Motion IIT JEE Physics Class 11 and 12

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UNIT AND DIMENSION & BASIC MATHEMATICS THEORY AND EXERCISE BOOKLET CONTENTS

S.NO.

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TOPIC

PAGE NO.

THEORY WITH SOLVED EXAMPLES ........................................ 3 – 23

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EXERCISE - I ......................................................................... 24 – 34

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EXERCISE - II ........................................................................ 35 – 36

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EXERCISE - III ........................................................................ 37 – 39

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ANSWER KEY ............................................................................40

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 www. motioniitjee.com , [email protected]

Page # 2

UNIT AND DIMENSION

394,50 - Rajeev Gandhi Nagar Kota, Ph. No. : 93141-87482, 0744-2209671 IVRS No : 0744-2439051, 52, 53, www. motioniitjee.com , [email protected]

Page # 3

UNIT AND DIMENSION 1.

PHYSICAL QUANTITY The quantites which can be measured by an instrument and by means of which we can describe the laws of physics are called physical quantities.

Types of physical quantities :

Fundamental 1.1

Derived

Supplementry

Fundamental Although the number of physical quantities that we measure is very large, we need only a limited number of units for expressing all the physical quantities since they are interrelated with one another. So, certain physical quantities have been chosen arbitrarily and their units are used for expressing all the physical quantities, such quantities are known as Fundamental, Absolute or Base Quantities (such as length, time and mass in mechanics) (i) All other quantites may be expressed in terms of fundamental quantities. (ii) They are independent of each other and cannot be obtained from one another. An international body named General Conference on Weights and Measures chose seven physical quantities as fundamental : (1) length

(2) mass

(3) time

(5) thermodynamic temperature

(4) electric current, (6) amount of substance

(7) luminous intensity. Note : These are also called as absolute or base quantities. In mechanics, we treat length, mass and time as the three basic or fundamental quantities. 1.2

Derived : Physical quantities which can be expressed as combination of base quantities are called as derived quantities.

For example : Speed, velocity, acceleration, force, momentum, pressure, energy etc. dis tan ce length = time time

Ex.1

Speed =

1.3

Supplementary : Beside the seven fundamental physical quantities two supplementary quantities are also defined, they are : (1) Plane angle

(2) Solid angle.

Note : The supplementary quantities have only units but no dimensions. 2.

MAGNITUDE : Magnitude of physical quantity = (numerical value) × (unit) Magnitude of a physical quantity is always constant. It is independent of the type of unit. ⇒ numerical value ∝ or

Ex.2

1 unit

n1u1 = n2u2 = constant

Length of a metal rod bar is unchanged whether it is measured as 2 metre or 200 cm. Observe the change in the Numerical value (from 2 to 200) as unit is changed from metre to cm.

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Page # 4 3.

UNIT AND DIMENSION

UNIT : Measurement of any physical quantity is expressed in terms of an internationally accepted certain basic reference standard called unit. The units for the fundamental or base quantities are called fundamental or base unit. Other physical quantities are expressed as combination of these base units and hence, called derived units. A complete set of units, both fundamental and derived is called a system of unit.

3.1.

Principle systems of Unit There are various system in use over the world : CGS, FPS, SI (MKS) etc Table 1 : Units of some physical quantities in different systems. System Physica l Qua ntity

CGS (Ga ussia n)

MKS (SI)

FPS (British)

Length

centimeter

meter

foot

Mass

gram

kilogram

pound

Time

second

second

second

Force

dyne

newton → N

poundal

W ork or Energy

erg

joule → J

ft-poundal

Power

erg/s

watt → W

ft-poundal/s

Fundame ntal

De rived

3.2

Supplementary units : (1) Plane angle : radian (rad) (2) Solid angle : steradian (sr)

*

The SI system is at present widely used throughout the world. In IIT JEE only SI system is followed.

3.3

Definitions of some important SI Units (i) Metre : 1 m = 1,650, 763.73 wavelengths in vaccum, of radiation corresponding to organ-red light of krypton-86. (ii) Second : 1 s = 9,192, 631,770 time periods of a particular from Ceasium - 133 atom. (iii) Kilogram : 1kg = mass of 1 litre volume of water at 4°C (iv) Ampere : It is the current which when flows through two infinitely long straight conductors of negligible cross-section placed at a distance of one metre in vacuum produces a force of 2 × 10–7 N/m between them. (v) Kelvin : 1 K = 1/273.16 part of the thermodynamic temperature of triple point of water. (vi) Mole : It is the amount of substance of a system which contains as many elementary particles (atoms, molecules, ions etc.) as there are atoms in 12g of carbon - 12. 1   2 m of a (vii) Candela : It is luminous intensity in a perpendicular direction of a surface of   600000  black body at the temperature of freezing point under a pressure of 1.013 × 105 N/m2.

(viii) Radian : It is the plane angle between two radiia of a circle which cut-off on the circumference, an arc equal in length to the radius. (ix) Steradian : The steradian is the solid angle which having its vertex at the centre of the sphere, cut-off an area of the surface of sphere equal to that of a square with sides of length equal to the radius of the sphere.

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Page # 5

UNIT AND DIMENSION Ex.3

Find the SI unit of speed, acceleration

Sol.

speed =

meter(m) dis tan ce = = m/s sec ond(s) time

acceleration = = 4.

(called as meter per second)

velocity displacement / time = time time

displacement ( time) 2

=

meter

= m/s2 (called as meter per second square)

sec ond2

S I PREFIXES The magnitudes of physical quantities vary order a wide range. The CGPM recommended standard prefixes for magnitude too large or too small to be expressed more compactly for certain power of 10. Power of 10 18

10 15 10 12 10 9 10 6 10 3 10 2 10 1 10

5.

Prefix

Symbol

exa peta tera giga mega kilo hecto deca

Power of 10

E P T G M k h da

–1

10 –2 10 –3 10 –6 10 –9 10 –12 10 –15 10 –18 10

Prefix

Symbol

deci centi milli micro nano pico femto atto

d c m µ

n p f a

GENERAL GUIDELINES FOR USING SYMBOLS FOR SI UNITS, SOME OTHER UNITS, SOME OTHER UNITS, AND SI PREFIXES (a) Symbols for units of physical quantities are printed/written in Roman (upright type), and not in italics For example : 1 N is correct but 1 N is incorrect (b)

(i) Unit is never written with capital initial letter even if it is named after a scientist.

For example : SI unit of force is newton (correct) Newton (incorrect) (ii)

For a unit named after a scientist, the symbol is a capital letter. But for other units, the symbol is NOT a capital letter.

For example : force

→

newton (N)

energy

→

joule (J)

electric current

→

ampere (A)

temperature

→

kelvin (K)

frequency

→

hertz (Hz)

length

→

meter (m)

mass

→

kilogram (kg)

luminous intensity

→

candela (cd)

time

→

second (s)

For example :

Note : (c)

The single exception is L, for the unit litre.

Symbols for units do not contain any final full stop all the end of recommended letter and remain unaltered in the plural, using only singular form of the unit.

