetic and Y is paramagnetic

(b)

54. The magnetising field required to be applied in (c)

opposite direction to reduce residual magnetism to zero is called

(d)

51. The variation of magnetic susceptibility ( c) with absolute temperature T for a ferromagnetic is given in figure, by +ve

(a) short and wide (b) tall and narrow (c) tall and wide (d) short and narrow

c

(a)

(b)

(0, 0)

T

(0, 0)

T

Round II Only One Correct Option 1. The susceptibility of a paramagnetic material is K at 27 °C. At what temperature will its susceptibility be K/2? (a) 600° C (c) 54° C

(b) 287° C (d) 327° C

55. The hysteresis cycle for the material of a transformer core is

+ve

c

(a) coercivity (b) retentivity (c) hysteresis (d) None of the above

(Mixed Bag) 2. Force between two identical short bar magnets whose centres are r metre apart is 8.1 N, when their axes are along the same line. If separation is increased to 3 r and the axis are rearranged perpendicularly, the force between them would become (a) 2.4 N (c) 0.1 N

(b) 1.2 N (d) 1.15 N

930 JEE Main Physics 3. A bar magnet has a magnetic moment equal to

5 ´ 10-5 Wb-m. It is suspended in a magnetic field which has a magnetic induction B equal to 8 p ´ 10-4 T. The magnet vibrates with a period of vibration equal to 15 s. The moment of inertia of magnet is (a) 4.54 × 10 4 kg-m2 (c) 4.54 × 10 -4 kg-m2

(b) 4.54 × 10 -5 kg-m2 (d) 4.54 × 10 5 kg-m2

4. In a vibration magnetometer, the time period of a bar magnet oscillating in horizontal component of earth’s magnetic field is 2 s. When a magnet is brought near and parallel to it, the time period reduces to 1 s. The ratio F / H of the fields, F due to magnet and H, the horizontal component will be 1 (b) 3

(a) 3

1 (c) 3

(d) 3

is 0.6 kg and the density is 7.8 ´ 103 kg m–3. If the area of hysteresis loop of alternating magnetising field of frequency 50 Hz is 0.722 MKS units, then hysteresis loss per second will be (a) 27.77 × J (c) 27.77 × 10–4 J

10–5

value of 4 p2 I / T2 for a short bar magnet is observed as 36 ´ 10-4 . In the experiment with deflection magnetometer with the same magnet, the value of 4 pd3/ 2 m 0 is observed as 108 /36. The magnetic moment of the magnet used is (b) 100 A-m (c) 200 A-m

(d) 1000 A-m

m2 is pivoted about a vertical diameter in a uniform horizontal magnetic field and carries a current of 2A. When the coil is held with its plane is N-S of 2A. When the coil is held with its plane in N-S direction, it experience a couple of 0.04 N-m; and when its plane is E-W, the corresponding couple is 0.03 N-m. The magnetic induction is

7. A coil of 50 turns and area 1.25 ×

(a) 0.2 T

(b) 0.3 T

10–3

(c) 0.4 T

(d) 0.5 T

8. Two short bar magnets of equal dipole moment M are fastened perpendicularly at their centres, figure. The magnitude of resultant of two magnetic field at a distance d from the centre on the bisector of the right angle is (a) (c)

m0 2 2 M 4 p d3

(b)

m0 M 4 p d3

(d)

(b) 2 vibs-m–1 (d) 20 vibs-m–1

10. A long magnet is placed vertically with its S-pole resting on the table. A neutral point is obtained 10 cm from the pole due geographic north of it. If H = 3.2 ´ 10–5 T , then the pole strength of magnet is (a) 8 ab-A-cm–1 (c) 32 ab-A-cm–1

(b) 16 ab-A-cm–1 (d) 64 ab-A-cm–1

meridian and the apparent dip is d1. It is then set in a vertical plane perpendicular to the geographic meridian. The apparent dip angle is d2 . The declination q at the place is (a) q = tan -1 (tan d1 tan d2 ) æ tan d1 ö (c) q = tan -1 ç ÷ è tan d2 ø (d) q = tan -1 (tan d1 - tan d2 )

12. A dip needle vibrates in the vertical plane perpendicular to magnetic meridian. The time period of vibration is found to be 2 s. The same needle is then allowed to vibrate in the horizontal plane and time period is again found to be 2 s. Then the angle of dip is (a) 0º (c) 45º

(b) 30º (d) 90º

13. A bar magnet of length 10 cm and having pole

strength equal to 10–3 Wb is kept in a magnetic field having magnetic induction B equal to 4 p ´ 10-3 T It makes an angle of 30° with the direction of magnetic induction. The value of the torque acting on the magnet is (a) 0.5 Nm (c) p ´ 10 -5 Nm

S

(b) 2p ´ 10 -5 Nm (d) 0.5 ´ 10 –5 Nm

14. A deflection magnetometer is adjusted in the usual Q

N d P

m 0 2M 4 p d3 m0 4p

(a) 10 vibs-m–1 (c) 4 vibs-m–1

(b) q = tan -1 (tan d1 + tan d2 )

(b) 2.777 × J (d) 27.77 × 10–6 J

6. In an experiment with vibration magnetometer, the

(a) 50 A-m

makes 10 oscillations per min under the action of earth’s magnetic field alone. When a bar magnet is placed at some distance along the axis of the needle, it makes 14 oscillations per min. If the bar magnet is turned so that its poles interchange their positions, then the new frequency of oscillation of the needle is

11. The plane of a dip circle is set in the geographic

5. The mass of a specimen of a ferromagnetic material

10–5

9. The magnetic needle of an oscillation magnetometer

2M d3

N

S

way. When a magnet is introduced, the deflection observed is q and the period of oscillation of the needle in the magnetometer is T. When the magnet is removed, the period of oscillation is T0 . Find the relation between T and T0 is (a) T2 = T02 cos q

(b) T = T0 cos q

T (c) T = 0 cos q

(d) T 2 =

T02 cos q

Magnetostatics 15. Two magnets of equal mass are

(a) 75º

(b) 60º

22. At the magnetic north pole of the earth, the value of

H

joined at 90° to each other as N1 shown in figure. Magnet N1S1 has a magnetic moment 3 times that of N2 S2 . The arrangement is pivoted so that it is free to rotate in horizontal plane. When in S2 equilibrium, what angle should N1S1 make with magnetic meridian? (c) 30º

N2 θ

S1

(d) 45º

centre by a thread. Its upper end is now loaded with a mass of 50 mg, and the needle becomes horizontal. If the strength of each pole is 98.1 ab-amp-cm and g = 981 cms –2 , then the vertical component of earth’s magnetic induction is (b) 0.25 G (d) 0.05 G

on the table. A neutral point was found at 20 cm from the pole. What is the pole strength if the vertical component of earth’s field is 0.4 × 10–4 Wbm–2? (b) 8 A-m (d) None of these

18. A magnet makes 5 oscillations per min in –4

B = 0.3 ´ 10 T . By what amount should the field be increased so that number of oscillations is 10 in the same time? 10–4

(a) 0.3 × T (c) 0.9 × 10–4 T

10–4

(b) 0.6 × T (d) 1.2 × 10–4 T

and 100 ab-amp-cm are placed with their axes in the same vertical line, with similar poles facing each other. Each magnet has a length of 1 cm. When separation between the nearer poles is 1 cm, the weight of upper magnet is supported by the repulsive force between the magnets. If g = 1000 cms –2 , then the mass of upper magnet is (b) 55 g

(c) 45 g

(d) 77.5 g

20. A magnet 20 cm long with its poles concentrated at its ends is placed vertically with its north pole on the table. At a point due 20 cm south (magnetic) of the pole, a neutral point is obtained. If H = 0.3 G, then the pole strength of the magnet is approximately (a) 185 ab-amp-cm (c) 18.5 ab-amp-cm

(b) 185 amp-m (d) 18.5 amp-m

21. A magnetic dipole is placed at right angles to the direction of lines of force of magnetic induction B. If it is rotated through an angle of 180°, then the work done is (a) MB

(b) 2 MB

(c) –2 MB

23. The period of oscillation of a bar magnet in a vibration magnetometer is 2 s. The period of oscillation of another bar magnet whose magnetic moment is 4 times that of Ist magnet is (a) 4 s (c) 2 s

(b) 1 s (d) 0.5 s

24. A steel wire of length l has a magnetic moment M. It is bent at its middle point at an angle of 60°. Then the magnetic moment of new shape of wire will be (a) M / 2

(b) M /2

(c) M

(d) 2 M

axis of 2 cm long bar magnet at large distances x and 3 x from the centre on opposite sides. The ratio of magnetic fields at A and B will be approximately equal to (a) 27 : 1 (c) 9 : 1

(b) 1 : 27 (d) 1 : 9

26. At a certain place, a magnet makes 30 oscillations per min. At another place where the magnetic field is double, its time period will be (a) 4 s (c) 1/2 s

(b) 2 s (d) 2 s

27. At a certain location in Africa, a compass points 12°

19. Two short magnets with pole strengths of 900 ab amp-cm

(a) 100 g

(b) maximum, minimum (d) minimum, minimum

25. The points A and B are situated perpendicular to the

17. A very long magnet is placed vertically with one pole

(a) 16 A-m (c) 4 A-m

the horizontal component of earth’s magnetic field and angle of dip are respectively (a) zero, maximum (c) maximum, maximum

16. A uniform magnetic needle is suspended from its

(a) 0.50 G (c) 0.005 G

931

(d) zero

west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60° above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. Specify the direction and magnitude of the earth’s field at the location. (a) 32° west of geographical meridian and 3.2 ´ 10 -4 T (b) 12° west of geographical meridian and 0.32 ´ 10 -4 T (c) 12° east of geographical meridian and 0.32 ´ 10 -4 T (d) 32° east of geographical meridian and 3.2 ´ 10 -4 T

More Than One Correct Option 28. A horizontal circular loop carries a current that looks anti-clockwise when viewed from above. It is replaced by an equivalent magnetic dipole N-S. Which of the following is true? (a) (b) (c) (d)

The line N-S should be along a diameter of the loop The line N-S should be perpendicular to the plane of the loop South pole should be below the loop North pole should be below the loop

932 JEE Main Physics 29. S is the surface of a lump of magnetic material.

33. What is represented by F in the relation F = H tan q? (a) F is magnetic field of earth (b) F is magnetic field of bar magnet (c) F is magnetic field due to current in coil of tangent galvanometer (d) None of the above

[NCERT Exemplar]

(a) Lines of B are necessarily continuous across S (b) Some lines of B must be discontinuous across S (c) Lines of H are necessarily continuous across S (d) Lines of H cannot all be continuous across S

30. Current flows through a straight cylinderical

34. Value of H on earth’s surface is (a) 0.32 G (b) 0.32 T (c) 0.32 × 10–4 G (d) 0.32 ×10–5 T

conductor of radius r. The current is distributed uniformly over its cross-section. The magnetic field at a distance x from the axis of the conductor has magnitude B (a) B = 0 at the axis 1 (c) B µ for n > r n

(b) B µ x for 0 £ x £ r

35. In the use of tangent galvanometer, H is (a) earth’s magnetic field (b) horizontal component of earth’s magnetic field (c) vertical component of earth’s magnetic field (d) None of the above

(d) B is maximum for n = r

31. A long solenoid has 1000 turs per metre and carries a current of 1 A. It has a soft iron core of m r = 1000. The core is heated beyond the Curie temperature Tc . [NCERT Exemplar]

(a) The H field in the solenoid is (nearly) unchanged but the B field decreases drastically (b) The H and B fields in the solenoid are nearly unchanged (c) The magnetisation in the core reverses direction (d) The magnetisation in the core diminishes by a factor of about 10 8

32. Essential difference between electrostatic shielding by a conducting shell and magnetostatic shielding is due to [NCERT Exemplar] (a) electrostatic field lines can end on charges and conductors have free charges (b) lines of B can also end but conductors cannot end them (c) lines of B cannot end on any material and perfect shielding is not possible (d) shells of high permeability materials can be used to divert lines of B from the interior region

Passage I A tangent galvanometer is used for detection and measurement of low electric currents. It is based on tangent law in magnetism, according to which F = G tan q, where q is angle with H made by a magnet suspended freely under the combined effect of m 2 pnI H and ( F ^ H ). Now, F = 0 , where n is 4p r number of turns in the coil of radius, r carrying current I. From, F = H tan q , we get 2 rH tan q = K tan q I= m0 n reduction

factor

36. At a particle place, V = H . The angle of dip is (a) 45º (c) 0º

of

tangent

(b) 90º (d) None of these

37. Magnetic field on the surface of earth is of the order of (a) 10– 4 T (c) 10–5 T

(b) 10–4 G (d) 10–5 G

38. What is the order of magnetic declination at a place on earth? (a) 20º East (c) 20º West

Comprehension Based Questions

where K is called galvanometer.

Passage II Magnetic field of earth is identical to magnetic field of a giant magnet held 20° west of geographic N-S at the centre of earth. At equator, horizontal component of earth is 0.32 G. Vertical component can be calculated from the relation V = H tan d, where d is angle of dip at the place. The value of d = 0° at equator and d = 90° at poles.

(b) 10º East (d) 10º West

Matching Type 39. Match the following column I with column II Column I Column II I. Magnetic moment A. [ML0 T –2 A –1] II. Permeability B. Vector III. Intensity of magnetisation C. Nm 3/Wb IV. Magnetic induction D. Scalar Code (a) I-A, II-B, III-C, IV-D (b) I-C, II-D, III-B, IV-A (c) I-D, II-C, III-A, IV-B (d) I-B, II-A, III-B, IV-D

Magnetostatics

933

43. Assertion An iron bar magnet is heated to 1000°C and

Assertion and Reason Directions

Question No. 40 to 48 are Assertion-Reason type. Each of these contains two Statements: Statement I (Assertion), Statement II (Reason). Each of these questions also has four alternative choice, only one of which is correct. You have to select the correct choices from the codes (a), (b), (c) and (d) given below (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 correct explanation of the Assertion. (c) If Assertion is true but Reason is false. (d) If Assertion is false but the Reason is true.

40. Assertion If a bar magnet is split parallel to its length in n parts, then magnetic moment of each part becomes 1/n times. Reason In this case the pole strength of each part remains as before but length becomes 1/n times.

41. Assertion The earth’s magnetic field undergoes a change with time. Reason The permeability of a ferro-magnetic material is depend upon the magnetic field.

42. Assertion Time period of oscillation of two magnets when like poles are in same direction (in a vibration magnetometer) is smaller, than the period of vibration when like poles are in opposite direction. Reason Moment of inertia increases in same position.

then cooled in a magnetic field in free space. It does not retain its magnetism. Reason Only ferro-magnets can show hysteresis.

44. Assertion The true geographic north direction cannot be found by using a compass needle. Reason The magnetic meridian of the earth is along the axis of rotation of the earth.

45. Assertion When a magnet is brought near iron nails, translatory force as well as a torque act on it. Reason The field due to a magnet is generally uniform.

46. Assertion Magnetic dipole possesses maximum potential energy when magnetic moment and magnetic field are parallel to each other. Reason Current loop is treated as a magnetic dipole.

47. Assertion A magnetic needle suspended by a silk thread is vibrating in the earth’s magnetic field. If the temperature of the needle is increased by 100°C, then the magnetic needle stops vibrating. Reason Time period of needle increases.

48. Assertion Susceptibility is defined as the ratio of intensity of magnetisation I to magnetic intensity H. Reason Greater the value of susceptibility smaller value of intensity of magnetisation I.

Previous Years’ Questions 49. A charge Q is uniformly distributed over the surface of

50. A horizontal straight wire 20 m long extending from

non-conducting disc of radius R. The disc rotates about an axis perpendicular to its plane and passing through its centre with an angular velocity w . As a result of this rotation, a magnetic field of induction B is obtained at the centre of the disc. If we keep both the amount of charge placed on the disc and its angular velocity to be constant and vary the radius of the disc then the variation of the magnetic induction at the centre of the disc will be represented by the figure.

east to west is falling with a speed of 5.0 m/s, at right angles to the horizontal component of the earth’s magnetic field 0.30 ´ 10–4 Wb / m 2 . The instantaneous value of the emf induced in the wire will be

[AIEEE 2012] B

(b)

(b) 3 mV

(c) 4.5 mV

(d) 1.5 mV

51. The magnitude of the magnetic field ( B) due to loop ABCD at the origin (O) is (a) zero (c)

B

(a)

[AIEEE 2011]

(a) 6.0 mV

m 0I é b - a ù 4 p êë ab úû

[AIEEE 2010]

m I ( b - a) (b) 0 24 ab m 0I é p ù (d) 2 ( b - a ) + ( a + b) ú 4 p êë 3 û

52. Due to the presence of the current I1 at the origin [AIEEE 2010]

R

R

B

B

(c)

(d) R

R

(a) The forces on AB and DC are zero (b) The forces on AD and BC are zero (c) The magnitude of the net force on the loop is given by m 0 I1 é p ù 2 ( b - a ) + ( a + b) ú 4 p êë 3 û (d) The magnitude of the net force on the loop is given by m 0 I1 ( b - a) 24 ab

934 JEE Main Physics 53. A bar magnet having a magnetic moment of

2 ´ 104 JT –1 is free to rotate in a horizontal plane. A horizontal magnetic field B = 6 ´ 10– 4 T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction 60° from the field is [Kerala CET 2009] (a) 2 J (c) 12 J

(b) 0.6 J (d) 6 J

54. If a diamagnetic substance is brought near the north or the south pole of a bar magnet, it is [Karnataka CET 2009]

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

61. Needles N1, N2 and N 3 are made of a ferro-magnetic,

a para-magnetic and a dia-magnetic substance respectively. A magnet when brought close to them will [AIEEE 2006] (a) attract N1 strongly, N2 weakly and repel N3 weakly (b) attract N1 strongly, but repel N2 and N3 weakly (c) attract all three of them (d) attract N1 and N2 strongly but repel N3

62. Permanent magnet has properties retentivity and coercivity respectively (a) high-high (c) low-high

[UP SEE 2006]

(b) low-low (d) high-low

attracted by both the poles repelled by both the poles repelled by the north pole and attracted by the south pole attracted by the north pole and repelled by the south pole

63. A magnetic needle suspended by a silk thread is

55. A thin bar magnet of length 2L is bent at the

[BVP Engg. 2006]

mid-point so that the angle between them is 60°. The new length of the magnet is [BVP Engg. 2008] (a) 2 L

(b) 3 L

(c) 2 L

(d) L

56. In a certain place, the horizontal component of

1 magnetic field is times the vertical component. 3 The angle of dip at this place is [Kerala CET 2008] (b) p/3 (d) p/6

(a) zero (c) p/2

associated with a ferromagnetic material? [Kerala CET 2008]

(a) It is strongly attracted by a magnet (b) It tends to move from a region of strong magnetic field to a region of weak magnetic field (c) Its origin is the spin of electrons (d) Above the curie temperature, it exhibits paramagnetic properties

58. Relative permittivity and permeability of a material are e r and m r , respectively. Which of the following values of these quantities are allowed for a diamagnetic material? [AIEEE 2008] (b) e r = 0.5, m r = 0.5 (d) e r = 0.5, m r = 1.5

59. The magnetic flux linked with the coil varies with

time as f = 3 t2 + 4 t + 9. The magnitude of the induced emf at 2 s is [BVP Engg. 2007] (a) 9 V

(b) 16 V

(c) 3 V

(d) 4 V

60. Nickel shows ferro-magnetic property at room temperature. If the temperature is increased beyond Curie temperature, then it will show [UP SEE 2007] (a) para-magnetism (c) no magnetic property

(a) the time period decreases (b) the time period remains unchanged (c) the time period increases (d) the needle stops vibrating

64. The magnetic moment m of a revolving electron around the nucleus varies with principal quantum number n as [DCE 2005] (a) m µ n

(b) m µ

1 n

(c) m µ n2

(d) m µ

1 n2

65. If the magnetic dipole moment of an atom off

57. Which of the following characteristic is not

(a) e r = 1.5, m r = 0.5 (c) e r = 1.5, m r = 1.5

vibrating in the earth’s magnetic field, if the temperature of the needle is increased by 500°C, then

(b) anti-ferromagnetism (d) dia-magnetism

dia-magnetic material, para-magnetic material and ferro-magnetic material are denoted by m d , m p and m f respectively, then [MHT CET 2005] (a) m p = 0 and m d ¹ 0 (c) m d ¹ 0 and m f ¹ 0

(b) m d ¹ 0 and m p ¹ 0 (d) m d = 0 and m p ¹ 0

66. The magnetic flux linked with a circuit of resistance 100 W increases from 10 to 60 Wb. The amount of induced charge that flows in the circuit is (in coulomb) [BVP Engg. 2005] (a) 0.5 (c) 50

(b) 5 (d) 100

67. Two short bar magnets of length 1 cm each have

magnetic moments 1.20 Am 2 and 1.00 Am 2 respectively. They are placed on a horizontal table parallel to each other with their N poles pointing towards the south. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the mid-point O of the line joining their centres is close to (Horizontal component of the earth’s magnetic induction is 3.6 ´ 10-5 Wb/m 2 ) [JEE Main 2013]

(a) 3.6 ´ 10 -5 Wb/m2

(b) 2.56 ´ 10 -4 Wb/m2

(c) 3.50 ´ 10 -4 Wb/m2

(d) 5.80 ´ 10 -4 Wb/m2

Magnetostatics

935

Answers Round I 1. 11. 21. 31. 41. 51.

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

2. 12. 22. 32. 42. 52.

(c) (c) (c) (d) (c) (d)

3. 13. 23. 33. 43. 53.

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

4. 14. 24. 34. 44. 54.

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

5. 15. 25. 35. 45. 55.

(a) (a) (b) (a) (a) (b)

6. 16. 26. 36. 46.

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

7. 17. 27. 37. 47.

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

8. 18. 28. 38. 48.

(a) (b) (b) (b) (b)

(a) (a) (d) (d) (d)

9. 19. 29. 39. 49.

10. 20. 30. 40. 50.

(b) (b) (b) (b) (b)

Round II 1. 11. 21. 31. 41. 51. 61.

(d) (c) (d) (a,d) (a) (c) (a)

2. 12. 22. 32. 42. 52. 62.

(c) (c) (a) (a,c,d) (c) (d) (a)

3. 13. 23. 33. 43. 53. 63.

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

4. 14. 24. 34. 44. 54. 64.

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

5. 15. 25. 35. 45. 55. 65.

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

6. 16. 26. 36. 46. 56. 66.

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

7. 17. 27. 37. 47. 57. 67.

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

8. 18. 28. 38. 48. 58.

(a) (c) (c) (c) (c) (a)

9. 19. 29. 39. 49. 59.

(b) (b) (a,b) (b) (c) (b)

10. 20. 30. 40. 50. 60.

(c) (a) (a to d) (a) (c) (a)

the Guidance Round I 1. If we cut along the axis of magnet of length l, then new pole m strength, m¢ = and new length l ¢ = l. 2 m ml M = \ New magnetic moment, M ¢ = ´ l = 2 2 2

3. As, W = BM cos 60° = \

MB or MB = 2 W 2

t = MB sin q = (2 W ) sin 60° = 2W

3 =W 3 2

4. Here, 2l = 3 cm,d1 = 24 cm and d 2 = 48 cm 3

S

N

S

N

As, the magnet is short,

If we cut perpendicular to the axis of magnet, then new pole l strength, m¢ = m and new length, l ¢ = 2

5. As, t = MB sin q or

\ New magnetic moment, l ml M M¢ = m ´ = = 2 2 2

6. As, F =

2. Here, r = 40 cm = 0.4 m q = 0°

(an axial line) –5

As, Þ or

V = 2.4 ´ 10 J/A-m; M = ? m M cos q V= 0 4p r2 2.4 ´ 10 –5 = 10 -7

M ´1 (0.4) 2

M = 38.4 Am2

B1 d 32 æ 48 cmö = =ç ÷ =8 B2 d13 è 24 cm ø

dt = MB cos q. It will be maximum, when dq

q = 0° m 0 6 M1M2 6 M1M2 × = 4p r4 r4

\In CGS system,

6 ´ 800 ´ 400 m0 = 12 dyne = 1= 20 ´ 20 ´ 20 ´ 20 4p

7. Hole reduces the effective length of the magnet and hence magnetic moment reduces.

8. As, magnetic moments are directed along SN, angle between M and M is q = 120°. \Resultant magnetic moment = M 2 + M 2 + 2 M cos120° = M 2 + M 2 + 2 M 2 ( -1 / 2) = M

936 JEE Main Physics 9. Given, uniform magnetic field

15. Horizontal component of earth’s magnetic field is given by, BH = B cos q B 0.3 3 cos d = H = = B 0.5 5 1 5 = sec d = cos q 3

B = 0.25 T The magnitude of torque t = 4.5 ´ 10 - 2 J

or

M

\ 30°

B

2

æ5ö tan d = sec2 d - 1 = ç ÷ - 1 è3ø

Now, Angle between magnetic moment and magnetic field q = 30° Torque experienced on a magnet placed in external magnetic field t = M ´B (Q A ´ B = AB sin q) t = MB sin q 4.5 ´ 10 - 2 = M ´ 0.25 ´ sin 30°

25 4 -1 = 9 3 -1 4 d = tan 3 =

\

16. In Fig. (a), at neutral point P, BH =

4.5 ´ 10- 2 M= 0.25 ´ sin 30° 4.5 ´ 10- 2 ´ 2 = 0.25 ´ 1

m0 æ M ö ç ÷ 4 p è d3 ø

N

1ö æ çQ sin 30° = ÷ è 2ø

n

W

= 0.36 J/T Thus, the magnitude of magnetic moment of the magnet is 0.36 J/T.

\ In Fig. (c),

M =0 M¢ = M - M = 0

In Fig. (d),

M ¢ = M 2 + M 2 + 2 MM cos 60° = 3 M

11. Given, number of turns n = 800

M = nIA

P

S (a)

E

S (b)

In Fig. (b) Net magnetic induction at P = resultant of

m0 2M = 2 BH 4 p d3

along horizontal and BH along vertical Þ Net magnetic induction at P. = (2 BH ) 2 + (BH ) 2 = 5 BH

17. From, B =

Current through solenoid I = 3 A As a current passes through a solenoid, a magnetic field is produced. By the use of Maxwell’s right hand grip rule, the magnetic field is along the axis of the solenoid. Using the formula of magnetic moment

n

W

s

Area of cross-section of solenoid A = 2.5 ´ 10 –4 m2

s E

P

10. As, M¢ = M2 + M2 = 2 M. As magnetic moments are in a closed loop in Fig. (b)

N

m 0i , 2r 4 p ´ 10 -7 ´ i 2 ´ 5 ´ 10 -2 7 ´ 10 -5 = 5.6 A i= 4 p ´ 10 -6

7 ´ 10 -5 = \

18. Points of zero magnetic field i. e. , neutral points lie on

= 800 ´ 3 ´ 2.5 ´ 10 –4 = 0.6 J/T along the axis of the solenoid

12. For each half pole strength m becomes half i. e. , M = m ´ 2l becomes half and volume V = a ´ 2l also becomes half therefore, l = W / V , remains constant.

equatorial line of magnet i. e. , along east and west.

19. Since the axis of the magnetic dipole placed at the centre of earth makes an angle 11.3° with the axis of earth, two possibilities arises as shown in Fig. (a) and (b). Hence the declination varies between 11.3°W to 11.3°E. N

13. In case of a toroid, the magnetic field is only confined inside the body of toroid in the form of concentric magnetic lines of force and there is no magnetic field outside the body of toroid. Thus the magnetic moment of toroid is zero. Hence, option (c) is correct.

14. At magnetic equator, the angle of dip is 0°. Then, the horizontal component, BH = B cos q = 0

N

S

S

11.3°

11.3°

W

E N S (a)

W

E N S (b)

Magnetostatics 20. Dip needle at neutral point will be horizontal at magnetic equator where angle of dip is zero degree. 2 l = 20 cm

21. Here, Þ

l = 10 cm,d = 40 cm 2 Md m As, neutral point, H = B = 0 2 2 2 4 p (d - l ) Þ

3.2 ´ 10 –5 =

10 -7 ´ 2 M (0.4) 15 ´ 15 ´ 10 –4

3.2 ´ 15 ´ 15 ´ 10 –4 ´ 10 -5 =9 M= 0.8 ´ 10 –7 9 M m= = = 45 A-m 2 l 0.2

\ \

29. Here, A = 0.5 m2,B = 2 T , q = 30° \

R=

H B0 = = 2 B0 cos d cos 45°

24. According to Gauss's law for electrostatic field,

ò E. ds = Q /e 0 It does not contradict for electrostatic fields as the electric field lines do not form a continuous closed path. According to Gauss's law in magnetic form, ò E. ds = 0 It contradicts for magnetic field because there is a magnetic field inside the solenoid and no field outside the solenoid carrying current but the magnetic field lines form the closed paths.

25. For a diamagnetic substance, I is negative and -I µ H. Therefore, the variation is represented by OC or OD. As magnetisation is small, so OC is the better choice.

26. Here, \ From,

V = (10 ´ 0.5 ´ 0.2) cm3 = 1 cm3 = 10 -6 m3 H = 0.5 ´ 10 4 Am–1,M = 5 Am2, B = ? M 5 I= = -6 = 5 ´ 10 6 Am V 10 B = m 0(I + H) B = 4 p ´ 10 –7 (5 ´ 10 6 + 0.5+10 4) = 6.28 T

27. When space inside the toroid is filled with air, B0 = m 0H When filled with tungsten, B = mH = m 0m rH = m 0(1 + c m) H Percentage increase in magnetic field/induction (B - B0) ´ 100 m 0 c mH ´ 100 = = = c m ´ 100 B0 m 0H = 6.8 ´ 10 –5 ´ 100 = 0.0068%

28. Here, B = 1.7 ´10 –5 T ,H = ? Now,

H=

B 1.7 ´ 10 –5 = 13.53 Am–1 = m 0 4 p ´ 10 -7

3 Wb 2 2

\ Moment of inertia I becomes,

1 æ 1ö 1 therefore ç ÷ = 4 è 4ø 64

Magnetic moment M becomes 1/4th. Now, as T = 2p

I , MH

\ T becomes 1/4 th

31. Here, n1 = 40 ,n2 = 20 ,H1 = 36 ´10 -6 T, H2 = ? \

H2 n22 (20) 2 1 = = = , H1 n12 ( 40) 2 4

or,

H2 =

23. For, H = R cos d \

f = BA cos q = 2 ´ 0.5 cos 30° =

30. Mass becomes 1/4 and length becomes 1/4.

22. In a permanent magnet at room temperature domains of a magnet are partially alingned due to thermal agitation.

937

32. As,

36 ´ 10 -6 = 9 ´ 10 -6 T 4

M1 T22 + T12 n22 + n12 4 2 + 12 2 160 = = 5: 4 = = = M2 T22 - T12 n12 - n22 12 2 - 4 2 128

33. Here, n1 = 10 oscillations per min d1 = 45°, T1 = 0.707 CGS units n2 = ?, d2 = 60° , R2 = 0.5 CGS units \ Þ or,

n2 H2 R2 cos d2 = = n1 H1 R1 cos d1 n2 1 0.5 cos 60° 0.5 ´ 1/ 2 = = = 10 0.707 cos 45° 2 0.5 ´ 2 ´ 1 / 2 n2 =

10 = 7.07 2

34. Given, number of turns of circular coil n = 16 Radius of circular coil r = 10 cm = 0.1 m Current, I = 0.75 A Magnetic field, B = 5.0 ´ 10 - 2 T Frequency, f = 2/s Magnetic moment of the coil, M = nIA = 16 ´ 0.75 ´ p (0.1) 2 = 16 ´ 0.75 ´ 3.14 ´ 0.1 ´ 0.1 = 0.377 J/T Frequency of oscillation of the coil 1 M ´B f= I 2p where I = Moment of inertia of the coil. Squaring on both the sides, we get 1 MB f2= × 4p 2 I MB 0.377 ´ 5 ´ 10 - 2 Þ = I= 4p 2f 2 4 ´ 3.14 ´ 3.14 ´ 2 ´ 2 = 1.2 ´ 10 - 4 kg-m 2 Thus, the moment of inertia of the coil is 1.2 ´ 10- 4 kg-m 2.

938 JEE Main Physics 35. As, T = 2p

I MBH

(BH ) 2 T1 = (BH )1 T2

Þ

42. When mass is quadrupled, i. e. , made 4 times. I becomes four times. As, T µ I

(B ) Þ T2 = T1 H 1 (BH ) 2 1 Here, n = 30 oscillations/min = oscillations/s 2 1 T1 = = 2 s \ n1 \

T2 = 2

\ T becomes twice, i. e. , motion remains SHM with time period = 2 T

43. When two identical bar magnets are held perpendicular to each other. M1 = M 2 + M 2 = M 2 ,I1 = I

\

BH 1 =2´ = 2s 2 BH 2

T1 = 25 / 4 s, T2 = ?

and

and M2 = M (as one magnet is removed) I2 = I1 / 2

36. When polarity is reversed, net magnetic moment 2 M - M = M, decreases. Therefore, time period of oscillation increases, i. e. ,T2 > T1.

1 M 2 æ 1 ö T2 I2 M1 = × =ç ÷ è 2ø 2 M T1 M2 I1

37. In a tangent galvanometer, applying tangent law, B = H tan q Þ

B = 0.34 ´ 10

–4

tan 30° = 0.34 ´ 10

–4

=

1 ´ 3

38. From, T = 2p

I I Þ 4 = 2p MB MB

field. t 2 = 3 2 = 2p

\ Dividing

= cos d =

galvanometers.

\

2 rH 2r H tan q1 = tan q2 m 0n1 m 0n2 n1 tan q1 tan 60° = = = 3 n2 tan q2 tan 45°

45. As, \

40. Here, l1 = 8 Am-1; B1 = 0.6 T, t1 = 4 K; l2 = ?; B2 = 0.2 T, t 2 = 16 K B (magnetic field induction) According to Curie law, I µ t ( in kelvin) \ or

l2 B2 l2 0.2 4 1 = ´ = ´ = l1 B1 l1 0.6 16 12 I2 = I1 ´

H = 3 ´ 10 -5 T, i = ?, q = 45° m ni From, F = 0 = H tan q , 2r 2 rH tan q 15 ´ 10 -2 ´ 3 ´ 10 -5 tan 45° i= =2´ m 0n ( 4p ´ 10 -7) ´ 25 = 0.29 A

R cos d = cos d R 1 Þ d = 60° 2

tan q2 d13 1 r3 = 3 = = tan q1 d 2 [r (3)1/3 ]3 3 tan q2 =

1 3 1 tan 60° = = tan q1 = 3 3 3 3

q2 = 30°

46. In a paramagnetic material, I µ H. Therefore, the graph between H and I is a straight line represented by choice (c) in figure.

47. As, work done, W = MB (cos q1 - cos q2) = 0

1 8 2 = = Am-1 12 12 3

41. Given, n = 25,r = 15 cm = 15 ´10 -2 m,

we get

Þ

I MH

H t1 1 we get, = R t2 2

39. In series, same current flows through two tangent i=

1 1 = 25 / 4 ´ 1/ 4 = 2 s 2 2 1/ 4

I , where R resultant intensity of earth’s MR

44. As, t1 = 3 = 2p

When it is cut into two equal parts in length, mass of each (length) 2 1 becomes th and M part becomes 1/2, l = mass 12 8 1 becomes . 2 1æ I ö 1 I /8 = ç2p T ¢ = 2p ÷ = T =2s MB ø 2 (M / 2) B 2 è

i. e. ,

1 21/ 4

T2 = T1 ´

B = 1.96 ´ 10 –5 T

1/ 2

(As there no change in angle between M and B and when loop in rotated by 30°) M1 = 3 M and M2 = M

48.

T2 = ? In pole combination, T2 M1 3M = = 3 T1 M2 M or

T2 = 3 T1 = 3 3 s

Magnetostatics

939

49. For diamagnetic substances, the magnetic susceptibility is

52. Copper is a diamagnetic material, therefore its rod align itself

negative and it is independent of temperature. Therefore, choice (d) is correct in figure.

where magnetic field is weaker and perpendicular to the direction of magnetic field at that field.

50. Because of large permeability of soft iron, magnetic lines of

53. As, m r < 1for substance X, it must be diamagnetic and m r > 1for

force prefer to pass through it. Concentration of lines in soft iron bar increases as shown in Fig. (b).

51. As temperature of a ferro-magnetic material is raised, its susceptibility c remains constant first and then decreases as shown in Fig. (a).

substance Y so it is must be paramagnetic.

54. This magnetic field is a measure of coercivity of the material. 55. Transformer core is of soft iron which has retentivity and small coercivity. Therefore, its hysteresis loop is tall and narrow.

Round II 1. For a paramagnetic material, K µ

1 T

K2 T1 = K1 T2

\

K / 2 27 + 273 = K T2

or Þ

T2 = 600 K = 600 - 273 = 327°C

4. As,

Fµ

i. e. ,

1 r4

\

3. Here, magnetic moment is given in weber-metre, which is the unit of m 0M. \

m 0M = 5 ´ 10 -5 Wb - m 5 ´ 10 -5 Am2 m0

or

M=

Also,

B = 8 p ´ 10 -4 = m 0H 8 p ´ 10 m0

H=

Now,

I T = 2p MH

\

I=

MHT 2 5 ´ 10 -5 ´ 8 p ´ 10 -4 ´ 15 2 = 4 p2 4 p 2m 20 =

or,

I=

1= 2 p

Þ

= 4.54 ´ 10 5 kg m2

…(ii)

V = volume of the material = mass/density A = area of hysteresis loop v = frequency of alternate magnetic field applied and t = time for which field is applied

then, energy loss in the material in t second is æ mö E = VAvt = ç ÷ Avt èd ø 0.6 = ´ 0.722 ´ 50 ´ 1 7.8 ´ 10 3 = 2.77 ´ 10 3 = 27.7 ´ 10 –4 J

6. In a vibration magnetometer, T = 2p

I MH

1 = MH = 36 ´ 10 -4 T2 In a deflection magnetometer, m 2M H= 0 3 4p d

\

4 p2

4 pd 3 M 10 8 = = 2m0 36 H

5 ´ 10 -5 ´ 8 p ´ 10 -4 ´ 225 4 p 2( 4p ´ 10 -7) 2 2250 ´ 10 -9 p (16 p 2) ´ 10 -14

1 M (F + H)

2 F+H F = = +1 1 H H F F +1= 4 Þ = 4 -1 = 3 H H

-4

\

…(i)

Dividing Eq. (i) by Eq. (ii), we get

5. If,

1 1 time. when, r becomes thrice, F becomes 4 time i. e. , 81 (3) 8.1 Therefore, F ¢ = = 0.1N 81

I I \ 2 = 2p MH MH

When an external magnet is brought near and parallel to H and the time period reduces to 1 s, net field must be (F + H).

2. When axes are in the same line, m 6 M1M2 F= 0 4 p r4

T = 2p

Multiplying Eq. (i) and Eq. (ii), we get 10 8 M 2 = 36 ´ 10 -4 ´ = 10 4 36 or

M = 10 2 = 100 A-m

…(i)

…(ii)

940 JEE Main Physics 7. As, M = niA = 50 ´ 2 ´1.25 ´10 –3 = 0.125 Am2

10. As, the magnet is long, we assume that the upper north pole

If normal to the face of the coil makes an angle q with the magnetic induction B, then in 1st case, torque = MB cos q = 0.04 and in second case, Torque = MB sin q = 0.03 \

MB = (0.04) 2 + (0.03) 2 = 0.05 B=

produces no effect. But due to south pole of the magnet is equal and opposite to horizontal component of earth’s magnetic field, i. e. , æm öm B= ç 0 ÷ 2 =H è4pø r m0 In CGS system, =1 4p \

0.05 0.05 = = 0.4 T M 0.125

8. Resolving the magnetic moments along OP

m = 3.2 ´ 10 –5 ´ 10 4 (gauss) 10 2 m = 32 ab-amp-cm -1

11. Let, qbe the declination at the place. As it is clear from figure. Geographic meridian

co s

δ

and perpendicular to OP, figure we find that component OP perpendicular OP cancel out. Resultant magnetic moment along OP is = M cos 45°+ M cos 45°.

1´

H

M sin 45°

δ

O

nδ si

co s

45

°

45°

2M

V

M sin 45° Perpendicular to Geographic meridian

2M = 2M 2 The point P lies on axial line of magnetic of moment = 2 M = 2M cos 45° =

B=

\

m 0 2 ( 2 M) 4p d3

60 =6s 10 I T1 = 2p =6 Þ MH 60 30 In the second case, T2 = s = 14 7 If B magnetic induction due to external magnet, then I 30 T2 = 2p = M (H + B) 7

\ …(i)

Dividing Eq. (i) by Eq. (ii), we get 6 H+B B = = 1+ 30 / 7 H H or

B æ7ö or ç ÷ = 1+ è5ø H

tan d1 sin q = = tan q tan d2 cos q æ tan d1 ö q = tan -1 ç ÷ è tan d2 ø

Þ

12. As, T1 = 2p …(ii)

V H cos q V V tan d2 = = H cos (90° - q) H sin q tan d1 =

and

9. In the first case, T1 =

2

Magnetic meridian

H

H

M

I MV I MH

and

T2 = 2p

\

T2 V = = tan q T1 H 2

or

B 49 24 = -1 = H 25 25

Þ

2 æT ö æ2ö tan q = ç 2 ÷ = ç ÷ = 1 è2ø è T1 ø

q = 45°

If n is number of vibrations/min in the third case when polarity of external magnet is reversed, then

13. In SI the unit of pole strength is Amp-metre. Here, the pole

60 I = 2p n M (H - B)

m 0m = 10 -3 Wb

T3 =

Dividing Eq. (i) by Eq. (iii), we get 6 H -B B 24 = = 1- = 160 / n H H 25 n 1 = or n = 2 vibrations per minute 10 5

strength is given in weber, which is the unit of (m 0m) and

…(iii) or,

m=

10 -3 m0

Magnetic moment of magnet, M = m ´2 l =

10 -4 10 -3 (0.1) = m m0

Magnetostatics \ Torque,

t = MB sin q æ10 -4 ö =ç ÷ ( 4p ´ 10 -3) sin 30° è m0 ø

Þ

=

14. In the usual setting of deflecting magnetometer, field due to magnet (F) and horizontal component (H) of earth’s field are perpendicular to each other. Therefore, the net field on the magnetic needle is F 2 + H 2. I …(i) T = 2p \ 2 M F + H2 When magnet is removed, I …(ii) T0 = 2p MH F Also, = tan q H Dividing Eq. (i) by Eq. (ii), we get T = T0

Þ \

18. As, or

m= B2 n22 10 2 = = =4 B1 n12 5 2

B2 = 4 B1 = 4 ´ 0.3 ´ 10 –4 T = 1.2 ´ 10 –4 T

\Increase in field = B2 - B1 = 0.9 ´ 10 –4 T

19. In CGS system,

m0 =1 4p

In equilibrium, net repulsion due to magnetic interaction = weight of upper magnet. Therefore, it is clear from figure. S = –900 1 cm N = +900

F 2 + H2 H

1 cm

2

H tan 2 q + H 2 S = +100

H 2

H sec q

1 cm

T2 = cos q or 22 = cos q T0 Þ

T

2

= T02 cos q

15. In equilibrium, the resultant magnetic moment will be along magnetic meridian. Let N1S1 makes an angle q with this resultant 1 M M tan q = 2 = = \ M1 3M 3 \

0.4 ´ 10 –4 = 16 A-m 25 ´ 10 -7

H

= =

m0 m ´ =V 4 p d2 m 10 -7 ´ = 0.4 ´ 10 –4 (20 / 100) 2

17. At neutral point,

-4

1 10 ( 4p ´ 10 -3) ´ = 0.5 Nm 2 4 p ´ 10 -7

941

N = –100

900 (100) 900 ( -100) 900 (100) 900 ( -100) + =m´g 12 22 22 32 1ö æ1 1 1 Þ 900 ´ 100 ç 2 - 2 - 2 + 2 ÷ = m ´ 1000 è1 2 2 3 ø æ 11ö 90 ç ÷ = m è18 ø

Þ Þ

q = 30°

m = 55 g

20. NS is a magnet held vertically with its north pole on the table.

16. As shown in figure, mV

P is neutral point, where NP = 20 cm shown in figure. Clearly. S

O

mg

20

mV

Strength of each pole, m = 98.1 ab - amp - cm, g = 981 cms–2, V =?

V=

θ 20 cm

P

If, H = 0.3, then clearly, m = 185 ab-amp-cm

At equilibrium, mV ´ 2 l = Mg ´ l Mg or V= 2m or

N

cm

20 cm

Mass, M = 50 , mg = 50 ´ 10 –3 g

√2

2l

21. Here, q1 = 90° , q2 = 270°, \ W = -MB [cos 270° - cos 90° ] = zero -3

50 ´ 10 ´ 981 = 0.25 G 2 ´ 98.1

22. At magnetic north of earth, H = 0 and d = 90° i. e. , maximum

942 JEE Main Physics 1 ; when M becomes 4 times, then T becomes its M half. Therefore new T = 1s. M A Pole strength m = . When the wire is l bent at its middle point O at 60°, then θ it is clear from the figure. l /2

23. As, T µ

28. Field due to circular loop carrying current is perpendicular to

24.

29. The lines of magnetic field induction B are necessarily

60° + q + q = 180°

θ

60° O

2 q = 180° - 60° = 120° Þ q = 60° QOAB is an equilateral triangle. or

l /2

B

\ AB = 2l ¢ = l / 2 New magnetic moment M ¢ = m (2l ¢ ) =

ml M = 2 2

25. On equitorial line, magnetic field due to magnet varies inversely as cube of the distance, therefore 3

B1 æ 3 x ö = ç ÷ = 27 : 1 B2 è x ø I 26. As, T = 2 p MB 1 time \ T= 2

continuous across the surface S of a lump of magnetic material. Outside the lump of magnetic material, H = B / m 0 and inside the lump of magnetic material, H = B / m 0 m r where m r is the relative permeability of material. Thus the lines of H cannot all be continuous across surface S.

30. Magnetic field induction at a point inside the cylindrical conductor is, F µ x. Magnetic field induction at a point outside the cylindrical 1 conductor is, F µ . x Magnetic field induction at a point on the axis of cylindrical conductor is, zero.

31. The primary origin of magnetism lies in the fact that the electrons are revolving and spinning about nucleus of an atom, which gives rise to currents and hence to magnetism.

33. F is the magnetic field strength at the centre of the circular coil of tangent galvanometer carrying current.

34. On earth’s surface, H = 0.32 G 35. H represents horizontal component of earth’s magnetic field.

60 =2s 30 2 New T = = 2s 2

36. From, V = H tan d,

Initial time period = \

the plane of the loop. As, current is anti-clockwise. So, N pole lies above the loop and south pole lies below the loop.

V H V tan d = = 1 H tan d =

27. Given, angle of declination

Þ

q = 12° west

d = 45°

37. On the surface of earth, B » 10 -5 T

GN

MN

(from question)

38. Magnetic declination is of the order of 20° west. 12°

40. The magnet is cut parallel to its length of its each part will be

GW

GE MS GS

Angle of dip d = 60° Horizontal component of earth’s magnetic field H =0.16 G Let the magnitude of earth’s magnetic field at that place is R. Using the formula, H = R cos d H 0.16 0.16 ´ 2 or R= = = cos d cos 60° 1 = 0.32 G = 0.32 ´ 10 - 4 T The earth’s magnetic field lies in a vertical plane 12° west of geographical meridian at an angle 60° above the horizontal.

of the same length as previous but the pole strength of each part, 1 = ´ initial pole strength n m \ m¢ = n Therefore, the magnetic moment m 1 M = m¢ ´ l ¢ = ´ l = (ml ) = n n n Hence, the magnetic moment of each part becomes 1/n times.

41. Earth’s magnetic field suffers a change in magnitude and direction with change of time. Studies of magnetic rock, give the idea that the direction of magnetic field is reversed. Also, the permeability of ferromagnetic material depends on magnetic fields as they strongly magnetised by relatively weak magnetising field in the same sense as the magnetising field. It also can be verified by hysteresis curve that permeability is greater for weaker fields.

Magnetostatics 42. Case I When the like poles of two magnets are placed in

T ¢ = 2p

I1 + I2 (M1 + M2) B

l1 O

=

It is clear from Eqs. (i) and (ii) that, T ‘ < T “. (770ºC), the iron behaves as paramagnetic more over when a magnet is heated, it losses its magnetism. Thus, iron bar magnet on heating to 1000ºC and then cooled, will not retain its magnetism.

45. The field due to a magnet is non-uniform. Therefore, it exerts

to a straight wire on AD and BC is parallel to elementary length of the loop.

53. Hence,

W = MB (cos q1 - cos q2)

= 2 ´ 10 4 ´ 6 ´ 10 -4 (cos 0° - cos 60° )

æ 1ö = 12 ç1 - ÷ = 6 J è 2ø

54. A diamagnetic substance is repelled by both the poles of the magnet.

55. On bending the magnet, the length of the magnet AC = AB + BC æ qö æ qö = L sin ç ÷ + L sin ç ÷ è2ø è2ø

56.

needle will be completely destroyed hence, the needle will stop vibration.

Thus, it is obvious that greater the value of susceptibility of a material greater will be the value of intensity of magnetisation i.e., more easily it can be magnetised.

49. Davisson and Germer experimentally established wave nature of electron by observing diffraction pattern while bombarding electrons on Ni-crystal.

50. As, induced emf, e = BH lv = 0.30 ´ 10 –4 ´ 20 ´ 5.0 mV = 3 mV

51. Net magnetic field due to loop ABCD at O is

1 =L 2 V V From, tan d = = = 3 H V/ 3 = 2L ´

47. We know that, at the high temperature, the magnetisation of

I cm = Þ cm µ I H

L

æ qö = 2 L sin ç ÷ ³ 2 L sin 30° è2ø

46. We know that, magnetic dipole posses maximum potential

48. From the relation, susceptibility of the material

M = 2 ´ 10 4 JT –1, B = 6 ´ 10 -4 T q1 = 0° , q2 = 60° , W = ?

both, a net force and a torque on the nails which will translate and also rotate the nails before striking to north pole of magnet with their induced south poles and vice-versa. energy when magnetic moment and magnetic field are anti-parallel. A current loop can be treated as a magnetic dipole. When the current loop has an area, A and carries a current, I then the magnetic dipole moment is given by M = I A.

m 0I m I m I - 0 = 0 ( b - a) 24 a 24 b 24 ab

52. The forces on AD and BC are zero because magnetic field due

44. A compass is simply a needle shaped magnet that mounted so it can rotate freely about a vertical axis. When it is held in a horizontal plane, the north pole end of the needle points, generally, towards the geomagnetic north pole (really a south magnetic pole remember). Thus, true geographic north direction cannot be found by using a compass needle. Now, vertical plane passing through the magnetic axis of earth’s magnet is called magnetic meridian.

C

B = BAB + BBC + BCD + BOA m I p m I p =0 + 0 + +0 - 0 ´ 4 pa 6 pb 6

...(ii)

43. At 1000ºC, which is quite above Curie’s temperature for iron

I

30° b

opposite direction then period of vibration is expressed as I1 + I2 (M1 - M2) B

B

D

...(i)

Case II When the likes poles of two magnets are placed in T ¢¢ = 2p

A

a

same direction then the time period of vibration is expressed as

943

Þ

d=

A

θ/2 θ/2

B

L

C

p 3

57. A ferromagnetic material cannot move from a region of strong magnetic field to a region of weak magnetic field.

58. For a diamagnetic material 0 < m r < 1 and e r or K > 1 59.

So, option (a) is correct. df As emf, e = = 6t + 4 + 0 dt Now at, t = 2 s, e = 6 ´ 2 + 4 = 16 V

60. Nickel exhibits ferromagnetism because of a quantum physics effect called exchange coupling in which the electron spins of one atoms interact with those of neighbouring atoms. The result is alignment of the magnetic dipole moments of the atoms, inspite of the randomizing tendency of atomic collisions. This persistent alignment is what gives ferromagnetic materials their permanent magnetism.

944 JEE Main Physics If, the temperature of a ferromagnetic material is raised above a certain critical value, called the Curie temperature, the exchange coupling ceases to be effective. Most such materials then become simply paramagnetic; that is, the dipoles still tend to align with an external field but much more weakly, and thermal agitation can now more easily disrupt the alignment.

61. A magnet will attract ferro-magnetic needle N1 strongly. It will attract paramagnetic needle, N2 weakly and repel diamagnetic needle, N3 weakly.

66. The amount of induced charge is given by q=

1 1 Df = (60 - 10) R 100

(where f1 = 100 Wb, f2 = 60 Wb,R = 100 W) 50 = = 0.5C 100

67.

Bnet = B1 + B2 + BH Bnet =

62. The materials for a permanent magnet should have high retentivity (so that the magnet is strong) and high coercivity (so that the magnetism is not wiped out by stray magnetic fields). As, the materials in this case is never put to cyclic changes of magnetisation, hence hysteresis is immaterial.

=

m 0 (M1 + M2) + BH 4p r3 10 -7 (1.2 + 1) ´ 3.6 ´ 10 -5 (0.1)3

= 2.56 ´ 10 -4 Wb/m2

63. When the magnetic needle is heated to such a high

64.

temperature, then the magnetic needle losses its magnetism, so, the restoring torque no longer acts and hence needle stops vibrating. neh From the knowledge of theory, m = \ m µn 4 pm

N BH B1 B2

S N

O

65. According to electron theory of magnetism, an atom of a diamagnetic material has no intrinsic dipole moment, whereas atom of a paramagnetic material has some intrinsic dipole moment i.e., m d = 0 and m p ¹ 0

S N

S

21

Electromagnetic Induction and Alternating Current JEE Main MILESTONE

< < < < < < < <
0, dt

so induced current is negative. S

(a)

(b)

i θ

Sample Problem 5 A bar magnet is brought near a B (Decreasing)

q > 90° , fB < 0,

solenoid as shown in figure. Will the solenoid attract or repel the magnet?

dfB > 0, dt

S

so induced current is negative.

21.3 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. Lenz’s law is not an independent principle. It gives the same result as the sign rules we introduced in connection with Faraday’s law, but is often easier to use. This law states that “The direction of any magnetic induction effect is such as to oppose the cause of the effect.” Later, we will see that Lenz’s law is directly related to energy conservation.

N

Interpret 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. S

N

N

S

S

N

S

N

Electromagnetic Induction and Alternating Current To apply Lenz’s law, you can remember RIN or Ä IN (when the loop lies on the plane of paper).

Note

Induced current

21.4 Motional Electromotive Force So far we 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.

i (Increasing)

RIN In RIN, R stands for right, I stands for increasing and N for north pole (anti-clockwise). It means, if a loop is placed on the right side of a straight current carrying conductor and the current i in the conductor is increasing, then induced current in the loop is anti-clockwis e ( N ) .

949

Increasing

Ä IN In Ä IN suppose the magnetic field in the loop is perpendicular to paper inwards ( Ä) and this field is increasing,

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

then induced current in the loop is anti-clockwise ( N ) .

++ ++

Sample Problem 6 A circular loop is placed near a current Fe

carrying conductor as shown. Find the direction of induced current, if the current in the wire is decreasing.

–

l

v

Fm l (Decreasing)

–– ––

Interpret Here, we apply RIN. In this case, loop is placed to the right of current carrying wire (not to the left as it appears, because if you move in the direction of current, loop lies to the right). Now, the Induced current current is decreasing, therefore, induced current in the loop is clockwise ( ) .

Sample

Problem 7 A current carrying straight wire passes inside a triangular coil as shown in figure. 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.

Fm = – e (v ´ B ) 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 until 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

i

Interpret Magnetic field lines round the current carrying wire are as shown in figure. Since, the lines are tangential to the loop ( q = 90° ) the flux passing through the loop is zero, whether the current is increased or decreased. Change in flux is zero. Therefore, induced current in the loop will be zero.

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 DV = El = Blv \

DV = 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 the conductor as long as the conductor continues to move through the uniform magnetic field.” Now, suppose the moving rod slides along a stationary U-shaped conductor, forming a complete circuit. No magnetic force acts on the charges in the stationary

950 JEE Main Physics U-shaped conductor, but there is an electric field caused by the charge accumulations at a and b. Under the action of this field a counter clockwise current is established around this complete circuit. The moving rod becomes a source of electromotive force. Within it, positive charge moves from lower to higher potential and in the remainder of the circuit, charge moves from higher to lower potential. We call this a motional electromagnetic force denoted by e, we can write,

3. Motional emf induced in a rotating bar. A conducting rod of length l rotates with a constant angular speed w 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 length dr at a distance r from O. This segment has a velocity, × × × × × × ×B P l dr × × × × × v× × O w × × × × × × ×

a

v=r w The induced emf in this segment is, de = Bvdr = B ( r w)dr Because every segment of the rod is moving perpendicular to B, an emf de of the same form is generated across each. Summing the emfs induced across all segments, which are in series, gives the total emf across the rod.

v

b

e = Bvl If R is the resistance of the circuit, then current in the circuit is, e Bvl i= = R R

\

e=

Note 1. The direction of motional emf or current can be given by right hand rule. Stretch your right hand. The stretched fingers point in the direction of magnetic field. Induced current (Upper side of palm)

\

l

ò 0 de l

=

ò 0Br wdr

=

B wl 2 2

e=

B wl 2 2

From right hand rule, we can see that P is at higher potential than O. Thus, V P – VO =

Stretched fingers (B) Velocity of conductor (Thumb)

Thumb is along the velocity of conductor. The upper side of palm is at higher potential and lower side on lower potential. If the circuit is closed the induced current within the conductor is along perpendicular to palm upwards. 2. In the figure shown, 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 i = R+ r i

a

B wl 2 2

Sample Problem 8 Two parallel rails with negligible resistance are 10.0 cm apart. They are connected by a 5.0 W resistor. The circuit also contains two metal rods having resistances of 10.0 W and 15.0 W 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 W resistor. a

c

e

5.0 Ω

4.0 m/s

2.0 m/s

e = Bvl l

⇒

v

R

R

r

b 10.0 Ω

d

f 15.0 Ω

b

or

i =

Bvl R+ r

Interpret In the figures, R = 5.0 W, r1 = 10 W, r2 = 15 W, e1 = 4 ´ 10 –3 V and e2 = 2 ´ 10 –3 V.

Electromagnetic Induction and Alternating Current i

e2

r1

r1

⇒

R

R

+ i2

r2

e1

e2

(a)

i1

r2

(b) i ′2 r1

R

r2

i ′1

e1 i′

(c)

From figure (b), Net resistance of the circuit Rr1 10 ´ 5 55 = r2 + = 15 + = W R + r1 3 10 + 5 \ Current,

i=

e2 Net resistance

=

2 ´ 10 –3 6 = ´ 10 –3 A 55 55 /3

\ Current through R, æ r ö i1 = ç 1 ÷ i è R + r1 ø

4 4 ´ 10 –3 A = mA 55 55

From figure (c) Net resistance of the circuit Rr2 = r1 + R + r2 = 10 + \

Current i¢ = =

5 ´ 15 55 = W 5 + 15 4

e1 Net resistance –3

4 ´ 10 55 / 4

16 ´ 10 –3 A 55 æ r ö \ Current through R, i1¢ = ç 2 ÷ i ¢ è R + r2 ø æ 15 ö æ 16 ö –3 =ç ÷ ç ÷ ´ 10 A è15 + 5 ø è 55 ø =

=

21.5 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 selfinductance of a circuit which is of great importance in our proceeding discussions. It can be defined in following two ways :

First Definition Suppose a circuit includes a coil with N turns of wire. It carries a current i. The total flux (NfB) linked with the coil is directly proportional to the current (i) in the coil, i.e., NfB µ i

æ 10 ö æ 6 –3 ö =ç ÷ ç ´ 10 ÷ A ø è10 + 5 ø è 55 =

951

12 mA 55

From superposition principle net current through 5.0 W resistor is, 8 i1¢ - i1 = mA from d to c. 55

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, NfB NfB = Li or L = i From here, we can define self-inductance (L ) of any circuit as, the total flux per unit current. The SI unit of selfinductance is henry (H).

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 eµ 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

952 JEE Main Physics This equation states that, the self-inductance of a circuit is the magnitude of self induced emf per unit rate of change of current.

The induced emf is produced whenever there is a change in the current in the inductor. This emf always acts to oppose this change.

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,

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.

Thus, an inductor is a circuit element which opposes the change in current through it. It may be a circular coil, solenoid etc.

From figure (a) The current is constant, and there is no self induced emf. Hence, Vab = 0. di is positive. dt The induced emf e 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.

From figure (b) The current is increasing, so

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.

di is dt negative. The self induced emf e opposes this decrease and Vab is negative.

From figure (c) The current is decreasing and

This is analogous to a battery with b, as its positive terminal. In each case, we can write the PD, Vab as, Vab = – e = L

The circuit 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. i

i

a

b

Potential Difference Across an Inductor

a

b

R Vab = iR (a)

We can find the direction of self-induced emf across an inductor from Lenz’s law.

L di Vab = L dt (b)

i (Increasing)

i (Constant) a

di dt

b a

b

+

–

di = 0 dt e=0 Vab = 0

e di > 0 dt Vab > 0

(a)

(b)

Note

Kirchhoff’s second law with an inductor In Kirchhoff’s second law (loop rule), 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. For example in the loop shown in figure, Kirchhoff’s second law gives the equation. di =0 E – iR – L dt R

i (Decreasing) a

b

–

+

H i

e di < 0 dt Vab < 0 (c)

Drop = iR

L L

H Drop = L

E

di dt

Electromagnetic Induction and Alternating Current Sample Problem 9 The inductor shown in figure has inductance 0.54 H and carries a current in the direction shown di that is decreasing at a uniform rate = – 0.03 A/s., The self dt induced emf is

Method of Finding Self-Inductance of a Circuit We use the equation, L = NfB / i to calculate the inductance of given circuit. A good approach for calculating the self-inductance of a circuit consists of the following steps :

i a

953

b

(a) Assume that there is a current i flowing through the circuit (we can call the circuit an inductor).

L

(a) 1.62 ´ 10 -2 V

(b) Determine the magnetic field B produced by the current.

(b) 0.62 ´ 10 -2 V (c) 2.62 ´ 10 -2 V

(c) Obtain the magnetic flux fB.

(d) 1.05 ´ 10 -3 V

(d) With the flux known, the self-inductance can be found from L = NfB / i.

Interpret (a) Self induced emf, e= –L

To demonstrate this procedure we now calculate the selfinductance of two inductors.

di = (– 0.54) (– 0.03) V dt

=1.62 ´ 10 –2 V

Inductance of a Solenoid

Sample Problem 10 In the circuit diagram shown in figure, R = 10 W, L = 5H, E = 20 V, i = 2 A. This current is decreasing at a rate of –1.0 A/s. Find Vab at this instant. a

R

E

L

i

(a) 20 V (c) 40 V

b

(b) 35 V (d) 25 V di dt

Va – iR – VL – E = Vb

\

Vab = Va – Vb = E + iR + VL

L=

= 20 + ( 2) ( 10) – 5 = 35 V

\

Note As the current is decreasing the inductor can be replaced by a di = 5 V in such a manner that this emf supports dt 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 source of emf, e = L ×

e= L a

R

di dt = 5V

E = 20 V i

Now, we can find V ab .

N is the number of turns per unit length. l

The magnetic flux through each turn is, NS fB = BS = m 0 i l Here, S is the cross-sectional area of the solenoid. Now,

= (5) (– 1.0) = – 5 V Now,

We can assume that the interior magnetic field due to a current i is uniform and given by equation, æNö B = m 0ni = m 0 ç ÷ i è l ø where, n =

Interpret (b) PD across inductor, VL = L

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.

NfB N æ m 0NSi ö m 0N 2S = ÷= ç i i è l ø l L=

This result shows that L depends on dimensions ( S, l ) and is proportional to the square of the number of turns. L µ N2 Because N = nl, we can also express the result in the form, L = m0

b

m 0N 2S l

or

(nl ) 2 S = m 0n2Sl = m 0n2V l

L = m 0n2V

Here, V = Sl is the volume of the solenoid.

954 JEE Main Physics Energy Stored in an Inductor

Sample Problem 12 (a) A toroidal solenoid with an air

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

core has an average radius of 15cm, area of a cross-section 12 cm 2 and 1200 turns. Obtain the self-inductance of the toroid. Ignore field variation across the cross-section of the toroid. (b) A second coil of 300 turns is wound closely on the toroid above. If the current in the primary coil is increased from zero to 2.0 A in 0.05 s, obtain the induced emf in the second coil.

Interpret (a) Total number of turns, N = 1200;

An increasing current in an inductor causes an emf between its terminals.

Area of cross-section of solenoid, A = 12 cm2 = 12 ´ 10 -4 m2

The work done per unit time is power. dW di P= = – ei = – Li dt dt

Therefore, the number of turns per unit length 1200 4000 -1 N m n= = = p 2pr 2p ´ 15 ´ 10 -2

From,

dW = – dU dW dU =– dt dt dU di or dU = Li di = Li dt dt

or we have,

The total energy U supplied while the current increases from zero to a final value i is, i 1 U = L ò idi = Li2 0 2 1 U = Li2 \ 2

Sample Problem 11 A long solenoid having 200 turns per cm carries a current of 1.5A. At the centre of the solenoid is placed a coil of 100 turns of cross-sectional area 314 . ´ 10 -4 m 2 having its axis parallel to the field produced by the solenoid. When the direction of current in the solenoid is reversed within 0.05 s, the induced emf in the coil is (a) 0.48 V (c) 0.0048 V

(b) 0.048 V (d) 48 V

and average radius of toroidal solenoid, r = 15 cm = 15 ´ 10 -2 m

Suppose that a current i is passed through the solenoid Magnetic field produced along the axis of the solenoid. 4000 B = m 0ni = 4 p ´ 10 -7 ´ ´ i = i ´ 1.6 ´ 10 -3 T p Magnetic flux linked with the solenoid itself f = B ´ area of cross-section × total number of turns = i ´ 1.6 ´ 10 -3 ´ 12 ´ 10 -4 ´ 1200 = i ´ 2.304 ´ 10 –3 Wb Now, L =

f i ´ 2.304 ´ 10 –3 = i i = 2.304 ´ 10 –3 H

(b) Magnetic flux linked with the second coil, f 2 = B ´ A ´ N2, where N2 is total number of turns of the second coil. \

f = i ´ 1.6 ´ 10 –3 ´ 12 ´ 10 -4 ´ 300 = 5.76 ´ 10 –4 ´ i Wb

Now, initial magnetic flux linked with the second coil, fi = 7.76 ´ 10 –4 ´ 0 = 0 Final magnetic flux linked with the second coil, f f = 5.76 ´ 10 –4 ´ final current

Interpret (c) The magnetic field is B = m0 nl

Given

n = 200 ´ 10 -2,

I = 1.5 A

B = 4p ´ 10 -7 ´ 200 ´ 10 -2 ´ 15 . Also flux

= 5.76 ´ 10 –4 ´ 2.0 = 1.152 ´ 10 –3 Wb Therefore, total change in flux, df = f f - fi = 1.152 ´ 10 –3 - 0 = 1.152 ´ 10 –3 Wb

B = 3.8 ´ 10 -2 W /m2

Time taken, dt = 0.05 s

f = B× A

Now,

f = 3.8 ´ 10 -2 ´ 314 . ´ 10 -4 = 1.2 ´ 10 -5 Wb. When the current in the solenoid is reversed, change in magnetic flux df = 2 ´ (1. 2 ´ 10 -5) = 2.4 ´ 10 -5 Wb From Faraday’s law, induced emf æ 2.4 ´ 10 -5 ö - df e= = 100 ´ ç ÷ = 0.048 V dt è 0.05 ø

e=

df 1.152 ´ 10 –3 = = 0.0234V dt 0.05

Sample Problem 13 What will happen to the inductance of a solenoid when the number of turns and the length are doubled keeping the area of cross-section same? L 2 (c) 2L (a)

(b) L (d) 4L

Electromagnetic Induction and Alternating Current Interpret (c) In case of B = m 0ni , f = B (nlS) = m 0n 2lSi and hence f L = = m 0n 2lS i = m0

a

solenoid

N2 S l

as

Nö æ ç as n = ÷ è lø

So, (a) when N and l are doubled,

21.6 Mutual Inductance 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.

(2N) 2 L¢ = m 0 S 2l Nl = 2m 0 S = 2L l i.e., inductance of the solenoid will be doubled.

Sample Problem 14 Calculate (i) 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 cm 2 and (ii) Calculate the self induced emf in the solenoid, if the current through it is decreasing at the rate of 50.0 A / s. (a) 1.81 ´ 10 -4 H, 9.05 mV

Consider two neighbouring coils of wire as shown in figure. 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 has mutual inductance ( M ). It also have two definitions as under

(b) 1.21 ´ 10 -3 H, 8.05 mV (c) 1.33 ´ 10

-2

H, 5.03 mV

(d) 2.35 ´ 10

-5

H, 4.26 mV

955

1

2

Interpret (a) (i) The inductance of a solenoid is given by, L=

m 0N 2S l

Substituting the values we have, L =

(ii) As, Here, \

( 4p ´10 –7 ) ( 300 ) 2 ( 4.00 ´ 10 –4 ) H ( 25.0 ´ 10 –2)

= 1.81 ´ 10 –4 H di e = –L dt di = – 50.0 A / s dt = 9.05 ´ 10

or

V

e = 9.05 mV

Sample Problem 15 What inductance would be needed to store 1.0kWh of energy in a coil carrying a 200 A current? (1kWh = 3.6 ´ 106 J) (a) 215 H (c) 180 H

(b) 106 H (d) 120 H

As, \

U = 1 kWh = 3.6 ´ 10 6 J L =

2U i2

2 (3.6 ´ 10 6) = = 180 H (200) 2

First definition

N 2 fB2 µ i1 or

N 2fB2 = Mi1

Here, the proportionality constant M is known as the mutual inductance M of the two circuits. N 2fB2 Thus, M= i1 From this expression, M can be defined as the total flux N 2f B2 linked with circuit 2 per unit current in circuit 1.

Second definition

Interpret (c) We have, i = 200 A and

i1

Suppose the circuit 1 has a current i1 flowing in it. Then total flux N 2fB2 linked with circuit 2 is proportional to the current in 1. Thus,

e = – (1.81 ´ 10 –4) ( – 50.0) –3

i1

1 2ö æ çQ U = Li ÷ è 2 ø

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 µ di1 / dt or

e2 = - Mdi1 / dt

956 JEE Main Physics 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. From the above equation, - e2 M= di1/dt

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, æN ö B = m 0 ç 1 ÷ i1 è l1 ø

This equation states that, the mutual inductance of two

circuits is the magnitude of induced emf e2 per unit rate of change of current di1 /dt. Note down the following points regarding the mutual inductance 1. The SI unit of mutual inductance is henry (H). 2. M depends upon closeness of the two circuits, their orientations and sizes and the number of turns etc.

and is directed parallel to the solenoid’s axis. The magnetic flux fB2 through the surrounding coil is, therefore, m Ni fB2 = B (pR12 ) = 0 1 1 pR12 l1 Now,

As,

e2 = – M (di1 / dt ) e1 = – M (di2 / dt ) N2fB2 M12 = i1

and

M21 =

N1fB1 i2

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) Suppose a current i1 flows through the primary circuit. (c) Determine the magnetic field B produced by the current i1. (d) Obtain the magnetic flux fB2 . (e) With the flux known, the mutual inductance can be found from, N f M = 2 B2 i1

Mutual Inductance of a Solenoid Surrounded by a Coil l1 R2 R1

N 2fB2 i1

æ N ö æm N i ö = ç 2 ÷ ç 0 1 1 ÷ pR12 è i1 ø è l1 ø

3. Reciprocity theorem M21 = M12 = M and

M=

\

=

m 0N1N 2pR12 l1

M=

m 0N1N 2pR12 l1

Notice that M is independent of the radius R2 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 fB1 is so complicated. However, since M12 = M21, we do know the result of this calculation.

Coefficient of Coupling of the Two Circuits Let us now calculate mutual inductance between two circuits in terms of the self-inductance of each circuit alone. Let us first consider a case when the total flux associated with one coil links with the other, i.e., a case of maximum flux linkage. Consider two coils placed adjacent to each other, N 2fB2 N1fB1 and M21 = M12 = i1 i2 Similarly,

L1 =

and

L2 =

N1fB1 i1 N 2fB2 i2

If all the flux of coil 2 links coil 1 and vice-versa then, Figure shows, a coil of N 2 turns and radius R2 surrounding a long solenoid of length l1, radius R1 and number of turns N1.

fB2 = fB1 Since, M12 = M21 = M, hence, we have

Electromagnetic Induction and Alternating Current M12 M21 = M 2 = \

In parallel

N1N 2fB1 fB2 i1i2

= L1L2 i

Mmax = L1L2

fB1 = K 2fB2

and

=

i1i2

of the coils and their relative closeness having value between 0 and 1.

In series If several inductances are in series so that there are no interactions through mutual inductance. L2

L3

d

b

⇒

L i

di dt di Vc – Vd = L2 dt di Vd – Vb = L3 dt Va – Vc = L1

From figure (a), i = i1 + i2 + i3 di di1 di2 di3 = + + dt dt dt dt di Va – Vb Va – Vb Va – Vb = + + dt L1 L2 L3 di Va – Vb = dt L

Here, L = equivalent inductance. From Eqs. (i) and (ii), we have L = L1 + L2 + L3

…(ii)

term becomes important. This mutual interaction may increase or decrease, the flux due to the self-induction. The equivalent inductance of the pair of coils in series is, L = L1 + L2 ± 2M

(b) 10 × 10 -3 Wb (d) 2 × 10 -7 Wb of the

coil, coil,

Total magnetic flux linked with the coil, f = Li = 10 ´ 10 -3 ´ 4 ´ 10 -3 = 4 ´ 10 -5 Wb Number of turns in the coil, N = 200 Therefore, magnetic flux through the cross-section of the coil (flux-linked with each turn) di dt

= …(i)

f 4 ´ 10 -5 = = 2 ´ 10 -7 Wb N 200

Sample Problem 17 A small square loop of wire of side l

From figure (b), di dt

…(i)

From Eqs. (i) and (ii), we get 1 1 1 1 = + + L L1 L2 L3

Adding all these equations, we get

Va – Vb = L

b

(b)

Interpret (d) Here, self-inductance Current through L = 10 mH = 10 ´ 10 –3H; i = 4 mA = 4 ´ 10 -3 A.

From figure (a),

Va – Vb = (L1 + L2 + L3 )

b a

(a)

(a) 4 × 10 -5 Wb (c) 4 × 10 -3 Wb

b

(b)

and

L

Sample Problem 16 The self-inductance of a coil having 200 turns is 10 mH. The magnetic flux through the cross-section of the coil corresponding to current is 4 mA, is

(a) a

L3

i

Note If the flux from one inductance links another, mutual inductance

Combination of Inductances

c

i3

(K £ 1)

Here, K = K1K 2 is a number, depending on the geometry

L1

L2

From figure (b),

M = K L1L2

a i

i2

or

N1N 2K1K 2fB1 fB2

= K1K 2L1L2 or

L1

or

fB2 = K1fB1

M21 M12 = M 2

\

i1

a

This is the maximum possible value of M as the total flux associated with one coil links with the other. In general only a fraction K 2 (< 1) of fB2 passes through the coil 1. Similarly a fraction K1 (< 1) of fB1 passes through coil 2. Hence,

957

…(ii)

is placed inside a large square loop of wire of side L (>l). The loops are coplanar and their centres coincide. What is the mutual inductance of the system? 2m 0 l 2 p L 2m L2 (c) 0 pl

(a)

2 2m 0 l 2 pL -2 2m 0L2 (d) pl (b)

958 JEE Main Physics Interpret (b) Considering the larger loop to be made up of four rods each of length L, the field at the centre, i.e., at a distance (L/2) from each rod, will be

Interpret (b)As inductances obey laws similar to grouping of resistances, L1 + L2 = 10 H L1L2 = 2.4 H (L1 + L2)

and I I

L

Substituting the value of (L1 + L2) from first expression into second, L1 L2 = (2.4) (L1 + L2 ) = 2.4 ´ 10 = 24 So that, (L1 - L2) 2 = (L1 + L2) 2 - 4L1L2

m B=4´ 0 4p m0 B=4´ 4p

i.e.,

i [sin a + sin b ] d i ´ 2 sin 45° L /2

m0 8 2 i 4p L So, the flux linked with smaller loop m 8 2 2 f 2 = B1S 2 = 0 li 4p L i.e.,

B1 =

and hence,

M=

(b) 2 ´ 10 -4 H

(c) 3 ´ 10 -5 H

(d) 1. 5 ´ 10 -3 H

and as

L1 + L2 = 10 H, L1 = 6 H L2 = 4 H

Sample

Problem 20 Two different coils have self-inductances L1 = 8 mH and L2 = 2 mH. At a certain instant the current in the two coils is increasing at the same constant rate and the power supplied to the two coils is the same. Find the ratio of energies stored in the two coils at that instant. 1 2 1 (c) 8

1 4 1 (d) 16 (b)

(a)

Sample Problem 18 A straight solenoid has 50 turns per (a) 5 ´ 10 -4 H

L1 - L2 = [(10) 2 - 4 ´ 24]1/ 2 = 2 H

and

f1 2 2 m 0 l 2 = i pL

cm in primary and 200 turns in the secondary. The area of crosssection of the solenoid is 4 cm 2. The mutual inductance is

i.e.,

e =L

Interpret (b) As, So,

Interpret (a) 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 = m 0n1i1

e1 L1 8 mH = = =4 e 2 L2 2 mH

As

P = ei = constant

So,

i1 e 2 1 = = i2 e1 4

As energy, stored in coil, U =

The magnetic flux through the secondary of N2 turns each of area S is given as, = m 0n1N2i1S Nf M= 2 2 i1 = m 0n1N2S Substituting the values,

æ 50 ö M = ( 4p ´ 10 –7) ç –2 ÷ (200) ( 4 ´ 10 –4) è10 ø

(a) 2 H, 8 H (c) 3 H, 7 H

(b) 4 H, 6 H (d) 5 H, 5 H

2

8 æ 1ö 1 ç ÷ = 2 è 4ø 4

Sample Problem 21 An inductor L = 20 mm, a resistor R = 100 W and a battery E = 10 V are connected in series. The time elapsed before the current reaches 99% of the maximum value is (a) 0.37 ms (c) 0.51 ms

inductors is 2.4 H when connected in parallel and 10 H when connected in series. What is the value of inductances of the individual inductors?

1 2 Li 2

2

=

= 5.0 ´ 10 –4 H

Sample Problem 19 The equivalent inductance of two

ù é di êë as dt = constant úû (given)

U1 L1 æ i1 ö = ç ÷ U2 L2 è i2 ø

So,

N2f2 = N2(BS) \

di dt

(b) 0.92 ms (d) 0.62 ms

Interpret (b) The time constant, l = The maximum current i = As,

L 20 = = 0.20 ms R 100

E 10 = = 0.10 A R 100 i = i0 (1 - e- t /l )

0.99 i = i0 (1 - e- t /l )

Electromagnetic Induction and Alternating Current e- t /l = 0.01

or

t = - ln (0.01) l t = 0.20 ms ln (100) = 0.92 ms

or or

Sample Problem 22 An LR circuit having L = 4 H, d R = 1W and E = 6 V is switched on at t = 0. The power dissipated in joule heating at t = 4 s is (a) 1210 W (c) 110 W

(b) 120 W (d) 100 W

Interpret (a) The time constant of the circuit is L 4 = =4s R 1 The current at t = 4 s is therefore E æ 1ö i = (1 - e- t / l ) = 6 ç1 - ÷ è R eø l=

\

di

i

t

ò 0 E - iR = ò 0

959

dt L Rt

or

i=

– E (1 - e L ) R

By letting E / R = i 0 and L / R = t L, the above expression reduces to, i = i 0 (1 – e–t / tL ) Here, i 0 = E / R is the current at t = ¥. It is also called the steady state current or the maximum current in the circuit. L And t L = is called time constant of the L-R circuit. At a R time equal to one time constant the current has risen to (1 – 1/ e) or about 63% of its final value i 0. The i-t graph is as shown in figure. i

= 6A ´ 6.3 A = 3.8 A The power dissipated in Joule heating

i0 = E/R

= i2 R = (3.8 A) 2 ´ 10 W = 140 W

21.7 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 S1, we connect R and L in series with constant emf E. Let i be the current at some time t after switch S1 is closed and di / dt be its rate of change at that time. Applying Kirchhoff’s loop rule strating at the negative terminal and proceeding counterclockwise around the loop E

0.63 i0

tL

t

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. 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 (= i2R) plus the rate at which energy is stored in the inductor (= iVbc = Li di / dt ) or

d dt

di æ 1 2ö ç Li ÷ = Li . è2 ø dt

Thus, S1

Ei = i 2R + Li

di dt

Decay of Current S2 i a

or

R

b

L

E - Vab - Vbc = 0 di E - iR - L =0 dt

c

Now suppose switch S1 in the circuit shown in figure has been closed for a long time and that the current has reached its steady state value i 0. Resetting our stopwatch to redefine the initial time, we close switch S2 at time t = 0 and at the same time we should open the switch S1 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

960 JEE Main Physics rule and to find current in the circuit at time t, let us draw the circuit once more.

Rate of increase of current, di i0 –t / tL = e dt tL di i0 E /R E = = = dt tL L /R L

At t = 0 , i a

i b

R

c

L

Applying loop rule we have, (as Va = Vc )

(Va – Vb ) + (Vb – Vc ) = 0 æ di ö iR + L ç ÷ = 0 è dt ø

or

Note Don’t bother about the sign of \ i

òi

\

0

= (0.37) (400) = 148 A /s (c) The steady state current in the circuit, E 200 i0 = = = 10A R 20

di . dt

di R = – dt i L di R t = – ò dt i L 0

21.8 Oscillations in L-C Circuit

i = i0e–t / tL

\

L where, t L = , is the time for current to decrease to 1/e or R about 37% of its original value. The i-t graph is as shown in figure. i

i0

0.37 i0

tL

t

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 or or

Substituting the value, we have di 200 = = 400 A /s dt 0.5 (b) At t = tL , di = (400) e–1 dt

dU d æ 1 2ö æ di ö i2R = – =– ç Li ÷ = Li ç – ÷ ø è dt ø dt dt è 2

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. i b

t=0 + C

–

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.

Interpret (a) This is the case of growth of current in an L-R circuit. Hence, current at time t is given by, i = i0 (1 – e–t / tL )

c

+ q0

L

æ –di ö i2R = Li ç ÷ è dt ø

Sample Problem 23 A coil of resistance 20 W and

t=t

–

q

L

a S

d S

(a)

(b)

Let us now derive an equation for the oscillations in an L-C circuit.

From figure (a), A capacitor is charged to a PD V0 = q0C Here, q0 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.

Electromagnetic Induction and Alternating Current From figure (b), When the switch is closed, the capacitor starts discharging. Let at time t charge on the capacitor is q ( < q0 ) 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.

d 2q æ 1 ö = –ç ÷q è LC ø dt 2

or

The general solution of Eq. (ii), is q = q0 cos (wt ± f )

difference across inductor,

For example in our case f = 0 as q = q0 at t = 0.

Vb – Va = Vc – Vd

Hence,

q æ di ö =Lç ÷ è dt ø C

…(i)

Now, as the charge is decreasing, di d 2q æ –dq ö =– 2 i=ç ÷ or è dt ø dt dt Substituting in Eq. (i), we get æ d 2q ö q = –L ç 2 ÷ C è dt ø

q = q0 cos wt

The oscillations of the L-C circuit are an electromagnetic analog to the mechanical oscillations of a block-spring system.

Mass Spring System

Inductor-Capacitor Circuit

1.

Displacement (x )

Charge (q )

2.

Velocity (v )

Current ( i )

Acceleration (a )

di Rate of change of current æç ö÷ è dt ø

3. 4.

d2x dt

2

= – w2 x, where w =

k m

d 2q dt 2

= – w2q , where w =

1 LC

x = A sin (wt ± f) or x = A cos (wt ± f) dx = w A 2 – x2 v= dt dv = – w2 x a= dt 1 Kinetic energy = mv 2 2 1 Potential energy = kx2 2

q = q 0 sin (wt ± f) or q = q 0 cos (wt ± f) dq = w q 02 – q 2 i = dt di Rate of change of current = = – w2q dt 1 Magnetic energy = Li 2 2

10.

1 1 1 1 2 mv 2 + kx2 = constant = kA 2 = mvmax 2 2 2 2

11.

|vmax| = Aw

1 q 02 1 2 1 2 1q2 = Limax = constant = Li + 2 C 2 2 2 C imax = q 0 w

12.

|amax| = w2 A

5. 6. 7. 8. 9.

…(iv)

Thus, we can say that charge in the circuit oscillates simple harmonically with angular frequency given by Eq. (iii). Thus, w 1 1 , f = w= = 2 p 2 p LC LC 1 and T = = 2p LC f

Shows a comparison of oscillations of a mass-spring system and an L-C circuit. S. No.

…(ii)

This is the standard equation of simple harmonic motion æ d 2x ö ç 2 = – w2x ÷ . è dt ø 1 …(iii) Here, w= LC

The potential difference across capacitor = potential

\

961

Potential energy =

æ di ö = w2q 0 ç ÷ è dt ømax

13.

1 k

C

14.

m

L

1q2 2 C

962 JEE Main Physics q

A graphical description of the energy transfer between the inductor and the capacitor in an L-C circuit is shown in figure. The right side of the figure shows the analogous energy transfer in the oscillating block-spring system.

q0 t

i

i=0 C ++++ t=0

i0

+q0

––––

k

L

E –q0

t

v=0 m

(a)

0

T

3T 2

2T

A

x=0

S

T 2

UC i = imax C t =T 4

2 qmax 2C

vmax q=0

L

m

B (b) S

UL x=0 0

i = imax B

C t = 3T 4

q=0

vmax

L

m

x=0 i=0

t =T

––––

v=0 L

E –q0

m (e) x=0

S

T 4

T 2

3T 4

T

charged to 300 V. It is then connected across a10 mH inductor. The resistance in the circuit is negligible.

S

+q0

2 Limax 2 t

Sample Problem 24 A capacitor of capacitance 25 mF is

(d)

C ++++

t

A

(a) Find the frequency of oscillation of the circuit. (b) Find the potential difference across capacitor and magnitude of circuit current1.2 ms after the inductor and capacitor are connected. (c) Find the magnetic energy and electric energy at t = 0 and t =1.2 ms.

Interpret (a) The frequency of oscillation of the circuit is,

Note

di In L-C oscillations, q, i and all oscillate simple harmonically with dt same angular frequency w . But the phase difference between q di p di is , while that between i and is p. and i or between i and dt 2 dt Their amplitudes are q 0 , q 0 w and w2q 0 respectively. So, now suppose

1 2p LC Substituting the given values, we have 1 f= = 318.3 Hz 2p ( 10 ´ 10 –3) ( 25 ´ 10 –6) f=

(b) Charge across the capacitor at time t will be, q = q 0 cos wt

q = q 0 cos, wt then and

dq i = = – q 0 w sin wt dt di = – q 0 w2 cos wt dt

Similarly potential energy across capacitor (UC ) and across inductor (U L ) also oscillate with double the frequency 2w but not simple harmonically. The different graphs are as shown in figure.

and Here,

i = – q 0 w sin wt

q 0 = CV0 = (25 ´ 10 –6) (300) = 7.5 ´ 10 –3 C

Now, charge in the capacitor after t =1.2 ´ 10 –3 s is q = (7.5 ´ 10 –3) cos (2p ´ 318.3) (1.2 ´ 10–3 ) C = –5.53 ´ 10 –3 C

Electromagnetic Induction and Alternating Current \ PD across capacitor, |q | 5.53 ´ 10–3 = 221.2 V V= = C 25 ´ 10–6 The magnitude of current in the circuit at t = 1.2 ´ 10

–3

s is,

|i | = q 0 w sin wt

= (7.5 ´ 10 –3) (2p ) (318.3) sin(2p ´ 318.3) (1.2 ´ 10 –3) A

= 10.13 A (c) At t = 0 Current in the circuit is zero. Hence, UL = 0 Charge in the capacitor is maximum. q 02

Hence,

UC =

1 2 C

or

UC =

1 (7.5 ´ 10 –3) 2 =1.125 J ´ 2 (25 ´ 10 –6)

\Total energy E = UL + UC = 1.125 J At t = 1.2 ms, 1 1 UL = Li 2 = (10 ´ 10 -3)( 10.13) 2 2 2 = 0.513 J \ UC = E – UL =1.125 – 0.513 = 0.612 J Otherwise UC can be calculated as, UC =

1 q 2 1 (5.53 ´ 10 –3) 2 = ´ 2 C 2 (25 ´ 10 –6)

= 0.612 J

21.9 Some Applications of Electromagnetic Induction (i) Eddy Currents

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. 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 machinary. These slots intercept the conducting paths and decreases the magnitudes of the induced currents.

(ii) 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.

(iii) 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. Suppose at time t = 0, plane of coil is perpendicular to the magnetic field.

ω

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.

v

963

v

The flux linked with it at any time t will be given by

f = NBA cos wt

F F

\ 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

(N = number of turns in the coil) df e=– = NBA w sin wt dt

or

e = e0 sin wt

where,

e0 = NBA w

964 JEE Main Physics Sample Problem 25 A circular loop of radius a having n turns is kept in a horizontal plane. A uniform magnetic field B exists in a vertical direction as shown in the figure. Find the emf induced in loop if the loop is rotated with a uniform angular velocity w about × × × × × × × ×

is called angular frequency of AC. 2p Also, w= = 2 pn T where, T is the time period or period of AC. It is equal to the time taken by the AC to go through one complete cycle of variation. Y AC

× × × × × × × × × × × × × × × × × × × × × × × ×

O

× × × × × × × ×

(i) an axis passing through the centre and perpendicular to the plane of the loop. (ii) a diameter. (a) pna2 Bw sin wt , no emf will be induced in the coil

t 2T

T

X

Y' AC

(b) Large emf will be induced in the coil, pna2Bw sin wt (c) No emf will be induced in the coil, pna2Bw sin wt (d) None of the above

T'

Interpret (a)

2T '

t 4T ' X

3T '

Y'

(i) The emf induces when there is change of flux. As in this case, there is no change of flux, hence no emf will be induced in the coil. (ii) If the loop is rotated about a diameter, there will be change of flux with time. In this case, emf will be induced in the coil. The area of the loop is A = pa 2. If the normal of the loop makes an angle q = 0° with the magnetic field at t = 0, this angle will becomes q = wt at time t. The flux of the magnetic field at this time is

The terms used of AC hold equally for alternating emf which may be represented by V = V0 sin wt or

B = V0 cos wt Y AC t

× × × × × × × ×

X

× × × × × × × ×

Y' × × × × × × × × × × × × × × × ×

f = nBpa2 cos q f = nBpa2 cos wt The induced emf is e=

df = pna2Bw sin wt dt

21.10 Alternating Current Most of the electric power generated and used in the world is in the form of AC i. e. , alternating current. The magnitude of an alternating current changes continuously with time and its direction is reversed periodically. It is represented by i = i0 sin wt or

i = i0 cos wt

Here, i is instantaneous value of current i. e. , magnitude of current at any instant of time t and i0 is the peak value or maximum value of AC. It is also called amplitude of AC, w

21.11 Peak and Root Mean Square Value of Alternating Current and EMF Mean Value or Average Value or Peak Value The steady current, which when passes through a circuit for half the time period of alternating current, sends the same amount of charge is as done by the alternating current in the same time through the same circuit, is called mean or average value of alternating current. It is denoted by im or iav 2i Thus, im or iav = 0 = 03 . 6 i0 p Thus, mean or average value of alternating current during a half cycle is 0.636 times (or 63.6% of) its peak value (i0 ). Similarly, mean or average value of alternating emf 2V0 = 0636 . Vm or Vav = p

965

Electromagnetic Induction and Alternating Current Note During the next half cycle, the mean value of alternating current and emf will be equal in magnitude but opposite in direction, the average of which over a complete cycle is always zero.

RMS Value The steady current, which when passes through a resistance for a given time will produce the same amount of heat as the alternating current does in the same resistance and in the same time, is called rms value of alternating current. It is denoted by i . i0 irms or iv = 0 = 0707 2 where, i0 = peak value of alternating current

Sample Problem 26 If a domestic appliance draws 2.5 A from a 220 V, 60 Hz power supply, find (i) the average current (ii) the average of the square of the current (iii) the current amplitude (iv) the supply voltage amplitude (a) 2A, 2.5A, 3.5A, 311 V (b) Zero, 2.5A, 3.5A, 311 V (c) Zero, 2.5A, 3.5A, 310 V (d) None of these

Interpret (b) (i) The average of sinusoidal AC values any whole number of cycles is zero. (ii) Rms value of current = i rms = 2.5A

(iii) irms =

Similarly, rms value of alternating emf V Vrms = 0 = 0707 . V0 2

(iv) Vrms = 220 V =

even for a complete cycle of AC.

Vm = 2 (Vrms) = 2 (220 V) = 311 V

1. The average value of sin wt, cos wt, sin2 wt, cos 2 wt, etc, is zero because it is positive for half of the time and negative for rest half of the time. Thus, (sin wt) = (cos wt) = (sin2 wt) = (cos 2 wt) = 0 \ If i = i 0 sin wt (i ) = (i 0 sin wt) = i 0 (sin wt) = 0 1 2. The average value of sin2 wt is 2 1 or (sin2 wt) = (cos 2 wt) = 2 i 2 = i 20 sin2 wt \ If then,

21.12 Current and Potential Relations We will consider now AC circuit containing pure resistor, inductor and capacitor.

Resistor in an AC Circuit R

i20 2

3. Like SHM, general expression of current voltage in an sinusoidal AC are, i = i 0 sin( wt ± f), V = V 0 sin( wt ± f) or i = i 0 cos( wt ± f) and V = V 0 cos( wt ± f) 4. The ratio,

V0 2

\ Supply voltage amplitude

Important Points

(i 2 ) = (i 20 sin2 wt) = i 20 (sin2 wt) =

i0 2

\ Current amplitude = 2 irms = 2(2.5A) = 3.5A

Note The rms of virtual value of alternating current and emf is same

Then,

2 (iav ) = (irsm) 2 = 6.25A 2

\

V / 2 p rms value = 0 = = 1.11 is known as form average value 2V 0 /p 2 2

factor. The different values i 0 i av and i rms are shown in figure.

If an AC circuit, fed by an alternating emf V = V0 sin wt contains pure resistance R, then current V V0 i= = sin wt = i0 sin wt R R We see from Fig. (a) that voltage and current are in phase if only resistance is in the circuit, i, VR V0 i0

VR = V0 sin ωt iR = i0 sin ωt

i

V0 i0

t

i0 irms = 0.707i0 iav = 0.637 i0

ωt

t

(a)

(b)

The corresponding phasor diagram is also shown. That is in a purely resistive AC circuit the current and voltage are in same phase.

966 JEE Main Physics Capacitor in an AC Circuit, Reactance of Capacitor

Impedance

If an AC circuit, fed by an alternating emf V = V0 sin wt pö æ contains pure capacitance C, the current i = i0 sin ç wt + ÷, è 2ø

According to Ohm’s law I =

For a pure resistor, where Z = R I =

C

Above equation shows that effective AC resistance, i. e. , capacitive reactance of capacitor is X C = 1/wC

V × R

where I and V are the rms or effective values. The quantity Z is called impedance. Since the phase affects the impedance and because the contributions of capacitors and inductors differ in phase from resistive components by 90 degrees, a process like vector addition (phasors) is used to develop expression for impedance.

Impedance Combination

It unit is ohm. We see from figure that current leads the voltage by 90° or p or the potential drop across the capacitor lags the 2 current passing it by 90°, y

EMF

π /2 π

ωt

π 2π

3 π/2

Z1 + Z2 = (R1 + jX1 ) + (R2 + jX 2 ) = (R1 + R2 ) + j ( X1 + X 2 ) = Req + jX eq X eq 2 2 , f = tan-1 | Z | = Req + X eq Req

π/2

iv

Z2

Z1

Ev

Current

E or i 0

V Z

For parallel set up

Zeq =

y'

The phasor diagram shows that in a purely capacitive circuit, current phasor leads the voltage phasor by 90°.

Inductor in AC Circuit, Reactance of Inductor If an AC circuit, fed by an alternating emf V = V0 sin wt, pö æ contains pure inductance, then current i = i0 sin ç wt - ÷ è 2ø V0 where, i0 = × wL Above equation shows that effective AC resistance, i. e. , inductive reactance of inductor is X L = wL Thus, unit of X L is ohm. Thus, we see from Fig. (a) that voltage across the inductor leads the current passing through it by 90°. Ev

y

0

Zeq = Req + jX eq = | Z | e jf The complex impedance of the parallel circuit takes the form Zeq =

Z1 Z2 (R + jX1 ) (R2 + jX 2 ) = 1 Z1 + Z2 (R1 + R2 ) + j ( X1 + X 2 )

= Req + jX eq = | Z | e jf

Sample Problem 27 A resistor of 200W and capacitor of 15mF are connected in series to a 220 V, 50 Hz AC source. The current in the circuit is [NCERT] (a) 0.755 A (c) 15 A

(b) 75 A (d) 1.5 A

Interpret (a) Given, 200 Ω

15 µ C

Current π /2

π 3π/2

ωt

0

π/2

2π

220 V

y'

iv

Phasor diagram shows that V L leads the current i by 90°.

Z2

Z1 Z2 Z1 + Z2

EMF E or i

Z1

1 1 1 = + Zeq Z1 Z2

R = 200 W, C = 15 mF = 15 ´ 10 -6F, V = 220 V, f = 50 Hz

Electromagnetic Induction and Alternating Current The impedance is

967

Series R-L Circuit

Z = R 2 + XC2 = R 2 + (2pfC) 2 -6 2

2

Z = (200) + (2 ´ 314 . ´ 50 ´ 10 ) = (200) 2 + (212) 2 = 2915 . W

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.

The current in the circuit is V 220 I= = = 0 .755A Z 291.5

Sample Problem 28 Calculate the capacitive reactance

VR

of a 5mF capacitor for a frequency of 106 Hz. (a) 0.032 W

(b) 0.05 W

(c) 0.1 W

(d) 0.2 W

Interpret (a) Capacitive reactance XC =

1 1 1 7 = = = = 0.032 W wC 2pfC 2p ´ 10 6 ´ 5 ´ 10 -6 10 ´ 22

VL

Suppose in phasor diagram current is taken along positive x-direction and V L along positive y-direction. So, we can write

Important Points Table Circuit Elements with AC Circuit Elements

Amplitude Relation

Circuit Quantity

V 0 = i 0R

R

in phase with i

Capacitor

V0 = i 0 X C

1 XC = wC

lags i by 90°

Inductor

V0 = i 0 X L

X L = wL

leads i by 90°

Resistor

Phase of V

1. The potential of point a with respect to point b is given by

θ

1 ö æ ç as X C = ÷ = iZ è wC ø

V = VR + jV L = iR + j (iX L )

Here, Z = R + jX L = R + j (wL ) 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 | = R2 + (wL ) 2

b

i

di Vl = + L , the negative of the induced emf. This expression gives dt the correct sign of VL in all cases.

2. With increase in frequency inductive reactance ( X L = wL ) i .e ., opposite of a given coil to ACincreases linearly with frequency. So, if w ® 0 , X L ® 0 and if w ® ¥, X L ® ¥. This is why an inductor is called low pass filter and as for DC w = 0,i .e ., X L = 0, the opposition of an inductance to DC is zero. 1 1 3. As, X C = , with increase in frequency X C decreases = wC 2 pfC non-linearly, i .e ., the opposition of a capacitor to AC decreases with increase in frequency so, if w ® 0, X C ® ¥ and if w ® ¥, X C ® 0. This is why a capacitor is called high pass filter and as for DC w ® 0, X C ® ¥, the opposition of a capacitor to DC is infinite,i .e .,a capacitor acts as open circuit in DC circuits in steady state.

4. As shown in figure, the graphs of R , X L and X C as functions of angular frequency w .

q = tan-1

| V L| | V R|

æX ö æ wL ö = tan-1 ç L ÷ or q = tan-1 ç ÷ è R ø è R ø

Series R-C Circuit Potential differences across a capacitor in AC lags in phase by 90° with the current in the circuit.

VR

VC

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 VR

R, X

θ

XL XC

x

VR

i

The potential difference leads the current by an angle,

L a

y

V

VL

R

VC

i

ω

V

y

x

968 JEE Main Physics V = VR - jVC = iR - j (iX C ) 1 ö æ i ö æ = iR - j ç ÷ = iZ ç as X C = ÷ è wC ø è wC ø æ 1 ö Here, impedance is, Z = R - j ç ÷ è wC ø

VS2 = (V L - VC ) 2 + VR2

æ 1 ö The modulus of impedance is, | Z | = R + ç ÷ è wC ø 2

2

and the potential difference lags the current by an angle, V æ 1/wC ö æX ö q = tan-1 C = tan-1 ç C ÷ = tan-1 ç ÷ è R ø è R ø VR æ 1 ö q = tan-1 ç ÷ è wRC ø

or

21.13 L-C-R Series Circuit As shown in the figure, inductance L, capacitance C and resistance R are connected IS together in series and supplied C with an alternating voltage. In VC such an arrangement the same L circuit current flows through VL VS all the components of the circuit and VR, V L and VC R VR indicate the voltages across the resistor, inductor and capacitor. Fig. (b) shows the circuit conditions when the inductive reactance ( X L ) is greater than the capacitive reactance ( X C ). In this case, since both L and C carry the same current and X L is greater than X C , it follows that V L must be greater than VC (V L = I S X L and VC = I S X C ) Remember that VC and V L are in anti-phase to each other due to their 90° leading and lagging relationship with the circuit current I S . As V L and VC directly oppose each other, a resulting voltage is created, which will be the difference between VC and V L. This is called the reactive voltage. VL VR

VC

IS

Phasors for V L and VC are in anti-phase.

(a)

VL VS

VL-VC θ

VR

VC

(b)

IS

The phasors for (V L - VC ), VR and VS in the figure form right angle triangle, a number of properties and values in the circuit can be calculated using Pythagoras theorem or some basic trignometry. Like

V L is greater than VC , so, the circuit behaves like an inductor.

VS = (V L - VC ) 2 + VR2

or

Total circuit impedance ( Z ) is Z = ( X L - X C ) 2 + R2 The phase angle between (V L - VC ) and VR can be found using trignometry tan q = opposite ¸ adjacent \

tan q =

(V L - VC ) (V - VC ) × \ q = tan-1 L VR VR

Also Ohm’s law states that R (or X ) =

V I

If (V L - VC ) and VR are each divided by the current I S , this allows the phase angle q to be found using the resistances and reactances. æ X - XC ö q = tan-1 ç L ÷ è ø R

When VC is larger than VL the circuit is capacitive VL The figure shows the phasor diagram for a L-C-R series circuit in which X C is greater than X L, showing that when VC exceeds V L, θ the resultant reactive voltage is now V –V L C given by (VC - V L ) and VS is the phasor sum of (VC - V L ) and VR × V

VR

IS

VS

C

The phase angle q now shows that the circuit current (I s ) leads supply voltage (VS ) by between 0° and 90°. The overall circuit is now capacitive. VS = (VC - V L ) 2 + VR2

When X L and X C are equal the circuit is purely resistive The figure shows the situation which VL must occur at some particular frequency, when X L and X C are equal. VS=VR The opposing and equal voltages VC IS and V L now completely cancel each θ = 0° other out. The supply voltage and the VS=VR= 0V circuit current must now be in phase so VC the circuit is apparently entirely resistive L and C have completely disappeared. This special case is called series resonance circuit.

Electromagnetic Induction and Alternating Current

21.14 Q-Factor (Quality Factor) The Q-factor or quality factor of a resonant L-C-R circuit is defined as ratio of the voltage drop across inductor (or capacitor) to applied voltage. Thus, Q=

voltage across L (or C ) applied voltage

Q=

1 L R C

Important Points 1. Let us take the most general case of a series L-C-R circuit in an AC | Z | = R2 + ( X L - X C )2 XL = XC 1 wL = wC 1 w= LC

If or or

f =

or

969

Sample Problem 29 A 25 mF capacitor, 0.1 H inductor and a 25.0 W resistor are connected series with a source, whose emf is given by V = 310 cos 314 t volt What is the frequency of the emf? What is the reactance of the circuit ? What is the impedance of the circuit ? What is the current in the circuit ? What is the phase angle of the current by which it leads or lags the applied emf ? (f) What is the expression for the instantaneous value of current in the circuit ? (g) What are the effective voltages across the capacitor, inductor and the resistor ? (h) What value of inductance will the impedance of the circuit minimum ? (a) (b) (c) (d) (e)

Interpret Here, L = 0.1H; C = 25 mF = 25 ´ 10 -6F; R = 25 W Also, the emf of the source is given by V = 310 cos 314 t The instantaneous value of alternative emf is given by V = V0 cos 2pft Comparing the Eqs. (i) and (ii), we have 314 314 (a) Frequency, f = = = 50 cycle s-1 2p 2 ´ 314 .

1 2p LC

The modulus of impedance | Z| = R And if the current is the phase with voltagei .e ,ifV = V 0 sin wt, then i = i 0 sin wt

... (i) …(ii)

(b) Inductive reactance, XL = 2pfL = 2 ´ 314 . ´ 50 ´ 0.1= 314 . W Capacitive reactance,

V V i0 = 0 = 0 | Z| R

where,

XC =

Such a condition is known as resonance and frequency known as reasonance frequency and is given by, 1 f = 2p LC

2. Response curves of series circuit The impedance of an 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 R Z j

XC XL– X

Net reactance of the circuit, XC - XL = 127.4 - 31.4 = 96 W As, XC > XL , the net reactance is capacitive.

= (25) 2 + (96) 2 = 99. 2 W Also, from Eqs. (i) and (ii), V0 = 310 V Therefore, virtual voltage, Vrms = (d) Hence, virtual current, irms =

Z

XL

= 127.4 W

(c) Impedance of the circuit, Z = R 2 + ( XC - XL) 2

The current in such a case is maximum.

i

1 1 = 2pfC 2 ´ 314 . ´ 50 ´ 25 ´ 10 -6

log ω

V0 310 = = 219. 2 V 2 2

VV 219. 2 = = 2. 21 A Z 99. 2

(e) The phase angle q is given by X - XL 96 tan q = C = = 3. 84 R 25 7. 54 ´ p q = 75. 4° = = 1. 316 rad 180 As XC > XL , the current leads the voltage by the angle q as obtained above.

970 JEE Main Physics (f) The instantaneous current is given by V i = i0 cos(2pft - f) = 0 cos(314 t - 1. 316) Z 310 = cos(314 t - 1. 316) 99. 2 or

i = 3.125 cos (314 t - 1. 316)

(g) Voltage across capacitor = irms XC = 2. 21 ´ 127. 4 = 281. 6 V Voltage across inductor = irms XL = 2. 21 ´ 31. 4 = 69. 4 V Voltage across resistor = irms R = 2. 21 ´ 25 = 55. 25 V (h) For impedance of the circuit to be minimum, 1 wL = wC Therefore, the required value of inductance, 1 1 1 L= 2 = = = 0. 405 H w C (2pf ) 2C (2p ´ 50) 2 ´ 25 ´ 10 -6

Sample Problem 30 When an alternating voltage of 220 V is applied across a device X, a current of 0.5 A flows through the circuit and is in phase with the applied voltage. When the same voltage is applied across another device Y, the same current again flows through the circuit, but it lags behind the applied voltage by p /2 radian. Calculate the current flowing in the circuit when same voltage is applied across the series combination of X and Y. (a) 0.5 A (c) 2.5 A

(b) 0.3 A (d) 4.5 A

Interpret (b) The current and voltage are in phase with each other, when alternating voltage is applied across a resistor. Hence, the device X is resistor. V 220 Obviously, R = rms = = 440 W irms 0.5 The current lags behind the voltage by phase angle p /2, when alternating voltage is applied across an inductor. Hence, the device Y is an inductor. V 220 Obviously, XL = rms = = 440 W Irms 0. 5 Vrms = 220 V ; R = 440 W; XL = 440 W If Z is impedance of L-R circuit, then

(a) 11 . ´ 10 -1H

(b) 11 . ´ 10 -2H

(c) 5. 5 ´ 120 -5H

(d) 6. 7 ´ 10 -7H

Interpret (b) Since voltage applied across the inductance leads the current by 45° (and not by 90°), the given inductance is not a pure inductance. The given inductance behaves as a series combination of L and R. If Z is impedance of the given inductance and q is a phase angle by which applied voltage leads the current, then Z = R 2 + XL2 and

XL2 + XL2: = 100 2XL = 100

or

or XL = 70. 71 W If L is self-inductance of the coil, then XL = 2pfL 70. 71 X or L= L = = 1.1254 ´ 10 -2H 2pf 2p ´ 1000

Sample Problem 32 A 100 mF capacitor in series with a 40 W resistance is connected to a 110 V-60 Hz supply. (i) What is the maximum current in the circuit? (ii) What is the phase lag between the current maximum and voltage maximum? (a) 1.5 A, 33° 33¢ (c) 3.2 A, 33° 33¢

(b) 1.5 A, 60° (d) 3.2 A, 60°

Interpret (c) Here, C = 100 mF = 10 -4F ; R = 40 W , Vrms = 110 V; f = 60 Hz; (i) As, irms =

Vrms 1 R + 2 2 wC 2

=

Z = R 2 + XL2 = 440 2 + 440 2 = 400 2 W

Sample Problem 31 An inductance coil has a reactance of100 W. When AC signal of frequency 100 Hz is applied to the coil, the applied voltage leads the current by 45°. Calculate the self-inductance of the coil.

…(ii)

Given, Z = 100 W, f = 45° From Eq. (ii), we have X tan 45° = L R XL or R = XL or 1= R Substituting for R in Eq. (i), we have

Here,

Therefore, current in the L-R circuit, E 220 irms = V = = 0. 3535 A Z 440 2

…(i)

X tan q = L L

2

R +

1 (2pfC) 2

110 1 40 2 + (2p ´ 60 ´ 10 -4) 2 =

Now,

Vrms

=

=

100 1600 + 703. 62

100 = 2.292 A 48

irms = 2i0 = 2 ´ 2.292 = 3.24 A

(ii) For C-R circuit, 1 /wC 1 1 tan q = = = R 2pfCR 2p ´ 60 ´ 10 -4 ´ 40 = 0.6631 q = 33°33¢

(emf lags behind the current)

Electromagnetic Induction and Alternating Current Sample Problem 33 A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series L-C-R circuit in which R = 3W, L = 25.48 mH and C = 796 mF. Then, the phase difference between the voltage across the source and the [NCERT Exemplar] current is (a) 53.1° (c) 28.3°

(b) 42.4° (d) 0°

Interpret (a) As, XL = 2pfL = 2 ´ 3.14 ´ 50 ´ 25.48 ´ 10 -3 = 8W XC = =

1 2pfC 1 = 4W 2 ´ 314 . ´ 50 ´ 796 ´ 10 -6

XC - XL R -1 æ 4 - 8 ö f = tan ç ÷ = - 53.1° è 3 ø

Phase difference f = tan -1

(a) 0.65 A, 1.0 H , 5 mF , 10.1mF (b) 1.65 A, 1.0 H, 5mF, 10.1mF (c) 0.65 A, 2.1 H , 5 mF , 10.1mF (d) None of the above

Interpret (a) (i) Here, VR = irmsR where, irms is the rms value of current in the circuit. V 65 \ irms = R = = 0.65 A R 100 (ii) VL = irms ´ XL V or XL = L irms XL =

Now,

XL = wL = 2pfL or L = L=

(iii) XC =

1 (a) 4 1 (c) 3

1 (b) 2

fµ

i. e. ,

1 1 = wC 2pfC 1 C= = 5 ´ 10 -6 = 5 mF 2p ´ 50 ´ 638.46

XC = \

1 2p LC

C¢ =

or

1 1 1 = = time 2 4 C

=

Sample Problem 35 A series L-C-R circuit is connected to an AC source of 220 V and 50 Hz shown in figure. If the readings of the three voltmeters V1, V2 and V3 are 65 V, 415 V and 204 V respectively, calculate (i) (ii) (ii) (iv)

313.85 = 1.0 H 2p ´ 50

(iv) Let C be the capacitance of capacitor that will produce resonance with inductor L = 1.0 H. Then 1 f= 2p (LC ¢ )

(d) None of these

Interpret (b) Resonance frequency, f =

the current in the circuit the value of inductor the value of the capacitor C and the value of C (for the same L) required to produce resonance.

1 4p 2f 2L 1 4p 2 ´ (50) 2 ´ 1.0

= 10.1 ´ 10 -5 F = 10.1mF

Parallel Resonant Circuit Figure shows a parallel resonant circuit in which resistor R and inductor L have been connected in series and this combination is connected in parallel with the capacitor C. To this combination, an alternating source of V = V0 sin wt is applied. R

V1 iC

100 Ω

L C

i

200 V 50 Hz

V2 L V3

XL 2pf

VC 415 = = 638.46 W irms 0.65

Sample Problem 34 Resonance frequency of a circuit is f. If the capacitance is made 4 times the initial, value then find the resonance frequency.

204 = 313.85 W 0.65

\

\

Since f is negative, the current in the circuit lags the voltage across the source.

971

V =V0 sin ωt

From the figure, i = iL + iC

972 JEE Main Physics V V V = + Z R + jwL - j/wC

or

Sample Problem 37 A pure inductor of 25 mH is connected to a source of 220 V. Frequency of the source is 50 Hz. The rms current is [NCERT]

V wCV = R + j wL j

(a) 14 A (c) 28 A

V = + j (wC )V R + j wL

Interpret (c) The inductive reactance XL = 2prL = 2 ´ 3.14 ´ 50 ´ 25 ´ 10 -3 W

1 1 = + j wC Z R + j wL

\

= 7.85 W The rms current in the circuit is V 220 V I= = = 28 A XL 7.85 W

1 is known as admittance (Y ). Thus, Z 1 R - j wL + j wC Y = = 2 Z R + w2L2 \

Y =

R2 + (wCR2 + w3L2C - wL ) 2 R2 + w2L2

The admittance will be minimum when wCR2 + w3L2C - wL = 0 or

w=

\

f =

Sample Problem 38 A capacitor of capacitance 240 pF is connected in parallel with a coil having inductance of 1.6 ´ 10 -2H and resistance 20 W. Calculate (i) the resonance frequency and (ii) the circuit impedance at resonance. (a) 3.2 ´ 10 6 Hz, 7.96 ´ 10 4 W

R2 1 - 2 LC L

w 1 = 2p 2p

(b) 7.96 ´ 10 4 Hz, 3.2 ´ 10 6 W (c) 4.5 ´ 10 7 Hz, 3.8 ´ 10 9 W

2

1 R - 2 LC L

is known as resonance frequency. At resonance frequency admittance is minimum of the impedance is maximum.

(d) 3.8 ´ 10 9 Hz, 4.5 ´ 10 7 W

Interpret (b) (i) The resonance frequency of a rejector L-C-R circuit is given by,

Thus, the parallel circuit does not allow this frequency from the source to pass in the circuit. Due to this reason, the circuit with such a frequency is known as rejector circuit. we have, dynamic resistance Zmax =

1 Ymax

(b) 20 A (d) 30 A

f=

1 2p

1 R2 LC L2

=

1 2p

1 (20) 2 (1.6 ´ 10 -2) (250 ´ 10 -12) (1.6 ´ 10 -2) 2

= 7.96 ´ 10 4 Hz

L = CR

(ii) The circuit impedance at resonance is given by Z=

V0 V CR = 0 \ Peak current through the supply = L/CR L

= 3.2 ´ 10 6 W

V0 The peak current through capacitor = = V0wC 1/wC Hence,

Q-factor =

V0wC wL = V0CR/L R

This is basically the measure of current magnification.

Sample Problem 36 A light bulb is rated at 100 W for a 220 V supply. The peak voltage of the source is (a) 100 V (c) 311 V

(b) 110 V (d) 210 V

21.15 Power in an AC Circuit The power is defined as the rate at which work is being done in the circuit. In AC circuit, the current and emf are not necessarily in the same phase, therefore, we write V = V0 sin wt, i = i0 sin(wt + q) The instantaneous power P = Vi = V0 sin wt, i = i0 sin(wt + q)

Interpret (c) The peak voltage of the source is

Pav = Vrmsirms cos q

Vm = 2V Given, \

V = 220 V Vm = 2 ´ 220 = 311V

1.6 ´ 10 -2 L = CR (250 ´ 10 -12)(20)

\

Pav =

V0 i0 cos q 2 2

Electromagnetic Induction and Alternating Current where, cos q =

Resistance (R) is called the power factor of Impedance ( Z )

Case V. When AC circuit contains resistance and inductance both

AC circuit.

Now,

Note If R = 0, cos q = 0 and Pav = 0, i .e ., in resistanceless circuit the

tan q =

and

power loss is zero. Such a circuit is called the wattless circuit and the current flowing is called the wattless current.

wL R

cos q =

R2 + ω2 L2 ωL

R

Different Cases Case I When AC circuit contains ohmic resistance. In this

=

case f = 0,\cos f = 1

= Vrms

R

2 Vrms R 2 (R + wL2 )

1 R2 + ω L – ω C

1

L

R

Then,

tan q =

and

cos q =

Case III When AC circuit contains only inductance p 2

p cos f = cos = 0 2

\

\

Case IV When AC circuit contains resistance and

wL -

1 wC

R R 1 ö æ R2 + ç wL ÷ è wC ø

Pav = Vrms ´ irms ´

capacitance both 1 1/wC = wCR R R cos q = 1 R2 + 2 2 wC

Then,

tan q =

and

R2 +

θ

\

or

=

1 2 2

ωC

Pav = Vrms ´

R

Vrms 1 R + 2 2 wC 2

1 w2C 2 ´

R 1 R + 2 2 wC 2

æ çQ irms è =

2 Vrms R 1 ö æ 2 çR + 2 2 ÷ è wC ø

R 1 ö æ R2 + ç wL ÷ è wC ø

2 Vrms R

1 ö æ R2 + ç wL ÷ è wC ø

2

2

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 (as R . (d) the pointer of the meter is stuck by some mechanical defect.

49. An emf of 15 V is applied in a circuit coil containing 5 H inductance and 10 W, the ratio of currents at time t = ¥ and t = 1 s is 1/2

(a)

e

1/2

e

(b)

-1

(c) 1 - e

-1

e

2

2

e -1

50. A 60 mF capacitor is connected to a 110 V, 60 Hz AC supply. Determine the rms value of the current in the circuit. [NCERT] (b) 2.1 A (d) 3.5 A

51. In step-up transformer, relation between number of turns in primary (N p ) and number of turns in secondary ( N s ) coils is (a) Ns > Np

(b) Np > Ns (c) Ns = Np

(d) N = 2Ns

52. The turns ratio of transformer is given as 2 : 3. If the current passing through the primary coil is 3 A. Find the current through the load resistance.

(b) 20 : 1 (d) 400 : 1

58. A low-loss transformer has 230 V applied to the primary and gives 4.6 V in the secondary. Secondary is connected to a load, which draws 5 A of current. The current (in ampere) in the primary is (a) 0.1 (c) 10

(b) 1.0 (d) 250

59. The armature of a DC motor has resistance of 20W. It draws a current of 1.5 A when run by 220 V of DC. The value of peak emf induced in it will be (a) 150 V (c) 190 V

(d) e -1

(a) 2.5 A (c) 3.1 A

(a) 1 : 20 (c) 1 : 400

(b) 170 V (d) 180 V

60. In an induction coil, the coefficient of mutual inductance is 4H. If current of 5A in the primary coil is cut-off i 1/1500 s, the emf at the terminals of the secondary coil will be (a) 15 kV (c) 10 kV

(b) 60 kV (d) 30 kV

61. In an ideal transformer, the voltage is stepped-down from 11 kV to 220 V. If the primary current be 100 A, the current in the secondary should be (a) 5 kA (c) 0.5 kA

(b) 1 kA (d) 0.1 kA

62. A transformer is used to light 140 W, 24 V lamp from

53. The number of turns in the primary coil of a

240 V AC mains. The current in the mains is 0.7 A. The efficiency of transformer is nearest to

transformer is 200 and the number of turns in secondary coil is 10. If 240 V AC is applied to the primary, the output from secondary will be

63. The reduce the resonant frequency in an L-C-R series

(a) 4.5 A

(a) 48 V

(b) 1.5 A

(b) 24 V

(c) 2 A

(c) 12 V

(d) 1 A

(d) 6 V

54. The primary winding of a transformer has 200 turns and its secondary winding has 50 turns, If the current in the secondary winding is 40 A, the current in the primary is (a) 10 A

(b) 80 A

(c) 160 A

(d) 800 A

55. The number of turns in a secondary coil is twice the number of turns in primary. A leclanche cell of 1.5 V is connected across the primary. The voltage across secondary is (a) 1.5 V (c) 240 V

(b) 3.0 V (d) zero

provide a potential difference of 2400 V. If the primary coil has 75 turns, the number of turns in the secondary coil is (b) 1200

(c) 1500

(b) 80% (d) 60%

circuit with a generator (a) (b) (c) (d)

[NCERT Exemplar]

the generator frequency should be reduced another capacitor should be added in parallel to the first the iron core of the inductor should be removed dielectric in the capacitor should be removed

64. The armature of a shunt wound motor can with stand current up to 8A before it overheats and it damaged. If the armature resistance is 0.5 W, minimum back emf that must be motor is connected to a 120 V line is (a) 120 V (c) 124 V

(b) 116 V (d) 4 V

65. A transformer is having 2100 turns in primary and

56. A step-up transformer is used on a 120 V line to

(a) 150

(a) 90% (c) 70%

(d) 1575

4200 turns in secondary. An AC source of 120 V, 10 A is connected to its primary. The secondary voltage and current are (a) 240 V, 5 A (c) 240 V, 10 A

(b) 120 V, 10 A (d) 120 V, 20 A

Electromagnetic Induction and Alternating Current 66. Which of the following combinations should be selected for better tuning of an L-C-R circuit used for communication ? [NCERT Exemplar] (a) (b) (c) (d)

R = 20 W, L = 1.5 H, C = 35 mF R = 25 W, L = 2.5 H, C = 45 mF R = 15 W, L = 3.5 H, C = 30 mF R = 25 W, L = 1. 5 H, C = 45 mF

develops a back emf of 210 V when the motor is running at fall speed, l the current in the armature is (a) 5 A (c) 120 A

from 220 V mains. If main current is 0.5 A, efficiency of transformer is

designed to operate at 220 V mains. At full speed, it

Round Only One Correct Option 1. The network shown in figure is part of a complete

(a) 90% (c) 96%

1Ω

(a) 20 V (c) 10 V

+

15 V

(b) 95% (d) 99%

(Mixed Bag) to the plane of the ring and the rails. When the speed of the ring is v, the current is the section PQ is

circuit. If a certain instant, the current i is 5 A, and is decreasing at a rate of 103 As -1, then ( VB - V A ) is –

(b) 10 A (d) 110 A

68. A transformer is used to light a 100 W-110 V lamp

67. A motor having an armature of resistance 2 W is

A

P×

×

×Q

×

×

×

×

×

×

B 5 mH

(b) 15 V (d) 5 V

2. A radio can tune over the frequency range of a portion of MW broadcast bond; (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 220 mH,what must be the range of its variable capacitor? [NCERT] [Hint For tuning the natural frequency i. e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radiowave.] (a) 87.8 to 198 pF (c) 63 to 168 pF

(b) 99 to 190 pF (d) 44 to 208 pF

3. A conducting wire frame is placed in a magnetic field, which is directed into the paper, figure. The magnetic field is increasing at a constant rate. The directions of induced currents in wire AB and CD are C A

2Rrv R 8Brv (d) R

(a) zero (c)

(b)

4Rrv R

5. A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the [NCERT] other coil? (a) 30 Wb (c) 23 Wb

(b) 33 Wb (d) 42 Wb

6. An irregular closed loop carrying a current has a shape such that the entire loop cannot lie in a single plane. If this is placed in a uniform magnetic field, the force acting on the loop (a) (b) (c) (d)

must be zero can never be zero may be zero will be zero only for one particular direction of the magnetic field

7. Predict the direction of induced current in the

B D

(a) A to B and C to D (c) A to B and D to C

983

situations described by the following figures r

(b) B to A and C to D (d) B to A and D to C

4. A vertical ring of radius r and resistance R falls vertically. It is in contact with two vertical rail which are joined at the top, see the figure. The rails are without friction and resistance. There is a horizontal uniform magnetic field of magnitude B perpendicular

q

p

S (i)

N

984 JEE Main Physics 9. In an L-R circuit shown in above figure switch S is S p

N x

q

r

y

closed at time t = 0. If e denotes the induced emf across inductor and i, the current in the circuit at any time t, then which of the following graphs, figure shows the variation of e with i? e

z

(ii)

e

(a)

y

(b)

Common axis x

i

i

e

z

e

(c)

(d)

(iii) Tapping key just closed y

i

i

10. Current growth in two L-R circuits (ii) and (iii) is as

Common axis

shown in figure (i). Let L1, L2 , R1 and R2 be the corresponding values in two circuits. Then

x z

(I)

1

(iv) Rheostat setting being changed

2 t

x

y r

(Tapping key just released) (v)

L1

V (ii)

(a) L1 > L2 (c) R1 > R2

R1 S

L2

R2

V (iii) S

(b) L1 < L2 (d) R1 = R2

11. A solenoid has 2000 turns wound over a length of Current (I ) decreasing at a steady rate (vi)

(a) only VI (c) (ii), (iii) and (iv)

(b) (i), (ii) and (iv) (d) All of the above

8. In the circuit shown in figure switch S is closed at time t = 0. The charge which passes through the battery in one time constant is L

EL (a) eR 2 eR 2 E (c) L

R

E

S

(b)

eL ER

æ Lö (d) E ç ÷ è Rø

0.30 m. The area of its cross-section is 1.2 ´ 10-3m 2 . Around its central section, a coil of 300 turns is wound. If an initial current of 2 A in the solenoid is reversed in 0.25 s, then the emf induced in the coil is equal to (a) 6 ´ 10 -4 V (c) 6 ´ 10 -2 V

(b) 4.8 ´ 10 -2 V (d) 48 kV

12. A thin semicircular conducting

×

ring of radius R is falling with × its plane vertical in a horizontal magnetic induction B, figure. At × M the position MNQ, the speed of × the ring is v. The potential difference developed across the ring is (a) zero 1 (b) BvpR2 , and M is at a higher potential 2 (c) pRBv, and Q is at a higher potential (d) 2RBv, and Q is at a higher potential

×

N

×B

×

×

× R

× v ×

× ×

× Q

× ×

985

Electromagnetic Induction and Alternating Current 13. Two parallel wires A1L and B1M placed at a distance

w are connected by a resistor R and placed in a magnetic field B which is perpendicular to the plane containing the wires (see figure). Another wire CD now connects the two wires perpendicularly and made to slide with velocity v through distance l. The power developed is A1

C

B2 l 2 v 2 R B2 w 2 v 2 (d) R

lv R Bwv (c) R

B

(b)

14. Three identical coils A, B and C are placed with their planes parallel to one another. Coils A and C carry current as shown in figure. Coils B and C are fixed in position and coil A is moved towards B. Then, current induced in B is in

B

C

18. The rails of a railway track insulated from each other and the ground are connected to a millivoltmeter. Find the reading of voltmeter, when a train travels with a speed of 180 km/h along the track. Given that the vertical component of earth magnetic field is 0.2 ´ 10-4 Wb/m2 and the rails are separated by 1m (b) 10 -2 V

resistance 1W is moved with a constant velocity v, in a uniform magnetic field of induction B = 2 Wbm -2 , as shown in figure. The magnetic field lines are perpendicular to the plane of the loop (directed into the paper). The loop is connected to network ABCD of resistors each of value 3W. The resistance of the lead wires SB and RD are negligible. The speed of the loop so, to have a steady current of 1 mA in the loop is B ×

×

×

×

×

×

×

×

×

×

×

×

×

×

3Ω

3Ω

S

P

C

A Q

R

(d) 1 V

a time varying current flows as shown in figure. The i ratio of current, 1 at any time t is i2 i1

15. A square metal wire loop PQRS of side 10 cm and

×

(c) 10 -3 V

19. Two inductors L1 and L2 are connected in parallel and

(a) clockwise current (b) anti-clockwise current (c) no current is induced in B (d) current is induced only when both coils move

×

(b) B to A and C to D (d) B to A and D to C

(a) G shows no deflection (b) G shows deflection on one side (c) Deflection of G to the left and right has constant amplitude (d) Deflection of G to the left and right has decreasing amplitude

(a) 10 -4 V

A

D

C

while it oscillates, the magnet moves in and out of the coil C connected to a galvanometer G. Then as the magnet oscillates.

M

D

(a) B

A

17. A magnet is suspended lengthwise from a spring and

B B1

s

(a) A to B and C to D (c) A to B and D to C

v

w

R

through its center. At the instant shown, what are the directions of the induced currents

n

L

l

16. The magnet in figure rotates shown on a pivot

3Ω

3Ω D

(a) 2 ms -1

(b) 2 ´ 10 -2 ms -1

(c) 20 ms -1

(d) 200 ms -1

i

L1 i

i2 L2

(a)

L2 L1

(b)

L1 L2

(c)

L22 ( L1 + L2 ) 2

20. A conducting rod PQ of length L = 1.0 m is moving with a uniform speed v = 2.0 ms -1 in a uniform magnetic field B = 4.0 T directed into the paper. A capacitor of capacity C = 10 mF is connected as shown in figure. Then, (a) (b) (c) (d)

(d)

×

×

×

A

×

× B

×

L21 ( L1 + L2 ) 2 ×

×

P ×

× v

×

×

×

×

Q ×

×

qA = - 80 mC and qB = + 80 mC qA = + 80 mC and qB = - 80 mC qA = 0 = qc charge stored in the capacitor increases exponentially with time

986 JEE Main Physics 21. In the circuit shown in

×

×

×

×

A H K figure, a conducting were HE × × × is moved with a constant × R C V speed v towards left. The × × × complete circuit is placed in a × D B E uniform magnetic field B × × × perpendicular to the plane of × circuit inwards. the current in HKDE is

(a) anti-clockwise (c) alternating

25. When an AC source of emf e = E0 sin(100 t) is

connected across a circuit, the phase difference between emf ( e) and current ( i) in the circuit is p observed to be , as shown in figure. If the circuit 4 consist possibly only of RC or LC in series, find the relationship between the two elements i

(b) clockwise (d) zero

e

φ

22. A square loop of side a placed in the same plane as a a long straight wire V i carrying a current i. The 1 centre of the loop is at a r distance r from the wire, where r >> a1 (see the figure). The loop is moved away from the wire with a constant velocity v. The induced emf in the loop is (a)

m 0 iav 2 pr

(b) m 0 ia3v m ia2 v (d) 0 2 2 pr

m iv (c) 0 2p

23. Some magnetic flux is changed

(a) 4 (c) 2

26. A long solenoid with 15 turns per cm has a small loop

of area 2.0 cm 2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing? [NCERT] (a) 7.5 ´ 106 V

(b) 8.5 ´ 106 V

(c) 7.5 ´ 10 4 V

(d) 7.5 ´ 105 V

4

0.1

t (s)

l = 1.0 m is situated in a uniform magnetic field B = 2 T. Perpendicular to the plane of loop. Resistance of connector is R = 2 W. Two resistance of 6 W and 3W are connected as shown in figure. The external force required to keep the connector moving with a constant velocity v = 2 ms -1 is B 6Ω

(b) 8 (d) 6

24. In an L-R circuit connected to a battery, the rate at which energy is stored in the inductor is plotted against time during the growth of current in the circuit. Which of the following, figure best represents the resulting curve ? dV —– dt

dV —– dt

(a)

(b) time

time

dV —– dt

dV —– dt

(c)

(a) 2 N

time

3Ω

v

(b) 1 N (c) 4 N

28. A copper rod of mass m

(d) 6 N B

slides under gravity on two smooth parallel rails l vT distance apart and set at an angle q to the horizontal. At the bottom, the rails are R θ joined by a resistance R, figure. There is a uniform magnetic field B perpendicular to the plane of the rails. The terminal velocity of the rod is mgR tan q B2 l 2 mgR sin q (c) B2 l 2 (a)

(d) time

(b) R = 1 kW, C = 10 mF (d) R = 1 kW, L = 10 H

27. A rectangular loop with a sliding connector of length

i (A)

from a coil of resistance 10 W. As a result, an induced current is developed in it, which varies with time as shown in figure. The magnitude of change in flux through the coil in weber is

(a) R = 1 kW, C = 5 mF (c) R = 1 kW, C = 1H

mgR cot q B2 l 2 mgR cos q (d) B2 l 2 (b)

Electromagnetic Induction and Alternating Current 29. Two coils A and B have coefficient of mutual

A

inductance M = 2H. The magnetic flux passing through coil A changes by 4 Wb in 10 s due to change in current in B. Then (a) (b) (c) (d)

change in current in B in this time interval is 0.5 A change in current in B in this time interval is 8 A the change in current in B in this time interval is 2 A a change in current of 1 A in coil A will produce a change in flux passing through B by 4 Wb

30. Two concentric and coplanar circular coils have radii a and b as shown in figure. Resistance of the inner coil is R. Current in the other coil is increased from 0 to i, then the total charge circulating the inner coil is

D

3Ω

2Ω R B

C

1 (a) A 242 1 (c) A 55

1 (b) A 220 1 (d) A 440

33. In the circuit shown, the coil has inductance and a

(a)

m o iab 2R

(b)

m o iapb2 2ab

(c)

m o ib 2pR

(d)

m o ia2 2Rb

b

resistance. When X is joined to Y, the time constant is t during growth of current. When the steady state is reached, heat is produced in the coil at a rate P. X is now joined to Z,

Y X

31. The current i in a coil varies with time as shown in the figure. The variation of induced emf with time would be i

0

T/4 T/2 3T/4 T

emf

(a) 0

t

(b) 0

emf

(c) 0

T/2 3T/4 T T/4

t

Z

(a) the total heat produced in the coil is P t 1 (b) the total heat produced in the coil is P t 2 (c) the total heat produced in the coil is 2P t (d) the data given is not sufficient to reach a conclusion

34. A small magnet M is allowed to fall through a fixed horizontal conducting ring R. Let g be the acceleration due to gravity. The acceleration of M will be

emf

T/4 T/2 3T/2 T

T/4 T/2 3T/2 T

t

M

emf

t

(d) 0

987

R

T/4 T/2 3T/2 T

t

32. A rectangular loop with a sliding connector of length 10 cm is situated in uniform magnetic field perpendicular to plane of loop. The magnetic induction is 0.1 T and resistance of connector (R) is 1W. The sides AB and CD have resistances 2 W and 3 W respectively. Find the current in the connector during its motion with constant velocity 1 m/s

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

< g when it is above R and moving towards R > g when it is above R and moving towards R < g when it is below R and moving away from R > g when it is below R and moving away from R

35. In a closed loop, which has some inductance but negligible resistance, uniform but time varying magnetic field is applied directed into the plane of the loop. Variation of field with time is shown. Initially current in the loop was zero. Then

988 JEE Main Physics (a) the magnetic field is constant

B

(b) the magnetic field is in the same plane as the circular coil and it may or may not vary

0

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

2

(c) the magnetic field has a perpendicular (to the plane of the coil component whose magnitude is decreasing suitably

4 t (s)

emf induced in the loop is zero at t = 2 s current in the loop will be maximum at t = 2 s direction of emf in the loop will change at t = 2 s None of the above

More Than One Correct Option 36. An emf is produced in a coil, which is not connected to an external voltage source. This can be due to [NCERT Exemplar]

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

the coil being in a time varying magnetic field the coil moving in a time varying magnetic field the coil moving in a constant magnetic field the coil is stationary in external spatially varying magnetic field, which does not change with time

(d) there is a constant magnetic field in the perpendicular (to the plane of the coil) direction

40. For an LCR circuit, the power transferred from the driving source to the driven oscillator is P = I 2 Z cos f. (a) Here, the power factor cos f ³ 0, P ³ 0 (b) The driving force can give no energy to the oscillator ( P = 0 ) in some cases (c) The driving force cannot syphon out (P < 0) the energy out of oscillator (d) The driving force can take away energy out of the oscillator

Comprehension Based Questions Passage I

37. The conductor ABCDE has the shape shown in figure. It lies in the y-z plane, with A and E on the y - axis. When it moves with a velocity v in a magnetic field B, an emf e is induced between A and E z B a O x

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

e e e e

E

C

A

a λ

D

= 0, if v is in y-direction and B is in the x-direction = 2Bav , if v is in the y-direction and B is in x-direction = Blv , if v is in z-direction and B is in x-direction = Blv , if v is in x-direction

38. An L-C circuit has capacitance C1 = C and inductance C and L2 = 2 L and a 2 third circuit has C3 = 2 C and L3 = L / 2. All the three capacitors are charged to the same potential V, and then made to oscillate. Then L1 = i. A second circuit has C2 =

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

maximum current is greatest in second circuit maximum current is greatest in third circuit maximum current is greatest in first circuit angular frequency of oscillation is same for all the three circuits

39. A circular coil expands radially in a region of magnetic field and no electromotive force is produced [NCERT Exemplar] in the coil. This can be because

A transformer is based on the principle of mutual induction. Input is supplied to primary coil and output is taken across the secondary coil of the np I p E transformer. It is found that s = , when = E p ns Is there is no energy loss. The efficiency of a transformer is given by Poutput EI h= = s s × Pinput EpEp

41. How much current is drawn by the primary coil of a transformer which steps down 220 V to 44 V to operator a device with an impedance of 880 W, (a) 1 A (c) 0.01 A

(b) 0.1 A (d) 0.02 A

42. A 110 volt AC is connected to a transformer of ratio 10. If resistance of secondary is 550 W, current through secondary will be (a) 10 A

(b) 2 A

(c) zero

(d) 55 A

43. A battery of 10 V is connected to primary of a transformer of ratio 20. The output across secondary is (a) 20 V (b) 5 V (c) 10 V (d) zero

Electromagnetic Induction and Alternating Current Passage II The amount of magnetic flux f linked with an area A held in a magnetic field of intensity B is f = B × A. An emf is induced in a coil when amount of magnetic flux - df linked with the coil changes as e = × Minus sign dt indicates that induced emf opposes the change in magnetic flux responsible for its production.

44. A circular coil of diameter 21 cm is held in a magnetic

field of induction 10-4 T. The magnitude of magnetic flux linked with the coil when the plane of the coil makes an angle of 30° with the field is (a) 3. 1 ´ 10 -6 Wb

(b) 1.414 Wb

(c) 1.73 ´ 10 -6 Wb

(d) 14.14 Wb

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

primary winding is not completely linked with secondary secondary winding is not completely linked with primary neither primary nor secondary None of the above

-2

(b) [ML2 T -2 A -1 ]

-2 -1

(d) [M-1 L2 T -1A2 ]

(c) [ML A T ]

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

Column II

Condenser Inductor Energy dissipation is due to A transformer

A. B. C. D.

increases AC reduces AC is conductor for DC resistance only

I-B, II-B, C III-D, IV- A,B I-D, II-C,D, III-B, IV-B,C I-A, II-B,C, III-D, IV-B I-C, II-B, III-D, IV-A 1

2

between two infinitely long current carrying wires in the same direction. Magnitude of currents in both the wires are same. Now match the following two columns.

I. Loop is moved towards right II. Loop is moved towards left III. Wire-1 is moved towards left IV. Wire-2 is moved towards right

II. Current i 2 is decreased

Column II A. Loops will attract each other B. Loops will repel each other C. Current i1 will increase D. Current i 2 will increase

50. Assertion Inductance coil are made of copper.

48. A square loop is symmetrically placed

Column I

Column I I. Current i1 is increased

Directions Questions No. 50 to 58 are Assertion-Reason type. Each of these contains two Statements: Statement I (Assertion), Statement II (Reason). Each of these questions also has four alternative choice, only one of which is correct. You have to select the correct choices from the codes (a), (b), (c) and (d) given ahead (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 correct explanation of the Assertion (c) If Assertion is true but Reason is false (d) If Assertion is false but the Reason is true

47. Match the following Column I to Column II. I. II. III. IV.

are shown in figure currents in them are in the same directions. Now match the following two columns.

Assertion and Reason

Matching Type Column I

49. Two coaxial identical circular current carrying loops

(a) I-A, II-B, III-C,D, IV-D (b) I-A, II-B, III-C, IV-D (c) I-A, II-B,C, III-D, IV-A,D (d) I-C, II-A,C, III-B, IV-C,D

46. The dimension of magnetic flux is (a) [M2 L2 T -2 A]

(a) I-A, II-B, III-A,B, IV-B,D (b) I-A, II-C, III-B, IV-D (c) I-A,B, II-B, III-C, IV-D (d) I-D, II-B,C, III-B,D, IV-D

III. Loop-1 is moved towards loop-2 IV. Loop-2 is moved away from loop-1

45. Magnetic flux produced in the

989

Column II A. Induced current in the loop is clockwise B. Induced current in the loop is anti-clockwise C. Induced current in the loop is zero D. Induced current in the loop is non-zero

Reason Induced current is more in wire having less resistance.

51. Assertion Two identical heaters are connected to two different sources one DC and other AC having same potential difference across their terminals. The heat produced in heater supplied with AC source is greater. Reason The net impedance of an AC source is greater than resistance.

52. Assertion The armature current in DC motor maximum when the motor has just started. E-e , where Reason Armature current is given by i = Ra e = the back emf and Ra = resistance of armature.

990 JEE Main Physics 53. Assertion If a variable frequency AC source is

56. Assertion The energy stored in the inductor

connected to a capacitor, then displacement current in it increases with increase in frequency. Reason The increase in frequency results in an increase of impedance.

of 2 H, when a current of 10 A flows through it is 100 J. Reason Energy stored in an inductor is directly proportional to its inductance.

54. Assertion The mutual inductance of two coils is

57. Assertion In series L-C-R circuit resonance can take

doubled if the self-inductance of the primary or secondary coil is doubled. Reason Mutual inductance is proportional to the self-inductance of primary and secondary coils.

55. Assertion In a series R-L-C circuit the voltage across resistor, inductor and capacitor are 8 V, 16 V and 10 V respectively. The resultant emf the circuit is 10 V. Reason Resultant emf of the circuit is given by the relation E = VR2 + ( VL - VC )2

place. Reason Resonance takes place if inductive and capacitive reactances are equal and opposite.

58. Assertion Making or breaking of current in a coil produces no momentary current in the neighbouring coil of another circuit. Reason At the time of making or breaking of current changes.

Previous Years’ Questions 59. In the given circuit, the AC source has w = 100 rad/s

62. A series of R-C circuit is connected to AC voltage

Considering the inductor and capacitor to be ideal. [IIT JEE 2012] The correct choice (s) is/are

source. Consider two cases : (A) when C is without a dielectric medium and (B) when C is filled with dielectric of constant 4. The current I R through the resistor and voltage Vc across the capacitor are compared in the two cases. Which of the following is is/are true? [IIT JEE 2011]

100 µF 100 Ω

0.5 H

50 Ω

I

(a) IRA > IRB

(b) IRA < IRB

VCA

(d) VCA < VCB

(c) 20 V

>

VCB

63. A rectangular loop has a sliding connector PQ of

(a) the current through the circuit 1 is 0.3 A (b) The current through the circuit I is 0.3 2 A (c) The voltage across 100 W resistor/s = 10 2 V (d) The voltage across 50 W resistor/s = 10 V

length l and resistance RW and it is moving with the speed v as shown. The set-up is placed in the uniform magnetic field going into the plane of the paper. The three currents I1, I2 and I are [AIEEE 2010] p

60. A fully charge capacitor C with initial charge q0 is

connected to a coil of salt inductance L at t = 0. The time at which the equation is stored equally between the electric and magnetic field is [AIEEE 2011] (a) p LC

(b)

(c) 2 p LC

(d)

p 4

RΩ

RΩ

v I

LC i1

LC

61. In AC series circuit, the resistance, inductive reactance and capacitive reactance are 3 W, 10 W and 14 W respectively. The impedance of the circuit is [Orissa JEE 2011]

(a) 5 W (b) 4 W (c) 7 W (d) 10 W

I

Blv Blv ,I = 6R 3R Blv Blv (b) I1 = - I2 = ,I= R R Blv 2Blv (c) I1 = I2 = ,I= 3R 3R Blv (d) I1 = I2 = I = R (a) I1 = I2 =

RΩ i2

Electromagnetic Induction and Alternating Current 64. An AC voltage source of variable angular frequency w

67. An

AC voltage source has an output of DV = (200 V) sin 2pft. This source is connected to a 100 W resistor. Rms current in the resistance is

and fixed amplitude V connected in series with a capacitance C and an electric bulbs of resistance R [IIT JEE 2010] (inductance zero) when w is increased (a) (b) (c) (d)

The bulb glows dimmer The bulb glows brighter Total impedence of the circuit is unchanged Total impedence of the circuit increases

[Kerala CET 2008]

(a) 1.41 A

(d) 0.71 A

[Kerala CET 2008]

(a) 48.8 mH (c) 187.5 mH

1 (Wbm -2 ) in such a way p that its axis makes an angle of 60° with B. The magnetic flux linked with the disc is [Kerala CET 2008]

magnetic field of induction

(d) 0.8 p mV

resistances R1 = 2 W and R2 = 2 W are connected to a battery of emf 12 V as shown in figure. The internal of the battery is negligible. The switch S is closed at t = 0. The potential drop across L as a function of time is

(b) 200 mH (d) 320 mH

69. A circular disc of radius 0.2 m is placed in a uniform

66. An inductor of inductance L = 4400 mH and resistor of

(a) 0.01 Wb (c) 0.06 Wb

(b) 0.02 Wb (d) 0.08 Wb

70. A coil of inductance 300 mH and resistance 2 W is

connected to a source of voltage 2 V. The current reaches half of its steady state value in

[AIEEE 2009]

[Kerala CET 2008]

L

(a) 0.05 s (c) 0.15 s

R1 S

(c) 3.41 A

the self-inductance of similar coil of 800 turns is

magnetic field 0.04 T with its plane perpendicular to the magnetic field. The radius of the loop starts sinking at 2 mms -1. The induced emf in the loop when the radius is 2 cm is [Kerala CET 2009]

E

(b) 2.41 A

68. If the self-inductance of 500 turn coil is 125 mH, then

65. A conducting circular loop is placed in a uniform

(a) 1.6 p mV (b) 3.2 p mV (c) 4.8 p mV

991

R2

(b) 0.1 s (d) 0.3 s

71. The flux linked with a circuit is given by f = t 3 + 3 t - 7. The graph between time (x-axis) and induced emf ( y-axis) will be

12 -3t (a) e V t (b) 6(1 - e - t /0.2 ) V

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

(c) 12e -5 t V (d) 6 e - 5 t V

a straight line through the origin straight line with positive intercept straight line with negative intercept parabola not through the origin

Answers Round I 1. 11. 21. 31. 41. 51. 61.

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

2. 12. 22. 32. 42. 52. 62.

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

3. 13. 23. 33. 43. 53. 63.

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

4. 14. 24. 34. 44. 54. 64.

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

5. 15. 25. 35. 45. 55. 65.

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

6. 16. 26. 36. 46. 56. 66.

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

7. 17. 27. 37. 47. 57. 67.

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

8. 18. 28. 38. 48. 58. 68.

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

9. 19. 29. 39. 49. 59.

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

10. 20. 30. 40. 50. 60.

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

Round II 1. 11. 21. 31. 41. 51. 61. 71.

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

2. 12. 22. 32. 42. 52. 62.

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

3. 13. 23. 33. 43. 53. 63.

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

4. 14. 24. 34. 44. 54. 64.

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

5. 15. 25. 35. 45. 55. 65.

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

6. 16. 26. 36. 46. 56. 66.

(a) (a) (c) (a,b,c) (b) (b) (c)

7. 17. 27. 37. 47. 57. 67.

(a) (d) (a) (a,c,d) (a) (a) (a)

8. 18. 28. 38. 48. 58. 68.

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

9. 19. 29. 39. 49. 59. 69.

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

10. 20. 30. 40. 50. 60. 70.

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

the Guidance Round I f =?

1. On rotating the magnet, no change in magnetic flux is linked with the coil. Therefore, induced emf current is zero.

2. For r ³ a, ò E × dl =

df dB d=A dt dt

f = B × A = B0 (2$i + 3$j + 4k$ ) × L2k$

As,

f = 4 B0L2 Wb

\

7. Using Fleming’s right hand rule, the direction of magnetic induction B in the region P is downward into the paper.

E

8. When the magnet is allowed to fall vertically along the axis of loop with its north pole towards the ring the upper face of the ring will become north pole in an attempt to choose the approaching north pole of the magnet.

a dB 2 dt r=a

r

Therefore, the accleration in the magnet is less than g.

9. Here, l = 50 m, v = 360 kmh-1 = 100 ms-1 Þ

E(2pr) = pa2

dB dt

B = 2 ´ 10 -4 Wbm-2 \Potential difference e = Blv = 2 ´ 10 -4 ´ 50 ´ 100 = 1 V

a2 dB E= 2r dt

\

\ Induced electric field, E µ

10. Here, B = B0( $i + k$ ) Area vector of ABCD = L2k$ Area vector of DEFA = L2 $i

1 r

3. Speed of jet plane v = 1800 km/h = 1800 ´

5 = 500 m/s 18

l = Distance between the ends of the wings = 25 m

Total area vector, A = L2 ( $i + k$ ) Total magnetic flux, f = B × A = B0( $i + k$ ) × L2 = B0L2 (1 + 1) = 2B0L2 Wb

The magnitude of magnetic field B = 5 ´ 10 - 4 T

y

Angle of dip d = 30° Use the formula of motional emf,

C (0,L,0)

e = BVvl (0,L,L) E

e = B sin d vl (where, BV = vertical component of the earth’s magnetic field. \

A (0,0,0)

e = 5 ´ 10 - 4 sin 30° ´ 500 ´ 25 = 3.1 V

4. Here, l = 36 m, v = 400 kmh-1 v=

400 ´ 1000 1000 = ms-1 60 ´ 60 9

B = V = 4 ´ 10 As

^i

BV = B sin d)

Thus, the voltage difference developed between the ends is 3.1 V.

-5

T

e = Blv = 4 ´ 10 -5 ´ 36 ´

z

6. Here, A = L2 k$ and B = B0 (2$i + 3$j 4k$ ) tesla

x

11. Induced current are clockwise. Therefore, induced magnetic field is into the plane of the paper. As it opposes the increasing inducing field, the inducing field must be out of the plane of the paper.

12. Induces charge doesn’t depend upon the speed of magnet.

1000 = 0.16 V 9

induced in the loop is e = BLv.

B (0,0,0)

F (0,0,L)

13. Here, v = 180 kmh-1 =

5. As magnetic flux linked with the loop is changing, emf

C (L,L,0) ^k

180 ´ 1000 = 50 ms-1, 60 ´ 60

l = 1 m, B = 0.2 ´ 10 -4 Wbm-2 As

e = Blv = 0.2 ´ 10 -4 ´ 1 ´ 50 = 10 -3 V = 1 mV

14. Though emf is induced in the copper ring. But there is no induced current because of cut in the ring. Hence, nothing opposes the free fall of the magnet. Therefore, a = g .

Electromagnetic Induction and Alternating Current 15. While moving due north, the truck intercepts vertical component of earth’s field. e = Blv = (90 ´ 10 -6)2.5 ´ 30

As

= 6.75 ´ 10

-23

, V = 6.75 mV

According to Lenz’s law, west end of the axle will be positive.

16. As there is no change in magnetic flux associated with the

17.

circuit, no current is induced in the circuit. The ammeter A shows no deflection. BA( cos 0° - cos 90° ) - df As q = = R R = =

2

Bpr (1 - 0) Bpr = R R

2

. 2 ´ 3143 ´ (10 -1) 2 0.01

= 6.286 C = 6.3 C

18. Coil A must be carrying a constant current in counter clockwise direction. That is why when A moves towards B, current induced in B is in counter clockwise direction, as per Lenz’s law. The current in Bwould stop when A stops moving.

19. q =

qf NA(B2 - B1) Npr 2(B2 - B1) = = R R R =

1000 ´ p ´ 10 -4 ´ (0.012 - 0) (200 + 400)

25. As L =

m 0N 2A l A®

\

\

M=

Number of turns = 20 Resistance of closed-loop = 10 W Angular speed w = 50 rad/s Magnitude of magnetic field B = 3 ´ 10 - 2 T Induced emf produced in the coil e = NBAw sin wt For maximum emf, sin wt = 1 \ Maximum emf e0 = NBAw = 20 ´ 3 ´ 10 - 2 ´ 3.14 (0.08) 2 ´ 50 e0 = 0.603 V e 0.603 Maximum current in the coil I0 = 0 = = 0.0603 A R 10 Average induced emf 1 2p 1 2p eav = ò edt = ò NBAw sin wt dt T 0 T 0 2p

eav =

1 ´ 0.5 ´ 25 ´ 10 -4 cos60° - 0 0.2

e = 3.12 ´ 10 3 V

eav

28. From f = Mi 10 -3 ´ 200 10 M1 f1 = = 0.625 = = -3 M2 f2 0.8 ´ 10 ´ 400 16

due to the resistance of the ring all energy discipates. NBA (cos q2 - cos q1) Dt

29. From e = Ldl / dt , L =

= - 800 ´ 4 ´ 10 -5 ´ 0.05 (cos 90° - cos 0° ) = - 800 ´ 4 ´ 10

-5

´ 0.05 (0 - 1)

= + 800 ´ 4 ´ 10

-5

´ 0.05 V

= 0.016 V

24. Mutual inductance between coils is M = K L1L2 Þ

NBA [cos 2p - cos 0° ] T NBA = [1 - 1] = 0 T

For full cycle average emf, eav = 0 Average power loss due to heating EI 0.603 ´ 0.0603 = 00 = = 0.018 W 2 2

22. Emf is induced in the ring and it opposes the motion. Then, 23. We have, e = -

1 é cos wt ù × NABw êë w úû T 0

=

in A (at t = 0) must be clockwise. Therefore, when coil A is made to rotate about a vertical axis, it would induce current in coil B in clockwise direction, as per Lenz’s law.

=

f 0.4 = = 0.2 H l 2

27. Given, radius of coil = 8 cm = 0.08 m

20. When current in coil B (at t = 0) is counterclockwise, constant

21.

2 ´2 ´ 4 times = 8 times 2

26. \ f = Mi

= 6.3 ´ 10 -6 C = 6.3 mC

df (NBA cos q - 0) e= = dt t

M = 1 2 ´ 10 -3 ´ 8 ´ 10 -3 k = 1 = 4 ´ 10 -3 = 4 mH

993

30. From L = When,

edt 8 ´ 0.05 = = 0.2 H dI 2

m 0N 2A m 0m rN 2A = l l m r = 1000 and N becomes 2

æ 1ö \L becomes 1000 ´ ç ÷ = 10 times è10 ø i.e.,

L = 10 ´ 0.1 = 1H

1 10

994 JEE Main Physics dI (2 - 3) = = - 10 3 As-1 dt 10 -3 dI e = -L dt 5 5 = - L( -10 3), L = 3 H = 5 mH 10

31. L = ?, e = 5 V, As \

41. Length of rod l = 1m Angular frequency of rod w = 400 rad/s Magnetic field B = 0.5 T The linear velocity of fixed end = 0 The linear velocity of other end = lw 0 + lw lw Average linear velocity v= = 2 2

32. The self-inductance L of a solenoid of length l and area of cross section A with fixed number of turns is

Obviously, L increases when l decreases and A increases.

l

m N 2A m 0N 2( pr 2) 33. L = 0 = l l

Fixed

4p ´ 10 -7 ´ (500) 2 ´ p ´ (0.025) 2 1

By using the formula of motional emf, Blw e = Bvl = ×l 2

= 4 ´ 10 ´ 10 7 ´ (500) 2 ´ (0.025) 2 = 6.25 ´ 10 -4 H

34.

[from Eq. (i)]

L2 N22 = L1 N12

e=

35. e =

L2 = L1

N22 æ 500 ö = 1.5 ç ÷ = 375 mH è 100 ø N12

MdI 20 = 0.09 ´ = 300 V dt 0.006

Thus, the emf developed between the centre and ring is 100 V.

42. From

43. Here, v = 50 Hz, Iv = 5 A, I = ? t =

37. As M =

1 R

L 1 = C 10

f m N 2Ai m 0N 2i 38. B = = 0 = A LA L di d 39. As e = - M = M (i0 sin wt) dt dt \

From I = I0 sin w t

2 10 3 = = 25 -5 40 32 ´ 10

m 0N1N2A , therefore, M becomes 4 times. l

e = Mi0 cos wt ( w) emax = Mi0 ´ 1 ´ w

= 5 2 sin 100 p ´

40.

\

L ¢ = L1 + L2 = L + L + 2L

45. As M µ N1N2, therefore M remains the same. 46.

1 1 1 2 L = + = or Lp = Lp L L L 2 where L is inductance of each part = \

10 , dt 30 d= = 10 -3 s 30 ´ 10 3

30 ´ 10 8 = 3 ´

1 3 3 p A = 5 2 sin = 5 2 ´ =5 300 3 2 2

44. Ignoring mutual induction, resultant, inductance

= 0.005 ´ 10 ´ 100 p = 5p di As e = M dt

1 s 300

I0 = 2 Iv = 2 = ´ 5 A

Q-factor of this circuit, Q=

e = L dI /dt dI e 90 = = = 450 As-1 dt L 0.2

36. Given, L = 2 H, C = 32 mF, R = 10 W Resonant angular frequency 1 1 wr = = = 125 rad/s LC 2 ´ 32 ´ 10 -6

0.5 ´ 1 ´ 400 ´ 1 2

e = 100 V

2

\

…(i)

ω

m N 2A L= 0 l

=

(Qv = rw)

Lp =

1.8 ´ 10 -4 = 0.9 ´ 10 -4 H 2 L 0.9 ´ 10 -4 = = 0.45 ´ 10 -4 H 2 2

Resistance of each part, r = 6 / 2 = 3W 1 1 1 2 Now, = + = rp 3 3 3 \

rp = 3 / 2W

Electromagnetic Induction and Alternating Current Time constant of circuit =

Lp rp

=

0.45 ´ 10 -4 3 /2

t = 3 ´ 10 Steady current, i =

-5

48. The voltmeter connected to AC mains is calibrated to read root mean square value or virtual value of AC voltage,

From i = i0(1 \

R - t e L)

-10 æ ´ 1ö ÷ = i0 æç1 - 1 ö÷ = i0 çç1 - e 5 ÷ è e2 ø è ø

i0 e2 = 2 i e -1

50. Given, capacitance of the capacitor C = 60 mF = 60 ´ 10 - 6 F Vrms = 110 V Frequency of AC supply f = 60 Hz Capacitive reactance 1 1 XC = = = 44.23 W 2pfC 2 ´ 3.14 ´ 60 ´ 60 ´ 10 - 6 The rms value of the current in the circuit V Irms = rms XC 110 = = 2.49 A 44.23

51. The number of turns in secondary coil is greater than the number turns in the primary coil.

52. We have, NsIs = NpIp Þ

NS IP = NP IS

Þ

2 3 = 3 Is

Þ

Is =

3 ´3 9 = = 4.5 A 2 2

53. E s =

ns 10 Ep = ´ 240 = 12 V np 200

54. ip =

ns 50 is = ´ 40 = 10 A np 200

55. The voltage across secondary is zero , as transformer does not 56.

58.

work on DC supply. ns E s 2400 = = = 20 np E p 120 ns = 20 np = 20 ´ 75 = 1500

ip is ip is

=

ns np

= 20 : 1

E s ip E 4.6 = Þ s ´ is = ´ 5 = 0.1 A E p is Ep 230 Frequency is not affected by transformer.

59. Input power P1 = 220 ´ 1.5 = 330 W 2

æ3ö Loss of power i 2R = ç ÷ ´ 20 = 45 W è2ø

< v2 > .

49. Here, i = i0 at t = ¥. Let i be the current at t = 1 s

1 20

\

XL + Xg = 0 or XL = - Xg .

i. e. ,

=

As

V 12 = =8 A rp 3 / 2

passive load, total reactance must vanish,

np ns

s

47. For delivering maximum power from the generator to the

i. e. ,

57.

995

Output power, P0 = 330 - 45 = 285 W P 285 \ Peak emf induced, V0 = 0 = = 190 V i 1.5 Ldi 5 60. e = =4´ = 30000 V = 30 kV dt 1 / 1500

61. is =

E pip Es

=

1100 ´ 100 = 500 A = 0.5 kA 220

62. Pi = 240 ´ 0.7 = 168 W , P0 = 140 W h=

P0 140 ´ 100 = ´ 100 » 80% Pi 168

63. Resonant frequency in an L-C-R series circuit is v r =

1 2p LC

The reduce v r ; c can be increased, by adding another capacitor in parallel to the first, E -V 64. From R= i 120 - V 0.5 = 8 V = 116 V n 4200 65. E s = s E p = ´ 120 = 240 V np 2100 is =

ns 2100 ip = ´ 10 = 5 A np 4200

66. The L-C-R circuit used for communication should possess high quality factor (Q-factor) of resonance, which is given by Q=

67.

1 R

L C

To make Q high; R should be low; L should be high and C should be low. E-e As, i= R Þ

68. As, h =

i=

220 - 210 10 = =5A 2 2

output power 100 = = 90.0% input power 220 ´ 0.5

996 JEE Main Physics

Round II 1. Moving from A to B -3

3

VB - VA = [5 ´ 10 ( -10 ) + 15 + 1 ´ 5] = 15 volt

2. Given, minimum frequency f1 = 800 kHz = 8 ´ 105 Hz Inductance L = 200 mH = 200 ´ 10 - 6 H = 2 ´ 10 - 4 H Maximum frequency f2 = 1200 kHz = 12 ´ 10 5 Hz For tuning, the natural frequency is equal to the frequency of oscillations that means it is the case of resonance. 1 Frequency of oscillations f = 2p LC 1 For capacitance C1, f1 = 2p LC1 C1 =

1 1 = 2 2 4p f1 L 4 ´ 3.14 ´ 3.14 ´ (8 ´ 105 ) 2 ´ 2 ´ 10 - 4 = 197.7 ´ 10

- 12

C2 =

Resistance of each half of ring = R / 2 As the two halves are in parallel, therefore, equivalent resistance = R/4 B (2r)v \Current in the section = R/4 8Brv I= R

5. Given, mutual inductance of coil. M = 1.5 H Current change in coil dI = 20 - 0 = 20 A Time taken in change dt = 0.5 s dI df Induced emf in the coil e = M = dt dt df = M × dI = 1.5 ´ 20

or

df = 30 Wb

F

Thus, the change of flux linkage is 30 Wb

= 197.7 pF For capacitance C 2, f2 =

through the two semicircular portions in parallel. Induced emf = B (2r)v.

6. A closed current carrying loop of any irregular shape and

1 2p LC 2

1 1 = 4p 2f22L 4 ´ 3.14 ´ 3.14 ´ (12 ´ 105 ) 2 ´ 2 ´ 10 - 4 = 87.8 ´ 10- 12 F

even not lying in a single plane, placed in a uniform magnetic field shall experience no net force. Therefore, force acting on the loop must be zero.

7. The magnetic field lines due to the current carrying wire are in the plane of the loop. Hence, no induced current is produced in the loop (because no flux lines crosses the area of loop).

= 87.8 pF Thus, the range of capacitor is 87.8 pF to 197.7 pF

3. As the magnetic field directed into the paper is increasing at a constant rate, therefore, induced current should produce a magnetic field directed out of the paper. Thus current in both the loops must be anti-clock wise. As area of loop on right side is more, therefore, induced emf on right side of loop will be more compared to the emf induced on the left side of the loop C A

B D

df dB ù é Qe== -A êë dt dt úû

4. When a ring movies in a magnetic field in a direction perpendicular to its plane, we replace the ring by a diameter (2r) perpendicular to the direction of motion. The emf is induced across this diameter. Current flow in the ring will be

8. In L-R circuit, the growing current at time t is given by i = i0[1 - e- t /t ]

where, i0 =

E L and t = R R

\ Charge passed through the battery in one time constant is t

t

0

0

q = ò idt = ò i0(1 - e- t / t)dt t

é i e- t / t ù -1 q = i0 t - ê 0 ú = i0 t + i0 t[ e - 1] ë -2 / t û 0 i t = i0 t - i0 t + 0 e i0 t (E / R) (L / R) El = = q= e e eR 2

9. In L-R circuit, current at any time t is given by tö E æç E E - t 1 - eL ÷÷ = - e L R çè R R ø R

i=

R

R

…(i)

R

di E - L t æ R ö E - L t = e ç ÷= e èLø L dt R R

Induced emf = L

- t di = Ee L dt

…(ii)

Electromagnetic Induction and Alternating Current

From Eq. (i),

iR = E - Ee

R - t L

17. As a given pole (N or S) of suspended magnet goes into the

Using Eq. (ii), iR = E - e or e = E - iR Therefore, graph between e and i is a straight line with negative slope and positive intercept.

10. As is clear for Fig. (i), steady state current for t = both the circuits is same. Therefore, V V or R1 = R2 = R1 R2

coil and comes about of its, current is induced in the coil in two opposite directions. Therefore, galvanometer deflection goes to left and right both. As amplitude of oscillation of magnet goes on decreasing, so does the amplitude of deflection.

18. Given, dBv = 0.2 ´ 10 -4 Wb/m2 We know that, v = 180 km/h e = Bvv l \ e = 2 ´ 10 -5 ´ 50 ´ 1 = 2 ´ 10 -5 ´ 50 = 1 ´ 10 -3 V = 10 -3 V

Again, from the same figure, we observe that t1 < t 2 L1 L2 < R1 R2

\

11.

19. As the inductors are in parallel, therefore, induced emf across the two inductors is the same i. e. , e1 = e2 æ di1 ö æ di ö L2ç ÷ = L2ç 2 ÷ è dt ø è dt ø

As R1 = R2, therefore, L1 < L2 Mdi æ m 0N1N2A ö di e= =ç ÷ è ø dt dt l =

4p ´ 10

-7

-3

´ 2000 ´ 300 ´ 1.2 ´ 10 ( 4) 0.3 ´ 0.25

= 4.8 ´ 10 -2 V

12. The emf induced in a conductor does not depend on its shape, but only on its end points, M and Q in this case. Thus the conductor is equivalent to an imaginary straight conductor of l = MQ = 2R. Therefore, potential difference developed across the ring = Blv = B (2R) v. And the direction of induced current is from Q to M. Therefore, Q is at higher potential.

13. Induced emf e = Blv = BWv Power developed =

997

e2 B2W 2v 2 = R R

Integrating both sides w.r.t. t , we get L1i1 = L2i2 i1 L2 = \ i2 L1

20. Motional emf across PQ V = Blv = 4(1) (2) = 8 volt This is the potential to which the capacitor is charged. As

q = CV

\

q = (10 ´ 10 -6)8 = 8 ´ 10 -5 C = 80mC

As magnetic force on electron in the conducting rod PQ is towards Q, therefore, A is positively charged and B is negatively charged i. e. ,

q A = + 80m C

and

qB = - 80m C

14. As coil A is moved closer to B, field due to A intercepting B is

21. Potential difference across the capacitor = emf induced across

increasing. Induced current in B must oppose this increase. Hence the current in B must be anti-clockwise.

HE = Blv which is constant. Therefore, charge stored in the capacitor is constant. Hence current in the circuitHKDE is zero.

15. Wheatstone bridge is balanced. Current through AC is zero.

22. Magnetic field intensity at a distance r from the straight wire

Effect resistance R of bridge is 1 1 1 1 = + = , R = 3W R 6 6 3

carrying current is

Total resistance = 1 + 3 = 4W

As area of loop,

Induced emf

and magnetic flux f = BA m iaz \ f= 0 2 pr

e = iR = Blv \

v=

iR 1 ´ 10 -3 ´ 4 = Bl 2 ´ 0.1

= 2 ´ 10

-2

ms

m 0i 2 pr A = a2 B=

The induced emf in the loop is

-1

e=

df d m 0ia2 = dt dt 2pr

e=

m 0ia2 dr m 0ia2v = 2pr 2 dt 2 pr 2

16. In the rotation of magnet, N pole moves closer to coil CD and S pole moves closer to coil AB. As per Lenz’s law, N pole should develop at the end corresponding to C. Induced current flows from C to D. Again S pole should develop at the end corresponding to B. Therefore, induced current in the coil flows from A to B.

where, v =

dr is velocity. dt

998 JEE Main Physics This acts as a cell of emf 4 V and internal resistance 2W, 6W and 3W resistors are in parallel.

23. Here, R = 10 W, As is known,

\

df |dq| = = |i dt| = area under i - t graphs. R df ( 4) (0.1) = = 0.2 df = 0.2R R 2

1 1 1 1+ 2 3 1 = + = = = Rp 6 3 6 6 2

\

Rp = 2 W

= 0.2 ´ 10 = 2 Wb 1 2

24. Energy stored in an inductor L carrying current i is U = Li 2

4V

Rate at which energy is stored dU 1 æ di ö æ di ö = = L2i ç ÷ = Lt ç ÷ è dt ø ø è dt 2 dt

4V

6Ω

3Ω

2Ω

2Ω 2Ω

At t = 0 , i = 0 , \

dU =0 dt

t = ¥ , i = i0 (constant), di \ =0 dt p 25. As the current (i) leads the voltage by , it is R-C circuit. 4 XC Hence, tan f = R p 1 Þ tan = 4 C/wR At

Þ

CwR = 1, w = 100 rad/s 1 1 -1 s CR = = w 100

As Þ

\Current through the connector (i) E 4 = = =1A Rp + r 2 + 2 Magnetic force on the connector = Bil = 2(1) (1) = 2 N Therefore, to keep the connector moving with a constant velocity, a force of 2 N has to be applied to the right side.

28. Terminal velocity of the rod is attained when magnetic force on the rod (Bil) balances the component of weight of the rod (mg sin q), figure. i. e. ,

Thus, from all the given options only (a) is correct answer. Area of small loop A = 2 cm 2 = 2 ´ 10 -4 m 2

Let e be the induced emf,

29. Here, M = 2H, df = 4 Wb, dt = 10 s

According to Faraday’s law, df d (BA) e= = dt dt dB d or e=A =A dt dt

(Q f = BA) (m 0 nI)

(Q Magnetic field inside the solenoid B = m 0 nI) dI or e = Am 0n dt e = 2 ´ 10 -4 ´ 4 ´ 3.14 ´ 10 -7 ´ 1500 ´ 20 (Q m 0 = 4p ´ 10 -7) e = 7.5 ´ 10 6 V Thus, the induced emf in the loop is 7.5 ´ 10 V

f = Mi

As

or Also,

df = M di df 4 di = = =2 A M 2 df = M(di) = 2(1) = 2 Wb m 0i ö 2 ÷ pa . è 2 pb ø

30. Initial flux with inner coil, when i = i is æç æm iö \Change in flux, df = ç 0 ÷ pa2 è 2 pb ø As

6

e = Blv = 2(1)(2) = 4 V

θ

Bl (BlvT ) = mg sin q R mg R sin q vT = B2l 2

dI 4 - 2 2 = = = 20 A/s dt 0.1 0.1

27. Motion emf induced in the connector

R

Bl ( e) = mg sin q R

26. Given, number of turns n = 15 per cm = 1500 per metre

Change in current

Bil = mg sin q æ eö B ç ÷ l = mg sin q èRø

sin θ mg

dq =

df R

\Total charge circulating the inner coil is 2 m a2 æ m i ö pa =ç 0 ÷ = 0 è 2 pb ø R 2Rb

Fm

Electromagnetic Induction and Alternating Current 31. e = -L

di dt

39. When a circular coil expands radially in a region of magnetic field, induced emf, developed is

T During 0 to , 4 So, T T For to , 4 2 For

di = constant dt e = - ve di =0 dt e=0 di = constant dt e = + ve

T 3T to , 2 4

e = B ´ l ´ v = B ´ rate of change of area Here, magnetic field B is in a plane perpendicular to the plane of circular coil. As e = 0, magnetic field must be in the plane of circular coil so that its component perpendicular to plane of coil is zero. Further, if the magnetic field has a component perpendicular to the plane of the coil, whose magnitude is decreasing suitably so that magnetic flux linked with the coil df stays constant then e = = 0. dt

1 V 100 2 ´ 3 11 Net resistance = 1 + = W 2+3 5

32. e = Bvl = (0.1) (0.1) =

\

æ 1 ö ÷ ç 1 A i = ç 100 ÷ = 11 220 ÷ ç è 5 ø p R 1 1 U = Li02 = ( tR) 2 2

40. Power transferred from the driving source to the driven oscillator is P = l 2 Z cos f. Therefore, power factor cos f ³ 0 and P ³ 0. For wattless component the driving force shall give no energy to the oscillator (P = 0, when f = 90°). Further, the driving force cannot syphon out energy out of the oscillator, i. e. ,P cannot be negative.

41. Here, ip = ? E p = 220 V , E s = 44 V, Rs = 880 W is =

33. P = (i0) 2 × R, i. e. ,(i0) 2 =

æP ö 1 ç ÷ = Pt èRø 9

be in such a direction that it attracts the magnet.

\

42. Here,

df æ dB ö = - A × ç ÷ = - A (slope of B-t graph) è dt ø dt

36. An emf is produced in a coil without connecting it to an

37. e = 0 , when conductor moves along its length. In (c) and (d),

38.

conductor moves at right angle to its length and B is perpendicular to that, therefore, e = Blv. 1 Angular frequency, w = × As L1C1 = L2 C 2 = L3C3 , therefore, LC angular frequency of oscillation is same for all the three circuits. From conservation of mechanical energy, 1 2 1 Li2 = CV 2 2 2 C i22 = V 2 L C As V is constant, therefore, i0 µ L C As is maximum for 3rd circuit, therefore, maximum L current (i0) is greatest for 3rd circuit.

E p = 110 V, K =

ns = 10 np

Rs = 550 W, is = ? n E s = E p ´ s = 110 ´ 10 = 1100 V np

At t = 2, slope is zero and it changes its sign. external voltage source only when amount of magnetic flux linked with the coil is changing with time. Choices (a) and (b) are correct. Choice (c) will be correct when the coil is entering or leaving the constant magnetic field partially.

Es 44 1 = = A Rs 880 20

E pip = E sis Ei 44 1 ip = s s = ´ = 0.01 A Ep 220 20

As

34. In both the cases (a) and (c) the induced current in the ring will 35. e = -

999

is =

43. K =

E s 110 = =2 A Rs 550

ns = 2. As a transformer does not work on battery, output np

voltage across secondary is zero.