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Page # 6

UNIT AND DIMENSION

For example :

(d)

Quantity

Correct

Incorrect

25 centimeters

25 cm

25 c m 25 cms.

Use of solidus ( / ) is recommended only for indicating a division of one letter unit symbol by another unit symbol. Not more than one solidus is used.

For example : Correct m/s

2

m/s/s 2

N s / m/ m

J/K mol

J / K / mol

kg/m s

kg / m / s

N s/m

(e)

Incorrect

Prefix symbols are printed in roman (upright) type without spacing between the prefix symbol and the unit symbol. Thus certain approved prefixes written very close to the unit symbol are used to indicate decimal fractions or multiples of a SI unit, when it is inconveniently small or large.

For example megawatt

1 MW = 10 6 W

centimetre

1 cm = 10 –2 m

kilometre

1 km = 10 3 m

millivolt

1 mV = 10 –3 V

kilowatt-hour

1 kW h = 10 3 W h = 3.6 M J = 3.6 × 10 6 J

microampere

1 µ A = 1 0 –6 A

angstrom

1 Å = 0 .1nm = = 1 0 –9

– 10

nanosecond

1 ns = 10

picofarad

1 pF = 10 –12 F

microsecond

1 µs = 1 0 – 6 s

gigahertz

1 GHz = 10 9 Hz

micron

1 µm = 1 0 –6 m

m

s

The unit 'fermi', equal to a femtometre or 10–15 m has been used as the convenient length unit in nuclear studies. (f)

When a prefix is placed before the symbol of a unit, the combination of prefix and symbol is considered as a new symbol, for the unit, which can be raised to a positive or negative power without using brackets. These can be combined with other unit symbols to form compound unit.

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Page # 7

UNIT AND DIMENSION

For example :

Correct

Quantity

(g)

Incorrect

cm3

(cm)3 = (0.01 m)3 = (10–2m)3 = 10–6 m3

mA2

(mA)2 = (0.001 A)2 = (10–3A)2 = 10–6 A2

0.01 m3 or 10–2 m3 0.001 A2

A prefix is never used alone. It is always attached to a unit symbol and written or fixed before the unit symbol.

For example : 103/m3 = 1000/m3 or 1000 m–3, but not k/m3 or k m–3. (h)

Prefix symbol is written very close to the unit symbol without spacing between them, while unit symbols are written separately with spacing with units are multiplied together.

For example : Quantity

Correct

Incorrect

1 metre per second

1 milli per second

1 ms

1 millisecond

1 metre second

1Cm

1 coulomb metre

1 centimetre

1 cm

1 centimetre

1 coulomb metre

1 ms

(i)

–1

The use of double profixes is avoided when single prefixes are available.

For example :

Quantity

Incorrect

10–9 m

1 nm (nanometre)

1 mµ m (milli micrometre)

10–6 m

1µm (micron)

1 m m m (milli millimetre)

10–12

1 pF (picofarad)

1 µ µ F (micro microfarad)

1 GW (giga watt)

1 kM W (kilo megawatt)

F

109 F

(j)

Correct

The use of a combination of unit and the symbols for unit is avoided when the physical quantity is expressed by combining two or more units.

Quantity

Correct

Incorrect

joule per mole Kelvin

J/mol K or J mol–1 K–1

Joule / mole K or J/mol Kelvin or J/mole K

newton metre second

Nms

newton m second or N m second or N metre s or newton metre s

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Page # 8

UNIT AND DIMENSION

5.1. Characteristics of base units or standards : (A) Well defined (B) Accessibility (C) Invariability (D) Convenience in use 5.2

Some special types of units : 1. 1 Micron (1µ) = 10–4 cm = 10–6 m (length) 2. 1 Angstrom(1 Å) = 10–8 cm = 10–10m (length) 3. 1 fermi (1f) = 10–13 cm = 10–15 m (length) 4. 1 inch = 2.54 cm (length) 5. 1 mile = 5280 feet = 1.609 km (length) 6. 1 atmosphere = 105 N/m2 = 76 torr = 76 mm of Hg pressure (pressure) –3 3 3 7. 1 litre = 10 m = 1000 cm (volume) 8. 1 carat = 0.0002 kg (weight) 9. 1 pound (Ib) = 0.4536 kg (weight)

6.

DIMENSIONS Dimensions of a physical quantity are the power to which the fundamental quantities must be raised to represent the given physical quantity. For example,

density =

mass mass = (length )3 volume

or density = (mass) (length)–3 ...(i) Thus, the dimensions of density are 1 in mass and –3 in length. The dimensions of all other fundamental quantities are zero. For convenience, the fundamental quantities are represented by one letter symbols. Generally mass is denoted by M, length by L, time by T and electric current by A. The thermodynamic temperature, the amount of substance and the luminous intensity are denoted by the symbols of their units K, mol and cd respectively. The physical quantity that is expressed in terms of the base quantities is enclosed in square brackets. [sinθ] = [cosθ] = [tanθ] = [ex] = [M0L0T0] 7.

DIMENSIONAL FORMULA It is an expression which shows how and which of the fundalmental units are required to represent the unit of physical quantity. Different quantities with units. symbol and dimensional formula. Quantity Displacement Area Volume

s A V

Symbol

Formula

×b
×b×h

Velocity

v

v=

Momentum

p

p = mv

Acceleration

a

a=

Force Impulse Work

F W

F = ma F × t F.d

∆s ∆t ∆v ∆t

S.I. Unit Metre or m (Metre)2 or m2 (Metre)3 or m3

D.F. M0LT0 M0L2T0 M0L3T0

m/s

M0LT–1

kgm/s

MLT–1

m/s2

M0LT–2

Newton or N N.sec N.m

MLT–2 MLT–1 ML2T–2

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Page # 9

UNIT AND DIMENSION

Energy

KE or U

K.E. =

1 mv 2 2

Joule or J

ML2T–2

watt or W

ML2T–3

kg/m3 Pascal or Pa N.m.

ML–3T0 ML–1T–2 ML2T–2

P.E. = mgh W t

Power

P

P=

Density Pressure Torque

d P τ

d = mass/volume P = F/A τ=r×F

Angular displacement θ

θ=

arc radius

radian or rad

M0L0T0

Angular velocity

ω

ω=

θ t

rad/sec

M0L0T–1

Angular acceleration

α

α=

∆ω ∆t

rad/sec2

M0L0T–2

Moment of Inertia

I

I = mr2

kg-m2

ML2T0

Frequency

v or f

f=

hertz or Hz

M0L0T–1

Stress

-

F/A

N/m2

ML–1T–2

Strain

-

∆
∆A ∆V ; ;
A V

-

M0L0T0

Youngs modulus

Y

Y=

N/m2

ML–1T–2

T

F W or
A

N J ; m m2

ML0T–2 ML0T–2

1 T

F/A ∆
/


(Bulk modulus of rigidity)

Surface tension

Force constant (spring)

k

F = kx

N/m

Coefficient of viscosity

η

 dv  F = η A  dx 

kg/ms(poise in C.G.S.) ML–1T–1

G

F=

Gravitation constant

Gm1 m 2 r2

Vg =

Gravitational potential Vg

PE m

N − m2 kg 2

J kg

M–1L3T–2

M0L2T–2

Temperature Heat

θ Q

Q = m × S × ∆t

Kelvin or K M0L0T0θ+1 Joule or Calorie ML2T–2

Specific heat

S

Q = m × S × ∆t

Joule kg .Kelvin

M0L2T–2θ–1

Latent heat

L

Q = mL

Joule kg

M0L2T–2

Coefficient of thermal

K

Q=

Joule m sec K

MLT–3θ–1

KA( θ1 − θ2 )t d

conductivity

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Page # 10

UNIT AND DIMENSION

Universal gas constant

R

PV = nRT

Joule mol.K

ML2T–2θ–1

Mechanical equivalent

J

W = JH

-

M0L0T0

Coulomb or C

M0L0TA

Ampere or A

M0L0T0A

of heat Charge

Q or q

I=

Q t

Current

I

-

Electric permittivity

ε0

ε0 =

1 q1q2 . 4πF r 2

Electric potential

V

V=

∆W q

Joule/coul

ML2T–3A–1

Intensity of electric field

E

E=

F q

N/coul.

MLT–3A–1

Capacitance

C

Q = CV

Farad

M–1L–2T4A2

Dielectric constant

εr

εr =

ε ε0

-

Resistance

R

V = IR

Ohm

ML2T–3A–2

Conductance

S

S=

1 R

Mho

M–1L–2T–3A2

Specific resistance

ρ

ρ=

RA


s

σ=

1 ρ

B

F = qvBsinθ

(coul.)2 N.m 2

or

C2 N − m2

M–1L–3T4A2

M0L0T0

or relative permittivity

Ohm × meter

ML3T–3A–2

or resistivity Conductivity or

Mho/meter

M–1L–3T3A2

specific conductance Magnetic induction

Tesla or weber/m2

MT–2A–1

or F = BIL dφ dt

Magnetic flux

φ

e=

Magnetic intensity

H

B=µH

Magnetic permeability

µ0

B=

L

e = L.

Electric dipole moment

p

Magnetic dipole moment

M

µ 0 Idl sin θ 4π r 2

Weber

ML2T–2A–1

A/m

M0L–1T0A

N amp 2

MLT–2A–2

Henery

ML2T–2A–2

p = q × 2


C.m.

M0LTA

M = NIA

amp.m2

M0L2AT0

of free space or medium Coefficient of self or

dI dt

Mutual inductance

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Page # 11

UNIT AND DIMENSION

8.

USE OF DIMENSIONS

8.1

Theory of dimensions have following main uses : Conversion of units : This is based on the fact that the product of the numerical value (n) and its corresponding unit (u) is a constant, i.e., n[u] = constant or n1[u1] = n2 [u2] Suppose the dimensions of a physical quantity are a in mass, b in length and c in time. If the fundamental units in one system are M1, L1 and T1 and in the other system are M2, L2 and T2 respectively. Then we can write.

n1[M1a Lb1 T1c ] = n 2 [Ma2 Lb2 T2c ]

...(i)

Here n1 and n2 are the numerical values in two system of units respectively. Using Eq. (i), we can convert the numerical value of a physical quantity from one system of units into the other system. Ex.4 The value of gravitation constant is G = 6.67 × 10–11 Nm2/kg2 in SI units. Convert it into CGS system of units. Sol. The dimensional formula of G is [M–1 L3 T–2]. Using equation number (i), i.e.,

n1[M1–1 L31 T1–2 ] = n 2 [M2–1 L32 T2–2 ] –1

3

M  L   T  n 2 = n1  1   1   1   M2   L 2   T2 

–2

Here, n1 = 6.67 × 10–11 M1 = 1 kg, M2 = 1 g = 10–3 kg L1 = 1 m, Substituting in the above equation, we get –1

n2 = 6.67 × 10

–11

3

L2 = 1cm = 10–2 m,

 1kg   1m  1s   –3   – 2     10 kg  10 m  1s 

T1 = T2 = 1s

–2

or n2 = 6.67 × 10–8 Thus, value of G in CGS system of units is 6.67 × 10–8 dyne cm2/g2. 8.2 To check the dimensional correctness of a given physical equation : Every physical equation should be dimensionally balanced. This is called the 'Principle of Homogeneity'. The dimensions of each term on both sides of an equation must be the same. On this basis we can judge whether a given equation is correct or not. But a dimensionally correct equation may or may not be physically correct. Ex.5 Show that the expression of the time period T of a simple pendulum of length l given by T =

2π

Sol.

l is dimensionally correct. g T = 2π

Dimensionally [T ] =

l g [L] [LT – 2 ]

= [T ]

As in the above equation, the dimensions of both sides are same. The given formula is dimensionally correct.

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Page # 12 8.3

UNIT AND DIMENSION

Principle of Homogeneity of Dimensions. This principle states that the dimensions of all the terms in a physical expression should be same. For example, in the physical expression s = ut +

1 2 1 at , the dimensions of s, ut and at2 all are same. 2 2

The physical quantities separated by the symbols +, –, =, >, < etc., have the same dimensions.

Note :

Ex.6 The velocity v of a particle depends upon the time t according to the equation v = a + bt +

c . d+ t

Write the dimensions of a, b, c and d. Sol. From principle of homogeneity [a] = [v] [a] = [LT–1]

or

Ans.

[bt] = [v] [b] =

or

[b] = [LT–2]

Similarly,

[d] = [t] = [T]

Further,

8.4

[v ] [LT –1] = [ t] [T ]

or

Ans.

[c ] = [v ] [d + t ]

or

[c] = [v] [d + t]

or

[c] = [LT–1] [T]

or

[c] = [L]

Ans.

To establish the relation among various physical quantities : If we know the factors on which a given physical quantity may depend, we can find a formula relating the quantity with those factors. Let us take an example.

Ex.7 The frequency (f) of a stretched string depends upon the tension F (dimensions of force), length l of the string and the mass per unit length µ of string. Derive the formula for frequency. Sol. Suppose, that the frequency f depends on the tension raised to the power a, length raised to the power b and mass per unit length raised to the power c. Then.

f ∝ [F]a [l ]b [µ]c or

f = k [F]a [l ]b [µ]c

Here, k is a dimensionless constant. Thus, [f ] = [F]0 [l]b [µ]c or

[M0 L0 T–1] = [MLT–2]a [L]b [ML–1]c

or

[M0L0T–1] = [Ma + c La + b – c T–2a]

For dimensional balance, the dimension on both sides should be same. Thus, and

a+c=0

...(ii)

a+b–c=0

...(iii)

– 2a = – 1

...(iv)

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Page # 13

UNIT AND DIMENSION

Solving these three equations, we get

a=

1 1 , c=– and b = – 1 2 2

Substituting these values in Eq. (i), we get

f = k(F)1/ 2 (l ) –1(µ ) –1/ 2 or

f =

k F l µ

Experimentally, the value of k is found to be Hence,

8.5

f =

1 2

1 F 2l µ

Limitations of Dimensional Analysis The method of dimensions has the following limitations : (i) By this method the value of dimensionless constant can not be calculated. (ii) By this method the equation containing trigonometrical, exponential and logarithmic terms cannot be analysed. (iii) If a physical quantity depends on more than three factors, then relation among them cannot be established because we can have only three equations by equalising the powers of M, L and T.