44. A = pr 2 = p(10.5 ´ 10 -2) 2 m2, B = 10 -4 T q = 90° - 30° = 60° f = BA cos q = 10 -4 ´ p(10.5 ´ 10 -2) 2 cos 60° f = 1.73 ´ 10 -6 Wb

45. Magnetic flux produced in primary winding because of this few magnetic lives of force completely their path in air only, to minimize this loss secondary winding is kept inside the primary winding.

46. The dimension of magnetic flux is [ML2T -2A -1] 50. Since, copper consists of a very small ohmic resistance so, inductance coils are made of copper. A large induced current is produced in such an inductance due to change in flux,d which offers a pretty opposition to the flow of current.

1000 JEE Main Physics 51. For the case of DC, the frequency is zero and the net

58. Before making current in a coil, the current is zero and before

impedance is equal to the resistance. For the case of AC, the impedance of the AC circuit is given by

breaking the current is maximum. In other words, it is constant in both the cases. Obviously, on making or breaking the current in a circuit, the current starts changing. The changing current produces changing magnetic field, which in turn produces induced current in the neighbouring coil of the circuit.

Z = R 2 + w2L2

52.

where, R = resistance, w = angular frequency and L = inductance. E-e From the relation, i= Ra

…(i)

59. We have, XL = wL = 10 ´ 0.5 = 50 W X1 =

When the motor is started e = 0 E Hence, Eq, (i) becomes istart = Ra

1 1 = = 100 W wC 100 ´ 100 ´ 10 -6 100 Ω

It is obvious that current is maximum when the motor has just started.

I1 =

1 1 XC = Þ XC µ f 2pfC

their mutual inductance is given by M = k L1L2

…(i)

It is clear from the relation, if self-inductances of primary and secondary coil are doubled the mutual inductance of the coils will be doubled.

56. The energy stored in the inductor is given by V=

1 2 1 Li0 = ´ 2 ´ (10) 2 = 100 J 2 2

It is obvious that energy stored is directly proportional to its inductance.

57. Resonance in L-C-R series circuit takes place when inductive reactance and capacitive reactance are equal and opposite i. e. , XL = XC or w0L = w0C 1 or w0 = C 1 or f0 = 2p LC In the words we can say that at resonant frequency 1 ö æ ÷ resonance can take place. ç f0 = è 2p LC ø

V

π/4

I2

Phase difference between I1 and V R 100 cos f1 = 1 = , Z1 100 2 f= I1 and V

p 4

50 Ω

50 Ω

I

Z2 = 50 2 , I2 =

E = VR2 + (VL - VC ) 2

E = 10 V

π/4

20 3 ´ = 10 2 2 2

=

55. The resultant emf in the L-C-R circuit is given by E = (8) 2 + (16 - 10) 2 = 64 + 36

20 1 = 100 2 5 2

1 V across 100 W = ´ 100 5 2

Current now increase in the capacitor with decrease of impedance of resistance. This is equal to displacement current between the plates of capacitor.

54. It two coils of inductance L1 and L2 are joined together, then

I1

Z1 = 100 2 ,

53. The impedance of a capacitor is given by

Thus, the impedance of the capacitor decreases with increase in the frequency f of a source.

100 Ω

I

V across

20 2 = 50 2 5 2

2 ´ 50 50 2 20 = = 10 2 2 p p f2 = I2 log V by 4 4

50 W =

I = I1 = I2 Inet = I12 + I22 I= =

4 1 + 25 ´ 2 25 ´ 2 5 1 = = 0.316 50 10

60. In the L-C oscillation, energy in transferred C to L or L to C, the maximum energy in L is = maximum energy in C is

1 2 LImax 2

2 q max 2C

Electromagnetic Induction and Alternating Current Equal equation will be, when 1 2 1 1 2 LI = × LImax 2 2 2 1 I= Imax 2

65. f = BA cos q = B( pr 2) cos 0° | e| =

1 Imax 2 p 2p p or wt = t= 4 T 4 T 1 p t= \ t = 2p LC = LC 8 8 4

61. Given, resistance R = 3 W

= 3.2p ´ 10 -6 V = 3.2pmV

66. In the R-L circuit, growth of current in given by I=

VL = L

XL = 10 W Capacitive reactantce, XC = 14 W The impedence of the series L-C-R circuit is

VL

Z = R 2 + ( XC - XL) 2 = (3) 2 + (14 - 10) 2 Z =5 W 1 ( wC) 2 4C

C

R

-R2t ù E é L ú 1 e ê R2 ê úû ë

Potential drop across L is

Inductive reactance,

62. As Z = R 2 +

df dr = Bp (2r) dt dt

= 0.04 ´ p(2 ´ 2 ´ 10 -2) ´ 2 ´ 10 -3

I = Imax ´ sin wt =

or

1001

67. I0 =

R2t R ö L æ ç- 2÷

æ -E ö dI = Lç ÷e dt è R2 ø

RT - 2 = Ee L

è

Lø

= 12e-2t /0.4 = 12e-5t V

V0 200 = =2 A R 100 2 I = 1.14 A Irms = 0 = 2 2

68. As L µ n2 \

L2 n22 = L1 n12 2

2

æn ö æ 800 ö L2 = ç 2 ÷ ´ L1 = ç ÷ ´ 125 mH = 320 mH è 500 ø è n1 ø Case A

Case B

V Z V B In = Z¢ VRA < VRB IRA =

Z¢ < Z

=

I1

I2

e 2vl B = 3R /2 3R R

I

R = 2W, E = 2 V E 2 I0 = = = 1 A R 2 Rt æ - ö I = I0 çç1 - e L ÷÷ ø è

From

R

-2 ö -2 1 æ t t= 1 I0 = I0 çç1 - e 0.3 ÷÷ = e 0.3 2 2 è ø

R

e = vlB

I1 = I2 =

64. As, Irms =

1 ( p ´ 0.2 ´ 0.2) cos 60° = 0.02 Wb p

70. Here, L = 300 mH = 300 ´ 10 -3 H [Q VR2 + VC2 = V02]

63. The circuit can be reduced as I=

1 Wbm-2, q = 60° , f = ? p

f = BA cos q ( pr 2) cos q

IRA < IRB

VCA > VCB

\

69. Here, r = 0.2 m, B =

1 vl B = 2 3R

Vrms 1 R + 2 2 wC 2

When wincreases, Irms increases, so the bulb glows brighter.

2t log e e = log e 1 - log e 2 = 0 - 0.6931 0.3 0.6931 ´ 0.3 t= = 0.1 s 2

71. Here, f = t 3 + 3t - 7 As, e =

dx -df = - (t 3 + 3t - 7) = - (3t 2 + 3) Þ -3 (t 2 + 1) dt dt

Clearly, the graph between e and t (along x-axis) will be a parabola. At t = 0 , e = - 3 ¹ 0 . \The parabola would not pass through the origin.

Electromagnetic 22 Waves JEE Main MILESTONE Electromagnetic Waves Characteristics of EM Waves Transverse Nature of EM Waves Electromagnetic Spectrum

Various Electromagnetic Radiation Applications of EM Waves

22.1 Electromagnetic Waves The velocity of electromagnetic waves in free space is given by E 1 c= 0 = B0 m 0e 0 where, m (= 1. 257 ´ 10-6 TmA -1 ) and e 0 (= 8.854 ´ 10-12 C2N-1m-2 ) are respectively the absolute permeability and absolute permittivity of free space. The velocity of electromagnetic waves in free space (vacuum) is equal to velocity of light in vacuum, i.e., 3 ´ 108 ms-1.

Conduction Current It is a current in the electric circuit which arises due to the flow of electrons in the connecting wires of the circuit, in a definite closed path. When a capacitor is connected to the battery, it starts storing the charge, due to conduction current. When the capacitor gets fully charged, the conduction current becomes zero in the circuit. Conduction current exists even if the flow of electrons is at uniform rate.

Displacement Current It is that current which comes into play in the region, whenever the electric field and hence, the electric flux is changing with it. 1. Displacement current was predicted by Maxwell. 2. The displacement current is given by the relation. iD = e0

df E dt

where, e0 = absolute permittivity of space, df E / dt = rate of change of electric flux. 3. In case of a steady electric flux linked with a region, the displacement current is zero.

In 1865, Maxwell predicted the existence of electromagnetic waves on the basis of his equations. According to him, an accelerated charge produces a sinusoidal time varying magnetic field, which in turn produces a sinusoidal time varying electric field. The two fields so produced are mutually perpendicular and are sources to each other. The mutually perpendicular time varying electric and magnetic fields constitute electromagnetic waves which can propagate through empty space.

Electromagnetic Waves Sample Problem 1 What is an instantaneous displacement current of 1.0 A current in the space between the parallel plates of mF capacitor? 5

(a) 10 Vs

-1

6

(b) 10 Vs

(c) 10 -6 Vs-1

(iii) This law also predicts that the isolated magnetic monopole does not exist. Mathematically, ò B × ds = 0 s

-1

(d) 10 7 Vs-1

3. Faraday’s law of electromagnetic induction (i) This law gives a relation between electric field and a changing magnetic flux.

df d V Interpret (b) As, id = e 0 E = e 0A æç ö÷ dt dt è d ø dV e A dV id = 0 ´ =C d dt dt . dV id 10 6 = = = 10 Vs-1 dt C 10 -6

or or

Thus, an instantaneous displacement current of 1.0 A can be set up by changing the potential difference across the parallel plates of capacitor at the rate of10 6 Vs-1.

(ii) This law tells that the changing magnetic field is the source of electric field. Mathematically, ò E × dl = -

(a) 2.4 A (c) 4.4 A

(b) 3.4 A (d) 5.2 A

(i) This law states that the magnetic field can be produced by a conduction current as well as by displacement current. (ii) This law also states that the conduction current and displacement current together have a property of continuity. (iii) At an instant, in a circuit, the conduction current is equal to displacement current. Mathematically,

Interpret (c) We have,

æ

ò E × dl = m0 çè ic + e0

Displacement current = Conduction current V 220 \ id = = = 4.4 A XC 50

Maxwell’s Equations Maxwell in 1862, gave the basic laws of electricity and magnetism in the form of four fundamental equations, which are known as Maxwell’s equations. In the absence of any dielectric and magnetic material, the Maxwell’s equations are based on experimental observations followed by all electromagnetic phenomena, may be stated in the integral form as below

dfB dt

4. Ampere-Maxwell's law

Sample Problem 2 In a electric circuit, a capacitor of reactance 50 W is connected across the source of 220 V. Find the displacement current.

df E ö ÷ dt ø

These equations are collectively called Maxwell’s equations.

Sample Problem 3 Electromagnetic waves travel in a medium with a speed of 2 ´ 10 8 ms-1. The relative permeability of the medium is 1. The relative permittivity is (a) 2.45 (c) 1.25

(b) 2.25 (d) 3.25

Interpret (b) Given, v = 2 ´ 10 8 ms-1 and m r = 1 The speed of EM waves in a medium is given by v =

1. Gauss’s law for electrostatics (i) This law gives the total electric flux in terms of charge enclosed by the closed surface. (ii) This law states that the electric lines of force start from positive charge and end at negative charge i.e., the electric lines of force do not form a continuous closed path. q Mathematically, ò E × ds = s e0 2. Gauss’s law for magnetism (i) This law shows that the number of magnetic lines of force entering a closed surface is equal to number of magnetic lines of force leaving that closed surface. (ii) This law tells that the magnetic lines of force form a continuous closed path.

1003

1 me

...(i)

where m and e are absolute permeability and absolute permittivity of the medium. Now,

m = m0 mr

and

e = e0 er

\ Eq. (i) becomes v=

1 1 1 = ´ m 0 m r e0 er m 0 e0 m r er

v=

c m r er

er = =

c2 v 2m r (3 ´ 10 8) 2 = 2.25 (2 ´ 10 8) 2 ´ 1

1004 JEE Main Physics Speed of Electromagnetic Waves A changing magnetic field produces a changing electric field (third concept of Maxwell) and a changing electric field produces a changing magnetic field as predicted by Maxwell. These induced electric and magnetic fields are in phase i.e., at any point the two fields reach their maximum at the same time (as shown in figure). It may be noted that the two fields are perpendicular to each other and are perpendicular to the direction of propagation i.e., electromagnetic waves are transverse in nature. Y E

3. The rate of flow of energy crossing a unit area in an electromagnetic wave is described by the vector S called the, Poynting vector, which is defined by the expression,

S=

1 E´B m0

Its magnitude S is related to the rate at which energy is transported by a wave across a unit area at any instant. Since, E and B are mutually perpendicular, hence

| E ´ B | = EB Thus,

S=

EB m0

The SI unit of poynting vector S is Js-1m-2 or Wm -2. v X

H Z

The speed of EM waves in vacuum is c =

1 . m 0 e0

Note The direction of Poynting vector S of an electromagnetic wave at any point gives the wave’s direction of travel and the direction of energy transport at that point. Further B = E/c , therefore, S can also be written as, S =

In isotropic medium, their speed

v=

1 1 = me m 0 m r e 0e r

v=

1 1 c × = m rer m 0e 0 n

Here, n = m r e r = refractive index of medium

E2 EB = m0c m0

This relation shows that the value of electric vector at any instant in the EM wave is about 377 times, the value of magnetic vector. It is because of this reason that while discussing the behaviour of light as EM wave, we prefer the use of electric vector. Average of poynting vector is given by I = S av =

cB02 E 0 B0 1 = e0 E 02c = 2m 0 2 2m 0

m r = relative permeability of medium e r = relative permittivity of medium

22.2 Characteristics of EM Waves 1. The electromagnetic waves are transverse in nature whose speed is same as that speed of light. 2. The two fields E and B have the same frequency of oscillation and they are in phase with each other. In keeping with these features in mind, we can assume that if the electromagnetic wave is travelling along positive direction along x-axis, the electric field is oscillating parallel to the y-axis and that the magnetic field is then oscillating parallel to the z-axis. Then, we can write the electric and magnetic fields as sinusoidal functions of position x and time t.

Ey = E0 sin(kx - wt ) Bz = B0 sin(kx - wt ) In this, E0 and B0 are the amplitudes of the fields. E 1 Further, c = 0 = = speed of light in vacuum. B0 e0 m 0

22.3 Transverse Nature of Eelctromagnetic Waves According to Maxwell electromagnetic waves conerst of time varying electric and magnetic fields, which are perpendicular to each other, as well as direction of wave propagation. y Let us consider a plane of EM wave travelling in the W x-direction and a V rectangular parallelopiped OPQRSUVW placed with its O edges parallel to the three axes. The electric and U magnetic fields z sinusoidally with x and t only are independent of y and z.

Q R

P

x

S

If the rectangular parallelopiped does not enclosed any charge, then according to Gauss’s law, the total electric flux across it must be zero i.e., ò E × ds = 0.

Electromagnetic Waves

ò

or

ò

E × ds +

PQRS

+

OWVU

ò

E × ds +

OUSP

ò

E × ds +

Sample Problem 5 If the earth receives 2 cal min -1 cm -2

E × ds

QWVR

ò

ò

E × ds +

SRVU

E × ds = 0

...(i)

ò

E × ds +

ò

E × ds +

QWVR

and

ò

E × ds = 0

...(ii)

ò

E × ds = 0

...(iii)

OPQW

From Eqs. (i), (ii) and (iii), we get E × ds +

PQRS

Interpret (a) From Poynting vector, we know that S=E´H = EH sin 90° = EH

ò

E × ds = 0

...(iv)

OWVU

E × ds = E xs and

PQRS

S = 2 cal min -1 cm–2

Given,

2 ´ 418 . ´ 10 4 -2 -1 Jm s 60 As S represents energy flux per unit area per second, we have =

When E x and E x ¢ are the x-component of E on faces PQRS and OWVU and s is the area of each face, then

ò

(b) 377 (d) 200

OUSP

SRVU

ò

solar energy. Then, the amplitudes of electric field of radiation is (V/m) (a) 102.3 (c) 150

OPQW

Since electric field E does not depend on y and z. \

1005

ò

EH =

E × ds = - E x ¢ s

EH =

OWVU

From Eq. (iv), we get

Þ

Ex s - Ex¢ s = 0

EH ´

2 ´ 418 . ´ 10 4 = 1400 60 m0 = 377 e0

E = 1400 ´ 377 H

i. e. ,

Ex = Ex¢

\

or

Ex = Ex¢ = 0

Amplitude of electric field is

The possibility, E x = E x ¢ predicts that electric field (E) remains the same at different points on x-axis i. e. , field is static. But static field cannot propagate a wave of finite wavelength then, E x = E x ¢ = 0 i. e. , no component of E is parallel to the direction of wave propagation. In other words, the electric field is perpendicular to the direcction of propagation of EM wave. By preceeding the same manner holds time to the magnetic field i.e., the magnetic field also perpendicular to direction of wave propagation since both the electric and magnetic fields are perpendicular to the direction of wave propagation. Therefore, EM waves are transverse in nature.

E = 1400 ´ 377 = 726.5 V /m E 0 = E 2 = 102.3 V /m

Energy Density of Electromagnetic Waves Consider a plane electromagnetic wave propagating along X-axis. The electric and magnetic fields in a plane electromagnetic wave can be given by E = E0 sin(kx - wt ) and B = B0 sin (kx - wt ). In any small volume dV, the energy of the electric field is 1 U E = e 0E 2dV and the energy of the magnetic field in 2 1 2 volume dV is U B = B dV 2m 0 (a) The average energy density of electric field is,

Sample

Problem

4 If

absolute permittivity and permeability of free space are e 0 = 8.85 ´ 10 -12 C 2N -1 m -2 and m 0 = 4p ´ 10 -7 Ns2C -2. Then, the velocity of electromagnetic waves is (a) 3 ´ 10 8 ms-1

(b) 3 ´ 10 5 ms-1

(c) 3.3 ´ 10 5 ms-1

(d) Zero

Interpret (a) Given, e 0 = 8.85 ´ 10 and We have,

m 0 = 4p ´ 10 c= =

-7

-12 2 -1 -2

CN m

-2 2

NC s

1 m0 e0

1 1 1 e 0E 2 = e 0 (E0 / 2 ) 2 = e 0E02 2 2 4

(b) The average energy density of magnetic field is,

uB =

B2 B2 (B / 2 ) 2 = 0 = 0 2m 0 2m 0 4m 0

(c) uE = uB, using the relation, 1 E = c= m 0 e0 B (d) Total average energy density i = uE + uB = 2uE = 2uB =

1 4p ´ 10 -7 ´ 8.85 ´ 10 -12

= 3 ´ 10 8 ms-1

uE =

1 1 B02 e0 E02 = 2 2 m0

The units of uE and uB are Jm-3.

1006 JEE Main Physics (e) Intensity

Sample Problem 7 A plane electromagnetic wave of

The energy crossing per unit area per unit time, perpendicular to the direction of propagation of EM wave is called intensity. total EM energy i.e., Intensity, I = surface area ´ time = uav

1 1 B02c W ´ c = e0 E02c = 2 2 m 0 m2

Radiation Pressure If a portion of electromagnetic wave of energy U is propagating with speed c, then linear momentum of U electromagnetic wave is given by p = . c The pressure exerted on the surface is defined as force per unit area F/A. Thus, Pressure, p =

1 æ dp ö 1 d æ U ö 1 (dU / dt ) F = ç ÷= ç ÷= A A è dt ø A dt è c ø c A

(dU / dt ) is the rate at which energy is A arriving at the surface per unit area, which is the magnitude of the poynting vector.

In this expression,

Thus the radiation pressure p exerted on the perfectly S absorbing surface, p = c If the surface is perfect reflector and incidence is normal, then, the momentum transported to the surface in time t is 2U 2S and the radiation pressure will be, p = . given by, p = c c

frequency 25 MHz travels in free space along the x-direction. At a particular point in space and time, E = 6.3j Vm -1, the value of [NCERT] magnetic field B at that point is (in Tesla) (a) 6.3 kT

(b) 6.3 ´ 10 –8 kT

(c) 2.1 ´ 10 –8 kT

(d) 2.1kT

Interpret (c) The magnitudes of the electric and magnetic fields in the electromagnetic wave are related as E B= c where, c is speed of light E = 6.3 V /m

Given,

c = 3 ´ 10 8 m/s B=

\

To find the direction, we note that E is along y-direction and the wave propagates along x-axis. therefore B should be in a direction perpendicular to both x and y-axes. Using vector algebra, E ´ B should be along x-direction. Since, ( + j ) ´ ( + k) = i, B is along the z-direction. Thus, B = 2.1 ´ 10 –8 kT

Sample Problem 8 The magnetic field in a plane electromagnetic wave is given by By = 2 ´ 10 -7 sin (0.5 ´ 103 x + 1.5 ´ 1011t)T. The frequency of the wave is (in GHz)

(b) 4.5 ´ 10 -12 J

(c) 5 ´ 10 -13 J

(d) 3.5 ´ 10 -10 J

Interpret (a)

The average value (energy/volume) is given by, 1 uav = e 0E 02 2

of

é æ x t öù By = B0 sin ê2p ç + ÷ ú è l T øû ë l=

We get,

density

Total volume of the cylinder V = A l \Total energy contained in the cylinder, æ1 ö U = (uav ) (V ) = ç e 0E 02 ÷ ( Al) è2 ø Substituting the values, we have 1 U = ´ (8.86 ´ 10 -12)(50) 2(10 ´ 10 -4)(50 ´ 10 -2)J 2 = 5.5 ´ 10 -12J

2p = 1.26 cm 0.5 ´ 10 3

1 = n = (1.5 ´ 10 11)/2p = 23.9 GHz T

and

energy

(b) 2.39 GHz (d) 23.9 GHz

Interpret (d) Comparing the given equation with

wave is given by, E = (50 N -1) sin w (t - x / c ). Find the energy

(a) 5.5 ´ 10 -12 J

[NCERT]

(a) 1.26 GHz (c) 12.6 GHz

Sample Problem 6 The electric field of an electromagnetic

contained in a cylinder of cross-section 10 cm 2 and length 50 cm along the x-axis.

6.3 = 2.1 ´ 10 –8 T 3 ´ 10 8

Sample Problem 9 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 cm 2, then the average force exerted on the surface during a 30 minute time span is [NCERT] (a) 1.2 ´ 10 -6 N (b) 2.16 ´ 10 -3 N (c) 2.16 ´ 10 -6 N (d) 1.2 ´ 10 -3 N

Interpret (a) The total energy falling on the surface is U = (18 W / cm2) ´ (20 cm2) ´ (30 ´ 60)

U = 6.48 ´ 10 5 J

Electromagnetic Waves Therefore, the total momentum delivered (for complete absorption) is U 6.48 ´ 10 5 = 2.16 ´ 10 –3 kg - m/s p= = c 3 ´ 10 8

Sample Problem 12 In a plane electromagnetic wave propagating in space has an electric field of amplitude 9 ´ 103Vm -1, then the amplitude of the magnetic field is (a) 2.7 ´ 10 12T

The average force exerted on the surface is p 2.16 ´ 10 -3 = 1.2 ´ 10 –6 N F= = t 0.18 ´ 10 4

(b) 9.0 ´ 10 –3T (c) 3.0 ´ 10 –4T (d) 3.0 ´ 10 –5T

Sample Problem 10 The average energy density of an -1

electromagnetic wave given by E = (50NC ) sin ( wt - kx) will be nearly (c) 10 -6 Jm-3

(d) 10 -5 Jm-3

1 1 e 0E 02 = ´ (8.85 ´ 10 –12) ´ (50) 2 » 10 -8 Jm-3 2 2

Sample Problem 11 The rms value of the electric field of the light coming from the sun is 720 NC -1. The average total energy density of the electromagnetic wave is (b) 6.37 × 10 -9 Jm -3 (d) 3.3× 10 -3 Jm -3

In the order of increasing wavelength, these waves are (i) Gamma rays, (ii) X-rays, (iii) Ultraviolet rays, (iv) Visible light, (v) Infrared waves, (vi) Microwaves and (vii) Radio waves The figure illustrates the general spectrum of the electromagnetic radiations, in which the wavelength is expressed in metre.

Red

700

Wavelength (nm) 600 500

400

Violet

= 8.85 ´ 10 –12 ´ (720) 2 = 4.58 ´ 10 –6 Jm-3

Indigo

2 Interpret (a)Total average energy = e0Erms

Blue

(a) 4.58 × 10 -6 Jm -3 (c) 81.35 × 10 -12 Jm -3

The array obtained on arranging all the electromagnetic waves in an order on the basis of their wavelength is called the electromagnetic spectrum.

Green

uav =

E 0 9 ´ 10 3 = 3 ´ 10 -5 T = c 3 ´ 10 8

22.4 Electromagnetic Spectrum

Interpret (a) Average energy density of electromagnetic wave

Yellow

(b) 10 -7 Jm-3

Interpret (d) Magnetic field, B0 =

Orange

(a) 10 -8 Jm-3

1007

Visible spectrum Wavelength (m) 108

107

106

105

104

Long waves 10

102

103 104 105

103

102

10

1

10–1 10–2 10–3 10–4 10–5 10–6 10–7 10–8 10–9 10–10 10–11 10–12 10–13 10–14 10–15 10–16

Radio waves 106 107

108 109

Infrared

Ultraviolet

X-rays

Gamma rays

1010 1011 1012 1013 1014 1015 1016 1017 1018 1019 1020 1021 1022 1023 1024 1025 Frequency (Hz)

22.5 Various Electromagnetic Radiations Gamma rays They were discovered by Becquerel and Curie in 1896. Their wavelength is of the order of 10-14 to 10-10m. The main sources are the natural and artificial radioactive substances. These rays affect the photographic plate. These rays are mainly used in the treatment of cancer disease.

X-rays They were discovered by Roentgen in 1895. Their wavelength is of the order of 10-12 m to 10-8m. X-rays are produced when highly energetic cathode rays are stopped

by a metal target of high melting point. They affect the photographic plate and can penetrate through the transparent materials. They are mainly used in detecting the fracture of bones, hidden bullet, needle, costly material, etc., inside the body and also used in the study of crystal structure.

Ultraviolet rays They were discovered by Ritter in 1801. Their wavelength is of the order 10-9 m to 4 ´ 10-7m . In the radiations received from sun, major part is that of the ultraviolet radiation. Its other sources are the electric discharge tube, carbon arc etc. These radiations are

1008 JEE Main Physics mainly used in excitation of photoelectric effect and to kill the bacteria of many diseases.

2. Infrared radiations are used (i) in green houses to keep the plants warm (ii) in revealing the secret writings on the ancient walls

Visible light This was first studied in 1666 by Newton. The radiations in the range of wavelength from 4 ´ 10-7m to

(iii) for looking through haze, fog and mist during war time, as these radiations can pass through them.

7 ´ 10-7m fall in the visible region. The wavelength of the light of violet colour is the shortest and that of red colour is the longest. Visible light is obtained from the glowing bodies, while they are white hot. The light obtained from the electric bulbs, sodium lamp, fluorescent tube is the visible light.

4. Ultraviolet radiations are used (i) in preserving the food stuffs. (ii) in the detection of invisible writing, forged documents, finger prints in forensic laboratory. (iii) Ultraviolet radiations are also used for knowing the structure of the molecules and arrangement of electrons in the external shells.

Thermal or Infrared waves They were discovered by Herchell in 1800. Their wavelength is of the order of 7 ´ 10-7m to 10-3 m. A body on being heated, emits out the

5. X-rays have many applications These rays provide us valuable information

infrared waves. These radiations have the maximum heating effect. The glass absorbs these radiations, therefore for the study of these radiations rock salt prism is used instead of a glass prism. These waves are mainly used for therapeutic purpose by the doctors because of their heating effect.

(i) about the structure of atomic nuclei (ii) in the study of crystal structure (iii) in the fracture of bones etc. 6. g-rays were used (i) in treatment of cancer and tumours

Microwaves They were discovered by Hertz in 1888. Their

(ii) to produce nuclear reactions.

wavelength is in the range of nearly 10-4 m to 1m. These waves are produced by the spark discharge or magnetron valve. They are detected by the crystal or semiconductor detector. These waves are used mainly in radar and long distance communication.

Check Point 1 1. Why are microwaves used in RADAR? 2. Is displacement current like conduction current, a source of magnetic field?

Radiowaves They were first discovered in 1895 by Marconi. Their wavelength is in the range of 0.1m to 10 m. They can

3. Radiowaves and gamma rays both are transverse in nature and

be obtained by the flow of high frequency alternating current in an electric conductor. These waves are detected by the tank circuit in a radio receiver or transmitter.

4. The wavelength of electromagnetic radiation is doubled. What

5

electromagnetic in character and have the same speed in vacuum. In what respect are they different? will happen to the energy of photon?

22.6 Applications of Electromagnetic Waves

5. Radiowaves diffract pronouncedly around buildings; while light waves which are also electromagnetic waves, do not. Why?

1. Radio and microwave radiations are used in radio and TV communication system. Microwave radiations are mainly used in radar and TV communication. S.N.

Name

Frequency range (Hz)

Wavelength range (m) Electromagnetic Spectrum -14

-10

Production

1.

Gamma (g ) rays

5 ´ 10

2.

X-rays

3 ´ 10 to 1 ´ 10

3.

Ultraviolet rays (UV)

8 ´ 10 to 8 ´ 10

4.

Visible light

4 ´ 10 to 8 ´ 10

4 ´ 10

5.

3 ´ 1011 to 4 ´ 1014

8 ´ 10-9 to 3 ´ 10-3

Excitation of atoms and molecules.

6.

Thermal or Infrared rays (IR) Microwaves

3 ´ 108 to 3 ´ 1011

10-3 to 1

Klystron value or magnetron valve

7.

Radiowaves

3 ´ 10 to 3 ´ 10

22

to 5 ´ 10

18

21

16

14

16

14

3

14

11

0.6 ´ 10 -13

10

-8

to 3 ´ 10 -9

-7

10

to 4 ´ 10

-7

to 8 ´ 10

to 10

5

Nuclear Origin Bombardment of high Z target by electrons

-7

4 ´ 10

-3

to 10

Excitation of atoms and spark. Excitation of atoms, spark and arc flame.

Oscillating circuits

Start Practice for

JEE Main Round I

(Topically Divided Problems)

Displacement Current and Maxwell’s Equations 1. Consider the following two statements regarding a linearly polarized plane electromagnetic wave. (i) Electric field and the magnetic field have equal average values. (ii) Electric energy and the magnetic energy have equal average values. (a) (i) is true (c) Both are true

(b) (ii) is true (d) Both are false

2. 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 [NCERT Exampler] lies in (a) visible (c) ultraviolet region

(b) infrared region (d) microwave region

3. The Maxwell’s four equations are written as (i) ò E × ds = q/ e 0

òS B × ds = 0 d (iii) ò E × dl = B × ds S dt òs (ii)

d E × ds dt òs The equation which have sources of E and B are (b) (i), (ii) (d) (i) and (iv) only

4. The magnetic field between the plates of a capacitor where r > R is given by (where r is the distance from the axis of plates and R is the radius of each plate of capacitor) m 0 iD 2pR2 m i (c) 0 D 2pr (a)

point between the capacitor plates indicates in figure express B in terms of the rate of change of the electric field strength i. e., dE / dt between the plates (a)

(b)

m 0 iD 2pR

(d) zero

m 0i 2pr

(c) zero

e 0m 0 r dE / dt 2 m i (d) 0 2p (b)

6. A linearly polarized electromagnetic wave given as E r = - E0 $i cos ( kz - wt ) is incident normally on a perfectly reflecting infinite wall at z = a. Assuming that the material of the wall is optically inactive, the [NCERT Exampler] reflected wave will be given as (a) Er = - E0 $i cos(kz + wt ) (b) Er = E0 $i cos(kz - wt ) (c) Er = - E0 $i cos(kz - wt ) (d) Er = - E0 $i sin (kz - wt )

7. The charge of a parallel plate capacitor is varying as q = q0 sin 2pft. The plates are very large and close together (area = A, separation = d). Neglecting edge effects, the displacement current through the capacitor is (a)

(iv) ò B × dl = m 0 I + m 0 e 0 (a) (i), (ii), (iii) (c) (i) and (iii) only

5. An expression for the magnetic field strength B at the

d Ae 0

(c) 2pfq0 cos 2pft

d sin 2pft e0 2pfq0 (d) cos 2pft e0

(b)

Electromagnetic Waves and Their Properties 8. A charged particle oscillates about its mean

equilibrium position with a frequency of 10 9 Hz. The electromagnetic waves produced will have (a) frequency of 10 9 Hz (b) frequency of 2 ´ 10 9 Hz (c) wavelength of 0-3 m (d) fall in the region of radiowaves

1010 JEE Main Physics 9. An electromagnetic wave of frequency n = 3 MHz passes from vacuum into a dielectric medium with permittivity e = 4.0, then (a) (b) (c) (d)

wavelength is 2 times and frequency becomes half wavelength is half and frequency remains unchanged wavelength and frequency both remains unchanged None of the above

10. In a plane electromagnetic wave, the electric field

oscillates sinusoidally of a frequency of 3 ´ 1010 Hz and amplitude 50 Vm -1. What is the wavelength of wave ? (a) 0.01 m (c) 0.1 m

(b) 0.05 m (d) 0.25 m

11. A radiowave has a maximum magnetic field induction of 10-4 T on arrival at a receiving antenna. The maximum electric field intensity of such a wave is (a) zero

(b) 3 ´ 10 4 Vm-1

(c) 5.8 ´ 10 -4 T

(d) 3.0 ´ 10 -5 T

12. If a free electron is placed in the path of a plane electromagnetic wave, it will start moving along (a) centre of earth (c) magnetic field

(b) equator of earth (d) electric field

13. A radio can tune into any station in the 7.5 MHz to 12 MHz band. The corresponding wavelength band of it is [NCERT] (a) 45 m to 65 m (c) 10 m to 25 m

(b) 25 m to 40 m (d) 45 m to 75 m

14. A radiowave of frequency 90 MHz enters a ferrite rod. If e r = 103 and m r = 10, then the velocity and wavelength of the wave in ferrite are (a) 3 ´ 10 8 ms -1; 333 . ´ 10 -2 m 6

-1

-2

8

-1

-1

(b) 3 ´ 10 ms ; 333 . ´ 10 m (c) 3 ´ 10 ms ; 333 . ´ 10 m (d) 3 ´ 107 ms -1; 333 . ´ 10 -3m

15. A cube of edge a has its edges parallel to x, y and z-axis of rectangular coordinate system. A uniform electric field E is parallel to y-axis. and a uniform magnetic field is E parallel to x-axis. The rate at which flows through each face of the cube is (a)

a2 EB parallel to x-y plane and zero in others 2m 0

(b)

a2 EB parallel to x-y plane and zero in others m0

a2 EB from all faces (c) 2m 0 (d)

a2 EB parallel to y-z faces and zero in others 2m 0

16. If e0 and m 0 represent the permittivity and

permeability of vacuum and e and m represent the permittivity and permeability of medium, then refractive index of the medium is given by (a)

m 0eo me

(b)

me m 0e0

(c)

m m 0e0

(d)

m 0eo m

17. The electric field intensity produced by the radiations coming from 100 W bulb at a 3 m distance is E. The electric field intensity produced by the radiations coming from 50 W bulb at the same [NCERT Exemplar] distance is (a) E/2

(b) 2 E

(c) E 2

(d)

2E

18. An electric field E and magnetic field B exist in a region. If these fields are not perpendicular to each other, then the electromagnetic wave (a) (b) (c) (d)

will not pass through the region will pass through region may pass through the region nothing is definite

19. The oscillating electric and magnetic field vectors of electromagnetic wave are oriented along (a) (b) (c) (d)

the same direction and in phase the same direction but have a phase difference of 90° mutually perpendicular directions and are in phase mutually perpendicular directions but has a phase difference of 90°

20. An electromagnetic wave, going through vacuum is described by E = E0 sin( kx - wt). Which of the following is independent of wavelength? (b) w

(a) k

(c) k / w

(d) kw

21. In a plane electromagnetic wave, the electric field

oscillates sinusoidally at a frequency of 20 . ´ 1010 Hz -1 and amplitude 48 Vm . The wavelength of the wave is [NCERT Exemplar] (a) 24 ´ 10 -10 m 8

(c) 416 . ´ 10 m

(b) 15 . ´ 10 -2 m (d) 3 ´ 10 8 m

22. The amplitude of electric field in a parallel beam of light of intensity 4 Wm -2 is (a) 40.5 NC-1

(b) 45.5 NC-1

(c) 50.5 NC-1

(d) 55.5 NC-1

23. A laser beam is sent to the moon and reflected back to earth by a mirror placed on the moon by an astronaut. If the moon is 38400 km from the earth, how long does it take the light to make the round trip? (a) 5 min (c) 2.5 s

(b) 2.5 min (d) 500 s

Electromagnetic Waves 24. Given the wave function (in SI units) for a wave to be y (x , t) = 103 sin p(3 ´ 106 x - 9 ´ 1014 t). The speed of wave is

32. An electromagnetic wave with poynting vector 6 Wm -2 is absorved by a surface area 12 m 2 . The force on the surface is

(a) 9 ´ 1014 ms -1

(b) 3 ´ 10 8 ms -1

(a) 24 ´ 10 -8 N

(b) 30 ´ 10 -5 N

(c) 3 ´ 106 ms -1

(d) 3 ´ 107 ms -1

(c) 2.4 ´ 10 -12 N

(d) 24 ´ 10 8 N

25. An EM wave radiates outwards from a dipole antenna, with E0 as the amplitude of its electric field vector. The electric field E0 which transports significant energy from the source falls off as [NCERT Exemplar]

1 r3 1 (c) r (a)

(b)

1 r2

26. An earth orbiting statellite has solar energy

collecting panel with total area 5 m 2 . If solar radiations are perpendicular and completely absorbed, the average force associated with the radiation pressure is (Solar constant = 1.4 kWm -2 ). -4

(a) 2.33 ´ 10 N

(b) 2.33 ´ 10 N

(c) 2.33 ´ 10 -5 N

(d) 2.33 ´ 10 -6 N

27. Which of the following has zero average value in a plane electromagnetic wave? (a) Kinetic energy (c) Electric field

(b) Magnetic field (d) Both (b) and (c)

28. 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? [NCERT] (a) 53 N/C (c) 93 N/C

(b) 123 N/C (d) 153 N/C

in any medium is expressed as (a)

(b)

1 me

(c)

m /e

(d)

m0 e

30. The sum radiates electromagnetic energy at the rate of 4.0 ´ 1026 W . Its radius is 7 ´ 108 m. The intensity of sun light at the solar surface is -8

-2

(a) 4.54 ´ 10 Wm

(b) 5.09 ´ 10

(c) 3.00 ´ 10 8 Wm-2

(d) 10 8 Wm-2

8

-2

Wm

31. The sun delivers 103 Wm -2 of electromagnetic flux to the earth’s surface. The radiation force on the roof of dimensions 10 m ´ 20 m is (a) 6.67 ´ 10 -6 N (c) 6.67 ´ 10 -3 N

of 2.0 ´ 108 ms -1. The relative permeability of the medium is 1.0. The relative permittivity is (a) 9.25 (c) 8.62

(b) 4.25 (d) 2.25

880 kHz and a power of 10 kW. The number of photons per second emitted are (a) 1.171 ´ 1025 (c) 1.71 ´ 1031

(b) 1.715 ´ 1030 (d) 1.025 ´ 1030

Electromagnetic Spectrum and its Uses 35. If, vs , vx and vm are the speeds of gamma rays, X-rays and microwaves respectively in vacuum, then (a) v s > v x > v m (c) v s > v x < v m

(b) v s < v x < v m (d) v s = v x = v m

36. The wavelength of X-rays lies between (a) (b) (c) (d)

maximum to finite limits minimum to certain limits minimum to infinite limits infinite to finite limits

37. Hydrogen atom does not emit X-rays because (a) (b) (c) (d)

it has single electron it has no neutron it has single neutron its energy levels are too close to each other

38. The correct sequence of the increasing wavelength of

29. According to Maxwell’s equation, the velocity of light 1 m 0e0

33. Electromagnetic waves travel in a medium at a speed

34. The radio transmitter operates at an frequency of

(d) remains constant

-3

1011

(b) 6.67 ´ 10 -4 N (d) 2.35 ´ 10 -4 N

the given radiation sources is (a) radioactive sources, X-ray tube, crystal oscillator, sodium vapour lamp (b) radioactive source, X-ray tube, sodium vapour lamp, crystal oscillator (c) X-ray tube, radioactive source, crystal oscillator, sodium vapour lamp (d) X-ray tube, crystal oscillator, radioactive source, sodium vapour lamp

39. The voltage applied across an X-ray tube is nearly equal to (a) 10 V (c) 1000 V

(b) 100 V (d) 10000 V

40. X-rays are produced by jumping of (a) (b) (c) (d)

electrons from lower to higher energy orbit of atom electrons from higher to lower energy orbit of atom protons from lower to higher energy orbit of nucleus proton from higher to lower energy orbit of nucleus

1012 JEE Main Physics 41. The wavelength of infrared rays is of the order of (a) 5 ´ 10 -7 m (c) 10

-7

to 10

(b) 10 -3 m

-3

11

(d) 10 to 10

travels in a free space along the x-direction. At the 9

42. Which of the following rays is emitted by a human body? (a) X-rays

(b) UV rays (c) Visible rays (d) IR rays

43. Molybdenum is used as a target element for the production of X-rays because it is (a) (b) (c) (d)

51. A plane electromagnetic wave of frequency 25 MHz particular point in space and time, E = 63 . ^j V/m . What is B at that point? (a) 2.1 ´ 10 -8 k$ T (c) 5.0 ´ 10 -6 k$ T

(b) 3.1 ´ 10 -5 k$ T (d) 0

52. The waves which have revolutionized telecommunication in more recent time, are

light and can easily deflect electrons light and can absorb electrons a heavy element with a high melting point an element having high thermal conductivity

53. The density of air at the top of mesosphere in

44. X-ray are not used for radar purposes, because they

comparison to that of near the earth’s surface is

are not (a) (b) (c) (d)

reflected by target partly absorbed by target electromagnetic waves completely absorbed by target

45. Which of the following electromagnetic waves have the longest wavelength? (a) Heat waves (c) Radiowaves

(b) Light waves (d) Ultraviolet waves

46. The shortest wavelength of X-rays emitted from an X-rays tube depends upon (a) (b) (c) (d)

nature of the gas in the tube voltage applied to tube current in the tube nature of target of the tube

47. If a source is transmitting electromagnetic wave of

frequency 8.2 ´ 106 Hz, then wavelength of the electromagnetic waves transmitted from the source will be (a) 36.6 m (c) 42.3 m

(b) 40.5 m (d) 50.9 m

48. If the wavelength of light is 4000 Å, then the number of wavelength in 1 mm length will be (a) 2.5 ´ 105

(b) 0.25 ´ 10 4

(c) 2.5 ´ 10 4

(d) 0

Earth’s Atmosphere and Propagation of Electromagnetic Waves 49. If, the earth did not have atmosphere, its surface temperature on a day time would be (a) higher (c) same as now

(b) power (d) not sure

50. The temperature variation in the region of stratosphere lies from (a) 290 K to 220 K (c) 220 K to 380 K

(b) 220 K to 280 K (d) 1180 K to 700 K

(a) microwaves (c) light waves

(b) radiowaves (d) TV waves

(a) 10 -3 times

(b) 10 -5 times

(c) 103 times

(d) 105 times

54. 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 [NCERT] wavelength? (a) 20 m (c) 10 m

(b) 15 m (d) 5 m

55. The atmosphere between the heights of 50 km and 80 km, is called (a) mesosphere (c) ionosphere

(b) ozonosphere (d) troposphere

56. The small ozone layer on top of the atmosphere is crucial for human survival because it (a) has ions (c) ionosphere

(b) reflects radio signals (d) troposphere

57. Television signals reach us only through the ground waves. The range R related with the transmitter height h is in proportion to (b) h1/2 (d) h-1

(a) h (c) h-1/2

58. Clouds are contained in a layer from the earth’s surface, which is called (a) troposphere (c) mesosphere

(b) stratosphere (d) ionosphere

59. Height h of transmitting antenna when R is radius of earth to have range, d is (a) d2 / 2R

(b)

2dR

(c) 2d2 / R

(d) 2R2 /d

60. Ozone layer blocks the radiations of wavelength (a) less than 3 ´ 10 -7 m

(b) equal to 3 ´ 10 -7 m

(c) more than 3 ´ 10 -7 m

(d) All of these

61. The ozone layer of the atmosphere lies in the region called (a) troposphere (c) mesosphere

(b) stratosphere (d) ionosphere

Electromagnetic Waves 62. The atmosphere above the height of 80 km is called (a) stratosphere (c) mesosphere

70. The mean electric energy density between the plates of a charged capacitor is (here, q = charge on the capacitor and A = area of the capacitor plate)

(b) troposphere (d) ionosphere

63. A TV tower has a height of 100 m. How much

(a)

(b)

population is covered by the TV broadcast if, the average population density around the tower is 1000 km -2 ? (radius of the earth = 637 . ´ 106 m)

q2 2e 0 A2

(c)

q2 2 e0 A

(d) 0

(a) 4 lakh (c) 40000

(b) 4 billion (d) 40 lakh

monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in [NCERT Exemplar] (b) infrared region (d) microwave region

65. The electric field intensity produced by the radiations coming from 100 W bulb at a distance 3m is E 0 . The electric field intensity produce d by the radiations coming from 50 W bulb at the same distance is [NCERT] E 2 E (c) 2 (a)

2E

hydrogen in inter stellar space is due to the interaction called the hyperfine interaction in atomic hydrogen, the energy of the emitted wave is nearly [NCERT]

(b) 1 J (d) 10 -24 J

67. In which one of the following regions of the electromagnetic spectrum will the vibrational motion of the molecules give rise to absorption (a) ultraviolet (c) infrared

(b) microwaves (d) radiowaves

68. In X-ray tube, the accelerating potential applied at the anode is V volt. The minimum wavelength of the emitted X-rays will be eV (a) h ev (c) ch

h (b) eV hc (d) eV

69. A plane electromagnetic wave is incident on a material surface. The wave delivers momentum p and energy, E. (a) p ¹ 0, E ¹ 0 (c) p = 0, E ¹ 0

(b) p = 0, E = 0 (d) p ¹ 0, E = 0

(a) 1.00 ´ 10 -25 Wh (c) 2.09 ´ 10 -23 Wh

(b) 1.83 ´ 10 -22 Wh (d) 4.05 ´ 10 -22 Wh

72. In an electromagnetic wave, the amplitude of electric field is 10 V/m. The frequency of wave is 5 ´ 1014 Hz. The wave is propagating along the z-axis, then the average density of electric field is (a) 2.21 ´ 10 -10 Jm-3 (c) 8.1 ´ 10 -12 Jm-3

(b) 3.25 ´ 10 -9 Jm-3 (d) 6.25 ´ 10 -3 Jm-3

73. The pressure exerted by an electromagnetic wave of

intensity I (watt/m 2 ) on a non-reflecting surface is (c = velocity of light) I c (d) I c2

(a)

66. The radiowave (wave length 21 cm) is emitted by

(a) 10 -17 J (c) 7 ´ 10 -8 J

energy of photon in watt hour for electromagnetic waves of wavelength 3000 Å, is (Given, h = 6.6 ´ 10-34 Js)

I c2 (c) Ic

(b) 2 E (d)

q 2e 0 A2

71. The

64. One requires 11 eV of energy to dissociate a carbon

(a) visible region (c) ultraviolet region

1013

(b)

74. When a plane electromagnetic wave travels in vacuum, the average electric energy density is given by (here E0 is the amplitude of the electric field of the wave) is 1 e 0 E20 4 (c) e 0 E20

1 e 0 E20 2 (d) 2 e 0 E20 (b)

(a)

75. In an apparatus, the electric field was formed to oscillate with an amplitude of 18 Vm -1. The magnitude of the oscillating magnetic field will be (a) 9 ´ 10 -9 T (c) 6 ´ 10

-8

(b) 11 ´ 10 -11 T (d) 4 ´ 10 -8 T

T

76. An electromagnetic wave going through vacuum is described by E = E0 sin( kx - wt); B = B0 sin( kx - wt). Which of the following is true? (a) E0 k = B0 w (c) E0 B0 = wk

(b) E0 k = 2 B0 w (d) E0 w = B0 k

77. Light waves travel in vacuum along the y-axis. Which of the following may represent the wavefronts? (a) x = constant (c) z = constant

(b) y = constant (d) x + y + z = constant

78. An electromagnetic wave travels along z-axis. Which of the following pairs of space and time varying fields would generate such a wave? (a) Ex, B y

(b) Ez, Bx

(c) Ey , Bz

(d) Ey , Bx

1014 JEE Main Physics 79. The electromagnetic waves travel in a medium which

80. If c is the speed of electromagnetic waves in a

has relative permeability by 1.3 and relative permittivity 2.14. Then, the speed of the electromagnetic wave in the medium will be

vacuum, its speed in a medium of dielectric constant k and relative permeability m is

(a) 1.8 ´ 107 ms -1 6

(c) 2.5 ´ 10 ms

-1

(b) 1. 8 ´ 10 8 ms -1 (d) 5.7 ´ 10 9 ms -1

Round II Only One Correct Option 1. If an electromagnetic wave is propagating in a medium with permittivity e and permeabillity m, then m is the e (a) (b) (c) (d)

intrinsic impedance of the medium square of the refractive index of the medium refractive index of the medium energy density of the medium

2. A capacitor having a capacity of 2pF. Electric field across the capacitor is changing with a value of 1012 Vs -1. The displacement current is (a) 2 A

(b) 4 A

(a) v =

(c) 6 A

(c) v =

(b) 106 Vs -1

(c) 1 Vs -1

(d) 01 . Vs -1

4. The fact that radiosignals reach the earth from outside the atmosphere, was discovered accidently by

1 m k

(d) v = c m k

energy. How much force is exerted on the plate if is exposed to sunlight for 20 min ? (a) 4 ´ 10 -7 N

(b) 5 ´ 10 -7 N

(c) 3 ´ 10 -7 N

(d) 6 ´ 10 -7 N

8. The maximum distance upto which TV transmission from a TV tower of height h can be received is proportional to (a) h1/2

(b) h

(d) h2

(c) h3/2

9. Radiowaves diffract around building although light waves do not. The reason is that radiowaves (a) (b) (c) (d)

3. Instantaneous displacement current of 1.0 A in the (a) 10 -6 Vs -1

(b) v =

(Mixed Bag)

(d) 10 A

space between the parallel plate of 1mF capacitor can be estabilished by changing potential difference of

c mc k m c

travel with speed target than c have much larger wavelength than light carry news are not electromagnetic waves

10. Out of the following electromagnetic radiation, which has the shortest wavelength? (a) Radiowaves (c) Ultraviolet

(b) infrared (d) X-rays

11. An electromagnetic wave going through vacuum is described by E = E0 sin( kx - wt); B = B0 sin( kx - wt). Which of the following equation is true?

(a) K.G. Jansky (c) Aryabhatta

(b) Millikan (d) Prof. Kanu

5. The curve drawn between velocity and frequency of a photon in vacuum will be (a) straight line parallel to frequency axis (b) straight line parallel to velocity axis (c) straight line passing through origin and making an angle of 45° with frequency axis (d) hyperbola

6. A radar sends the waves towards a distant object and receives the signal reflected by object. These waves are (a) sound waves (c) radiowaves

(b) light waves (d) microwaves

7. The average energy flux of sunlight is 1.0 kWm -2 . This energy light is falling normally on the plate surface of area 10 cm 2 which completely absorbs the

(a) E0 k = B0 w (c) E0 B0 = wk

(b) E0 w = B0 k (d) None of these

12. The electric field of plane electromagnetic wave in vacuum is represented by E x = 0; ®

®

. cos[2p ´ 108 ( t - x / c]; E z = 0 E y = 05

What is the direction electromagnetic waves? (a) Along x-z direction (c) Along x-direction

of

propagation

of

(b) along y-direction (d) a long yz-direction

13. A circular ring of radius r is placed in a homogeneous magnetic field perpendicular to the plane of the ring. The field B changes with time according to the equation, B = kt, where k is a constant and t is the time. The electric field in the ring is (a)

kr 4

(b)

kr 3

(c)

kr 2

(d)

k 2r

Electromagnetic Waves 14. A large parallel plate capacitor, whose plates have

an area of 1 m 2 and are separated from each other by 1 mm, is being charged at a rate of 25 Vs -1. If the dielectric between the plates has the dielectric constant 10, then the displacement current at this instant is (a) 25 mA (c) 2.2 mA

(b) 11 mA (d) 1.1 mA

with time having an amplitude 1 Vm -1. The frequency of wave is 05 . ´ 1015 Hz. The wave is propagation along X-axis. What is the average energy density of magnetic field? (b) 2.2 ´ 10 -12 J m-3 (d) 4.4 ´ 10 -12 Jm-3

reflected by the surface, find the pressure exerted on the surface. (b) 6.67 ´ 10 -9 Nm-2 (d) 9.87 ´ 10 -8 Nm-2

17. The speed of electromagnetic wave in a medium of dielectric constant 2.25 and relative permittivity, m is 8

(a) 1 ´ 10 ms

-1

(c) 4 ´ 10 8 ms -1

8

(b) 2.5 ´ 10 ms

-1

(d) 3 ´ 10 8 ms -1

18. The dielectric constant of air is 1.006. The speed of electromagnetic wave travelling a ´ 108 ms-1, where a is about (a) 2.8 (c) 3

in

air

is

(b) 4.5 (d) 3.2

(a) 4.3 ´ 106 (c) 5.6 ´ 107

(b) 4.5 ´ 107 (d) 5.8 ´ 106

electromagnetic oscillations differ respectively from those of the corresponding electric vector by (a) 0 and

p 2

(c) 0 and 0

(b) 0 and ¥ (d)

p p and 2 2

amplitude of an electromagnetic wave in vacuum is doubled with no other changes made to the wave. As a result of this doubling of the amplitude, which of the following statement is true? (a) (b) (c) (d)

The speed of wave propagation changes only The frequency of the wave changes only The wavelength of the wave changes only None of the above

25. The wave of wavelength 5900 Å emitted by any atom or molecule must have some finite total length which is known as the coherence length. For sodium light, this length is 2.4 cm. The number of oscillations in this length will be (a) 4.068 ´ 10 8

(b) 4.068 ´ 107

(c) 4.068 ´ 106

(d) 4.068 ´ 105

26. A radiation of 200 W is incident on a surface which is

19. The flood light is covered with a filter that transmits red light. The electric field of the increasing beam is represented by a sinusoidal plane wave V/m. The Ex = 36 sin (1.20 ´ 10 z - 3.6 ´ 1015 t) 2 average intensity of beam in W/m will be (a) 6.88 (c) 1.72

of 3.9 ´ 1026 W . Its radius is 6.96 ´ 108 m . The intensity of sunlight (in Wm -2 ) at the solar surface will be

24. The

16. If the intensity of the incident radio wave of 1 W/m 2 is (a) 5.67 ´ 10 -9 Nm-2 (c) 8. 57 ´ 10 -9 Nm-2

22. The sun radiates electromagnetic energy at the rate

23. The phase oriented magnetic vector associated with

15. In a plane electromagnetic wave electric field varies

(a) 1.1 ´ 10 -12 Jm-3 (c) 3.3 ´ 10 -12 Jm-3

1015

(b) 3.44 (d) 0.86

60% reflecting and 40% absorbing. The total force on the surface is (a) 107 . ´ 10 -6 N (b) 13 . ´ 10 -6 N (c) 107 . ´ 10 -7 N (d) 103 . ´ 10 -7 N

27. The sun delivers 104 Wm -2 of electromagnetic flux to

20. A plane electromagnetic wave of wave intensity 2

-2

the earth’s surface. The total power that in incident on a roof of dimensions 10 m2 will be

strikes a small mirror of area 30 cm , held 6 Wm perpendicular to the approaching wave. The momentum transferred in kg ms -1 by the wave to the mirror each second will be

28. The electric field (in NC-1) in an electromagnetic

21. In an electromagnetic wave, the electric and

wave is given by E = 50 sin w( t - x / c). The energy stored in a cylinder of cross-section 10 cm 2 and length 100 cm along the x-axis will be

(a) 1. 2 ´ 10 -10 (c) 3.6 ´ 10 -8

(b) 2.4 ´ 10 -9 (d) 4.8 ´ 10 -7

magnetizing fields are 100 Vm -1 and 0.265 Am -1. The maximum energy flow is -2

(a) 26.5 Wm (c) 68.2 Wm-2

-2

(b) 70.5 Wm (d) 0

(a) 10 4 W (c) 106 W

(a) 5.5 ´ 10 -12 J (b) 1.1 ´ 10 -11 J (c) 2.2 ´ 10 -11 J

(d) 1.65 ´ 10 -11 J

(b) 105 W (d) 107 W

1016 JEE Main Physics 29. A plane electromagnetic wave of intensity 10 Wm -2 2

strikes a small mirror of area 20 cm , held perpendicular to the approaching wave. The radiation force on the mirror will (a) 6.6 ´ 10 -11 N (c) 133 . ´ 10

-10

N

(b) 133 . ´ 10 -11 N

(d) 6.6 ´ 10

-11

N

30. Electric fields induced by changing magnetic fields are (a) conservative (b) non-conservative (c) may be conservative or non-conservative depending on the conditions (d) Nothing can be said

31. Assume that a lamp radiates power P uniformly in all directions. What is the magnitude of electric field strength at a distance r from the lamp? (a) (c)

P pce 0 r2

(b)

P 2pe 0 r2 c

P 2pcer2

average power output of 800 W. The maximum value of electric field at a distance 4.0 m from the source is (b) 57.8 Vm-1 (d) 54.77 Vm-1

33. The electric field for a plane electromagnetic wave travelling in the positive z-direction is represented by which one of the following? (a) k$1 E0 e i( kz - wt + f) (c) i$ E e i( kz + wt + f) 1 0

(b) $i1 E0 e i( kx - wt + f) (d) k$ E e i( kz + wt + f) 1 0

34. The magnetic field between the plates of radius 12 cm separated by distance of 4 mm of a parallel plate capacitor of capacitance 100 pF along the axis of plates having conduction current of 0.15 A is (a) zero

(b) 1.5 T

(c) 15 T

(d) 0. 15 T

35. A lamp emits monochromatic green light uniformly in all directions. The lamp is 3% efficient in converting electric power to electromagnetic waves and consume 100 W of power. The amplitude of the electric field associated with the electromagnetic radiation at a distance of 5 m from the lamp will be (a) 2.68 Vm-1

-1

(c) 2.01 Vm

xö æ (c) Ex = 33 cos p ´ 1011 çt - ÷ and è cø æ B y = 11 ´ 10 -7 cos p ´ 1011 çt è

xö ÷ cø

xö æ (d) Ey = 66 cos 2p ´ 1011 çt - ÷ and è cø xö æ Bz = 2.2 ´ 10 -7 cos 2p ´ 1011 çt - ÷ è cø

suited to observed particle of radius 3 ´ 10-4 cm is of the order of

32. A point source of electromagnetic radiation has an (a) 64.7 Vm-1 (c) 56.72 Vm-1

xö æ (b) Ey = 11 cos 2p ´ 1011 çt - ÷ and è cø xö æ B y = 11 ´ 10 -7 cos 2p ´ 1011 çt - ÷ è cø

37. The frequency of electromagnetic waves which is best

P pe 0 cr2

(d)

xö æ (a) Ey = 33 cos p ´ 1011 çt - ÷ and è cø xö æ Bz = 1.1 ´ 10 -7 cos p ´ 1011 çt - ÷ è cø

(b) 3.15 Vm-1 (d) 0

36. A plane electromagnetic waves travelling along the x-direction has a wavelength of 3 mm. The variation in the electric field occurs in the y-direction with an amplitude 66 Vm -1. The equation for the electric and magnetic fields as a function of x and t are respectively

(b) 1016 Hz (d) 1011 Hz

(a) 1015 Hz (c) 1020 Hz

38. Radiowaves received by a radio telescope from distant aparts, may have a wavelength of about 8 0.20 m. If the speed of the wave is 3 ´ 10 ms -1, then frequency of the wave will be (a) 1.5 ´ 10 9 Hz 3

(c) 1.5 ´ 10 Hz

(b) 1.5 ´ 10 8 Hz (d) 135 Hz

39. A TV tower has a height of 100 m. The average

population density around the tower is 1000 km -2 and radius of the earth is 6.37 ´ 106 m . The population covered is (a) 3 ´ 105

(b) 4 ´ 106

(c) 6 ´ 105

(d) 8 ´ 102

40. In a plane of electromagnetic wave, the electric field

oscillates sinusoidally at a frequency of 2 ´ 1010 Hz and amplitude 48 Vm -1. The amplitude of oscillating magnetic field will be (a) 16 ´ 10 -8 Wbm-2

(b) 12 ´ 10 -8 Wbm-2

(c) 18 ´ 10 -5 Wbm-2

(d) 2.0 ´ 10 -6 Wbm-2

41. What should be the height of transmitting antenna, if the TV telecast is to cover a radius of 128 km? (a) 1280 m (c) 1024 m

(b) 1000 m (d) 1332 m

42. In an electromagnetic wave, the amplitude of electric field is 1 Vm -1. The frequency of wave is 5 ´ 1014 Hz. The wave is propagating along z-axis. The average energy density of electric field in J/m 3 will be (a) 3.2 ´ 10 -12 Jm-3 (c) 5.2 ´ 10 -13 Jm-3

(b) 2.2 ´ 10 -12 Jm-3 (d) 7.2 ´ 10 -12 Jm-3

Electromagnetic Waves 43. A point source of electromagnetic radiation has an average power output of 800 W. The maximum value of electric field at a distance 3.5 m from the source will be (b) 16.15 Vm-1 (d) 32.5 Vm-1

(a) 62.6 Vm-1 (c) 8.08 Vm-1

(a) will have frequency of 10 9 Hz (b) will have frequency of 2 ´ 10 9 Hz (c) will have a wavelength of 0.3 m (d) fall in the region of radiowaves

49. An EM wave of intensity I falls on a surface kept in vacuum and exerts radiation pressure p on it . Which of the followings are true? [NCERT Exemplar]

More Than One Correct Option 44. An electromagnetic wave travels in vacuum along z direction: E = ( E1i$ + E2 $j ) cos ( kz - wt). Choose the correct options from the following [NCERT Exemplar] (a) The

associated magnetic field is given 1 $ B = ( E1i + E2 $j ) cos ( kz - wt ) c (b) The associated magnetic field is given 1 B = ( E1i$ - E2 $j ) cos ( kz - wt ) c (c) The given electromagnetic field is circularly polarised (d) The given electromagnetic wave is plane polarised 10

oscillates sinusoidally at a frequency of 2 ´ 10 and amplitude 54 V.

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

as

Radiation pressure is I/c if the wave is totally absorbed Radiation pressure is I/c if the wave is totally reflected Radiation pressure is 2I/c if the wave is totally reflected Radiation pressure is in the range I/c < p < 2I/c for the real surfaces

50. Which of the following pairs of space and time as

varying E and B fields would generate a plane electromagnetic wave travelling along the z-direction? (a) Ex, Bz (c) Ex, B y

45. In a plane electromagnetic wave, the electric field Hz

(a) The amplitude of oscillating magnetic field will be 18 ´ 10 -8 Wbm-2 (b) The amplitude of oscillating magnetic field will be 18 ´ 10 -7 Wbm-2 (c) The wavelength of electromagnetic wave is 1.5 m (d) The wavelength of electromagnetic wave is 1.5 cm

(b) Ey , Bz (d) Ey , Bx

51. If m 0 , e0 are the absolute permeability and

permitivity respectively, of space, m r , e r are the relative permeability and permittivity respectively of mediums and m, e are the absolute permeability and permittivity of medium respectively, then the refractive index of the medium is (a)

me m 0e0

(b)

m r er m 0e0

(c)

m r er

(d)

me m 0er

46. Which of the following Maxwell’s equations have sources of E and B ? q e0

(a)

òs

(b)

òs B. ds = q

E.ds =

d (c) ò E. ds = s dt (d)

1017

52. Which of the following statements is correct in relation to electromagnetic waves in an isotropic medium?

òs B. ds d

òs E. dI = m 0 e 0 dt òs E. ds

47. The electromagnetic wave travelling along z-axis is given as E = E 0 cos (kz - wt). Choose the correct options from the following 1$ k´E c (b) The electromagnetic field can be written in terms of the associated magnetic field as E = c ( B ´ k$ ) (c) k$ × E = 0, k$ × B = 0, (d) k$ ´ E = 0, k$ ´ B = 0

(a) Energy due to electric field is equal to that due to magnetic field (b) Electric field E and magnetic field B are in phase (c) For given amplitude of E, the intensity increases as the first power of frequency f (d) For the cylindrical wavefronts, the amplitude of the 1 waves varies in proportion to 2 , where r is the radius of r the wavefront

(a) The associated magnetic field is given as B =

48. A charged particle oscillates about its mean

equilibrium position with a frequency of 109 Hz. The electromagnetic waves produced [NCERT Exemplar]

Comprehension Based Questions Passage I The magnetic field in a plane electromagnetic wave is given by By = 2 ´ 10-7 sin ( 0.5 ´ 103 x + 1.5 ´ 1011 t) T It is propagating in space.

53. What is the frequency of this electromagnetic waves ? (a) 5.0 ´ 1010 Hz (c) 1.9 ´ 1010 Hz

(b) 2.4 ´ 1010 Hz (d) 1.3 ´ 1010 Hz

1018 JEE Main Physics 54. Expression for the electric field is 3

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

-1

11

(a) Ey = 60 sin( 0.5 ´ 10 x + 1.5 ´ 10 t ) Vm

(b) Ey = 600 sin( 0.5 ´ 103 x + 1.5 ´ 1011t ) Vm-1 (c) Ez = 60 sin( 0.5 ´ 103 x + 1.5 ´ 1011t ) Vm-1 (d) Ez = 600 sin( 0.5 ´ 103 x + 1.5 ´ 1011t ) Vm-1

55. Intensity of the electromagnetic wave is about -2

(a) 3 Wm

-2

(b) 4 Wm

-2

63. Match the following Column I with Column II. Column I -2

(d) 6 Wm

(c) 5 Wm

56. Maximum value of electric field is (a) 6 ´ 10 -16 Vm-1

(b) 60 Vm-1

(c) 600 Vm-1

(d) 6000 Vm-1

I. II. III. IV.

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

57. What is the wavelength of this electromagnetic waves?

(a) 0.6 ´ 10 -2 m (c) 1.9 ´ 10 -2 m

I-A, II-B, III-C, IV-D I-D, II-C, III-B, IV-A I-D, II-B, III-A, IV-C I-C, II-A, III-B, IV-D

(b) 1.3 ´ 10 -2 m (d) 2.5 ´ 10 -2 m

Ultraviolet Infrared X-rays Microwaves

58. The total electromagnetic power of sun falling on the earth (radius of the earth = 6400 km) is (a) 5.1 ´ 1017 W

(b) 6.7 ´ 1015 W

(c) 5 ´ 1018 W

(d) 5.1 ´ 1019 W

59. The total electromagnetic power of the sun is (b) 5.6 ´ 1026 W (d) 5.8 ´ 1032 W

60. Total electromagnetic power that is incident on a roof of dimensions (8 m ´ 10 m ) on the surface of the earth is (a) 8 ´ 10 4 W

(b) 8 ´ 105 W

(c) 8 ´ 10 9 W

(d) 2.56 ´ 10 4 W

61. The radiation force on the roof is

(b) 4.07 ´ 10 -6 N (d) 6.00 ´ 10 -5 N

Radar system Roengton Heat radiation To destroy living tissues

64. Match the following Column I with Column II. Column I

The sun delivers 10-3 Wm -2 of electromagnetic flux to the earth’s surface, which is at a distance of 1.5 ´ 1011 m from the sun. The whole incident electromagnetic flux is absorbed by the earth, then answer the following questions.

(a) 3.05 ´ 10 -5 N (c) 2.67 ´ 10 -4 N

A. B. C. D.

I-D, II-C, III-B, IV-A I-A, II-B, III-C, IV-D I-B, II-A, III-D, IV-C I-C, II-B, III-D, IV-A

Passage II

(a) 5.6 ´ 1028 W (c) 5.6 ´ 1022 W

Column II

Column II

I.

g-ray

A.

6 ´ 10-9 to 4 ´ 10-7 m

II. III.

X-rays UV-rays

B. C.

10-3 to 0.3 m 1 ´ 10-13 to 3 ´ 10-8 m

IV.

Microwaves

D.

0. 6 ´ 10-14 to 1 ´ 10-10 m

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

I-A, II-B, III-C, IV-D I-D, II-A, III-B, IV-C I-D, II-C, III-A, IV-B I-B, II-A, III-D, IV-C

Assertion and Reason Directions

Question No. 65 to 79 are Assertion-Reason type. Each of these contains two Statements: Statement I (Assertion), Statement II (Reason). Each of these questions also has four alternative choice, only one of which is correct. You have to select the correct choices from the codes (a), (b), (c) and (d) given below (a) If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion (b) If both Assertion and Reason are true but Reason is not the correct explanation of the Assertion (c) If Assertion is true but Reason is false (d) If Assertion is false but the Reason is true

65. Assertion Environment damage has increased the

Matching Type Column II

amount of ozone in the atmosphere. Reason Increase the ozone increases the amount of ultraviolet radiation on the earth.

A.

Gauss’s law

66. Assertion The changing electric field produces a

II.

dfE ö æ ò B × dl = m0 çè I + E0 dt ÷ø

B.

III.

ò E × dl = - dt ò sB × ds

d

C.

IV.

1 ò E × ds = e0

D.

Faraday’s laws of electromagnetic induction Ampere Maxwell’s law Ampere’s circuit law

62. Match the following Column I with Column II. Column I I.

ò B × dl = m0 I

ò s r ds

magnetic field. Reason A changing magnetic field produces an electric field.

67. Assertion Short wave band are used for transmission of radiowaves to a large distance. Reason Short waves are reflected from atmosphere.

Electromagnetic Waves 68. Assertion In an electromagnetic waves, the direction

1019

electron while falling on a metal of high atomic number.

of the magnetic field induction B is parallel to the electric field E. Reason Electric field vector E and magnetic field vector B, have the same frequency.

74. Assertion Microwaves are better carrier of signals

69. Assertion Magnetic field lines cannot start from a

75. Assertion X-rays astromy is possible only from

point n or end at a point. Reason The line integral of magnetic field induction over a closed path is not zero.

70. Assertion Electromagnetic waves are transverse in nature. Reason The electric and magnetic fields of an EM wave are perpendicular to the direction of wave propagation.

71. Assertion Out of the following radiations, microwave, ultraviolet and X-rays, microwaves has the shortest wavelength. Reason The microwave do not deviate from the obstacles in their path while going from one location to another.

72. Assertion The light can travel in vacuum but sound cannot do so. Reason Light is an electromagnetic wave and sound is a mechanical wave.

73. Assertion g-rays are more energetic than X-rays. Reason g -rays are of nuclear origin but X-rays are produced due to sudden deceleration of high energy

than optical waves. Reason Microwaves move faster than optical waves. satellites orbiting the earth. Reason Efficiency of X-rays telescope is larger as compared to any other telescope.

76. Assertion The electromagnetic wave is transverse in nature. Reason The wave propagates in straight line.

77. Assertion Television signals are received through sky-wave propagation. Reason The ionosphere reflects electromagnetic waves of frequencies greater than a certain critical frequency. Ultraviolet radiations being higher frequency waves are dangerous to human being. Reason Ultraviolet radiations are absorbed by the atmosphere.

78. Assertion

79. Assertion If earth did not have atmosphere, its average surface temperature would be lower than what is now. Reason Green house effect of the atmosphere would be absent if earth did not have atmosphere.

Previous Years’ Questions 80. An electromagnetic wave in vacuum has the electric and magnetic fields E and B, which are mutually perpendicular to each other. The direction of polarization is given by X and that of wave $ then propagation by k, [AIEEE 2012] $ || B ´ E (b) X || E and (a) X || B and k $ ||E ´ B (d) X || E and (c) X || E and k

$ ||E ´ B k $ ||B ´ E k

81. The transverse displacement y ( x, t) of a wave on a string is given by y( x, t) = e- i (ax

2

+ bt 2 + 2 ab xt)

represents a

. This

[AIEEE 2011]

b a

(a) wave moving in x-direction with speed

(a) Cosmic rays (c) b-rays

(b) Gamma rays (d) X-rays

83. The oscillating electric and magnetic vectors of an electromagnetic wave are oriented along [Karnataka CET 2010]

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

the same direction but differ in phase by 90° the same direction and are in phase mutually perpendicular direction and are in phase mutually perpendicular direction and differ in phase by 90°

84. An electromagnetic wave propagating along north has its electric field vector upwards. Its magnetic field vector point towards [Orissa JEE 2010]

(b) standing wave of frequency

b 1 (c) standing wave of frequency b

(a) North

(b) East

(c) West

(d) Downwards

85. The

(d) wave moving in + x-direction with speed

a b

82. Which of the following is not electromagnetic waves? [Orissa JEE 2011]

electric and magnetic fields of an electromagnetic wave are [Karnataka CET 2010] (a) in phase and parallel to each other (b) in opposite phase and parallel to each other (c) in opposite phase and parallel to each other (d) in phase and perpendicular to each other

1020 JEE Main Physics Answers Round I 1. 11. 21. 31. 41. 51. 61. 71.

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

2. 12. 22. 32. 42. 52. 62. 72.

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

3. 13. 23. 33. 43. 53. 63. 73.

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

4. 14. 24. 34. 44. 54. 64. 74.

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

5. 15. 25. 35. 45. 55. 65. 75.

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

6. 16. 26. 36. 46. 56. 66. 76.

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

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

8. 18. 28. 38. 48. 58. 68. 78.

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

9. 19. 29. 39. 49. 59. 69. 79.

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

10. 20. 30. 40. 50. 60. 70. 80.

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

(a) (a) (c) (a) (a,b,c) (b) (a) (b)

8. 18. 28. 38. 48. 58. 68. 78.

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

9. 19. 29. 39. 49. 59. 69. 79.

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

10. 20. 30. 40. 50. 60. 70. 80.

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

7. 17. 27. 37. 47. 57. 67. 77.

Round II 1. 11. 21. 31. 41. 51. 61. 71. 81.

(a) (a) (a) (c) (a) (a,c) (c) (a) (a)

2. 12. 22. 32. 42. 52. 62. 72. 82.

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

3. 13. 23. 33. 43. 53. 63. 73. 83.

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

4. 14. 24. 34. 44. 54. 64. 74. 84.

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

5. 15. 25. 35. 45. 55. 65. 75. 85.

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

6. 16. 26. 36. 46. 56. 66. 76.

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

7. 17. 27. 37. 47. 57. 67. 77.

the Guidance Round I 1. Electric and magnetic fields and energies have equal average values.

2. E = 11eV = 11 ´ 16 . ´ 10 n=

or

=

11 ´ 1.6 ´ 10 h

-19

J = hn

-19

11 ´ 1.6 ´ 10 -19 6.62 ´ 10 -34

= 2.65 ´ 10 15 Hz This frequency radiation belongs to ultraviolet region.

3. Equation (i) is Gauss’s law and equation (iv) is Ampere’s law Here, two equations are Maxwell’s equations.

4. Using Ampere circuit law,

ò B × dl = m 0iD B ´ 2pr = a 0iD

or or m0 4p m = 0 2p

5. As, B =

2iD m 0 = r 4p 2iD m 0 = r 4p

B = m 0iD / 2pr df ´ e0 E dt 2 df ´ e0 E r dt

m e pr 2dE m 0 e 0r dE = 0 0 = 2prdt 2 dt

6. When a wave is reflected from denser medium, the reflected wave is without change in type of wave but with a change in phase by 180° or p radian. Therefore, for the reflected wave we use z = - z , $i = - $i and additional phase of p in the incident wave. The incident EM wave is, E r = E 0 $i cos(kz - wt ) The reflected EM wave is E r = E 0 ( - $i ) cos[ k (- z ) - wt + p ]= - E 0 $i cos[ -( kz + wt )+ p ] = E $i cos[ - ( kz + wt )] [Q cos( q + p ) = - cos q] 0

= E 0 $i cos ( kz + wt )

7. As, i =

[Q cos( - q) = cos q]

dq d = (q 0 sin 2pft ) = q 02pf cos 2pft . dt dt

8. We know that the speed of all electromagnetic waves is the same in vacuum.

9. We know that wavelength is constant.

1 and frequency remains 2

10. Here, n = 3 ´ 1010 Hz, c = 3 ´ 10 8m/s As wavelength, l =

c 3 ´ 10 8 = 0.01m = n 3 ´ 10 10

11. As, E 0 = cB0 = 3 ´ 10 8 ´ 10 -4 = 3 ´ 10 4 Vm-1

Electromagnetic Waves 12. The electron placed in the path of electromagnetic wave will experience force due to electric field vector and not due to magnetic field vector.

13. Given, frequency f1 = 7.5 MHz

I=

22. As, or

1 e 0E 02c 2

E0 =

2´4 2I = e 0c (8.85 ´ 10 -12) ´ (3 ´ 10 8) = 55.5 NC-1

Frequency f2 = 12 MHz Speed of EM wave c = 3 ´ 10 8 m/s

23. As, t =

Wavelength corresponding to frequency f1 l=

c 3 ´ 10 8 = f1 7.5 ´ 10 6

3000 = = 4.0 m 7.5 Wavelength corresponding to frequency f2 l=

c 3 ´ 10 8 300 = = 25 m = 6 f2 12 ´ 10 12

Thus, the corresponding wavelength band is 25 m to 40 m.

14. As, v ferrite = and l ferrite =

c 3 ´ 10 8 = 3 ´ 10 6 ms-1 = m r er 10 ´ 10 33 v ferrite 3 ´ 10 6 = 3.33 ´ 10 -2 m = n 90 ´ 10 6

2s 2 ´ 38400 ´ 1000 = 2.5 s = c 3 ´ 10 8

24. We have, y ( x, t ) = 103 sin p(3 ´ 10 6 x - 9 ´ 1014t) = 10 3 sin 3 ´ 10 6 p ( x - 3 ´ 10 8t ) Comparing it with the relation, 2p y ( x, t ) = a sin ( x - ct ); l We note that, c = 3 ´ 10 8 ms-1

25. From a diode antenna, the electromagnetic waves are radiated outwards. The amplitude of electric field vector (E 0) which transports significant energy from the source falls off intensity inversely as the distance (r) from the antenna, i.e., E 0 µ1/ r .

26. As, Power = I ´ area = (1.4 ´ 103) ´ 5

15. Energy flowing per sec per unit area from a face is =

1 [E ´ B]. It will be in the negative z-direction. It shows m0

that the energy will be flowing in faces parallel to xy-plane and is zero in all other faces. Total energy flowing per EBa2 1 second from a face in xy-Plane = (EB sin 90° ) a2 = m0 m0

16.

1/ m 0e 0 c me Refractive index = 0 = = m 0e 0 c 1 / me

17. Electric field intensity on a surface due to incident radiation is, E =

U P U = , where = P = power At A t

\ E µ P (for the given area of the surface) E E ¢ P ¢ 50 1 Hence, = = = or E ¢ = 2 E P 100 2

18. The electromagnetic wave being packets of energy moving with speed of light may pass through the region.

19. In electric and magnetic waves are mutually perpendicular directions and they are in same phase.

Force,

1 v where v is the velocity of electromagnetic wave, which is independent of wavelength of wave but depends upon the nature of medium of propagation of wave.

21. As, l = c / n = 3 ´ 10 8 / 2 ´ 1010 = 1.5 ´ 10 -2 m

Power solar constant ´ area = c speed of light

=

1.4 ´ 10 3 ´ 5 = 2.33 ´ 10 -5 N 3 ´ 10 8

or magnetic field is zero.

28. Given, magnetic field part of harmonic electromagnetic wave B0 = 510 nT E Speed of light in vacuum c = 0 B0 where, E 0 is the electric part of the wave E0 3 ´ 10 8 = 510 ´ 10 - 9 E 0 = 153 N/C

or

Thus, the amplitude of the electric field part of wave is 153 N/C.

29. Velocity of light in a medium, c=

and wt = q0 or w = q0 / t k / w = t / x = 1/ ( x / t) =

F=

27. In electromagnetic wave, the average value of electric field

20. As, kx = q or k = q / x \

1021

30. As, Intensity = =

1 1 = m 0 e 0m r e r me

Power 4.0 ´ 10 26 = Area 4pr 2 4 ´ 10 26 22 4´ ´ (7 ´ 10 8) 2 7

= 4.54 ´ 10 8 Wm-2

1022 JEE Main Physics Total power Solar constant ´ area = Velocity of light Velocity of light

31. Radiation force = =

32.

3

10 ´ (10 ´ 20) = 6.67 ´ 10 -4 N 3 ´ 10 8

53. The density of air in mesosphere with height decreases from 1/103 to 1/105 times that due to the surface of the earth.

54. As we know that the direction of electromagnetic wave is perpendicular to both electric and magnetic fields. Here, electromagnetic wave is travelling in Z-direction, then electric and magnetic fields are in X-Y direction and are perpendicular to each other.

1 As, F = PA, But, P = c IA 6 ´ 12 So, F= = 24 ´ 10 -8 N = c 3 ´ 10 8

Frequency of waves f = 30 MHz = 30 ´ 10 6 Hz Speed c = 3 ´ 10 8 m/s

33. Here, v = 2 ´ 10 8 ms-1, m r = 1 and c = 3 ´ 10 8ms-1

c=fl Wavelength of electromagnetic waves Using the formula,

Speed of electromagnetic wave in the medium is 1 1 1 1 v= = = ´ me m 0m r ( e 0 e 0) m 0e 0 m r er er =

\

(3 ´ 10 8) 2 c2 = 2.25 = 2 (2 ´ 10 8) 2 ´ 1 v

l=

Thus, the wavelength of electromagnetic waves is 10 m.

57. Range, R = 2hr , where r is the radius of earth. So, R µ h1/ 2 .

34. Number of photons, 10 ´ 10 3 P = 1.171 ´ 10 31 n= = hn 6.6 ´ 10 -34 ´ 880 ´ 10 3

59. Distance, d = 2hR or h = d 2/2R 60. Ozone layer blocks the high energy radiations like UV (3 ´ 10 -7m ).

36. According to Daun-Hunt law, the wavelength of X-rays lies between minimum to certain limit.

62. The earth’s atmosphere above the height of 80 km up to 400 km is called ionosphere.

38. Radioactive source, X-ray tube, sodium vapour lamp, crystal oscillator.

39.

63. Distance, d = 2hR

hc 6.6 ´ 10 -34 ´ 3 ´ 10 8 As, V = » 10000 V = el 1.6 ´ 10 -19 ´ 10 -10

Population covered = pd 2 ´ population density = 3.114 ´ (2 ´ 0.1 ´ 6.37 ´ 10 3) ´ 1000 » 40 lakh

40. X-rays are produced when there is vacancy for the electron on inner complete orbits of an atom and jump of electrons takes place from higher orbit to lower orbit.

41. The

wavelengths of 8 ´ 10 -7 m and 10 -3 m.

infrared

rays

lie

64. It lies in ultraviolet region. 65. The electric field intensity produce by radiations coming from 50 W bulb at the same distance is 2E .

between

6.6 ´ 10 -34 ´ 3 ´ 10 8 = 0.94 ´ 10 -24 J 21 ´ 10 -2 = 10 -24 J

66. As, E = hc / l =

42. Generally, temperature of human body is 37°C, corresponding to which intrared and microwave radiations are emitted from the human body.

67. Molecular spectra due to vibrational motion lie in the

44. X-rays being of high energy radiations, penetrate the target and hence these are not reflected back.

68. 46. As, eV = hc / l or

47.

51.

h = hc / eV i. e, l µ1/ V

c 3 ´ 10 8 Here, l = = = 36.6 m v 8.2 ´ 10 6

48. As n =

c 3 ´ 10 8 300 = = 10 m = 6 f 30 ´ 10 30

microwave region of EM spectrum due to Kirchhoff’s law in spectroscopy the same will be absorbed. hc hc As, eV = or l = l eV

69. When plane electromagnetic wave is incident on the material surface the wave delivers some momentum and every to the surface and hence, P ¹ 0 and E ¹ 0.

70. Electric energy stored in charge capacitor

-3

10 m 1 mm x = = = 0.25 ´ 10 4 l 4000 Å 4000 ´ 10 -10

E 6.3 As, E = = = 2.1 ´ 10 8 T c 3 ´ 10 8 Since Bis perpendicular to the direction of propagation of EM waves as well as the electric field.

U=

1 q2 1 q2 1 q = = ´ Ad 2 C 2 e 0 A / d 2 e 0 A2

\ Mean electric energy density between the plates of a U 1 q2 charge capacitor, u = = Ad 2 e 0 A2

71. Here,

l = 3000 Å = 3 ´ 10 -7m

Electromagnetic Waves 71. As E =

hc 6.6 ´ 10 -34 ´ 34 ´ 3 ´ 10 8 = l 3 ´ 10 -7 = 6.6 ´ 10 =

-19

75. Here, E 0 = 18 Vm-1, B0 = ? B0 = E 0 /c = 18 / 3 ´ 10 8 = 6 ´ 10 -8 T

J ( Ws)

76. As,

-19

6.6 ´ 10 60 ´ 60

= 1.83 ´ 10

Wh

-22

77. As, velocity of light is perpendicular to the wavefront and

Wh

light is travelling in vacuum along the y-axis. Then, the wavefront is represented by y = constant.

The average density due to electric field 2

1 1 æE ö 2 uE = e 0 E rms = e0 ç 0 ÷ 2 2 è 2ø 1 1 = e 0E 02 = ´ (8.85 ´ 10 -12) ´ (10) 2 4 4 = 2.21 ´ 10 -10 Jm-3

or Now, pressure,

travelling in z-direction. E, B and k form a right handed ^

^

^

^

system k is along z-axis. (As, i ´ j = k). ^

Þ

79. As, v =

F´s F´c = A´I A

^

^

E x i + By j = c k 3 ´ 10 8 c = £~ - 1.8 ´ 10 8 ms-1 me 0 1.3 ´ 2.14

80. Speed of light in vacuum, c =

F I = A c F I p= = A c

1 m 0e 0

...(i)

and in another medium v=

74. The average electric energy density of the plane electromagnetic wave travelling in vacuum is 2

uE =

78. E x and By would generate a plane electromagnetic wave

i. e. ,E is along x-axis and B is along y-axis.

energy As, intensity, I = area ´ time =

2p E0 and w = 2pn = c also k = l B0

Their relation gives, E 0k = B0 w

72. Here, E 0 = 10 Vm-1, n = 5 ´ 1014 Hz

73.

1023

1 1 1 æE ö 2 e 0 E rms = e 0 ç 0 ÷ = e 0E 02 4 2 2 è 2ø

1 me

…(ii)

Dividing Eq. (i) by (ii), we get c v= mc where m = ratio of two permittivities and K = dielectric constant (ratio of two permittivities)

Round II m = has the dimensions of resistance, hence it is called e the intrinsic impedance of the medium. dQ d dV As, i = = 2 ´ 10 -12 ´ 10 12 = 2 A = (CV ) = C dt dt dt

1. As,

2.

3. As,

Q CV æV ö or iD = C ç ÷ where, iD = displacement current = èt ø t t

or

V iD 1.0 = = -6 Vs-1 = 10 6 Vs-1 t C 10

4. K.G. Jansky discovered accidently the radio signals coming from outside the atmosphere and reaching the earth.

5. Velocity of photon in vacuum is constant for all frequencies. 6. Now a days microwaves are used to locate the flying objects by radar.