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Page # 14

UNIT AND DIMENSION

BASIC MATHEMATICS 9.

MENSURATION FORMULAS : r : radius ; V = Volume (a) Circle

d = diameter ; S.A = surface area

Perameter : 2πr = πd,

Area

: πr2 =

1 2 πd 4

(b) Sphere Surface area = 4πr2 = πd2 , Volume =

4 3 1 πr = πd3 3 6

(c) Spherical Shell (Hollow sphere) Surface area = 4πr2 = πd2 Volume of material used = (4πr2)(dr), dr = thickness (d) Cylinder Lateral area = 2πrh V = πr2h Total area = 2πrh + 2πr2 = 2πr (h + r) (e) Cone Lateral area = πr

r 2 + h2

h = height

1 2 2  2  Total area = π r  r + h + r  V = πr h   3

(f) Ellipse Circumference ≈ 2π

a2 + b2 2

area = πab a = semi major axis

b a

b = semi minor axis

a

(g) Parallelogram A = bh = ab sin θ

h θ

b

a = side ; h = height ; b = base θ = angle between sides a and b

b

(h) Trapezoid

h

h area = (a + b) 2

a a and b parallel sides h = height

(i) Triangle

bh ab sin γ = s(s − a)(s − b)(s − c ) = 2 2 a, b, c sides are opposite to angles α, β , γ area =

b = base ; h = height s=

1 (a + b + c ) 2

α

c

b γ

β

a

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Page # 15

UNIT AND DIMENSION

h

(j) Rectangular container lateral area = 2(
b + bh + h
)

b
side
, b, h Mathematics is the language of physics. It becomes easier to describe, understand and apply the physical principles, if one has a good knowledge of mathematics. V =
bh

10.

LOGARITHMS : (ii) If ex = y, then x = loge y = ln y (iii) If 10x = y, then x = log10y (i) e ≈ 2.7183 (iv) log10y = 0.4343 loge y = 2.303 log10 y (v) log (ab) = log (a) + log (b) a (vi) log  = log (a) – log (b) b

(vii) log an = n log (a)

11.

TRIGONOMETRIC PROPERTIES :

(i)

Measurement of angle & relationship between degrees & radian In navigation and astronomy, angles are measured in degrees, but in calculus it is best to use units called radians because of they simplify later calculations. B Let ACB be a central angle in circle of radius r, as in figure. Then the angle ACB or θ is defined in radius as Arc length AB θ= ⇒ θ= Radius r

If

θ

C

r

A

r = 1 then θ = AB

The radian measure for a circle of unit radius of angle ABC is defined to be the length of the circular arc AB. since the circumference of the cirlce is 2π and one complete revolution of a cicle is 360°, the relation between radians and degrees is given by the following equation. π radians = 180°

ANGLE CONVERSION FORMULAS 1 degree =

π 180°

(≈ 0.02) radian

1 radian ≈ 57 degrees Ex.15 Covert 45° to radians :

Convert

π rad to degrees : 6

45 •

Degrees to radians : multiply by

π 180°

Radians to degrees : multiply by

180° π

π π = rad 180 4

π 180 • = 30° 6 π

Ex.16 Convert 30º to radians : Sol.

30 º×

π π = rad 6 180 º

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Page # 16

UNIT AND DIMENSION

Ex.17 Convert Sol.

π rad to degrees. 3

π 180 × = 60 3 π

Standard values π rad 6

(1) 30° = (4) 90° =

(2) 45° =

π rad 2

(7) 150° =

π rad 4

(3) 60° =

(5) 120° =

5π rad 6

2π rad 3

π rad 3

(6) 135° =

(8) 180° = π rad

3π rad 4

(9) 360° = 2π rad

(Check these values yourself to see that the satisfy the conversion formulaes) (ii) Measurement of positive & Negative Angles :

y y

Positive measure

Negative Measure

x

x

An angle in the xy-plane is said to be in standard position if its vertex lies at the origin and its initial ray lies along the positive x-axis (Fig). Angles measured counterclockwise from the positive x-axis are assigned positive measures; angles measured clockwise are assigned negative measures. y

y

y

y –

x

3π

x

x

5π 2

x

3π – 4

9π 4

(iii) Six Basic Trigonometric Functions :

y

O

r

θ x

y

oppsite side

hy po te

nu se

P(x,y)

adjacent side P(x,y)

The trigonometric fucntion of a general angle θ are defined in tems of x, y and r.

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Page # 17

UNIT AND DIMENSION

Sine :

sin θ =

opp y = hyp r

hyp r Cosecant : cos ecθ = opp = y

Cosine:

cos θ =

adj x = hyp r

Secant :

opp y Tangent: tan θ = adj = x

sec θ =

hyp r = adj x

adj x Cotangent: cot θ = opp = y

VALUES OF TRIGONOMETRIC FUNCTIONS If the circle in (Fig. above) has radius r = 1, the equations defining sin θ and cos θ become cosθ = x,

sinθ = y

We can then calculate the values of the cosine and sine directly from the coordinates of P. Ex.18 Find the six trigonometric ratios from given fig. (see above) Sol.

opp 4 sinθ = hyp = 5

5

adj 3 cosθ = hyp = 5

4

θ

opp 4 tanθ = adj = 3 sec θ =

cot θ =

hyp 5 = opp 3

3

adj 3 = opp 4

cosec θ =

hyp 5 = opp 4

Ex.19 Find the sine and cosine of angle θ shown in the unit circle if coordinate of point p are as shown. y    – 1, 3   2 2    3 2

1

θ

x

1 2

Sol.

cos θ = x-coordinate of P = –

12.

1 2

sin θ = y-coordinate of P =

3 2

Values of sin θ, cos θ and tan θ for some standard angles.

Degree

0

30

37

Radians

0

π/6

37π / 180

sin θ

0

1/2

3/5

1/ 2

4/5

cos θ

1

3 /2

4/5

1/ 2

3/5

tan θ

0

1/ 3

3/4

1

4/3

45

π/ 4

53

60

90

120

135

53π / 180

π/3

π/2

2π / 3

3π / 4

π

3 /2

1

3 /2

1/ 2

0

1/2

0

–1/2

– 1/ 2

–1

3

∞

– 3

–1

0

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180

Page # 18

UNIT AND DIMENSION

A useful rule for remembering when the basic trigonometric funcions are positive and negative is the CAST rule. If you are not very enthusiastic about CAST. You can remember it as ASTC (After school to college) y nd nd II Quadrant I Quadrant

A all positive

S sin positive

x T Tan positive IIInd Quadrant

C cos positive IVnd Quadrant

The CAST rule RULES FOR FINDING TRIGONOMETRIC RATIO OF ANGLES GREATER THAN 90°. Step 1 →

Identify the quadrant in which angle lies.

Step 2 → (a) If angle = (nπ ± θ)

where n is an integer. Then

π   (b) If angle = ( 2n + 1) + θ where n is in interger. Then 2   π   trigonometric function of ( 2n + 1) ± θ = complimentry trignometric function of θ and 2   sign will be decided by CAST Rule.

Ex.20 Evaluate sin 120° sin 120° = sin (90° + 30°) = cos 30° =

3 2

Aliter sin 120° = sin (180° – 60°) = sin 60° =

3 2

Sol.

Ex.21 Evaluate cos 210° Sol.

cos 210° = cos (180° + 30°) = – cos 30° = –

Ex.22 tan 210° = tan (180° + 30°) = tan 30° = + 13.

3 2 1 3

IMPORTANT FORMULAS (i) sin2θ + cos2θ = 1 (ii) 1 + tan2θ = sec2θ 2 2 (iii) 1 + cot θ = cosec θ (iv) sin2θ = 2 sin θ cos θ (v) cos 2θ = 2 cos2θ – 1 = 1 – 2 sin2θ = cos2θ – sin2θ (vi) sin (A ± B) = sin A cos B ± cos A sin B (vii) cos (A ± B) = cos A cos B ∓ sin A sin B C+D C–D (viii) sin C + sin D = 2 sin  cos   2   2 

(x) cos C + cos D = 2 cos (xii) tan 2θ =

2 tan θ 2

1 – tan θ

C+D C–D cos 2 2

C–D C+D  cos  (ix) sinC – sin D = 2 sin 2    2 

(xi) cos C – cos D = 2 sin (xiii)

D–C C+D sin 2 2

tan A ± tan B tan(A ± B) = 1 ∓ tan A tan B

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Page # 19

UNIT AND DIMENSION (xiv) sin(90° + θ) = cos θ (xvi) tan (90° + θ) = – cot θ (xviii) cos(90° – θ) = sinθ (xx) sin(180° – θ) = sin θ (xxii) tan (180° + θ) = tan θ (xxiv) cos (–θ) = cos θ

•

sin A sin B sin C = = a b c

Sine Rule

•

(xv) cos (90° + θ) = – sin θ (xvii) sin(90° – θ) = cos θ (xix) cos (180° – θ) = – cos θ (xxi) cos (180° + θ) = – cos θ (xxiii) sin(– θ) = – sin θ (xxv) tan (–θ) = – tan θ

Cosine rule

B c

a

a2 = b2 + c2 – 2bc cos A

C

A b

4 Ex.23

3

90°

53°

37° x Find x :

S

o

l

.

14.

sin 90° sin 53° = x 4 x= 5

SMALL ANGLE APPROXIMATION It is a useful simplification which is only approximately true for finite angles. It involves linerarization of the trigonometric functions so that, when the angle θ is measured in radians. sin θ ~ θ cosθ ~ 1 or cos θ ~ 1 –

θ2 for the second - order approximation 2

tan θ ~ θ Geometric justification

tan

Object

arc θ

θ

D

arc

tan

d

•

•

Small angle approximation. The value of the small angle X in radians is approximately equal to its tangent. When one angle of a right triangle is small, is hypotenuse in approximately equal in length to the leg adjacent to the small angle, so the cosine is approximately 1. The short leg is approximately equal to the arc from the long leg to the hypotenuse, so the sine and tangent are both approximated by the value of the angle in radians.

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Page # 20 15.

UNIT AND DIMENSION

BINOMIAL THEOREM : (1 ± x)n = 1 ± nx +

n(n – 1)x 2 ........... 2!

(1 ± x)–n = 1 ∓ nx +

n(n + 1) 2 x ......... 2!

If x 0 ⇒ The line will pass through (0, 1)

(0,1) θ

tan θ =

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3 2

x

1 th if x 4

Page # 22

UNIT AND DIMENSION

Ex.29 Draw the graph for the equation : 2y + 4x + 2 = 0 ⇒ y = – 2x – 1 Sol. 2y + 4x + 2 = 0 m = – 2 < 0 i.e., θ > 90° c = – 1 i.e., line will pass through (0, –1)

tanθ = –2

θ

(0,–1)

: (i) If c = 0 line will pass through origin. (ii) y = c will be a line parallel to x axis.

(0,c)

(0,0) (iii) x = c will be a line perpendicular to y axis

(c,0) (0,0)

(ii)

Parabola A general quadratic equation represents a parabola. y = ax2 + bx + c a≠0 if a > 0 ; It will be a opening upwards parabola. if a < 0 ; It will be a opening downwards parabola. if c = 0 ; It will pass through origin. y ∝ x2 or y = 2x2, etc. represents a parabola passing through origin as shown in figure shown. y y

x ∝ y2

y ∝ x2

x

x

2

k = 1/2mv

(i) e.g.

(ii)

y = 4 x2 + 3x

e.g.

k=

1 mv2 2

k y=4x2+3x v 2

y=–4x +3x

Note : That in the parabola y = 2x2 or y ∝ x2, if x is doubled, y will beome four times. Graph x ∝ y2 or x = 4 y2 is again a parabola passing through origin as shown in figure shown. In this cae if y is doubled, x will become four times. y = x2 + 4 or x = y2 – 6 will represent a parabola but not passing through origin. In the first equation (y = x2 + 4), if x doubled, y will not become four times.

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Page # 23

UNIT AND DIMENSION 17.

SIMILAR TRIANGLE Two given triangle are said to be similar if

(1)

All respective angle are same or

(2)

All respective side ratio are same.

P A

R C Q As example, ABC, PQR are two triangle as shown in figure.

B

If they are similar triangle then (1)

∠A=∠P ∠B=∠Q ∠C=∠R

OR AB BC AC = = PQ QR PR

(2)

A

P 5 Ex.30

3

B

Q

x

O

6 Find x : Sol.

By similar triangle concept AB OB = PQ OQ 5 6 = 3 x

⇒

x=

18 5

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Page # 24

UNIT AND DIMENSION

(ONLY ONE OPTION IS CORRECT)

Exercise - I SECTION A : UNITS

1. Which of the following sets cannot enter into the list of fundamental quantities in any system of units ? (A) length, mass and velocity (B) length, time and velocity (C) mass, time and velocity (D) length, time and mass Sol.

2. Which of the following is not the name of a physical quantity ? (A) kilogram (B) impulse (C) energy (D) density Sol.

3. Light year is the unit of (A) speed (B) mass (C) distance (D) time Sol.

4. PARSEC is a unit of (A) Time (C) Distance Sol.

(B) Angle (D) Velocity

5. Which of the following is not the unit of time (A) solar day (B) parallactic second (C) leap year (D) lunar month Sol.