7. Here, energy flux = 1.0 kWm-2 = 1.0 ´ 103 Wm-2 Area of plate, A = 10 cm2 = 10 ´ 10 -4 m2 = 1.0 ´ 10 -3 m2

and Time,

t = 20 min = 20 ´ 60 s = 1200 s

Now total power = flux ´ area ´ time = 10 3 ´ 10 -4 ´ 20 ´ 60

\

= 120 watt Power 120 Force = = = 4 ´ 10 -7 N Velocity 3 ´ 10 8

8. Distance, d = 2hR or d µ h1/ 2 9. Diffraction takes places when the wavelength of wave is comparable with the size of the obstacle in path. The wavelength of radio waves is greater than the wavelength of light waves. Therefore, radio waves are diffracted around building.

10. The X-rays has the shortest wavelength along the following radiations.

11. Use method of dimensions. Equating the dimensions of two sides, concludes that the relation (a) is dimensionally correct.

1024 JEE Main Physics 12. Equation second shows that the electromagnetic wave 13.

travels along the positive x-axis. df As, ò E × dl = - B dt d or E ´ 2pr = (kt ´ pr 2) = kpr 2 dt kr or E= 2

14. As, C =

= 8.85 ´ 10 Now,

= (Q f = BA)

24. Velocity

of an electromagnetic wave 1 8 -1 c= = 3 ´ 10 ms is independent of amplitude of m 0e 0

F

electromagnetic, wave frequency and wavelength of electromagnetic wave.

= 2.2 ´ 10 -6 A = 2.2 mA

15. In an electromagnetic wave, the average energy density of magnetic field mB = average energy density of electric field 1 vE = e 0E 02 4 1 = ´ (8.85 ´ 10 -12) ´ 12 4

25. Number of oscillations in coherence length =

26. Here, Ftotal = Fre + Fabs 1.2P 0.4P 1.6P + = c c c 1.6 ´ 200 = 1.07 ´ 10 -6 N = 3 ´ 10 8

=

2I 2 ´1 = 6.67 ´ 10 -9Nm-2 = c 3 ´ 10 8

27. Total power = Solar constant ´ Area

17. The speed of electromagnetic waves in a medium is,

= 10 4 ´ (10 ´ 10) = 10 6 W

c 1 1 3 ´ 10 8 v= = = = me m 0 e r e 0m r m 0e 0 4 ´ 2.25

28. Energy contained in a cylinder,

= 1 ´ 10 8ms-1

U = average energy density ´ volume 1 = e 0E 02 ´ Al 2 1 = ´ (8.85 ´ 10 -12) ´ (50) 2 ´ (10 ´ 10 -4) ´ 1 2

18. Ths speed of electromagnetic waves in vacuum, 1 = 3 ´ 10 -8 ms-1 m 0e 0

Air acts almost as vacuum, So,

8

= 1.1 ´ 10 -11 J

8

3 ´ 10 = a ´ 10 Þ a = 3

29. Radiation force = momentum transferred per sec by

1 2

19. Average intensity, Iav = e 0E 02 ´ c

electromagnetic wave to the mirror

8.85 ´ 10 -12 ´ (36) 2 ´ 3 ´ 10 8 = 2

=

20. Momentum transferred in one second by electromagnetic

30. The electric field induced by changing magnetic field

wave to the mirror is 2S av A 2 ´ 6 ´ (30 ´ 10 -4) = c 3 ´ 10 8

31.

= 1.2 ´ 10 -10 kg ms-1

depends upon the rate of change of magnetic flux. Hence it is non-conservative. pressure p As, intensity, I = = arc 4 pr 2 = average energy density ´ velocity 1 = e 0E 02c 2

-1

21. Here, an amplitude of electric field, E 0 = 100 Vm , amplitude of magnetizing field, H0 = 0.265 Am-1. We know that the maximum rate of energy flow S = E 0 ´ H0 = 100 ´ 0.265 = 26.5 Wm-2

2S av A 2 ´ (10) ´ (20 ´ 10 -4) = c 3 ´ 10 8

= 1.33 ´ 10 -10 N

= 1.72 Wm -2

p=

l 0.024 = l 5900 ´ 10 -10

= 4.068 ´ 10 6

= 2.21 ´ 10 -12 Jm-3

c=

3.9 ´ 10 26 = 5.6 ´ 10 7 Wm-2 22 8 2 4´ ´ (6.96 ´ 10 ) 7

23. The electric and magnetic field vectors are in the same phase

dQ d dV i = (CV ) = C = 8.85 ´ 10 -8 ´ 25 dt dt dt

16. Here, pressure p =

P 3.9 ´ 10 26 = A 4 pr 2

in electromagnetic wave but their orientation is perpendicular to each other as well as perpendicular to the direction of propagation of electromagnetic wave.

e 0KA (8.85 ´ 10 -12) ´ 10 ´ 1 = d 10 -3 -8

22. Intensity =

\

E0 =

2p p = 2 4pe 0r c 2pe 0r 2c

Electromagnetic Waves 32. Intensity of electromagnetic wave is I = or

E0 =

A = p (d) 2 = p (2hR) = 2phR

Pav E2 = 0 2 2m 0 c 4p ´ r

= 2 ´ 3.14 ´ 100 ´ 6.37 ´ 10 6 Average populations density = 100 km-2

m 0 cPav ( 4p ´ 10 -7) ´ (3 ´ 10 8) ´ 800 = 2p ´ ( 4) 2 2 pr 2

= 1000 ´ (1000) -2 = 10 -3 m-2 \Population covered = 2 ´ 3.14 ´ 100 ´ 6.37 ´ 10 6 ´ 10 -3

= 54.77 Vm-1

= 4 ´ 10 6

33. The required wave in z-direction, is E z = E 0 sin (kx - wt + f) L E 2 = E 0 × ei ( kx- wt + f ) L

Þ

34. As B µ r , since the point is on the axis, where r = 0 , so B = 0. 35.

P 1 Average intensity = = e 0E 02c 2 2 4 pR or

E c

40. As B = =

3 2 ´ 3.14 ´ 25 ´ 8.85 ´ 10 -12 ´ 3 ´ 10 8

48 = 16 ´ 10 -8 Wbm-2 3 ´ 10 8

41. The height, h of the transmitting antenna is given by h =

42. The average energy density ue is given by 1 1 1 e 0E 2 = e 0(E 0 / 2) 2 = e 0E 2 2 2 4 1 -12 2 = ´ (8.85 ´ 10 ) ´ 1 4 = 2.2 ´ 10 -12 Jm-3

ue =

= 2.68 Vm-1

36. Here, l = 3 mm = 3 ´ 10 -3m, E 0 = 66 Vm-1 \

B0 =

E0 66 = = 2.2 ´ 10 -7 T c 3 ´ 10 8

As, electromagnetic wave is propagating along x-axis and electric field oscillation is along y-direction, the magnetic field oscillation is along z-direction using the relation for harmonic wave 2p E y = E 0 cos ( ct - x) l 2 pc (t - x / c) E y = E 0 cos l \

E y = 66 cos

2p ´ 3 ´ 10 8 (t - x / c) 3 ´ 10 -3

43. The intensity of electromagnetic induction, I=

Bz = B0 cos

\

E = (E1$i + E 2$j ) cos (kz - wt )

37. If l is radius of the particle, then l = 3 ´ 10 -4 ´ 10 -2 m Frequency of electromagnetic wave ,

Then, to observe the particle, the frequency of wave should be maximum than10 14 Hz i. e. ,10 15 Hz.

38. As q = n =

c 3 ´ 10 8 = = 1.5 ´ 10 8 Hz l 0.2

39. The distance d upto which TV transmission can be viewed is given by d = 2hR Area in which TV transmission can be viewed

4p ´ 10 -7 ´ 3 ´ 10 8 ´ 800 2 ´ 3.14 ´ (3.5) 2

44. In electromagnetic wave, the electric field vector is given as,

2 pc (t - x / c) l

c 3 ´ 10 8 = 10 14 = l 3 ´ 10 -6

æ m cP ö E m = ç 0 2av ÷ = è 2 pr ø = 62.6 Vm-1

= 2.2 ´ 10 -7 cos 2p ´ 10 11(t - x / c)

n=

Pav E2 = m 2 2mc 4 pr

where, E n = amplitude of electric vector oscillator

= 66 cos 2p ´ 10 11(t - x / c) and

r2 2Re

where, Re = radius of the earth. and radius to covered, (128 ´ 10 3) 2 = 1280 m h= 2 ´ (6.4 ´ 10 6)

P E0 = 2pR 2e 0 c =

1025

45.

In electromagnetic wave, the associated magnetic field vector, E (E $i + E 2$j ) B= = 1 cos (kz - wt ) c c As E and B are perpendicular to each other and the propagation of electromagnetic wave is perpendicular to E as well as B, so the given electromagnetic wave is plane polarised. E 54 As, B0 = 0 = = 18 ´ 10 -8 T c 3 ´ 10 8 \

l=

c 3 ´ 10 8 = 1.5 ´ 10 -2 m = 1.5 cm = n 2 ´ 10 10

46. The equations (a) and (c) are related to source of electric field and equation (d) with source of magnetic field.

47. Suppose an electromagnetic wave is travelling along negative z-direction. Its electric field is given by E = E 0 cos (kz - wt ) which is perpendicular to z-axis. It acts along negative y-derection.

1026 JEE Main Physics The associated magnetic field B in electromagnetic wave is along x-axis, i.e., along k$ ´ E. E 1 B0 = 0 \ B = (k$ ´ E) c c

As

54. As, electromagnetic wave is propagating along x-direction and magnetic field vector is along y-direction, hence electric field vector is along z-direction since electromagnetic wave is of transverse nature. Therefore, æ 2 px 2 pt ö + E 0 = E 0 sinç ÷ è l T ø

The associated electric field can be written in terms of $. magnetic field as E = c (B ´ k)

48. Here, n = 10 9Hz, l = c / v = 3 ´ 10 8 / 10 9 = 0.3 m This wavelength radiation ( = 0.3 m) or frequency radiation 10 9 Hz falls in the region of radio waves.

\

55. Intensity of electromagnetic wave is I=

49. Radiation pressure (p) is the force exerted by electromagnetic wave on unit area of the surface, i.e., rate of change of momentum per unit area of the surface. intensity I Momentum per time per unit area = = speed of wave c Change in momentum per unit time per unit area = DI / c = radiation pressure (p), i.e., p = DI / c. Momentum of incident wave per unit time per unit area I / c . When wave is fully absorbed by the surface, the momentum of the reflected wave per unit time per unit area = 0. Radiation pressure ( p) = change in momentum per unit time DI I I per unit area = = -0 = c c c When wave is totally reflected, then momentum of the reflected wave per unit time per unit area = - I / c . Radiation pressure ( p) =

I æ I ö 2I - ç- ÷ = c è cø c

E z = 60 sin(0.5 ´ 10 3 x + 1.5 ´ 10 11t ) Vm-1 1 B02 1 (2 ´ 10 -7) 2 ´ (3 ´ 10 8) c= ´ 2 m0 2 ( 4p ´ 10 -7)

» 5 Wm-2

56. As, B0 = 2 ´ 10 -7 T; \

E 0 = cB0 = (3 ´ 10 8) ´ (2 ´ 10 -7) = 60 Vm-1

57. Given By = 2 ´ 10 -7 sin (0.5 ´ 103 x + 1.5 ´ 1011 t) æ 2px 2pvt ö Compare it with, By = B0 sin ç + ÷ è l l ø æ 2 p 2 pt ö = B0 sinç + ÷ è l T ø 2p we get, = 0.5 ´ 10 3 l 2 or l= = 1.26 ´ 10 -2 m 0.5 ´ 10 3 » 1.3 ´ 10 -2 m

58. Total electromagnetic power of the sun falling on earth = electromagnetic flux ´ area of the earth

Here, p lies between I / c .

= 10 3 ´ 4p ´ (6.4 ´ 10 6) 2

50. The electromagnetic waves are produced due to sinusoidal variation of electric field vector and magnetic field vector perpendicular to each other as well as perpendicular to the direction of propagation of waves. Since electromagnetic waves are propagating along z-direction therefore E and B should be either along x-axis and y-axis along y-and x-axis respectively. c v

51. As, m = =

1/ m 0e 0 me

=

me m rm 0 e 0 e r = = m r er m 0e 0 m 0e 0

52. In electromagnetic waves average energy due to electric field is equal to that due to magnetic field. Also the electric vector E and magnetic vector B show sinusoidal variation with same frequency is same phase. E and B are in one plane which in perpendicular to the propagation of electromagnetic waves.

53. From the wave equation, clearly, or

n=

2 = 1.5 ´ 10 11 T

1 1.5 ´ 10 11 = T 2p

» 2.4 ´ 10 10 Hz

= 5.1 ´ 10 17 W

59. Total electromagnetic power of the sun = 10 3 ´ 4p ´ (1.5 ´ 10 11) 2 = 5.6 ´ 10 26 W

60. Total electromagnetic power incident on roof on the earth = 10 3 ´ area of roof = 10 3 ´ (8 ´ 10) = 8 ´ 10 4 W

61. Radiation force on the roof =

Total power on roof 8 ´ 10 4 = Velocity of light 3 ´ 10 8

= 2.67 ´ 10 -4 N

65. Assertion is true false but Reason is because increase isozone does not increase the ultraviolet relation on the earth.

66. The changing electric field produces a magnetic field. It is accounted by Ampere-Maxwells law.

68. In an electromagnetic wave, the direction of B ^ E. 69. The magnetic lines of force form a closed path, hence the line integral of B over a closed path is zero.

Electromagnetic Waves 71. Both energy of X-rays is maximum as compared microwave and ultraviolet rays.

73. Both A and R are true, but R is not correct explanation of A. In fact, the energy of g-rays is more than X-rays because the frequency of g-rays is higher than that of X-rays and E = hn.

74. The optical waves used in optical fibre communication and better carrier of signals than microwaves. The speed of microwave and optical wave is the same in the vacuum.

1027

78. Ultraviolet radiations are electromagnetic waves. The wavelength of these waves ranges between 4000 Å to 100 Å that is of smaller wavelength and higher frequency. They are absorbed by ozone layer of stratosphere in atmosphere. They cause skin diseases and they are harmful to eye and may cause permanent blindness.

79. Earth is heated by sun’s infrared radiation. The earth also

radio waves, but it absorbs X-rays. Thus, X-rays telescope cannot be used on surface of earth.

emits radiation most in infrared region. These radiations are reflected back by heavy gases like CO2 is atmosphere. These back radiation keep the earth’s surface warm at night. This phenomenon is called green house effect. When the atmosphere were absent then temperature of earth falls.

76. The electromagnetic wave contains sinusoidally time varying

80. In an electromagnetic wave, the direction of propagation of

electric and magnetic fields which act perpendicularly to each other as well as at right angle (90°) to the direction of propagation of waves, so it is quite clear that electromagnetic waves are transverse in nature. The two components of field may be represented as xö æ E = E 0 sin wçt - ÷ è vø

wave, electric field and magnetic field are mutually perpendicular to each other i. e. , the wave propagates perpendicular to E and B or along E ´ B. While polarisation of wave takes place parallel to electric field vector.

75. The earth’s atmosphere is transparent to visible light and

xö æ B = B0 sin wçt - ÷ è vø

77. In sky wave propagation the radio waves which have frequency between 2 MHz to 30 MHz, are reflected back to the ground by the ionosphere. But radio waves having frequency greater than 30 MHz cannot be reflected by the ionosphere because at this frequency they penetrates the ionosphere. It makes the sky wave propagation less reliable for propagation of TV signal having frequency greater than 30 MHz. Critical frequency is defined as the higher frequency that is returned to the earth by the ionosphere. Thus, above this frequency a wave whether it is electromagnetic will penetrate the ionosphere and is not reflected by it.

81. Given, y( x, t) = e-( ax = e-(

2

+ bt 2 + 2 ab xt )

ax +

bt )2

It is a function of type, y = f ( wt + kx) \ y( x, t ) represents wave travelling along x-direction. speed of wave =

w = k

b = a

b a

82. b-rays are beams of fast moving electrons. 83. E and B are mutually perpendicular to each other and are in phase i. e. , they become zero and minimum and the same place and at the same time.

84. The magnetic field vector points towards east. 85. It is in the phase and perpendicular to each other.

23

Ray Optics and Optical Instruments JEE Main MILESTONE Reflection of Light Spherical Mirrors Refraction of Light Refraction from a Spherical Surface

Lens Deviation by Prism Dispersion by a Prism Optical Instruments

23.1 Reflection of Light When a beam of light is incident on a polished interface, it is thrown back in same medium. This phenomenon is called reflection. In reflection, the frequency, speed and wavelength do not change, but a phase change may occur depending on the nature of reflecting surface. Experimentally, it is found that the rays corresponding to the incident and reflected waves make equal angles with the normal to the surface. Thus, the two laws of reflection can be summarized as under. 1. Ð i = Ðr 2. Incident ray, reflected ray and normal lie on the same plane

ra en t

ed

id

ra

y

i r

Re fle ct

In c

y

Normal

Note The above two laws of reflection can be applied to the reflecting surfaces which are not even horizontal.

Reflection from Plane Mirror (Surface) In case of reflection from plane surface such as plane mirror (i) The image is always erect, virtual and of exactly the same size as the object. The image is formed as much behind the mirror as the object is in from of it. (ii) The image is laterally inverted. (iii) If keeping the incident ray fixed, the plane mirror is rotated through an angle q, the reflected ray turns through double the angle i.e., 2q in that very direction. (iv) If the object is fixed and the mirror moves relative to the object with a speed v, the image moves with a speed 2v relative to the object.

The light is that form of energy which makes objects visible to our eyes. The branch of physics, which deals with nature of light, its sources, properties, measurement, effects and vision is called optics. For the sake of convenience, study of optics is generally divided into two parts namely (i) geometrical optics or ray optics, and (ii) wave optics. This chapter deals with the geometrical optics.

Ray Optics and Optical Instruments (v) If the mirror is fixed and the object moves relative to the mirror with a speed v, the image also moves with the same speed v relative to the mirror.

Sample Problem 3

Two plane mirrors are inclined at 30º as shown in figure. A light ray is incident at angle 45º. Find total deviation produced by combination of mirror after two successive reflection.

(vi) Deviation suffered by a light ray incident at an angle Ð i is given by d = (180 - 2i) (vii) If there are two mirrors inclined at an angle d, the total number of images formed for an object kept 2p 2p between the two is equal to or æç - 1ö÷ , which ever è q ø q is odd. (viii) The minimum size of a mirror required to see the full image of a person, is half the height of the person. (ix) If a plane mirror is rotated about an axis perpendicular to plane of mirror, then reflected ray image do not rotate.

M2

45°

30°

(a) 60° ­ (c) 50° ­

Interpret (a) Deviation at mirrorM1, d1 = 180° – 2 ´ 45° = 90° ­ Deviation at mirror M2, d2 = 180° – 2 ´ 15° = 150° ­ Total deviation d = d2 – d1 = 150° – 90° = 60° ­ M2

on the wall of a room in which an observer at the centre of room may see the full image of the wall of height h behind him is h 3 2h (c) 3

45° d1

(d) h

M1M2 x = h 2x + y

A

A′ M1 y

h

O

x

x

O′

A

B′

hx M1M2 = (2x + y) x = y , then M1M2 =

h 3

Sample Problem 2 Find velocity of image when object and mirror both are moving toward each other with velocity 2 ms-1 and 3 ms-1 respectively. (a) 8 ms-1

(b) - 8 ms-1

(c) - 5 ms-1

(d) 5 ms-1

Interpret (a) Here,

Incident light

Incident light

B

If

(i) Concave mirror If reflection takes place from the inner surface, the mirror is called concave [Fig. (a)]. (ii) Convex mirror If reflection takes place from the outer surface, the mirror is called convex [Fig. (b)].

M2

Size of mirror,

23.2 Spherical Mirrors Mirrors having their reflecting surface spherical are called spherical mirrors. Spherical mirrors are of two types

h1d1 = h2d 2

vOM = vIM vO – v M = (vI – v M )

Þ

15°

30°

Interpret (a) From D O ¢ M1M2 and DO ¢ AB, we get

Þ

d2

h 2

(b)

M1

(b) 58° ­ (d) 68° ­

Sample Problem 1 The minimum size of the mirror fixed

(a)

1029

( +2ms-1) – (–3ms-1) = vI + ( -3 ms-1) vI = 8 ms-1

C P

+ ve

P

C

B (a) Concave mirror

(b) Convex mirror

Definitions of Some Terms Related to Spherical Mirrors Centre and radius of curvature The centre of curvature and radius of curvature of a mirror are the centre and radius of the sphere of which the mirror is a part. In the given figure, AC is the radius of curvature and C, the centre of curvature.

A

C

P

1030 JEE Main Physics Pole Pole of the spherical mirror is mid-point of

its

Concave

Convex

reflecting surface. In figure it is shown by P.

Principal axis The principal axis of a spherical mirror is the

C

line joining the pole and centre of curvature. In the figure, PC is principal axis.

C

2. A ray parallel to principal axis is reflected through the focus, and vice-versa. Also, mutually parallel rays after reflection intersect on the focal plane.

Principal focus Principal focus is a point on the principal axis of the mirror at which the light rays coming parallel to principal axis actually meet after reflection or appear to meet.

F C

P

F

F

P F

C

3. The light coming through the focus of mirror or coming towards focus, becomes parallel to principal axis. (a) Concave mirror

(b) Convex mirror

For concave mirror focus, is infront of the mirror, while for convex mirror, focus is behind the mirror. Focus of concave mirror is real, while focus of convex mirror is virtual.

F

F

Focal length The distance between pole and focus of a spherical mirror is called its focal length. It is represented by f. R i.e., f = 2

Sign Convention for Mirrors According to the sign convention, (i) Origin should be placed at the pole (P). (ii) All distances should be measured from the pole (P). (iii) Object distance is denoted byu , image distance byv, focal length byf and radius of curvature by R. (iv) Distance measured in the direction of incident ray are taken as positive, while in the direction opposite of incident ray are taken negative.

Ray Tracing In geometrical optics, to locate the image of an object. Tracing of a ray as it reflects or refracts, is very important. 1. A ray going through centre of curvature is reflected back along the same direction.

Table 23.1 Image Formation by Concave Mirror S. No. 1.

Position of Object

Ray Diagram

Details of Image

M

At infinity

P

F

C

Real, inverted, very small (m – 1) between 2F and infinity

M

Between F and C O C

P

F

M′

5.

At F

Real, inverted, very large [m ® (¥)] at infiinity

M O

6.

P

F

C

Virtual, erect, large in size (m > + 1) behind the mirror

Between F and P M O C

P

F

M

Table 23.2 Image Formation by Concave Mirror S. No. 1.

Position of Object

Ray Diagram

At infinity

Details of Image Virtual, erect, very small (0 < m < < + 1) at F

M

I P

F

C

M'

2.

In front of mirror

Virtual, erect, diminished (m < + 1) between P and F

M

F O

P M'

I

C

1032 JEE Main Physics Sample Problem 4 An object is placed at 10 cm in front of a concave mirror of radius of curvature 15 cm. Which one of the following statement regarding the nature of image is true? (a) Image is small, imaginary and inverted (b) Image is magnified, real and inverted (c) Image is magnified, false and straight (d) Image is small, imaginary and straight

(a) – 5 cm (c) 6 cm

(b) 5 cm (d) – 6 cm

Interpret (a) The focal length, F = - f

Interpret (b) Given focal length, f=

Sample Problem 5 An object of length 2.5 cm is placed at 1.5f from a concave mirror, where f is the focal length of the mirror. The length of the object is perpendicular to the principal axis. Find the length of the image. Is the image erect or inverted?

u = -1.5f

and

-15 = - 7.5 cm 2

f

Object O O

I

C

1.5 f

Image

From mirror formula,

The object distance, u = -10 cm, then from mirror formula 1 1 1 + = v u f 1 1 1 + = v -10 - 7.5 1 1 1 =+ v 7.5 10 10 ´ 7.5 v= = - 30 cm -2.5

Þ Þ

The image is 30 cm from the mirror on the same side as the object. v ( -30) Also magnification, m= - == -3 u ( -10) Hence, image is magnified, real and inverted.

Mirror Formula The relation among object distance (u), image distance (v) and the focal length ( f ) of mirror (of any time) can be established as, 1 1 1 = + f v u

we have, 1 1 1 1 1 1 + = or + =u v f -15 . f v f 1 1 1 1 = - =v 15 . f f 3f m=

or

h2 = - 2 or h2 = - 2h1 = - 5 cm h1

The image is 5 cm long. The minus sign shows that it is inverted.

Sample Problem 6 A concave mirror of focal length 10 cm and a convex mirror of focal length 15 cm are placed facing each other 40 cm apart. A point object is placed between the mirrors, on their common axis and 15 cm from the concave mirror. Find the position of the image produced by the successive reflections, first at concave mirror and then at convex mirror. (b) + 10 cm

(a) 6 cm

f = + 15

I v f f -v = = = O u f -u f

P2 I2

O = size of object perpendicular to principal axis. 2

æ f - vö dv v2 æ f ö = = =ç ÷ ÷ =ç du u2 è f - u ø è f ø

2

2

æ f - vö AI v2 æ f ö = =ç ÷ ÷ =ç AO u2 è f - u ø è f ø

2

where, AI = area of image and AO = area of object.

(d) + 30 cm

I1

O 15cm

M

P1 15cm

M¢

Given, So,

Areal magnification, mar =

f = – 10 40 cm

M

Axial magnification,

(c) + 15 cm

Interpret (a) According to given problem, for concave mirror.

where, I = size of image perpendicular to principal axis

max

3f v = = -2 . f u 15

Now,

Lateral magnification, m=

F

P

u = -15 cm and f = - 10 cm 1 1 1 + = i. e. , v = - 30 cm v -15 -10

i.e., concave mirror will form real, inverted and enlarged image I1 of object O at a distance 30 cm from it, i.e., at a distance 40 - 30 = 10 cm from convex mirror. For convex mirror, the image I1 will act as an object and so for it u = -10 cm and f = + 15 cm.

Ray Optics and Optical Instruments 1 1 1 + = i. e. , v = + 6 v -10 15

n e1

So, final image I2 is formed at a distance 6 cm behind the convex mirror and is virtual as shown in above figure.

Sample Problem 7 Lower half of concave mirror’s reflecting surface is covered with an opaque (non-reflective) material. The intensity of an image of an object placed in front [NCERT] of the mirror is

1033

µ1 µ2 e2

(ii) Now, we can write Snell’s law as, m sin i = constant

...(i)

For two media, m1 sin i1 = m 2 sin i2

...(ii)

(iii) Snell’s can be written as

Object

1m 2

Image

Concave mirror

(a) twice the original image (b) half the original image (c) same as the original image (d) one-fourth the original image

Interpret (b) One may think that the image will now show only half of the object, but according to the laws of reflection to be true for all points of the remaining part of the mirror, the image will be that of the whole object. However, as the area of the reflecting surface has been reduced, the intensity of the image will be low (in this case half).

Check Point 1 1. Does the mirror formula hold good for a plane mirror? 2. An object is placed between two plane parallel mirrors. Why do the distant images get fainter and fainter?

3. Why are mirrors used in search-lights parabolic and not concave spherical?

=

sin i1 v1 l1 m1 = = = sin i2 v2 l 2 m 2

Here, v1 is the speed of light in medium 1 and v2 in medium 2. Similarly, l1and l 2 are the corresponding wavelengths. n i1

i1 Rarer Denser

1 2 i2

i2

i1 > i2 v2 < v1 m2 > m 1 l2 < l1

i1 < i2 v2 > v1 m2 < m1 l2 > l1

If m 2 > m1, then v1 > v2 and l1 > l 2, i.e., in a rarer medium, speed and hence, wavelength of light is more. (iv) 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 speed of light in vacuum c

m=

4. If you were driving a car, what type of mirror would you prefer to use for observing traffic at your back?

23.3 Refraction of Light

(i) If medium 1 is a vacuum (or in practice air) we refer 1 m 2 as the absolute refractive index of medium 2 and denote it by m 2 or simply m (if no other medium is there).

speed of light in medium

=

v

Sample Problem 8 Light is incident from air on oil at an angle of 30º. After moving through oil-1, oil-2, and glass it enters water. If the refractive indices of glass and water are 1.5 and 1.3, respectively, find the angle which the ray makes with normal in water.

When light passes from one medium, say air, to another medium, say glass, a part is reflected back into the first medium and the rest passes into the second medium. When it passes into the second medium, it either bends towards the normal or away from the normal. This phenomenon is known as refraction.

Laws of Refraction (Snell’s Law)

Denser Rarer

1 2

30° Air Oil-1 Oil-2 Glass Water r

æ 1ö (a) sin -1 ç ÷ è 2.6 ø

æ 3 ö (b) sin -1 ç ÷ è 2.6 ø

æ 1ö (c) sin -1 ç ÷ è 3.6 ø

(d) sin -1 (2.6)

1034 JEE Main Physics Interpret (a) As we know,

Apparent Shift of an Object due to Refraction

m sin i = constant Þ

m air sin i(air) = m(glass) sin r(glass) sin r(glass)

Again,

m = air sin iair m glass

...(i)

m glass sin iglass = m water sin iwater

...(ii)

From Eqs. (i) and (ii), we get sin 30° = 13 . sin r sin r =

1 1 æ 1ö = , r = sin -1 ç ÷ è 2.6 ø . 2 ´ 13 2.6

Due to bending of light at the interface of two different media, the image formation due to refraction creates an illusion of shifting of the object position. Consider an object O in medium. After refraction, the ray at the interface bends. The bent ray, when it falls on our eyes, is perceived as coming from I. For nearly normal incident rays, and will be very small.

Rarer µ2

Sample Problem 9 A ray of light is incident on a transparent glass slab of refractive index 1.62. If the reflected and refracted rays are mutually perpendicular, what is the angle of incidence? i

A B I

r

θ2

θ1

Denser µ1

O 90°

r´

tan q1 = sin q1 = (a) 58.3° (c) 60º

Similarly,

(b) 85.3º (d) 65º

sin q2 =

Interpret (a) Let the angle of incidence angle of reflection and angles of refraction be i, r and r¢, respectively.

Þ

Now, as per the question 90° - r + 90° - r ¢ = 90° Þ

r ¢ = 90° - i

(because i = r)

Þ

In case of reflection, according to Snell’s law, we have sin i = m sin (90°-i)

Þ

tan i = m

or

-1

i = tan [m ] = tan -1(162 . ) = 58.3°

Sample Problem 10 Refractive index of glass with respect to water is 1.125. If the absolute refractive index of glass is 1.5, find the absolute index of water. (a) 1.33 (c) 0.33

(b) 2.33 (d) 0.44

Interpret (a) Here, the refractive index of glass with respect to water i.e., w mg = 1125 and absolute refractive index of glass . mg = 15 . . We know that, a mg a mw = w mg =

1.5 = 1.33 1.125

AB Image distance from the refracting surface sin q1 m = 1m 2 = 2 sin q2 m1 AB OB

AB m 2 = BI m1 BI Apparent depth = OB Real depth m = 2 m1

1sin i = m sin r ¢ or

AB Object distance from the refracting surface

So, Shift = Real depth–Apparent depth = Real depth

Case I If

m1 < m 2

Shift becomes negative, image distance > object distance, i.e., image is further from the refracting surface.

Case II If

m1 < m 2

Shift becomes positive, image distance < object distance, i.e., image is closer to the refracting surface.

Case III If m 2 = 1 or m1 = m 1ö æ Shift = Real depth ç1 - ÷ è mø

Ray Optics and Optical Instruments

Refraction through a Glass Slab

From the figure, AB =

Let a glass slab of thickness t, and refractive index m be taken, an object is placed at O.

Since,

C

C

M

E

E

I1 O I A

A

B

B O

d = t [sin i - cos i tan r] sin i sin i or sin r = m= m sin r

Further,

P

M

I1

Normal shift

\

(as, AC = t)

d = AB sin (i - r ) t [sin i cos r - cos i sin r] = cos r

N

N

AC t = cos r cos r

1035

tan r =

sin i 2

m - sin2 i

The expression for d now is µ D

D

t

F

Plane surface CD forms its image (virtual) at I1. This image acts as object for EF, which finally forms the image (virtual) at I. Distance OI is called normal shift and its value is

Proof Let,

OA = x (Refraction from CD)

AI1 = mx BI t BI = 1 = x + m m

(Refraction from EF)

OI = ( AB + OA) - BI

\

æ m - 1ö Note For small angles of incidence d = ti ç ÷. è m ø

Sample Problem 11 A vessel has a concave mirror of focal length 30 cm placed at the bottom. It is filled with water upto 20 cm. The position of the image of the sun from the surface of water is

1ö æ OI = ç1 - ÷ t è mø then,

æ ö cos i ÷t sin i d = ç 1ç ÷ m 2 - sin2 i ø è

F

t

tö æ 1ö æ = (t + x ) - ç x + ÷ = ç1 - ÷ t è mø è mø

(a) 7.5 cm above surface of water (b) 7.5 cm below surface of water (c) 15 cm above surface of water (d) 15 cm below surface of water Real depth m 20 ´ 3 = = 15 cm 4

Interpret (a) Apparent depth =

The ray of light which would have meet line AB at O will now meet this line at I after two times refraction from the slab. Hence, 1ö æ OI = ç1 - ÷ t è mø

15 cm

In the following figure, ray MA is parallel to ray BN. But the emergent ray is displaced laterally by a distance d which depends upon m, t and i and its value is given by

Shift in position = 20 - 15 = 5 cm

20 cm

Lateral Shift

æ ö cos i ÷ sin i. d = t ç1 ç 2 2 ÷ m - sin i ø è N

µ i

r i-r

A i M

d

D

t

B C

Apparent,

R = 60 - 5 = 55 cm

Apparent,

f = 27.5 cm

\ Image of the sun is formed (27.5 - 20 = 7.5 cm) above the surface of water.

Sample Problem 12 A cylindrical vessel having height and diameter = 30 cm is placed on a horizontal surface. A point P is 5 cm from the centre. Upto what height should water be filled so that the particle P is visible, is? (a) 30 cm (c) 13.3 cm

(b) 15 cm (d) 26.7 cm

P

1036 JEE Main Physics Interpret (d) On filling water in the vessel, the ray diagram is as follows From the figure, S

x

45° x

4√2

r

t1 t + 2 m1 m 2

\

36 5 3 = + 7 5 /3 m 2

or

3 36 15 = -3 = m2 7 7

\

m2 =

3

r √23

P

Interpret (a) Apparent depth (AD) =

7 = 1.4 5

L

20–x

tan r =

20 - x 3 = 30 - x 23

10 23 - 3 = 30 - x 23 10 23 48 30 - x = = = 26.7 cm 1.8 23 - 3

Total Internal Reflection Whenever a ray of light goes from a denser medium to a rarer medium, it bends away from the normal. As angle of incidence in denser medium increases, angle of refraction also increases in rarer medium. The angle of incidence ( Ði ) in denser medium for which the angle of refraction ( Ðr) in rarer medium is 90° is called the critical angle ( ÐC ). m sinC = rarer sin 90° m denser

Sample Problem 13 A fish in an aquarium, approaches the left wall at a rate of 3 ms-1, and observes a fly approaching it at 8 ms-1. If the refractive index of water is (4/3), find the actual velocity of the fly.

x

Þ y

(b) 2.75 ms-1 (d) 4.75 ms-1

Interpret (a) For the fish, apparent distance of the fly from the wall of the aquarium is mx. If x is actual distance, then apparent d (mx) velocity will be be = dt (v app) fly = m v fly Now, the fish observes the velocity of the fly to be 8 ms-1. Therefore, apparent relative velocity = 8 ms-1 v fish + (v app) fly = 8 ms-1 Þ

sinC =

mr md

æm ö C = sin-1 ç r ÷ è md ø

3 + mv fly = 8 3 v fly = 5 ´ = 3.75 ms-1 4

This phenomenon is called total internal reflection. For total internal reflection to take place following set of conditions must be obeyed. (i) The ray must travel from denser medium to rarer medium. (ii) The angle of incidence( Ði ) must be greater than critical angle( ÐC ).

Sample Problem 15 An isotropic point source (bulb) is placed at a depth (h) below the water surface. A floating opaque disc is placed on the surface of water, so that the bulb is not visible from the surface. What is the minimum radius of the disc? Take refractive index of water = m. Interpret As shown in figure, light from bulb will not emerge out of the water, if at the edge of disc. i>C

Sample Problem 14 A layer of oil 3 cm thick is flowing on a layer of coloured water 5 cm thick. Refractive index of coloured water is 5/3 and the apparent depth of the two liquids appears to be 36/7 cm. What is the refractive index of oil? (a) 1.4 (c) 3

mr md

Now, if the angle of incidence ( Ði ) in the rarer medium is greater than the critical angle ( ÐC ), then, the ray instead of suffering refraction is reflected back in the same (denser) medium.

µx

(a) 3.75 ms-1 (c) 0.75 ms-1

Þ

=

(b) 2.4 (d) 2

sin i > sin C

...(i)

Now, if R is the radius of disc and h is the depth of bulb from it R sin i = 2 R + h2 and

sin C =

1 m

Ray Optics and Optical Instruments R

A

C

h

i>C

1037

23.4 Refraction from a Spherical Surface Spherical surfaces are of two types (i) Convex

m

(ii) Concave 1 I

O

So, Eq. (i) becomes,

2

P

R 2

R +h

2

>

1 m 1

h R> m -1

or

O

Applications of Total Internal Reflection 1. Diamonds Diamonds are known for their spectacular brilliance. Their brilliance is mainly due to the total internal reflection of light inside them. The critical angle for diamond-air interface (@ 24.4° ) is very small, therefore once light enters a diamond, it is very likely to undergo total internal reflection inside it. By cutting the diamond suitably, multiple total internal reflections can be made which are responsible for the shine of diamond. 2. Optical fibres Optical fibres too make use of the phenomena of total internal reflection. Optical fibres are fabricated with high quality composite glass/quartz fibres. Each fibre consists of a core and cladding. The refractive index of the material of the core is higher than that of cladding. When a signal in the form of light is directed at one end of the fibre at a suitable angle, it undergoes repeated total internal reflection along the length of the fibre and finally comes out at the other end. Since, light undergoes total internal reflection at each stage there is no appreciable loss in intensity of light. Optical fibres are used for transmission of optical signals.

I

For both surfaces refraction formula is given by 1 m -1 1m 2 - =1 2 v u R is refractive index of second medium with respect to first. If m1 and m 2 are refractive indices of first and second medium with respect to air, then m 2 m1 m 2 - m1 = v u R

1m 2

Sample Problem 16 A linear object of length 4 cm is placed at 30 cm from the plane surface of hemispherical glass of radius 10 cm. The hemispherical glass is surrounded by water. Find the final position and size of the image. B

4 cm A

30 cm

(a) 5.3 cm

(b) 4.3 cm

(c) 5 cm

(d) 2.3 cm

Interpret (a) For 1st surface, m1 =

4 3 , m 2 = , u = -20 cm 3 2

R = + 10 cm

and B¢¢

Low m

B¢

st surface

B

5.3cm 5.3cm 4cm A¢¢

A¢

A

u

v

Higher m

u¢ v¢

Light undergoes successive total internal reflections as it moves through an optical fibre.

2

P

Using,

m 2 m1 (m 2 - m1) = v m R

1038 JEE Main Physics Þ

(3 /2) ( 4 /3) (3 /2 - 4 /3) = v ( -20) 10

Þ

v = - 30 cm A¢ B¢ m1v A¢ B¢ ( 4 /3) ( -30) = = Þ AB m 2u ( 4cm) (3 /2) ( -20)

Using,

Some Definitions Relating Lenses Optical centre The optical centre is a point within or outside the lens, at which incident rays refract without deviation in its path. P

Þ

A¢ B¢ = 5.3 cm A¢ B¢ behaves as the object for plane surface 3 4 and R = ¥ , u¢ = - 40 m1 = , m 2 = 2 3 m1 m 2 ( 4 /3) (3 /2) = Þ = Þ v ¢ u¢ v¢ ( -40) Solving it, we will get, v¢ = - 35.4 cm A¢ B¢¢ (m1v ¢ ) Now, using, = A¢ B¢ (m 2u¢ )

P Q

Q O R

O Optical centre

Optical centre

R

S

S

Principal axis The straight line passing through the optical centre of lens is called principal axis of lens.

A¢¢ B¢¢ (3 / 2) ( -35.4) = Þ A¢¢ B¢¢ = 5.3 cm (5.3) ( 4 / 3) ( -40) The final images in all the above cases are shown in figure.

Principal focus Lens has two principal foci. (i) First principal focus It is a point on the principal axis of lens, the rays starting from which (convex lens) or appear to converge at which (concave lens) become parallel to principal axis after refraction.

23.5 Lens Lens is a transparent medium bounded by two curved surfaces. Lenses are of two types (i) Convex or convergent lens (ii) Concave or divergent lens

O F1

O

F1

f1

(i) Convex or Convergent Lens The transparent medium bounded by two bulging surfaces is called convex lens. It is of three types (as shown).

f1

(ii) Second principal focus It is the point on the principal axis at which the rays coming parallel to the principal axis converge (convex lens) or appear to diverge (concave lens) after refraction from the lens.

O

F2 O

F2

(a) Double-convex lens

(b) Plano-convex lens

(b) Concavo-convex lens

(ii) Concave or Divergent Lens The transparent medium bounded by two hollow surfaces is called concave lens. It is of three types (as shown).

f2 f2

Both the foci of convex lens are real, while that of concave lens are virtual.

Focal length The distance between focus and optical centre of lens is called focal length of lens.

Laws of Formation of Images by Lens (i) The rays coming parallel to principal axis of lens pass through the focus after refraction. (ii) The rays coming from the focus of lens go parallel to the principal axis of lens after refraction. (a) Double-convex lens

(b) Plano-convex lens

(b) Concavo-convex lens

(iii) The rays of light passing through optical centre go straight after refraction without changing their path.

Ray Optics and Optical Instruments Table 23.3 S. No. 1.

Position of Object

Formation of Image by a Convex Lens

Ray Diagram

Position of Image

At infinity

2.

Beyond 2 F1

Nature and Size of Image

At the principal focus (F2 ) or in the focal plane

Real, inverted and extremely diminished

Between F2 and 2 F2

Real, inverted and diminished

At 2 F2

Real, inverted and of same size as the object

Beyond 2 F2

Real, inverted and highly magnified

At infinity

Real, inverted and highly magnified

On the same side as the object

Virtual, erect and magnified

2F2

F2 2F2

1039

F1

A B′ 2F2

F2 B 2F 1

F1 A′

3.

At 2 F1

A F2

B 2F1

F1

2F2 B′ A′

4.

Between F1 and 2 F1

A F2 2F1 B

F1

2F2 B′ A′

5.

At F1

2F2

6.

F1

F2

2F2

Between F1 and optical centre F2 2F1

F1

2F2

1040 JEE Main Physics

Hot Spot

Lens Maker’sFormula

In figure, an object is placed on principal axis of lens, the two spherical

Sample Problem 17 The radii of curvature of the

surfaces of the lens have their centres at C1 and C2 .

faces of a double convex lens are 10 cm and 15 cm. Its focal length is 12 cm.The refractive index of glass lens is

A

G

µ1

D O

p

C2

(a) 1.33 (c) 1.5

E I

C1

Interpret (c) Given, f = + 12cm, R1 = + 10 cm, R2 = - 15cm . Refractive index of air is taken as unity.

B

The general equation of refraction at a spherical surface is m2 m1 m2 - m1 = v u R

…(i)

For the first refraction, the object at O, the image is at O1 and the centre of curvature is at C1. When u, v and R denote their X-coordinate. m2 m1 m2 - m1 …(ii) = v1 u R1 For the second refraction at AEB, the light goes from the medium m2 to medium m1. Applying Eq. (i), we get m1 m2 m1 - m2 …(iii) = v v1 R2 Adding Eqs. (ii) and (ii), we get ö æ1 1 1 æ m2 1ö - = ç - 1÷ ç - ÷ v u è m1 ø è R1 R2 ø

æ 1 1 1ö çQ - = ÷ …(iv) è v u fø

If the refractive index of material of the lens m. It is placed in air,

m2 = m and m1 = 1

\Eq. (iv), becomes,

1 = (m - 1) f

æ1 1ö ç - ÷ R R è 1 2ø

…(v)

From the Lens maker’s formula, we have æ1 1 1ö = (m - 1) ç - ÷ f è R1 R2 ø Putting the values in above expression, 1 1 ö æ 1 = (m - 1) ç ÷ è10 -15 ø 12 1 1ö æ 1 Þ = (m - 1) ç + ÷ è10 15 ø 12 æ 3 + 2 ö (m - 1) = (m - 1) ç ÷= è 30 ø 6 6 Þ = m -1 12 1 3 Þ m = + 1 = = 1.5 2 2

Sample Problem 18 A biconvex lens has a radius of curvature of magnitude 20 cm, which one of the following options describe best the image formed of an object height 2 cm placed 30 cm from the lens? (given refractive index of lens m = 3 / 2). (a) Real, inverted, height = 4 cm (b) Virtual, upright, height = 1cm (c) Real, inverted, height = 1cm (d) Virtual, inverted, height = 0.3 cm

This equation is known as Lens maker’s formula. and

(b) 1.05 (d) 1.0

1 1 1 - = =P v u f

is known as Lens’s equation. where, P = power of lens

Cartesian Sign Convection 1. All figures are draw with light travelling from left to right. 2. All distances are measured from a reference surface, such as a wavefront or a refracting surface. Distances to the left of the surface are negative. 3. The refractive power of a surface that makes light rays more convergent is positive. The focal length of such a surface is positive. 4. The distance of a real object is negative. 5. The distance of a real image is positive. 6. Heights above the optical axis are positive. 7. Angles measured clockwise from the optic axis are negative.

Interpret (c) Given, R1 = + 26 cm, R2 = -20 cm and m = Using Lens maker formula, æ 1 1 ö æ3 ö æ 1 1 1ö 1 = (m - 1) ç - ÷ = ç - 1÷ ç ÷= f è R1 R2 ø è 2 ø è 20 20 ø 20 Þ f = 20 cm Here, u = -30 cm, f = 20 cm, v = ? 1 1 1 Using lens formula, = f v u 1 1 1 or = 20 v -30 1 1 = Þ v 60 or

v = 60 cm

3 2

Ray Optics and Optical Instruments m=

Magnification,

v hi = u ho

Substituting the values, we get 3 / 2 -1 fwater = fair æ3 /2 ö 1 ÷ ç è 4 /3 ø

60 h = i -30 2 m

or

= 4 fair = 4 ´ 10 = 40 cm

Þ hi = - 4 cm It means the image is real, inverted and of height (h) = 4 cm,

Sample Problem 21 An object is placed at a distance of

Sample Problem 19 A magician during a show makes a glass lens with m = 147 . disappear in a trough of liquid. The refractive index of the liquid is (a) 1.84 (c) 1.47

1041

(b) 1.0 (d) 1.33

10 cm to the left on the axis of a convex lens L1 of focal length 20 cm. A second convex lens L2 of focal length 10 cm is placed co axially to the right of the lens L1 at a distance of 5 cm from it. Find the position of the final image and its magnification. 2 cm on the right of the second lens, 3.33 3 2 (b) 16 cm on the right of the second lens, 1.33 3 2 (c) 16 cm on the right of the first lens, 1.33 3 (d) None of the above

(a) 16

Interpret (c) From Lens maker’s formula, we have æ1 1 1ö = (m 2 - m1) ç - ÷ f è R1 R2 ø In order to make the lens disappear the refractive index of the liquid 1 must be equal to 1.47. This means m1 = m 2 this given = 0 or f ® ¥. f The lens in the liquid will act like a plane sheet of glass.

Interpret (b) Here, for1st lens, u1 = -10 cm and f1 = 20 cm 1 1 1 - = v1 u1 f1

Note From the refractive index value, it can be concluded that liquid is glycerine.

Formation of Image by Concave Lens

Þ

1 1 1 = v1 20 10

The image formed is always virtual, erect and diminished and lies between the lens and F2 for all positions of the object.

Þ

v1 = - 20 cm L1

L2

O1

O2

Object F1 2F2

O′

2F1

O

F2 f1

Image

f2 5 cm

Sample Problem 20 The focal length of convex lens is 10 cm in air. Find its focal length in water. (Given, m g =3/2 and mw = 4/3) (a) 10 cm (c) 30 cm

(b) 20 cm (d) 40 cm

Interpret (d) As, and

1 fwater

Here,

æ1 1 1ö = (mg - 1) ç - ÷ fair è R1 R2 ø

...(i)

æ mg öæ1 1ö =ç - 1÷ ç - ÷ è mw ø è R1 R2 ø

...(ii)

Dividing Eq. (i) by (ii), we get fwater æ mg -1 ö ÷÷ = çç fair è mg /mw - 1ø

i.e., the image is virtual and hence lies on the same side of the object. This will behave as an object for the second lens. 1 1 1 For 2nd lens, = v 2 u2 f2 u 2 = - (20 + 5) , f2 = 10 cm

1 1 1 + = v 2 25 10 Þ

v2 =

50 2 = 16 cm 3 3

2 i.e., final image is at a distance of16 cm on the right of the second 3 lens. The magnification of the image is given by, m=

4 v1 v 2 20 50 . = = = 133 u1 u 2 10 3 ´ 25 5

1042 JEE Main Physics Sample Problem 22 In the given diagram, the position of the image formed by the lens combination is f = +10

–10

1 1 1 - = v3 ¥ 30

Þ

[NCERT]

Þ v3 = 30 cm The final image is formed 30 cm to the right of the third lens.

+ 30 cm

Magnification of Lens

30 cm 5 cm

The lateral, transverse or linear magnification produced by a lens is defined by Height of image I m= = Height of object O

10 cm

(a) 30 cm to the left of the second lens (b) 30 cm to the right of the third lens (c) 15 cm to the left of the second lens (d) 15 cm to the right of the third lens

A real image I I ¢ of an object OO¢ formed by a convex lens is shown in figure.

Interpret (b) Image formed by the first lens 1 1 1 - = v1 u1 f1 1 1 1 = v1 -30 10

Þ Þ

O′ I

O

P

v1 = 15 cm

I′

The image formed by the first lens serves as the object for the second. This is at a distance of (15 - 5) cm = 10 cm to the right of the second lens. Though the image is real, it serves as a virtual object for the second lens, which means that the rays appear to come from it for the second lens 1 1 -1 = Þ v2 = ¥ v 2 10 10 The virtual image is formed at an infinite distance to the left of the second lens. This acts as an object for the third lens. 1 1 2 - = v3 u3 f3

u

v

Height of image I I ¢ v = = Height of image OO ¢ u Substituting v and u with proper sign, we get II ¢ v -I = = OO ¢ O -u I v or =m= O u v Thus, m= u

Important points 1. Power of lens, P = Þ

1 f ( in m)

P = P1 + P2 + ... =

i =1

100 P = f ( in cm)

Magnification of combination, n

Power of convex lens is positive and of concave lens is negative.

2. If distance of an object from first focus of lens is a1 and distance of image from second focus is a2, then its focal length.

M = m1 ´ m2 ´ ... = P m i =1

4. If two lenses of focal lengths f1 and f2 are separated by a distance x, then its equivalent focal length

f = a1a2 This is Newton’s formula.

Combination of Thin lenses 3. If two or more lenses are placed in contact, then equivalent focal length of the combination. n 1 1 1 1 = + + ... = å f f1 f2 f i =1 i

Power of combination,

n

åPi

x f1

f2

Ray Optics and Optical Instruments 1 1 1 x = + F f1 f2 f12 f

1043

f side focal length of the combination is . But on joining two parts in 2 opposite sense the net focal length becomes ¥ (or net power = 0).

Power of combination,

f

P = P1 + P2 - x PP 12

f

f 2

∞

Total magnification remains unchanged i.e., m = m1 ´ m2

5. If a lens is made of a number of layers of different refractive indices, then number of images of an object formed by the lens is equal to number of different media.

f (a)

(b)

(c)

(d)

6. Cutting of a lens (i) If a symmetrical convex lens of focal length f is cut into two parts along its optical axis, then focal length of each part (a plano-convex lens) is 2f. However, if the two parts are joined as shown in figure, the focal length of combination is again f. f

2f

2f

f

f

7. Silvering of a lens (i) Let a plano-convex lens is having a curved surface of radius of curvature R and has refractive index m. If its plane surface is silvered, it behaves as a concave mirror of focal length, R f = 2( m - 1) (ii) If the curved surface of plano-convex lens is silvered, then it behaves as a concave mirror of focal length, R f = 2m

(a)

(b)

(c)

(d)

(ii) If a symmetrical convex lens of focal length f is cut into two parts along the principal axis, then focal length of each part remains changed at f. If these two parts are joined with curved ends on one

(iii) If one surface of a symmetrical double convex lens is silvered, then the lens behaves as a concave mirror of focal length, R f =2(2m - 1)

1044 JEE Main Physics Table 23.4 Difference between Lens and Mirror S. No. Nature of Lens/mirror Focal Length (f )

Power 1 1 PL = , PM = f f

Converging/Diverging

1.

Convex lens

+ ve

+ ve

converging

2.

Concave mirror

– ve

+ ve

converging

3.

Concave lens

– ve

– ve

diverging

4.

Convex mirror

+ ve

– ve

diverging

Sample Problem 23 A convergent lens of 6D is combined with a diverging lens of –2 D. Find the power and focal length of the combination. (a) 26 cm (c) 30 cm

Interpret (d) Here, and

(b) 20 cm (d) 25 cm P1 = 6D

Interpret (d) Here, f1 = 10 cm, f2 = - 6cm, F = ? Use the formula,

1 1 1 = + F f1 f2 =

1 1 1 - =10 6 15

F = -15 cm

P2 = - 2D

Using the formula, P = P1 + P2 = 6 - 2 = 4D \

Ray Diagram

f = 1 / P = 1 / 4 m = 25cm

Check Point 2 1. Is the ratio of frequencies of ultraviolet rays and infrared rays

Sample Problem 24 A convex lens of 10 cm focal length is combined with a concave lens of 6 cm focal length. Find the focal length of the combination. (a) –15 cm (b) 15 cm (c) 10 cm (d) –10 cm

in glass, more than, less than or equal to one ?

2. Can convergent lens in one medium behave as a divergent lens in some other medium ?

3. Why an air bubble inside a transparent liquid behaves like a divergent lens ?

4. The sun is seen a little before it rises and for a short while after it sets. Explain why ?

Ray Optics and Optical Instruments

1045

23.6 Deviation by Prism

Sample Problem 25 The angle of minimum deviation for a glass prism with m = 3 equals the refracting angle of the

A prism is a homogeneous, transparent medium bounded by two plane surfaces inclined at an angle A with each other. These surfaces are called as refracting surfaces and the angle between them is called angle of prism A.

prism. What is the angle of the prism?

P

i

M

A O

δm N

r1 r2

(a) 60° (c) 45°

(b) 30° (d) 90°

Interpret (a) Given, A = dm æ A + dm ö sinç ÷ è 2 ø m= Aö æ ç sin ÷ è 2ø

Using, e

O Q

\

R

Figure shows the refraction of monochromatic light through a prism. Here, i and e represent the angle of incidence and angle of emergence respectively, r1 and r2 are two angles of refraction. If m is the refractive index of the material of the prism, then sin i sin e m= = sin r1 sin r2 The angle between the incident ray and the emergent ray is known as the angle of deviation d. For refraction through a prism it is found that i + e = A + d and r1 + r2 = A

i=e

or

sin A 3= = Aö æ ç sin ÷ è 2ø

\

cos

A A × cos 2 2 æ Aö sin ç ÷ è2ø

2 sin

A 3 = 2 2 A = 30° or A = 60° 2

\

23.7 Dispersion by a Prism

Minimum Deviation It is found that the angle of δ deviation d varies with the angle of incidence i of the ray incident on the first refracting face of the δm prism. The variation is shown in figure and for one angle of incidence it has a minimum value dmin. At this value

æ A + dm ö sinç ÷ è 2 ø 3= Aö æ ç sin ÷ è 2ø

Dispersion of light is the phenomenon of splitting of white light into its constituent wavelengths on passing through a dispersive medium, e. g., prism. Cause of dispersion is the variation of refractive index of prism with wavelength. As, lV > l R, hence, mV > m R and consequently, dV > dR.

i=e r1 = r2

A

i1

dR dv

t ligh ite h W

Vio let

It therefore, follows that r1 = r2 = r A r= 2 Further at d = dm = (i + i) - A A + dm i= 2 æ A + dm ö sinç ÷ è 2 ø sin i or m = m= Aö sin r æ ç sin ÷ è 2ø Note For thin prism, d m = (m - 1)A.

Red Screen

R V

Angular Dispersion It is the angular separation between the two extreme rays. Angular dispersion, q = dV - dR = (mV - m R) A Dispersive Power The dispersive power of a prism material is measured by the ratio of angular dispersion to the mean deviation suffered by light beam. \ Dispersive power, w=

dV - dR mV - m R = d m -1

1046 JEE Main Physics where m is the mean value of refractive index of prism. The dispersive power of a prism depends only on its material and is independent of angle of prism, angle of incidence or size of the prism.

Interpret (a) We know that dispersion produced by a thin prism, q = (mV - mR) A Here,

mV = 1.68, m R = 1.56 A = 18° q = (1.68 – 1.56) ´ 18° = 2.16°

and \

Dispersive power is a unitless and dimensionless term. Dispersive power of a flint glass prism is more than that of a crown glass.

Sample Problem 27 Calculate the dispersive power for crown glass from the given data

Dispersion without Deviation (Direct Vision Prism)

and

mV = 1.523

1. To produce dispersion without mean deviation, we use a combination of two prisms of different materials such that Flint

V R

A R V Crown

A¢

(a) 0.01639

mR = 1.5145 (b) 1.05639

(c) 0.05639

(d) 2.05639

Interpret (a) Here, mV = 1.523 and mR = 1.145 Mean refractive index, m =

Dispersive power is given by, (m - mR) w= V (m - 1) =

æ m - 1ö A¢ = ç ÷A è m ¢ - 1ø 2. Net dispersion caused

= ( mV - m R) A + ( m ¢V - m ¢R ) A¢ = ( m - 1) A (w - w¢ ) = d (w - w¢ )

Deviation without Dispersion (Achromatic Prism) 1. To produce deviation without dispersion, we use a combination of two prisms of different materials such [m - m R] that A¢ = V A [m ¢V - m ¢R ] Flint

1.523 +1.5145 = 1.51875 2

1.523 –1.5145 = 0.01639 (1.51875 –1)

Sample Problem 28 A prism of crown glass with refracting angle of 5º and mean refractive index = 1.51 is combined with a flint glass prism of refractive index = 1.65 to produce no deviation. Find the angle of flint glass. (a) 3.92º (c) 5.32º

(b) 4.68º (d) 7.28º

Interpret (a) Let A¢ be the angle of flint glass prism. A = 5° and m = 1.51for crown glass prism. Deviation produced by flint glass,

Here,

d¢ = (m - 1) A = (1.51 – 1) ´ 5 = 2.55° For no deviation, or

A

d¢ = d 0.65 A¢ = 2.55 2.55 A¢ = = 3.92° 0.65

R V

A¢

Crown

w 2. Resultant deviation produced = d é1 - ù êë w¢ úû

Sample Problem 26 Find the dispersion produced by a thin prism of 18º having refracting index for red light = 1.56 and refractive index for violet light = 1.68. (a) 2.16º (b) 1.16º (c) 3.16º (d) 2.10º

23.8 Optical Instruments Optical instrument is a device which is made from proper combination of mirrors, prisms and lenses. The principle of working of optical instruments depends on laws of reflection and refraction of light.

Microscope It is an optical instrument which forms a magnified image of a small nearby object and thus, increases the visual angle subtended by the image at the eye so that the object is seen to be bigger and distinct.

Ray Optics and Optical Instruments

1047

Simple Microscope

Telescope

A simple microscope is a convex lens of short focal length which is fixed in a frame provided with handle.

Telescope is an optical instrument which increases, the visual angle at the eye by forming the image of a distant object at the least distance of distinct vision, so that the object is seen distinct and bigger.

A¢

L A

A¢¢

β

α

B¢

O

B

F

F

u D

Magnification of simple microscope (a) When final image is formed at least distance of distinct vision,

M = 1+

D f

Astronomical Telescope It consists of two converging lenses placed coaxially. The one facing the distant object is called the objective and has a large aperture and large focal length. The other is called the eyepiece, as the eye is placed closed to it. The eyepiece tube can slide within the objective tube, so that the separation between the objective and the eye-piece may be varied. Objective lens

D (b) For relaxed eye, M = f where, D = least distance of distinct vision.

Eyepiece

Parallel rays from object at ∞ α

A¢¢

Compound Microscope

B¢¢

Figure shows a simplified version of a compound microscope. It consists of two converging lenses arranged coaxially. The one facing the object is called objective and the one close to eye is called eye piece. The objective has a smaller aperture and smaller focal length than those of the eyepiece. uo Objective lens

fo

Eyepiece Fe h¢

h Fo

D

ue

(a) For relaxed eye, M¥ = -

fo fe

In this position, length of telescope L¥ = fo + fe (b) When final image is formed at least distance of distinct vision

MD = -

θ

fo æ fe ö ç1 + ÷ fe è Dø

Length of telescope, vo

LD = f o + ue

ue D

f o = focal length of objective lens and f e = focal length of eyepiece

Magnification of compound microscope (a) For relaxed eye, M¥ = -

A′

Magnification of astronomical telescope

h¢¢

u

o′ Fe

Fe β

α

O

vo æ D ö ç ÷ uo è fe ø

In this position, length of microscope

L¥ = vo + f e (b) When final image is formed at least distance of distinct vision.

v æ Dö MD = - o ç1 + ÷ uo è fe ø Length of microscope,

Terrestrial Telescope In an astronomical telescope, the final image is inverted with respect to the object. To remove this difficulty, a convex lens of focal length f is included between the objective and the eyepiece in such a way that the focal plane of the objective is a distance 2f away from this lens. B¢¢ B¢¢

Parallel rays from object a at ∞

A¢ A¢¢

LD = vo + ue

B¢

vo = distance of first image from object lens, uo = distance of object from objective lens and f e = focal length of eyepiece

β

A¢¢ L fo

2f

2f

1048 JEE Main Physics Given,

Magnification of terrestrial telescope,

2a = Diameter = 100 inch = 254 cm. l = 6000 Å = 6 ´ 10 -5 cm, then

(a) For relaxed eye, M¥ = fo / fe

Dq =

In this position, length of telescope,

= 2.9 ´ 10 –7 rad

L¥ = f o + 4 f + f e (b) When final image is formed at least distance of distinct vision,

f æ f ö Mo = o ç1 + e ÷ fe è Dø L¥ = f o + 4 f + ue f o = focal length of objective lens and

microscope of focal length 20 cm. Find the angular magnification produced if the image is formed at 30 cm from the lens.

(iii) Galilean telescope

(b) 2.05 (d) 1.5

Interpret (a) Given, f = + 30 cm v = - 30 cm

and

f e = focal length of eyepiece

1 1 1 - = , we have v u f 1 1 1 = -30 uo 20

Using the formula, we have,

A simple model of Galilean telescope is shown in figure. A convergent lens is used as the objective and a divergent lens as the eyepiece. fo

a

Sample Problem 30 An object is seen through a simple

(a) 2.08 (c) 3.08

Length of telescope,

Parallel rays from object at ∞

0.61 ´ 6 ´ 10 –5 127

B¢¢ ue A¢¢

β

a

E

β

A¢

B¢

Magnification of Galilean telescope. f (a) For relaxed eye, M¥ = o fe In this position, length of telescope

L¥ = f o - f e (b) When final image is formed at least distance of distinct f æ f ö vision, MD = o ç1 - e ÷ fe è Dø

uo = 12cm D 25 The angular magnification, M = = = 2.08 uo 12

Sample Problem 31 A galilean telescope is 27 cm long when focussed to form an image at infinity. If the objective has a focal length of 30 cm, what is the focal length of the eyepiece? (a) 3 cm (c) 2 cm

Interpret (a) Given, fo = + 3cm Length of telescope is given 27 cm Therefore, ue = + 3cm For the final image at infinity, the intermediate image should lie at first focus of eyepiece of the Galilean telescope,

Sample Problem 29 Light of wavelength 6000Å is coming from a star. The limit of resolution of a telescope whose objective has a diameter of 100 inch is

Objective

Eyepiece

F1

27 cm 30 cm

3 cm

fe = - 3cm

Length of telescope

LD = f o - ue

(b) –3 cm (d) –2 cm

Check Point 3 1. When does a ray incident on a prism deviate away from the base?

2. Why are ‘holes’ (rings) observed sometimes round the sun or moon?

(a) 2.9 ´ 10 –7 rad

(b) 1.5 ´ 10 –5 rad

3. Refractive index of glass for lights of yellow, green and red

(c) 5.3 ´ 10 –4 rad

(d) 3.3 ´ 10 –6 rad

colours are my , mg and m r respectively. Rearrange these symbols in an increasing order of value.

Interpret (a) Resolving power of telescope is given by 0.61 l Dq = a where, l is wavelength, a is radius of the lens.

4. What should be the position of a object relative to a biconvex lens so that it behaves like magnifying lens?

5. If a telescope is inverted, will it serve as a microscope?

WORKED OUT Examples Example 1

A beaker containing liquid is placed on a table underneath a microsope which can be moved along a vertical scale. The microscope is focussed, through the liquid, on a mark on the table and the reading on the scale is a. It is next focused on the upper surface of the liquid and the reading is b. More liquid is added and the observations are repeated, the corresponding reading being c and d. The refractive index if the liquid is d -b d -c -b+ a d -c-b+ a (c) d -b (a)

Solution

b -d d -c-b+ a d -b (d) d + c-b-a (b)

Using Eqs. (iii) and (iv), we obtain sin i m 2 = sin i m1 m tan i = 2 m1

Þ

= m(d - c - b + a) Since, the liquid is added in second case, h2 - h1 = d - b d -b m= d -c-b+ a

A given ray of light suffers Q minimum deviation in an equilateral prism P R P. Additional prism Q and R of identical shape and of the same material as P are now added as shown in the figure. The ray will now suffer

Solution

No deviations occur on interfaces 2 and 3 as there is no change in medium. However, deviation at interface 4 is same as it was on interface 2 with only prism P. 1

Example 2

(c) tan -1 (sin r)

(d) tan -1(sin i)

i r

Þ

sin r ¢ = cos r sin i m 2 = cos r m1

Example 4 The sun (diameter D) subtends an angle of q rad at the pole of a concave mirror of focal length f. The diameter of the image of the sun formed by the mirror is

r¢ Rarer

(a) f q

Solution

(b) f q

(c) 2 f q

... (i)

(d) D q

Since, the sun is at very large distance, u = ¥ Image of the sun

θ d d

...(ii) 1 1 1 + = ¥ v f

f

...(iii)

Þ

…(iv)

Þ d = (2a)v 2 Putting, 2a = q and v = f , we obtain d = f q

According to the law of refraction i =r

4 R

Denser

From the given condition,r + r ¢ = 90° Solution of Eqs. (i) and (ii) yields,

Q 3

2 P

Solution Applying Snell’s law for refraction, sin i m = 2 sin r¢ m1

µ2

Since, at the time of total internal reflection, m sin qc = 2 , using Eq. (v), we obtain m1

(a) greater deviations (b) no deviation (c) same deviation as before (d) total internal reflection

Since, h2 ³ h1 the difference of the depths = h2 - h1

(b) sin -1 (cot i)

r′

Example 3

In first case, the real depth h1 = m (d - c ). Similarly, in the second case, the real depth h2 = m (d - c)

(a) sin -1 (tan r)

... (v)

µ1

qc = sin -1 (tan r)

The real depth = (Refractive index) apparent depth

A ray of light from a denser medium strikes a rarer medium at angle of incidence i. The reflected and refracted rays make an angle of 90° with each other. The angles of reflection and refraction are r and r¢, respectively.The critical angle is

i r

d

1050 JEE Main Physics Example 5 A ray enters a glass sphere of refractive index m = 3 at an angle of incidence 60° a ray is reflected and refracted at the farther surface of the sphere. The angle between the reflected and refracted rays at this surface is (a) 50° (c) 90°

(b) 60° (d) 40°

Solution

Solution

One image will be real and the other will be virtual. Since, they are of the same size, one will have magnification m and the other -m. 1 1 1 + = u1 u1m f or

1æ 1ö 1 ç1 + ÷ = è u1 mø f

and

1 1 1 = u2 u2m f

or

1 æ 1ö 1 ç1 - ÷ = u2 è m ø f

Refraction at P, 60º

P r1 r2 r′2

Q α

i2

... (i)

... (ii)

From Eqs. (i) and (ii), we get u1 + u2 =2 f

sin 60° = 3 sin r1 Þ Since

f=

or

sin r1 = (1/2) Þ r1 = 30° r2 = r1

Example 8

A ray of light passes through four transparent media with refractive indices m1, m 2, m3 and m 4 as shown in the figure. The surfaces of all media are parallel. If the emergent ray CD is parallel to the incident ray AB, we must have

\

r2 = 30° sin r2 1 Refraction at Q = = sin i1 3 Putting r2 = 30° , we obtain i2 = 60°

µ1

Reflection at Q, r2¢ = r2 = 30°

\

u1 + u2 2

µ2

D

µ3 C

B

a = 180° - (r2¢ + i2) 180° - (30° + 60° ) = 90°

µ4

A

Example 6

A soldier directs a laser beam on an enemy by reflecting the beam from a mirror. If the mirror is rotated by an angle q, by what angle will reflected beam rotate? (a) q/ 2 (c) 2 q

Solution

(b) q (d) None of these Let M1OM2 be the

initial position of the mirror. The mirror is rotated through an angle qto the position M1¢OM2¢ . PO is the

N′

M

Solution

Considering Snell’s law m sin q = constant, and i1 = i2

(given). N

Q′

µ1

Q

µ3

µ4 i4

θ

i3

M2 i1

When an object is at distances of u1 and u 2 from the poles of a concave mirror, images of the same size are formed. The focal length of the mirror is

i2

sin i1 m 2 sin i2 , = sin i2 m1 sin i3 =

Example 7

(b) u1 - u2 u - u2 (d) 1 2

µ2

M′2

ÐPOQ = 2i and ÐPOQ ¢ = ÐPON ¢ + ÐN ¢ OQ = 2i - 2q. \ The reflected beam rotates through an angle 2 q.

(a) u1 + u2 u + u2 (c) 1 2

(b) m 2 = m 4 (d) m 2 = m3

θ

P

incident light OQ on the initial 1 reflected ray and OQ¢ is the reflected M′1 ray after rotating the mirror by angle q. If i = initial incidence angle, then

(a) m1 = m3 (c) m 4 = m1

\

m3 sin i3 m 4 = , m 2 sin i4 m3

sin i1 m 4 = sin i4 m1

Since,

i1 = i4 ,

Therefore

m 4 = m1

Start Practice for

JEE Main Round I

(Topically Divided Problems)

Reflection of Light

6. For a convex mirror, the variation of u versus v is

1. A dentist has a small mirror of focal length 16 mm. He views the cavity in the tooth of a patient by holding the mirror at a distance of 8 mm from the cavity. The magnification is (a) 1

(b) 1.5

(c) 2

(d) 3

given by v

(a)

2. Given width of aperture = 3 mm and l = 500 nm. For

mirror is 2r. An isotopic point source of light is placed exactly mid ways between the mirror and the screen. Assume that mirror reflects 100% of incident light. Then the ratio of illuminance on the screen with and without the mirror is (b) 2 : 1

(c) 10 : 9

(d) 9 : 1

4. An object is placed a symmetrically between two

u

v

(c)

(d) u

O

u

O

7. A fish is a little away below the surface of a lake. If the critical angle is 49°, then the fish could see things above water surface within an angular range of q° where Air Water

plane mirrors inclined at an angle of 72°. The number of image formed is (a) 5 (c) 2

u

O

v

(b) 18 mm (d) 18 light years

3. The separation between the screen and a plane

(a) 10 : 1

(b) O

what distance ray optics is good approximation? (a) 18 m (c) 18 Å

v

θ

(b) 4 (d) infinite

49°

5. From a spherical mirror, the graph of 1 / v versus 1 / u is given by

(a) q = 49°

1 v

1 v

(a) 1 u

1 u

O

1 v

1 v

(c) 1 u

1 ° 4

(d) q = 90°

20 cm. A second car 2 m broad and 1.6 m height is 6 cm away from the first car. The position of the second car as seen in the mirror of the first car is (a) 19.35 cm (c) 21.48 cm

(b) 17.45 cm (d) 15.49 cm

9. A convex mirror forms an image one-fourth the size of the object. If object is at a distance of 0.5 m from the mirror, the focal length of mirror is

(d)

O

(c) q = 24

8. A car is fitted with a convex mirror of focal length

(b)

O

(b) q = 98°

O

1 u

(a) 0.17 m (c) 0.4 m

(b) – 1.5 m (d) – 0.4 m

1052 JEE Main Physics 10. A person of 6 feet in length can see his full size erect

18. Sun subtends an angle of 0.5° at the centre of

image in a mirror 2 feet in height. This mirror has to be

curvature of a concave mirror of radius of curvature 15 m. The diameter of the image of the sun formed by the mirror is

(a) plane or convex (c) necessarily convex

(b) plane or concave (d) necessarily concave

11. A point object is placed at a distance of 30 cm from a convex mirror of a focal length 30 cm. The image will form at (a) (b) (c) (d)

infinity pole 15 cm behind the mirror no image will be formed

12. A plane mirror is reflecting a ray of incident light is rotated through an angle of about an axis through the point of incidence in the plane of the mirror perpendicular to the plane of incident, then (a) (b) (c) (d)

the reflected ray rotates through an angle 2 q the reflected ray rotates through an angle of q the reflected ray does not rotate None of the above

13. With a concave mirror, an object is placed at a distance x1 from the principal focus, on the principal axis. The image is formed at a distance x2 from the principal focus. The focal length of the mirror is (a) x 1x 2 (c)

x1 x2

(b) (d)

x 1 + x2 2 x1x2

0.2 m. How far should the mirror be held from his face in order to give an image of two fold magnification? (b) 0.2 m (d) 0.4 m

15. To focal length of a concave mirror is 12 cm. Where should an object of length 4 cm be placed so that an image 1 cm long is formed? (a) 48 cm (c) - 60 cm

(b) 3 cm (d) 15 cm

16. The focal length of a concave mirror is 20 cm. Where an object must be placed to form an image magnified to times when the image is real? (a) - 30 cm from the mirror (b) 10 cm from the mirror (c) 20 cm from the mirror (d) 15 cm from the mirror

17. A spherical mirror forms diminished virtual image of magnification 1/3. Focal length is 18 cm. The distance of the object is (a) 18 cm (c) 48 cm

(b) 7.55 cm (d) 6.55 cm

19. A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved? [NCERT] (a) 54 cm (c) 28 cm

(b) 27 cm (d) 475 cm

20. An object 5 cm tall is placed 1 m from a concave spherical mirror which has a radius of curvature of 20 cm. The size of the image is (b) - 0.55 cm (d) 0.60 cm

(a) 0.11 cm (c) 0.55 cm

21. An convex mirror of radius of curvature 1.6 m has an object placed at a distance of 1 m from it. The image is formed at a distance of (a) (b) (c) (d)

8/13 m in front of the mirror 8/13 m behind the mirror 4/9 m in front of the mirror 4/9 m behind the mirror

22. A short linear object of length b lies along the axis of a concave mirror. The size of the image is equal to

14. A man has a concave shaving mirror or focal length

(a) - 0.1 m (c) 0.3 m

(a) 8.55 cm (c) 6.55 cm

(b) - 36 cm (d) infinite

æu - f ö (a) b ç ÷ è f ø

1 /2

æu - f ö (c) b ç ÷ è f ø

æ f ö (b) b ç ÷ èu - f ø

1 /2

æ f ö (d) b ç ÷ èf - uø

23. 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. The total deviation of the ray is (a) 2 q (c) 360° - 2 q

(b) 240° - 2 q (d) 180° - q

24. A plane mirror is approaching you at 10 cms -1. Your image shall approach you with a speed of (a) + 10 cms -1

(b) -10 cms -1

(c) + 20 cms -1

(d) - 20 cms -1

25. A candle is placed before a thick plane mirror. When looked obliquely in the mirror, a number of images are seen from the surfaces of the plane mirror. Then (a) (b) (c) (d)

first image is brightest second image is brightest third image is brightest all images beyond second are brighter

Ray option and Optical Instruments 26. An object is approaching a plane mirror at 10 cms -1. A stationary observer sees the image. At what speed will the image approach the stationary observer? (a) 10 cms -1

(b) 5 cms -1

(c) 20 cms -1

(d) 15 cms -1

v

mirror. If you stand behind the object, 30 cm from the mirror and look at its image, for what distance must you focus your eyes? (a) 20 cm

(b) 60 cm

(c) 80 cm

(d) 40 cm

(a) upright and real (c) inverted and virtual

reflected solar beam when the light is incident at an angle of a = 40° to the vertical. At what angle b to the horizontal should a plane mirror be placed? (b) 20°

(c) 50°

(d) 40°

30. The sun (diameter d) subtends an angle q radian at the pole of a concave mirror of focal length f . The diameter of the image of sun formed by mirror (a) qf

(b)

q f 2

(c) 2 qf

(d)

u

u

O

v

v

(c)

(d) u

O

O

u

35. If the space between the lenses in the

(b) upright and virtual (d) inverted and real

29. It is necessary to illuminate the bottom of a well by

(a) 70°

(b) O

28. When a convergent beam of light is incident on a plane mirror, the image formed is

v

(a)

27. A small object is placed 10 cm in front of a plane

1053

q f p

31. At what angle should a ray of light be incident on the

combination shows were filled with water, what would happen to the focal length and power of the lens combination? Focal length Power (a) (b) (c) (d)

Decreased Decreased Increased Increased

increased unchanged unchanged decreased

36. Two convex lenses placed in contact form the image of a distance object at P. If the lens B is moved to the right, the image will [NCERT] A B

face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.

P

[NCERT]

(a) 16°

(b) 29°

(c) 45°

(d) 58°

32. A spherical mirror forms an image of magnification m = ± 3. The object distance, if focal length of mirror is 24 cm, may be (a) 32 cm, 24 cm (c) 32 cm only

(b) 32 cm, 16 cm (d) 16 cm only

Refraction of Light 33. How will the image formed by a convex lens be affected, if the central portion of the lens is wrapped in blank paper, as shown in the figure. (a) No image will be formed (b) Full image will be formed but is less bright (c) Full image will be formed but without the central portion (d) Two image will be formed, one due to each exposed half

34. The distance v of the real image formed by a convex lens is measured for various object distance u. A graph is plotted between v and u. Which one of the following graphs is correct?

(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 length of the lenses

37. A layered lens as shown in figure is made of two types of transparent materials indicated by different shades. A point object is placed on its axis. The object will form (a) 1 image (c) 3 images

(b) 2 images (d) 9 images

38. A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.) [NCERT] (a) 4.6 m2 2

(c) 5.6 m

(b) 3.2m (d) 2.6 m2

1054 JEE Main Physics 39. The relation between n1 and n2 , if the behaviour of light ray is as shown in the figure.

44. A concave lens of focal length 20 cm produces an image half in size of the real object. The distance of the real object is (a) 20 cm (c) 10 cm

n1 n2

Lens

(a) n2 > n1 (c) n1 > n2

(b) n1 >> n2 (d) n1 = n2

40. A convex lens A of focal length 20 cm and a concave lens B of focal length 56 cm are kept along the same axis with the distance d between them. If a parallel beam of light falling on A leaves B as a parallel beam, beam then distance, d in cm, will be (a) 25 (c) 30

(b) 36 (d) 50

41. As shown in figure, the liquids L1, L2 and L3 have

refractive indices 1.55, 1.50 and 1.20 respectively. Therefore, the arrangement corresponds to

L1

(a) biconvex lens (c) concavo-convex lens

L2

L3

(b) biconcave lens (d) convexo-concave lens

(b) 30 cm (d) 60 cm

45. An object 15 cm high is placed 10 cm from the optical centre of a thin lens. Its image is formed 25 cm from the optical centre in the same side of the lens as the object. The height of the image is (a) 2.5 cm (c) 16.7 cm

(b) 0.2 cm (d) 37.5 cm

46. One surface of a lens is convex and the other is concave. If the radii of curvature are r1 and r2 respectively, the lens will be convex, if (a) r1 > r2 (c) r1 < r2

(b) r1 = r2 (d) r1 = 1 / r2

47. A lens for ms a virtual image 4 cm away from it when an object is placed 10 cm away from it. The lens is a……lens of focal length…… (a) (b) (c) (d)

concave, 6.67 cm concave, 2.86 cm convex, 2.86 cm may be concave or convex, 6.67 cm

1 m forms a real, 3 inverted image twice in size of he object. The distance of the object form the lens is

48. A convex lens of focal length

(a) 0.5 m (c) 0.33 m

(b) 0.166 m (d) 1 m

42. You are given four sources of light each one providing

49. The radius of curvature of the curved surface of a

a light of a single colour, blue, green and yellow. Suppose the angle of refraction for a beam of yellow light corresponding to a particular angle of incidence at the interface of two media is 90°. Which of the following statement is correct if the source of yellow light is replaced with that of other lights without changing the angle of incidence? [NCERT Exemplar]

plano-convex lens is 20 cm. If the refractive index of the material of the lens be 1.5, it will [NCERT Exemplar]

(a) The beam of red light would undergo total internal reflection (b) The beam of red light would bend towards normal while it gets refracted through the second medium (c) The beam of blue light would undergo total internal reflection (d) The beam of green light would bend away from the normal as it gets refracted through the second medium

43. A convex lens of focal length f produces a virtual image n times the size of the object. Then the distance of the object from the lens is (a) ( n - 1) f æ n - 1ö (c) ç ÷f è n ø

(b) ( n + 1) f æ n + 1ö (d) ç ÷f è n ø

(a) act as a convex lens only for the objects that lie on its curved side (b) act as a concave lens for the objects that lie on its curved side (c) act as a convex lens irrespective of the side on which the object lies (d) act as a concave lens irrespective of side on which the object lies

50. Consider an equiconvex lens of radius of curvature R and focal length f . If f > R, the refractive index m of the material of the lens (a) (b) (c) (d)

is greater than zero but less than 1.5 is greater than 1.5 but less than 2.0 is greater than one but less than 1.5 None of the above

51. A convex lens for ms an image of an object placed 20 cm away from it at a distance of 20 cm on the other side of the lens. If the object is moved 5 cm towards the lens, the image will move

1055

Ray option and Optical Instruments (a) (b) (c) (d)

5 cm towards the lens 5 cm away from the lens 10 cm towards the lens 10 cm away from the lens

i

(a)

shown if figure. If the space between them is filled with water, its power will

(b)

2

r

52. A convex lens is placed in contact with a mirror as

i

1

ir

1

(c)

1 2

r

1

(d)

2

2

59. What is the relation between refractive indices m1,m2 , (a) (b) (c) (d)

decrease increase remain unchanged increase or decrease depending on the focal length

and m 3 if the behaviour of light rays is as shown in figure.

µ1

53. The power of a thin convex lens ( a n g = 1.5) is + 5.0 D.

When it is placed in a liquid of refractive index a ne , then it behaves as a concave lens of focal length 100 cm. The refractive index of the liquid a nl will be (a) 5 / 3

(b) 4/3

(c)

3

(d) 5/4

54. A concave lens with unequal radii of curvature made of glass (m g = 15 . ) has focal length of 40 cm. If it is immersed in a liquid of refractive index m = 2, then (a) (b) (c) (d)

it behaves like a convex lens of 80 cm focal length it behaves like a concave lens of 20 cm focal length its focal length becomes 60 cm nothing can be said

55. An object approaches a convergent lens from the left of the lens with a uniform speed 5 m/s and stops at the focus. The image [NCERT Exemplar] (a) moves away from the lens with an uniform speed 5 m/s (b) moves away from the lens with an uniform acceleration (c) moves away from the lens with a non-uniform acceleration (d) moves towards the lens with a non-uniform acceleration

(a) m3 < m2 , m2 = m1 (c) m3 < m2 < m1

(c) 20 cm

æn ö (a) l1 ç 1 ÷ è n2 ø

contact has a power of + 2 D. The convex lens has power + 5D. What is the ratio of the dispersive powers of the convergent and divergent lenses? (b) 3 : 5

(c) 5 : 2

æn ö (b) l1 ç 2 ÷ è n1 ø

æ n - n1 ö (d) l1 ç 2 ÷ è n1 ø

(c) l1

61. The optical density of turpentine is higher than that of water while its mass density is lower. Figure shows a layer of turpentine floating over water in a container. For which one of the four rays incident on turpentine in figure the path shown is correct? [NCERT Exemplar] 1

2

3

4

Air

A

Turpentine B

Water

(d) 15 cm

57. An achromatic convergent doublet of two lenses in

(a) 2 : 5

(b) m2 < m1, m2 = m3 (d) m3 > m2 > m1

medium of refractive index, n1 enters a denser medium of refractive index, n2. The wavelength in the second medium is

by a convex lens when the object is 10 cm away from it. A real image twice as long as the object will be formed when it is placed at a distance……from the lens. (b) 30 cm

µ2

60. Monochromatic light of wavelength, l1 travelling in

56. A virtual image twice as long as the object is formed

(a) 40 cm

µ3

(d) 5 : 3

58. There are certain material developed in laboratories which have a negative refractive index (figure). A ray incident from air (medium 1) into such a medium (medium 2) shall follow a path given by [NCERT Exemplar]

(a) 1

(b) 2

(c) 3

(d) 4

62. What is the angle of incidence for an equilateral prism of refractive index 3 so that the ray is parallel to the base inside the prism? (a) 30° (c) 60°

(b) 45° (d) Either 30° or 60°

63. A car is moving with at a constant speed of 60 kmh -1 on a straight road. Looking at the rear view mirror, the driver finds that the car following him is at distance of 100 m and is approaching with a speed of 5 km h -1. In order to keep track of the car in the rear,

1056 JEE Main Physics the driver begins to glance alternatively at the rear and side mirror of his car after every 2 s till the other car overtakes. If the two cars were maintaining their speeds, which of the following statement (s) is/are correct? [NCERT Exemplar] (a) The speed of the car in the rear is 65 km h-1 (b) In the side mirror the car in the rear would appear to approach with a speed of 5 km h-1 to the driver of the leading car (c) In the rear view mirror, the speed of the approaching car would appear to decrease as the distance between the cars decreases (d) In the side mirror, the speed of the approaching car would appear to increase as the distance between the cars decreases

64. A ray of light travelling in glass

Water

æm = 3 ö is incident on a ÷ ç è 2ø horizontal glass air surface at the critical angle q c. If thin layer 4 of water æçm = ö÷ is now poured è 3ø

Glass R

R

(b) 45° (d) 180°

65. Light is incident from a medium X at an angle of incident i and is refracted into a medium Y at angle of refraction r. The graph sin i versus sin r is shown in figure. Which of the following conclusions would fit the situation? 1. Speed of light in medium Y is 3 times 0.2 that in medium X. sin r 2. Speed of light in 30° sin i medium Y is 1/ 3 times 0 0.4 0.2 that in medium X. 3. Total internal reflection will occur above a certain i value. (a) 2 and 3 (c) 2 only

(b) 1 and 3 (d) 3 only

66. The direction of ray of

(b) 2

(c) 3

Later on, it was found that a housefly was sitting on the objective lens of the telescope, in photograph, (a) (b) (c) (d)

there is a reduction in the intensity of the image there is an increase in the intensity of the image the image of housefly reduced the image of the housefly will be enlarged

68. The saparation as between two microscopic particles is measured PA and PB by two different lights of wavelength 2000 Å and 3000 Å respectively, then (a) PA < PB (c) PA = PB

(b) PA > PB (d) PA < 3 /2PB

69. The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eyepiece is found to be 20 cm. The focal lengths of the lenses are (a) 18 cm, 2 cm (c) 10 cm, 10 cm

(b) 11 cm, 9 cm (d) 15 cm, 5 cm

1 cm. 3.8 1 The focal length of objective lens is cm. What is the 4 magnification of eyepiece? the distance of object from objective lens is

(a) 5 (c) 100

(b) 10 (d) 200

71. The focal lengths of the objective and eyelenses of a microscope are 1.6 cm and 2.5 cm respectively. The distance between the two lenses is 21.7 cm. If the final image is formed at infinity, the distance between the object the objective lens is (a) 1.8 cm

(b) 1.70 cm (c) 1.65 cm

(d) 1.75 cm

72. Two points, separated by a distance of 0.1 mm, can just be inspected on a microscope when light of wavelength 6000Å is used. If the light of wavelength 4800Å is used, the limit of resolution is (a) 0.8 mm (c) 0.1 mm

(b) 0.08 mm (d) 0.04 mm

73. The diameter of moon is 3.5 ´ 103 km and its distance

1

light incident on a concave mirror is shown 2 by PQ while directions in which the ray would travel after reflection is shown by four rays marked 1, 2, 3 and 4 (figure). Which of the P 3 four rays correctly shows the direction of reflected ray? (a) 1

67. A photograph of the moon was taken with telescope.

70. In compound microscope, magnifying power is 95 and

on the glass air surface, the angle at which the ray emerges into air at the water-air surface is (a) 60° (c) 90°

Optical Instrument

Q

F

4

from the earth is 3.8 ´ 105 km. The focal length of the objective and eyepiece are 4 m and 10 cm respectively. The limit of resolution of diameter of the image of the moon will be approximately (a) 2° (c) 40°

(b) 21° (d) 50°

74. With diaphragm the camera lens set at f / 2, the correct exposure times is 1/100 s. Then with diaphragm set at f /8, the correct exposure time is (d) 4

(a) 1/100 s (c) 1/200 s

(b) 1/400 s (d) 16/100 s

Ray option and Optical Instruments 75. An object is viewed through a compound microscope and appears in focus when it is 5 mm away from the objective lens. When a sheet of transparent material 3 mm thick is placed between the objective and the microscope, the objective lens has to be moved 1 mm to bring to object back into the focus. The refractive index of the transparent material is (a) 1.5

(b) 1.6

(c) 1.8

(d) 2.0

76. A hypermetropic person having near point at a distance of 0.75 m puts on spectacles of power 2.5 D. The near point now is at (a) 0.75 m

(b) 0.83 m

(c) 0.36 cm

(d) 0.26 m

77. An astronomical telescope has a converging eyepiece of focal length 5 cm and objective of focal length 80 cm. When the final image is formed at the least distance of distinct vision (25 cm), the separation between the two lenses is (a) 75.0 cm (c) 84.2 cm

(b) 80.0 cm (d) 85.0 cm

(b) 2 cm (d) 0.5 ´ 10 -2 cm

84. A ray of light incident at an angle q on a refracting face of a prism emerges from the other face normally. If the angle of the prism is 5° and the prism is made of a material of refractive index 1.5, the angle of incidence is [NCERT Exemplar] (a) 7.5°

(b) 5°

astronomical telescope are respectively 2cm and 5cm. Final image is formed at (1) least distance of distinct vision (2) infinity. Magnifying powers in two cases will be

(c) 15°

(d) 2.5°

85. The refractive index of a prism for a monochromatic wave is 2 and its refracting angle is 60°. For minimum deviation, the angle of incidence will be (a) 30°

(b) 45°

(c) 60°

(d) 75°

86. It is desired to make a converging achromatic combination of mean focal length 50 cm by using two lenses of materials A and B. If the dispersive powers of A and B are in ratio 1 : 2, the focal lengths of the convex and the concave lenses are respectively (a) 25 cm and 50 cm (c) 50 cm and 100 cm

78. The focal length of objective and eye lens of an

(a) - 48, - 40 (c) - 40, + 48

(a) 200 cm (c) 0.5 cm

1057

(b) 50 cm and 25 cm (d) 100 cm and 50 cm

87. Two parallel light rays are incident at one surface of a prism of refractive index 1.5 as shown in figure. The angle between the emergent rays is nearly 30°

(b) - 40, 48 (d) - 48, + 40

i r

θ

θ θ

79. A man's near point is 0.5 m and far point is 3 m. Power spectacle lenses repaired for (i) reading purpose (ii) seeing distant object, respectively. (a) - 2 D and + 3 D (b) + 2 D and – 3 D (c) + 2 D and 0.33 D (d) – 2 D and + 0.33 D

80. A hypermetropic person has to use a lens of power + 5 D to normalise his vision. The near point of the hypermetropic eye is (a) 1 m

(b) 1.5 m

(c) 0.5 m

(d) 0.66 m

81. The focal length of the objective and the eyepiece of a microscope are 4 mm and 25 mm respectively. If the final image is formed at infinity and the length of the tube is 16 cm, then the magnifying power of microscope will be (a) – 337.5

(b) –3.75

(c) 3.375

(d) 33.75

82. A simple microscope consists of a concave lens of power – 10D and a convex lens of power + 20D in contact. If the image is formed at infinity, then the magnifying power when D = 25 cm is (a) 2.5

(b) 3.5

(c) 2.0

(d) 3.0

83. The magnifying power of an astronomical telescope is 10 and the focal length of its eye-piece is 20 cm. The focal length of its objective will be

(a) 19°

(b) 37°

(c) 45°

(d) 49°

88. The refractive index of the material of a prism is 2 and the angle of prism is 30°. One of its refracting faces is polished, the incident beam of light will retrace back for angle of incidence (a) 0°

(b) 45°

(c) 60°

(d) 90°

89. The cross-section of a glass prism has the form of an isoceles triangle. One of the refracting faces is silvered. A ray of light falls normally on the other refracting face. After being reflected twice, it emerges through the base of the prism perpendicular to it. The angles of the prism are (a) 54°, 54°, 72° (c) 45°, 45°, 90°

(b) 72°, 72°, 36° (d) 57°, 57°, 76°

90. The maximum refractive index of a prism which permits the passage of light through it, when the refracting angle of the prism is 90°, is (a)

3 3 (c) 2

(b) 3 (d) 2

2

1058 JEE Main Physics

Round 1. Two point sources A and B of luminous intensities 1 cd and 16 cd respectively are placed 100 cm apart. A grease spot screen is placed between the two sources. For the grease spot to become indistinguishable from both the sides, it should be placed at (a) 80 cm from 16 cd lamp and 20 cm from 1 cd (b) 20 cm from the 16 cd and 80 cm from 1 cd 400 100 (c) cm from 16 cd and cm from 1 cd 3 3 100 400 cm from 16 cd and cm from 1 cd (d) 3 3

2. A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. What happens as the [NCERT] needle is moved farther from the mirror? (a) (b) (c) (d)

Image goes on increasing Image goes on decreasing Image will be unchanged Image may change or change

(Mixed Bag) (a) Light of l = 400 nm undergoes (b) Light of l = 500 nm undergoes total internal reflection (c) Neither of the two wavelengths undergoes total internal reflection (d) Both wavelengths undergoes total internal reflection

7. An object is placed 30 cm to the left of a diverging lens whose focal length is of magnitude 20 cm. Which one of the following correctly states the nature and position of the virtual image formed? Nature of image Distance from lens (a) (b) (c) (d) (e)

inverted enlarged erect, diminished inverted, enlarged erect, diminished inverted, enlarged

8. A ray of light passes through four transparent

3. When an object is kept at a distance of 30 cm from a

medium with refractive indices m1, m2 , m 3 and m4 as shown in the figure. The surfaces of all media are parallel. If the emergent ray CD is parallel to the incident ray AB. We must have

concave mirror, the image is formed at a distance of 10 cm. If the object is moved with a speed of 9 ms -1, the speed with which images moves, is (a) 0.1 ms -1 (b) 1 ms -1

µ1 µ2 µ3

(d) 9 ms -1

(c) 3 ms -1

B

4. When a glass slab is placed on a cross made on a sheet, the cross appears raised by 1 cm. The thickness of the glass is 3 cm. The critical angle for glass is (a) sin -1( 0.33)

(b) sin -1( 0.5 )

(c) sin -1( 0.67)

(d) sin -1( 3 / 2 )

eye-piece as thin lenses of focal lengths 1 cm and 5 cm respectively. The distance between the objective and the eye-piece is 20 cm. The distance at which the object must be placed infront of the objective if the final image is located at 25 cm from the eyepiece, is numerically (c) 95/89 cm

6. Parallel beam containing light of l = 400 nm and 500 nm is incident on a prism as shown in figure. The refractive index m of the prism is given by the relation, 0 . 8 ´ 10-14 m( l) = 1.20 + l2

µ4 D C

A

(a) m1 = m2

(b) m = m3

(c) m3 = m 4

(d) m3 = m1

9. A diminished image of an object is to be obtained on a

5. A compound microscope has an objective and

(a) 95/6 cm (b) 5 cm

60 cm to the right 12 cm to the left 60 cm to the left 12 cm to the right 12 cm to the left

(d) 25/6 cm θ θ

Which of the following statement is correct?

screen 1.0 m away from it. This can be achieved by approximately placing (a) (b) (c) (d)

a convex mirror of suitable focal length a convex mirror of suitable focal length a convex lens of focal length less than 0.25 m a concave lens of suitable focal length

10. A ray light passes through an equilateral prism such that the angle of incidence and the angle of emergence are both equal to 3/4th of the angle of prism. The angle of minimum deviation is (a) 15° (c) 45°

(b) 30° (d) 60°

11. A Cassegrain telescope uses two mirrors as shown in figure. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be? [NCERT]

Ray option and Optical Instruments

Objective mirror

surface, so that it undergoes a total internal reflection. How much time would it take to traverse the straight fibre of light 1 km? Air

Secondary mirror Eye piece

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

235 mm away from small mirror 285 mm away from small mirror 305 mm away from small mirror 315 mm away from small mirror

12. A thin equiconvex lens of refractive index 3/2 and radius of curvature 30 m is put in water (refractive 4 index = ). Its focal length is 3 (a) 0.15 m (c) 0.45 m

(b) 0.30 m (d) 1.20 m

13. A 16 cm long image of an object is formed by a convex lens on a screen. On moving the lens towards the screen, without changing the position of the object and the screen, 9 cm long image is formed again on the screen. The size of the object is (a) 9 cm (c) 12 cm

1059

(b) 11 cm (d) 13 cm

Air

(a) 3.33 ms (c) 3.85 ms

Glass

60°

(b) 5.77 ms (d) 6.67 ms

18. A thin lens has focal length, f1 and its aparture has

diameter d. It forms an image of intensity I. Now the d is central part of the aparture upto diameter 2 blocked by an opaque paper. The focal length and image intensity will be change to I 4 f I (c) and 2 2 (a) f and

3I 4 3f I and (d) 4 2 (b) f and

19. Two similar plano-convex lenses are combined together in three different ways as shown in figure. The ratio of focal lengths in three cases will be

14. Two lenses, one concave and the other convex of same power are placed such that their principal axes coincide. If the separation between the lenses is x, then (a) (b) (c) (d)

real image is formed for x = 0 only real image is formed for all values of x system will behave like a glass plate for x = 0 virtual image is formed for all values of xother than zero

15. A ray of light falls on a transparent glass slab with refractive index (relative to air) of 1.62. The angle of incidence for which the reflected and refracted rays are mutually perpendicular is (a) tan -1(162 . ) -1

(c) cos (162 . )

(b) sin -1(162 . )

(d) None of these

16. A double convex lens made out of glass (refractive index, m = 15 . ) has both radii of curvature of magnitudes 20 cm. Incident light rays parallel to the axis of this lens will converge at a distance, d such that (a) d = 10 cm (c) d = 40 cm

20 cm 3 (d) d = 20 cm

(b) d =

17. A light ray from air is incident as shown in figure at one end of the glass fibre (refractive index, m = 1.5) making an incidence angle of 60° on the lateral

(a) 2 : 1 : 1 (c) 2 : 2 : 1

(b) 1 : 2 : 2 (d) 1 : 1 : 1

20. A glass convex lens (m g = 1.5) has a focal length of

8 cm when placed in air. What would be the focal length of the lens when it is immersed in water (m w = 1.33) ? (a) 32 cm (c) 4 cm

(b) 16 cm (d) 2 cm

21. A beam of electrons is used in a YDSE experiment to slit width is d, when the velocity of electrons is increased, then (a) (b) (c) (d)

no interference is observed fringe width increases fringe width decreases fringe width remains same

22. In the Young’s double slit experiment, the central maxima in observed to be I 0 . If one of the slits is covered, then the intensity at the central maxima will become I0 2 I0 (c) 4 (a)

(b)

I0 2

(d) I0

1060 JEE Main Physics 23. When an angle of incidence on a material is 60°, the reflected light is completely polarized. The velocity of the refracted ray inside the material in ms -1 is (a) 3 ´ 10 8 (c) 0.5 ´ 10

(b)

Water

6 cm h2

Glass

3 ´ 10 8

(d) 1.5 ´ 10 8

8

4 cm h1

24. A convex lens and a concave lens each having same focal length of 25 cm are put in contact of a combination of lenses. The power of the combination is (a) zero (c) 100

(b) infinite (d) 10

length 20 cm is 0.08. The longitudinal chromatic aberration in of the lens is (b) 1.6 cm (d) 0.16 cm

26. A rectangular glass slab ABCD of refractive index m1,

is immersed in water of refractive index m2 (m1 > m2 ). A ray of light is incident at the surface AB of the slab as shown. The maximum value of the angle of incidence a max , such that the ray comes out only from the other surface CD is given by

µ1

αmax

31. How many images are formed by the lens shown, if an object is kept on its axis? (a) (b) (c) (d)

µ1

1 2 3 4

1 cm O

C

20 cm 40 cm

(a) 32 cm

æ 1 öù (c) sin ê a1 × cos ç sin -1 ÷ ú è a2 ø û ë é -1 a1 ù (d) ê sin ú a2 û ë

(b) 0.6 cm

(c) 6 cm

(d) 0.5 cm

33. A thin plano-convex lens focal length f is split into

27. A ray of light makes an angle of 10° with the horizontal above it and strikes a plane mirror which is inclined at an angle q to the horizontal. The angle q for which the reflected ray becomes vertical is (c) 100°

µ2

µ=1.33

µ=1

-1 é

(b) 80°

(b) 666.64 m (d) 576.64 m

the position and nature of image.

éæm ö æ m öù (a) sin -1 ê ç 1 ÷ cos ç sin -1 2 ÷ ú è m1 ø û ë è m2 ø æa ö (b) sin -1 ç 2 ÷ è a1 ø

(a) 50°

(a) 566.64 m (c) 586.45 m

µ2 C

B

water. Given that m for water is 4/3 and the velocity of light in vacuum is 3 ´ 1010 cms -1. Calculate equivalent optical path.

32. For a optical arrangement shown in the figure. Find

D

A

(b) 8.0 cm (d) 5 cm

30. The light takes in travelling a distance of 500 m in

25. The dispersive power of the material of lens of focal (a) 0.08 cm (c) 0.8 cm

Coin

(a) 7.0 cm (c) 10 cm

two halves. One of the halves is shifted along the optical axis. The separation between object and image plane 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.

O

(d) 40°

28. A telescope has an objective of focal length 50 cm and an eyepiece of focal length 5 cm. The least distance of distinct vision is 25 cm. The telescope is focussed for distinct vision on a scale 200 cm away. The separation between the objective and eyepiece is (a) 74 cm

(b) 75 cm

(c) 60 cm

(d) 71 cm

29. A 4 cm thick layer of water covers a 6 cm thick glass slab. A coin is placed at the bottom of the slab and is being observed from the air side along the normal to the surface. Find the apparent position of the coin from

1.8 m

(a) 0.1 m

(b) 0.4 m

(c) 0.9 m

(d) 1 m

34. A plano-convex lens has a thickness of 4 cm. When placed on a horizontal table, with the curved surface in contact with it, the apparent depth of the bottom most point of the lens is found to be 3 cm. If the lens is inverted such that the plane face is in contact with the table, the apparent depth of the centre of the

Ray option and Optical Instruments plane face is found to be 25/8 cm. Find the focal length of the lens. Assume thickness to be negligible (a) 85 cm (c) 75 cm

(b) 59 cm (d) 7.5 cm

in between an object and a screen. The distance between object and screen is x. If numerical value of magnification produced by lens is m, focal length of lens is mx ( m + 1)2

(b)

( m + 1)2 (c) x m

mx ( m - 1)2

medium at angle of incidence i. The reflected and refracted rays make an angle of 90° with each other. The angles of reflection and refraction are r and r' respectively. The critical angle is (a) sin -1( tan r ¢ )

(b) sin -1( tan r )

(c) tan -1( tan r ¢ )

(d) tan -1(tan i )

37. If eye is kept at a depth h inside water of refractive index and viewed outside, then the diameter of the circle through which the outer objects become visible, will be

(c)

(b)

2

m -1 2h

(d)

2

m -1

h m2 + 1 h m2

38. A ray of light is incident at 60° on one face of a prism which has angle 30°. The angle between the emergent ray and incident ray is 30°. What is the angle between the ray and the face from which its emerges? (a) 0°

(b) 30°

(c) 60°

39. A small coin is resting on the

(d) 90° A

bottom of the beaker filled with a liquid. Aray of light from the coin travels upto the 4 cm surface of the liquid and moves along the surface O (figure). How fast is light travelling in the liquid?

(a) 0.88 m (c) 0.75 m

(b) 0.90 m (d) 0.63 m

magnifying power for relaxed eye is 25. If the focal length of eyelens is 5 cm, then the object distance for objective lens will be

36. A ray of light from a denser medium strikes a rarer

(a)

a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose? [NCERT]

42. The length of the compound microscope is 14 cm. The

( m - 1)2 (d) x m

h

(b) 4° (d) 2.6°

41. The image of a small electric bulb fixed on the wall of

35. A convex lens of focal length, f is placed somewhere

(a)

(a) 3° (c) 5.33°

1061

3 cm

(a) 2.4 ´ 10 8 m/s

(b) 1.8 ´ 10 8 m/s

(c) 3.0 ´ 10 8 m/s

(d) 5.0 ´ 105 m/s

B

40. A thin prism P1 with angle 4° and made from glass of

refractive index 1.54 is combined with another thin prism P2 made from glass of refractive index 1.72 to produce dispersion without deviation. The angle of prism P2 is

(a) 2.4 cm (c) 1.5 cm

(b) 2.1 cm (d) 1.8 cm

43. A simple telescope, constiting of an objective of focal length 60 cm and a single eyelens of focal length 60 cm and a single eyelens of focal length 5 cm is focussed on a distant object is such a way that parallel rays come out from the eyelens. If the object subtends an angle 2° at the objective, the angular width of the image (a) 10°

(b) 24°

(c) 35°

(d) 48°

44. 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) 21 cm (c) 15 cm

(b) 11 cm (d) 17 cm

45. P is a point on the axis of a concave mirror. The image of P formed by the mirror, coincides with P. A rectangular glass slab of thickness t and refractive index m is now introduced between P and the mirror. For image of P to coincide with P again, the mirror must be moved (a) towards P by (m - 1)t 1ö æ (c) towards P by t ç1 - ÷ è mø

(b) away from P by (m - 1)t 1ö æ (d) away from P t ç1 - ÷ è mø

46. A point object is placed at a distance of 25 cm from a convex lens of focal length 20 cm. If a glass slab of thickness, t and refractive index, 1.5 is inserted between the lens and the object, the image is formed at infinity. The thickness, t is (a) 15 cm

(b) 5 cm

(c) 10 cm

(d) 20 cm

47. A plano-convex lens fits exactly into a plano-concave lens. Their plane surfaces are parallel to each other. If the lenses are made of different materials of refractive indices m1 and m2 and R is the radius of curvature of the curved surface of the lenses, then focal length of the combination is

1062 JEE Main Physics R 2 (m1 + m2 ) R (c) (m1 - m2 )

R 2 (m1 - m2 ) 2R (d) (m1 + m2 )

(a)

53. A point object is placed at the centre of a glass sphere

(b)

of radius, 6 cm and refractive index, 1.5. The distance of the virtual image from the surface of the sphere is

48. One of the refracting surfaces of a prism of angle 30° is silvered. A ray of light incident at an angle of 60° retraces its path. The refractive index of the material of prism is (a)

(b) 3/2

3

(c) 2

(d)

2

49. The focal length of objective and eyepiece of a microscope are 1 cm and 5 cm respectively. If the magnifying power for relaxed eye is 45, then of the tube is (a) 9 cm

(b) 15 cm

(c) 12 cm

(d) 6 cm

50. A glass prism ABC (refractive index 1.5), immersed in water (refractive index 4/3). A ray of light is incident normally on face AB. If it is totally reflected at face AC, then

(a) 2 cm (c) 6 cm

(b) 4 cm (d) 12 cm

54. In order of obtain a real image of magnification 2, using a converging lens of focal length 20 cm, where should an object be placed (a) 50 cm (c) -50 cm

(b) 30 cm (d) -30 cm

55. A plano-convex lens of refractive index 1.5 and radius of curvature 30 cm is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens an object be placed in order to have a real image of the size of the object? (a) 20 cm (c) 60 cm

(b) 30 cm (d) 80 cm

56. Double-convex lenses are to be manufactured from a B

θ

glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required, if the focal length is to be 20 cm?

A

[NCERT]

(a) 18 m (c) 17 cm

C

(a) sin q ³ (c) sin q =

8 9 3 2

(b) sin q ³ (d)

(c)

2h m -1

(b)

are fV , fG and fR respectively. Which of the following is the true relationship?

2 8 < sin q < 3 9

of a tank containing a liquid of refractive index m. P is a small object at a height h above the mirror. An observer O vertically above P outside the liquid sees P and its image in a mirror. The apparent distance between these two will be (a) 2mh

57. The focal lengths for violet, green and red light rays

2 3

51. A plane mirror as placed at the bottom

(a) fR < fG < fV (c) fG < fR < fV O

P h

2h m

lenses is + 2D. The power of convex lens is + 5D. The ratio of dispersive power of convex and concave lens will be (a) 5 : 3

1ö æ (d) hç1 + ÷ è mø

and the same light takes t2 second to travel 10x cm in a medium. Critical angle for corresponding medium will be -1 æ

t ö (b) sin ç 2 ÷ è10t1 ø

æ10t ö (c) sin -1 ç 1 ÷ è t2 ø

æ t ö (d) sin -1 ç 1 ÷ è10t2 ø

(b) 3 : 5

(c) 2 : 5

(d) 5 : 2

59. The radius of the convex surface of plano-convex lens is 20 cm and the refractive index of the material of the lens is 1.5. The focal length of the lens is (a) 30 cm

ö (a) sin ç ÷ è t1 ø

(b) fV < fG < fR (d) fG < fV < fR

58. The power of an achromatic convergent lens of two

52. Light takes t1 second to travel a distance x in vacuum

-1 æ 10t2

(b) 22 cm (d) 26 cm

(b) 50 cm

(c) 20 cm

(d) 40 cm

60. A combination of two thin convex lenses of focal length 0.3 m and 0.1 m will have minimum spherical and chromatic aberrations if the distance between them is (a) 0.1 m

(b) 0.2 m

(c) 0.3 m

(d) 0.4 m

61. In a given direction, the intensities of the scattered light by scattering substance for two beams of light are in the ratio of 256 : 81. The ratio of the frequency of the first beam to the frequency of the second because (a) 4 : 3

(b) 16 : 9

(c) 2 : 1

(d) 3 : 4

Ray option and Optical Instruments 62. An astronomical telescope has an angular magnification of magnitude 5 for distant object. The separation between the objective and the eyepiece is 36 cm and the final image is formed at ¥. The focal length f0 of the objective and the focal length fe of the eyepiece are (a) f0 = 30 cm and fe = 6 cm (b) f0 = 15 cm and fe = 12 cm (c) f0 = 8.5 cm and fe = 12.9 cm (d) f0 = 40 cm and fe = 11 cm

near point of distinct vision is about 25 cm in front of the eye. The cornea of the eye provides a converging power of about 40 D, and the least converging power of the eye lens behind the cornea is about 20 D. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye lens) of [NCERT] a normal eye. (a) 10 to14 D (c) 28 to 32 D

(b) 20 to 24 D (d) 14 to 18 D

64. A beam of parallel rays is brought to a focus by a plano-convex lens. A thin concave lens of the same focal length is joined to the first lens. The effect of this is (a) the focal points shifts away from the lens by a small distance (b) the focus remains undisturbed (c) the focus shifts to infinity (d) the focal points shifts towards the lens by a small distance

θ 45°

66. A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a [NCERT] parallel beam of light. (a) 10° (c) 23°

(b) 20° (d) 18°

V

(a)

(b)

V R

R (d) V

R

R

68. Following figure shows the

V

R

multiple reflections of a light T ray along a glass corridor Q S where the walls are either P parallel or perpendicular to one another. If the angle of incidence at point P is 30°, what are the angle of reflection of the light ray at points Q, R, S and T, respectively? (a) 30°, 30°, 30°, 30° (c) 30°, 60°, 60°, 30°

(b) 30°, 60°, 30°, 60° (d) 60°, 60°, 60°, 60°

69. When the rectangular metal tank is filled to the top with an unknown liquid, as 3 cm observer with eyes level with the top of the tank can just 4 cm see the corner E; a ray that refracts towards the observer at the top surface of the liquid is shown. The refractive index of the liquid will be (b) 1.4

(c) 1.6

(d) 1.9

70. A concave mirror and a converging lens (glass with

shown in the figure. A ray incident normally to one face is totally reflected, if, q = 45°. The index of refraction of glass is less than 1.41 equal to 1.41 greater than 1.41 None of the above

dispersion of white light by a prism?

(a) 1.2

65. A triangular prism of glass is

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

67. Which of the following diagrams shows correctly the

(c)

63. For a normal eye, the far point is at infinity and the

1063

m = 1.5) both have a focal length of 3 cm when in air. 4 When they are in water æçm = ö÷, their new focal è 3ø length are (a) fLens = 12 cm, fMirror = 3 cm (b) fLens = 3 cm, fMirror = 12 cm (c) fLens = 3 cm, fMirror = 3 cm (d) fLens = 12 cm, fMirror = 12 cm

71. A ray of light strikes a plane mirror M at an angle of 45° as shown in the figure. After reflection, the ray passes through a prism of refractive index 1.5 whose apex angle is 4°. The total angle through which the ray is deviated is (a) 90° (c) 92°

(b) 91° (d) 93°

45°

4°

1064 JEE Main Physics More Than One Correct Option 72. A point object is at 30 cm from a convex glass lens

æm = 3 ö of focal length 20 cm. The final image of ÷ ç s è 2ø object will be formed at infinity if (a) another concave lens of focal length 60 cm is placed in contact with the previous lens (b) another convex lens of focal length 60 cm is placed at a distance of 30 cm from the first lens (c) the whole system is immersed in a liquid of refractive index 4/3 (d) the whole system is immersed in a liquid of refractive index 9/8

73. Consider an extended object immersed in water

76. A ray of light travelling in a transparent medium falls on a surface separating the medium from air, at an angle of incidence of 45°. The ray undergoes total internal reflection. If n is the refractive index of the medium with respect to air, select the possible values of n from the following (a) 1.3 (c) 1.5

(b) 1.4 (d) 1.60

77. A diminished image of an object is to be obtained on a screen 1.0 m from it. This can be achieved by appropriately placing (a) (b) (c) (d)

A concave mirror of suitable focal length A convex mirror of suitable focal length A convex lens of focal length less than 0.25 m A concave lens of suitable focal length

contained in a plane trough. When seen from close to the edge of the trough the object looks distorted because [NCERT Exemplar]

78. A planet is observed by an astronomical refracting

(a) the apparent depth of the points close to the edge are nearer the surface to the water compared to the points away from the edge. (b) the angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air (c) some of the points of the object far away from the edge may not be visible because of total internal reflection (d) water in a trough acts as a lens and magnifies the object

(a) The distance between the objective and the eyepiece is 16.02 m (b) The angular magnification of the planet is -800 (c) The image of the planet is inverted (d) The objective is larger than the eyepiece

74. There are three optical media, 1,2 and 3 with their refractive indices m1 > m2 > m 3. (TIR– total internal reflection) (a) When a ray of light travels from 3 to 1 no TIR will take place (b) Critical angle between 1 and 2 is less than the critical angle between 1 and 3 (c) Critical angle between 1 and 2 is more than the critical angle between 1 and 3 (d) Chances of TIR are more when ray of light travels from 1 to 3 compare to the case when it travel from 1 to 2

75. Parallel rays of light are falling on convex spherical surface of radius of curvature R = 20 cm as shown. Refractive index of the medium is m = 15 . . After refraction from the spherical surface parallel rays µ=1.5

(a) actually meet at some point (b) appears to meet after extending the refracted rays backwards (c) meet (or appears to meet) at a distance of 30 cm from the spherical surface (d) meet (or appears to meet) at a distance of 60 cm from the spherical surface

telescope having an objective of focal length 16 m and an eyepiece of focal length 2 cm

79. A converging lens is used to form an image on a screen. When the upper half of the lens is covered by an opaque screen (a) (b) (c) (d)

Half the image will disappear Complete image will be formed Intensity of the image will increase Intensity of the image will decrease A

B

[NCERT Exemplar] D

C

80. A rectangular black of glass ABCD has a refractive index 1.6. A pin is placed midway on the face AB (figure). When observed from the face AD, the pin shall (a) (b) (c) (d)

appear to be near A appear to be near D appear to be at the center of AD not be seen at all

81. Between the primary and secondary rainbows, there is a dark band known as Alexander’s dark band. This is because [NCERT Exemplar] (a) light scattered into this region interfere destructively (b) there is no light scattered into this region (c) light is absorbed in this region (d) angle made at the eye by the scattered rays with respect to the incident light of the sun lies between approximately 42° and 50°

Ray option and Optical Instruments 82. A convex lens forms a real image of a point object placed on its principal axis. If the upper half of the lens is painted black, [NCERT Exemplar] (a) the image will be shifted upward (b) the image will be shifted downward (c) the intensity of the image will decrease (d) the image will not be shifted

index m of glass is 3/2, is (a) 41.8° (c) 30°

(b) 60° (d) 44.3°

is 49°. Its value for red colour would be

can be brougth closer to the eye than the normal near point. This results in [NCERT Exemplar] (a) a larger angle to be subtended by the object at the eye and hence viewed in greater detail (b) the formation of a virtual erect image (c) increase in the field of view (d) infinite magnification at the near point

84. An astronomical refractive telescope has an objective of focal length 20 m and an eyepiece of focal length [NCERT Exemplar] 2 cm. The length of the telescope tube is 20.02 m The magnification is 1000 The image formed is inverted An obdective of a larger aperture will increase the brigthtness and reduce chromatic aberration of the image

85. If a hollow convex shaped glass is filled with water and surrounding is glass. The lens will act as (a) (b) (c) (d)

88. Critical angle for glass air-interface where refractive

89. Critical angle for air water interface for violet colour

83. A magnifying glass is used, as the object to be viewed

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

1065

convex lens concave lens glass plate convex mirror

(a) 49° (c) 48°

(b) 50° (d) cannot say

90. A point source of light is held at a depth h below the surface of water. If C is critical angle of air water interface, the diameter of circle of light coming from water surface would be (a) 2 h tan C (c) h sin C

(b) h tan C (d) h /sin C

91. Choose the correct statement. (a) The speed of light in the meta-material is v = c |m | c (b) The speed of light in the meta-material is v = |m | (c) The speed of light in the meta-material is v = c (d) The wavelength of the light in the metal-material ( l m ) is given by l m = l ab |a |, where l air is the wavelength of the light in air.

92. For light incident from air on a metal-material the appropriate ray diagram is θ1 Air (a) Metamaterial θ2

θ1

(b)

Air Metamaterial

θ2

Comprehension Based Questions Passage I Total internal reflection is the phenomenon of reflection of light into denser medium at the interface of denser medium with a rarer medium. Light must travel from denser to rarer and angle incidence denser medium must be greater than critical angle (C) for the pair of media in contact. We can show that 1 m= sin C

86. Critical angle for water air interface is 48.6°. What is the refractive index of water? (a) 1

(b) 3/2

(c) 4/3

(d) 3/4

87. Light is travelling from air to water at Ði = 50°, which is greater than critical angle for air water interface. What fraction of light will be totally reflected? (a) 100% (c) None of them

(b) 50% (d) Cannot say

(c)

θ1 Air Metamaterial θ 2

θ1

(d)

Air Metamaterial θ2

93. The following statements carefully to mark correct option are options given below (a) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1 (b) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1 (c) Statement 1 is true, Statement 2 is false (d) Statement 1 is false, Statement 2 is true Statement 1 The formula connecting u, v and f, for a spherical mirror is valid only for mirrors whose sizes are very small compared to their radii of curvature.

Statement 2 Laws of reflection are strictly valid for plane surface, but not for large spherical surfaces.

1066 JEE Main Physics Matching Type 94. An optical component and an object S placed along optical axis are given in Column I. The distance between the object and the component can be varied. The properties of images are given in Column II. Match all the properties of images from Column II with the appropriate components given in Column I. Column I S

I.

S

II.

Column II

A.

Real image

B.

Virtual image

(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 correct explanation of the Assertion (c) If Assertion is true but Reason is false (d) If Assertion is false but the Reason is true

96. Assertion Convergent lens property of converging remain same in mediums. Reason Property of lens whether the ray is diverging or converging depends on the surrounding medium.

97. Assertion A concave mirror and convex lens both have the same focal length in air. When they are submerged in water, they will still have the same focal length. Reason The refractive index of water is greater than the refractive index of air.

98. Assertion The focal length of the objective of the S

III.

C.

Magnified image

telescope is larger than that of eyepiece. Reason The resolving power of telescope increases when the aperture of objective is small. A short sighted person cannot see objects clearly when placed beyond 50 cm. He should use a concave lens of power 2 D. Reason Concave lens should form image of an object at infinity placed at a distance of 50 cm.

99. Assertion S

IV.

D.

Image at infinity

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

I-A, B, C, D; II-B; III-A, B, C, D; IV-A, B, C, I-A, B II-B; C; III-A, B, C IV-D I-B, C; II-A, B; III-C, D; IV-A, C, D I-A, B, C; II-B, C; III-D; IV-A, B

95. A simple telescope used to view distant object has eyepiece and objective lenses of focal lengths fe and fo , respectively. Then

II. III.

Column I Intensity of light received (A) by lens Angular magnification (B) Length of telescope (C)

IV.

Sharpness of image

I.

Column II Radius of aperture

Dispersion of lens Focal length of objective lens and eyepiece lens (D) Spherical aberration

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

I-A; II-C; III-C; IV-A, B; C I-A; II-B; III-C; IV-D I-D; II-B; C; III-A; IV-C; D I-C, D; II-B; C; D; III-A; B; IV-A

Assertion and Reason Direction

Question No. 96 to 105 are Assertion-Reason type. Each of these contains two Statements : Statement I (Assertion), Statement II (Reason). Each of these questions also has four alternative choice, only one of which is correct. You have to select the correct choices from the codes (a), (b) (c) and (d) given below

By roughening the surface of a glass sheet its transparency can be reduced. Reason Glass sheet with rough surface absorbs more light.

100. Assertion

The refractive index of diamond is 6 and that of liquid is 3. If the light travels from diamond to the liquid , it will initially reflected when the angle of incidence is 30°. 1 Reason m = where m is the refractive index of sin C diamond with respect to liquid.

101. Assertion

102. Assertion The colour of the green flower seen through red glass appears to be dark. Reason Red glass transmits only red light.

103. Assertion The stars twinkle while the planets is do not. Reason The stars are much brgger in size than planets.

104. Assertion Owls can move freely during night. Reason They have large number of rods on their retina.

105. Assertion The focal length of the mirror is f and distance of the object from the focus is u, the magnification of the mirror is f/u. Size of image Reason Magnification = Size of object

Ray option and Optical Instruments

1067

Previous Years’ Questions 106. Let the xz-plane be the boundary between two

110. The initial shape of the wavefront of the beam is

transparent media. Medium 1 in z ³ 0 has a refractive index of 2 and medium 2 with z < 0 has a refractive index 3. A ray of light in medium 1 given

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

^

by the vector A = 6 3 ^i + 8 3 ^j - 10 k is incident on the plane of separation. The angle of refraction in medium 2 is [AIEEE 2011] (a) 45°

(b) 60°

(c) 75

111. The speed of light in the medium is (a) (b) (c) (d)

(d) 30°

107. This question has a paragraph followed by two statements, Statement 1 and Statement 2. Of the given four alternatives after the statements, choose the one that describes the statements. A thin air film is formed by putting the convex surface of a plano-convex lens over a plane glass plate. With monochromatic light, this film gives an interference pattern due to light reflected from the top (convex) surface and he bottom (glass plate) [AIEEE 2011] surface of the film. Statement 1 When light reflects from the air-glass plate interface, the reflected wave suffers a phase change of p. Statement 2 The centre of the interference pattern is dark. (a) Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation of Statement 1. (b) Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation of Statement 1. (c) Statement 1 is false, Statement 2 is true. (d) Statement 1 is true, Statement 2 is false.

108. A car is fitted with a convex side-view mirror of focal length 20 cm. A second car 2.8 m behind the first car is overtaking the first car at a relative speed of 15m/s. The speed of the image of he second car as seen in the [AIEEE 2011] mirror of the first one is (a)

1 m/s 15

(b) 10 m/s

(c) 15 m/s

(d)

1 m/s 10

Direction (Q. No. 109-111) Based on the following paragraph. An initially parallel cylindrical beam travels in a medium of refractive index, m( I ) = m 0 + m 2 I , where m 0 and m 2 are positive constants and I is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius.

109. As the beam enters the medium, it will (a) (b) (c) (d)

[AIEEE 2010]

diverge converge diverge near the axis and converge near the periphery travel as a cylindrical beam

convex [AIEEE 2010] concave convex near the axis and concave near the periphery planar [AIEEE 2010]

minimum on the axis of the beam the same everywhere in the beam directly proportional to the intensity maximum on the axis of the beam

112. An object 2.4 m infront of a lens forms a sharp image on a film 12 cm behind the lens. A glass plate 1 cm thick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object shifted to be [AIEEE 2012] in sharp focus on film? (a) 7.2

(b) 2.4

(c) 3.2

(d) 5.6

113. A transparent solid cylinder rod has a refractive

2 . It is surrounded by air. A light ray is 3 incident at the mid point of one end of the rod as [AIEEE 2009] shown in the figure.

index of

θ

The incident angle q for which the light ray grazes along the wall of the rod is [AIEEE 2009] æ 1ö (a) sin -1 ç ÷ è 2ø

æ 3ö (b) sin -1 ç ÷ è 2 ø

æ 2 ö (c) sin -1 ç ÷ è 3ø

æ 1 ö (d) sin -1 ç ÷ è 3ø

114. A thin lens of glass (m = 1.5) of focal length + 10 cm is immersed in water (m = 1.33). The new focal length is [AIEEE 2009]

(a) 12 cm (c) 40 cm

(b) 20 cm (d) 40 cm

115. A

student performed the experiment of determination of focal length of a concave mirror by u-v method using an optical bench of length 1.5 m. The focal length of the mirror used is 24 cm. The maximum error in the location of the image can be 0.2 cm. The 5 sets of (u,v) values recorded by the student (in cm) are (42,56) (48,48) (60,40) (63,33) (78,39). The data set (s) that cannot come from experiment and is (are) recorded is (are) [AIEEE 2009] (a) (42,56) (c) (66,33)

(b) (48,48) (d) (78,39)

1068 JEE Main Physics 116. A ray OP of monochromatic light is incident on the

118. In an optics experiment, with the position of the

face AB of prism ABCD near vertex. B at an incident angle of 60° (see figure). If the refractive index of the material of the prism is 3, which of the following is [IIT JEE 2010] (are) correct?

object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance u and the image distance v, from the lens, is plotted using the same scale for the two axes. A straight line passing through origin and making an angle of 45° with the x-axis meets the experimental [AIEEE 2009] curve at P. The coordinate of P will be

B

O

C

60° 135° P

90°

75°

A

D

(a) The ray gets totally internally reflected at face CD (b) The ray comes out through face AD (c) The angle between the incident ray and the emergent ray is 90° (d) The angle between the incident ray and the emergent ray is 120°

æf f ö (a) ç , ÷ è 2 2ø

(b) (f, f)

(c) (4f, 4f)

(d) (2f, 2f)

119. The graph between angle of deviation ( d) and angle of incidence ( i) for triangular prism is represented by [JEE main 2013] δ

(a)

t

δ

(d)

(c)

1 (d) m +1

(c) (m + 1)

O

t

δ

is m times the focal length of the lens. The linear magnification produced by the lens is (b) 1/m

(b) O

117. The distance between an object and a divergent lens

(a) m

δ

O

O

t

t

Answers Round I 1. 11. 21. 31. 41. 51. 61. 71. 81.

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

2. 12. 22. 32. 42. 52. 62. 72. 82.

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

3. 13. 23. 33. 43. 53. 63. 73. 83.

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

4. 14. 24. 34. 44. 54. 64. 74. 84.

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

5. 15. 25. 35. 45. 55. 65. 75. 85.

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

6. 16. 26. 36. 46. 56. 66. 76. 86.

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

7. 17. 27. 37. 47. 57. 67. 77. 87.

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

8. 18. 28. 38. 48. 58. 68. 78. 88.

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

9. 19. 29. 39. 49. 59. 69. 79. 89.

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

10. 20. 30. 40. 50. 60. 70. 80. 90.

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

Round II 1. (a) 11. (d) 21. (c) 31. (a) 41. (c) 51. (b) 61. (a) 71. (c) 81. (a,d) 91. (b) 101. (a) 111. (a)

2. (b) 12. (d) 22. (c) 32. (b) 42. (d) 52. (c) 62. (a) 72. (a,d) 82. (c,d) 92. (c) 102. (a) 112. (d)

3. (b) 13. (c) 23. (b) 33. (b) 43. (b) 53. (c) 63. (b) 73. (a,b,c) 83. (a,b) 93. (c) 103. (b) 113. (d)

4. (c) 14. (b,c ) 24. (a) 34. (c) 44. (b) 54. (b) 64. (c) 74. (a,d) 84. (a,b,c) 94. (a) 104. (c) 114. (d)

5. (c) 15. (a) 25. (b) 35. (a) 45. (c) 55. (a) 65. (c) 75. (a,d) 85. (b) 95. (a) 105. (a) 115. (a,b,c)

6. (a) 16. (d) 26. (a) 36. (b) 46. (a) 56. (b) 66. (a) 76. (c,d) 86. (c) 96. (d) 106. (a) 116. (d)

7. (b) 17. (c) 27. (d) 37. (c) 47. (c) 57. (b) 67. (c) 77. (a,c) 87. (c) 97. (d) 107. (b) 117. (d)

8. (a) 9. (c) 10. (b) 18. (b) 19. (d) 20. (a) 28. (d) 29. (a) 30. (b) 38. (d) 39. (b) 40. (a) 48. (a) 49. (b) 50. (a,b) 58. (b) 59. (d) 60. (b) 68. (c) 69. (a) 70. (a) 78. (a,d) 79. (b,c) 80. (a,d) 88. (a) 89. (c) 90. (a) 98. (d) 99. (a) 100. (c) 108. (a) 109. (b) 110. (d) 118. (d) 119. (c

the Guidance Round I 1. We have, m = 2. As, DF =

f -16 -16 = = =2 f - u -16 - ( -8) -8

d2 l

or

3 ´ 10 -3 ´ 3 ´ 10 -3 90 m = 18 m = DF = 5 500 ´ 10 -9

3.

I I I As, E1 = 2 , and E 2 = 2 + 2 r 9r r \

E1 I 9r 2 = ´ E 2 r 2 10I

or

E 2 10 = E1 9 n=

4. Here, Note that

360° =5 72°

1 1 1 + = -30 v 30 1 2 1 = = v 30 15

v = 15 cm The distance is behind the mirror. or

12. By keeping the incident ray is fixed, if plane mirror rotates through an angle qreflected ray rotates through an angle 2q. θ 2θ

1 1 1 + = u v f 1 1 1 + = f - x1 f - x2 f

Now, Þ

1 1 1 + = u v f 1 1 1 =- + v u f

f - x2 + f - x1 1 = ( f - x1) ( f - x2) f

or or

Now, compare it with equation, y = mx + c. Therefore, graph is a straight line having negative slope.

f 2 - fx2 - fx1 + x1x2 = 2f 2 - f ( x1 + x2) f 2 = x1x2

or

f = x1x2

6. Think in terms of rectangular hyperbola.

or

7. The angular range is clearly twice the critical angle.

This is Newton’s mirror formula. f As, m = f -u

8. From the mirror formula,

or or or

9. As, or or

1 1 1 + = v -600 20 1 1 1 = + v 20 600 1 31 = v 600 600 v= cm = 19.35 cm 31 f 1 f = = f - u 4 f - ( -0.5) 4f = f + 0.5 or 3f = 0.5 0.5 f= m = 0.17 m 3

10. The image is erect and diminished. So, the mirror is necessarily convex.

θ θ

13. As, u = f - x1 and v = f - x2

360 is odd when object is placed asymmetrically. q

5. We have, or

11. We have,

14.

Þ

2=

-0.2 -0.2 - u

or

2=

0.2 0.2 + u

or

0.4 + 2u = 0.2 2u = 0.2 - 0.4 = - 0.2

or

u = - 0.1m

or

15.

f As, m = f -u Þ or or

-

1 -12 12 = = 4 -12 - u 12 + u

12 + u = - 48 u = - 60 cm

1070 JEE Main Physics 16. As, m =

f f -u

20. Here, f = -10 cm, O = 5 cm,

Þ

-2 =

-20 -20 - u

Now, as,

Þ

-2 =

20 20 + u

Þ

or

20 + u = - 10 u = - 30 cm

or

21.

17. Clearly, the given mirror is a convex mirror. As

f m= f -u

Þ

1 18 = 3 18 - u

or

3 ´ 18 = 18 - u

or or

18.

u = - 100 cm, I = ? I f m= = O f -u I=

-10 -10 ´5 = ´ 5 cm -10 - ( -100) 90

= - 0.55 cm 1.6 Here, f = m = 0.8 m, u = - 1 m 2 1 1 1 10 18 9 we have, = = + 1= = v 0.8 -1 8 8 4 4 or v= m 9 (Positive sign indicates that the distance is behind the mirror)

u = - 2 ´ 18 cm u = - 36 cm

1 p We have, q = ´ rad 2 180 diameter of image As, =q focal length 1 15 p or diameter of image = ´ ´ ´ 100 cm = 6.55 cm. 2 180 2

æ f ö ÷b è f -uø

22. Length of image = ç

23. Total deviation = (180° - 2a) + (180° - 2b) = 360° - 2( a + b) But, 90° - a + 90° - b + q = 180° q=a+b

or

19. Given, radius of curvature of concave mirror, R = - 36 m (For concave mirror radius of curvature is taken as negative) R 36 \ Focal length f = = = - 18 cm 2 2 u = - 27 cm (Object distance is always taken as negative)

90–β β

α α θ

Distance of object

O = 2.5 cm Use the mirror formula 1 1 1 1 1 1 = = + Þ 18 v 27 f v u 1 1 1 1 -3 + 2 =+ = =v 18 27 54 54 Height of object

C

F 36 cm

β

90–α

\ Total deviation = 360° - 2q

P

24. O

I x

x

O

I

Distance of screen from mirror v = - 54 cm Let the size of image be I. By using the formula of magnification for mirror v I m== u O - ( -54) I = - 27 2.5 I = - 5 cm The negative sign shows that the image is formed in front of the mirror and it is inverted. Thus, the screen should be placed at a distance 54 cm and the size of image is 5 cm, real, inverted and magnified in nature.

x– v

x– v

It is clear from figure, the new distance is 2x - 2v. The distance of image from object is reduced by an amount 2v in one second. Thus, the required speed is 2 ´ 10 = 20 cms-1 in direction your own speed i. e. , positive.

25. The first images is due to reflection from the front surface i. e. , in unpolished surface of the mirror. So, only a small fraction is the incident light energy is reflected. The second image is due to reflection from polished surface. So, a major portion of light is reflected. Thus, the second image is the brightest.

Ray Optics and Optical Instruments sin ic =

Plane mirror

26. 10 cm O' x Observer x + y – 10

I'

I y

\

or

27. Clearly, the distance of image from observer is 40 cm. 10 cm

10 cm

r1 + ic = A

r1 + 41° = 60° Þ r1 = 19° sin i1 Using the formula, m= sin r1

It is clear from figure the distance of image with reference to observer reduces by 10 cm in one second. Thus, required speed is 20 - 10 = 10 cms-1.

O

sin i1 = 1.524 sin 19° = 1.524 ´ 0.3256 i1 = sin - 1 (0.4962)

or

I

30 cm

32.

Observer

i1 = 29° 75 ¢ Thus, the angle should be 29° 75¢. f As, magnification, m = f -u If, f = - 24 and

28. Clearly, the image is upright and real.

or or

-24 - u = - 8 u + 24 = 8

i = 70°

u + 24 = - 8

40°

u = - 32 cm

or

20° 70°

33. Only the light-gathering maker reduced.

70°

30.

m = + 3, then -24 3= -24 - u

or u = 8 - 24 cm = - 16 cm Now, if, m = - 3, then -24 -3 = -24 - u

29. Clearly, i + r = i + i = 140° or

1 1 = = 0.6561 m 1.524

ic = 41° For a prism r1 + r2 = A here r2 = ic

10 cm

O

1071

34. Think in terms of rectangular hyperbola.

Clearly, the plane mirror makes an angle of 20° with vertical and 70° with horizontal. Thus b is 70°. I f As, = d u

æ1 1ö - ÷ (Lens macer formula è R1 R2 ø

35. As, P = (m - 1) ç

1 m decreases, P decreases and f increases because, P = ) f

36. Power of the system decreases due to separation between the lenses. So, the focal length increases.

d

37. Note that two refractive indices are involved. Therefore, two images will be formed. I

38. Let the bulb is placed at point O AB = AC = r

or

31. Angle of prism,

I=

d f or I = qf u

A = 60°

A

Refractive index of prism m = 1.524 Let i be the angle of incidence. The critical angle is ic because it just suffers total internal refraction, so we use critical angle.

If the light falls at an angle of incidence equal to critical angle ic , then only a circular area is formed because if angle of incidence is less than the critical angle it will refract into air and when angle of incidence is greater than critical angle then it will be reflected back in water.

60° i1

B

r1

ic C

C

ic

A

ic ic

B

ic

O

The source of light is 80 cm below the surface of water i. e. AO = 80 cm, mw = 1.33

1072 JEE Main Physics Using the formula for critical angle, 1 sin ic = mw 1 sin ic = = 0.75 1.33 ic = 48.6° AB In DOAB tan ic = AO r or tan ic = l

44. As, m =

45.

\ 1 f

or

d 1 1 + =0 f1 f2 f1f2

47.

d = - 36 cm

41. As seen from a relatively rarer medium (L2 or L3) the interface

42. Here, for yellow light, r = 90° when i = C. As i is kept same, C

43.

must be smaller for total internal reflection. From 1 m= , C will be smaller, when m is larger. Out of given sin C colours, m is largest for blue colour. Critical angle will be smallest for blue colour.Therefore, blue light would undergo total internal reflection. f As, n = f +u f f +u= n f æ1 - n ö or u= -f =ç ÷f è n ø n æ n - 1ö or u = -ç ÷f è n ø \

|u| =

n -1 f n

1 1 > r1 r2

r1 < r2 1 1 1 As, = - (Lens formula) f v u 1 1 1 or = f -4 -10 1 1 1 or = f 10 4 1 2 -5 3 or = =f 20 20 20 or f =cm = - 6.67 cm 3 or

1 1 d = 20 56 20( -56) 56 - 20 d = 20 ´ 56 -20 ´ 56

L1 L2 is concave and L2 L3 is convex. The divergence produced by concave surface is much smaller than the convergence due to the convex surface. Hence, the arrangement corresponds to concavo-convex lens.

1ö r2 ø

æ 1 1ö For lens to be concave, ç - ÷ > 0 è r1 r2 ø

behaves like a diverging lens.

Þ

æ1 è r1

I = 15 ´ 2.5 cm = 37.5 cm

46. As, = (m - 1) ç - ÷ (Lens maker formula)

39. When convex lens is surrounded by denser medium, it

Þ

u = - 20 cm I v As, = = magnification O u I -35 or = 15 -10 or

A = 2.58 m2

d 1 1 + = f1 f2 f1f2

u = - 40 + 20

or

A = 3.14 ´ (90.7) 2 = 25865.36 cm 2

\

-20 + u = - 40

or

r = l tan ic = 80 tan 48.6

40. As, power of combination, P =

1 -20 = 2 -20 + u

Þ

r = 80 ´ 1.1345 = 90.7 cm Area of circular surface of water, through which light will emerge A = pr 2

f f +u

48.

The negative sign indicates that the lens is concave. 1 f 3 As, m = = -2 = 1 f +u +u 3 2 1 - - 2u = Þ 3 3 1 2 or -2u = + = 1 3 3 1 or u = - m = - 0.5 m 2

49. Here, m =1.5 If object lies on place side; R1 = ¥ , R2 = - 20 cm æ1 1ö 1 = (m - 1) ç - ÷ f è R1 R2 ø æ1 1ö 1 = (1.5 - 1) ç + ÷ = è ¥ 21ø 40 f = + 40 cm. The lens behaves as convex.

Ray Optics and Optical Instruments If object lies on its curved side R1 = 20 cm, R2 = ¥ æ1 1ö 1 1 æ 1 1ö - ÷= = (m - 1) ç - ÷ = (1.5 - 1) ç ø è f¢ 20 ¥ 40 è R1 R2 ø f ¢ = 40 cm .

O

50.

\

54. We have, -

O

or

R f= 2(m - 1)

Now,

f >R

\

R >R 2(m - 1)

or

1 >1 2(m - 1)

or or or

or

f = 80 cm So, it behaves like a convex lens of focal length 80 cm.

or

55. When an object approaches a convergent lens from the left

2(m - 1) < 1 1 m -1< 2 1ö æ m < ç1 + ÷ è 2ø

56.

m < 1.5

of the lens with a uniform speed of 5 m/s, the image moves away from the lens with a non-uniform acceleration. For example, if f = 20 m and v = -50 m ; - 45 m , - 40 m and -35 m; we get v = 33.3 m; 36 m ; 40 m and 46.7 m. Clearly, image moves away from the lens with a non-uniform acceleration. f As, m = f +u

or

u = - 30 cm

57. The condition for achromatism is w1P1 + w2P2 = 0

Now,

æ2ö We have, From lens maker formula,5 = (1.5 - 1) ç ÷ èRø

\

...(i) …(ii)

w1P1 = - w2P2 w1 P =- 2 w2 P1

Þ

positive power of the convex lens, negative power of the plano-concave lens of water and zero power of the plane mirror. Clearly, the power of the system decreases.

-7.5 + 5n = 0.5n

-40 - 2u = 20 or -2u = 20 + 40

or

52. The power of the given system is a combination of the

or

f = 20 cm 20 -2 = 20 + u

Again, or

f f - 10

2f - 20 = f

or

v = 30 cm The change in image distance is (30 - 20) cm i. e. , 10 cm and away from the lens.

Dividing Eq. (i) by Eq. (ii), we get 0.5n -5 = 1.5 - n

+2 =

Now,

or

æ1.5 ö æ 1 ö -1 = ç - 1÷ ç ÷ øèRø è n

0.5 æ 1 ö 1 =÷ ç2 è 20 ø f 1 1 = f 80

or

Now, u = - 15 cm, v = ? 1 1 1 = v -15 10 1 1 1 or + = v 15 10 1 1 1 or = v 10 15 1 3 -2 1 or = = v 30 30

and

1 æ1.5 ö æ 1 ö =ç - 1÷ ç ÷ ø è 20 ø f è 2

Now,

51. Clearly, 2f = 20 cm or f = 10 cm

53.

æ1 1 1ö = (1.5 - 1) ç - ÷ 40 è R1 R2 ø

(from lens maker formula) 1 1 1 =Þ R1 R2 20

1 æ2ö We have, = (m - 1) ç ÷ (By lens maker formula) èRø f

or

-7.5 = - 4.5n 75 5 = an e = n = 45 3

or

The lens behaves as convex.

1073

or

P1 + P2 = 2D 5 + P2 = 2 or P2 = - 3D w1 -3 3 == w2 5 5

58. According to Snell’s law, m=

sin i sin i or sin r = m sin r

As m is negative; sin r is negative. Therefore, r is negative.

(in magnitude)

1074 JEE Main Physics 59. The central ray goes undeviated. So, m 2 = m1.

65. We have,

Also, m3 < m 2. n1 =

60. As,

c nl l = = v1 nl1 l1

and

c nl l n2 = = = v 2 nl 2 l 2

Now,

æn ö n1 l 2 or l 2 = ç 1 ÷ l1 = n2 l1 è n2 ø

or

1

2

3

m= 3

or

So, speed of light in Y is 3 times lowered.

66. In question figure, PQ is a ray of light passing through focus, and falling on the surface of a concave mirror. On reflection, from the mirror, the ray becomes parallel to principal axis of the mirror.

61. In figure, the path shown for the ray 2 is correct. The ray suffers two refractions : At A, ray goes from air to turpentine, bending towards normal. At B, ray goes from turpentine to water (i. e. , From denser to rarer medium), bending away from normal.

sin r 1 = tan 30° = sin i 3 sin i = 3 sin r

67. Because aperture decreases. 68. The resolving limit (minimum separation), P µl PA 2000 = PB 3000

Þ

4

PA 2 = PB 3

Þ A

Air

Þ Turpentine

B

62. We have,

Þ

Water

æ 60° - dm ö sinç ÷ ø è 2 m= 3= æ 60° ö sinç ÷ è 2 ø

where, m = refractive index 3 æ 60° + dm ö = sinç Þ ÷ ø è 2 2 æ 60° + dm ö sin 60° = sinç ÷ ø è 2 60° + dm or = 60° 2 A + dm 60° + 60° or = = 60° dm = 60° ; i = 2 2

\

69.

64. We have, or

mg sin qc = m1 sin 90° mg sin qc = 1

When water is poured, mw sin r = m s sin qc or

mw sin r = 1

Again,

m a sin q = mw sin r

or

m a sin q = 1

or

sin q = 1

or

q = 90°

PA < PB

f As, M = o fe 9=

fo fe

fo = 9fe

or

L = fo + fe

Also,

20 = fo + fe 20 = 9fe + fe

or or or

20 = 10 fe fe = 2 cm

\

fo = 9 ´ 2 cm = 18 cm

or

70. For the objective, 1 1 1 = v o -1 /3.8 1 / 4

63. As is known, in the side mirror, the speed of approaching car would appear to increase as the distance between the cars decreases.

3 PA = 2 PB 2 PA = PB 3

or or or Now,

Again,

or

1 + 3.8 = 4 vo 1 1 = 0.2 = vo 5 v o = 5 cm 5 Mo = = - 19 1 3.8 M = Mo ´ Me -95 = - 19 ´ Me 95 Me = =5 19

Ray Optics and Optical Instruments 1 f

71. As, L = v o + fe Þ v o = L - fe v o = 19.2 cm 1 1 1 \From bus formula, = 19.2 uo 1.6 or

-

1 10 10 = uo 16 192

or

-

1 120 - 10 110 = = uo 192 192 uo = -

72. We have,

192 cm = - 1.75 cm 110

l 2 4800 = = 0.8 l1 6000

77.

3.5 ´ 10 3 3.5 180° rad = Here, a = ´ 5 3.8 ´ 100 p 3.8 ´ 10

Also,

=

3.5 ´ 180 ´ 7° 38 ´ 100 ´ 22

M=

fo 400 = = 40 fe 10

\ Limit of resolution, b =

78.

40 ´ 35 ´ 180 ´ 7° 35 ´ 100 ´ 22

= 21.1° » 21°

74. As, t µ

1 1 -0.75 u 1 100 25 or =u 75 10 1 4 5 or =- u 3 2 1 -8 - 15 23 or = =u 6 6 6 or u=m = - 0.26 m 23 f D 5 ´ 25 125 Separation = fo + e = 80 + = 80 + fe + D 5 + 25 30 Þ

New resolution limit = 0.8 ´ 0.1mm = 0.08 mm

73.

79. Four reading purposes,

2

f d2

\

1 1 f2 or µ µ4 100 æ f ö 2 100 ç ÷ è2ø tµ

f2 æfö ç ÷ è8ø

2

or t µ 64

Dividing Eq. (ii) by Eq. (i), 64 = 16 4 16 t= s 100

10t = or æ

1ö m

75. As, lateral shift = t ç1 - ÷ è ø Þ or or or

1ö æ 1 = 3ç1 - ÷ è mø 1 1 =13 m 1 1 2 =1- = m 3 3 3 m = = 1.5 2

2.5 =

= 84.16 cm = 84.2 cm fo æ f ö Case I As, M = - ç1 + e ÷ fe è Dø 200 æ 5ö M=÷ ç1 + 5 è 25 ø 1ö 6 æ M = - 40 ç1 + ÷ = - 40 ´ = - 48 è 5ø 5 fo 200 Case II M= == - 40 fe 5

When diaphragm of the camera-lens is set at f /2 with 1 exposure time = s, then 100

Again,

1 u

76. As, = - (lens formula)

or

or

1 v

1075

u = - 25 cm, v = - 50 cm, f = ? 1 1 1 1 1 1 = - =+ = f v u 50 25 50 100 P= = + 2D f

For distinct vision,

…(i)

…(ii)

80.

f ¢ = distance of far point = - 3 m 1 1 P = = - D = - 0.3 D f¢ 3 1 1 We have, f = = m = 20 cm P 5 1 1 1 Now, (from lens formula) - = v u 20 1 1 1 or = v -25 20 1 1 1 or = v 20 25 1 5-4 1 1 or or = = v 100 v 100 d = 100 cm = 1 m

or

81. When final image is formed at ¥, then M= Now, \

vo uo

æ d ö vo ç ÷= è fe ø fo

æd ö ç ÷ è fe ø

v o = 16 - fe = 16 - 2.5 = 13.5 cm 13.5 25 M= ´ = - 337.5 -0.4 2.5

1076 JEE Main Physics 82. Power of combination, P = P1 + P2 = + 20 - 10 = + 10D 1 1 = m = 10 cm P 10 D 25 For image at infinity, M = = = 2.5 f 10 f f As, M = o or , 10 = o , Þ fo = 200 cm 20 fe f=

83.

r1

As, From

85. As,

Þ

or

Þ dm = 30° \ Angle of incidence, A + dm 60 + 30 90 i= = = = 45° 2 2 2

86. We have, ratio of dispersive powers,

Now, or Now, Þ

w1 1 = w2 2 f1 1 w =- 1 =f2 2 w2

f2 = - 2 ´ 25 cm = - 50 cm

\ Required angle = 2 ´ 18.6 = 37.2° \ or Now, or or or

30° = r1 + 0° r1 = 30° sin i = 2 sin 30° 1 sin i = 2 ´ 2 1 sin i = 2 i = 45°

89.

30°

i

a 25

b A i i

90°–i

90°–2i i'

d i'

90°–θ

θ

θ l

From Dabc, A + 90° + (90° - i) = 180° i=A

or

Now, complementary angle at point d , q = 2i \

q = 2A

clearly, option (b) satisfies this condition.

90. For passing the ray from prism, A 2 æ 90° ö m £ cosecç ÷ è 2 ø m £ cosec

f2 = - 2f1 1 1 1 (for the combination of lenses) = + F f1 f2 1 1 1 = + 50 f1 -2f1

A = r1 + r2

88. As,

= 1.5 ´ sin 5° = 1.5 ´ 0.087 = 0.1305 i1 = sin -1 (0.1325) = 7.5° æ A + dm ö sinç ÷ è 2 ø m= A sin 2 æ 60° + dm ö sinç ÷ ø è 2 (as m = 2) 2= 60° sin 2 1 æ 60° + dm ö = sinç ÷ ø è 2 2 æ 60° + dm ö sin 45° = sinç ÷ ø è 2

Again,

\ i = 30° 1 sin 30° (ii) (By Snell's law) = 1.5 sin r 1.5 or sin r = = 0.75 2 or r = 48.6° (iii) q = r - i = 18.6°

N2

i2 = 0° \ r2 = 0° r1 + r2 = A r1 = A - r2 = 5 - 0 = 5° sin i1 m= , sin i1 = m sin r1 sin r1

f1 = 25 cm

(i) Angle between any two lines is the same as the angle between their perpendiculars.

90º

As the ray emerges from the other face of prism normally,

or

87. Following arguments lead us easily to the right choice.

A

i1

-2 + 1 1 = 2f1 -2f1

2f1 = 50

or

84. Here,A = 5° , m =1.5, i1 = ?

N1

50 f =

or

m£ 2 \

m max = 2

30°

Ray Optics and Optical Instruments

Round II 1. We have, or or

1 16 = 2 x (100 - x) 2 1 4 or 5x = 100 = x 100 - x x = 20 cm

mirror f = + 15 cm (Focal length of convex mirror is taken as positive) u = - 12 cm Size of object O = 4.5 cm Using the mirror formula, 1 1 1 = + f v u 1 1 1 1 1 1 4+5 9 = Þ = + = = 15 v 12 v 15 12 60 60 Distance of image from the mirror v = 6.7 cm

F

As the needle moves away from the mirror, the image also moves away from the mirror (as u ® ¥ , v ® f ) and the size of image goes on decreasing. 1 1 1 The mirror formula is, + = u v f du dv (for concave mirror) - 2 - 2 =0 u v dv du or - 2 = 2 v u or

dv v du 10 ´ 10 =- 2 = ´ 9 ms-1 = - 1 ms-1 dt u dt 30 ´ 30 1ö m 1 1 1- = m 3 1 1 2 2 = 1 - = or m = m 3 3 3 1 3 = sin ic 2

æ2ö ic = sin -1ç ÷ è3ø

or

ic = sin -1(0.67)

C

Using the formula of magnification, v I m== u O - 6.7 I = -12 4.5 Size of image I = 2.5 cm As I is positive, so image is erect and virtual. Magnification m is given by I 2.5 25 5 m= = = = O 4.5 45 9

2

or

or

The positive sign shows that the image is formed behind the mirror.

3.

2 3

5. For the eyepiece,

2. Given, focal length of convex

Distance of object

sin ic =

or

or or Now,

Now, or or \

v e = - 25 cm, fe = 5 cm 1 1 1 =ue 25 5 1 -1 - 5 =ue 25 25 6

ue = -

25 6 120 - 25 95 = cm = cm 6 6 1 1 1 = 95 / 6 uo 1 v o = L - |ue| = 20 -

1 6 = -1 uo 95 95 cm 89 95 |uo| = cm 89 uo = -

m1 = 1.20 +

0.8 ´ 10 -14 ( 400 ´ 10 -9) 2

or

m1 = 1.20 +

0.8 ´ 10 -14 400 ´ 400 ´ 10 -18

or

m1 = 1.20 +

0.8 16

or

m1 = 1.20 + 0.05

6. We have,

or or Þ Again,

m1 = 1.25 1 sin ic = = 0.8 1.25 ic = 53.13° m 2 = 1.20 +

0.8 ´ 10 -14 (500 ´ 10 -9) 2

æ

4. We have, m = 1 = 3ç1 - ÷ è ø or or Now,

θ θ

or

m 2 = 1.20 +

0.8 25

1077

1078 JEE Main Physics or or or Þ Now,

m 2 = 1.20 + 0.32 m 2 = 1.232 1 sin ic = = 0.81 1.232

A = 60° 3 i = i ¢ = and A = 45° 4

10. Given,

A

ic = sin -1 0.81= 54.26° sin q = 0.8 or q = 53.13°

This angle is clearly greater than critical angle corresponding to wavelength 400 nm. So, light of 400 nm wavelength undergoes total internal reflection.

i

P

δm i

180–δm

180°

θ

i'

7. When an object is placed between 2f and f (focal length) of the diverging lens, the image is virtual, erect and diminished as shown in the graph. To calculate the distance of the image from the lens, we apply

\ or

i + i¢ = A + d 90° = 60° + d

\ d = 30° Note that, i = i ¢ is the condition for minimum deviation. Hence, d = 30° = dmin 20 cm

10 cm

11. Given, distance between objective mirror and another mirror

1 1 1 = f v u 1 1 1 (as m = - 30 = + -20 v 30

Þ cm) Þ

v=-

(20) (30) 20 + 30

Thus, distance = -12 cm ( to the left of the diverging lens)

8. As, there is no deflection between mediums 1 and 2. Therefore, m1 = m 2 reduced size can be formed by a concave mirror of a convex lens. u = 2f + x, then 1 1 1 As, mirror equation is, + = u v f 1 1 1 Þ + = 2f + x v f Let

Þ

1 1 1 f+x = = v f 2f + x f (2f + x)

Þ

v=

f (2f + x) f+x

It is given that, u + v = 1.0 m é f (2f + x ) f ù 2f + x + = (2f + x ) ê1 + ú < 1.0 m f+x f + xû ë

or

Radius of curvature of small mirror = R2 = 140 mm \ Focal length of small mirror,

f2 =

140 = 70 mm 2

The image of an object placed at infinity, formed by the objective mirror, will act as a virtual object for small mirror. So, the object distance for small mirror u = f1 - d i.e.,

9. Image can be formed on the screen if it is real. Real image of

or

d = 20 mm Radius of curvature of objective mirror = R1 = 220 mm 220 \ Focal length of objective mirror, f1 = = 110 mm 2

(2f + x ) 2 < 1.0 m f+x (2f + x ) 2 < ( f + x )

This will be true only when, f < 0.25 m

u = 110 - 20 = 90 mm 1 1 1 Using mirror formula, + = v u f2 1 1 9 -7 2 1 1 1 = = = - = v f2 u 70 90 630 630 v = 315 mm or v = 31.5 cm Thus, the final image is formed at 315 mm away from small mirror. 1 f

æ m1 ö æ 1 1ö - 1÷ ç - ÷ è m 2 ø è R1 R2 ø

12. As, = ç Þ or or or

1 æ 3 /2 öæ 1 1 ö + =ç - 1÷ ç ÷ ø è è 4 /3 0.3 0.3 ø f 1 æ9 öæ 2 ö = ç - 1÷ ç ÷ f è 8 ø è 0.3 ø 1 1 2 = ´ f 8 0.3 f = 1.20 m

13. As, y = y1 ´ y 2 = 16 ´ 9 = 4 ´ 3 = 12 cm

Ray Optics and Optical Instruments 1 f

14. As, = Þ

x x 1 1 1 1 + = - + f1 f2 f1f2 f1 f1 f12 1 x = f f12

20. As,

15. We know that, m=

(Snell’s law)

r = 90° - i sin i m= = tan i sin(90° - i)

or Þ

i = tan -1(m) = tan -1(1.62)

or

Þ

16. Use, d = f ; in lens maker formula, we get

17. Given, a = 1.5, Ði = 60° and n = 1km = 103 m When the total internal reflection just takes place from lateral surface, i = C i. e,60° = C 1 Þ sin C = m sin 60° =

1 3 = m 2

2 m= 3

Þ

The time taken by light to travel some distance in a medium, 2 ´ 10 3 mx 3 = 3.8528 ms t= = c 3 ´ 10 8

22. When one of the slit is covered, amplitude becomes half. Therefore, the intensity of central maximum will becomes I0 I = 0 2 4 (2)

23. Here ip = 60° , v = ?

Þ

pr 2 Ar 2 2 2 I2 æ A2 ö 4 = 4 pr - pr = 3 pr = 3 =ç ÷ = 2 2 I1 è A1 ø pr 4 pr 4 pr 2 4

Þ

I2 3 = I1 4

Þ

I2 =

c = tan ip = tan 60° = 3 v c 3 ´ 10 8 v= = = 3 ´ 10 8 ms-1 3 3 100 100 P1 = = = + 4D f1 25 100 100 P2 = = = - 4D f2 -25 m=

24. Here, and

For the combination, P = P1 + P2 = 4 - 4 = 0 D

25. Here, f = 20 cm, w = 0.08 Longitudinal chromatic aberration = wf = 0.08 ´ 20 = 1.6 cm

26. The ray comes out from CD, means rays after refraction from AB get, total internal reflected at AD D

A µ1 αmax

18. As, I µ A2, where I = intensity and A = amplitude 2

50 50 ´ fa = ´ 8 ´ 1.33 17 17

electron ( p = mv) increases. The de-Broglie wavelengths ( l = h / mv) decreases. The fringe width is, b = lD / d , so b µ l. Hence, if l decreases then b decreases.

f = 20 cm

Þ

fw =

~ - 32 cm

1 2 2 1 = (m - 1) = (1.5 - 1) ´ = f R 20 20 Þ

fw (1.5 - 1) 0.5 = ´ 1.33 = fa æ 1.5 ö 0.17 - 1÷ ç è1.33 ø

21. When velocity of an electron increases, the moments of

sin i sin r

i + r = 90°

and

fw ( a mg - 1) = fa (w mg - 1)

Þ

Þ f > 0 for every x. For x = 0 , f = ¥ Hence, for x = 0 , system Snell’s law will behave like a glass plate.

1079

B

C

2

3 I1 and the focal length remains unchanged. 4

19. In every condition, two plano-convex lenses are placed, closed to each other i. e. , distance between is assumed to be zero. 1 1 1 = + i. e. , F f1 f2

According to Snell’s law m1 sin a max = m2 sin r1 Þ

ém ù a max = sin -1ê 1 sin r1ú ëm2 û

Also,

r1 + r2 = 90°

Þ

r1 = 90° - r2 = 90 - C

Þ

æ 1 ö r1 = 90° - sin -1ç ÷ è 2m1 ø

Þ

æm ö r1 = 90° - sin -1ç 2 ÷ è m1 ø

…(i)

…(ii)

1080 JEE Main Physics Then, from Eqs. (i) and (ii), we get a max

ém æ m öù = sin -1ê 1 sinç90° - sin -1 2 ÷ ú m1 ø û è ëm2 ém æ m öù = sin -1ê 1 cosç sin -1 2 ÷ ú m1 ø û è ëm2

27. From figure, we see that q + q + 10° = 90° Vertical RR θ

θ

10°

IR Horizontal line

Here, a = 1.8 m and

q = 40°

28. Here, fo = 50 cm, fe = 5 cm, D = 25 cm and uo = 200 cm \ Separation between the objective and eyepiece lens is, u f feD L= oo + uo - fo ( fe + D) =

or h2 = 0.6 cm The positive sign shows that the image is erect.

33. This is a modified displacement method problem.

2q + 10 = 90 2q = 80

Þ

h2 1(32) =1 1.33( -40)

\

Plane mirror

Þ Þ

Given, m1 = 1, and m 2 = 1.33 Applying equation for refraction through spherical surface, we get, m 2 m 2 m 2 - m1 = v u R 1.33 1 1.33 = v -40 -20 After solving v = - 32 cm. h mv The magnification is, m = 2 = 1 h1 m 2 u

200 ´ 50 5 ´ 25 + = 71 cm (200 - 50) (5 + 25)

f=

4 cm

4 cm

In this case refraction of the rays starting from t 0 takes d actual m 4 or 3= m 4 or m= 3

place from a plane surface. So, we can use, d app =

1 ö 1 ö æ æ s = 4ç1 ÷ ÷ + 6ç1 è 3 / 2ø è 4 / 3ø = 3. 0 cm h = h1 + h2 - s = 4 + 6 - 3 = 7.0 cm

30. We know that, m=

velocity of light in vacuum velocity of light in water

4 3 ´ 10 10 = 3 velocity of light in water 10

Velocity of light in water = 2.25 ´ 10 cms Time taken =

As shown in Fig. (b). In this case refraction takes place from a spherical surface. Hence, applying m 2 m1 m 2 - m1 = v u R we have,

-1

500 ´ 100 = 2.22 ´ 10 -6 s 2.25 ´ 10 10

Equivalent optical path = m ´ distance travelled in water 4 = ´ 500 = 666.64 m. 3

31. Only one converging point is found by this lens. Therefore, only one image is formed.

or \

1 4 /3 1- 4 /3 = ( -25 / 8) -4 -R 1 1 8 1 = = 3R 3 25 75 R = 25 cm

Now, to find the focal length we will use the lens maker formula æ1 1 1ö = (m - 1) ç - ÷ f è R1 R2 ø 1 ö 1 æ 4 öæ 1 = ç - 1÷ ç ÷= è 3 ø è ¥ -25 ø 75

32. According to Cartesian sign convention, u = - 40 cm, R = - 20 cm

O (b)

O (a)

æ æ 1ö 1ö s = h1ç1 - ÷ + h2ç1 - ÷ è m2ø è m1 ø

Thus,

a2 - d 2 = 0.4 m 4a

34. As shown in Fig. (a)

29. Using equation, the total apparent shift is

or

d = 0.6 m

Solving, we get, \

a+d 2 = a-d 1

\

f = 75 cm

Ray Optics and Optical Instruments 35. Here, x = u + v

…(i) m=

As,

f f -v = f +u f

39. From figure, OA = 4 cm, AB = 3 cm \ Now,

and image is real, magnification is negative. f -(m + 1) f -m = ,u = \ f +u m

From,

f -v f v = (m + 1) f

OB = 4 2 + 3 2 = 5 cm m=

1 1 OB 5 = = = sin C AB / OB AB 3

m=

-m =

From Þ Put in Eq. (i)

Solving, we get

40. For dispersion without deviation,

Þ Þ

1 1 sin i = = m sin r ¢/ sin i sin r ¢

and the distance of image from the lens is (3 - u) m i.e., v = (3 - u) m From Lens formula, 1 1 1 - = v u f 1 1 1 = (3 - u) - u f

Denser i r B r' Rarer

or

1 1 1 + = (3 - u) u f

or

u + 3 -u 1 = u (3 - u) f

ÐCBD = 90° \ or \

90° - r + 90° - r ¢ = 90° r ¢ = 90° - r sin i sin i sin C = = sin(90° - r) cos r =

sin i = tan i cos i

3f = 3u - u 2 u 2 - 3u + 3f = 0 (Q i = r)

Now,

C = sin -1(tan i) = sin -1(tan r).

37. Let, r be the radius of circle through which other objects become visible. The rays of light must be incident at critical angle C. 1 r Now, sin C = = 2 m r + h2 or or Þ Diameter,

m 2r 2 = r 2 + h 2 (m 2 - 1)r 2 = h 2 r= 2r =

- ( -3) ± 9 - 4 ´ (3 f ) 2 + 3 ± 9 - 2f u= 2

u=

Condition for image to be formed real on the screen. or

9 - 12 f ³ 0 9 ³ 12f or f £ 0.75 m

Thus, the maximum possible focal length of the lens required for this purpose is 0.75 m.

42. As, L¥ = v o + fe h

m2 -1 2h m2 -1

38. Here, i1 = 60° , A = 30° and d = 30° i1 + i2 = A + d, i2 = 0 Hence, angle between the rays and the face from which it emerges = 90° - 0° = 90°

As,

A (m ¢ - 1) = A¢ (m - 1) 4 (1.72 - 1) 0.72 = = A1 (1.54 - 1) 0.54 4 ´ 0.54 A1 = = 3° 0.72

41. Suppose the object is placed at u metre in front of the lens

As is clear as shown in figure

O

c c 3 ´ 10 8 = = c /v m 5 /3 = 1.8 ´ 10 8 m/ s

-(m + 1) x= f + (m + 1) f m mx f= (m + 1) 2

36. At the critical angle, sin C =

1081

Þ 14 = v o + 5 Þ v o = 9 cm Magnifying power of microscope for relaxed eye v D Now, magnification, m = o × uo fe Þ or

9 25 ´ 40 5 9 ´ 25 9 40 = = cm = 1.8 cm 5 ´ 25 5 25 =

1082 JEE Main Physics 43. As, m =

fo fe

48. As, it is clear from figure, Angle subtended by an image Angle subtended by an object fo a = fe b

We know that m = \ Þ

b ´ fo 6002 ´ 60 a= = = 2 ´ 12 = 24° fe 5

44. As, a convex lens alone can form a real image as well as a virtual image, therefore, the lens in the present question is a convex lens. Let, f be the focal length of the lens and m be the magnification produced.

As, \ or \

As \

\

In the second case, when image is virtual. From

u = - 6 cm, v = ( -6 m ) cm 1 1 1 - = v u f 1 6 1 1 1 + = or 1 - = m f -6m 6 f

From, ...(ii)

45. When a slab of thickness, t is introduced between P and the

Length of the tube, v o + fe = 10 + 5 = 15 cm

50. For total internal reflection at AC-face sin i ³

direction through the same distance.

46. In the situation given, the image will be formed at infinity, if the object is at focus of the lens i. e. , at 20 cm from the lens. Hence, shift in position of object 1ö æ x = 25 - 20 = ç1 - ÷t è mø 1 ö æ 5 = ç1 ÷t è 1.5 ø t = 15 cm

47. The focal length of combination of two lenses is 1 1 1 = + F f1 f2 1 1ö æ 1 1ö æ 1 - ÷ = (m1 - 1) ç + ÷ + (m 2 - 1) ç ø è è -R ¥ ø F ¥ R m1 - 1 m 2 - 1 R R R 1 m1 - m 2 or F = × = m1 - m 2 F R

fo = 1 cm 1 1 1 = v o uo fo 1 9 1 + = ; v o = 10 cm vo vo 1

mirror, the apparent position of P shifts towards the mirror by tö æ çt - ÷ × Hence, the mirror must be moved in the same è mø

=

æDö ç ÷ è fe ø 25 v v -45 = - o ´ Þ o =9 5 uo uo vo uo

For objective lens, image is real. v \ v o = + v o and uo = - o 9 Given,

Add. Eq. (i) and Eq. (ii), we have 22 22 or f = 2= = 11 cm f 2

\

i2 = 0 , and r2 = 0 r1 + r2 = A r1 + 0 = 30° r1 = 30° sin i1 sin 60° 3 /2 m= = = = 3 sin r1 sin 30° 1/ 2

M=-

…(i)

30° 60°

49. For the relaxed eye, magnifying power is

In the first case, when image is real, u = - 16 cm, v = (m ´ 16) cm 1 1 1 - = v u f 1 16 1 1 1 = + = or 1 + m f 16 m 16 f

A

A = 30° , i = 60° As the ray retraces its path on reflection at the silvered face, therefore,

mw mg

sin q ³

4 3 ´ 1.5

sin q ³

8 9

51. Note that image formation by a mirror does not depend on the medium. As, P is at a height h above the mirror, image of P will be at a depth h below the mirror. If d is depth of liquid in the tank, apparent depth of P , d -h x1 = m \Apparent distance between P and its image d + h d - h 2h = x2 - x1 = = m m m

52. As, C = Now, \

x 10 x , and v = t1 t2 sin C ¢ =

1 v 10 x t1 = = ´ t2 x m c

æ10t ö C ¢ = sin -1ç 1 ÷ è t2 ø

Ray Optics and Optical Instruments - 1 1m 2 1 = - in given case, R v u medium (1) is glass and (2) is air 1 -1 1 1 1 gm a - 1 gm a 1.5 = - g Þ = \ R v u -6 1.5v -6 1 - 1.5 1 1.5 = + Þ -6 6 v 0.5 1 1 Þ = + 6 v 4 1 1 1 2 1 = - ==Þ v 12 4 12 6

53. By refraction formula,

Þ

1m 2

Þ

1 F

60. For minimum spherical and chromatic aberration d = F1 - F2 = 0.3 - 0.1 = 0.2 m

61. According to Rayleigh scattering, the intensity of scattered light, I µ

1 l4

I µ f4 \

or

f1 æ I1 ö =ç ÷ f2 è I2 ø

æ 256 ö =ç ÷ è 81 ø

|m| =

1/ 4

æ 44 ö =ç 4÷ è3 ø

1/ 4

=

4 = 4 :3 3

30 R = = 10 cm 2m 2 ´ 1.5

Þ

In order to have a real image of the same size of the object must be placed at centre of curvature i. e. ,u = 2f Hence, u = 2 ´ 10 = 20 cm

power of eye lens = 20 D

To observe the objects at infinity, the eye uses its least converging power means power is maximum, = 40 + 20 = 60 D The distance between cornea = focal length of eye lens 100 100 5 f= = = cm P 60 3 To focus objects at the near point on the retina 5 u = - 25 cm, v = cm 3 1 1 1 1´ 3 1 15 + 1 16 Using Lens formula, = - = + = = f v u 5 25 25 25 25 cm Þ f= 16 1 100 ´ 16 Power of lens = = = 64 D f 25 i.e.,

(Q both faces have same radius of curvature) For double convex lenses R1 = R, R2 = - R (For double convex lens, one radius is taken positive and other negative)

R1

R2

æ1 1 1ö = (1.55 - 1) ç + ÷ 20 è R1 R2 ø 1 2 = 0.55 ´ R 20

\

R = 0.55 ´ 2 ´ 20 = 22 cm

58. As, P1 + P2 = 2D and P1 = 5D, so P2 = - 3D For a cromatic combination, w1 æ -P2 ö æ -3 ö 3 =ç ÷ = -ç ÷ = è 5 ø 5 w2 è P1 ø

fe = 6 cm, fo + 6 = 36 fo = 36 - 6 = 30 cm

63. Given, the power of cornea = 40 D and least converging

mg = 1.55

Thus, the required radius of curvature is 22 cm. 1 1 As, F µ Þ m µ (Q f = 2 R) m -1 r

6fe = 36 cm

\ Þ

56. Given, the refractive index of glass with respect to air

æ1 1 a 1ö = ( mg - 1) ç - ÷ f è R1 R2 ø

…(i)

5fe + fe = 36 On putting the fe in Eq. (ii), we have

55. Focal length of the system (Concave mirror)

Focal length of lens, f = + 20 cm Using the Lens maker’s formula

fo =5 fe

and length of telescope …(ii) l = fo + fe = 36 cm From Eq. (i) fo = 5fe , putting the value of fo in Eq. (ii), we get

u = - 30 cm

a

1/ 2

4

62. We have, magnification,

f u+ f f 20 -2 = = u + f u + 20

F=

I1 æ f1 ö =ç ÷ I2 è f2 ø

or

m=

Þ

57.

1 1ö - ÷ Þ F = 40 cm è 20 ¥ ø

æ 59. As, = (15 . - 1) ç

v = - 6 cm

54. For real image, m = - 2 \

1083

64.

Power of eye lens = 64 - 40 = 24 D

Thus, the range of accommodation of the eye lens is 20 D to 24 D. 1 1 1 As, = + F f1 f2 Þ

f1 = - f2 F=¥

65. For total internal reflection, q>C

Þ sin q > sin C Þ sin q >

1 2

1084 JEE Main Physics 1 1 Þm> 2 Þm> sin 45° sin q

or

m>

or

m > 1.41

68. R 60°

66. Given, radius of curvature of concave mirror, R = - 36 m

30° 30° 30° 30°

(For concave mirror radius of curvature is taken as negative) R 36 = = - 18 cm \ Focal length f = 2 2

30° 30°

T

p

From the figure, the required angle of reflections are 30°, 60°, 60°, 30°

69. Light ray is going from liquid (Densor) to air (Reson) and angle of refraction is 90°, so angle of incidence must be 4 equal to critical angle, there fore sin C = 5

20 cm

10 cm

90°

u = - 27 cm (Object distance is always taken as negative)

Distance of object Height of object

C

5 cm 3 cm

O = 2.5 cm 4 cm

Also, C

F 36 cm

P

m=

1 5 = = 1.2 sin C 4

70. Focal length of lens will increase by four times i. e. , 12 cm while focal length of mirror will not affected by medium.