6. Which of the following system of units is NOT based on the unit of mass, length and time alone (A) FPS (B) SI (C) CGS (D) MKS Sol.

7. The SI unit of the universal gravitational constant G is (B) Nm2kg–2 (A) Nm kg–2 (C) Nm2 kg–1 (D) Nmkg–1 Sol.

8. The SI unit of the universal gas constant R is : (B) watt K–1 mol–1 (A) erg K–1 mol–1 –1 –1 (C) newton K mol (D) joule K–1 mol–1 Sol.

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Page # 25

UNIT AND DIMENSION 9. The unit of impulse is the same as that of : (A) moment force (B) linear momentum (C) rate of change of linear momentum (D) force Sol.

13. One watt-hour is equivalent to (A) 6.3 × 103 Joule (B) 6.3 × 10–7 Joule (C) 3.6 × 103 Joule (D) 3.6 × 10–3 Joule Sol.

10. Which of the following is not the unit of energy? (A) watt-hour (B) electron-volt (C) N × m (D) kg × m/sec2 Sol.

SECTION : B DIMENSIONS 14. What are the dimensions of lenth in force × displacement/time (A) –2 (B) 0 (C) 2 (D) none of these Sol.

11. A physical quantity is measured and the result is expressed as nu where u is the unit used and n is the numerical value. If the result is expressed in various units then (A) n ∝ size of u (B) n ∝ u2 (C) n ∝ √u (D) n ∝ 1/u Sol.

12. If the unit of length is micrometer and the unit of time is microsecond, the unit of velcoity will be : (A) 100 m/s (B) 10 m/s (C) micrometers (D) m/s Sol.

15. The angular frequency is measured in rad s–1. Its dimension in length are : (A) – 2 (B) –1 (C) 0 (D) 2 Sol.

16. A dimensionless quantity : (A) never has a unit (B) always has a unit (C) may have a unit (D) does not exit Sol.

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Page # 26

UNIT AND DIMENSION

17. A unitless quantity : (A) never has a nonzero dimension (B) always has a nonzero dimension (C) may have a nonzero dimension (D) does not exit Sol.

18. If a and b are two physical quantities having different dimensions then which of the following can denote a new physical quantity (A) a + b (B) a – b (E) sin (a/b) (C) a/b (D) ea/b Sol.

19. Two physical quantities whose dimensions are not same, cannot be : (A) multiplied with each other (B) divided (C) added or substracted in the same expression (D) added together Sol.

Sol.

21. The dimensions of universal gravitational constant are (A) M–1 L3 T–2 (B) M–1 L3 T–1 –1 –1 –2 (C) M L T (D) M–2 L2 T–2 Sol.

22. The SI unit of Stefan's constant is : (B) J s m–1 K–1 (A) Ws–1 m–2 K–4 –1 –2 –1 (C) J s m K (D) W m–2 K–4 Sol.

23. What are the dimensions of Boltzmann's constant? (B) ML2T–2K–1 (A) MLT–2K–1 0 –2 (C) M LT (D) M0L2T–2K–1 Sol. 20. Choose the correct statement(s) : (A) All quantities may be represented dimensionally in terms of the base quantities. (B) A base q uant i t y cannot b e re pres ente d dimensionally in terms of the rest of the base quantities. (C) The dimension of a base quantity in other base quantities is always zero. (D) The dimension of a derived quantity is never zero in any base quantity.

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Page # 27

UNIT AND DIMENSION 24. Planck's constant has the dimensions of : (A) force (B) energy (C) linear momentum (D) angular momentum Sol.

25. The velocity 'v' (in cm/s) of a particle is given in terms of time 't' (in s) by the equation b t+c The dimensions of a, b and c are a b c a (A) L2 T LT2 (B) LT2 (C) LT–2 L T (D) L Sol.

v = at +

b LT LT

c L T2

26. The position of a particle at time 't' is given by the relation

27. The time dependence of a physical quantity ? P = P0exp(–α t2) where α is a constant and t is time The constant α (A) will be dimensionless (B) will have dimensions of T–2 (C) will have dimensions as that of P (D) will have dimensions equal to the dimension of P multiplied by T–2 Sol.

28. Force F is given in terms of time t and distance x by F = A sin C t + B cos D x A C and are given by Then the dimensions of B D –2 0 0 –1 –2 (A) MLT , M L T (B) MLT , M0L–1T0 0 0 0 0 1 –1 (C) M L T , M L T (D) M0L1T–1, M0L0T0 Sol.

V0 [1 – e – αt ] α where V0 is a constant and α > 0. The dimensions of V0 and α are respectively. (A) M0L1T0 and T–1 (B) M0L1T0 and T–2 (C) M0L1T–1 and T–1 (D) M0L1T–1 and T–2 Sol.

x(t) =

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Page # 28

UNIT AND DIMENSION

29. The Van der Waal equation for 1 mole of a real gas is a    P + 2  ( V – b) = RT  V  where P is the pressure, V is the volume, T is the absolute temperature, R is the molar gas constant and a, b are Van dar Waal constants. The dimensions of a are the same as those of (A) PV (B) PV2 (C) P2V (D) P/V Sol.

30. In above question 29, the dimensions of b are the same as those of (A) P (B) V (C) PV (D) nRT Sol.

31. In above question 29, the dimensions of nRT are the same as those of (A) energy (B) force (C) pressure (D) specific heat Sol.

33. Which pair of following quantities has dimensions different from each other. (A) Impulse and linear momentum (B) Plank's constant and angular momentum (C) Moment of inertia and moment of force (D) Young's modulus and pressure Sol.

34. A pair of physical quantities having the same dimensional formula is : (A) angular momentum and torque (B) torque and energy (C) force and power (D) power and angular momentum Sol.

35. If force (F) is given by F = Pt–1 + α t, where t is time. The unit of P is same as that of (A) velocity (B) displacement (C) acceleration (D) momentum Sol.

32. In above question 29, the dimensional formula for ab is (B) ML4T–2 (C) ML6T–2 (D) ML8T–2 (A) ML2T–2 Sol.

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Page # 29

UNIT AND DIMENSION 36. The product of energy and time is called action. The dimensional formula for action is same as that for (A) power (B) angular energy (C) force × velocity (D) impulse × distance Sol.

37. Dimensions of pressure are the same as that of (A) force per unit volume (B) energy per unit volume (C) force (D) energy Sol.

39. In the above question dimensions of same as those of (A) wave velocity (C) wave amplitude Sol.

b are the c

(B) wavelength (D) wave frequency

40. What is the physical quantity whose dimensions are M L2 T–2 ? (A) kinetic energy (B) pressure (C) momentum (D) power Sol.

41. Which one of the following has the dimensions of ML–1T–2 ? (A) torque (B) surface tension (C) viscosity (D) stress Sol.

38. When a wave traverses a medium, the displacement of a particle located at x at time t is given by y = a sin (bt – cx) where a, b and c are constants of the wave. The dimensions of b are the same as those of (A) wave velocity (B) amplitude (C) wavelength (D) wave frequency Sol.