Use the mirror formula 1 1 1 1 1 1 = = + Þ 18 v 27 f v u 1 1 1 1 -3 + 2 =+ = =v 18 27 54 54 Distance of screen from mirror v = - 54 cm Let the size of image be I. By using the formula of magnification for mirror v I m== u O - ( -54) I = - 27 2.5 I = - 5 cm The negative sign shows that the image is formed in front of the mirror and it is inverted. Thus, the screen should be placed at a distance 54 cm and the size of image is 5 cm, real, inverted and magnified in nature. If we move the object near to the mirror (as u ® f , v ® ¥) the screen should be moved away from mirror. As the distance of object is less than focal length, (u < f ) no screen is required, because the image formed is virtual.

67. Because in dispersion of white light, the rays of different colours are not parallel to each other. Also deviation takes place in same direction.

71. As, dnet = dmirror + dprism = (180 - 2i) + (m - 1) A = (180 - 2 ´ 45) + (1.5 - 1) ´ 4 = 92°

72. Final image is formed at infinity if the combined focal length of the two lenses (in contact) becomes 30 cm 1 1 1 Thus, = + 30 20 f i. e. , when another concave lens of focal length 60 cm is kept in contact with the first lens. Similarly, let m be the refractive index of a liquid in which focal length of the given lens becomes 30 cm. Then 1 æ3 ö æ 1 1ö …(i) = ç - 1÷ ç - ÷ 20 è 2 ø è R1 R2 ø 1 æ3 /2 ö æ 1 1ö =ç - 1÷ ç - ÷ ø è R1 R2 ø 30 è m

…(ii)

From Eqs. (i) and (ii), we get m=

9 8

73. Here, an extended object lies immersed in water contained in a plane trough. When seen from close to the edge of the trough, the object looks distorted on account of refraction of light from denser to rarer medium. Therefore, apparent depths of the points close to the edge and nearer to the surface of water is more compared to points away from the edge.Further, the angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air. Again, some of the points of the object, far away from the edge may not be visible because of total internal reflection.

Ray Optics and Optical Instruments 74. Total internal reflection takes place when ray of light travels

82. By painting black the upper half of the lens, intensity of

from denser to rarer medium. Further, Since,

75.

76.

image will reduce but its position will not be shifted.

m m sin q12 = 2 and sin q13 = 3 m1 m1 m3 m3 > m1 m1

83. A magnifying glass is used as the object to be viewed can be

q12 > q13 Smaller the value of critical angle more the change of total internal reflection. m m m - m1 Using, 2 - 1 = 2 ,we get v u R 1.5 1.0 1.5 - 1.0 = v ¥ 20 or v = 60 cm 1 1 1 As, m ³ > > ³ 2 = 1.414 sin C sin 45° 1 / 2

brought closer to the eye than the normal near point. This results in a larger angle to be subtended by the object at the eye and hence viewed in greater detail. Also, it results in the formation of a virtual, erect image.

84. Here, fo = 20m and fe = 2 cm = 0.02 m In normal adjustment, length L = fo + fe = 20.02 m f 20 Magnification = o = = 1000 fe 0.02

77. A concave mirror of suitable length and a convex of focal length £ 0.25 m .

of

telescope

tube

The image formed is inverted ( w.r. t. the object) 1 f

æ mL öé1 1ù - 1÷ ê - ú è m m ø ë R1 R2 û

85. Use, = ç

\ Possible values of m are 1.5 and 1.6.

86.

Since (mL < m m) because lens is of water and mL = mw 1 1 1 4 As, m = = = = × sin C sin 48.6 0.75 3

87. Light cannot undergoes total internal reflection when it is

78. Tube length = fo + fe f Angular magnification = o fe

88.

fo + fe = 16 + 0.02 = 16.02 m f 16 Angular magnification = o = = - 800 0.02 fe Then,

travelling from air to water, i. e. , from rarer to denser medium. 1 1 2 sin C = = = = 0.6667 m 3 /2 3 C = sin -1(0.6667) = 41.8°

Objective lens is larger than eyepiece in aperture focal length.

79. Due to blocking only intensity is going to decrease. Each part of reflection mean by put it.

89. From As

m=

1 1 , sin C = m sin C

mv > m r \ Cv < C r

90. As shown in figure, B

O

80. In figure a pin is held at L, mid point of AB. When seen from face AD (so long i < C) image of L appears to be at L¢ , close to A. 1 1 From sin C = = = 0.625 m 1.6 C = sin -1 (0.625) = 38.7°

1085

A L′

L

B h

C

iC TIR

So when angle of incidence D C becomes greater than C( = 38.7° ), the rays starting from L will undergo total internal reflection and pin shall not be seen at all.

81. Alexander ’s dark band between the primary and secondary rainbows is because light scattered into this region interferes destructively. Further, primary rainbow subtends an angle of 41°-43° at the eye of the observer w.r.t. the incident light, and secondary rainbow subtends an angle of about 51° to 54° at the eye of the observer w.r.t the incident light. Therefore, the region between the angles of 41° to 51° is dark.

91.

When angle of incidence is slightly greater than C, light undergoes total internal reflection. \Diameter of circle of light coming from water surface = 2r = 2(OB) = 2OS tan C = 2h tan C c As, m = , for meta material v 1 v= (m)

92. Meta material has a negative refractive index \ Þ m 2 is negative \ q2 is negative.

sin q2 =

m1 sin q1 m2

1086 JEE Main Physics 93. It is only valid for paraxial rays. 96. In air or water, a convex lens made of glass behaves as a convergent lens but when it is placed in carbon disulfide, it behaves as a divergent lens. Therefore, when a convergent lens is placed inside a transparent medium of refractive index greater than that of material of lens, it behaves as a divergent lens. It simply concludes that property of a lens whether the ray is diverging or converging depends on the surrounding medium.

\

æ1 1 1ö = (w ng - 1) ç - ÷ fw è R1 R2 ø

As, w ng < ang , hence focal length of lens in water increase. The refractive index of water is 4/3 and that of air is 1. Hence, mw > m a .

98.

f The magnifying power telescope in relaxed state is, m = o fe So, for high magnification, the focal length of objective length should be larger than or eyepiece. d Resolving power of a telescope = × 1.22l For high resolving power, diameter (d) of objective should be higher.

99. We know that power of lens is a reciprocal of its focal length, 1 1 = = 2D 50 f 100 Since, lens is concave hence, its power will be 2D. If the objective is placed at infinity, then

103. The stars twinkle while the planes do not. it is due to variation in density of atmospheric layer. As, the stars are very far and giving light continuously to us. So, the light coming from stars is found to change their intensity continuously. Hence, they are seen twinkling. Also stars are much bigger in size than planets but it has nothing to deal with twinkling phenomenon.

104. Owls can move freely during night, because they have large

105.

number of cones on their retina which helps them to see in night. I f f Magnification produced by mirror, m = = = O f -u x where, x is distance from focus.

106. As, refractive index for z > 0 and z £ 0 is different xy-plane should be boundary between two media. Angle of incidence, cos i = \

Thus, concave lens will form an image of the object at infinity at a distance of 50 cm.

100. When glass surface is made rough, then light incident on it is scattered in different directions. Due to which its transparency decreases.

101.

There is no effect of roughness on absorption of light. 1 Refraction index of diamond w.r.t. liquid l m d = sin C \

6 1 = 3 sin C

A2 2

An +

A42

1 u

108. As, + Þ

=

1 2

sin i 3 = Þ r = 45° sin r 2

1 1 = v f 1 du 1 dv - 2 =0 u dt v 2 dt

Þ

dv -v 2 æ du ö = 2 ç ÷ dt u è dt ø

But

v f = u u-f 2

\

+

A22

i = 60°

From Snell’s law,

hence, P =

u = ¥ , v = ?, f = 50 cm 1 1 1 From the formula, - = v u f 1 1 1 - = v ¥ -50 v = - 50 cm

C = 45°

of white light. Thus, when green flower is seen through red glass its absorbs the green colour, so it appears to be dark.

does not change as, f = R / 2, but for the lens by lens maker formula,

and

1 = sin 45° 2

102. Red glass transmits only red light and absorbs all the colours

97. If mirror is placed in a medium other than air its focal length æ1 1 1ö = ( a ng - 1) ç - ÷ fa è R1 R2 ø

sin C =

or

2

æ f ö æ du ö æ ö 1 dv 0.2 ms-1 =-ç ÷ ç ÷=ç ÷ ´ 15 = dt 15 è u - f ø è dt ø è -28 - 0.2 ø

109. As, intensity is maximum at axis, \ m will be maximum and speed will be minimum on the axis of the beam. \ Beam will converge.

110. For a parallel cylindrical beam, wavefront will be planar. 111. Speed of light in the medium is minimum on the axis of the beam.

Ray Optics and Optical Instruments 112. Shift in image position due to glass plate, Now,

For focal length of the lens, 1 1 1 1 1 = - = f v u 12 -240 1 20 + 1 or = f 240 240 f= cm Þ 21

f = + 10 cm, m = 1.5, m ¢ = 1.35 f ¢ m ¢ (m - 1) f¢ = (m - m ¢ ) 10 ´ 1.33 ´ 0.5 f¢ = = 40 cm 0.15

Using \ 1 v

115. As, + or

1 1 1 = f v u 1 1 1 3 21 = - = u v f 35 240 1 48 ´ 3 - 7 ´ 21 1 = =u 1680 560

1 1 = u f 1 1 1 + = -|v| |u| -| f | v=

Þ

u = - 5.6 m

which is not in the permissible limit. So, (66, 33) is incorrect recorded. 2

θ

r

m= Þ At interface 2, Þ

For

90° i

sin q = m sin r (90° - r) = C

|u| = 78,| f | = 24 cm (78) (24) |v| = = 32 cm 78 - 24

which is not in the permissible limit. So, (78, 39) is incorrect recorded.

sin q sin r

sin(90° - r) = sin C 1 cos r = m

Þ

|u|| f | |u| - | f |

|u| = 66 cm,| f | = 24cm 66 ´ 24 |v| = = 36 cm 66 - 24

For

113. At interface 1,

116. On refraction at face AB, …(i)

sin 60° = 3 sin r1

Þ r = 30° From Eq. (i), we see that 1 2 sin q = sin 30° = 3 3 1 ö æ \ q = sin -1ç ÷ è 3ø 3 2

(By Snell’s law)

r1 = 30 °

So, O B 60°

1ö æ çQ sin C = ÷ è 2ø

P

C 60°

1 3 cos r = = 2 2/ 3

Þ

114. As, sin C =

æ 1 ö q = sin -1ç ÷ è 3ø

Here,

the image on the film.

Þ

sin q m 2 = sin r m1

\

Now, to get back image on the film, lens has to form image 1ö 35 æ at ç12 - ÷ cm = cm such that the glass plate will shift è 3ø 3

Þ

1 2 2 1 ´ Þ sin q = 3 2

sin r = sin(90° - C) = cos C =

1ö 1 ö 1 æ æ S = ç1 - ÷t = ç1 ÷ ´ 1 cm = cm è 1.5 ø è mø 3

As

1087

45°

90°

75°

A

D

This shows that the refracted ray is parallel to side B of prism. For side CD angle of incidence will be 45°, which can be concluded from quadrilateral PBCQ. By refraction of face CD, 3 sin 45° = 1 sin r2,

i θ

So,

sin r2 =

3 2

(By Snell’s law)

1088 JEE Main Physics Which is impossible. So, there will be TIR at face CD.

Let, the coordinates of P be ( x, x), then u = - x

Now, the angle of incidence at AD will be 30°. This, angle of emergence will be 60°. Thus, angle between incident and emergent will be 90°.

and

v=x 1 1 1 = f x -x 2 = x

From Eq. (i),

117. Here, u = - mf , u = v , f = - f Using lens formula or or

1 1 1 - = , we have v u f

or x = 2f \Coordinate of point P are (2f , 2f ).

1 1æ 1 ö -1 1 (m + 1) = (m + 1) = - ç1 + ÷ = v fè m ø mf u

119. We know that the angle of deviation depends upon the angle of incidence. If we determine experimentally, the angles of deviation corresponding to different angles of incidence and then plot i ( on- x - axis) and d( o - y - axis), we get a curve as

v 1 = u m+1

Linear magnification =

v 1 = u (m + 1)

shown in figure. y

118. The lens formula is

δ

1 1 1 = f v u

...(i)

δ δm

The graph between u and v will be curve as shown in figure. u i1

(0, 0)

v

i

i2

x

Clearly if angle of incidence is gradually increased, from a small value, the angle of deviation first decreases, becomes minimum for a particular angle of incidence and then begins to increase.

24 Wave Optics JEE Main MILESTONE <
v1, the angle of refraction will now be greater than angle of incidence, and if ic is critical angle, we have n sin ic = 2 n1 Thus, if i = ic then, sin r = 1 and r = 90°. Obviously, for i > ic there cannot be any refracted wave.

(b) Path difference = 2l cos q 2l cos q = nl

For x to be minimum, n = 1 cos q =

1 2

q = 60° , x = D tan 60° = 3D P sθ co 2λ θ 2λ

x O D

1092 JEE Main Physics

24.3 Interference of Light When two light waves of exactly equal frequency having phase difference which is constant with respect to time travelled in same direction and overlap each other, then intensity is not uniform in space.

where the two waves meet in opposite phase, i. e. , the intensity of light is minimum is called the destructive interference. Second wave

Resultant wave

First wave t

y

At some points, the intensity of light is maximum (more than the sum of individual intensities of those waves), while at some points, the intensity of light is minimum (less than the difference of individual intensities of those waves).

(a) Constructive interference

First wave

Thus, formation of maximum intensity at some points and minimum intensity at some other points by the two identical light waves travelling in same direction is called the interference of light.

Resultant wave

t

y

The interference at the points where the two waves meet in same phase, i. e. , the intensity of light is maximum, is called the constructive interference while at the points

Second wave (b) Destructive interference

Hot Spot

Conditionsof Maxima and Minima

Let y 1 = A1 sin wt and y 2 = A2 sin(wt + f) be two simple harmonic waves of same frequency travelling in the w same direction, A1 and A2 are their amplitudes, f is the initial phase difference between them and is the 2p common frequency of the two waves. By the principle of superposition, the resultant displacement is y = y1 + y2 = A1 sin wt + A2 sin( wt + f) From this expression, the resultant amplitude is given by

For destructive interference (minimum intensity) The intensity I is minimum, when cos f = - 1 i. e., when phase difference, f = (2n - 1)p; n = 1, 2, . . . etc. n = 1 for first order minima, n = 2 stands for second order minima l Path difference = (2n - 1) 2

A 2 = A12 + A22 + 2A1A2 cos f The resultant intensity I is given by I µ A2 I = I1 + I2 + 2 I1 I2 cos f

Þ

Thus, the resultant amplitude (or the resultant intensity) at a point depends on the phase difference at that point between the two superposing waves. (i) For constructive interference (maximum intensity) The intensity I is maximum, when cos f = + 1, i. e., when phase difference f = 2np ; n = 0, 1, 2, ... etc. n = 0 stands for zero order maxima n = 1 stands for 1st order maxima n = 2 stands for IInd order maxima Path difference = nl So,

I

max

= A21 + A22 + 2A1A2 = ( A1 + A2 )2

I min = A 21 + A 22 - 2 A1A2 = ( A1 - A2 )2

So,

Imax = 4A2

Imax = 0 – 5π – 4π – 3π – 2π

φ

O

π

2π

3π

4π

5π

Figure shows the variation of intensity I with the phase difference f due to the superposition of two waves of equal amplitudes. The graph is called the intensity distribution curve. when, and when

(ii)

π

f = 0, ± 2p, ± 4p . . . . . , Imax = 4A2 f = ± p, ± 3, ± 5p, . . . , Imin = 0

Wave Optics

1093

Sample Problem 2 Light waves from two coherent source

Sample Problem 3 Light waves from two coherent

as of having intensity ratio 81 : 1 produce interference. Then, the ratio of maxima and minima in the interference pattern will be

sources having intensities I and 2I cross each other at a point with a phase difference of 60°. The intensity at the point will be

(a)

18 23

(b)

16 25

(c)

25 16

(d)

23 18

(a) 4.414 I

(d) 6.441 I

We know that, the amplitude A of resultant wave is A = a2 + b 2 + 2 ab cos f A2 = a2 + b 2 + 2 ab cos f

A1 = 9A2 Imax ( A1 + A2) 2 = Imin ( A1 - A2) 2

\

(c) 4 I

Interpret (a) Here, I1 = I , I2 = 2 I and f = 60°

I A2 81 Interpret (c) Given, 1 = 12 = I2 A2 1 A1 9 \ = A2 1 or

(b) 5.455 I

...(i)

We also know that, I µ A2 \ Required intensity,

From Eq. (i), we get Imax (9A2 + A2) 2 (10) 2 25 = = = Imin (9A2 - A2) 2 (8) 2 16

I = I2 + I2 + 2 I1I2 cos f = I + 2I + 2 I ´ 2I cos 60° = 4. 414 I

Young’s Double Slit Experiment One of the first to demonstrate, the interference of light was Thomas Young in 1801. This experiment is a demonstration that matter and energy can display characteristics of both waves and particles and demonstrates the fundamentally probabilistic nature of quantum mechanical phenomena. Young’s experimental arrangement for double slit experiment consists of two narrow slits S1 and S2 on which the light beam was incident from another slit S. A screen is placed in the path of light emerging out of the slits S1 and S2 on which alternate coloured fringes are observed when the distance between the two slits was increased, it was observed that the bands or fringes disappeared. Therefore, an optimum distance between them is required to observe the well defined fringes pattern. Moreover, with white light, only a few coloured fringes were obtained, thus, a well defined interference pattern could be observed only with monochromatic light. With monochromatic light, we get alternate dark and bright fringes. Former is known as minima and later as maxima of the interference pattern. A′ A S1

Light S

Screen

D S2

B′ Young’s Double Slit Experiment

B

The observed interference of light waves can be explained on the basics of Huygen’s principle. Each of the primary wavefront reaches the slits S1 and S2 simultaneously and secondary wavefronts originate. The two sets of the secondary wavefronts originating at S1 and S2 superimpose is the gap between the slits and the screen will produce the phenomena of interference. At the point where a crest wave from S1, intersects with another crest wave from S2 a double crest is formed. In a similar way, a double trough is formed. The crests and troughs differ from each other in that the displacement amplitudes in the two are opposite to each other. Since, intensity of light is proportional to the square of amplitude, so both double crest or double trough corresponds to maximum intensity. However, at the points where a crest wavefront intersects with a trough wavefront, the wave amplitude is reduced to minimum (zero). It is clear that the location of double crests as well as the location of minimum amplitude lie on the straight lines which seems to be originating from the mid-point of the slits S1 and S2. The lines of maximum amplitudes are called antinodal lines and that of minimum amplitude are called nodal lines. The points where the nodal lines meet the screen is the location of interference minimum and the point where the antinodal lines meet the screen is the location of interference maximum.

Mathematical Derivation Let S1 and S2 be section of two coherent sources of monochromatic light illuminated by a monochromatic point source having wavelength l. Let the separation between the slits be d ×YY ¢ is the screen on which the light transmitted through the two slits is received. Let the

1094 JEE Main Physics distance of the screen YY ¢ Y from the slits be D. P Assuming d x (green) x (blue) < x (green) x (blue)/x (green) = 5400/4360

29. In a double slit interference experiment, the distance between the slits is 0.05 cm and screen is 2 m away from the slits. The wavelength of light is 8.0 ´ 10-5 cm . The distance between successive fringes is (a) 0.24 cm (c) 1.28 cm

(b) 3.2 cm (d) 0.32 cm

30. Two light rays having the same wavelength l in vacuum are in phase initially. Then the first ray travels a path L1 through a medium of refractive index n1, while the second ray travels a path of length L2 through a medium of refractive index n2 . The two waves are then combined to observe interference. The phase difference the two waves is 2p ( L2 - L1 ) l 2p (c) ( n2 L1 - n1L2 ) l (a)

2p ( n1L1 - n2 L2 ) l 2p æ L1 L ö (d) ç - 2÷ l è n1 n2 ø (b)

31. In Young’s experiment, the wavelength of red light is d

d

24. The correct formula for fringe visibility is Imax - Imin Imax + Imin I (c) V = max Imin

(a) V =

Imax + Imin Imax - Imin Imin (d) V = Imax (b) V =

25. Two coherent waves are represented by y1 = a1 cos wt and y2 = a2 sin wt, superimposed on each other. The resultant intensity is proportional to (a) ( a1 + a2 ) (c) ( a21 + a22 )

(b) ( a1 - a2 ) (d) ( a12 - a22 )

26. The maximum intensity of fringes in Young’s

7.8 ´ 10-5 cm and that of blue light 5.2 ´ 10-5 cm . The value of n for which ( n + 1)th blue light band coincides with nth red bond is (a) 4 (c) 3

(b) 2 (d) 1

32. We shift Young’s double slit experiment from air to water. Assuming that water is still and clear, it can be perdicted that the fringe pattern will (a) remain unchanged (c) shrink

(b) disappear (d) be enlarged

33. In Young’s double slit experiment, the separation between slit is halved and the distance between the slits and screen is doubled. The fringe width is (a) unchanged (c) double

(b) halved (d) quardrupled

experiment is I. If one of the slit is closed, then the intensity at that place becomes I 0 . Which of the following relation is true?

34. In an experiment, the two slits are 0.5 mm apart and

27. The equations of two interfering waves are

the fringes are observed to 100 cm from the plane of the slits. The distance of the 11th bright fringe from the 1st bright fringe is 9.72 mm. The wavelength is

(a) I = I0 (c) I = 4 I0

(b) I = 2I0 (d) I = 0

y1 = b cos wt and y2 = b cos( wt + f). For destructive interference the path difference is (a) 0° (c) 180°

(b) 360° (d) 720°

28. The Young’s double slit experiment is performed with blue and with green light of wavelengths 4360 Å and 5460 Å respectively. If, x is the distance of 4th maxima from the central one, then

(a) 4.86 ´ 10 -5 cm (b) 5.72 ´ 10 -4 cm (c) 5.87 ´ 10 -4 cm (d) 3.25 ´ 10 -4 cm

35. Figure shows a standard two slit arrangement with slits S1, S2, P1, P2 are the two minima points on either side of P.

Wave Optics

42. In Young’s double slit experiment, the seventh

Screen S1

maximum with wavelength l1 is at a distance d1 and the same maximum with wavelength l2 is at a distance d2 . Then, d1/ d2 =

P1 S

P S2

P2

S3 Second screen

S4

At P2 on the screen, there is a hole and behind P2 is a second 2-slit arrangement with slits S3, S4 and a second screen behind them. [NCERT Exemplar] (a) There would be no interference pattern on the second screen but it would be lighted (b) The second screen would be totally dark (c) There would be a single bright point on the second screen (d) There would be a regular two slit pattern on the second screen

36. Two coherent light sources S1 and S2 ( l = 6000 Å) are

1 mm apart from each other. The screen is placed at a distance of 25 cm from the sources. The width of the fringes on the screen should be (a) 0.015 cm (c) 0.01 cm

(b) 0.013 cm (d) 0.10 cm

37. Through quantum theory of light we can explain a number of phenomena observed with light, it is necessary to retain the wave nature of light to explain the phenomenon of (a) Photoelectric effect (c) Compton effect

(b) Diffraction (d) Black body radiation

38. If the intensities of the two interfering beams in Young’s double slit experiment be I1 and I2 , then the contrast between the maximum and minimum intensity is good when (a) I1 is much greater than I2 (b) I1 is much smaller than I2 (c) I1 = I2 (d) Either I1 = 0 or I2 = 0

39. The fringe width at a distance of 50 cm from the slits in Young’s experiment for light of wavelength 6000 Å is 0.048 cm. The fringe width at the same distance for l = 5000 Å, will be (a) 0.04 cm (c) 0.14 cm

(b) 0.4 cm (d) 0.45 cm

40. Two waves originating from sources S1 and S2

having zero phase difference and common wavelength l will show complete destructive interference at a point P, if, ( S1 P - S2 P) = (a) 5l

3l (b) 4

4l (c) 2

11l (d) 2

41. Two coherent sources of intensities I1 and I2 produce

an interference pattern. The maximum intensity in the interference pattern will be (a) I1 + I2

(b) I21 + I22

(c) ( I 1 + I2 )2

(d) ( I1 +

1107

I2 )2

(a)

l1 l2

(b)

l2 l1

(c)

l21 l22

(d)

l22 l21

43. When monochromatic light is replaced by white light in Fresnel’s biprism arrangement, the central fringe is (a) coloured (c) dark

(b) white (d) None of these

44. In the setup shown in figure, the two slits, S1 and are not S2 equidistant from the slit S S. The central fringe at O is, then (a) (b) (c) (d)

S1 O S2

always bright always dark Either dark or bright depending on the position of S Neither dark nor bright

45. In a double slit interference experiment,the distance between the slits is 0.05 cm and screen is 2 m away from the slits. The wavelength of light is 6000 Å. The distance between the fringes is (a) 0.24 cm (c) 1.24 cm

(b) 0.12 cm (d) 2.28 cm

46. The separation between successive fringes in a double slit arrangement is x. If the whole arrangement is dipped under water, what will be the new fringe separation? [The wavelength of light being used is 5000 Å] (a) 1.5 x

(b) x

(c) 0.75 x

(d) 2 x

47. In a Young’s experiment, two coherent sources are placed 0.90 mm aprt and the fringes are observed one metre away. If it produces the second dark fringe at a distance of 1 mm from the central fringe, the wavelength of monochromatic light used will be (a) 60 ´ 10 -4 cm

(b) 10 ´ 10 -4 cm

(c) 10 ´ 10 -5 cm

(d) 6 ´ 10 -5 cm

48. In the Young’s double slit experiment, the interference pattern is found to have an intensity ratio between bright and dark fringes as 9. This implies that (a) the intensities at the screen due to two slits are 5 units and 4 units respectively (b) the intensities at the screen due to two slits are 4 units and 1 unit respectively (c) the amplitude ratio is 3 (d) the amplitude ratio is 2

1108 JEE Main Physics 49. Interference fringes are being produced on screen

56. In Young’s double slit experiment, the intensity on

XY by the slits S1 and S2 . In figure, the correct fringe locus is

screen at a point where path difference is l is K . What will be intensity at the point where path difference is l/4?

W1

S1 S

W3

P S2

Q

W2

(a) PQ (c) W3W4

50. Microwaves from a transmitter are directed normally towards a plane reflector. A detector moves along the normal to the reflector. Between positions of 14 successive maxima, the detector travels a distance of 0.14 m. The frequency of transmitter is (b) 1010 Hz (d) 6 ´ 1010 Hz

51. Two waves having the intensities in the ratio 9 : 1 produce interference. The ratio of maximum to minimum intensity is equal to (a) 10 : 8

(b) 9 : 1

(c) 4 : 1

(d) 2 : 1

52. In Young’s double slit experiment, phase difference between light waves reaching 3rd bright fringe from the central fringe when l = 5000 Å, is (a) 6p (c) 4p

(b) 2p (d) zero

53. In an interference pattern produced by two identical slits, the intensity at the slit of the central maximum is I. The intensity at the same spot when either of the slits is closed is I 0 . Therefore, (a) (b) (c) (d)

I = I0 I = 2I0 I = 4 I0 I and I0 are not related to each other

54. In a two slits experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by 5 ´ 10-2 m towards the slits, the change in fringe width is 3 ´ 10-5 m. If separation between the slits is 10-3m, the wavelength of light used is (a) 4500 Å (c) 5000 Å

(b) 3000 Å (d) 6000 Å

55. In Young’s double slits experiment, let S1 and S2 be

the two slits, and C be the centre of the screen. If ÐS1CS2 = q and l is the wavelength, the fringe width will be l q 2l (c) q (a)

(b) lq (d)

l 2q

(b) K/2

(c) K

(d) zero

57. In the Young’s double slit experiment, a mica slip of

W4

(b) W1W2 (d) XY

(a) 1.5 ´ 1010 Hz (c) 3 ´ 1010 Hz

(a) K/4

thickness t and refractive index m is introduced in the ray from first source S1. By how much distance fringes pattern will be displaced? d (m - 1)t D d (c) (m - 1) D

(a)

D (m - 1)t d D (d) (m - 1) d (b)

58. Oil floating on water appears coloured due to interference of light. What should be the order of magnitude of thickness of oil layer in order that this effect may be observed? (a) 1,000 Å (c) 10 Å

(b) 1 cm (d) 100 Å

and waves y1 = A1 sin( wt - b1) y2 = A2 sin( wt - b2 ) superimpose to form a resultant wave whose amplitude is

59. Two

(a)

A21 + A22 + 2 A1 A2 cos (b1 - b2 )

(b)

( A21 + A22 + 2 A1 A2 sin (b1 - b2 )

(c) A1 + A2 (d) | A1 + A2 |

60. In Young’s double slit experiment, 12 fringes are obtained in a certain segment of the screen when light of wavelength 600 nm, is used. If the wavelength of light is changed to 400 nm, number of fringes observed in the same segment of the screen is given by (a) 12 (c) 24

(b) 18 (d) 30

61. In Young’s double slit experiment, distance between two sources is 0.1 mm. The distance of screen from the sources is 20 cm. Wavelength of light used is 5460 Å. Then angular position of first dark fringe is (a) 0.08° (c) 0.20°

(b) 0.16° (d) 032°

62. Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is p/2 at point A and p at point B. Then the difference between the resultant intensities at A and B is (a) 2 I (c) 5 I

(b) 4 I (d) 7 I

Wave Optics 63. In a Young’s double slit experiment using red and blue lights of wavelengths 600 nm and 480 nm respectively, the value of n from which the nth red fringe coincides with ( n + 1) the blue fringes is

(a) 40 m (c) 47 m

1109

(b) 43 m (d) 38 m

71. Three waves of equal frequency having amplitudes

64. In Young’s double slit experiment, distance between

10 mm, 4 mm, 7mm arrive at a given point with p successive phase difference of , the amplitude of 2 the resulting wave (in mm) is given by

source is 1 mm and distance between the screen and source is 1m. If the fringe width on the screen is 0.06 cm, then l is

72. An astronaut floating freely in space decides to use

(a) 5 (c) 3

(a) 6000 Å (c) 1200 Å

(b) 4 (d) 2

(b) 4000 Å (d) 2400 Å

65. In Young’s double slit experiment, the central bright fringe can be identified (a) (b) (c) (d)

as it has greater intensity than the other bright fringes as it is wide than the other bright fringes as it is narrower than the other bright fringes by using while light instead of monochromatic light

66. Two slits, 4 mm apart are illuminated by light of wavelength 600 Å. What will be the fringe width on a screen placed 2 m from the slits? (a) 0.12 mm (c) 3.0 mm

(b) 0.3 mm (d) 4.0 mm

67. In a Young’s double slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case (a) there should be no interference fringe (b) there should be an interference pattern for red mixing with one for blue (c) there should be alternate interference patterns of red and blue (d) None of the above

68. Consider a point at the focal point of a convergent lens. Another convergent lens of short focal length is placed on the other side. What is the nature of the wavefront emerging from the final image? (b) Elliptical (d) Square

69. A parallel beam of light of wavelength 3141.59 Å is incident on a small aperture. After passing through the aperture, the beam is no longer parallel but diverges at 1° to the incident direction. What is the diameter of the aperture? (a) 180 m (c) 1.8 m

(b) 18 mm (d) 0.18 m

70. Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and [NCERT] wavelength 400 nm.

(b) 5 (d) 7

his flash light as a rocket. He shines a 10W light beam in a fixed direction so that he acquires momentum in the opposite direction. If his mass is 80 kg, how long must he need to reach a velocity of 1 ms -1? (a) 9 s (c) 2.4 ´ 106 s

(b) 2.4 ´ 103 s (d) 2.4 ´ 10 9 s

73. A beam of light consisting of two wavelengths 650 nm and 520 nm is used to illuminate the slit of a Young’s double slit experiment. Then the order of the bright fringe of the longer wavelength that coincide with a bright fringe of the shorter wavelength at the least distance from the central maximum is (a) 1 (c) 3

Diffraction of Light

(a) Cylindrical (c) Spherical

(a) 4 (c) 6

(b) 2 (d) 4

74. Two identical radiators have a separation of d = l/4 where l is the wavelength of the waves emitted by either source. The initial phase difference between the source is p/4. Then the intensity on the screen at a distant point situated at an angle, q = 30° from the radiators is (here, I 0 is intensity at that point due to one radiator alone) (a) I0 (c) 3I0

(b) 2I0 (d) 4 I0

75. In Young’s double slit experiment, the 8th maximum with wavelength l1 is at a distance, d1 from the central maximum and the 6th maximum with wavelength l2 is at a distance, d2 . Then, d1/ d2 is equal to (a)

4 æ l2 ö ç ÷ 3 è l1 ø

(b)

4 æ l1 ö ç ÷ 3 è l2 ø

(c)

3 æ l2 ö ç ÷ 4 è l1 ø

(d)

3 æ l1 ö ç ÷ 4 è l2 ø

76. Light of wavelength 500 nm is used to form interference pattern in Young’s double slit experiment. A uniform glass plate of refractive index 1.5 and thickness 0.1 mm is introduced in the path of one of the interfering beams. The number of fringes which will shift the cross wire due to this is (a) 100 (c) 300

(b) 200 (d) 400

1110 JEE Main Physics 77. Air has refractive index 1.003. The thickness of air column, which will have one more wavelength of yellow light (6000 Å) than in the same thickness of vacuum is (a) 2 mm (c) 2 m

(b) 2 cm (d) 2 km

78. The distance between the first and the sixth minima in the diffraction pattern of a single slit is 0.5 mm. The screen is 0.5 m away from the slit. If the wavelength of light used is 5000 Å. Then the slit width will be (a) 5 mm (c) 1.25 mm

(b) 2.5 mm (d) 1.0 mm

79. Plane microwaves are incident on a long slit having a width of 5 cm. The wavelength of the microwaves if the first minimum is formed at 30° is (a) 2.5 cm (c) 25 cm

(b) 2 cm (d) 2 mm

80. A plane wave of wavelength 6250 Å is incident

normally on a slit of width 2 ´ 10-2 cm. The width of the principal maximum on a screen distant 50 cm will be (a) 312.5 ´ 10 -3cm (c) 312 cm

(b) 312.5 ´ 10 -4 cm (d) 312.5 ´ 10 -5 cm

81. The main difference between the phenomena of interference and diffraction is that (a) diffraction is caused by reflected waves from a source whereas interference is caused due to refraction of waves from a source (b) diffraction is due to interaction of waves derived from the same source, whereas interference is that bending of light from the same wavefront (c) diffraction is due to interaction of light from wavefront, whereas the interference is the interaction of two waves derived from the same source (d) diffraction is due to interaction of light from the same wavefront whereas interference is the interaction of waves from two isolated sources

82. Light of wavelength 6000 Å is incident on a single slit. The first minimum of the diffraction pattern is obtained at 4 mm from the centre. The screen is at a distance of 2 m from the slit. The slit width will be (a) 0.3 mm (c) 0.15 mm

(b) 0.2 mm (d) 0.1 mm

83. The Fraunhofer diffraction pattern of a single slit is formed in the focal plane of a lens of focal length 1 m. The width of slit is 0.3 mm. If third minimum is formed at a distance of 5 mm from central maximum, then wavelength of light will be (a) 5000 Å (c) 7500 Å

(b) 2500 Å (d) 8500 Å

84. What should be refractive index of a transparent medium to be invisible in vacuum? (a) 1 (c) > 1

(b) < 1 (d) None of these

85. A slit 5 cm wide is irradiated normally with microwaves of wavelength 1.0 cm. Then the angular spread of the central maximum on either side of incident light is nearly (a) 1/5 rad (c) 5 rad

(b) 4 rad (d) 6 rad

86. Which of the following phenomena is not common to sound and light waves? (a) Interference (c) Coherence

(b) Diffraction (d) Polarisation

87. A beam of ordinary unpolarised light passes through a tourmaline crystal C1 and then it passes through another tourmaline crystal C2 , which is oriented such that its principal plane is parallel to that of C2 . The intensity of emergent light is I 0 . Now C2 is rotated by 60° about the ray. The emergent ray will have an intensity (a) 2I0

(b) I0 / 2

(c) I0 / 4

(d) I0 / 2

88. What is the Brewster’s angle for air to glass transition? (Refractive index of glass = 1.5) [NCERT] (a) 15° 27¢ (c) 50° 16 ¢

(b) 36° 27 ¢ (d) 56° 18 ¢

89. An unpolarised beam of intensity 2 a2 passes through a thin polaroid. Assuming zero absorption in the polaroid, the instensity of emergent plane polarised light is (a) 2a2

(b) a2

(c)

2 a2

(d)

a2 2

90. 80 g of impure sugar, when dissolved in a litre of water gives an optical rotation of 9.9°, when placed in a tube of length 20 cm. If, the specific rotation of sugar is 66°, then concentration of sugar solution will be (a) 80 gL-1

(b) 75 gL-1

-1

(d) 50 gL-1

(c) 65 gL

91. If, for a calcite crystal, m 0 and m e are the refractive indices of the crystal for O-ray and E-ray respectively, then, along the optic axis of the crystal (a) m 0 = m e (c) m e < m 0

(b) m e > m 0 (d) None of these

92. l a and l m are the wavelengths of a beam of light in air and medium respectively. If q is the polarising angle, the correct relation between l a , l m and q is (a) l a = l m tan2 q

(b) l m = l a tan2 q

(c) l a = l m cot q

(d) l m = l a cot q

Wave Optics 93. In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3. [NCERT] (a) 0.15° (c) 0.27°

(b) 0.30° (d) 0.45°

Round II Only One Correct Option 1. The ratio of intensities of two waves is 9 : 1. They are producing interference. The ratio of maximum and minimum intensities will be (a) 10 : 8

(b) 9 : 1

(c) 4 : 1

(d) 12 : 1

2. Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities are (a) 9I and I

(b) 5I and 2 I (c) 2 I and 9I

(d) 4I and 2 I

3. Interference was observed in interference chamber when air was present, now the chamber is evacuated and if, the same light is used, a careful observer will see

(a) interference in which width of the fringe will be slightly increased (b) interference with bright band (c) interference with dark band (d) All of the above

4. A stone thrown into still water, creates a circular wave pattern moving radially outwards. If, r is the distance measured from the centre of the pattern, the amplitude of the wave varies as (a) r -3/2

(b)

r -1/2 2

(c) r -1

(d) r1/3

5. Two coherent sources of intensities I1 and I2 produce

an interference pattern. The maximum intensity in the interference pattern will be (a) ( I 1 +

I2 )2

(b) I21 + I22 (d) ( I 1 + I2 )2

(c) I 1 + I2

6. If, an interference pattern has maximum and minimum intensities in 36 : 1 ratio, then what will be the ratio of amplitudes? (a) 4 : 5

(b) 7 : 5

(c) 6 : 5

(d) 3 : 4

7. The waves of wavelength 5900 Å emitted by any atom or molecule must have some finite total length which is known as coherent length. For sodium light, this length is 2.4 cm. The number of oscillations in this length will be

1111

94. The 6563 Å H a sign line emitted by hydrogen in a

star is found to be red-shifted by 15 Å. Estimate the speed with which the star is receding from the earth. [NCERT]

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

6.9 m/s approaching the earth 6.86 m/s receding the earth 7.9 m/s receding the earth 8.9 m/s receding the earth

(Mixed Bag) (a) 4.068 ´ 105

(b) 4.068 ´ 106

7

(d) 4.068 ´ 10 4

(c) 4.068 ´ 10

8. In a certain double slit experimental arrangement interference fringes of width 1.0 mm each are observed when light of wavelength 5000 Å is used. Keeping the setup unaltered, if the source is replaced by another source of wavelength 6000 Å, the fringe width will be (a) 1.2 mm (c) 1.8 mm

(b) 1.5 mm (d) 2.0 mm

9. A beam of circularly polarised light is completely absorbed by an object on which it falls. If, U represents absorbed energy and w represents angular frequency, then angular momentum transferred to the object is given by U w2 U (c) w

U 2w 2U (d) w

(a)

(b)

10. In

Young’s double-slit experiment using monochromatic light of wavelength l, the intensity of light at a point on the screen where path difference is l i.e., K units. What is the intensity of light at a point where path difference is l/3? [NCERT] K 4 K (c) 7

K 5 K (d) 2 (b)

(a)

11. In Young’s double slit experiment, we get 60 fringes in the field of view of monochromatic light of wavelength 4000 Å. If we use monochromatic light of wavelength 6000 Å, then the number of fringes obtained in the same field of view are (a) 60

(b) 90

(c) 40

(d) 1.5

12. The phenomenon which does not take place in sound waves is (a) scattering (c) interference

(b) diffraction (d) polarisation

1112 JEE Main Physics 13. Two Nicol prisms are first crossed and then one of them is rotated through 60°. The percentage of incident light transmitted is (a) 1.25 (c) 37.5

(b) 25.0 (d) 50

14. Find the thickness of a plate which will produce a change in optical path equal to half the wavelength l of the light passing through it normally. The refractive index of the plate m is equal to l (a) 4(m - 1) (c)

2l (b) 4(m - 1)

l (m - 1)

(d)

l 2(m - 1)

source falls on a single slit 1.00 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between the first dark fringes on either side of the central bright fringe is (b) 1.2 mm (d) 2.4 mm

16. Light of wavelength 2 ´ 10-3 m falls on a slit of width

4 ´ 10-3 m. The angular dispersion of the central maximum will be (a) 30° (c) 90°

(b) 60° (d) 180°

17. nth bright fringe if red light ( l1 = 7500 Å) coincides with ( n + 1)th bright fringe ( l2 = 6000 Å). The value of n, is (a) 4 (c) 3

of

green

light

(b) 5 (d) 2

18. In double-slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1°. What is the spacing between the two slits? [NCERT] (a) 3.44 ´ 10 -4 m

(b) 3.03 ´ 10 -4 m

(c) 4.03 ´ 10 -4 m

(d) 2.68 ´ 10 -4 m

19. A light of wavelength 5890 Å falls normally on a thin air film. The minimum thickness of the film such that the film appears dark in reflected light is (a) 2.945 ´ 10 -7 m

(b) 3.945 ´ 10 -7 m

(c) 4.95 ´ 10 -7 m

(d) 1.945 ´ 10 -7 m

20. In Young’s double slit experiment, a minimum is obtained when the phase difference of super imposing waves is (a) zero

(b) (2n - 1) p (c) np

(a) 3 cm (c) 2 cm

(b) 0.011 cm (d) 4 cm

22. The velocity of a moving galaxy is 300 km s -1 and the apparent change in wavelength of a spectral line emitted from the galaxy is observed as 0.5 nm. Then, the actual wavelength of the spectral line is (a) 3000 Å (c) 6000 Å (e) 5500 Å

(b) 5000 Å (d) 4500 Å

23. How fast a person should drive his car so that the red

15. A beam of light of wavelength 600 nm from a distant

(a) 1.2 cm (c) 2.4 cm

of prism is 1°. The fringe width with light of wavelength 6000 Å will be

(d) ( n + 1) p

21. In Fresnel’s biprism (m = 1.5) experiment the distance between source and biprism is 0.3 m and that between biprism and screen is 0.7 m and angle

signal of light appears green? (Wavelength for red colour = 6200 Å and wavelength for green colour = 5400 Å) (a) 1.5 ´ 10 8 m/s 7

(c) 3.9 ´ 10 m/s

(b) 7 ´ 107 m/s

(d) 2 ´ 10 8 m/s

24. In Young’s double slit experiment, if d, D and l represent, the distance between the slits, the distance of the screen from the slits and wavelength of light used respectively, then the bandwidth is inversely proportional to (a) l (c) D (e) D2

(b) d (d) l2

25. A star is moving towards the earth with a speed of

4.5 ´ 106 m/s. If the true wavelength of a certain line in the spectrum received from the star is 5890 Å. Its apparent wavelength will be about (c = 3 ´ 108 m/s) (a) 5890 Å (c) 5802 Å

(b) 5978 Å (d) 5896 Å

26. In Young’s two slit experiment the distance between the two coherent sources is 2 mm and the screen is at a distance of 1 m. If the fringe width is found to be 0.03 cm, then the wavelength of the light used is (a) 4000 Å (c) 5890 Å

(b) 5000 Å (d) 6000 Å

27. In Young’s experiment the wavelength of red light is 7.8 ´ 10-5 cm and that of blue light 5.2 ´ 10-2 cm . The value of n for which ( n + 1)th blue bright band coincides with nth red band is (a) 4 (c) 2

(b) 3 (d) 1

28. Two slits separated by a distance of 1 mm are illuminated with red light of wavelength 6.5 ´ 10-7 m . The interference fringes are observed on a screen place 1 m from the slits. The distance between the third dark fringe and the fifth bright fringe is equal to (a) 0.65 mm (c) 3.25 mm

(b) 1.63 mm (d) 4.88 mm

Wave Optics 29. In Young’s double slit experiment, the 7th maximum wavelength l1 is at a distance d1 and that with wavelength l2 is at a distance d2 . Then ( d1/ d2 ) is (b) ( l2 / l1 ) (d) ( l22 / l21 )

(a) ( l1 / l2 ) (c) ( l21 / l22 )

1113

d ( > > b) from the slits. At a point on the screen directly in front of one of the slits, certain wavelengths are missing, figure. Some of these missing wavelengths are

30. If white light is used in a biprism experiment, then (a) (b) (c) (d)

fringe pattern disappers all fringes will be coloured central fringe will be white while others will be coloured central fringe will be dark

31. Two waves of same frequency and same amplitude from two monochromatic source are allowed to superpose at a certain point. If in one case the phase difference is 0° and in other case is p /2, the ratio of the intensities in the two cases will be (a) 1 : 1 (c) 4 : 1

(b) 2 : 1 (d) None of these

32. Two polaroids are kept crossed to each other. Now one of them is rotated through an angle of 45°. The percentage of incident light now transmitted through the system is (a) 15%

(b) 25%

(c) 50%

(d) 60%

33. In Young’s double slit experiment, the two slits act as coherent sources of equal amplitude A and wavelength l. In another experiment with the same setup, the two slits are sources of equal amplitude A and wavelength l but are incoherent. The ratio of the intensity of light at the mid-point of the screen in the first case to that in the second case is (a) 2 : 1 (c) 3 : 4

(b) 1 : 2 (d) 4 : 3

34. If the two waves represented by y1 = 4 sin wt and

interfere at a point, the y2 = 3 sin( wt + p/3) amplitude of the resulting wave will be about (a) 7

(b) 5

(c) 6

(d) 3.5

35. In a biprism experiment, by using light of wavelength 5000 Å, 5 mm wide fringes are obtained on a screen 1.0 m away from the coherent sources. The separation between the two coherent sources is (a) 1.0 mm (c) 0.05 mm

(b) 0.1 mm (d) 0.01 mm

36. How will the diffraction pattern of single slit change when yellow light is replaced by blue light? The fringe will be (a) wider (c) brighter

(b) narrower (d) fainter

37. White light is used to illuminate the two slits in a Young’s double slit experiment. The separation between slits is b and the screen is at a distance

S1

d

P

b S2

b2 2b2 , d 3d 2b2 (c) l = 3d

(a) l =

b2 3b2 , 2d 2d 3b2 (d) l = 4d (b) l =

38. A beam of unpolarized light having flux 10-3 W falls normally on a polarizer of cross-sectional area 3 ´ 10-4 m 2 . The polarizer rotates with an angular frequency of 31.4 rads -1. The energy of light passing through the polarizer per revolution will be (a) 10 -4 J

(b) 10 -3 J

(c) 10 -2 J

(d) 10 -1 J

39. In a biprism experiment, 5th dark fringe is obtained at a point. If a thin transparent film is placed in the path of one of waves, then 7th bright fringes is obtained at the same point. The thickness of the film in terms of wavelength l and refractive index, m will be (a)

1.5l (m - 1)

(b) 1.5(m - 1) l

(c) 2.5(m - 1) l

(d)

2.5l (m - 1)

40. The ratio of intensities of successive maxima in the differaction pattern due to single slit is (a) 1 : 4 : 9 4 4 (c) 1 : 2 : 9 p 25 p2

(b) 1 : 2 : 3 4 9 (d) 1 : 2 : 2 p p

41. The equations of displacement of two waves are given as y1 = 10 sin(3 pt + p/3) y2 = 5(sin 3 pt + 3 cos 3 pt), then what is the ratio of their amplitude? (a) 1 : 2 (c) 1 : 1

(b) 2 : 1 (d) None of these

42. Light of wavelength l is incident on a slit width d.

The resulting diffraction pattern is observed on a screen at a distance D. The linear width of the principal maximum is equal to the width of the slit, if D equals d2 2l 2l2 (c) d (a)

d l 2l (d) d

(b)

1114 JEE Main Physics 43. A glass slab of thickness 8 cm contains the same number of waves as 10 cm of water when both are traversed by the same monochromatic light. If the refractive index of water is 4/3, the refractive index of glass is (a) 5/4 (c) 5/3

(b) 3/2 (d) 16/15

(a) 1.5 (c) 1.2

50. Two sources S1 and S2 of intensity I1 and I2 are

placed in front of a screen [Fig.(a)]. The pattern of intensity distribution seen in the central portion is given by Fig. (b) [NCERT Exemplar]

44. Fringes are obtained with the help of a biprism in

S1

the focal plane of an eyepiece distant 1 m from the slit. A convex lens produces images of the slit in two positions between biprism and eyepiece. The distances between two images of the slit in two positions are 4.05 ´ 10-3 m and 2.90 ´ 10-3 m respectively. The distance between the slits will be (a) (b) (c) (d)

x S2 Fig. (a)

3.43 ´10 -3 m 0.343 m 0.0343 m 43.3 m

d = 10-4 D (d = distance between slits, D = distance of screen from the slits). At a point P on the screen resultant intensity is equal to the intensity due to individual slit I 0 . Then the distance of point P from the central maximum is ( l = 6000 Å)

x=0 Fig. (b)

45. In Young’s double slit experiment

(a) 0.5 mm (c) 1 mm

(b) 2 mm (d) 4 mm

46. In Young’s double slit experiment, the intensity of the maxima is I. If the width of each slit is doubled, the intensity of the maxima will be (a) I/2 (c) 4 I

(b) 2 I (d) I

47. The maximum number of possible interference maxima for slit separation equal to twice the wavelength in Young’s double slit experiment is (a) Infinite (c) 3

(b) 5 (d) zero

48. If, I 0 is the intensity of the principal maximum in the

single slit diffraction pattern, then what will be its intensity when the slit width is doubled? (a)

(b) 1.6 (d) 1.3

I0 2

(b) I0

(c) 4 I0

(d) 2I0

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

S1 and S2 have the same intensities S1 and S2 have a constant phase difference S1 and S2 have the same phase S1 and S2 have the same wavelength

51. In Young’s double slit experiment, white light is used. The separation between the slits is b. The screen is at a distance d (d >> b) from the slits. Some wavelengths are missing exactly in front of one slit. These wavelengths are b2 d b2 (c) l = 3d

(a) l =

2b2 d 2b2 (d) l = 3d (b) l =

52. In the Young’s double slit experiment, the ratio of intensities of bright and dark fringes is 9. This means that (a) the intensities of individual sources are 5 and 4 units respectively (b) the intensities of individual sources are 4 and 1 units respectively (c) the ratio of their amplitudes is 3 (d) the ratio of their amplitudes is 2

53. Consider sunlight incident on a pinhole of width

More Than One Correct Option 49. A ray of light travelling in a transparent medium falls on a surface separating the medium from air, at an angle of incidence of 45°. The ray undergoes total internal reflection. If, m is refractive index of the medium w.r.t. air, select the possible values of m from the following

103 A. The image of the pinhole seen on a screen shall be [NCERT Exemplar] (a) (b) (c) (d)

a sharp white ring different from a geometrical image a diffused central spot, white in colour diffused coloured region around a sharp central white spot

Wave Optics 54. In an interference arrangement similar to Young’s double slit experiment, the slits S1 and S2 are illuminated with coherent microwave sources each of frequency 106 Hz. The sources are synchronized to have zero phase difference. The slits are separated by distance d = 150 m. The intensity I(q ) is measured as a function of q, where q is defined as shown. If I 0 is maximum intensity, then I(q ) for 0 £ q £ 90° is given by

(b) n = 7.5 ´ 1014 Hz (d) l = 9000 Å

56. In Young’s double slit experiment, the distance between the two slits is 0.1 mm and wavelength of light used is 4 ´ 10-7 m. If the width of fringe on screen is 4 mm, the distance between screen and slit is (c) 0.1 cm

(d) 1 m

57. Consider the diffraction pattern for a small pinhole. As the size of the hole is increased (a) (b) (c) (d)

[NCERT Exemplar]

the size decreases the intensity increases the size increases the intensity decreases

ratio of intensities of a bright band and a dark band is 16 : 1. The ratio of amplitudes of interfering waves is (b) 5/3

c

Medium-1 f

Y h

g

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

the same in medium-1 and medium-2 larger in medium-1 than in medium-2 larger in medium-2 than in medium-1 different at b and d

61. The phases of the light wave at c, d, e and f are fc , fd , fe and ff respectively. It is given that fc ¹ ff (a) (b) (c) (d)

fc cannot be equal to fd fd can be equal to fe ( fd - ff ) is equal to ( fc - fe ) ( fd - fc ) is not equal to ( ff - fe )

62. Light travels as a (a) (b) (c) (d)

parallel beam in each medium convergent beam in each medium divergent beam in each medium divergent beam in one medium and convergent beam in the other medium

Passage II

58. In Young’s double slit experiment, on interference,

(a) 16

a X

d

60. Speed of light is

of refractive index 1.5. Inside the medium, its frequency is n and its wavelength is l. Then,

(b) 1 cm

b

e

55. A light of wavelength 6000 Å in air enters a medium

(a) 0.1 mm

ef and gh represent wavefronts of the light wave in medium-2 after refraction.

Medium-2

(a) I( q) = I0 for q = 0° I (b) I( q) = 0 for q = 30° 2 I0 (c) I( q) = for q = 90° 4 (d) I( q) is constant for all values of q

(a) n = 5 ´ 1014 Hz (c) l = 4000 Å

1115

(c) 4

(d) 1/4

59. For light diverging from a point source [NCERT Exemplar]

(a) the wavefront is spherical (b) the intensity decreases in proportion to the distance squared (c) the wavefront is parabolic (d) the intensity at the wavefront does not depend on the distance

Comprehension Based Questions Passage I The figure shows a surface XY separating two transparent media, medium-1 and medium-2. The lines ab and cd represent wavefronts of a light wave travelling in medium-1 and incident on XY . The lines

In YDSE, intensity of light from two slits is proportional to their width, and also to the square of amplitudes of light from the slits, i. e., I1 w1 a2 = = I2 w2 b2 At points, where crest of one wave falls on crest of the other and through falls on trough, intensity of light is maximum. Interference is said to be constructive. At points, where crest of one wave falls on trough of the other, resultant intensity of light is minimum. The interference is said to be destructive. I max ( a + b)2 = I min ( a - b)2 For sustained interference, the two sources must be coherent. Two independent sources cannot be coherent. 63. The light waves from two coherent sources are represented by y1 = a1 sin wt and y2 = a2 sin( wt + p/2) The resultant amplitude will be (a) a1

(b) a2

(c) a1 + a2

(d)

a21 + a22

1116 JEE Main Physics 64. The ratio of intensities in the interference pattern in the above question will be (a) 9 : 1 (c) 2 : 1

Directions

(b) 4 : 1 (d) 1 : 4

65. The amplitudes of light waves from two slits are in the ratio 3 : 1. Widths of slits are in the ratio (a) 1 : 9 (c) 9 : 1

(b) 1 : 3 (d) 3 : 1

66. The widths of two slits in YDSE are in the ratio 1 : 4. The ratio of amplitudes of light waves from two slits will be (a) 9 : 1 (c) 1 : 2

(b) 4 : 1 (d) 1 : 2

y1 = a1 sin w1t and y2 = a2 sin( w1t + p/2). Will you observe interference pattern? Yes No Sometimes Cannot say

68. Match the following column I with column II.

III. IV.

system when the space between the lens and the glass plate is filled with a liquid of refractive index greater than that of glass, the central spot of the pattern is dark. Reason The refraction in Newton’s ring cases will be from denser to rarer medium and the two interfering rays are reflected under similar conditions.

71. Assertion To observe diffraction of light the size of

Matching Type

I. II.

Question No. 70 to 82 are Assertion-Reason type. Each of these contains two Statements : Statement I (Assertion), Statement II (Reason). Each of these questions also has four alternative choice, only one of which is correct. You have to select the correct choices from the codes (a), (b), (c) and (d) given below (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 correct explanation of the Assertion (c) If Assertion is true but Reason is false (d) If Assertion is false but the Reason is true

70. Assertion Newton’s rings are formed in the reflected

67. Two independent sources emit light waves given by (a) (b) (c) (d)

Assertion and Reason

Column II

Column I Polarization Change in the path of light on screen entering another medium Bending of light around corners Light added to light can produce darkness

A. B.

Refraction Diffraction

C.

Interference

D.

Wave nature of light

obstacle/aperture should be of the order of 10-7 m. Reason 10-7 m is the order of wavelength of visible light.

72. Assertion A famous painting was painted by not using brush strokes in the usual manner, but rather a myriad of small colour dots. In this painting the colour you see at any given place on the painting changes as you move away. Reason The angular separation of adjacent dots changes with the distance from the painting.

73. Assertion In Young’s experiment, the fringe width

Code

(c) I-B, II-A, III-A, B, IV-B, D

for dark fringes is same from that for white fringes. Reason In Young’s double slit experiment, when the fringes are performed with a source of white light, then only black and bright fringes are observed.

(d) I-C, II-B, III-B, C, IV-A, C

74. Assertion Thin films such as soap bubble or a thin

(a) I-A, II-B, III-C, D, IV-A, D (b) I-D, II-A, III-B, D, IV-C, D

69. Match the following column I with column II. Column I I. II. III. IV.

Polarization m = tani p Interference Focal length of a lens

A. B. C. D.

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

I-C, II-C, III-D, IV-A, B I-A, II-C, III-D, IV-B, I-A, II-B, III-C, IV-D I-D, II-C, III-A, IV-C, D

Column II Depends on nature of material Reciprocal of power of lens Brewster’s law Coherent sources

layer of oil on water show beautiful colours when illuminated by white light. Reason It happens due to the interference of light reflected from the upper surface of the thin film.

75. Assertion Corpuscular theory fails in explaining the velocities of light in air and water. Reason According to corpuscular theory, light should travel faster in denser media than in rarer media.

76. Assertion When a tiny circular obstacle is placed in the path of light from some distance, a bright spot is seen at the centre of shadow of the obstacle. Reason Destructive interference occurs at the centre of the shadow.

Wave Optics 77. Assertion Out of radio waves and microwaves, the former undergoes more diffraction. Reason Radio waves have greater compared to microwaves.

frequency

78. Assertion No diffraction is produced in sound waves near a very small opening. Reason For diffraction to take place the aperture of opening should be of the same order as wavelength of the waves.

79. Assertion In Young’s experiment, for two coherent sources,the resultant intensity given by 2 f I = 4 I 0 cos × 2 Reason Ratio of maximum and minimum intensity 2 I max ( I1 + I2 ) × = I min ( I1 - I2 )2

1117

80. Assertion The film which appears bright in reflected system will appear dark in the transmitted light and vice-versa. Reason The condition for film to appear bright or dark in reflected light are just reverse to those in the transmitted light.

81. Assertion For best contrast between maxima and minima in the interference pattern of Young’s double slit experiment, the intensity of light emerging out of the two slits should be equal. Reason The intensity of interference pattern is proportional to square of amplitude.

82. Assertion In Young’s double slit experiment, the fringes become indistinct if one of the slits is covered with cellophane paper. Reason The cellophane paper decreases the wavelength of light.

Previous Years’ Questions 83. In Young’s double slit experiment, one of the slit is

wider than other, so that amplitude of light from one slit is double of that from other slit. If, I m be the maximum intensity, the resultant intensity I when they interfere at phase difference f, is given [AIEEE 2012] by Im ( 4 + 5 cos f ) 9 fö I æ (b) m ç1 + cos2 ÷ 3 è 2ø fö I æ (c) m ç1 + 4 cos2 ÷ 5 è 2ø fö I æ (d) m ç1 + 8 cos2 ÷ 9 è 2ø (a)

86. As the beam enters in the medium, it will (a) (b) (c) (d)

87. A thin air film is formed by putting the convex surface of a pleno-convex lens over a plane glass plate. With monochromatic source of light, this film gives an interference pattern due to light reflected from the top (convex) surface and the bottom (glass plate) surface of the film. [AIEEE 2011] Statement I When light reflects from the air glass plate interface, the reflected want suffers a phase change of p.

Passage An initially parallel cylindrical beam travels in a medium of refractive index m( I ) = m 0 + m2 I , where m 0 and m2 are positive constants and I is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius.

84. The initial shape of the wavefront of the beam is

Statement II The centre of the interface pattern is dark.

88. Two coherent point sources S1 and S2 are separated by a small distance d as shown. The fringes obtained on the screen will be [JEE main 2013] Screen

[AIEEE 2010]

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

planar convex concave convex near the axis and concave near the periphery

85. The speed of light in the medium is (a) (b) (c) (d)

maximum on the axis of the beam minimum on the axis of the beam the same everywhere in the beam directly proportional to the intensity

[AIEEE 2010]

travels as a cylindrical beam diverge converge diverge near the axis and converge near the periphery

[AIEEE 2010]

d

D

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

points straight lines semi-circle concentric circles

1118 JEE Main Physics Answers Round I 1. 11. 21. 31. 41. 51. 61. 71. 81. 91.

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

2. 12. 22. 32. 42. 52. 62. 72. 82. 92.

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

3. 13. 23. 33. 43. 53. 63. 73. 83. 93.

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

4. 14. 24. 34. 44. 54. 64. 74. 84. 94.

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

5. 15. 25. 35. 45. 55. 65. 75. 85.

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

6. 16. 26. 36. 46. 56. 66. 76. 86.

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

7. 17. 27. 37. 47. 57. 67. 77. 87.

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

8. 18. 28. 38. 48. 58. 68. 78. 88.

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

9. 19. 29. 39. 49. 59. 69. 79. 89.

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

2. 12. 22. 32. 42. 52. 62. 72. 82.

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

3. 13. 23. 33. 43. 53. 63. 73. 83.

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

4. 14. 24. 34. 44. 54. 64. 74. 84.

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

5. 15. 25. 35. 45. 55. 65. 75. 85.

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

6. 16. 26. 36. 46. 56. 66. 76. 86.

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

7. 17. 27. 37. 47. 57. 67. 77. 87.

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

8. 18. 28. 38. 48. 58. 68. 78. 88.

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

9. 19. 29. 39. 49. 59. 69. 79.

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

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

10. 20. 30. 40. 50. 60. 70. 80. 90.

Round II 1. 11. 21. 31. 41. 51. 61. 71. 81.

(c) (c) (b) (b) (c) (a,c) (c) (a) (b)

10. 20. 30. 40. 50. 60. 70. 80.

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

the Guidance Round I 1. For a given time, optical path remain constant \

m1x1 = m 2x2

or1.5 ´ 2 = m 2 ´ 2.25 1.5 ´ 2 2 \ m2 = = 2.25 1.5 20 4 = = 15 3

2. According to Brewster's law, the light reflected from the top of

3.

glass slab gets polarised. The light refracted into the glass slab and the light emerging from the glass slab is only partially polarised. Therefore, when a polaroid is held in the path of emergent light at P, and rotated about an axis passing through the centre and perpendicular to plane of polaroid, the intensity of light shall go through a minimum but not zero for two orientations of the polaroid. 1 Using the relation, L µ 2 we get, r L1 r22 3 2 9 = = = L2 r12 4 2 16 \

L2 =

16 L1 9

16 7 L1 - L1 = L1 9 9 DL Increase in illumination = ´ 100 L1 DL = L2 - L1 =

Change in L,

=

4. Given,

separation = 0.28 ´ 10 -3 m.

7 ´ 100 = 78% 9

between

slits

d = 0.28

mm

Distance between screen and slit D = 1.4 m, Distance between central bright and fourth bright fringe x = 1.2 cm = 1. 2 ´ 10 - 2 m Number of fringes n = 4

Dl d 4 ´ 1.4 ´ l -2 1.2 ´ 10 = 0.28 ´ 10 - 3

For constructive interference x = n

Wavelength,

l=

1.2 ´ 10 - 2 ´ 0.28 ´ 10 - 3 4 ´ 1.4

l = 6 ´ 10 - 7 m

Wave Optics l = 600 ´ 10 - 9 m

or

10. In figure, a ray of light AB is incident from air onto glass slab

= 600 nm

[Q 1 nm = 10

-9

m]

The wavelength of light is 6 ´ 10 - 7 m.

5. Here, the width of the slit is 10 4 Å, i.e., 10000 Å. The

of width d at angle q. It is reflected partially at B and refracted at B along BC at Ðr . At C, the ray is partially reflected along CD and partially refracted (not shown). To calculate phase difference between rays reflected from B and C, we find

wavelength of (visible) sunlight varies from 4000 Å to 8000 Å. As width of slit a > l. (wavelength of light), therefore no diffraction occurs. The image seen throught the slit shall be a fine sharp slit white in colour at the centre.

A

q D E

B r

6. For refracted light (In this process wavelength and speed

d

changes but frequency remains the same) Wavelength of refracted light l ¢ =

r

l 589 ´ 10 - 9 = 1.33 m

C

= 4.42 ´ 10 - 7 m Velocity of refracted light v =

c 3 ´ 10 8 = = 2.25 ´ 10 8 m/s 1.33 m

D l d 1 and lµ m From Eqs. (i) and (ii), b=

\

bµ

…(i)

1 m

1 m

Angular fringe width, q0 = Dq (width Dy = b) b Dl 1 l q0 = = ´ = Þ D d D d q0 = 1° = p /180 rad l = 6 ´ 10

and \

d=

-7

m

l 180 = ´ 6 ´ 10 -7 q0 p

= 3.44 ´ 10 -5 m » 0.03 mm

9. As field of view is same in both cases n1b1 = n2b 2 or \

æ Dl ö æ Dl ö n1ç 1 ÷ = n2ç 2 ÷ è d ø è d ø

æn ö or l 2 = ç 1 ÷ l1 è n2 ø

æ 84 ö l 2 = ç ÷ ´ 4358 è 62 ø l 2 = 5904 Å

DT =

nd æ sin 2 q ö cç1 - 2 ÷ n ø è

1/ 2

=

1/ 2

sin 2 q ö n ´ Td æ ÷ ç1 l è n2 ø

2pDT 2p nd æ sin 2 q ö ç1 - 2 ÷ = T l è n ø

-1/ 2

-1/ 2

As reflection at C is from medium of higher refractive index, additional phase diff. of p is introduced. Hence required phase difference =

DY Dq = D

Now,

\

Phase difference = Df =

æyö Here, sin q = ç ÷ èDø So,

sin q sin q , sin r = n sin r

æ sin 2 q ö ÷ cos r = 1 - sin 2 r = ç1 n ø è

…(ii)

The refractive index of water is greater than air, therefore fringe width will decrease.

8.

Time difference, D T = time taken to travel BC in glass BC d / cosr nd = = = v c /n c cos r From Snell's law, n =

7. As we know,

b µl µ

1119

2p nd æ sin 2 q ö ç1 - 2 ÷ l è n ø

-1/ 2

+ p.

11. When two coherent light beams of intensities I1 and I2 superimpose, then maximum intensity is ( I1 + I2) 2 and minimum intensity is ( I1 - I2) 2. But when two incoherent light beams of intensities I1 and I2 superimpose, then maximum intensity is (I1 + I2) and minimum intensity is (I1 - I2). \ Imax = 5I and Imin = 3I

12. For incoherent waves, \

Imax = nI I 32 n = max = = 16 I 2

13. The distance between zeroth order maxima and second order minima is 3 b 3 y1 = + b = b = ´ 0.2 mm = 0.3 mm 2 2 2 \The distance of second maxima from point P is y = ( 4.8 + 0.3) mm = 5.1mm

1120 JEE Main Physics 14. At a point of maxima,

21. The position of 30th bright fringe, -2

Imax = 4I0 = 4 Wm

y30 =

I0 = 1 Wm-2

\

Now position shift of central fringe is 30 lD y0 = d D But we know, y 0 = (m - 1)t d 30 lD D = (m - 1)t d d

16. Path difference, Dx = S1S3 - S 2S3 = 0 2p Dx = 0 l

\

f=

\

I3 = I0 + I0 + 2 I0 + I0(cos 0° )

\

I3 = 4I0

The path difference at S 4 is xd Dx ¢ = S1S 4 - S 2S 4 = D

\ \

Þ

lD ö æ ÷ çHere, x = è 2d ø

d lD l ´ = D 2d 2 2p l f¢ = × =p l 2 =

\

\

23.

Imax = 4I + I + 2 2I1 ´ I = 9I

and

Imin = I1 + I2 - 2 2I1I2

Þ

Imin = I

b µ1/ d

I1 = I0 ´ 25% 25 I0 = I0 ´ = 100 4 The intensity of transmitted light from upper surface is 3I I I = I0 - 0 = 0 4 4 \The intensity of reflected light from lower surface is 3I 50 3I = 0 I2 = 0 ´ 4 100 8 2 Imax ( I1 + I2) \ = Imin ( I1 - I2) 2

\Phase difference = f = 90°

3ö ÷ 8ø

2

æ1 3ö ç ÷ 8ø è2

2

R = a12 + a22 + 2a1a2 cos f = a12 + a22 \Resultant intensity, I µ R 2 Þ I µ ( a12 + a22)

26. Consider the slit widths are equal, so they produces waves of equal intensity say, I¢. Resultant intensity at any point IR = 4I ¢ cos2 f , where f is the phase difference between the wave at the point observation. For maximum intensity, …(i) f = 0° Þ Imax = 4I ¢ = I If, one of slit is closed, resultant intensity at the same point will be I¢ only i. e. , I ¢ = I0

2

Imax = 2 Imin æ I 3I0 ö ç 0 ÷ 8 ø è 4

colours/wavelength/frequencies, therefore, there shall be no interference fringes.

…(ii)

Comparing Eqs. (i) and (ii), we get, I ¢ = I = 4I0 . l 2

27. For destructive interference, path difference = × 2p l × = p radian = 180°. l 2 D Distance of nth maxima, x = nl µ l d Þ l b < lg \ xblue < xgreen Therefore, phase difference =

28.

20. As light from two slits of YDSE is of different

y 2 = a2 sin wt

and

(as I1 = I2 = I)

3I0 ö ÷ 8 ø

Imax - Imin Imax + Imin

25. As, y1 = a1 cos wt = a1 sin( wt + 90° )

19. The intensity of light reflected from upper surface is

=

light is travelling in vacuum along the y-axis, therefore, the wavefront is represented by y = constant. lD As, b = d

24. Fringe visibility (V ) is given by, V =

\

æ1 ç + è2

30 l 30 ´ (6000 ´ 10 -10) = 0.5 = t (3.6 ´ 10 -5) m = 1.5

\

18. As, Imax = I1 + I2 + 2 I1I2

\

(m - 1) =

22. As, velocity of light is perpendicular to the wavefront, and

I4 = I0 + I0 + 2I0 cos p = 0 I 4I Ratio = 3 = 0 = ¥ I4 0

æ I0 ç + è 4

30 lD d

29. Distance between successive fringes = fringe width =b =

lD 8 ´ 10 -5 ´ 2 = 0.32 cm = 0.05 d

Wave Optics 30. Optical path for 1st ray = n1L1

42. As, d1 = 7λ1

Optical path for 2nd ray = n2L2 2π ∆x ∴ Phase difference, ∆φ = λ 2π = (n1L1 − n2L2) λ

31. Here, n1λ1 = n2λ 2 −5

n(7.8 × 10 ) = (n + 1) (5.2 × 10 ) n(2.6 × 10 −5) = (5.2 × 10 −5) ∴

33.

n =2 2D Here, β = , In water λ decreases. So, β also decreases, 2d i. e. , the fringe pattern shinks. λD The fringe width, β= 2d λ (2D) and β′ = d

∴

and

d 2 = 7λ 2

∴

d1 λ 1 = d2 λ 2

44. Path difference, x = ( SS1 + S1O) − ( SS 2 + S 2O). If, x = nλ , then central fringe at O will be bright. If, x = (2n − 1) λ / 2, the central fringe at O will be dark.

45. Distance between the fringes = fringe width ⇒

n = 10

nλ D We know that x11 − x1 = 2d ( xn) (2d) (0.972) × (5 × 10 −2) ∴ λ= = nD 10 × 100 = 4.86 × 10

−5

46. When the arrangement is dipped in water,

47. As, x = (2n − 1) ∴

48. As, or

source of fresh light for the slits S3 and S 4 . Therefore, there will be a regular two slit pattern of the second screen. Choice (d) is correct. λD Here, β = , Here, λ = 6000 Å = 6000 × 10 −8 cm 2d

or

2d = 1 mm = 0.1 cm

∴

β=

6000 × 10 −8 × 25 = 0.015 cm 0.1

37. Wave nature of light alone can explain the phenomenon like diffraction.

38. For maximum contrast, I1 = I2 39. As, β′ =

λ′ 5000 β= × 0.48 = 0.04 cm λ 6000

40. Hence, path difference, S1P − S 2P = odd integral multiple of

λD 2 d

λ=

2xd 2 × 10 −3 × 0.9 × 10 −3 = (2n − 1)D (2 × 2 − 1) × 1

∴

Imax ( a + b) 2 = = 9 or Imin ( a − b) 2

a+ b =3 a−b

3a − 3b = a + b 2a = 4b I1 a2 4b 2 = = 2 = 4 :1 I2 b 2 b

49. Interference fringes are bands on screen XY runnings parallel to the lengths of slits. Therefore, the locus of fringes is represented correctly by W3W4 .

50. Distance between two successive maxima is obtained as,

∴

51. As, ∴

λ=

0.14 m = 10 −2 m, 14

ν=

c 3 × 10 8 = = 3 × 1010 Hz λ 10 −2

I1 a2 9 = = I2 b 2 1 a 3 = b 1 2

λ / 2 = 11( λ / 2)

41. Let, I1 = a2 and I2 = b2 ∴

β x 3 = = x = 0.75x µ 4 /3 4

= 6 × 10 −7 m = 6 × 10 −5 cm

cm

D = 25 cm

λD 6 × 10 −7 × 2 = d 0.05 × 10 −2

= 24 × 10 −4 m = 0.24 cm

35. As there is a hole at minima point P2, the hole will act as a

and

x=β =

β′ =

xn = x11 − x1 = 9.72 mm i. e. ,

D d

43. At the centre, all colours meet in phase, hence central fringe

β′ = 4β

34. Here, 2d = 0.5mm = 5 × 10 −2cm and D = 100 cm

36.

D d

is white.

−5

32.

1121

Imax = ( a + b) 2 = a2 + b 2 + 2ab = I1 + I2 + 2 I1 I2 = ( I1 + I2) 2

⇒

Imax ( a + b) 2  3 + 1 = =  = 4 :1 Imin ( a − b) 2  3 − 1

52. Path difference of 3rd bright fringe from central fringe = 3λ, ∴Corresponding phase difference = 3(2π ) = 6π rad.

1122 JEE Main Physics 53. When one slit is closed, amplitude becomes half and intensity becomes 1/4th

or

54.

= I + 4I + 2 I ´ 4I cos p / 2 = 5I Resultant intensity Iq 2 = I1 + I2 + 2 I1I2 cos q2

lD As, b = d

= I + 4I + 2 I ´ 4I cos p lD ¢ b¢ = d l(D - D ¢ ) b - b¢ = d l ´ 5 ´ 10 -2 3 ´ 10 -5 = 10 -3

and \

= 5I - 4I = I \

Iq1 - Iq 2 = 5I - I = 4I

63. According to question, nlr = (n + 1) lb

3 ´ 10 -5 = 6 ´ 10 -7 m = 6000 Å 50 lD l l As, b = = = d d /D q or

55.

Resultant intensity Iq1 = I1 + I2 + 2 I1I2 cos q1

1 I 4 I = 4I0

I0 =

i. e. ,

62. Here, I1 = I, I2 = 4I, q1 = p / 2, q2 = p

l=

Þ

n + 1 l r 600 4 = = = n l b 480 5

Þ

1 4 = -1 Þ n = 4 n 5

64. From, b =

56. When path difference is l and Imax = 4I = K When path difference is \

l , phase difference, f = p / 2 4

IR = I1 + I2 + 2 I1I2 cos f K = I + I = 2I = 2

= 6 ´ 10 -7 m = 6000 Å

65. When white light is used instead of monochromatic light, the central bright fringe becomes white, while others are coloured. Hence, distinction is made.

66. As, b =

57. For a path difference (m - 1)t , the shift is D x = (m - 1)t d

lD 6000 ´ 10 -10 ´ 2 = d 4 ´ 10 -3

= 0.3 ´ 10 -3 m = 0.3 mm

67. There is no interference fringes.

58. Oil floating on water appears coloured only when thickness of oil layer » wavelength of light »1000 Å.

59. Amplitude A1 and A2 are added as vectors. Angle between these vectors is the phase difference (b1 - b 2) between them \

lD bd 0.06 ´ 10 -2 ´ 10 -3 = Þl = D 1 d

68. The nature of the wavefronts emerging from the final image is spherical.

69. From diffraction at a single slit, size of aperture, a =

R = A12 + A22 + 2A1A2 cos(b1 - b 2) or,

60. Here, n1 = 12, l1 = 600 nm

a=

\

n1l1 = n2l 2 n l 12 ´ 600 = 18 n2 = 1 1 = 400 l2 1 5

61. Here, d = 0.1mm = 10 -4 , D = 20 cm = m l = 5460 Å = 5.46 ´ 10

-7

m

Angular position of first dark fringe is, x l 5.46 ´ 10 -7 q= = = D 2d 2 ´ 10 -4 = 2.73 ´ 10 -3 rad = 2.73 ´ 10 -3 ´

180° p

~0.16 Å = 0.156°-

3141.59 ´ 10 -10 sin 1°

= 18 ´ 10 -6 m = 18 mm

n2 = ?, l 2 = 400 nm As,

l sin q

70. Given, aperture a = 4 mm = 4 ´ 10 - 3 m Wavelength

l = 400 nm = 400 ´ 10 - 9 m

Ray optics is good approximation upto a distance equal to Fresnel’s distance (ZF ). ZF =

a2 4 ´ 10 - 3 ´ 4 ´ 10 - 3 = l 400 ´ 10 - 9 ZF = 40 m

71. The amplitude of the waves are a1 = 10 mm, a2 = 4 mm and a3 = 7 mm and phase difference between Ist and IInd wave is p p and that between 2nd and 3rd wave is × Then, phase 2 2 difference between 1st and 3rd is p. Combining Ist with 3rd, their resultant amplitude is given by A12 = a12 + a22 + 2a1a2 cos f

Wave Optics A1 = 10 2 + 7 2 + 2 ´ 10 ´ 7 cos p

or

= 100 + 49 - 140 = 9 = 3 mm Now, combining this with 2nd wave, we have the resultant amplitude p A2 = A 12 + a 22 + 2A1a2 cos 2 A = 3 2 + 4 2 + 2 ´ 3 ´ 4 ´ cos 90°

or

= 9 + 16 = 25 = 5 mm

72. Let, it takes a time t sec for astronaut to acquire a velocity of 1 ms -1. Then, energy of photon, s = 10t

10t Momentum = = 80 ´ 1 c 80 ´ 1 ´ 3 ´ 10 8 t= 10 = 2.4 ´ 10 9 s

\ Þ

77. We have, mv = 1and m a = 1.003 lv m a = = 1.0003 l a mv

\

x = lvn = l a (n + 1) n + 1 lv Þ = = 1.0003 n la 1 1 1 + = 1.0003 or, = 0.0003 Þ n n 1 10 4 Þ n= = 0.0003 3 4 10 = 2 mm \ x = l an = 6000 ´ 10 -7 mm ´ 3 Now

78. Distance between first and sixth minima x=

d = 2.5 ´ 10 -3 m = 2.5 mm

n ´ 650 ´ 10 -19D (n + 1) ´ 520 ´ 10 -19 = ´D d d n=4

Þ

74. The intensity at a point on screen is given by I = 4I0 cos2( f/2) When f is the phase difference. In this problem f arises (i) p due to initial phase difference of and (ii) due to path 4 difference for the observation point situated at q = 30°. Thus, p 2p p 2p l f= + - (sin 30° ) (d sin q) = + l 4 4 l 4 p p p = + = 4 4 2 f p æpö Hence, = and I = 4I0 cos2ç ÷ = 2I0 è 4ø 2 4

75. Position of nth maxima from central maxima is given by xn =

8l1D d1 6l D and for 6th maxima x6 = 2 d2 x6 = x8 d1 n1l1 4 æ l1 ö = = ç ÷ d 2 n2l 2 3 è l 2 ø

76. The number of fringes shifting is decided by the extra path difference produced by introducing the glass plate. The extra path difference is, (m - 1)t = nl or (1.5 - 1) ´ 0.1 ´ 10 -3 = n ´ 500 ´ 10 -9 Þ

l=

Þ

n = 100

a sin q 5 sin 30° = = 2.5 cm n 1

80. Here, l = 6250 Å = 6250 ´ 10 -10 m a = 2 ´ 10 -2 cm = 2 ´ 10 -4 m D = 50 cm = 0.5 m 2lD \Width of central maxima = a 2 ´ 6250 ´ 10 -10 ´ 0.5 = 2 ´ 10 -4 = 312.5 ´ 10 -3 cm

81. In interference, we use two sources while in diffraction, we use light from two points of the same wavefront.

82. From, a sin q = nl, a

For 8th maxima x8 =

Þ

79. As, a sin q = nl

nl D d

Now,

5lD d

for 1st maxima, nlD 1 ´ 500 ´ 10 -10 ´ 2 \ d= = x 0.5 ´ 10 -3

73. As, 4b = (n + 1)b2 Þ

1123

x nlD 1 ´ 6000 ´ 10 -10 ´ 2 = nl or, a = = D x 4 ´ 10 -3 a = 3 ´ 10 -4 = 0.3 mm

\

83. As, a sin q = nl Þ or

ax (Hence, focal length is used as D) = 3l D ax 0.3 ´ 10 -3 ´ 5 ´ 10 -3 l= = 3D 3 ´1 = 5 ´ 10 7 m = 5000 Å

84. For, the light to be invisible in vacuum, m of medium must be equal to m of vacuum, which is 1.

85. Angular spread on either side is, q =

l 1 = rad a 5

1124 JEE Main Physics 86. Sound waves cannot be polarised as they are longitudinal. Light waves can be polarised as they are transverse.

87. Intensity of light from C 2 = I0 On rotating the crystal through 60°, I = I0 cos2 60°

(Malus law)

2

æ 1ö = I0 ç ÷ = I0 / 4 è2ø mg = 1.5

88. Given,

Let ip be the Brewster’s angle.

Distance between screen and slit D = 1m Wavelength of light l = 600 nm = 600 ´10 - 9 m 4 Refractive index of water mw = 3 Using the formula of angular width l q= D l¢ and q¢ = D l where, l¢ = m q¢ l ¢ l = = q l ml

Refractive index, m = tan ip tan ip = 1.5 ip = tan - 1 (1.5)

q 0.2 ´ 3 = = 0.15 m 4

\ Intensity of polarised light from first nicol prism 1 I = 0 = ´ 2 a2 = a2 2 2

94. Given, wavelength of Ha , l = 6563 Å = 6563 ´ 10 - 10 m Red- shift Dl = 15 Å

90. Here, q = 9.9° , l = 20 cm = 0.2 m and s = 66°

Since, the star is found to be red-shifted, hence star is receding away from earth and doppler’s shift is negative. l Dl = - v c 15 ´ 3 ´ 10 8 Dl. c v==6563 l v = - 6.86 ´ 10 5 m/s

q 9.9 c= = = 0.075 g cc-1= 75 gL-1. ls 2 ´ 66

91. Along the optic axis, m 0 = m e la and m = tan q m l l m = a = l a cot q tan q

92. We know, lm =

Negative sign shows that the star is receding (away) from earth.

93. Given, angular width q = 0.2°

Round II 1. As,

Imax Imin

\

æ ç ç =ç çç è

2

2 ö I1 ö æ 9 + 1÷ ç + 1÷ I2 ÷ ÷ =4 ç 1 ÷ =ç 9 ÷ 1 I1 - 1÷÷ - 1÷ ç ø è 1 I2 ø

Imax : Imin = 4 : 1

4. As, I æç = è

Imin = ( I1 - I2)

For maximum intensity, q = 0° Þ

2

= ( I - 4I ) 2 = I

P ö 1 µ A2 Þ A µ 2÷ ø r 4 pr

5. The resultant intensity, IR = I1 + I2 + 2 I1I2 cos q

2. We have, Imax = ( I1 + I2) 2 = ( I + 4I) 2 = 9I and

4ö æ çQ m = ÷ è 3ø

Thus, the angular fringe width is 0.15° as the apparatus is immersed in water.

89. The intensity of plane polarised light = 2a2.

\

q¢ =

or

ip = 56° 18 ¢

As,

…(ii)

Dividing Eq.(ii) from Eq.(i), we get

From the Brewster’s law, or

…(i)

6. As,

The possible beams are 9I and I.

3. The refractive index of air is slightly more than 1. When

Þ

chamber is evacuated, refractive index decreases and hence, the wavelength increases and fringe width also increases.

Þ

IR = I1 + I2 + 2 I1I2 = ( I1 + I2) 2 ö æ a1 ç + 1÷ ø è a2

2 2

æ a + a2 ö Imax 36 =ç 1 = ÷ = Imin æ a1 ö 2 è a1 - a2 ø 1 ç - 1÷ ø èa a1 + a2 6 = a1 - a2 1 6a1 - 6a2 = a1 + a2

Wave Optics Þ

6a1 - a1 = 6a2 + a2

Þ

5a1 = 7a2 a1 7 = a2 5

Þ

12. As sound waves are longitudinal, therefore, polarization of sound waves is not possible.

13. If, I0 is intensity of unpolarized light, I0 2 From 2nd Nicol prism, I2 = I1 cos2(90° - 60° ) then from 1st Nicol prism, I1 =

7. The number of oscillations in coherence light l 0.024 = = 40677.9 = 4.068 ´ 10 4 l 5900 ´ 10 -10

8. As, Þ

2

=

1 5000 b1 l1 or = = b 2 l 2 b 2 6000

Þ

6000 = 1.2 mm 5000

b2 =

9. Angular momentum

L=

3 I0 æ 3 ö ç ÷ = I0 2è 2 ø 8

I2 = 37.5 I0 nl (m - 1) l nl = 2 l t= 2 (m - 1)

14. From, (m - 1) t = nl, we get, t =

nh 2p

when,

U = nhn then

w = 2pn Þ

n=

w 2p

\

U=

nhw 2p

or

U = Lw L=

15. Given, l = 600 nm = 6 ´ 10 -7 m, a = 1mm = 10 -3m, D = 2m Distance between the first dark fringes on either side of central bright fringe = width of central maximum 2lD 2 ´ 6 ´ 10 -7 ´ 2 = a 10 -3 -4 = 24 ´ 10 m = 2.4 mm

U w

=

10. Given, wavelength of light = l

16. Angular dispersion of central maximum = angular dispersion

When the path difference is l ( x), the phase difference

of 1st minimum ( = 2q) l 2 ´ 10 -3 1 From, = sin q = = a 4 ´ 10 -3 2

2p 2p f= ×x = × l = 2p l l Resultant intensity IR = I1 + I2 + 2 I1I2 cos f IR = I + I + 2 II cos 2p = 2I + 2I = 4I = K

(Given) …(i) (QI1 = I2 = I)

When path difference is l/3, then Phase difference =

2p 1ù é êëQ cos 3 = - 2 úû [From Eq. (i)]

Thus, the intensity of light at a point of path difference K . 4

11. As, x = n1b1 = n2b2 = n1l1 = n2l2 n2 =

\

q = 30° 2q = 2 ´ 30° = 60°

17. As the two bright fringes coincide \

nl1 = (n + 1) l 2 n + 1 l1 7500 5 = = = n l 2 6000 4

2p l 2p × = l 3 3

In this condition, resultant intensity 2p æ 1ö IR¢ = I + I + 2 II cos = 2I + 2I ç - ÷ è 2ø 3 K IR¢ = I = 4

\

1125

n1l1 60 ´ 4000 = = 40 6000 l2

l is 3

1+

1 5 = Þn = 4 n 4

18. Given, wavelength of light l = 600 nm = 600 ´ 10 - 9 m Angular width of fringe q = 0.1° = Using the formula q =

0.1 p rad 180

l d

Spacing between the slits d =

l 600 ´ 10 - 9 ´ 180 = 0.1 ´ p q

d = 3.44 ´ 10 - 4 m Thus, the spacing between the two slits is 3.44 ´ 10 - 4 m.

1126 JEE Main Physics 27. As, x = (n + 1) lb = nlr

19. If, thin film appears dark then, 2mt cos r = nl

for normal incident

When

r = 0°

Then,

2mt = nl nl t= 2m

Þ

t min =

Þ

Þ Þ

l 5890 ´ 10 -10 = = 2.945 ´ 10 -7m 2m 2 ´1

28. Here, d = 1mm = 10 -3 m, l = 6.5 ´ 10 -7 m

20. In case of destructive interference (minima) phase difference is odd multiple of p.

21.

Now,

( a + b) l As, b = , 2a(m - 1) a where, a = distance between source and biprism = 0.3 m

and

b = distance between biprism and screen = 0.7 m a = angle of prism = 1° , m = 1.5, l = 6000 ´ 10 -10 m b=

Hence,

n + 1 l r 7.8 ´ 10 -5 3 = = = l b 5.2 ´ 10 -5 2 n 1 3 1+ = b 2 n =2

Þ

(0.3 + 0.7) ´ 6 ´ 10 -7 2 ´ 0.3 (1.5 - 1) ´ (1° ´ p / 180)

\

D = 1m D 1 x5 = nl = 5 ´ 6.5 ´ 10 -7 ´ -3 d 10 = 32.5 ´ 10 -4 m lD x3 = (2n - 1) 2d (2 ´ 3 - 1) ´ 6.5 ´ 10 -7 = 2 ´ 10 -3 = 16.25 ´ 10 -4 m x5 - x3 = (32.5 - 16.25)10 -4 m = 16.25 ´ 10 -4 m = 1.63 mm

= 1.14 ´ 10 -4 m

29. From, x = nl

= 0.011 cm

22. Here, Dl = 0.5 nm = 0.5 ´ 10 -9m

Þ

v = 300 kms-1 = 3000 ´ 10 3ms-1 As, Þ

Dl v = l c

v 23. Doppler shift, l¢ = l æç1 - ö÷

When f = 90°

cø

è

æ vö 5400 Å = 6200 ç1 - ÷ è cø é 54 ù v = ê1 ë 62 úû

Þ

IR ¢ = I + I + 2 II cos 90° = 2I

Now,

IR 4I = = 2 :1 IR ¢ 2I

polarised light from 1st polaroid = I0 / 2. On rotating through 45°, intensity of light from 2nd polaroid,

lD d

2

æ 4.5 ´ 10 6 ö v ÷ = 5802 Å 25. As, l¢ = l æç1 - ö÷ = 5890 ç1 5

Þl =

then,

32. If, I0 is intensity of unpolarised light, then intensity of

= 3.9 ´ 10 7 m/s

24. Fringe width, b =

spot will be white, while the surrounding fringes will be coloured.

When, f = 0° then IR = I + I + 2 II cos 0° = 4I

= 5000 Å

26. From, b =

\

31. From, IR = I1 + I2 + 2 I1I2 cos f

= 5 ´ 10 -7 cm

è

D D and d 2 = 7l 2 d d d1 l1 = d 2 l2

d1 = 7l1

30. When white light is used in a biprism experiment, central

Dl ´ c 0.5 ´ 10 -9 ´ 3 ´ 10 8 l= = v 300 ´ 10 3

i. e. ,

D d

cø

è

3 ´ 10

ø

lD , d

I æ 1 ö I0 æI ö I = ç 0 ÷ (cos 45° ) 2 = 0 ç ÷ = è2ø ø è 4 2 2 Þ

I = 25% of I0

33. When sources are coherent, then IR = I1 + I2 + 2 I1I2 cos f -2

b.d 0.3 ´ 10 ´ 2 ´ 10 = D 1

-3

= 6 ´ 10 -7 m = 6000 Å

At middle point of the screen, f = 0° then IR = I + I + 2 II cos 0° = 4I

Wave Optics When sources are incoherent, then,

Now, Þ

34. Given, y1 = 4 sin wt y 2 = 3 sin( wt + p / 3)

and

a = 4, b = 3, f = p / 3

\

R = a2 + b 2 + 2ab cos f

2

or,

(5000 ´ 10 -10) ´ 1.0 = d

-7

5 ´ 10 = 10 -4 m = 0.1mm 5 ´ 10 -3

36. Fringe width µ wavelength of light. Therefore, fringe will become narrower.

37.

æ b2 ö Distance, S 2 P = (d 2 + b 2)1/ 2 = d ç1 + 2 ÷ d ø è

and

1/ 2

a 10 = = 1: 1 b 10

\

42. The linear width of central principal maximum =

43. Let, the wavelength of monochromatic light in glass be lg cm, and in water be lw cm. \ Number of waves in 8 cm of glass =

æ b2 b2 ö ÷ =d + = d ç1 + 2 2d 2d ø è

waves in 10 cm of glass =

S1P = d

Þ

x=d +

b2 b2 -d = 2d 2d

For missing wavelengths, (2n - 1)

b2 l =x= 2 2d

b2 , d 2b2 n = 2, l = 3d

Now, \

For

Energy transmitted/revolution, æI ö = (IA)T = ç 0 A÷T è2 ø f0T 10 -3 ´ 0.2 = = 10 -4 J 2 2 l D 9 lD For 5th dark fringe, x1 = (2n - 1) = 2d 2d =

mg =

c c and mw = ng vw

mg

nw nlw 5 = = ng nlg 4

mw

=

8 , and number of lg

5 5 4 5 mw = ´ = 4 4 3 3

44. Distance between the slits, d = d1d 2 = 4.05 ´ 10 -3 ´ 2.90 ´ 10 -3

38. Here, w = 31.4 rads -1 \Time period of revolution, 2p 2 ´ 3.14 T= = = 0.2 s w 31.4

8 × lw 8 10 l 10 5 or, w = = = lg lw lg 8 4

mg =

n = 1, l =

For

2lD d

If, it is equal to width of slit (d), then 2lD d2 = d or, D = 2l d

\Path difference = S 2 P - S1P

39.

2

41. Here, a = 10 , b = 52 + (5 3) 2 = 10

5 ´ 10 -3 d=

2

4 4 æ 2 ö æ 2 ö æ 2 ö 1: ç ÷ : ç ÷ : ç ÷ = 1: 2 : è 3p ø è 5p ø è 7p ø 9p 25p 2

= 37 » 6 lD From, b = d Þ

2.5l (m - 1)

40. The ratio of intensities of successive maxima is

2

= 4 + 3 + 2 ´ 4 ´ 3 cos p / 3

35.

t=

or,

Here,

2

D 7 lD = d d D x2 - x1 = (m - 1) t d lD é D 9ù 7 - ú = (m - 1) t d êë d 2û

For 7th bright fringe, x2 = nl

IR ¢ = I1 + I2 = I + I = 2I IR 4I = =2 IR ¢ 2I

\

1127

= 3.427 ´ 10 -3 m

45. The resultant intensity at any point P is æ fö I = 4I0 cos2ç ÷ è2ø I0 = 4I0 cos2 f / 2

\ or \

cos

f 1 = 2 2 f p 2p = or f = 2 3 3

If, Dx is the corresponding value of path difference at P , then

1128 JEE Main Physics 2p ( Dx) l 2p 2p = Dx 3 l xd Dx = D 1 1 xd = 3 l D

51. Path difference between the rays reaching infront of slit S1 is

f=

Þ As, \ or

x=

S 2P - S1P = ( b 2 + d 2)1/ 2 - d S1

b

l 6 ´ 10 -7 = = 2 ´ 10 -3 m 3d / D 3 ´ 10 -4

S2

This is the distance of point P from central maximum.

46. As, Imax = I = I1 + I2 + 2 I1I2 When width of each slit is doubled, intensity from each slit becomes twice i. e. ,

æ b2 ö d ç1 + 2 ÷ d ø è

Þ

I ¢1 = 2I1 \

I ¢max = I ¢ = I ¢1 + I ¢2 = 2 I ¢1 + I ¢2

Þ

47. For maxima on the screen,

n = 1, 2........ , l =

For,

2l sin q = nl

The maximum value of sin q = + 1 \ n =2

48. The intensity of principal maximum in the single slit

æ a + a2 ö Imax =9Þç 1 ÷ =9 Imin è a1 - a2 ø a1 + a2 =3 a1 - a2 a1 3 + 1 a = Þ 1 =2 a2 3 - 1 a2

Þ Þ

diffraction pattern does not depend upon the slit-width.

49. For total internal reflection, i > C Therefore, C < 45°

or,

1 sin C 4 m> sin 45° m> 2 m=

The possible values of m are 1.5 and 1.6.

I1 : I2 = 4 : 1

Therefore,

53. As width of pinhole is 103 A =1000 Å and wavelength of sunlight ranges from 4000 Å to 8000 Å, therefore, sunlight is diffracted on passing through the pinhole. The image of the pinhole seen on the screen shall be different from a geometrical image. Infact, the image of pinhole will consist of a diffused coloured region around a sharp central white spot. c v

54. For microwave, l = =

3 ´ 10 8 = 300 m 10 6

50. From the pattern of intensity distribution seen in the central portion, we find that (i) as intensity of successive maxima is the same, S1 and S 2 have the same intensity. (ii) as width of successive maxima appears to increase slightly, S1, S 2 must have a constant phase difference. (iii) as minimum intensity is zero, S1, S 2 must have the same wavelength.

b2 b2 , .... so on d 3d

2

52. As,

\ Eq. (i) is satisfied by -2, - 1, 0 , 1, 2.

\

(By first approximately)

...(i)

2 sin q = n

As

(2n - 1)l 2

b 2 (2n - 1)l = 2d 2 b2 l= (2n - 1)d

Þ

= 2(I1 + I2 + 2 I1 + I2) = 2I

\

-d =

(Binomial expansion)

= 2I1 + 2I2 + 2 2I1 ´ 2I2

Given,

1/ 2

ö æ (2n - 1)l b2 d ç1 + + K÷ - d = 2 2d 2 ø è

Þ

I ¢2 = 2I2

d sin q = nl d = slit-width = 2l

d

For distructive interference at P (2n - 1)l S 2P - S1P = 2 (2n - 1) l 2 2 1/ 2 i. e. , (b + d ) - d = 2

x = 2 mm

and

P

P Y

S1 θ

d S2

θ ∆x D

Wave Optics Dx = d sin q 2p Phase difference, f = (Path difference) l 2p 2p = (d sin q) = (150 sin q) = p sin q 300 l As,

\

61. Points c and d are on same wavefront. i. e. , Also

fc = fd fe = ff

Thus,

fd - ff = fc - fe

62.

IR = I1 + I2 + 2 I1I2 cos f

Medium-1 Medium-2

Here, I1 = I2 and f = p sin q æ p sin q ö IR = 2I1[1 + cos( p sin q)] = 4I1 cos2ç ÷ è 2 ø

\

æ p sin q ö IR will be maximum when cos2ç ÷ =1 è 2 ø \

(IR) max = 4I1 = I0 2æp

I = I0 cos ç è

Hence,

sin q ö ÷ 2 ø

63. The resultant amplitude, R = a12 + a22 + 2a1a2 cos p / 2 = a12 + a22

64. In the interference pattern, lmax ( a + b) 2 (1 + 2) 2 = 9 :1 = = lmin ( a - b) 2 (1 - 2) 2

then I = I0 cos2 0° = I0

If, q = 0 ,

If, q = 30° , then I = I0 cos2( p / 4) = I0 / 2

2

65. As,

If, q = 90° , then I = I0 cos2( p / 2) = 0

66. As,

55. Given, la = 6000 Å, \

na =

In water,

3 ´ 10 8 c = 5 ´ 10 14 Hz = l a 6000 ´ 10 -10 lm =

l a 6000 Å . = m 15 .

n m = n a = 5 ´ 10 14 Hz lD d bd 4 ´ 10 -3 ´ 0.1 ´ 10 -3 = 1m D= = l 4 ´ 10 -7

56. From, b = \

1129

57. As size of hole (a) is increased, width of central maximum of æ 2 f lö diffraction pattern of hole ç = ÷ decreases. As the same è a ø

amount of light emergy is now distributed over a smaller area, the intensity increases. lmax ( a + b) 2 16 = = 1 lmin ( a - b) 2 a+ b 4 \ = a-b 1

58. As,

4a - 4b = a + b or

3a = 5b \

a 5 = b 3

59. For light diverging from a point source, the wavefront is (diverging) spherical wavefront. The intensity varies inversely as the area of the wavefront ( = 4pr 2) i.e., intensity decreases in proportion to the distance squared.

60. Obviously, medium-2 is denser because ray bends towards normal.

\

w1 a2 æ 3 ö = = ç ÷ = 9 :1 w2 b 2 è 1 ø l1 w1 1 a2 = = = l2 w2 4 b 2 a = 1: 2 b

67. Two independent sources cannot be coherent . Therefore, no interference pattern would be observed.

70. When the medium between plane-convex lens and plane glass is rarer than the medium of lens and glass, the central spot of Newton's ring is dark. The darkness of central spot is due to the phase change of p which is introduced between the rays reflected from denser to rarer and rarer to denser medium.

71. For diffraction to occur, the size of an obstacle/aperture is comparable to the wavelength of light wave. The order of wavelength of light wave is10 -7m, so diffraction occurs.

72. When we seen the painting which is painted by a myriad of small colour dots near our eyes, scientillating colour of dots are visible due to diffraction of light and when we go away from the painting our eyes blend the dots and we see the different colours. This is due to the change is angular separation of adjacent dots as the distance of the object changes.

73. In Young's experiment fringe width for dark and white fringes are same while in the same experiment, when a white light as a source is used, the central fringe is white around which few coloured fringes are observed on either side.

74. The beautiful colours are seen on account of interference of light reflected from the upper and the lower surfaces of the thin films. Since, condition for constructive and destructive interference depends upon the wavelength of light therefore, coloured interference fringes are observed.

1130 JEE Main Physics Imin = ( Ia - Ib ) 2 = 0 , or absolute dark.

75. According to Newton's to corpuscular theory of light, the light should travel faster in denser media than in rarer media. It is contrary to present theory of light which explains the light travels faster in air (rarer) and in water (denser).

\ It provides a better contrast.

82. When one of slits is covered with cellophane paper, the intensity of light emerging from the slit is decreased (because this medium is translucent). Now, the two interfering beam have different intensities or amplitudes. Hence, intensity at minima will not be zero and fringes will become indistinct.

76. The waves diffracted from the edges of circular obstacle, placed in the path of light, interfere constructively at the centre of the shadow resulting in the formation of a bright spot.

77. The waves which consist longer wavelength have more diffraction. Since radio waves have greater wavelength than microwaves, hence radio waves undergo more diffraction than microwaves.

83. Here, a1 = 2a2 Þ I1 = 4I2 = 4I0 As, Now,

78. The diffraction of sound is only possible when the size of

Therefore,

l1 = l2 = l0

I = I1 + I2 + 2 I1I2 cos f

Im [from Eq. (i)] (5 + 4 cos f) 9 I I æ fö = m [1 + 4(1 + cos f)] = m ç1 + 8 cos2 ÷ 9 9 è 2ø = 5I0 + 4I0 cos f =

79. For two coherent sources, Putting,

...(i)

= 4I0 + I0 + 2 4I0I0 cos f

opening should be of the same order as its wavelength and the wavelength of sound is of the order of 1.0 m, hence, for a very small opening no diffraction is produced in sound waves. l = l1 + l2 + l1 l2 cosf

Imax = ( I1 + I2) 2 = (3 I2) 2 = 9I2 = 9I0

84. Parallel cylindrical beam gives planar wavefront.

l = l0 + l0 + 2 l0 ´ l0 cos f

On simplifying the above expression, we get l = 2l0(1 + cos f) f ö æ = 2l0 ç1 + 2 cos2 - 1÷ è 2 ø = 2l0 ´ 2 cos2 Þ Also, and \

I = 4l0 cos2

f 2

f 2

86.

lmax = ( l1 + l2) lmin = ( l1 - l2)

2

x

Axis

2

2 lmax ( l1 + l2) = lmin ( l1 - l2) 2

80. For reflected system of the film, the maxima or constructive (2n - 1) l , while the maxima for 2 transmitted system of film is given by equation 2mt cos r = nl

interference is 2mt cos r =

where t is thickness of the film and r is angle of reflection. From these two equations, we can see that condition for maxima in reflected system and transmitted system are just opposite.

81. When intensity of light emerging from two slits is equal, the intensity at minima,

c c m v Since I is decreasing so m also decreasing and hence v increases.

85. As, m = Þ v =

Converge when it enter in the medium, when light is moving and as it enters the medium than along the axis, velocity is decreasing so as we move away from the centre i. e. , x in figure, the wave covers smaller distance and hence shape is convex.

87. Statement 1 When light reflects from denser used (glass) a phase shift of p is generated.

Statement 2 Central maxima or minima depends on thickness of the lens.

88. It will concentric circle.

Dual Nature of Radiation and 25 Matter JEE Main MILESTONE < <
W1 \

( f0 )2 > ( f0 )1

as

W = hf0

Here, f0 = threshold frequency.

Graph between V 0 and f Let us now plot a graph between the stopping potential V0 and the incident frequency f. The equation between them is, eV0 = hf - W or

2 (Slope)1 = (Slope)2 =

Let us plot a graph between maximum kinetic energy Kmax of photoelectrons and frequency f of incident light. The equation between Kmax and f is, Kmax

1

æ hö æW ö V0 = ç ÷ f - ç ÷ è eø è eø

Again comparing with y = mx + c, the graph between V0 and f is a straight line with positive slope constant) and negative intercept

h (a universal e

W (which depends on the e

metal). The corresponding graph is shown in the figure.

\

E (in eV ) =

12375 12375 = = 34.4eV l (in Å ) 2800

(a) Kmax = E - W = (4.4 – 2.3) eV = 2.1 eV (b) Kmax = eV0 \ 2.1 eV = eV0 or V0 = 2.1 V

Hertz and Lenard’s Observations The phenomena of photoelectric emission was discovered in 1887 by Heinrich Hertz (1857-1894), during his electromagnetic wave experiments. In his experimental investigation on the production of electromagnetic waves by means of spark discharge. Hertz observed that high voltage sparks across the detector loop were enhanced when the emitter plate was illuminated by ultraviolet light from an arc lamp. Wilhelm Hallwachs and Phillip Lenard investigated the phenomenon of photoelectric effect in detail during 1886-1902. Lenard observed that when ultraviolet radiations were allowed to fall on the emitter plate of an evacuated glass tube enclosing two electrodes (metal plates), current flows in the circuit. As soon as the ultraviolet radiations were stopped the current flow also stopped. These observations indicate that when ultraviolet radiations fall on the emitter plate, electrons are ejected from it which are attracted towards the positive collector plate by the electric field. The electrons flow through the evacuated glass tube, resulting in the current flow. Thus light falling on the surface of the emitter causes current in the external circuit. Hallwachs and Lenard studied how this photocurrent varied with collector plate potential and with frequency of incident light.

Dual Nature of Radiation and Matter Note • The electrons will be emitted from metal surface only, if the

frequency of incident light is greater than threshold frequency n0 . • The electrons are emitted instantaneously. The interaction between

photons and electrons is one to one. So, weak incident light very few photons arrive per unit time, but each one has enough energy to eject an electron instantaneously. • The work function f and threshold frequency n0 varies from metal to metal. • The frequency n and wavelength l are related as v = c /l. The energy hc of photon = hn = × l • For a given intensity I = nhn greater than frequency lesser will be number of photons. Hence, lesser will be photoelectric current i .e ., 1 photocurrent µ µ l. n

Sample Problem 2 In the above instance, what will be the maximum kinetic energy of the emitted photoelectrons? (a) 2.5 eV

(b) 1 eV

(c) 1.51 eV

(d) 3.51 eV

Interpret (c) Incident wavelength, l = 2600 Å \ incident frequency n= Then,

c 3 ´ 10 8 Hz = l 2600 ´ 10 -10

25.4 Matter Waves and Wave Nature of Particle In case of light some phenomenon like diffraction and interference can be explained on the basis of its wave character. However, the certain other phenomenon such as black body radiation and photoelectric effect can be explained only on the basis of its particle nature. Thus light is said to have a dual character. Such studies on light wave were made by Einstein in 1905. Louis-de-Broglie, in 1942 extended the idea of photons to material particles such as electron and he proposed that matter also has a dual character as wave and as particle.

de-Broglie Relation According to de-Broglie, a wave is associated with energy moving particle. These waves are called de-Broglie waves or matter waves. According to quantum theory, energy of photon E = hn

KEmax = hn - f0 6.63 ´ 10 -34 ´ 3 ´ 10 8 hn = 2600 ´ 10 -10

From Eqs. (i) and (ii), we get, hn = mc2

KEmax = hn - f0 = 4.78 eV - 3.27 eV = 1.51 eV (2.42 ´ 10 -19 J)

Sample Problem 3 In the above instance, what will be the maximum velocity of the photoelectrons? (Mass of the electron = 9.11 ´ 10 -31 kg) (b) 1.5 ´ ms-1 6

(c) 0.7289 ´ 10 ms

-1

6

(d) 0.56 ´ 10 ms

-1

1 2

2 Interpret (c) As, KEmax = mv max

\ As, v max =

2KEmax 2 ´ 2.42 ´ 10 -19 = m 9.11 ´ 10 -31 = 0.7289 ´ 10 6 ms-1

Check Point 1 1. Do X-rays show phenomenon of photoelectric effect? 2. It is harder to remove a free electron from copper than from sodium? Which metal has greater work function? Which has higher threshold wavelength?

3. Green light ejects electrons from a certain photosensitive surface, yellow light does not. Will (a) red light (a) violet light eject photoelectrons from the same substance?

4. A photon and an electron have same wavelength. Which particle is moving faster?

…(i)

If mass of the photon is taken as m, then as per Einstein’s equation …(ii) E = mc2

= 7.65 ´ 10 -19 = 4.78 eV

(a) 1 m ´ 6 ms-1

1135

h

c = mc2, l

where, l = wavelength of photon h l= mc de-Broglie asserted that the above equation is completely a general function and applies to photon as well as all other moving particles. So,

l=

h h = mv 2 mE

where, m is mass of particle and v is its velocity. (i) de-Broglie wavelength associated with charged particle h h h l= = = p 2mE 2mqV (ii) de-Broglie wavelength of a gas molecule h l= 3mkT where, T = absolute temperature k = Boltzmann’s constant = 1.38 ´ 10-23 J/K (iii) Ratio of wavelength of photon and electron The wavelength of photon of energy E is given by

1136 JEE Main Physics hc while the wavelength of an electron of kinetic E h energy K is given by l c = × Therefore for same 2mK energy, the ratio lp =

c 2mc 2 K = 2mK = le E E2

lp

l = 3800 Å. Ultraviolet light of l = 2600 Å is incident on silver surface. What will be the value of work function? (b) 6.5 ´ 10 -18 J

(c) 6.5 ´ 10 -19 J

(d) 5.5 ´ 10 -10 J

or

W = hn 0 =

m = 1. 675 ´ 10 -17 kg

Sample Problem 7 The de-Broglie wavelength associated with an electron moving with a speed of 5.4 ´ 106 m/ s is l e and for a ball of mass 150 g travelling at 30 m/s is l b , then which of the following relations between l e and l b is true? (a) l e < l b

(b) l b < l e

l=

hc 6.63 ´ 10 -34 ´ 3 ´ 10 8 = l0 3800 ´ 10 -10

h p

m = 9.11 ´ 10 -31 kg, v = 5.4 ´ 10 6 m/s p = mv = 9.11 ´ 10 -31 ´ 5.4 ´ 10 6

Sample Problem 5 The de-Broglie wavelength associated with an electron, accelerated through a potential difference of 100 volts is [NCERT]

Interpret

(d) 2l b = l e

where, h is Planck’s constant and p is momentum. For the electron,

= 5.23 ´ 10 -19 J (3.2 eV)

(a) 0.529 nm (b) 52.9 nm

(c) l e = l b

Interpret (b) We know that de-Broglie wavelength is given by

Interpret (a) Here, l 0 = 3800 Å Work function f

\

Note The particle with this mass could be a proton or a neutron.

Sample Problem 4 Photoelectric threshold of silver is (a) 5.23 ´ 10 -19 J

æ l ö æv ö Then, mass of particle m = me ç e ÷ ç e ÷ è løèvø ö æ 1ö æ 1 -31 ´ 10 -4 ÷ m = (9.11 ´ 10 kg) ´ ç ÷ ´ ç ø è 3 ø è1.813

(c) 0.123 nm (d) 1.23 nm

= 4.92 ´ 10 -24 kg - m/ s \

l=

(c) Accelerating potential V = 100 V

h 6.63 ´ 10 -34 = p 4.92 ´ 10 -24

l e = 0.135 nm

The de-Broglie wavelength, h 1.227 nm l= = p V 1.227 l= nm = 0.123 nm 100

m¢ = 0.150 kg, v¢ = 30 m/s

For the ball,

p¢ = m¢ v ¢ = 0.150 ´ 30 = 4.50 kg-m/s h 6. 63 ´ 10 -34 J - s = = 1.47 ´ 10 -34 m p¢ 4. 50 kg - m/ s

\

lb =

\

lb < le

Note The de-Broglie wavelength associated with an electron in this case is of the order of X-ray wavelengths.

Sample Problem 8 Calculate the de-Broglie wavelength

Sample Problem 6 A particle is moving three times as fast

of neutrons whose energy is 1 eV. (Given mass of electrons 1.67 ´ 10 -27 kg)

as an electron. The ratio of the de-Broglie wavelength of the particle to that of electron is1.813 ´ 10 -4 . The mass of particle is [NCERT]

(a) 1.675 ´ 10 -27 kg

(b) 3.25 ´ 10 -13 kg

(c) 4.75 ´ 10 -27 kg

(d) 16.73 ´ 10 -13 kg

Interpret (a) de-Broglie wavelength of a moving particle, having mass m and velocity v is h h l= = p mv h Mass = m = lv h For an electron, mass me = l en e v l \ = 3 and = 1.813 ´ 10 -4 ve le

(a) 0.2857Å (c) 0.3150 Å

(b) 0.2167 Å (d) 0.375 Å

Interpret (a) We know that, l=

h 6.6 ´ 10 -84 = 2mE 2 ´ 1.67 ´ 10 -27 ´ 1.6 ´ 10 -10

= 0.2857 ´ 10 -10 = 0.2857 Å

Sample Problem 9 A metal has a work function of 2.0 eV. It is illuminated by monochromatic light of wavelength 500 nm. Calculate (a) the threshold wavelength, (b) the maximum energy of photoelectrons, (c) the stopping potential. (Given, Planck’s constant, h = 6.6 ´ 10 -34 Js, charge on electron, e = 1.6 ´ 10 -19 C and 1 eV = 1.6 ´ 10 -19 C ).

Dual Nature of Radiation and Matter Interpret Here, h = 6.6 ´ 10 -34 Js, e = 1.6 ´ 10 -19 C, 1 eV = 1.6 ´ 10 -19 J f = 2.0 eV = 2.0 ´ 1.6 ´ 10 -19 = 3.2 ´ 10 -19 J, l = 500 nm = 500 ´ 10 -9 m (a) If l 0 is the threshold wavelength, then f=

hc 6.6 ´ 10 -34 ´ 3 ´ 10 8 hc or l 0 = = f l0 3.2 ´ 10 -19 = 618.75 ´ 10 -9 m = 618.75 nm

(b) The maximum energy of the photoelectrons, 1 hc 2 = hv - f = -f mv max l 2 6.6 ´ 10 -34 ´ 3 ´ 10 8 - 3.2 ´ 10 -19 = 500 ´ 10 -9 = 3.96 ´ 10 -19 - 3.2 ´ 10 -19 = 0.76 ´ 10 -19 J (c) The stopping potential is given by 1 2 eV0 = mv max = 0.76 ´ 10 -19 2 0.76 ´ 10 -19 0.76 ´ 10 -19 or = 0.475 V V0 = = e 1.6 ´ 10 -19

25.5 Davisson and Germer Experiment The wave nature of the material particles as predicted by de-Broglie wave confirmed by Davisson and Germer (1927) in United States and by GP Thomson (1928) Scotland. Experimental arrangement used by Davisson and Germer is as shown in the figure. Electrons from hot tungsten cathode (C ) are accelerated by a potential difference V between the cathode and anode ( A). A narrow hole in the anode renders the electrons into a fine beam of electrons

1137

and allows it to strike the nickel crystal. The electrons are scattered in all directions by the atoms in the crystal. The intensity of the electron beam scattered in a given direction is found by the use of a detector. By rotating the detector about an axis through the point O, the intensity of the scattered beam can be measured for different values of f, the angle between incident and the scattered direction of electron beam. Davisson and Germer, found that the intensity of scattered beam of electrons was not the same but different at different angles of scattering. The beam of electron was allowed to fall normally on the surface of nickel crystal. It is observed that below 44 V, the graph of electron intensity is smooth, at 44 V a bump begins to appear and continues to move upwards reaching a pronounced maximum at 54 V. Beyond 54 V the bump diminishes with increasing potential and almost vanish at around 68 V. Let f be angle of incidence and q the angle of scattering. Intensity is maximum for f = 50°, then from figure q + f + q = 180 ° Þ

2 q + f = 180 °

Þ

2 q = 180° - 50° = 130 °

or

q = 65°

Thus, for angles of incidence 50°, scattering angle relative to the set of Bragg’s plane is 65°.

40 V

44 V

48 V

54 V

60 V

64 V

– Electron gun C

+

A

Incident

Detector

Sc a be tte am red

2d sin q = l

Beam

θ

For most of the crystals, the spacing between atomic planes is about 1 Å, therefore the Bragg’s equation for maxima in the diffraction pattern becomes

f θ

O Bragg’s plane Nickel single crystal Davisson and Germer Experiment

l = 2 ´ 1 ´ sin 65° = 1.67 Å \ The de-Broglie wavelength of diffracted electrons is 1.67 Å. Now, for 54 V electron, the de-Broglie wavelength is 12.27 Å l= V 12.27 Å = 1.67 Å = 54

WORKED OUT Examples Example 1

What is the frequency and energy of a photon of wavelength 6000Å ? (Given h = 6.6 ´ 10 -34 Js, c = 3 ´ 10 8 ms-1) 4

(a) 15 ´ 10 Hz and 3.3 ´ 10 (b) 6 ´ 10

10

Hz and 4.4 ´ 10

-10

-19

J

J

(c) 5 ´ 10 12 Hz and 3.3 ´ 10 -19 J (d) 5 ´ 10 14 Hz and 3.3 ´ 10 -19 J

Solution

Frequency n =

c 3 ´ 10 8 = 5 ´ 10 14 Hz = l 6000 ´ 10 -10

and energy = hn = 6.6 ´ 10 -34 ´ 5 ´ 10 14 = 3.3 ´ 10 -19 J

Example 2

The work function of sodium is 2.3 eV.Calculate the maximum wavelength for the light that will cause photoelectrons to be emitted from the sodium. (a) 6400 Å

Solution

(b) 5000 Å

KEmax

(c) 5400 Å hc = hn - f 0 = - f0 l hc - f0 > 0 l

(d) 5200 Å

(Kinetic energy is always positive) ch l< f0

Þ l max =

In an experiment on photoelectric effect it was observed that for incident light of wavelength 1.98 ´ 10 -7 m, stopping potential is 2.5 V. What is the energy of photoelectrons with maximum speed, work function f 0 and threshold frequency? (a) 6.25 eV , 3.75 eV , 9.10 ´ 10 14Hz (b) 3.75 eV , 6.25 eV , 9.10 ´ 10 14Hz (c) 4.75 eV , 6.25 eV , 8.10 ´ 10 14Hz (d) None of the above KE max = eV = 2.5 eV

Energy of the incident light E= =

( f 0) = E - KE = 3.75 eV f Moreover f 0 = hn 0 Þ n 0 = 0 h 3.75 ´ 1.6 ´ 10 -19 \ = 9.10 ´ 10 14 Hz n0 = 6.6 ´ 10 -34

Example 4

Find the ratio of de-Broglie wavelength of proton and a-particle which has been accelerated through the same potential difference.

hc 6.6 ´ 10 -34 ´ 3 ´ 10 8 = l 1.98 ´ 10 -7 6.6 ´ 10 -34 ´ 3 ´ 10 8 eV = 6.25 eV 1.98 ´ 10 -7 ´ 1.6 ´ 10 -19

(b) 2 2

(a) 3 2

Solution

ma

(c) 2 3

(d) 2 5

lp 2maq aV h , = 2mpq pV 2mqV l a

l= mp

As,

=

lp 1 q 2 and a = Þ = 4 ´2 =2 2 4 la qp 1

Example 5

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 the photocell varies as follows. l

l

(a)

ch 3 ´ 10 8 ´ 6.62 ´ 10 -34 = 5400 Å = f0 2.3 ´ 1.6 ´ 10 -19

Example 3

Solution

Now, work function

(b) O

O

λ

l

λ

l

(c)

(d) O

O

λ

λ

Solution

On increasing wavelength of light the photoelectric current decreases and at a certain wavelength (cut-off) above which photoelectric current stops. Hence, graph (a) is correct.

Example 6

The ratio of de-Broglie wavelengths of molecules of hydrogen and helium which are at temperature 27°C and 127°C respectively is (a)

1 2

(b)

Solution

3 8

Wavelength l = =

(c)

8 3

(d) 1

h h = mv rms 3mKT

lH mHeTHe = = l He mHTH

4(273 + 127) 8 = 2(273 + 27) 3

Start Practice for

JEE Main Round I

(Topically Divided Problems)

Particle Nature of Light 1. A particle is dropped from a height H. The de-Broglie wavelength of the particle as a function of height is [NCERT Exemplar] proportional to (b) H1/ 2 (d) H-1/ 2

(a) H (c) H0

2. A parallel beam of light is incident normally on a

plane surface absorbing 40% of the light and reflecting the rest. If the incident beam carries 60 W of power, the force exerted by it on the surface is (a) 3.2 ´ 10 -8 N

(b) 3.2 ´ 10 -7 N

(c) 5.12 ´ 10 -7 N

(d) 5.12 ´ 10 -8 N

3. Calculate the energy of a photon with momentum 3.3 ´ 10-13 kg-ms -1, given Planck’s constant to be 6.6 ´ 10-34 Js (a) 7.3 ´ 10 4 J

(b) 9.9 ´ 10 -5 J

(c) 1.3 ´ 105 J

(d) 8.1 ´ 103 J

4. A particle of mass 1 mg has the same wavelength as an electron moving with a velocity of 3 ´ 106 ms -1. The velocity of the particle is (a) 3 ´ 10 -31 ms -1 (c) 2.7 ´ 10

-18

ms

(b) 2.7 ´ 10 -21 ms -1

-1

(d) 9 ´ 10 -2 ms -1

5. The energy that should be added to an electron to reduce its de-Broglie wavelength from 10-10 m to 0.5 ´ 10-10 m , will be (a) (b) (c) (d)

four times the initial energy thrice the initial energy equal to the initial energy twice the initial energy

6. The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly (a) 1.2 nm (b) 1.2 ´ 10 -3 nm

7. Photons absorbed in matter are converted to heat. A source emitting n photon/sec of frequency n is used to converting of ice at 0°C to water at 0°C. Then, the time, T taken for the conversion (a) decreases with increasing n, with v fixed (b) decreases with n fixed, v increasing (c) remains constant with n and n changing such that nn = constant (d) increases when the product nn increases

8. There are two sources of light each emitting with a power of 100 W. One emits X-rays of wavelength 1 nm and the other visible light of wavelength 500 nm. Find the ratio of number of photons of X-rays in the photons of visible light of the given wavelength? (a) 1 : 500 (c) 1 : 20

(b) 1 : 250 (d) 100

9. Consider a metal exposed to light of wavelength 600 nm. The maximum energy of the electron doubles when light of wavelength 400 nm is used. The work function in eV is [NCERT Exemplar] (a) 1.50 eV (c) 1.94 eV

(b) 1.02 eV (d) 2.76 eV

10. An important spectral emission line has a wavelength of 21 cm. The corresponding photon energy is (h = 6.62 ´ 10-34 Js and c = 3 ´ 108 cm -1) (a) 5.9 ´10 -8 eV

(c) 5.9 ´ 10

-6

eV

(b) 5.9 ´ 10 -4 eV

(d) 11.8 ´ 10 -6 eV

11. The energy of a photon of green light of wavelength 50000 Å is

(a) 3.459 ´ 10 -19 J (c) 4.132 ´ 10 -19 J

(b) 3.973 ´ 10 -19 J (d) 8453 ´ 10 -19 J

12. What will be the number of photons emitted per

second by a 10 W sodium vapour lamp assuming that 90% of the consumed energy is converted into light? [Wavelength of sodium light is 590 nm, h = 6.63 ´ 10-34 J-s]

(c) 1.2 ´ 10 -6 nm

(a) 0.267 ´ 1018

(b) 0.267 ´ 1019

(d) 1.2 ´ 101 nm

(c) 0.267 ´ 1020

(d) 0.267 ´ 1017

1140 JEE Main Physics 13. If the energy of photons corresponding to the wavelength of 6000 Å is 3.2 ´ 10-19 J, the photon energy for a wavelength of 4000 Å will be (a) 1.11 ´ 10 -19 J

(b) 2.22 ´ 10 -19 J

(c) 4.40 ´ 10 -19 J

(d) 4.80 ´ 10 -19 J

20. Represents a graph of kinetic energy of most energetic photoelectrons K max (in eV) and frequency n for a metal used as cathode in photoelectric experiment. The threshold frequency of light for the photoelectric emission from the metal is Kmax. (eV)

14. A radio transmitter operates at a frequency 880 kHz and a power of 10kW. The number of photons emitted per second is

3

(a) 1.72 ´ 1031

(b) 1.327 ´ 1025

2

(c) 1.327 ´ 1037

(d) 1.327 ´ 10 45

1 0

Emission of Electrons and Photoelectric Effect 15. The photoelectric threshold of Tungsten is 2300 Å. The energy of the electrons ejected from the surface by ultraviolet light of wavelength 1800 Å is ( h = 6.6 ´ 10-34 J-s) (a) 0.15 eV (c) 15 eV

(b) 1.5 eV (d) 150 eV

16. Consider a beam of electrons (each electron with energy E0 ) incident on a metal surface kept in an evacuated chamber. Then, [NCERT Exemplar] (a) no electrons will be emitted as only photons can emit electrons (b) electrons can be emitted but all with an energy, E0 (c) electons can be emitted with any energy, with a maximum of E0 - f ( f is the work function) (d) electrons can be emitted with any energy, with a maximum of E0

1015Hz

(a) 4 ´ 1014 Hz 14

(c) 2.0 ´ 10

(a) 3 ´ 10 -21

(b) 3.2 ´ 10 -19

(c) 3 ´ 10 -17

(d) 3 ´ 10 -15

18. Radiations of two photon’s energy, twice and ten times the work function of metal are incident on the metal surface successively. The ratio of maximum velocities of photoelectrons emitted in two cases is (a) 1 : 2 (c) 1 : 4

(b) 1 : 3 (d) 1 : 1

19. A proton, a neutron, an electron and an a-particle have same energy. Then their de-Broglie wavelengths compare as [NCERT Exemplar] (a) (b) (c) (d)

lp = ln > le > la la = lp > ln > le le = lp > ln > la le = lp > ln > la

(b) 3.5 ´ 1014 Hz (d) 2.7 ´ 1014 Hz

21. For a certain metal, v = 2 v0 and the electrons come out with a maximum velocity of 4 ´ 106 ms -1. If the value of v = 5 v0 , then maximum velocity of photelectrons will be (a) 2 ´ 107 ms -1

(b) 8 ´ 106 ms -1

-1

(d) 8 ´ 105 ms -1

6

(c) 2 ´ 10 ms

22. The wavelength of the photoelectric threshold for silver is l 0 . The energy of the electron ejected from the surface of silver by an incident light of wavelength l ( l < l 0 ) will be (a) hc ( l 0 - l ) (c)

17. Ultraviolet radiation of 6.2 eV falls on an aluminium surface (work function 4.2 eV). The kinetic energy in joule of the fastest electron emitted is approximately.

Hz

ν

h æ1 1ö ç ÷ c è l l0 ø

hc l0 - l æ l - lö (d) hc ç 0 ÷ è l0 l ø (b)

23. A metal surface is illuminated by a light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one-fourth of its original value, then the maximum kinetic energy of the emitted photoelectrons would become (a) (b) (c) (d)

four times the original value twice the original value 1/6th of the original value unchanged

24. The work function of a metal is 1eV. Light of wavelength 3000 Å is incident on this metal surface. The velocity of emitted photoelectrons will be (a) 10 ms -1 (c) 10 4 ms -1

(b) 103 ms -1 (d) 106 ms -1

25. In the photoelectric effect, the velocity of ejected electrons depends upon the nature of the target and (a) (b) (c) (d)

the frequency of the incident light the polarisation of the incident light the time for which the light has been incident the intensity of the incident light

Dual Nature of Radiation and Matter 26. Light of wavelength 4000 Å is incident on a metal plate whose work function is 2 eV. The maximum KE of the emitted photoelectron would be (a) 0.5 eV (c) 1.5 eV

(b) 1.1 eV (d) 2.0 eV

27. A photon of energy 3.4 eV is incident on a metal having work function 2 eV. The maximum KE of photoelectrons is equal to (a) 1.4 eV (c) 5.4 eV

(b) 1.7 eV (d) 6.8 eV

28. The frequency of the incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectron is (a) (b) (c) (d)

double the earlier value unchanged more than doubled less than doubled

with light of wavelength 332 nm. The retarding potential required to stop the escape of photoelectrons is (b) 2.66 eV (d) 4.81 eV

30. If the work function for a certain metal is 3.2 ´ 10

J and it is illuminated with light of frequency n = 8 ´ 1014 Hz, the maximum kinetic energy of the photoelectron would be (a) 2.1 ´ 10 -19 J

(b) 3.2 ´ 10 -19 J

(c) 5.3 ´ 10 -19 J

(d) 8.5 ´ 10 -19 J

31. Ultraviolet radiations of 6.2 eV falls on an aluminium surface. KE of fastest electron emitted is (work function = 4.2 eV) (b) 3.2 ´ 10 -19 J (d) 9 ´ 10 -32 J

32. Ultraviolet light of wavelength 300 nm and intensity

1.0 Wm -2 falls on the surface of a photosensitive material. If one percent of the incident photons produce photoelectrons, then the number of photoelectrons emitted from an area of 1.0 cm2 of the surface is nearly (a) 9.61 ´ 1014 s -1 (c) 1.51 ´ 1012 s -1

(b) 4.12 ´ 1013 s -1 (d) 2.13 ´ 1011 s -1

33. The photoelectric threshold wavelength for a metal surface is 6600 Å. The work function for this metal is (a) 0.87 eV (c) 18.7 eV

35. The

difference between kinetic energies of photoelectrons emitted from a surface by light of wavelength 2500 Å and 5000 Å will be (a) 1.61 eV (c) 3.96 eV

(b) 2.47 eV (d) 3.96 ´ 10 -19 eV

36. When a point source of light is 1 m away from a photoelectric cell, the photoelectric current is found to be I mA. If the same source is placed at 4 m from the same photoelectric cells, the photoelectric current (in mA) will be I 16 (c) 4 I (a)

I 4 (d) 16 I (b)

The photoelectrons are emitted when light of wavelength 2000 Å falls on it. The potential difference applied to stop the fastest photo electrons is (h = 4.14 ´ 10-15 eV) (a) 2.24 V (c) 4.8 V

(b) 1.2 V (d) 3.6 V

38. The work function of tungsten and sodium are 4.5 eV -19

(a) 3.2 ´ 10 -21 J (c) 7 ´ 10 -25 J

(b) 1.14 eV (d) 4.58 eV

37. The work function of a metallic surface is. 5.01 eV.

29. A metal surface of work function 1.07 eV is irradiated

(a) 1.07 eV (c) 3.7 eV

(a) 0.57 eV (c) 2.29 eV

1141

(b) 1.87 eV (d) 0.18 eV

34. Light of wavelength 4000 Å incident on a sodium surface for which the threshold wavelength of photoelectrons is 5420 Å. The work function of sodium is

and 2.3 eV respectively. If the threshold wavelength, l for sodium is 5460 Å, the value of l for tungsten is (a) 2791 Å (c) 1925 Å

(b) 3260 Å (d) 1000 Å

39. A photon of energy E ejects a photoelectrons from a metal surface whose work function is W0 . If this electron enters into a uniform magnetic field of induction B in a direction perpendicular to the field and describes a circular path of radius r, then the radius, r is given by (a) (c)

2m (W0 - E) eB 2m ( E - W0 ) eB

(b) (d)

2e ( E - W0 ) mB 2mW0 eB

40. The energy flux of sunlight reaching the surface of

the earth is 1.388 × 10 3 W/m2 . How many photons (nearly) per square metre are incident on the earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm. [NCERT] (a) 3.8 ´ 1021 photon / m2 -s (b) 4.1 ´ 1018 photon / m2 -s (c) 2.6 ´ 1019 photon / m2 -s (d) 1.9 ´ 1020 photon / m2 -s

41. The work function for the surface of Al is 4.2 eV. How much potential difference will be required to just stop the emission of maximum energy electrons emitted by light of 2000 Å ? (a) 1.51 V (c) 2.99 V

(b) 1.99 V (d) None of these

1142 JEE Main Physics 42. Two monochromatic beams A and B of equal intensity I, hit a screen. The number of photons hitting the screen by beam A is twice that by beam B. Then, what inference can you move about their frequencies ? (a) (b) (c) (d)

The frequency of beam B is twice that of A The frequency of beam B is half that of A The frequency of beam A is twice of B None of the above

a surface with a spacing d = 0.1 nm. The first maximum of intensity in the reflected beam occurs at q = 30°. The kinetic energy E of the beam (in eV) is (c) 0.21 eV

surface and electrons are ejected with kinetic energy E. If the KE is to be increased to 2 E, the wavelength must be changed to l¢ where l 2

(b) l ¢ = 2 l

l (c) < l¢ < l 2

(d) l ¢ > l

function W0 is l 0 . What is the threshold wavelength W for a metal whose work function is 0 ? 2 (b) 2l 0

(c)

l0 2

(d)

l0 4

46. The work function of a substance is 4.0 eV. The longest wavelength of photoelectric emission approximately. (a) 540 nm (b) 400 nm

light that can cause from this substance (c) 310 nm

is in an electric field v = v0 $i (v 0 > 0) E = E0 $i (E0 = constant > 0) field. It's de-Broglie wavelength at time t is given by (a)

l0 æ eE0 t ö ç1 + ÷ è m v0 ø

(c) l 0

æ eE0t ö (b) l 0 ç1 + ÷ è mv 0 ø (d) l 0t

51. If the mass of neutral = 1.7 ´ 10-27 kg, then the de-Broglie wavelength of neutral of energy 3 eV is (h = 6.6 ´ 10-34 J- s) (a) 1.6 ´ 10 -16 m

45. The threshold wavelength for a metal having work

(a) 4 l 0

(a) remains constant (b) increases with time (c) decreases with time (d) increases and decreases periodically

(d) 0.78 eV

44. Light of wavelength l strikes a photo sensitive

(a) l ¢ =

and is in a magnetic field B = B0 $j .Then it's [NCERT Exemplar] de-Broglie wavelength

50. An electron (mass m) with an initial velocity

43. A neutrons beam of energy E scatters from atoms on

(a) 10.2 eV (b) 5.02 eV

49. An electron is moving with an initial velocity v = v0 i$

(b) 1.6 ´ 10 -11 m (c) 1.4 ´ 10 -10 m (d) 1.4 ´ 10 -11 m

52. An electron (mass m) with an initial velocity v = v0 $i is

in an electric E = E0 $j . If l 0 = h / mv0 , it's de-Broglie [NCERT Exemplar] wavelength at time t is given by (b) l 0 1 +

(a) l 0 l0

(c) 1+

(d) 220 nm

e2 E 20t2 m2 v20

(d)

e2 E 2t2 m2 v20

l0 æ e2 E 2t2 ö ç1 + 2 02 ÷ m v0 ø è

53. An electron of mass m and charge e initially at rest

Wave Nature of Particle 47. An electron and photon have same wavelength. If E is the energy of photon and p is the momentum of E in SI unit is p

electron, then the magnitude of (a) 3.33 ´ 10 -9 (c) 1.1 ´ 10

-19

(b) 3.0 ´ 10 8 (d) 9 ´ 1016

48. The wavelength of de-Broglie wave associated with a thermal neutron of mass m at absolute temperature T is given by (Here, k is the Boltzmann constant) (a) (c)

h 2m kT h 3k mT

h m kT h (d) 2 m kT (b)

gets accelerated by a constant electric field E. The rate of change of de-Broglie wavelength of this electron at time t ignoring relativistic effect is -h e Et2 - mh (c) e Et2 (a)

- eEt E -h (d) eE

(b)

54. What should be the velocity of an electron so that its momentum becomes equal to that of a photon of wavelength 5200 Å ? (a) 103 ms -1 (b) 1.2 ´ 103 ms -1 (c) 1.4 ´ 103 ms -1 (d) 2.8 ´ 103 ms -1

Dual Nature of Radiation and Matter

Round II Only One Correct Option 1. The stopping potential V for photoelectric emission for a metal surface is plotted along Y -axis and frequency n of incident light along X-axis. A straight line is obtained as shown. Planck’s constant is given by slope of the line product of slope of the line and charge on the electron intercept along Y-axis divided by charge on the electron product of intercept along X-axis and mass of the electron (e) product of slope and mass of the electron

(Mixed Bag) 7. The de-Broglie wavelength L associated with an

elementary particle of linear momentum p is best represented by the graph L

radiation. The potential difference to stop the ejection is 2 V. If the incident light is changed to 3000 Å, then the potential required to stop the ejection of electrons will be (a) greater than 2V (c) ¥

(b) less than 2V (d) zero

3. An electron and a proton have the same de-Broglie wavelength. Then the kinetic energy of the electron is (a) (b) (c) (d)

zero infinity equal to kinetic energy of the proton greater than the kinetic energy of proton

4. Energy required to remove an electron from an aluminium surface is 4.2 eV. If light of wavelength 2000 Å falls on the surface, the velocity of fastest electrons ejected from the surface is (a) 2.5 ´ 1018 ms -1 (c) 6.7 ´ 1018 ms -1 (d) None of the above

produce photoelectric effect in certain metal is 200 nm. The maximum kinetic energy acquired by electron due to radiation of wavelength 100 nm will be (b) 6.2 eV (d) 200 eV

6. What is the de-Broglie wavelength (in Å) of the a-particle accelerated through a potential difference V? 0. 287 V 0.101 (c) V

p

p

L

L

(c)

(d) p

p

8. Maximum velocity of photoelectron emitted is

e ratio of electron is 1.76 ´ 1011 Ckg -1, m then stopping potential is given by 4.8 ms -1. The

(a) 5 ´ 10 -10 JC-1 (c) 7 ´ 1011 JC-1

(b) 3 ´ 10 -7 JC-1 (d) 2. 5 ´ 10 -2 JC-1

9. Two identical metal plates shown photoelectric effect by a light of wavelength l Å falls on plate A and l B on plate B ( l A = 2l B ). The maximum kinetic energy is (a) 2 KA = KB (c) KA = 2KB

KB 2 KB (d) KA = 2

(b) KA
m2 ) have the same de-broglie wavelength. Then

[NCERT Exemplar]

(a) their momenta are the same (b) their energies are the same (c) energy of A1 is less than the energy of A2 (d) energy of A1 is more than the energy of A2

Dual Nature of Radiation and Matter 25. In which of the following situations are heavier of the two particles has smaller de-Broglie wavelength? The two particles (a) move with same speed (b) move with same KE (c) move with same linear momentum (d) have fallen through the same height

26. The de-Broglie wavelength of a photon is twice the de-Broglie wavelength of an electron. The speed of c the electron is ve = . Then 100 Ee = 10 -4 Ep p (c) e = 10 -2 me c

(a)

Ee = 10 -2 Ep p (d) e = 10 -4 me c (b)

1145

emission of photoelectrons takes place. The maximum kinetic energy of the emitted photoelectrons is given by K max = hn - f 0 . If the frequency of the incident light is n 0 (called threshold frequency), the photoelectrons are emitted from metal without any kinetic energy. So, hn 0 = f 0

30. Stopping potential of emitted photoelectron is given by hv - f0 e hv (c) e

(b) hv - f0

(a)

(d)

f0 + hv e

31. The variation of maximum kinetic energy ( K max ) of

the emitted photoelectrons with frequency ( n) of the incident radiations can be represented by Kmax

Kmax

27. When photons of energy 4.25 eV strike the surface of a metal, the ejected photelectrons have a maximum kinetic energy E A eV and de-Broglie wavelength l A . The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is EB = ( E A - 1.50 ) eV. If the de-Broglie wavelength of these photelectrons is l B = 2l A , then (a) (b) (c) (d)

the work function of A is 2.25 eV the work function of B is 4.20 eV EA = 2.0 eV EB = 2. 75 eV

(a)

(b) ν

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

the work function of A is 1.50 eV the work function of B is 4.0 eV TA = 2. 00 eV All of the above

29. The maximum KE of photoelectrons ejected from a photometer when it is irradiated of wavelength 400 nm is 1 eV. If the threshold energy of the surface is 0.9 eV (a) the maximum KE of photoelectrons when it is irradiated with 500 nm photons will be 0.42 eV (b) the maximum KE in case (a) will be 1.425 eV (c) maximum KE will increase if the intensity of radiation is increased

Comprehension Based Questions Passage I According to Einstein, when a photon or light of frequency n or wavelength l is incident on photosensitive metal surface of work function f 0 , where f 0 < hn (here h is Planck’s constant), then the

Kmax

(c)

(d)

28. When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV and de-Broglie wavelength l A . The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is TB = ( TA - 150) eV. If the de-Broglie wavelength of these photoelectrons l B = 2l A , then

ν

Kmax

ν

ν

Passage II According to de-Broglie, a moving material particle exhibits dual nature ( i. e., a particle as well as a wave). He also predicted that a wave is associated with every moving material particle (which controls the particle) called matter wave and its wavelength is called de-Broglie wavelength given by h l= mv where h is Planck’s constant, m is the mass of the particle moving with velocity v. The existence of matter waves was firstly experimentally verified by Division and Germer using slow moving electrons which were accelerated with moderate accelerating potential.

32. An electron is accelerated under a potential difference of 64 V, the de-Broglie wavelength associated with electron is (use charge of electron 1.6 ´ 10-19 C, mass of electron 9.1 ´ 10-31 kg; h = 6.623 ´ 10-34 J-s). (a) (b) (c) (d)

1.53 Å 2.53 Å 3.35 Å 4.54 Å

1146 JEE Main Physics 33. If a-particle and proton have same momenta, the

35. Assertion Work function of copper is greater than the

ratio of de-Broglie wavelength of a-particle and proton is

work function of sodium, but both have same value of threshold frequency and threshold wavelength. Reason The frequency is inversely proportional to wavelength.

(a) 2

(b) 1

(c) 1/2

(d) 1/4

34. If a-particle and proton are accelerated through the same potential difference, then the ratio of de-Broglie wavelength of a-particle and proton is (b) 2 2 1 (d) 2

(a)

2 1 (c) 2 2

36. Assertion The de-Broglie wavelength of a molecule varies inversely as the square root of temperature. Reason The root mean square velocity of the molecule depends on the temperature.

37. Assertion Stopping potential is a measure of KE of photoelectron. Reason W = eVs =

Assertion and Reason Directions

Question Nos. 35 to 40 are Assertion-Reason type. Each of these contains two Statements : Statement I (Assertion), Statement II (Reason). Each of these questions also has four alternative choice, only one of which is correct. You have to select the correct choices from the codes (a), (b), (c) and (d) given below (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 correct explanation of the Assertion (c) If Assertion is true but Reason is false (d) If Assertion is false but the Reason is true

1 mv2 = KE 2

38. Assertion A photon has no rest mass, yet it carries definite momentum. Reason Momentum of photon is due to energy hence its equivalent mass. The relative velocity of two photons travelling in opposite direction is the velocity of light. Reason The rest mass of photon is zero.

39. Assertion

40. Assertion Photoelectric effect demonstrates the particle nature of light. Reason The number of photoelectrons is proportional to the frequency of light.

Previous Years’ Questions 41. Light of wavelength l falls on a metal having work function

hc . Photoelectric effect will take place only if l0 [DCE 2009]

(a) l ³ l 0 (c) l £ l 0

(b) l ³ 2 l 0 (d) l = 4 l 0

42. The surface of the metal is illuminated with the light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work [AIEEE 2009] function of metal is (a) 1.42 eV (c) 1.68 eV

(b) 1.51 eV (d) 3.0 eV

matter has a wave nature ?

(a) both (c) photon only

Outgoing electrons i

90°

d

[VIT EEE 2008]

(b) Electron diffraction (d) Photon diffraction

44. Out of a photon and an electron the equation E = pc, is valid for

Incoming electrons 90°–1

43. Which phenomenon best supports the theory that (a) Electron momentum (c) Photon momentum

Directions Questions Nos. 45 to 47 are based on the following paragraph. Wave property of electrons implies that they will show diffraction effects. Davission and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see figure), is obtained by requiring electron waves reflected from the planes of atom in a crystal interfere contruetively shown in the figure.

[BVP Engg. 2008]

(b) neither (d) electron only

Crystal plane

45. Electrons accelerated by potential V are diffracted from a crystal. If d = 1 Å and i = 30°, V should be about (h = 6.6 ´ 10-34 J-s, me = 9.1 ´ 10-31 kg, -19 C) [AIEEE 2008] e = 1.6 ´ 10 (a) 2000 V (b) 50 V

(c) 500 V

(d) 1000 V

Dual Nature of Radiation and Matter 46. If a strong diffraction peak is observed when

1147

Which of the following graphs can be expected to represent the number of electrons N detected as a function of the detector position y = 0 (corresponds to the middle of the slit) ?

electrons are incident at an angle i from the normal to the crystal planes with distance d between them (see figure), de-Broglie wavelength l dB of electrons can be calculated by the relationship (n is an integer)

y

y

[AIEEE 2008]

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

d sin i = nl dB 2d cos i = nl dB 2d sin i = nl dB d cos i = nl dB

(a)

d

N

(b)

d

N

y

47. In an experiment, electrons are made to pass through a narrow slit of width d comparable to their de-Broglie wavelength. They are detected on a screen at a distance D from the slit (see figure). [AIEEE 2008]

(c)

y

d

N

(d)

d

N

48. Photon of frequency n has a momentum associated with it. If c is the velocity of the light, the moments is

y=0

d

[AIEEE 2007]

v (a) c hv (c) 2 c

D

(b) hvc (d)

hv c

Answers Round I 1. 11. 21. 31. 41. 51.

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

2. 12. 22. 32. 42. 52.

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

3. 13. 23. 33. 43. 53.

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

4. 14. 24. 34. 44. 54.

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

5. 15. 25. 35. 45.

(b) (b) (a) (b) (b)

4. 14. 24. 34. 44.

(c) (c) (a,d) (c) (c)

5. 15. 25. 35. 45.

(b) (a) (a,b,d) (d) (b)

6. 16. 26. 36. 46.

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

7. 17. 27. 37. 47.

(a,b,c) (b) (a) (b) (b)

8. 18. 28. 38. 48.

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

9. 19. 29. 39. 49.

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

10. 20. 30. 40. 50.

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

Round II 1. 11. 21. 31. 41.

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

2. 12. 22. 32. 42.

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

3. 13. 23. 33. 43.

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

6. 16. 26. 36. 46.

(c) (c) (b,c) (b) (c)

7. 17. 27. 37. 47.

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

8. 18. 28. 38. 48.

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

9. 19. 29. 39.

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

10. 20. 30. 40.

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

the Guidance Round I 1. Velocity acquired by a particle while falling from a height H

8. Total E is constant. Let, n1 and n2 be the number of photons of

is, v = 2gH

X-rays and visible region

l=

h h or l µ H = mv m 2gH

1 2

Þ

2. Momentum of incident light per second E 60 p1 = = = 2 ´ 10 -7 c 3 ´ 10 8 Momentum of reflected light per second E 60 60 ´ = = 1.2 ´ 10 -7 p2 = 100 c 3 ´ 10 8

= (2 + 1.2) ´ 10 -7 = 3.2 ´ 10 -7 N

n1 l1 n 1 = = 1 : 500 Þ 1 = n2 l 2 n2 500

Þ

9. The maximum energy = hn - f æ1230 ö 1 æ1230 ö - f÷ = ç - f÷ ç è 600 ø 2 è 400 ø 1230 f= = 1. 02 eV Þ 1200 hn hc As, E = = (in eV) el l

Þ

\Force on the surface = change in momentum per second = p2 - ( - p1) = p2 + p1

10.

3. Given, E / c = 3.3 ´ 10 -13 kg ms-1;

=

E = 3.3 ´ 10 -13 ´ c = 3.3 ´ 10 -13 ´ 3 ´ 10 8

So,

= 9.9 ´ 10 -5 J

n1E1 = n2 E 2 hc hc = H2 n1 l1 l2

Then,

6.6 ´ 10 -34 ´ 3 ´ 10 8 = 5.9 ´ 10 -6 eV 1.6 ´ 10 -19 ´ 0.21

12. Energy of photon,

4. de-Broglie wavelength, l =

h mv

E=

hc 6.63 ´ 10 -34 ´ 3 ´ 10 8 6.63 ´ 3 = ´ 10 -18 = l 59 590 ´ 10 -9 90 ´ 10 = 9 W 100

As both particle and electrons having same wavelength therefore, their momentum will be equal to

Light energy produced per second =

Þ

\Number of photons emitted per sec 9 ´ 59 = - 18 = 2.67 ´ 10 19 6.63 ´ 3 ´ 10

mpv p = mev e

Þ

9.1 ´ 10 -31 ´ 3 ´ 10 6 mv vp = e e = mp 10 -6

Þ

v p = 2.7 ´ 10 -18 m/s h 2mE 1 lµ E l1 E = 2 l2 E1 l=

5. As, Þ Þ

13. As, E = hc / l E µ l / l;

or

E 2 = E1 ´ l1 / l 2

So,

= 3.2 ´ 10 -19 ´ 6000 / 4000 = 4.8 ´ 10 -19 J

14. Number of photons emitted per sec, P Power = Energy of photon hn 10000 = = 1.72 ´ 10 31 6.6 ´ 10 -34 ´ 880 ´ 10 3

E 10 -10 = 2 E1 0.5 ´ 10 -10

Þ Þ Therefore,

E 2 = 4 E1 added energy = E 2 - E1 = 3E1

6. Energy E = 1 MeV = 10 6 eV; hc = 1240 eV nm E=

n=

hc hc 1240 eV nm or l = = = 1.24 ´ 10 -3 nm l E 10 6 eV

7. Decreases with increasing n, with v fixed, decrease with n fixed, v increasing and nv is constant.

15. We know, E k = =

hc æ 1 1ö ç ÷ (in eV) e è l l0 ø 6.6 ´ 10 -34 ´ 3 ´ 10 8 1.6 ´ 10 -19

= 1.5 eV

æ 10 10 10 10 ö ç ÷ è1800 2300 ø

Dual Nature of Radiation and Matter 16. When a beam of electrons of energy E 0 is incident on a metal

1 2

24. We know, mv 2 =

surface kept in an evacuated chamber, electrons can be emitted with maximum energy E 0 (due to elastic collision ) and with any energy less than E 0 ,when part of incident energy of electron is used in liberating the electrons from the surface of metal.

=

1 2 1 mv 22 = 10 f0 - f0 = 9f0 2

\

of the incident light. hc As, maximum KE = - f0 l =

1 2

19. Kinetic energy of particle K = mv 2 or mv = 2mK h h de-Broglie wavelength l = = mv 2mK For the given value of K , l µ1 / m. 1 1 1 1 \ lp : ln : le : la = : : : mp mn me ma Since mp = mn , hence l p = l n As ma > mp , therefore l a > l p, As me < mn , therefore l e > l n, Hence, la < lp = ln < le

6.6 ´ 10 -34 ´ 3 ´ 10 8 1 ´ - 2 = 1.1 eV 400 ´ 10 -10 1.6 ´ 10 -19

27. Maximum KE = E - f0 = 3.4 - 2 = 1.4 eV 28. Let, E1 and E 2 be the KE of photoelectron for incident light of frequency n and 2n, respectively. hn = E1 + f0

Then

h2n = E 2 + f 2(E1 + f0) = E 2 + f0

and So,

or E 2 = 2E1 + f0 It means the KE of photoelectron becomes more than double.

29. Retarding potential Vs =

20. From graph, n = 1015 Hz Kmax = 3eV = 3 ´ 16 . ´ 10

or

J

= (10 - 7.3) ´ 10 14

30. Maximum KE = hn - f0 = 6.63 ´ 10 -34 ´ 8 ´ 10 14 - 3.2 ´ 10 -19 = 2.1 ´ 10 -19 J

31. KE of fastest electron = E - f0 = 6.2 - 4.2 = 2.0 eV

= 2.7 ´ 10 14 Hz

21. As,

So,

1 mv12 = 2hn 0 - hn 0 = hn 0 and 2 1 mv 22 = 5hn 0 - hn 0 = 4hn 0 2 1 1 mv 22 = 4 ´ mv12 2 2

or

22. As, E k =

hc f0 1240 ´ 10 -9 - 1.07 = le e 330 ´ 10 -9

= 3.73 - 1.07 = 2.66 V -19

Kmax = hn - hn 0 K 3 ´ 1.6 ´ 10 -19 n 0 = n - max = 10 15 h 6.6 ´ 10 -34

As,

2 ´ 3.14 ´ 1.6 ´ 10 -19 = 10 6ms-1 9.1 ´ 10 -31

25. The velocity of photoelectrons depends upon the frequency 26.

v1 1 f0 = = v2 9f0 3

6.63 ´ 10 -34 ´ 3 ´ 10 8 -1 (3 ´ 10 -7) ´ 1.6 ´ 10 -19

v=

or

18. We have, mv12 = 2f0 - f0 = f0 and

hc - f0 l

= 4.14 - 1 = 3.14 eV

17. We have, E k = E - f0 = 6.2 - 4.2 = 2.0 eV = 2.0 ´ 1.6 ´ 10 -19 = 3.2 ´ 10 -19 J

1149

v 2 = 2v1 = 2 ´ 4 ´ 10 6 = 8 ´ 10 6 ms-1 é l - lù hc hc = hc ê 0 ú l l0 ë l0l û

23. The maximum KE of the emitted photoelectrons is independent of the intensity of the incident light but depends upon the frequency of the incident light.

= 2 ´ 1.6 ´ 10 -19 = 3.2 ´ 10 -19 J

32. Energy incident over 1 cm2 = 1.0 ´ 10 -4 J; Energy required to produce photoelectrons = 1.0 ´ 10 -4 ´ 10 -2 = 10 -6 J As, number of photoelectrons ejected = number of photons which can produce photoelectrons = energy required for producing electron/energy of photon. =

10 -6 10 -6 ´ 300 ´ 10 -9 = 1.51 ´ 10 12 s-1 = hc / l 6.6 ´ 10 -34 ´ 3 ´ 10 8

33. As, f0 =

hc 6.6 ´ 10 -34 ´ 3 ´ 10 8 (in eV) = = 1.87 eV el 0 6600 ´ 10 -10 ´ 1.6 ´ 10 -19

1150 JEE Main Physics Number of photons incident on the earth’s surface

34. As, f0 = hc / l0 (in eV) -34

8

6.62 ´ 10 ´ 3 ´ 10 = 2.29 eV = 5420 ´ 10 -10 ´ 1.6 ´ 10 -19

35.

hc hc hc( l 2 - l1) (in eV) As, DE = = l1 l 2 l1l 2 =

n=

= 3.838 ´ 10 21 photon/m 2-s

36. Photoelectric current (I) µ Intensity of incident light and 1 (distance) 2 Iµ

hc - W0 l

W0 = 4.2 eV = 4.2 ´ 1.6 ´ 10 -19 = 6.72 ´ 10 -19 J

= 2.47 eV

So,

E k = hn - W0 =

41. As,

6.62 ´ 10 -34 ´ 3 ´ 10 8 ´ (5000 - 2500) ´ 10 -10 2500 ´ 5000 ´ 10 -20 ´ 1.6 ´ 10 -19

intensity µ

1.388 ´ 10 3 P = 3.838 ´ 10 21 = E 3.616 ´ 10 -19

1 (distance) 2

6.62 ´ 10 -34 ´ 3 ´ 10 8 - 6.72 ´ 10 -19 2000 ´ 10 -10

So,

Ek =

Þ

E k = 9.9 ´ 10 -19 - 6.72 ´ 10 -19

Þ

E k = 3.18 ´ 10 -19 J

Hence, stopping potential V0 =

2

I æ 1ö I¢ = I ç ÷ = è 4ø 16

Hence

37. Energy of the incident light, E =

38.

12375 = 618 . V 2000

According to relation, E = W0 + eV0 E - W0 6.18 eV - 5.01 eV = = 1.17 eV = 1.2 V Þ V0 = e e l hc (W0)T As, W0 = ; = Na l 0 (W0) Na lT or

lT =

42. As,

E = nAv n = nBvB

43. We have, 2d sin q = l and d = 10 -10 m \

p=

Also,

or

44. As, E = Þ

and a charged particle placed in uniform magnetic field experience of force, mv 2 F= r mv 2 evB = Þ r r=

mv 2m(E - W0) = eB eB

40. Given, energy per unit area per second, P = 1.388 ´ 103 W/m2 Let n be the number of photons incident on the earth per square metre. Wavelength of each photon = 550 nm = 550 ´ 10 -9 m. hc (where h is the planck’s Energy of each photon, E = l constant) =

6.63 ´ 10 -34 ´ 3 ´ 10 8 = 3.616 ´ 10 -19 J 550 ´ 10 -9

E=

6.6 ´ 10 -34 ´ 1.6 ´ 10 -19 2 ´ (1.7 ´ 10 -27) 6.6 ´ 1.6 ´ 10 -2 eV 2 ´ 1.7

= 20.5 ´ 10 -2 eV = 0.21 eV

2(E - W0) =v m

Þ

h 6.6 ´ 10 -34 = = 6.6 ´ 10 -21 kg-ms -1 -10 10 10 -10

=

39. We know that Einstein equation, 1 mv 2 2

nA v =2 = B nB vA

The frequency of beam B is twice that of A

l Na ´ (W0) Na 5460 ´ 2.3 = = 2791Å (W0)T 4.5

E = W0 +

Þ

E k 3.18 ´ 10 -19 = 1.99 V = e 1.6 ´ 10 -19

45.

hc hc - W0 and 2E = - W0 l l¢ W ö æ ç1 + 0 ÷ è l ¢ E + W0 E ø = Þ l¢ = l W0 ö æ l 2E + W0 ÷ ç2 + è E ø

W ö æ ç1 + 0 ÷ è E ø 1 Since, > W ö 2 æ ç2 + 0 ÷ è E ø l So, l¢ > 2 hc As, work function, W0 = l0

where l 0 is the threshold wavelength or 1 W0 µ l0 W W0 l ¢0 l ¢ or l 0 ¢ = 2l 0 \ 0 = 0 or = W0 W ¢0 l0 l0 2 hc 12400 46. As, l0 = = = 3100 Å = 310 nm 4 W0

1151

Dual Nature of Radiation and Matter h hc l l E hc / l = = c = 3 ´ 10 8 ms-1 p h/l

47. For electron, p = ; and for photon, E = \

48.

52. Initial de-Broglie wavelength of electron, l0 =

Force on electron in electric field , F = - eE = - eE 0 $j Acceleration of electron, a =

1 3 KE of thermal neutron, mv 2 = kT 2 2

The initial velocity of electron along x-axis v x0 = v 0 $i. Initial

h h l= = p 3kmT

So,

velocity of electron along y-axis, v y 0 = 0 . Velocity of electron after time t along x-axis, v x = v 0 $i

49. Here, v = v 0$i , B = B0$j. Force on moving electron due to magnetic field is F = - e ( v ´ B) = - e [v $i ´ B $j ] = - eV B k$ 0

50.

0

0

0

As this force is perpendicular to v and B , so the magnitude of v will not change, i.e., momentum ( = mv) will remain constant in magnitude. Hence de-Broglie wavelength l = h/mv remains constant. h ...(i) Initial de-Broglie wavelength of electron, l 0 = mv 0

(Q There is no acceleration of electron along x-axis) Velocity of electron after time t along y-axis, eE æ eE ö v y = 0 + ç - 0 $j ÷ t = - 0 t $j è m ø m Magnitude of velocity of electron after time t is æ - eE 0 ö | v | = v x2 + v y2 = v 02 + ç t÷ è m ø

Acceleration of electron a =

de-Broglie wavelength, l¢ =

F eE 0 $i = m m

Velocity of electron after time t, æ eE $i ö v = v 0 $i + ç 0 ÷ t è m ø

=

de-Broglie wavelength associated with electron at time t is l0 h h l= = = é ù é æ mv eE 0 ù eE 0 ö t mêv 0 ç1 + t ÷ ú mê1 + mv 0 úû mv 0 ø û ë ë è [From Eq. (i)]

51. As, E = 3 eV = 3 ´ 1.6 ´ 10

J

h l= 2mE =

1+

l0 2 2 2 2 2 e E 0 t (m v 0)

eE ; v = ?; t = t m eE v = u + at = 0 + t (from equation of motion) m

\

de-Broglie wavelength, l=

6.6 ´ 10 -34 2 ´ 1.7 ´ 10 -27 ´ 3 ´ 1.6 ´ 10 -19

h h h = = mv m( eEt / m) eEt

Rate of change of de-Broglie wavelength, -h dl h æ 1 ö = ç- ÷ = dt eE è t 2 ø eEt 2

54. As, mv =

= 1.65 ´ 10 -11 m

h h = mv mv 1 + e2E 2 t 2/ (m2 v 2) 0 0 0

53. Here, u = 0; a =

æ æ eE 0 ö $ eE 0 ö $ = çv 0 + t ÷ i = v 0 ç1 + t÷ i è è mv 0 ø mv 0 ø

-19

2

æ e2E 2 t 2 ö = v 0 1 + ç 20 2 ÷ è m v0 ø

Force on electron in electric field,F = - eE = - e [ -E 0 $i ] = eE 0 $i

\

eE F = - 0 $j m m

It is acting along negative y-axis.

mv = 3kmT

or

h mv 0

or

h l v=

(de-Broglie equation) 6.6 ´ 10 -34 h = ml 9.1 ´ 10 -31 ´ 5200 ´ 10 -10 = 1.4 ´ 10 3 ms-1

1152 JEE Main Physics

Round II h p

E k = eV = hn - f0

1. As,

V=

or

h f n- 0 e e

h e h = e ´ slope of straight line.

1 2

Vs =

or

2. According to Einstein’s photoelectric equation , E = W0 + Kmax hc é 1 1ù V0 = ê ú e ë l l0 û

It gives the value of l 0 and then we find the value of new stopping potential.

3.

Then, if l decreases V0 increases. 1 As, EK = mv 2 or mv = 2mEK 2 As per question; mpv p = mev e 2mpEK p = 2meEK e

or

EK e

or

EK p

=

mp me

>1

EK e > EK p 1 hc 2 As, mv = - f 0 (in eV) 2 l

or

4.

=

6.6 ´ 10 -34 ´ 3 ´ 10 8 - 4.2 2000 ´ 10 -10 ´ 1.6 ´ 10 -19

= 2 eV = 2 ´ 1.6 ´ 10 -19 J v = 2 ´ 2 ´ 1.6 ´ 10 -19 / 9.1 ´ 10 -31

\

=

5. Here, l0 = 200 min; l = 100 nm; hc = 1240 eV-nm e hc hc Maximum KE = (in eV) le l 0 e hc æ 1 1ö = ç ÷ e è l l0 ø 1 ö æ 1 = 1240 ç ÷ = 6.2 eV è100 200 ø

= =

h h = mv 2meh 6.6 ´ 10 -34 2 ´ ( 4 ´ 1.66 ´ 10 -27) ´ (2 ´ 1.6 ´ 10 -19) ´ V 0.101 V

2 2 mv m vm = 2e 2( e / m)

( 4.8) 2 = 7 ´ 10 11JC-1 2 ´ 1.76 ´ 10 -11

9. We know, KA =

hc - f0 lA hc - f0 lB hc KA 2lB 1 = < hc KB 2 lB KB =

and

\

KA < KB / 2

or

10. As,

lp la

=

h 2empV 2 ´ 2e 4mpV

=2 2 3 2

11. Kinetic energy of a particle at temperature T K is, E = kT The de-Broglie wavelength associated with it is, h h l= = 2mE 3 2m ´ kT 2 1 lµ i. e. , T \

300 1 l 927 27 + 373 = = = 1200 2 l 27 927 + 273

or

l 927 =

= 6.7 ´ 10 6 ms-1

6. As, l =

1 p

8. As, eVs = mv m2

Slope of straight line between V and n is

Þ

h p

7. As, l = or L = i. e. ,L µ × The curve (d) is correct.

12. As, l = where,

l 27 l = 2 2

h h 1 Þ lµ = 2m0E 2m0qV m0 m0 = mass of the charge

Þ

l1 M = l2 m

Þ

l 2 = l1

or

l2 = l

m M m M

(Q l1 = l )

13. Given, power of lamp, P = 100 W Wavelength of the sodium light, l = 589 nm = 589 ´ 10 –9 m Planck constant h = 6 .63 ´ 10 -34 J-s

Dual Nature of Radiation and Matter (i) Energy of each photon hc 6.63 ´ 10 -34 ´ 3 ´ 10 8 (Q c = 3 ´ 10 8 m/s) E= = l 589 ´ 10 –9 = 3.38 ´ 10 -19 J =

3.38 ´ 10 -19 eV 1.6 ´ 10 - 19

= 2.11 eV (ii) Let n photons are delivered per second. Power (From P = En) n= \ Energy of each photon 100 = = 3 ´ 10 20 photon/s 3.38 ´ 10 -19 = 3 ´ 10 20 photon/s are delivered

14. As, E = \ or Now,

hc 1 or E µ l l E 2 l1 = E1 l 2

hn - f0 =

1 2 mv max = eVs 2

Momentum of the reflected light = 0; as the light is completely absorbed. E E Force exerted by light, F = - 0 = c c F E /c Pressure on surface, p = = 4 pr 2 4 pr 2 =

= 2.46 - 1.36 = 1.10 eV 1 1 1 h 15. As, E e = mv 2 = (mv)v = æç ö÷ v 2 2 2 è lø hc and Ep = ; l Ee v = \ E p 2c or

Now,

pe = mv =

and

ph =

\

pe =1 ph

66 / (3 ´ 10 8) 4 ´ (22 / 7) ´ (0.10) 2

= 1.75 ´ 10 -6 Pa

18. Here, lth = 5200 Å, Thus, wavelength less than 5200 Å cannot produce the photoelectric effect. 1 l

\Energy of a photon = hn =

hc = hc n l

20. Velocity of electron, v = h / (ml) Let h = 6.6 ´ 10 -9 Js and m = 9 ´ 10 -31 kg.

f 0 = hn 2 - eVs = E s - eVs

h l

h l

16. KE of photoelectrons increases with increase in frequency of the incident light and is independent of the intensity of incident light. Photoelectrons are emitted if the wavelength of the incident light is less than threshold wavelength, as hc f0 = l0 Photoelectric emission is an instantaneous process photoelectrons may not be emitted from a gas with ultraviolet light if the work-function of that gas is larger than the energy of UV light.

17. Light falling per second on the surface of sphere E=

Momentum of the light falling per second on the surface of E sphere = c

19. As, wave number, v =

10000 l E 2 = E1 ´ 1 = 1.23 ´ = 2.46 eV 5000 l2

1153

66 ´ 100 = 66 W 100

(a) When l1 = 10 nm = 10 ´ 10 -9 m = 10 -8 kg v1 =

6.6 ´ 10 -34 2.2 = ´ 10 5 = 10 5 m/s (9 ´ 10 -31) ´ 10 -8 3

(b) When l 2 = 10 -1 nm = 10 -1 ´ 10 -9 m = 10 -10 kg. v2 =

6.6 ´ 10 -34 » 10 7 m/s (9 ´ 10 -31) ´ 10 -10

(c) When l3 = 10 -4 nm = 10 -4 ´ 10 -9 m = 10 -13 m v3 =

6.6 ´ 10 -34 » 10 10 m/s (9 ´ 10 -31) ´ 10 -13

(d) When l 4 = 10 -4 nm = 10 -4 ´ 10 -9 m = 10 -13 m v4 =

6.6 ´ 10 -34 » 10 12 m/s (9 ´ 10 -31) ´ 10 -15

As v3 and v 4 are greater than velocity of light ( = 3 ´ 10 8 m/ s), hence relativistic correction is needed for l = 10 -4 nm and l = 10 -6 nm.

21. The saturation current depends upon the intensity of the incident light and not on energy of the incident light. The intercept of straight line. In a graph between E k and n or negative energy axis given the value of would function of cathode metal. The point where the straight line out the frequency gives the value of threshold frequency whereas the slope of straight line can help to find the Planck’s constant.

22. As, E k = Kv 0 =

hc - f0 l

When, l will decease, v 0 and E k will decrease.

1154 JEE Main Physics h p

h l

24. As l = or p = \

p1 l 2 l = = = 1 or p1 = p2 p2 l1 l

E=

1 p2 1h 2 = 2 m 2m l2

2

TA = 2.0 eV f A = 4 - TA = 4 - 2 = 2.0 eV fB = 6 - TA = 6 - 2 = 4. 0 eV hn hc = 1 eV = - f0 = - f0 l 400 ´ 10 -9

On solving,

h 1 Þ l µ if n is same mv m h 1 so, l µ if, E is same. (i) Also l = 2mE m (ii) When two bodies fall from a certain height both acquire equal velocity (v) so, l will be less for heavier particle. h h 100h …(i) For electron, l e = = = me v e me( c / 100) mec

\

25. As, l =

26.

1 mev e2 or mev e= 2E e me 2 h h h or E e = 2 2 le = = mev e 2me E e l e me

29. E max

f 0 = 1.9 eV

As

hc = 400 ´ 10 -9 ´ (1 + 1.9)

Hence, E max =

\

Kinetic energy, E e =

For photon of wavelength l p , energy = E p =

l max =

…(ii)

30. Maximum kinetic energy

hc 2l2e me l emec 100h mec = ´ = = ´ = 100 E e 2le h2 h mec h

Kmax = hn - f0 = eV hn - f0 V= e

\

31. As, Kmax = hn - f0, when n = n 0 , Kmax = 0

Ee 1 = = 10 -2 E p 100

0 = hn 0 - f0

For electron, pe = mev e = me ´ c /100 pe 1 = = 10 -2 \ mec 100

27. As, l = or or

h l E ; so, B = A lA EB 2 mE

32. As, l =

12.27 Å 12. 27 = = 1.534 Å V 64 h p

EA or E A = 4EB EB

33. As, l = and pa = pp , so la = lp

EA = E A - 1.5 or E A = 2. 0 eV 4

34. As, qV = mv 2 or mv = 2qVm

2=

1 2

\

l=

or

lµ

fB = 4.70 - 0.50 = 4.20 eV

28. As, E = f 0 + (KE) max 4 = f A + TA

4.5 = fB + (TA - 1.5) or 6 = fB + TA From Eqs. (i) and (ii), fB - f A = 2 According to the de-Broglie hypothesis, h h lA = = mv 2mTA

and

f0 = hn 0

or

If n < n 0 , then Kmax is negative, i. e. , no photoelectric emission takes place. Thus, graph (c) is possible.

f A = 4.25 - 2.00 = 2.25 eV

\ and

hc 6.6 ´ 10 -24 ´ 3 ´ 10 8 = f0 1.9 ´ 1.6 ´ 10 -19

= 650 ´ 10 -9 m = 650 nm

hc hc = lp 2 le

Ep

\

400 ´ 10 -9 ´ 2.9 eV - 1.9 eV 500 ´ 10 -9

= 2.32 eV - 1.9 eV = 0.42 eV

(Q l p = 2l e) \

lB =

h 2mTB

1/ 2

1.5 æ 1ö ç ÷ =1è2ø TA

1 E m E µ \ 1 = 2 < 1 or E1 < E 2. 2 E 2 m1

So,

lA T T - 1.5 æ 1.5 ö = B = A = ç1 ÷ è lB TA TA TA ø

\

…(i) …(ii) …(iii)

\

h h = mv 2q Vm 1 qm

q p mp la = lp q ama =

1 1 1 ´ = 2 4 2 2

35. When work function of copper is greater than the work function of sodium, then f Cu > f Na (hn 0) Cu > (hn 0) Na c But we know that, n 0 = lc

…(i)

Dual Nature of Radiation and Matter Hence, Eq. (i) becomes, æ hc ö æ hc ö ç ÷ >ç ÷ è l 0 ø Cu è l 0 ø Na Þ

41. As, E k =

36. de-Broglie wavelength associated with gas molecules varies

v rms

3 RT = M

37. It is fact that, greater is KE of photoelectron, greater is the potential required to stop it. Hence, stopping potential is a measure of KE of photoelectron. It can be understood from the relation, eVs = KE or

Vs = KE (in eV)

l £ l0 hc 1240 eV nm As, f0 = - Ek = - 1.68 eV l 400 nm or

as 1 T Also, root mean square velocity of gas molecules is,

hc hc l l0

E k is positive, i. e. , photoelectric emission will take place if, hc hc ³ l l0

( l 0) Na > ( l 0) Cu

lµ

42.

= 3.1 - 1.68 = 1.42 eV

43. Matter has a wave nature that is best supported by the phenomenon of electron diffraction.

44. Relativistic energy is given by

hn E = mc = hn Þ m = 2 c Therefore, momentum of photon hn hn = mc = 2 ´ c = c c Thus, photon possessed momentum due to its equivalent mass even its rest mass is zero.

39. Velocity of first photon = u = c Velocity of second photon = v = - c Now, relative velocity of first photon with respect to second photon u -v c - ( - c) = = ( c) ( - c) uv 1- 2 1c c2 2c 2c 2c = = = =c c2 1 + 1 2 1+ 2 c Also the rest mass of photon is zero.