42. If force, acceleration and time are taken as fundamental quantities, then the dimensions of length will be : (B) F–1 A2 T–1 (C) FA2T (D) AT2 (A) FT2 Sol.

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Page # 30

UNIT AND DIMENSION

43. The dimensions ML–1T–2 can correspond to (A) moment of a force or torque (B) surface tension (C) pressure (D) co-efficient of viscosity

(useful relation are τ
= r × F , S = F/l, F = 6 π η r v,, where symbols have usual meaning) Sol.

44. Which of the following can be a set of fundamental quantities (A) length, velocity, time (B) momentum, mass, velocity (C) force, mass, velocity (D) momentum, time, frequency Sol.

47. In a certain system of units, 1 unit of time is 5 sec, 1 unit of mass is 20 kg and unit of length is 10m. In this system, one unit of power will correspond to (A) 16 watts (B) 1/16 watts (C) 25 watts (D) none of these Sol.

48. In a book, the answer for a particular question is expressed as b =

ma  2kl   1+  here m represents k  ma 

mass, a represents accelerations, l represents length. The unit of b should be (A) m/s (B) m/s2 (C) meter (D) /sec Sol.

45. If area (A) velocity (v) and density (ρ) are base units, then the dimensional formula of force can be represented as (A) Avρ (B) Av2ρ (C) Avρ2 (D) A2vρ Sol.

49. ρ = 2 g/cm3 convert it into MKS system kg kg (B) 2 × 103 3 (A) 2 × 10–3 3 m m kg kg (C) 4 × 103 3 (D) 2 × 106 3 m m Sol.

46. The pressure of 106 dyne/cm2 is equivalent to (A) 105 N/m2 (B) 106 N/m2 7 2 (C) 10 N/m (D) 108 N/m2 Sol.

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Page # 31

UNIT AND DIMENSION F sin(βt) (here V = velocity, F = force, t = V2 time) : Find the dimension of α and β (A) α = [M1L1T0], β = [T–1] (B) α = [M1L1T–1], β = [T1] (C) α = [M1L1T–1], β = [T–1] (D) α = [M1L–1T0], β = [T–1] Sol.

50. α =

51. Given that v is the speed, r is radius and g is acceleration due to gravity. Which of the following is dimension less (A)

v 2g r

(B) v2rg

(C) vr2g

(D)

v2 rg

Sol.

54. The velocity of water waves may dpend on their wavelength λ , the density of water ρ and the acceleration due to gravity g. The method of dimensions gives the relation between these quantities as (A) v2 = kλ–1 g–1 ρ–1 (B) v2 = k g λ (C) v2 = k g λ ρ (D) v2 = k λ3 g–1 ρ–1 where k i s a di mensi onl e ss constant Sol.

Sol.

52. If E, M, J and G denote energy, mass, angular momentum and gravitational constant respectively, then

EJ2

M5 G2 (A) length Sol.

has the dimensions of (B) angle

(C) mass

(D) time

55. If the unit of force is 1 kilonewton, the length is 1 km and time is 100 second, what will be the unit of mass : (A) 1000 kg (B) 10 kg (C) 10000 kg (D) 100 kg Sol.

53. The dimensions ML–1T–2 may correspond to (A) work done by a force (B) linear momentum (C) pressure (D) energy per unit volume

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Page # 32

UNIT AND DIMENSION

56. A body moving through air at a high speed 'v' experiences a retarding force 'F' given by F = K A d vx where 'A' is the surface area of the body, 'd' is the density of air and 'K' is a numerical constant. The value of 'x' is : (A) 1 (B) 2 (C) 3 (D) 4 Sol.

p

57. The velocity of a freely falling body changes as g hq where g is acceleration due to gravity and h is the height. The values of p and q are : (A) 1, (C)

1 2

1 ,1 2

(B)

1 1 , 2 2

59. The value of G = 6.67 × 10–11 N m2 (kg)–2. Its numerical value in CGS system will be : (A) 6.67 × 10–8 (B) 6.67 × 10–6 (C) 6.67 (D) 6.67 × 10–5 Sol.

60. The density of mercury is 13600 kg m–3. Its value of CGS system will be : (A) 13.6 g cm–3 (B) 1360 g cm–3 –3 (C) 136 g cm (D) 1.36 g cm–3 Sol.

(D) 1, 1

Sol.

61. If the acceleration due to gravity is 10 ms–2 and the units of length and time are changed to kilometre and hour, respectively, the numerical value of the acceleration is : (A) 360000 (B) 72000 (C) 36000 (D) 129600 Sol. 58. Choose the correct statement(s) : (A) A dimensionally correct equation must be correct. (B) A dimensionally correct equation may be correct. (C) A dimensionally incorrect equation may be correct. (D) A dimensionally incorrect equation may be incorrect. Sol.

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Page # 33

UNIT AND DIMENSION 62. If 'c' the velocity of light 'g' the acceleration due to gravity and 'P" the atmospheric pressure are fundamental units, then the dimensions of length will be (A) c/g (B) P × c × g (C) c/P (D) c2/g Sol.

63. The units of length, velocity and force are doubled. Which of the following is the correct change in the other units ? (A) unit of time is doubled (B) unit of mass is doubled (C) unit of momentum is doubled (D) unit of energy is doubled Sol.

65. If the units of M, L are doubled then the unit of kinetic energy will become (A) 2 times (B) 4 times (C) 8 times (D) 16 times Sol.

BASIC MATHEMATICS 66. The radius of two circles are r and 4r what will be the ratio of their Area and perimeter. Sol.

67. Internal radius of a ball is 3 cm and external radius is 4 cm. What will be the volume of the material used. Sol. 64. If the units of force and that of length are doubled, the unit of energy will be : (A) 1/4 times (B) 1/2 times (C) 2 times (D) 4 times Sol.

68. Binomial (a) (99)1/2 Sol.

(b) (120)1/2

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(c) (126)1/3

Page # 34

UNIT AND DIMENSION

4 x

2

71. 69. A normal human eye can see an object making an angle of 1.8° at the eye. What is the approximate height of object which can be seen by an eye placed at a distance of 1 m from the eye.

y

Find x and y :

2 3

Sol.

h 1.8°

1m Sol.

70. Draw graph for following equations : (ii) x = 4t – 3 (i) v = v0 – at (iii) x = 4 at2 (iv) v = – gt Sol.

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Page # 35

UNIT AND DIMENSION

Exercise - II 1*. Which of the following is not the unit length : (A) micron (B) light year (C) angstrom (D) radian Sol.

4. The equation of state for a real gas at high temperature is given by P =

nRT a where − 1/ 2 V − b T V( V + b)

n, P V & T are number of moles, pressure, volume & temperature respectively & R is the universal gas constant. Find the dimensions of constant ‘a’ in the above equation. Sol.