40. Photoelectric effect is bases upon quantum theory of light or particle nature of light. The number of photoelectrons emitted is proportional to intensity of incident light. It does not depend on frequency of light.

m0 c2

E=

1 - v 2 / c2

or

E2 =

m02c4 1 - v 2 / c2

or

E2 =

m02c6 c2 - v 2

38. Equivalent mass of photon (m) is given by 2

1155

(i)

Momentum is given by p= E2 =

or

m0v 1 - v 2 / c2 m02v 6 c2 - v 2

\

E 2 - p 2c2 = m02c4

or

E 2 = p 2c2 + m2c2

For photon, rest mass m0 = 0 , so, E = pc For electron m0 ¹ 0 , so, E ¹ pc

45. For constructive interference, and from Bragg’s equation; 2d sin i = nl =

h , on substituting values, we get 2 2mV

V = 50 volt

46. Expression is given by, 2d sin i = nldB 47. As, diffraction pattern has to be wider, then short width. So, (d) is correct option.

48. The momentum of the photon, p =

h hn = l c

Electronic 26 Devices JEE Main MILESTONE
> ih Conductivity, s = ne m ee, where m e is the mobility of electron.

1158 JEE Main Physics Donor Energy Levels in n-type Semiconductor

Acceptor Energy Levels in p-type Semiconductor

When the impurity of antimony is introduced, then each atom of germanium has an extra electron. The energy of these electrons is less than the least energy of conduction band and is greater than the highest value of energy of valence band. Thus,the electrons form a specific energy level below donor energy level.

When a small quantity of trivalent atom is introduced into pure germanium, the holes are formed. These holes form a specific energy level above the valence band, called acceptor energy level. Minimum acceptor energy level is 0.08 eV.

Conduction band Donor energy gap

0.01 eV

Valence band

Ef

Minimum donor energy level is 0.05 eV for Si and 0.01` eV for Ge. By giving this much amount of energy to the electron they become free and go to the conduction band. This energy level is very near to the conduction band.

(ii) p-type Semiconductors When a small quantity of a trivalent impurity such as indium (In). boron (B) aluminium (Al) etc., having three valence electrons is introduced into the pure germanium, then such type of semiconductors are called p-type or acceptor type semiconductors. The three valence electrons of indium atom form covalent bonds with the valence electrons of three neighbouring germanium atoms and their remains a lack of one electron (or deficiency of one electron). This lack of one electron is called the hole. This hole soon captures an electron from its neighbouring germanium atom and a hole is created in this neighbouring atom. This hole is equivalent to a positively charged particle. Since, positive holes are responsible to increase the conductivity in this crystal, the crystal so obtained is called the p-type crystal and the impurity atom is called acceptor impurity. Obviously, the density of holes is equal to the density of acceptor ions, i. e. , nh = na. For a p -type semiconductor nh >> ne; ih >> ie . Conductivity, s = nh m h e, where m h is the mobility of holes. Hence, the conductivity of semiconductor. s = e [ne m e + nh m h] +4 Ge +4 Ge

Acceptor energy level Er Valence band

0.08 eV

Sample Problem 1 Determine the number of donor atoms which have to be added to an intrinsic germanium semiconductor to produce an n-type semiconductor of conductivity 5 W-1 cm -1 given that the mobility of electrons in n-type Ge is 3900 cm 2V -1s -1. Neglect the contribution of holes to conductivity. Take charge on electron, e = 16 . ´ 10 -19 C. (a) 8. 013 ´ 10 21 m-3 (b) 6.2 ´ 10 20 m-3 (c) 5.3 ´ 10 19 m-3 (d) 4.8 ´ 10 18 m-3

Interpret (a) Here, e = 1.6 ´ 10 -19 C; s = 5 W -1 cm-1 = 500 W -1 m-1; m e = 3900 cm2 V -1s-1 = 0.39 m2V -1s-1 Now, s = ene m e (neglecting contribution of holes) s 500 \ ne = = em e 1.6 ´ 10 -19 ´ 0.39 = 8.013 ´ 10 21 m-3

Sample Problem 2 A pure Si crystal has 5 ´ 10 28atoms m –3. It is doped by 1 ppm concentration of pentavalent As. The number of holes is (approx). (Given ni = 1.5 ´ 1016 m -3) (a) 2.2 ´ 10 6 m-3

(b) 4.5 ´ 10 9 m-3

(c) 6.2 ´ 10 6 m-3

(d) 8.1´ 10 9 m-3

[NCERT]

Interpret (b) Note that thermally generated electrons (ni ~10 16 m-3 ) are negligibly small as compared to those produced +4 Ge

+3 In Hole

+4 Ge

Conduction band

by doping. Therefore, ne » N0 . Since, ne hh = ni2, the number of holes nh = (2.25 ´ 10 32) / (5 ´ 10 22) nh ~4.5 ´ 10 9 m–3

Electronic Devices

26.2 Semiconductor Diode When a p -type semiconductor is brought into a close contact with n-type semiconductor crystal, the resulting arrangement is called a p-n junction or junction diode. In the p -type semiconductor, the holes are the majority carriers and electrons are minority carriers whereas in n-type semiconductor, the electrons are majority carriers and holes are minority carriers. Junction p

(a)

1159

increases with the migration of more charge carriers across the junction. This electric field opposes further flow of electrons from the n-region to the p-region and that of holes from the p-region to n-region. This electric field sets a potential barrier VB at the junction which opposes further diffusion of free charge carriers into opposite regions. In the vicinity of the junction, a region is created, which is devoid of free charge carriers and has immobile ions. This region in which no free charge carriers are available is called a depletion region. It is like a no man land on a border. Figure represents the potential distribution near the junction. This potential acts as a barrier, hence, known as potential

n

Conduction band

barrier.

Fermi level Fermi level (b)

(c)

Valence band

Unneutralised acceptor atom – – p-type – –

+

–

+

Hole

–

Acceptor atom (neutral)

+

Unneutralised donor atom + n-type +

There are two methods of biasing the p-n junction.

1. Forward Biasing A p-n junction is said to be Electron

+

Extent of Extent of negative positive space space charge charge region region on p-type on n-type side side

Biasing of the p-n Junction

Donor atom (neutral)

Depletion layer

Flow of charges across semiconductor diode

On account of difference in concentration of charge carriers in the two sections of p-n junction, the electrons from n-region diffuse through the junction into p-region and the holes from p-region diffuse into n-region. Since, the hole is a vacancy of an electrons, when an electron from n-region diffuses into the p-region, the electron falls into the vacancy i. e. , it completes the covalent bond. (This process is called electron-hole recombination). Due to migration of charge carriers across the junction, the n-region of the junction will have its electrons neutralized by holes from the p-region, leaving only ionized donor atoms (positive charges) which are bound and cannot move. Similarly, the p-region of the junction will have ionized acceptor atoms (negative charges) which are immobile. The accumulation of electric charges of opposite polarities in the two regions of the junction gives rise to an electric field between these regions as if a fictitious battery is connected across the junction with its positive terminal connected to n-region and negative terminal connected to p-region. The strength of electric field across the junction

forward biased if the positive terminal of the external battery B is connected to p-side and the negative terminal to the n-side of p-n junction Fig. (a). VB

p +

+

++

+

n + + +

+

a Depletion Layer + – B + – b +

– B Forward biasing of p-n junction diode

The circuit diagram for forward biasing of p-n junction is shown in Fig. (b). In forward biasing, the forward bias voltage opposes the potential barrier VB. due to it, the potential barrier is considerably reduced, the depeletion region becomes thin. The majority carriers, electrons in the n-region are repelled by the negative potential due to battery B and move towards the p-n junction. Similarly, the majority carriers, holes in the p-region are repelled by the positive potential, towards the junction. The positive potential of p-region attracts the electrons from the n-region and negative potential of n-region attracts the holes from the p-region. On crossing the junction, the number of the electrons and holes will combine with each other. For each electron hole

1160 JEE Main Physics combination, a covalent bond near the positive terminal of the battery is broken and the liberated electron enters the positive terminal of the battery B through lead wires. This action results in a new hole, which under the force of applied voltage moves towards the p -n junction. At the other end, the electrons from the negative terminal of the battery enter the n-region to replace the electrons lost due to the combination with the holes at the junction. Thus, an electric current will flow due to migration of majority carriers across the p-n junction; which is called forward current. Since, the small increase in forward voltage shows the large increase in forward current hence, the

I-V characteristics of forward and reverse bias The following circuit diagram shows the arrangement for studying V -I characteristics of a p-n junction diode. (a) Forward bias Voltmeter (V)

p

n Milliammeter (mA) Switch

+

–

(b) Reverse bias

resistance of p-n junction is low to the flow of current

Voltmeter (V)

when forward biased.

2. Reverse Biasing A p-n junction is said to be reverse biased if the positive terminal of the external battery B is connected to n-side and the negative terminal to p-side of the p-n junction, Fig. (a). The circuit diagram for reverse biasing of pn junction is shown in Fig. (b). p +

VB

Switch –

+

(c) Typical V-I characteristics of a silicon-diode I (mA)

+ + +

+

n Microammeter (µA)

n +

+

p

+

a Depletion Layer – +

100 80 60 40 20 100 80 60 40 20

Forward bias

0.2 0.4 0.6 0.8 1.0

Vbr

B

Reverse bias b B Reverse biasing of p-n junction diode

In reverse biasing, the reverse bias voltage supports the potential barrier VB. Now the majority carriers are pulled away from the junction and the depletion region becomes thick. There is no conduction across the junction due to majority carriers. However, a few minority carriers (holes in n-section and electrons in p-section) of p-n junction diode cross the junction after being accelerated by high reverse bias voltage. they constitute a current that flows in the opposite direction. this is called reverse current of leakage current. Since, the large increase in reverse voltage shows small increase in reverse current, hence, the resistance of p-n junction

is high to the flow of current when reverse biased.

V (Volt)

10 20 30

I (µA)

Diode Resistance (i) Static or DC resistance of diode In forward biasing, the ratio of potential applied to the junction diode to the current corresponding to it, is called static or DC resistance of diode. Thus, rDC =

V i

Its unit is ohm (W ).

(ii) Dynamic or AC resistance of diode In forward bias, the reciprocal or the slope of characteristic curve is called dynamic or AC resistance of diode. Thus, Its unit is ohm (W) .

rAC =

DV Di

Electronic Devices

1161

Transformer

Important Points

A

1. The curves for charge density, electric field and potential barrier for a

x

Primary RL

Secondary

p-n junction are shown in figure. n p –

+

y

B Half-wave rectifier circuit

Charge density

Input Voltage at A

Electric field

Input AC time

Voltage across RL

Output voltage

2. Figure shows the characteristic curve plotted between the current i and the potential differenceV for the p-n junction diode. It may be mentioned here that in the forward bias, the current is of the order of milliampere while in the reverse bias, the saturated currenti s is of the order of microampere. +

i Forward bias

Zener breakdown voltage B

0 is

V

+

Zener break down Reverse bias

time Input AC voltage and output voltage waveforms from the rectifier circuit.

The no-load output DC voltage of an ideal half-wave rectifier is Vpeak Vrms = 2 Vpeak VDC = p where, VDC, Vav -the DC or average output voltage Vpeak -the peak value of the phase input voltage Vrms -the root-mean-square value of output voltage.

26.3 Diode as a Rectifier Junction diode allows current to pass only when it is forward biased. So, if an alternating voltage is applied across a diode, the current flows only in that part of the cycle, when the diode is forward biased. This property is used to rectify alternating voltages and the circuit used for this purpose is called a rectifier, and the process is known as rectification.

Half-Wave Rectifier In half-wave rectifier of the single phase supply, either the positive or negative half of the AC wave is passed, while the other half is blocked. Because only one-half of the input waveform reaches the output, mean voltage is lower. Half-wave rectifier requires a single diode in a single phase supply. Half-wave produce for more ripple than full-wave rectifiers, and much more filtering is needed to eliminate harmonics of the AC frequency from the output.

Full-Wave Rectifier Like the half wave circuit, a full-wave rectifier circuit produces an output voltage or current which is purely DC or has some specified DC component. Full wave rectifiers have some fundamental advantages over their half-wave rectifier counterparts. The average (DC) output voltage is higher than for half wave, the output of the full wave rectifier has much less ripple than that of the half wave rectifier producing a smoother output waveform. In full-wave rectifier circuit, two diodes are used, one for each half of the cycle. A multiple winding transformer is used whose secondary winding is split equally into two halves with a common centre trapped connection (C). This configuration results in each diode conducting in turn when its anode terminal is positive with respect to the transformer centre point C producing an output during both half cycles, twice that for the half-wave rectifier, so it is 100% efficient.

1162 JEE Main Physics Transformer

Important Points D1

A

C

AC input

even 1/2 cycle RL

Current flows when D1 conducts

odd 1/2 cycle

Current flows when D2 conducts

B

1. Various parameters for half-wave rectifier (a) i =

V 0 sin( wt + f) RL + RF

Where RF is forward resistance of junction. In forward biasing, RF > p. Hence, the fractional change in the æ Dn ö majority carriers ç ÷ would be much less than that in the minority è n ø æ Dp ö carriers ç ÷ × In general, the fractional change due to the photo è p ø effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the forward bias current. Hence, photodiodes are used in reverse bias.

–10 V 1 µA

(a) 10 W (c) 10 12 W

0.5 0.8 V(volt)

(b) 20 W (d) 10 7 W

Interpret (d) Considering the diode characteristics as a straight line between I = 10 mA to I = 20 mA passing through the origin, the resistance can be found using Ohm’s law. From the curve, V = - 10 V, I = - 1mA 10 V \ rrb = =1 ´ 10 7 W =10 7 W 1mA

26.5 Junction Transistor A transistor is a small device constructed from p-and n-type silicon (or germanium) semiconductors which can be used as a triode valve and transfers the resistance (because transistor = trans + resistance i. e. , transfer of resistance). In junction transistor, there are three semiconductor surfaces which are known as emitter, base and collector. The middle portion is called base, which is very thin of the order of 10-6 m. The other two are known as emitter and collector.

Electronic Devices Types of Transistors (i) p-n-p transistor The circuit of p-n-p transistor and symbol are shown in Fig. (a) and (b) respectively. Left handed emitter base (p-n) junction (forward biased) is of low resistance in p-n-p transistor and right handed base collector (n-p) junction (reverse biased) is of high resistance. p-n-p E p

B n

C p

C

E

E → Emitter B → Base C → Collector

(a)

ie = ib + ic

(b)

(ii) n-p-n transistor The circuit of n-p-n transistor and symbol are shown in Fig. (a) and (b) respectively. E n

n-p-n

B p

C n

E

C

B –

–

+

Due to this, the probability of electron-hole combination in base region is very small (» 5%). Most of the electrons (» 95%) cross into collector region, where they are swept away by the positive terminal of the battery Vcb connected to the collector. Corresponding to each electron that is swept by the collector and that enters the positive pole of the collector-base battery Vcb an electron enters the emitter from the negative pole of the emitter-base battery Vbe. Thus, in n-p-n transistors, the current is carried inside the transistor as well as external circuit by the electrons. If ie, ib and ic are respectively the emitter current, base current and collector current, then

B

+ –

+ –

+

(a)

1165

(b)

It may be pointed out that the arrows point in the direction of conventional current or hole current inspite of the fact that in the n-p-n transistor, the current is carried by electrons.

2. Action of p-n-p transistor The p-type emitter of a p -n- p transistor is forward biased by connecting it to positive pole of emitter base battery Vbe and the p-type collector is reverse biased by connecting it to the negative pole of the collector base battery Vcb as shown in figure.

In this left handed junction is forward biased and right handed junction is reverse biased.

ie

ic

p-n-p C

E

Transistor Action The action of both the types of transistors i. e. , n- p-n and p-n-p is similar, except that the majority and minority carriers in the two cases are of opposite nature.

1. Action of n-p-n transistor

Figure shows the biasing of an n-p-n transistor. The n-type emitter is forward biased by connecting it to negative pole of the battery Veb (emitter-base battery) and n-type collector is reverse biased by connecting it to the positive pole of the battery Vcb (collector-base battery). ie

ic

n-p-n c

Ib – + Vbe Forward Bias

B – + Vcb Reverse Bias

The majority carriers (which are electrons) in the emitter are repelled towards the base due to the forward bias. The base contains holes as majority carriers but their number density is small as it is doped lightly as compared to emitter or collector.

ib +

–

B +

Veb Forward Bias

–

Vcb Reverse Bias

In this case, majority carriers in emitter are holes and they are repelled towards the base due to the forward bias. As base is thin and lightly doped very small as compared to collector and emitter, it has a low number density of electrons. When holes enter the base region, then only about 5% electron-hole combination takes place. Most of the holes = 95% reach the collector under the influence of reverse bias. As one hole reaches the collector, an electron leaves the negative pole of collector-base battery VB and combines with it. At the same time, an electron is released from some covalent bond in the emitter, creating a hole in the emitter. The electron released, enters the positive pole of the emitter-base battery Veb. Thus, current in p-n-p transistor is carried by holes and at the same time their concentration is maintained as explained above. In this case also, ie = ib + ic

1166 JEE Main Physics Transistor as an Amplifier The amplifier is a device which amplifies the power of varying current or alternating current. In other words, the amplifier is a device which changes the weak varying alternating current or voltage into strong varying alternating current or voltage.

consequence, input characteristics for various values of VCE will give almost identical curves. Curves IB /µA 100 80 70

As an amplifier, a transistor can be used in the following three configurations :

60 40 20

(i) Common base amplifier (ii) Common emitter amplifier

Characteristics of a Transistor The input and output characteristics of an n-p-n transistors are as follows. To study the input characteristics of the transistor in CE configuration a curve is plotted between base-current IB against the base, emitter

0.2 0.4 0.6 0.8 1.0

VBE / V

Whereas the output characteristics is obtained by observing the variation of I C as VCE is varied, keeping I B constant. The plot of I C versus VCE for different fixed values of I B gives one output characteristics. Base current IB

Collector current (Ic) in mA

(iii) Common collection amplifier

VCE = 10 V

voltage VBE . The collector emitter voltage VCE is kept fixed while studying the dependence of I B on VBE . Since, the transistor is operated as an amplifier over large range of VCE , the reverse bias across the base collector junction is high most of the time. Since, the increase in VCE appears as increase in VCE , its effect on I B is negligible. As a

60 µA 50 µA 40 µA 30 µA 20 µA 10 µA

Collector to emitter voltage (VCE) in volts

Hot Spot TRANSISTOR AS AN AMPLIFIER (CE Configuration) Common emitter amplifier The given circuit diagram shows a n-p-n transistor which has been used in common-emitter mode. The input and output waveforms are shown along with the circuit. Current gain in CE configuration, at constant collector voltage ratio of change in collector current gain to the change in base current is called current gain. It is denoted by b. æ Di ö b = ç C ÷ = constant è DiB øVC

\

The value of b is always greater than1.

Power gain The ratio of change in output power to the change in input power is called power gain. \

P Power gain = out Pin

Also, power gain = b2 ´ Resistance gain

The relation between a and b is given by b =

Resistance gain The ratio of output (load) resistance to the input resistance is called resistance gain. \

R Resistance gain = out R in

Voltage gain The ratio of change in output voltage to the change in input voltage is called voltage gain. It is denoted by AV . Voltage gain, Þ

AV =

DVC Di R = c ´ out DVb Dib R in

A V = b ´Resistance gain

a b or a = 1- a 1+ b

Sample Problem 8 In a common emitter amplifier, the low resistance of the output circuit is 500 times the resistance of the input circuit. If a = 0.98, then find the voltage gain and the power gain. (a) 23500, 1300500 (c) 22505, 1300500

(b) 24500, 1200500 (d) 23500, 1200500

Interpret (b) Given, a = 0.98 and

Rout = 500 Rin

Electronic Devices We have, current gain, b=

iB =

a 0.98 = = 49 1 - a 1 - 0.98

\

= ( 49) (500) = 24500 æR ö Power gain = (b) 2 ç out ÷ è Rin ø

10 ´ 0.010 = 0.10 mA (V - VBE ) RB = BB , VBE = 0.6 V IB RB =

\

= ( 49) 2 (500) = 1200500

(2 - 0.6) = 14 k W 0.10

Sample Problem 10 In a CE transistor amplifier, the

Sample Problem 9 For a CE transistor amplifier, the audio signal voltage across the collector resistance of 2 k W is 2 V. Suppose the current amplification factor of the transistor is 100. The value of RB in series with VBB supply of 2V, if the DC base [NCERT] current has to be 10 times the signal current is (b) 14 k W (d) 54 k W

output resistance in 500 kW and current gain b = 49.If the power gain of the amplifier is 5 ´ 106 , the input resistance is (a) 240 W (c) 180 W

(b) 165 W (d) 290 W

Interpret (a) Given, 500 kW, b = 49 and P = 5 ´ 10 6 P = b2

We have,

Interpret (b) The output AC voltage is 2 V. So the AC collector

5 ´ 10 6 =

current. ic =

ic 1 mA = = 0.010 mA b 100

The DC base current has to be

æR ö Voltage gain = (b) ç out ÷ è Rin ø

(a) 4 k W (c) 28 k W

1167

2 = 1mA 2000

Ri =

\

The signal current through the base is therefore, given by

26.6 Transistor as an Oscillator When the key K is closed, collector current begins to increase. The magnetic flux linked with ticker coil L¢ and hence, with L also begins to increase. This supports the forward bias of the base-emitter circuit. As a result of this, the emitter current increases. Consequently, the collector current also continues to increase till it attains it saturation value.

R0 R2

( 49) 2 ´ 500 Ri 49 ´ 49 ´ 500 5 ´ 10 6

= 240 W

When the capacitor begins to discharge through inductance L, the emitter current and hence, the collector current begin to decrease. Consequently, the magnetic flux linked with the ticker coil L¢ and hence, with the tank coil L decreases. The forward bias of the emitter-base circuit is opposed thereby further reducing the emitter and the collector currents. This process continues till the collector current becomes zero. At this, the mutual induction has once again no role to play. The frequency of oscillations is given by 1 n= 2 p LC

L′ C B

n-p-n E

L

B2

+ –

Sample Problem 11 When a transistor is used in common emitter configuration, the following output characteristics are obtained. When VCE is 10 V and IC is 4 mA, then the value of b AC (current amplification factor) is

C B1 + –

Collector current (IC) in mA

L′′ K

When the collector current attains its saturation value, the magnetic flux linked with the ticker coil L¢ attains a steady value. So, mutual inductance has no role to play.

60 µA 50 µA 40 µA 30 µA 20 µA 10 µA

Collector to emitter voltage (in volts)

(a) 50

(b) 100

(c) 150

(d) 200

1168 JEE Main Physics Interpret (c) Current amplification factor (b) is

26.7 Logic Gates

æ DI ö b AC = ç C ÷ è DIB ø V

CE

For finding the value of b AC at the stated values, take any two characteristics of IB which lie above and below the given value of IC . Here, IC = 4 mA (choose characteristics for IB = 30 A and 20 m A) At VCE = 10 V, we read the two values of IC from the graph, then DIB = (30 - 20) mA = 10 mA DIC = ( 4. 5 - 3) mA = 1. 5 mA b AC = 1. 5 mA/10 mA = 150

\

Sample Problem 12 In the given figure, the VBB supply can be varied from 0 V to 5V. The Si transistor has bDC = 250 and RB = 100 k W, RC = 1k W, VCC = 5 V. Assume that, when the transistor is saturated, VCE = 0 V and VBE = 0.8 V. The minimum base current for which the transistor will reach saturation is [NCERT] IC C

RB IB Vi

VBB

RC

A logic gate is an elementary building block of a digital circuit. Most logic gates have two inputs and one output. At any given moment, every terminal is in one of the two binary conditions low (0) or high (1), represented by different voltage levels. The logic state of a terminal can, and generally does, change often, as the circuit processes data. In most logic gates, the low state in approximately zero volts (0V), while the high state is approximately five volts positive (+ 5V). There are seven basic logic gates. AND, OR, NOT, XOR, NAND, NOR and XNOR. The basic logic gates are of three types

The OR Gate The OR gate is a davice has two or more inputs and one output. this devices combines two inputs to give one output. The logic symbol of OR gate is

+

A

V0 –

B

E VCC

Y

The Boolean expression for OR gate is Y = A+ B This indicates Y equals A OR B.

(a) 10 mA (c) 30 mA

(b) 20 mA (d) 40 mA

Truth table for OR gate (Y = A + B) A

B

Y

Interpret (b) At saturation, VCE = 0 V

0

0

0

VBE = 0.8 V

0

1

1

VCE = VCC - IC RC 5V V 0IC = CC = = 5 mA RC 1 kW

1

0

1

1

1

1

\

and

5 mA I - 20 mA IB = C = b 250

Note The input voltage at which the transistor will go into saturation is given by

V1H = VBB = IBRB + VBE = 20 mA ´ 100 kW + 0. 8 V = 2. 8 V The value of input voltage below which the transistor remains cut- off is given by VIL = 0. 6 V, V1H = 2.8 V Between 0.0 V and 0.6 V, the transistor will be in the switched off state. Between 2.8 V, and 5 V it will be in switched on state. One more interesting observation is that the transistor is in active state when IB varies from 0.0 mA to 20 mA. In this range, IC = bIB is valid. In the saturation range, IC £ bIB.

The output of an OR gate assumes 1 if one of more inputs assume 1. The output is high when either of inputs A or B is high, but not if both A and B are high.

The AND Gate The AND gate a device has also two or more inputs and one output. The logic symbol of AND gate is given as under. The Boolean expression for AND gate is Y = A × B, this indicates Y equals to A AND B.

Truth table for AND gate ( Y = A × B ) A

B

Y

0 0 1 1

0 1 0 1

0 0 0 1

The output of an AND gate is 1 only, when all the inputs assume 1.

1169

Electronic Devices The logic symbol of NOR gate is shown as

The NOT Gate The NOT gate is a device which has only one input and only one output. The logic symbol of NOT gate is as shown in figure.

A

y=A+ B

B

The Boolean expression for NOR gate is Y = A + B, which A

Y

indicates that ‘A OR B are negated’

Truth table for NOR gate

The Boolean expression for NOT gate is Y = A, which indicates Y equals NOT A.

Truth Table for NOT gate (Y = A)

A

B

Y¢

Y

0

0

0

1

1

0

1

0

A

Y

0

1

1

0

0 1

1 0

1

1

1

0

The output of a NOT gate assumes 1, if input is 0 and vice-versa. These basic gates (OR, AND and NOT) can be combined in various ways to provide large number of complicated digital circuits. Few combinations of gates are given as under.

The NOR gate and NAND gate can be said to be universal gates since combinations of them can be used to accomplish any of the basic operations and can thus produce an inverter, an OR gate or an AND gate. The non-inverting gates do not have this versatility since, they can’t produce an invert.

Some Basic Functions of NOR Gate

The NAND Gate In this type of gate, the output of AND gate is fed to input of a NOT gate and final output is obtained at output of NOT gate. A

Y′

NOR A

NOR

A

⇒

NOR B

NOR

⇒ A

A

A + B = AB

Y

B

A+A=A

A

AND

AB

B

B

The logic symbol of NAND gate is shown as A Y B

A B

The Boolean expression of NAND gate is Y = A × B, which indicates A and B are negated.

A

B

Y¢

Y

0 0 1 1

0 0 0 1

1 1 1 0

A B

OR

A+ B

Which can be implemented by the gate arrangements shown. They can also be implemented using NAND gates only. A

y′

⇒

2. A Å B = ( A + B) ( AB) A OR B AND NOT A AND B.

In this type of gate, the output of OR gate is fed to input of the NOT gate and final output is obtained at output of the NOT gate. A

A+ B

Exclusive OR Gate

The NOR Gate

B

NOR

Logically, the exclusive OR (XOR) operation can be seen as either of the following operations 1. A Å B = AB + BA A AND NOT B OR B AND NOT A.

Truth table for NAND gate 0 1 0 1

A+ B NOR

y

AB y′′

B

AB + AB A ⊕ By

A B

A B

AB y′′

1170 JEE Main Physics Exclusive OR with NAND

Interpret

A B

(d) NOR and NAND gates are universal gates. Any digital circuit can be realised by repetitive use of these NOR and NAND gate.

A+B

B

Sample Problem 13 Identify the true statement for OR gate. A⊕ B A

A

A

AB

B

B

A

The implementation of the Exclusive OR ( XOR) operation with just NAND gates illustrates the function of NANDs as universal gates. A

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

Output y will be 1, when in put A or B or both are 1. Output y will be 0,when either of the inputs A or B is 1. OutputY will be 1, only when both the inputs A andBare 1. Output Y will be 1, only when either of the inputs A and B are 1.

Interpret (a) In an OR gate, output y is 1 only when either or both the Inputs A and B are 1. The truth table is shown below A 0 0 1 1

A A+ B (A+ B) (AB) B

B

A⊕ B

B 0 1 0 1

The XOR Gate The XOR gate can be obtained by combination of OR, AND and NOT gates as shown under Y′

A A

AB

B

Y 0 1 1 1

Y

1. A Å B = AB + B A A AND NOT B OR B AND NOT A

B

2. A Å B = ( A + B) ( AB)

Y ′′

The logic symbol of XOR gate is shown as

A OR B AND NOT A AND B

A

Y

B

Sample Problem 11 The combination of gates shown

The Boolean expression for XOR gate is

below yields.

Y = A B + AB = A Å B

A

Truth Table for XOR gate X B

(a) OR gate (c) XOR gate

Interpret

(b) NOT gate (d) NAND gate

(a) Truth table for given combination is A

B

X

0 0 1 1

0 1 0 1

0 1 1 1

This comes out to be truth table of OR gate.

A

B

Y

0 0 1 1

0 1 0 1

0 1 1 0

From truth table, we can conclude that if A and B are not indentical, output is 1 and when A and B are indentical, output is 0.

The XNOR Gate (exclusive-NOR) Gate If XOR gate with NOT gate, we get XNOR gate. The XOR gate followed by an inverter. Its output is “true” if the inputs are the same, and “false” if the inputs are different. A

Sample Problem 12 Any digital circuit can be realised by repetitive use of only (a) NOT gate

(b) OR gate

(c) AND gate (d) NOR gate

Y

B X NOR gate

Y =A Å B or A υ B

Electronic Devices

Some Useful Boolean Identities

Truth table for XNOR gate Inputs

(a) Commutative laws

Output

A

B

A XNOR B

0 0 1 1

0 1 0 1

1 0 0 1

(i) A + B = B + A;

Y′

(ii) A × B = B × A

(b) Associative laws (i) A + (B + C ) = ( A + B) + C; (ii) A × (B × C ) = ( A + B) × C

(c) Distributive laws

Sample Problem 14 In the following circuit, the output 1 for all possible inputs A and B is expressed by which of the truth table given below A

Y

(i) A × (B + C ) = A × B + A × C; (ii) ( A + B) × ( A + C ) = A + B × C

(d) Absorption laws (i) A + A × B = A; (ii) A × ( A + B) = A; (iii) A × ( A + B) = A × B

(e) Double complement function A = A, A + B = A + B ; A . B = A × B

B

(f) Boolean indentities

(a) A 0 0 1 1

B 0 1 0 1

Y 0 1 1 1

A

B

Y

0 0 1 1

0 1 0 0

1 0 1 1

(b)

A

B

Y

0 0 1 1

0 1 1 1

0 1 1 1

(ii) A + A × B = A + B

(i) A × ( A + B) = A × B

(iii) A + B × C = ( A + B) × ( A + C ) (iv) ( A + B) × ( A + C ) = A × C + A × B

(g) De-Morgan’s Theorem It states that the complement of the whole sum is equal to the product of individual complements and vice-versa. i.e., (ii) A × B = A + B

(i) A + B = A × B

(c)

(h) Basic OR and AND relations

OR (i) (ii) (iii) (iv)

A+ A+ A= A+

AND 0= A 1= 1 A= A A =1

A× 0 = 0 A× 1 = A A× A = A A× A = 0

26.8 Transistor as a Switch Vcc Load Output

Relay

Flywheel Diode

(d)

Interpret

1171

Ic

A

B

Y

1

1

1

0

1

0

1

1

0

0

1

1

(a) Boolean Expression

Y ¢ = A + B andY = Y = A + B = A + B

Truth table given below A

B

Y

0

0

0

1

0

1

0

1

1

1

1

1

RB + Vin Ω –

ia

β

Vce

R

The circuit resembles that of the Common-Emitter circuit. The difference this time is that to operate the transistor as a switch the transistor needs to be turned either fully ‘‘OFF’’ (Cut-off) fully ‘‘ON’’ (Saturated). An ideal transistor switch would have an infinite resistance when turned ‘OFF’ resulting in zero current flow and zero resistance, when turned ‘‘ON’’, resulting in maximum current flow. In practice, when turned ‘‘OFF’, small leakage currents flow through the transistor and when fully “ON” the device has a low resistance value causing a small saturation voltage (Vce ) across it. In both the cut-off and saturation regions, the power dissipated by the transistor is at its minimum.

WORKED OUT Examples Example 1

In a common emitter emplifier, the phase difference between the input signal voltage and output signal voltage is (a)

p 4

Solution

(d)

Mobility of electrons is 2 to 3 times larger than that of hole. Hence, m e > m h .

Example 5

(b) p

(c) 0

Solution

The voltage gain of the amplifier shown in figure

is

p 2

100 kW 1kW

In CE amplifier only a phase shift of p exists.

+ Vo

Example 2

In a full wave rectifier circuit operating from 50 Hz mains frequency, the fundamental frequency in the ripple would be (a) 50 Hz (c) 25 Hz

Solution

10 kW

(b) 100 Hz (d) 70.7 Hz (a) 10 (c) 9.6

fout = 2 finput = 2 ´ 50 = 100 Hz

Consider an n-p-n transistor amplifier in CE. configuration. The current gain in the transistor is 100. If the collector changes by 1 mA. What will be change in emitter current? (a) 1.1 mA (c) 0.01 mA

(b) 1.01 mA (d) 10 mA

Current gain b =100 and a =

Given \

b 100 = ; 1 + b 101

Dic 100 1mA = = Die 101 Die Die = 1.01 mA

Example 4

In a semiconducting material the mobilities of electron and hole are m e and m h respectively. Which of the following is true? (a) m e > m h (b) m e = m h (c) m e < m h (d) m e > 0; m h > 0

(b) 100 (d) 1000

Solution

Example 3

Solution

–

Voltage gain Av =

Example 6

Evaluate Y1 = AB + AB and Y2 = A + AB

(a) 0, A + B (c) A , B

Solution Put

Vo Rf 100 k W = = = 100 Vi Ri 1 kW

(b) 1, A + B (d) AB , AB

Y1 = AB + AB,

AB = X then Y1 = X + X = 1 = Y2 = A (1 + B) + A B = A + ( A + A) B = A + B

Example 7

A transistor has hFE = 95, find h FB

(a) 1.9 (c) 0.56

Solution

(b) 0.20 (d) 0.99 h FB = a , hFE = b a=

b 95 = b + 1 96

Start Practice for

JEE Main Round I

(Topically Divided Problems)

Semiconductors 1. In an n-type silicon, which of the following statements is true?

[NCERT]

(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

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

P is semiconductor and Q is conductor P is conductor and Q is semiconductor P is n-type semiconductor and Q is p-type semiconductor None of the above

7. The energy band diagrams for three semiconductor samples of silicon are as shown. We can then assert that

2. The correct relation between ne and nh in an intrinsic semiconductor at ordinary temperature, is (a) ne > nh (c) ne = nh

(b) ne < nh (d) ne = nh = 0

3. The relation between number of free electrons (n) is a semiconductor and temperature (T) is given by (a) n µ T

(b) n µ T 2

(c) n µ T

(d) n µ T 32/

4. The conductivity of a semiconductor increases with increase in temperature because

[NCERT Exemplar]

(a) number density of free current carriers increases (b) relaxation time increases (c) both number density of carriers and relaxation time increase (d) number density of carriers increases, relaxation is time decreases but effect of decreases in relaxation time is Imuch less than increase in number density

5. The forbidden energy band gap in conductors semiconductors and insulators are EG1, EG2 and EG3 respectively. The relation among them, is [NCERT] (a) EG1 < EG2 < EG3 (c) EG2 > EG1 < EG3

X

Y

8. The ratio of electron and hole current in a semiconductor is 7/4 and the ratio of drift velocities of electrons and holes is 5/4, then ratio of concentrations of electrons and holes will be (a) 5/7 (c) 25/49

(b) 7/5 (d) 49/25

9. In figure, V 0 is the potential barrier across a p-n junction, when no battery is connected across the junction [NCERT Exemplar] 1 2 3

(b) EG1 > EG2 > EG3 (d) EG3 > EG1 > EG2

6. Wire P and Q have the same resistance at a ordinary room temperature. When heated resistance of P increases and that of Q decreases. We conclude that

Z

(a) Sample X is undoped while samples Y and Z have been doped with a third group impurity respectively (b) Sample X is undoped while both samples Y and Z have been doped with a fifth group impurity (c) Sample X has been doped with equal amounts of third and fifth group impurities while samples Y and Z are undoped (d) Sample X is undoped while samples Y and Z have been doped with a fifth group and a third group impurity respectively

V0

1174 JEE Main Physics (a) 1 and 3 both correspond to forward bias of junction (b) 3 corresponds to forward bias of junction and 1 corresponds to reverse bias of junction (c) 1 corresponds to forward bias and 3 corresponds to reverse bias of junction (d) 3 and 1 both correspond to reverse bias of junction

10. A piece of copper and other of germanium are cooled from the room temperature to 80 K, then (a) resistance of each will increase (b) resistance of each will decrease (c) the resistance of copper will increase, while that of germanium will decrease (d) the resistance of copper will decrease, while that of germanium will increase

11. In figure assuming the diodes to be ideal, [NCERT Exemplar] –10V

A

R

D1

5 ´ 1028 atoms m -3, then the number of acceptor atoms in silicon per cubic centimetre will be (a) 2.5 ´ 1030 atoms cm-3 (b) 2.5 ´ 1035 atoms cm-3 (c) 10 . ´ 1013 atoms cm-3

16. When an electrical conductivity of semiconductor is

due to the breaking of its covalent bands, then the semiconductor is said to be (a) acceptor (b) donor

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 and 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

(d) extrinsic

17. For a transistor amplifier, the voltage gain

[NCERT]

(a) remains constant for all frequencies (b) is high and low frequencies and constant in the middle frequency range (c) is low at high and low frequencies and constant at mid frequencies (d) None of the above

Junction Diode are made of semiconductors having band gaps of 2.5 eV, 2 eV and 3 eV, respectively. Which one will be able to detect light of wavelength of 6000 Å ? (a) 1.05 eV

increases with the rise in its temperature decrease with the rise in its temperature decrease does not change with the rise in its temperature first increase and then decreases with the rise in its temperature

13. An n-type and a p-type silicon semiconductor can be obtained by doping pure silicon with sodium and magnesium phosphorus and boron respectively boron and phosphorus respectively indium and sodium respectively

(b) 0.05 eV

(c) 0.25 eV

(d) 2.06 eV

19. A sinusoidal voltage of peak value 200V is connected

to a diode and resistor R in the circuit figure, so that half-wave rectification occurs. If the forward resistance of the diode is negligible compared to R, the rms voltage (in volt) across R approximately

E0 = 200 volt

12. Electrical conductivity of a semiconductor

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

(c) intrinsic

18. Three photodiodes D1, D2 and D3 D2

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

(d) 10 . ´ 1015 atoms cm-3

(a) 200

R

(b) 100

(c)

200 2

(d) 280

20. Two identical p-n junction may be connected in series with a battery in three ways as shown in the adjoining figure. The potential drop across the p-n junctions are equal in p n

n p

p n

p n

14. The fermi level of an intrinsic semiconductor is pinned at the centre of the band gap. The probability of occupation of the highest electron state in valence band at room temperature, will be (a) zero (c) half

Circuit 1

Circuit 2 n p

n p

(b) between zero and half (d) one

15. A silicon specimen is made into a p-type

semiconductor by dopping, on an average, one indium atom per 5 ´ 107 silicon atoms. If the number density of atoms in the silicon specimen is

Circuit 3

(a) circuit 1 and circuit 2 (c) circuit 3 and circuit 1

(b) circuit 2 and circuit 3 (d) circuit 1 only

Electronic Devices 21. In the half. wave rectifier, circuit operating from 50

28. The value of current in the following diagram will be

Hz mains frequency, the fundamental frequency in the ripple would be (a) 25 Hz

(b) 50 Hz

(c) 70.7 Hz

(d) 100 Hz

22. A 220 V AC supply is connected between points A and B (figure). What will be the potential difference V across the capacitor? [NCERT Exemplar]

200 Ω

+ 5V

29. Write the name of the following gate that the circuit shown in figure. +5V

220V A.C.

C

V

D1 V0

B

(a) 220 V (c) 0 V

D2

(b) 110 V (d) 220 2 V

23. The correct curve between potential (V) and distance (d) near p-n junction is n

p

(a)

(b) d

V n

(c)

(d)

p

d

(b) OR gate (d) XOR gate

(a) the current in the reverse biased condition is generally very small (b) the current in the reverse biased condition is small but the forward biased current is independent of the bias voltage (c) the reverse biased current is strongly dependent on the applied bias voltage (d) the forward biased current is very small in small in comparison to reverse biased current

n

d

V

(a) AND gate (c) NOR gate

30. In a p-n junction diode

V

V

p

+ 3V

(b) 10 -2 A (d) 0.025 A

(a) zero (c) 10 A

A

p

1175

n d

31. p-n junction is said to be forward biased, when 24. Two amplifier are connected after the orher 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. [NCERT]

(a) 3.6 V (c) 2.0 V

(b) 4.2 V (d) 5.2 V

25. The value of ripple factor for full-wave rectifier is (a) 40.6%

(b) 48.2%

(c) 81.2%

(d) 121%

26. The average value of output direct current in a half wave rectifier is I (a) 0 p pI0 (c) 2

I (b) 0 2 2I0 (d) p

(a) the positive pole of the battery is joined to the n-semiconductor (b) the positive pole of the battery is joined to the n-semiconductor and p-semiconductor (c) the positive pole of the battery is connected to n-semiconductor and p-semiconductor (d) a mechanical force is applied in the forward direction

32. The reverse bias in a junction diode is changed from 8 V to 13V, then the value of the current changes from 40 m A to 60 mA. The resistance of junction diode will be (a) 2 ´ 105 W

(a) e -V /kT (b) eV /kT (c) ( e eV /kT - 1) (d) ( eV /kT - 1)

(d) 4 ´ 105 W

(c) 3 ´ 10 W

33. Consider the junction diode is ideal. The value of current in the figure is + 4V

27. For a junction diode, the ratio of forward current ( I f ) and reverse current is [I e = electronic charge, V = voltage applied across junction, k = Boltzmann constant, T = temperature in kelvin]

(b) 2.5 ´ 105 W

5

(a) zero

34. Hole is

p-n

(b) 10 -2 A

300 Ω

+ 1V

(c) 10 -1 A

(d) 10 -3 A [NCERT Exemplar]

(a) an anti-particle of electron (b) a vacancy created when an electron leaves a covalent bond (c) absence of free electrons (d) an artifically created particle

1176 JEE Main Physics 35. The output of the given circuit in figure [NCERT Exemplar]

39. In the circuit shown in figure. If the diode forward voltage drop is 0.3 V, the voltage difference between A and B is

(a) would be zero at all times (b) would be like a half-wave rectifier with positive cycle in output (c) would be like a half-wave rectifier with negative cycle in output (d) would be like that of a full-wave rectifier

36. In the case of forward biasing of p-n junction, which one of the following figures correctly depicts the direction of flow of carriers? – + p – + n – +

– + p – + n – +

5 kW

Vp

Vp

– + p – + n – +

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

1.3 V 2.3 V 0 0.5 V

5 kW B

Junction Transistor 40. In an n-p-n transistor, the collector current is 10 mA, if 90% of the electrons emitted reach the collector, the emitter current ( I E ) and base current (I B ) are given by (a) IE = - 1 mA, Ib = 9 mA (b) IE = 9 mA, IB = - 1 mA (c) IE = 1 mA, IB = 11 mA (d) IE = 11 mA, IB = 1 mA

41. The transfer ratio of the transistor is 50. The input

(b)

(c)

0.2 mA

[NCERT Exemplar]

Vm sin ωt

(a)

A

– + p – + n – +

resistance of the transistor when used in the CE configuration is 1 kW. The peak value for an AC input voltage of 0.01 V of collector current is (a) 500 mA

(b) 0.25 mA (c) 400 mA

(d) 0.01 mA

42. Current gain in common-emitter configuration is

(d)

more than 1, because Vp

Vp

37. A semiconductor device is connected in a series circuit with a battery and a resistance. A current is found to pass through the circuit. If the polarity of the battery is reversed, the current drops almost to zero. The device may be (a) (b) (c) (d)

a p-type semiconductor an n-type semiconductor a p-n junction an intrinsic semiconductor

(c) l c > l b

(d) l e > l b

43. Current gain common-base configuration is less than 1, because (a) l e < l b

(b) l b < l e

(c) l c < l e

(d) l e < l c

44. Three amplifier stages each with a gain of 10 are cascaded. The overall gain is (b) 30 (d) 100

45. A transistor has b = 40. A change in base current of [NCERT Exemplar]

C E

B

(b) l c < l e

(a) 10 (c) 1000

38. Truth table for the given circuit. A

(a) l c < l b

100 mA, produces change in collector current (a) 40 ´ 100 mA (c) 100 + 40 mA

(b) (100 - 40 mA) (d) 100 ´ /40 mA

46. Current gain of a transistor in common base mode is 0.95. Its value in common emitter mode is

D

A B 0 0 (a) 0 1 1 0 1 1

E 1 0 1 0

A B 0 0 (b) 0 1 1 0 1 1

E 1 0 0 1

A B 0 0 (c) 0 1 1 0 1 1

E 0 1 0 1

A B 0 0 (d) 0 1 1 0 1 1

E 0 1 1 0

(a) 0.95 (c) 19

(b) 1.5 (d) (19 ) -1

47. The current gain of a transistor in a common emitter configuration is 40. If the emitter current is 8.2 mA, then base current is (a) 0.02 mA (c) 2.0 mA

(b) 0.2 mA (d) 0.4 mA

48. In

a common emitter transistor amplifier b = 60, R0 = 5000 W and internal resistance of a transistor is 500 W. The voltage amplification of amplifier will be (a) 500

(b) 460

(c) 600

(d) 560

Electronic Devices 49. In a n-p-n transistor 1010 electrons enter the emitter -6

in 10 s. 4% of the electrons are lost in base. The current transfer ratio will be (a) 0.98 (c) 0.96

(b) 0.97 (d) 0.94

10 mA. If 95 percent of the electrons emitted reach the collector, which of the following statements are true? [NCERT Exemplar] The emitter current will be 8 mA The emitter current will be 10.53 mA The base current will be 5.53 mA The base current will be 2 mA

55. What is the name of the gate obtained by the A Y B

(a) NAND (b) NOR (c) NOT (d) XOR

56. A truth table is given below. Which of the following has this type of truth table?

51. A transistor has a base current of 1 mA and emitter current 90 mA. The collector current will be (a) 90 mA (c) 89 mA

(b) 1 mA (d) 91 mA

52. What is the value of A A in Boolean algebra? (a) zero (c) A

(b) NAND gate (d) NOR gate

combination shown in figure?

50. In a n-p-n transistor circuit, the collector current is

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

(a) AND gate (c) OR gate

1177

(b) 1 (one) (d) A

53. What is the output Y of the gate circuit shown in figure?

A

B

Y

0 1 0 1

0 0 1 1

1 0 0 0

(a) NOR gate (c) AND gate

(b) OR gate (d) NAND gate

57. Which of the following logic gate is represented by the combination of logic gates

A

y

A

B

(a) A × B

(b) A × B

(c) A × B

(d) A × B

Y B

54. Which gate is represented by the symbolic diagram given here?

(a) NAND gate (c) AND gate

(b) NOR gate (d) OR gate

A Y B

Round

II

Only One Correct Option

3. In a common base amplifier circuit, calculate the

1. Pure sodium (Na) is a good conductor of electricity because the 3s and 33p atomic bands overlap to form a partially filled conduction band. By contrast the ionic sodium chloride (NaCl) crystal is (a) Insulator (c) Semiconductor

(b) Conductor (d) None of these

2. For a transistor, the current amplification factor is 0.8. The transistor is connected in common emitter configuration. The change in the collector current when the base current changes by 6 mA is (a) 6 mA

(b) 4.8 mA

(c) 24 mA

(Mixed Bag)

(d) 8 mA

change in base current, if that in the emitter current is 2 mA and a = 0. 98. (a) 0.04 mA (b) 1.96 mA (c) 0.98 mA

(d) 2 mA

4. In case of n-p-n-transistors the collector current is always less than the emitter current because (a) collector side is reverse biased and emitter side is forward biased (b) after electrons are lost in the base and only remaining ones reach the collector (c) collector side is forward biased and emitter side is reverse biased (d) collector being reverse biased attracts less electrons

1178 JEE Main Physics 5. The input resistance of a common emitter transistor

11. The circuit shown in the figure contains two diodes

amplifier, if the output resistance is 500 kW, the current gain, a = 0.98 and power gain is 60625 . ´ 106 , is

each with a forward resistance of 50 W and with infinite backward resistance. If the battery is 6 V, the current through the 100 W resistance (in ampere) is

(a) 198 W

(b) 300 W

(c) 100 W

(d) 400 W

150 Ω

6. The following configuration of gate is equivalent to figure.

50 Ω

(a) NAND (c) OR

(b) XOR (d) None of these (a) zero (c) 0.03

load, R = 50 k W. The voltage amplification obtained from this triode will be (a) 30.42 (c) 28.18

100 Ω

6V

7. In a triode, g m = 2 ´ 10-3 ohm -1; m = 42, resistance (b) 29.57 (d) 27.15

(b) 0.02 (d) 0.036

12. A full wave rectifier circuit along with the input and

8. In the figure, potential difference between A and B is

output are shown in the figure, the contribution from the diode I is (are)

10 kΩ

V t

Input 10 kΩ

30 V

10 kΩ

Input

(a) zero (c) 10 V

(b) 5 V (d) 15 V

V

9. When A is the internal stage gain of an amplifier and b is the feedback ratio, then the amplitude becomes as oscillator if A 2 1 (b) b is negative and magnitude of b = A (c) b is negative and magnitude of b = A 1 (d) b is positive and magnitude of b = A

O

(a) b is negative and magnitude of b =

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

(b) A, C (d) A, B, C, D

C B A B

D

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

C

an OR gate and an AND gate respectively an AND gate and a NOT gate respectively an AND gate and an OR gate respectively an OR gate and a NOT gate respectively

14. In p-n junction, the barrier potential offers resistance

A

to

AND G1 Y

NAND G3 G2

t

A

0, 0 0, 1 1, 0 1, 1

B

D

figure, are equivalent to

G2 G1

B C Output

13. The combination of NAND gates shown here under

of inputs A, B, C are as follows A = B = C = 0 and A = B = 1, C = 0, then the logic states of output D are A

A

(a) C (c) B, D

10. For the given combination of gates, if the logic states

B C

Output

I

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

free electrons in n-region and holes in p-region free electrons in p-region and holes in n-region only free electrons in n-region only holes in p-region

Electronic Devices

1179

15. In the case of forward biasing of p-n junction, which

21. A semiconductor device is connected in a series

one of the following figures correctly depicts the direction of flow of carriers ?

circuit with a battery and a resistance. A current is found to pass through the circuit. If the polarity of the battery is reversed, the current drops almost to zero. The device may be

(a)

Vb – + p – + – +

Vb – + p – + n – +

n

(b)

VF

VF

(c)

Vb – + p – + n – +

Vb – + p – + – +

(d)

a p-type semiconductor an n-type semiconductor a p-n junction an intrinsic semiconductor

22. The correct relation between the two current gains a n

and b in a transistor is a 1+ a b (c) a = 1+ b

(a) b =

VF

VF

16. In an intrinsic semiconductor, the fermi level is (a) (b) (c) (d)

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

nearer to valency band than conduction band equidistance from conduction band and valency band nearer to conduction band than valency band bisecting the conduction band

17. In a common base amplifier circuit, calculate the

23. The diode used in the circuit shown in the figure has a constant voltage drop of 0.5 V at all currents and a maximum power rating of 100 mW. What should be the value of the resistor R, connected in series with the diode for obtaining maximum current ?

I

(b) 1.96 mA (d) 2 mA

18. Platinum and silicon are heated upto 250°C and after that cooled. In the process of cooling (a) resistance of platinum will increase and that of silicon will decrease (b) resistance of silicon will increase and that of platinum will decrease (c) resistance of both will increase (d) resistance of both will decrease

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

figure. n p n

RL Vout

Vin

C

D

zero same as the input half-wave rectified full wave rectified

20. Doping of a semiconductor (with small traces of impurity atoms) generally changes the resistivity as follows (a) (b) (c) (d)

(b) 5 W (d) 200 W

24. An n- p-n-transistor circuit is arranged as shown in

and C and the output is across B and D. Then the output is

A

1.5 V

(a) 1.5 W (c) 6.67 W

19. As shown is figure, the input is across the terminal A

B

0.5 V

R

change in base current if that in the emitter current is 2 mA and a = 0.98 (a) 0.04 mA (c) 0.98 mA

b 1- b 1+ b (d) a = b (b) a =

does not alter increases decreases may increase or decrease depending on the dopant

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

a common base amplifier circuit a common emitter amplifier circuit a common collector circuit None of the above

25. In p-type semiconductors, conduction is due to (a) (b) (c) (d)

greater number of holes and less number of electrons only electrons only holes greater number of electrons and less number of holes

26. The junction diode in the following circuit requires a minimum current of 1 mA to be above the knee point (0.7 V) of its I-V characteristic curve. The voltage across the junction diode is independent of current

1180 JEE Main Physics above the knee point, if VB = 4 V, then the maximum value of R so that the voltage is above knee point will be 0.7 V

R

33. The circuit shown in the figure contains two diodes each with a forward resistance of 30 W and with infinite backward resistance. If the battery is 3 V, the current through the 50 W resistance (in ampere) is 70 Ω 70 Ω

(a) 3.3 kW

(b) 4.0 kW

50 Ω

3V +

VB

(c) 4.7 kW

(d) 6.6 kW

27. In a common emitter amplifier, using output resistance of 5000 W and input resistance of 2000 W, if the peak value of input signal voltage is 10 mV and b = 50, then peak value of output voltage is (a) 5 ´ 10 -6 V

(b) 12.50 ´ 10 -6 V

(c) 1. 25 V

(d) 125.0 V

(a) zero

(b) 0.01

(c) 0.02

34. In the network shown, the current flowing through the battery of negligible internal resistance is 10 Ω

20 Ω

28. What is the output of the combination of the gates shown in the figure?

10 Ω

A

Y

(d) 0.03

(a) 0.10 A

3V

(b) 0.15 A

(c) 0.20 A

(d) 0.30 A

35. The output Y of the logic circuit shown in figure is

B

(a) A + A × B

(b) ( A + B) + ( A × B )

(c) ( A + B) × ( A × B)

(d) ( A + B) × ( A + B )

29. The length of germanium rod is 0.928 cm and its area of cross-section is 1 mm 2 . If for ni = 2.5 ´ 1019 m -3, m h = 0.19 m 2 V -1s -1 (a) 2.5 kW

(b) 4.0 kW

(c) 5.0 kW

germanium (d) 10.0 kW

best represented as A B C

Y

III II I

(a) A + A × C

(b) A + B × C

(c) A + B × C

(d) A + B + C

30. A junction diode is connected to a 10 V source and

103 W rheostate figure. The slope of load line on the characteristic curve of diode will be

More Than One Correct Option 36. For transistor action, which of the following statements are correct?

103 Ω

(a) 10 -2 AV -1

(b) 10 -3 AV -1

-4

(d) 10 -5 AV -1

(c) 10

AV

-1

31. If, a zener diode (VZ = 5 V and I Z = 10 mA) is

connected in series with a resistance and 20 V is applied across the combination, then the maximum resistance one can use without spoiling zener action is (a) 20 kW (c) 10 kW

(b) 15 kW (d) 1. 5 kW

32. If the output of a logic gate is 0 when all its inputs are at logic 1, then the gate is either (a) NAND or XNOR (c) XOR or NOR

(b) NOR or OR (d) AND or NOR

[NCERT]

(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

37. When an electric field is applied across a samiconductor

[NCERT Exemplar]

(a) electrons moved 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

1181

Electronic Devices 38. A transistor is used in common emitter mode as an amplifier, then (a) the base emitter junction is forward biased (b) the base emitter junction is reversed biased (c) the input signal is connected in series with the voltage applied to the base emitter junction (d) the input signal is connected in series with the voltage applied to bias the base collector junction

39. 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 [NCERT Exemplar] true? (a) (b) (c) (d)

Electrons crossover from emitter to collector Holes move from based to collector Electrons move from emitter to base Electrons from emitter move out of base without of base without of base without going to the collector

40. In a p-n- p transistor circuit, the collector current is 10 mA. If 90% of the holes emitted from emitter reach the collector; (a) (b) (c) (d)

the emitter current will be 9 mA the emitter current will be 11 mA the base current will be 1 mA the base current will be - 1 mA

The emitter current will be 8 mA The emitter current will be 10.53 mA The base current will be 0.53 mA The base current will be 2 mA

Comprehension Based Questions Passage The input and output resistances in a common-base amplifier circuits are 400 W and 400 kW respectively. The emitter current is 2 mA and current gain is 0.98.

44. Power gain of transistor is (a) 950

(b) 960

(c) 970

(d) 980

45. The collector current is (a) 1.84 mA (b) 1.96 mA (c) 1.2 mA

(d) 2.04 mA

46. The base current is (a) 0.012 mA (c) 0.032 mA

(b) 0.022 mA (d) 0.042 mA

47. Voltage gain of transistor is (a) 960

(b) 970

(c) 980

(d) 990

48. If peak of peak voltage of input AC source is 0.1 V. The peak of peak voltage of the output will be (a) 9.8 V

41. The breakdown in a reverse biased p-n junction diode is more likely to occur due to

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

[NCERT Exemplar]

(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 a depletion region if the doping concentration is large

(c) 980 V

(d) 970 V

49. For a CE transister 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. [NCERT] (a) 0.01 V and 10 mA (c) 0.001 V and 100 mA

(b) 0.1 V and 14 mA (d) 2.0 V and 10 mA

Matching Type 50. Match the following Column I with Column II

42. Figure shows the transfer characteristics of a base biased CE transistor. Which of the following statements are true? [NCERT Exemplar]

(b) 98 V

Column I

Column II

I.

n-p-n transistor

A.

II.

p-n-p transistor

B.

Vo

N C

E

0

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

At Vi At Vi At Vi At Vi

0.6 V

2V

V

= 0.4 V, transistor is in active state = 1V, it can be used as an amplifier = 0.5 V, It can be used as a switch turned off = 2.5 V, it can be used as a switch turned on

III.

Light emmiting diode

B

C

E

IV.

Zener diode

D. B

43. In a 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? [NCERT Exemplar]

C.

Code (a) I-A, II-B, III-C, IV-D (c) I-C, II-D, III-B, IV-A

(b) I-D, II-A, III-B, IV-C (d) I-B, II-A, III-C, IV-D

1182 JEE Main Physics Assertion and Reason Directions

Question No. 51 to 66 are Assertion-Reason type. Each of these contains two Statements : Statement I (Assertion), Statement II (Reason). Each of these questions also has four alternative choice, only one of which is correct. You have to select the correct choices from the codes (a), (b), (c) and (d) given below (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 correct explanation of the Assertion (c) If Assertion is true but Reason is false (d) If Assertion is false but the Reason is true

51. Assertion The resistivity of a semiconductor increases in temperature. Reason In a conducting solid, the rate of collisions between free electrons and ions increases with increase of temperature.

52. Assertion Zener diode works on a principle of breakdown voltage. Reason Current increases suddenly after breakdown voltage. P-n junction diode can be used even at ultrahigh frequencies Reason capacitive reactance of p-n diode increases as the frequency increases.

53. Assertion

54. Assertion In vacuum tubes (valves), vacuum is necessary for the movement of electrons between electrodes otherwise electrons collide with air particle and lose their energy. Reason In semiconductors devices, external heating or vacuum is not required.

55. Assertion De-morgan’s theorem A + B = A × B may be explained by the following circuit.

Y

A B

Reason A p-n photodiode detects wavelength l if, hc > E g. l

57. Assertion Thickness of depletion layer is fixed in all semiconductor devices. Reason No free charge carriers are available in depletion layer.

58. Assertion If forward current changed by 1.5 mA when forward voltage in semiconductor triode is changed from 0.5 V to 2 V, the forward resistance of diode will be 1 W. Reason The forward resistance is given by DV f Rf = DI f

59. Assertion The resistivity of a semiconductor increases with temperature. Reason The atoms of a semiconductor vibrate with larger amplitude at higher temperature thereby increasing its resistivity.

60. Assertion When base region has larger width, the collector current increases. Reason Electron hole combination in base results in increase or base current.

61. Assertion In a common base circuit, current gain is 0.95. If, base current is 60 mA. then emitter current is 1200 mA. I Reason Current gain in common base circuit is, a = C . IE

62. Assertion The value of current through p-n junction in the adjoining figure will be 10 mA. Reason In the above figure, p-side higher potential than n-side. p 5V

Y

=

n

300 Ω 2 V

63. Assertion NAND or NOR gates are called digital building blocks. Reason The repeated use of NAND or NOR gates can produce all the basics or completed gates.

Reason In the following circuit, for output 1, inputs ABC are 101. A B

64. Assertion The coordination number of face centred crystal is 8. Reason The coordination number is number of the closest neighbouring atoms in a crystal structure.

65. Assertion The logic gates NOT can be built using C

Y

56. Assertion A p-n photodiode is made from a semiconductor for which, E g = 2.8 eV . This photo diode will not detect the wavelength of 6000 nm.

diode. Reason The output voltage and the input voltage of the diode have 180° phase difference.

66. Assertion NOT gate is also called invertor. Reason NOT gate inverts the input signals.

Electronic Devices

1183

Previous Years’ Questions 67. The output of an OR gate is connected to both the

inputs of NAND gate. The combination will serve as a [IIT JEE 2011] (a) OR gate (c) NOR gate

(b) NOT gate (d) AND gate

71. A p-n junction ( D) shown in the figure can act as a rectifier. An alternating current source ( V ) is connected in the circuit. The current ( I ) in the resistor ( R) can be shown by [AIEEE 2009]

68. The combination of gates shown below yields

D

[AIEEE 2011]

V– V

R

A X

I

B

I

(a)

(a) OR gate (c) XOR gate

(b)

(b) NOT gate (d) NAND gate

t

69. If in a p-n junction diode, a square input signal of 10 V is applied as shown

t I

I

(c)

(d)

5V

t

t

RL

72. How many NAND gates are used in an OR gate ? [UP SEE 2009]

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

–5 V

Then the output signal across RL will be [AIEEE 2009] 10 V

(a)

73. In a transistor the base is (a) (b) (c) (d)

(b) –10 V

Y B

–5 V

70. The graph given below represents the I-V characteristics of a zener diode. Which part of the characteristics curve is most relevant for its [AMU Engg. 2009] operation as a voltage regulator. I (mA) a Forward bias

Reverse bias Vz e

d

c

b

(b) bc

(c) cd

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

(d) de

AND gate OR gate NOT gate NAND gate

75. A Si and a Ge diode has identical physical dimensions. The band gap in Si is larger than that in Ge. An identical reverse bias is applied across the diodes. [WB JEE 2008] (a) (b) (c) (d)

V (V)

I (µA)

(a) ab

[UP SEE 2009]

A

(d)

current

[ UP SEE 2009]

an insulator a conductor of low resistance a conductor of high resistance an extrinsic semiconductor

74. The figure show the symbol of a

5V

(c)

Four Two Three Five

The reverse current in Ge is larger than that in Si The reverse current in Si is larger than that in Ge The reverse current is identical in the two diodes The relative magnitude of the reverse currents cannot be determined from the given data only

Answers Round I 1. 11. 21. 31. 41. 51.

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

2. 12. 22. 32. 42. 52.

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

3. 13. 23. 33. 43. 53.

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

4. 14. 24. 34. 44. 54.

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

5. 15. 25. 35. 45. 55.

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

6. 16. 26. 36. 46. 56.

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

2. 12. 22. 32. 42. 52. 62. 72.

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

3. 13. 23. 33. 43. 53. 63. 73.

(a) (a) (b) (c) (b,c) (c) (a) (b)

4. 14. 24. 34. 44. 54. 64. 74.

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

5. 15. 25. 35. 45. 55. 65. 75.

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

6. 16. 26. 36. 46. 56. 66.

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

7. 17. 27. 37. 47. 57.

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

8. 18. 28. 38. 48.

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

9. 19. 29. 39. 49.

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

10. 20. 30. 40. 50.

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

9. 19. 29. 39. 49. 59. 69.

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

10. 20. 30. 40. 50. 60. 70.

(d) (c) (b) (b,c) (c) (c) (d)

Round II 1. 11. 21. 31. 41. 51. 61. 71.

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

7. 17. 27. 37. 47. 57. 67.

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

8. 18. 28. 38. 48. 58. 68.

(c) (b) (a) (a,c) (b) (a) (a)

the Guidance Round I 1. In an n-type semiconductor, it is obtained by doping the Ge or Si with pentavalent atoms, In n-type semiconductor, electrons are majority carriers and holes are minority carriers.

2. At ordinary temperature, ne = nh. (in intrinsic semiconductor) 3. For semiconductor, n = AT3/ 2e-Eg / 2kT n µ T3/ 2

So,

4. Knowledge based question. 5. In insulators, the forbidden energy gap is very large, in case

\

Ie Rev e = Ih Rhv h

or

ne Ie v h 7 4 7 = ´ = ´ = nh Ih v e 4 5 5

9. When p-n junction is forward biased, it opposes the potential barrier across junction. When p-n junction is reverse biased, it supports the potential barrier junction, resulting increases in potential barrier across the junction. Thus, option (b) is correct.

of semiconductor, it is moderate and in conductors the energy-gap is zero.

10. The temperature coefficient of resistance of copper is

6. Conductor has positive temperature coefficient of resistance

11. In the given circuit, p side p-n junction D1 is connected to

but semiconductor has negative temperature coefficient of resistance.

lower voltage and n-side of D1 to higher voltage. Therefore, D1 is reverse biased. The p-side of p-n junction D2 is at higher potential and n-side of D2 is at lower potential. Therefore, D2 is forward biased. Hence, current flows through the junction from B to A i.e., option (b) is correct.

7. In sample x, no impurity energy level seen, so it is undoped. In sample y, impurity energy level lies below the conduction band, so it is doped with fifth group impurity. In sample z, impurity energy level lies above the valence band, so it is doped with third group impurity.

8. As, I = nAev d or

I µ nv d

positive and germanium is negative.

12. Electrical conductivity of a semiconductor increases with rise in temperature because more covalent bonds will be broken with rise in temperature. Due to which more number of electrons and holes will be available for the conduction of electricity in a semiconductor.

Electronic Devices 13. Phosphorous is pentavalent and Boron is trivalent material.

25. The ripple factor for full wave rectifier is 0.482 which is 48.2%.

14. The probability of occupation of the highest electrons state in valence band at room temperature becomes half according to Fermi distribution.

26. The average value of output direct current in a half-wave rectifier is = (average value of current over a cycle)/2 = (2I0 /p ) /2 = I0 /p

15. Number density of atoms in silicon specimen = 5 ´ 10 28 2

7

atoms/m = 5 ´ 10 silicon atoms, so total number of indium atoms doped per atoms, so total number of indium atoms doped per cm3 of silicon will be n = 5 ´ 10 22/5 ´ 10 7 = 10 15 atoms cm-3 .

27. Current in junction diode, IF = I0( eeV / kT - 1) In forward biasing, V is positive and in reverse bias V is negative. Then, Ir = I0 IF I0( eeV / kT - 1) = Ir I0

16. Intrinsic semiconductor ® conductivity is due to the breaking the covalent band Extrinsic semiconductor ® conductivity is due to the breaking of covalent bond and excess of charge carriers due to impurity.

= ( eeV / kT - 1)

28. Here, p-n junction as forward biased with voltage

17. The voltage gain is low at high and low frequencies and constant at mid frequency.

\

18. Energy of incident light (photon), hv =

6.6 ´ 10 -34 ´ 3 ´ 10 8 = 2.06 eV 6 ´ 10 -7 ´ 1.6 ´ 10 -19

For the incident radiation to be detected by the photodiode, energy of incident radiation (photon) should be greater than the band gap. This is true only for D2. V0 200 = 2 2

20. In circuit 1, n is connected with n, which is not a series combination of p- n junction. In circuit 2, each p-n junction is forward biased, same current flows, giving same potential difference across junction. In circuit 2, each p-n junction is reverse biased, same leakage current will flow, giving equal potential difference across each p-n junction diode.

21. In half-wave rectifier, we get the output only in one-half cycle of input AC, therefore, the frequency of the ripple of the output is same as that of input AC i. e. , 50 Hz.

22. During positive half cycle, p-n junction conducts. Potential difference across C = peak voltage of the given AC voltage = V0 = Vrms = 2 = 220 2 V.

23. V -d curve near the junction will be as shown by curve (a). 24. Given, voltage gain of first amplifier, AV1 = 10 Voltage gain of second amplifier, AV2 = 20 Input voltage Vi = 0.01 V V Total voltage gain AV = o = AV1 ´ AV2 Vi \

Vo = 10 ´ 20 0.01 Vo = 2 V

=5 -3 =2 V 2 1 Current, I = = =`10 -2 A 200 100

30. The forward voltage, overcomes the barrier voltage. Due to which the forward current is high but depends upon the forward voltage applied. The reverse voltage supports the barrier voltage, due to which the reverse current is low.

31. For forward biasing of p-n junction, the positive terminal of

19. If half-wave rectifier the output voltage is the RMS voltage =

1185

32.

external battery is to be connected to p-type semiconductor and negative terminal of battery to the n-type semiconductor. DV (13 - 8) As, Re = = 2.5 ´ 10 5 W = DI (60 - 40) ´ 10 -6

33. As, If =

4 -1 1 = = 10 -2 A 300 100

35. During positive half cycle of input AC voltage, the p-n junction is forward biased. The resistance of p-n junction is low. The current in the circuit is maximum. In this situation, a maximum potential difference will appear across resistance connected in series of circuit. Due to it, there is no output voltage across p-n junction. During the negative half-cycle of input AC voltage the p-n junction is reverse biased. The resistance of p-n junction becomes high which will be more than resistance in series. Due to it, there will be voltage across p-n junction with negative cycle in output.

36. In forward biasing both positive and negative charge carriers move towards the junction.

37. When polarity of the battery is reversed, the p-n junction becomes reverse biased. So, therefore, no current flows.

38. In circuit, C = A × B and D = A × B E = C + D = ( A × B) + ( A + B)

1186 JEE Main Physics truth table of this arrangement of gates be as given below A

B

A

C = A× B

d = A×B

E = (C + D)

0 0 1 1

0 1 0 1

1 1 0 0

0 0 0 1

0 1 0 0

0 1 0 1

Ic =10 mA

50. Here,

95 95 Ie Þ 10 mA= Ie 100 100 100 ´ 10 2 Ie = = 10.53 mA Þ ic = = 10 -3 A 95 2 ´ 10 3

\

Ic =

ro

\Current gain

b=

39. Let V be the potential difference between A and B. Then V - 0.3 = (5 + 5) ´ 10 3 ´ (0.2 ´ 10 -3) = 2

100 =

V = 2 + 0.3 = 2.3 V 90 As, collector current, IC = ´ IE 100 where, IE = emitter current Þ 10 = 0.9 ´ IE Þ IE = 11mA Also, IE = IB + IC Base current IB = IE - IC = 11 - 10 = 1mA or

40.

Vi = Rbib ´ 1 ´ 10 3 ´ 10 -5 = 10 -2 V = 10 mV

52. When A = 1, and when then

Ic Ib V 0.01 Ib = i = 3 = 10 -5 A RI 10 Ic = 50 ´ 10 -5 A = 500 mA

... (i)

53. The input to AND gate will be A and B. So, the output is Y = A × B.

54. For this, the boolean expression is Y = A + B, which is for [from Eq. (i)]

Ic > 1or Ic > Ib Ib I As, a = c < I or Ic < Ie Ie

42. As, b = 43.

A A = 1.0 = 0 A = 0, A × A = 0.1 = 0

then

b=

then,

NOR gate.

55. Here, the output of AND gate is made as the input of NOT gate, we get NAND gate.

56. In NOR gate, Y = A + B 0 + 0 = 0 = 1, 1 + 0 = 1 = 0

i. e. ,

44. If K is a gain of one stage, then total gain of n stages

0 + 1= 1 = 0

= (K) n = 10 3 = 1000

45. We know that, b =

10 -3 ib

ib = 10 -5 A

Þ

41. Given, b = 50 , Ri = 1000 W , Vi = 0.01 V

and

iC ib

DIc DIb

DIc DbDIb = 40 ´ 100 mA a 0.95 0.95 As, b = = = 19 = 1 - a 1 - 0.95 0.05

1+ 1= 1 = 0

and A

or

46.

47. As, b =

48.

B

Ic Ie - Ib Ie I = = - 1or, e = 1 + b Ib Ib Ib Ib

8.2 8.2 Ie = = = 0.20 mA 1 + b 1 + 40 41 R 5000 As, AV = b o = 60 ´ = 600 Ri 500 or

Y

Ib =

Y=A+B According to Demorgan’s theorem, Y = A + B = A × B = AB This is the output equation of AND gate.

49. Number of electrons reaching the collector, 96 ´ 10 10 = 0.96 ´ 10 10 100 n ´e Emitter current, Ie = e t nc ´ e Collector current, Ie = t Thus, current transfer ratio, nc =

a=

Ic nc 0.96 ´ 10 10 = = = 0.96 Ie ne 10 10

57. From the figure, the truth table A

B

Y

0 0 1 1

0 1 0 1

0 0 0 1

i. e. , AND gate.

Electronic Devices

1187

Round II 1. In NaCl, the Na + and Cl- ions both have noble gas electronic

Current in the circuit I =

configuration corresponding to completely filled bands. Since, the bands do not overlap, there must be a gap between the filled bands and the empty bands above them. Thus, NaCl is an insulator.

Current in arm AB = I =

Also, Þ

b=

0.8 0.8 = =4 (1 - 0.8) 0.2

b=

Dic Dib

positive (ii) the fraction of the output voltage feedback i. e. , 1 b= i. e. , the reciprocal of the voltage gain without A feedback.

Dic = b ´ Dib = 4 ´ 6 = 24 mA

10. Here Boolean expression is, Y = ( A + B) × C. So, Y = A × B = 0.1 = 1 as the boolean expression of it is, Y = A × B.

Dib = Die - Dic = 2 - 1.96 = 0.04 mA

(for NAND gate)

4. We have, ie = 2b + ic Þ

11. In circuit the upper diode junction is forward biased and the

ic = ie - ib

5. Voltage gain, Av = b

lower diode junction is reverse biased. Thus, there will be no conduction across lower diode junction. Now the total resistance of circuit = 100 + 150 + 50 = 300 W 6 Current in 100 W = = 0.02 A 300

R2 R1

Also current gain, b =

a 0.98 = = 49 1 - a 1 - 0.98

æ 500 ´ 10 3 ö Av = 49 ç ÷ R1 ø è \

Power gain = 6.0625 ´ 10 6

12. The junction diode I will provide output when forward …(i)

biased. It will be so during negative half cycle of input AC voltage applied.

…(ii)

13. For first case, the Boolean expression is, Y = A × B = A + B hence for OR gate and for second case, the Boolean expression is Y = A × B = A × B, and hence for AND gate.

From (i) and (ii), we get 3

49 ´ 500 ´ 10 = 6.0625 ´ 10 6 R1

14. In p-n junction, the barrier potential offers resistance to free

49 ´ 5 ´ 10 5 6.0625 ´ 10 6

15. Positive charge (i. e. , holes) should move in the direction of

Þ

R1 =

\

R1 = 198W

6. Output of G1 = ( A + B)

electrons in n-region and holes in p-region. current and negative charge (i. e. , electrons) should move opposite to the direction of current.

16. In intrinsic semiconductor of Fermi level is near the middle

Output of G 2 = A × B Output of G3 = ( A + B) × A × B

7.

2 ´ 5 ´ 10 3 = 10 V 10 3

9. The conditions for a circuit to oscillate are (i) the feedback is

3. As, Dic = aDie Þ 0.92 ´ 2 = 1.96 \

2 10 3

Potential difference across A and B =

2. Here, a = 0.8 We have,

V 30 2 A = = R 15 ´ 10 3 10 3

Which give XOR gate mR mR As, A v = = rp + R (m/g m) + R =

42 ´ (50 ´ 10 3) 42 / (2 ´ 10 -3) + 50 ´ 10 3

= 29.57

8. Here p-n junction is forward biased. If p-n junction is ideal, its resistance is zero. The effective resistance across A and B 10 ´ 10 = = 5 kW 10 + 10

17.

of the forbidden gap. DI As, a = c Þ DIc = 0.98 ´ 2 = 1.96 mA DIc and DIb = DIe - DIc = 2 - 1.96 = 0.04 mA

18. The

temperature coefficient of resistance of silicon (i. e. , semiconductor) is negative and that of platinum (i. e. , conductor) is positive.

19. It is a circuit of full wave rectifier. 20. By doping a semiconductor with some impurity atoms increases the conductivity and hence decreases the resistivity (as r = 1/ s ) .

21. The reverse biasing of p-n junction supports the barrier voltage, due to which the current through the junction due to majority carriers becomes nearly zero.

1188 JEE Main Physics Ie Ib = +1 Ic Ic 1 1 1+ b b or a = = + 1= a b b 1+ b

32. If A = 1, B = 1 and Y = 0 , the gate can be NOR gate, NAND

22. As, Ie = Ib + Ic or or

gate or exclusive NOR gate (i. e. , XOR gate).

33. In the circuit, the upper diode D1 is reverse biased and the

23. Current through circuit, P 100 ´ 10 -3 = = 0.2 A 0.5 V Voltage drop across R = 1.5 - 0.5 = 1.0 V

lower diode D2 is forward biased. Thus there will be no current across upper diode junction. The effective circuit will be shown in figure.

I=

70 Ω

Hence, R = 1/0.2 = 5W input and output circuit. Thus the circuit is of n-p-n transistor with a common emitter amplifier mode.

25. In p-type semiconductor holes are majority carriers and electrons are minority carriers.

26. As, VB = Vknee + IR 4 = 0.7 + 10 -3R

or

R = 3.3 /10 -3 = 3.3 ´ 10 3 W R DV0 =b 0 Ri DVi

So, DV0 = DVi ´ b \

500 = 1250 mV = 1.25 V 2000

28. The input of OR gate is A and ( A × B)

R = 50 + 70 + 30 = 150 W V 3 Current in circuit, I = = = 0.02 A R 150 junction diode is reverse biased. So, effective resistance of circuit = 10 + 10 = 20 W 3 I= = 0.15 A 20 B ×C = Y ¢ At logic gate II, the Boolean expression is A + (B × C) = Y ¢ ¢ At logic gate III, the Boolean expression is A + (B × C) = Y

Y = A + ( A × B)

Hence,

29.

The resistance of circuit

35. At logic gate I, the Boolean expression is

R0 Ri

DV0 = 10 ´ 50 ´

+

34. The upper junction diode is forward biased and middle

or

27. Av =

36. For a transistor b =

rl L We have, R = = A ni e (m e + m h) A =

2.5 ´ 10 19

0.928 ´ 10 -2 ´ 1.6 ´ 10 -19(0.39 + 0.19) ´ 10 -6

or

30. If V is the voltage across the junction and I is the circuit current, then V + IR = E E V V E or I= - =- + R R R R 1 1 \Slope of load line = - = = 10 -3 AV -1 R 1000

31. Voltage available across load resistance R = 30 - 5 = 15 V Resistance of load,

= i.e.,

Rinput µ

IC IB IC b Vinput IB Vinput IC

×b

1 IC

Therefore Rinput is inversely proportional to the collector current. For high collector current, the Rinput should be small for which the base region must be very thin and lightly doped for a transistor action, the emitter junction is forward biased and collector junction is reverse biased.

37. When electric field is applied across a semiconductor, the

R 10 mA

R=

IB = Rinput =

= 4000 W

20 V

50 Ω

3V

24. Here, emitter is forward biased and is common between

5V

15 = 1.5 ´ 10 3 = 1.5 kW 10 ´ 10 -3

electrons in the conduction band get accelerated and acquired energy. They move from lower energy level to higher energy level. While the holes in valence band move from higher energy level to lower energy level, where they will be having more energy.

Electronic Devices 38. The circuit of n-p-n transistors as CE amplifier is as shown in figure. Then, the base emitter are forward biased and input signal is connected between base and emitter. C

1189

46. As, Ib = Ie - Ic = 2 - 1.06 = 0.04 mA 47. As, Av = a

Ie

R0 ( 400 ´ 10 3) = 0.98 ´ = 980 Ri 400

48. As, V0 = Vi ´ voltage gain = 0.1 ´ 980 = 98 V

Ib

49. Given, collector resistance Routput = 2 kW = 2000 W R

Output

Current amplification factor of the transistor b AC = 100 Voutput = 2 V

Audio signal voltage

Rinput = 1kW = 1000 W Voutput Routput Voltage gain AV = = b AC Vinput Rinput

Input (base) resistance

Input

Q

39. In an n-p-n transistor when base emitter junction is forward biased and collector base junction is reverse biased, the majority charge carriers electrons move from emitter to base. In base few electrons get neutralised by electron-hole combination and the remaining electrons cross over the collector.

Ie =

100 100 Ic = ´ 10 = 11mA 90 90

Ib = Ie - Ic = 11 - 10 = 1mA

41. In reverse biasing, the minority charge carriers will be accelerated due to reverse biasing, which on striking with atoms cause ionisation resulting secondary electrons and thus more number of charge carriers. When doping concentration is large, there will be large number of ions in the depletion region, which will give rise to a strong electric field.

Base (input) current IB =

43. Here, Ic = 10 mA, Ic = \

10 mA =

95 le 100

95 100 = 10.53 mA le or Ie = 10 ´ 100 95

Vinput

45. As, Ic = aIe = 0.98 ´ 2 = 1.96 mA

=

0.01 1000

= 10 ´ 10 -6 A = 10 mA

51. Resistivity of semiconductor decreases with temperature. The atoms of a semiconductor vibrate with large amplitudes at higher temperatures there by increasing conductivity not resistivity.

52. When the reverse voltage across the zener diode is equal to or more than the breakdown voltage, the reverse current increases sharply. Input (A)

Output (A)

53. As, capacitive resistance X c =

1 1 = wC 2pnC

54. In vacuum tubes, vacuum is necessary and the working of semiconductor devices is independent of heating or vacuum.

55. Assertion is true but reason is false A

AB

B Y = (AB).C

C

If A = 1, B = 0 , C = 1, then Y = 0

56. For detection of a particular wavelength ( l) by a p-n photo diode, energy of incident light > Eg Þ For Þ

44. Power amplification, ap = aAV = 0.98 ´ 980 = 960

2 = 0.01 V 100 (2000 /1000)

Rinput

42. From the given transfer characteristics of a base biased common emitter transistor, we note that (i) When Vi = 0.4 V, there is no collector current. The transistor circuit is in cut-off state. (ii) When Vi = 1 V (which is in between 0.6 V to 2 V), the transistor circuit is in active state and it used as an amplifier. (iii) When Vi = 0.5 V, there is no collector current. The transistor is in cut-off state. The transistor circuit can be used as a switch turned off. (iv) When Vi = 2.5 V, the collector current becomes maximum and transistor is in saturation state and can be used as switch turned on state.

b AC(Routput /Rinput )

=

Ic = 10 mA

40. Here,

Voutput

\ Input signal voltage Vinput =

i. e. ,

hc >l Eg

Eg = 2.8eV hc 6.6 ´ 10 -34 ´ 3 ´ 10 8 = 441.9 nm l= = Eg 2.8 ´ 1.6 ´ 10 -19 hc < 6000 nm, Eg

So diode will no wavelength of 6000 Å

1190 JEE Main Physics 57. Depletion layer of semiconductor is affected by various

68. Truth table for given combination is

factors e. g . , biasing, temperature condition etc. So, it cannot be a fixed layer. In depletion layer, negatively and positively immobile ions are present. Therefore, no free charge carriers are available in such layer.

58. As, Rf =

DVf (2 - 0.5) V = 10 3 W = 1kW = DIf 1.5 ´ 10 -3 A

A

B

X

0 0 1

0 1 1

0 1 1

This comes out to be truth table for OR gate.

69. For, Vi < 0, 5V

59. Assertion is not true as resistivity of a semiconductor decreases with increase of temperature. The atoms of a semiconductor vibrate with large amplitudes at higher temperatures thereby increasing its conductivity and not its resistivity.

VI

–5 V

60. When base region has larger width, electrons hole

RL

Vo

Fig. (i)

combination increases the base current. The output collector current decreases from Ie + Ib + Ic = constant. Vi

61. In a common base configuration,

I I -I I a = C = E B =1- B IE IE IE IB = 1 - a = 1 - 0.95 = 0.05 IE

or

62. The p-side of p-n junction is taken at higher potential than n-side so, p-n junction is forward biased. Taking its resistance to be zero and on applying Ohm’s law, we get V 5 -2 = = = 10 -2 A = 10 -2 ´ 10 3 mA = 10 mA R 300

63. NAND or NOR gates are called universal (digital) building blocks because using repeated order of these two types of gates we can produce all the basic gates namely, OR, AND or complex gates.

64. The co-ordination number of face centered crystal is 12. It is the number of the closest neighbouring atom in a crystal structure.

65. NOT gate inverts the signal applied to it. But in diode, the input and output are in same phase. Thus, NOT gate cannot be built by a diode.

66. NOT gate inverts the input signal i. e. , if input is 1, then output will be zero or vice-versa. Therefore, it is called as invertor. NOT gate inverts the input order means that for low input, it gives high output of for high input, it gives low output.

67. The output, Y ¢ = A + B A B

and

Vo

VI

Fig. (iii)

Fig. (ii)

I IE = B mA = 1200 mA 0.05

So,

Vo

The diode is reverse biased and hence offer infinite resistance, so, circuit would be like as shown in Fig. (ii) and Vo = 0. For Vi > 0 , the diode is forward biased and circuit would be as shown in Fig. (iii) and Vo = Vi .

70. If the reverse biased is greater than the V2, there is breakdown condition. In breakdown region i. e.; V1 > V2. For a long range of load, R2, the voltage remains the constant through the current may be large.

I (mA) a V2 d

c

b

V (V)

e I (µA)

71. The given circuit works as half-wave rectifier. In this circuit, we will get current through R when p-n junction diode is forward biased and there is no current when p-n junction is reversed biased. Thus the current through resistance will be shown by graph (b).

72. For this purpose we use NAND gate in manner as shown. The first two NAND gates are operated as NOT gates and their outputs are fed to the third. The resulting circuit is OR gate. A

A Y = (A . B) = A+ B

B

B

Here, three gates are used. Y′

Y =Y¢ = A + B

i. e. , output of a NOR gate.

Y

73. In a transistor the base is conductor of low resistance. 74. The given figure is the symbol of NAND gate. (i. e. , combination fo NOT and AND gates)

75. When diode is reverse biased, then the applied voltage supports the barrier voltage. Due to it, the reverse current is weak. It will be identical in two diodes.

Atoms, Molecules 27 and Nuclei JEE Main MILESTONE < < < < < < <