2. A particle is in a uni-directional potential field where the potential energy (U) of a particle depends on the x-coordinate given by Ux = k(1 – cos ax) & k and ‘a’ are constants. Find the physical dimensions of ‘a’ & k. Sol. 5. The distance moved by a particle in time t from centre of a ring under the influence of its gravity is given by x = a sinωt where a & ω are constants. If ω is found to depend on the radius of the ring (r), its mass (m) and universal gravitational constant (G), find using dimensional analysis an expression for ω in terms of r, m and G. Sol.

3. The time period (T) of a spring mass system depends upon mass (m) & spring constant (k) & length of the spring (l) [k =

Force ] . Find the relation among, length

(T), (m), (l) & (k) using dimensional method. Sol.

6. If the velocity of light c, Gravitational constant G & Plank’s constant h be chosen as fundamental units, find the dimension of mass, length & time in the new system. Sol.

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Page # 36

UNIT AND DIMENSION

7. A satellite is orbiting around a planet. Its orbital velocity (v0) is found to depend upon (a) Radius of orbit (R) (b) Mass of planet (M) (c) Universal gravitation constant (G) Using dimensional analysis find an expression relating orbital velocity (v0) to the above physical quantities. Sol.

10. Use the small angle approximations to find approximate values for (a) sin 8° and

(b) tan 5°

Sol.

8. The angle subtended by the moon's diameter at a point on the earth is about 0.50°. Use this and the act that the moon is about 384000 km away to find the approximate diameter of the moon.

θ

D

rm (A) 192000 km

(B) 3350 km

(C) 1600 km

(D) 1920 km

Sol.

9. Use the approximation (1 + x)n ≈ 1 + nx, |x| 0 =

Function defined for x ≥ 0

…..

3

d  1  d −1 1 ( x ) = ( −1)x − 2 = − 2  = dx  x  dx x

Function defined for x ≥ 0 (b)

3

1 −4 / 5 x 5

derivative not defined at x = 0

Rule No.3 The Constant Multiple Rule If u is a differentiable function of x, and c is a constant, then In particular, if n is a positive integer, then

d du (cu) = c dx dx

d (cx n ) = cn x n−1 dx

Ex.34 The derivative formula d (3x 2 ) = 3 (2x ) = 6x dx

says that if we rescale the graph of y = x2 by multiplying each y-coordinate by 3, then we multiply the slope at each point by 3. Ex.35 A useful special case The derivative of the negative of a differentiable function is the negative of the function’s derivative. Rule 3 with c = – 1 gives. d d d d ( −u) = ( −1.u) = −1 . (u) = − (u) dx dx dx dx

Rule No.4 The Sum Rule The derivative of the sum of two differentiable functions is the sum of their derivatives. If u and v are differentiable functions of x, then their sum u + v is differentiable at every point where u and v are both differentiable functions in their derivatives. d d du dv du dv (u − v ) = [u + ( −1) v ] = + ( −1) = − dx dx dx dx dx dx

The sum Rule also extends to sums of more than two functions, as long as there are only finite , u2, ........ un are differentiable at x, then so if u1 + u2 + ....... + un, then 1 f u

n

c t i o

n

s

i n

t h

e

s u

m

.

I f

u

du du du d (u1 + u 2 + ...... + un ) = 1 + 2 + ........ + n dx dx dx dx

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VECTOR & CALCULUS

Page # 25

Ex.36 (a) y = x4 + 12x

(b) y = x3 +

4 2 x – 5x + 1 3

dy d 3 d 4 2 d d = x + (5 x ) + (1)  x − dx dx dx  3  dx dx

dy d 4 d = (x ) + (12 x ) dx dx dx

= 4x3 + 12

= 3x2 +

4 . 2x – 5 + 0 3

8 x−5 3 Notice that we can differentiate any polynomial term by term, the way we differentiated the polynomials in above example.

= 3x2 +

Rule No. 5 The Product Rule If u and v are differentiable at x, then if their product uv is considered, then

d dv du (uv ) = u +v . dx dx dx

The derivative of the product uv is u times the derivative of v plus v times the derivative of u. In prime notation (uv)’ = uv’ + vu’. While the derivative of the sum of two functions is the sum of their derivatives, the derivative of the product of two functions is not the product of their derivatives. For instance, d d 2 (x . x) = ( x ) = 2x, dx dx

while d ( x ). d ( x ) = 1.1 = 1 , which is wrong dx dx 2

+ 1) (x3 + 3)

Ex.37

F

Sol.

Using the product Rule with u = x2 + 1 and v = x3 + 3, we find

i n

d

t h

e

d

e

r

i v

a

t i v

e

s

o

f

y

=

(

x

d [( x 2 + 1)( x 3 + 3)] = (x2 + 1) (3x2) + (x3 + 3) (2x) dx

= 3x4 + 3x2 + 2x4 + 6x = 5x4 + 3x2 + 6x Example can be done as well (perhaps better) by multiplying out the original expression for y and differentiating the resulting polynomial. We now check : y = (x2 + 1) (x3 + 3) = x5 + x3 + 3x2 + 3 dy = 5x4 + 3x2 + 6x dx

This is in agreement with our first calculation. There are times, however, when the product Rule must be used. In the following examples. We have only numerical values to work with. Ex.38 Let y = uv be the product of the functions u and v. Find y’(2) if u(2) = 3, u’(2) = – 4, v(2) = 1, and v’(2) = 2. Sol.

From the Product Rule, in the form y’ = (uv)’ = uv’ + vu’, we have y’(2) = u(2) v’(2) + v(2) u’(2) = (3) (2) + (1) (–4) = 6 – 4 = 2

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VECTOR & CALCULUS

Page # 26

Rule No.6 The Quotient Rule If u and v are differentiable at x, and v(x) ≠ 0, then the quotient u/v is differentiable at x, d u and  = dx  v 

v

du dv −u dx dx v2

Just as the derivative of the product of two differentiable functions is not the product of their derivatives, the derivative of the quotient of two functions is not the quotient of their derivatives. Ex.39 Find the derivative of y = Sol.

t2 − 1 t2 + 1

We apply the Quotient Rule with u = t2 – 1 and v = t2 + 1

dy ( t 2 + 1) 2t − ( t 2 − 1). 2t = dt ( t 2 + 1)2 =

2t 3 + 2t − 2t 3 + 2t 2

( t + 1)

2

 d  u  v( du / dt ) − u(dv / dt )   As   =  dt v2 v  

=

4t 2

( t + 1)2

Rule No. 7 Derivative Of Sine Function d (sin x ) = cos x dx

Ex.40 (a) y = x2 – sin x :

(b) y = x2 sin x :

dy d = 2x − (sin x ) = 2x – cos x dx dx

dy d = x2 (sin x ) + 2x sin x dx dx

Difference Rule

Product Rule

= x2cosx + 2x sinx dy sin x = (c) y = : dx x =

x.

d (sin x ) − sin x .1 dx x2

Quotient Rule

x cos x − sin x x2

Rules No.8 Derivative Of Cosine Function d (cos x ) = − sin x dx

Ex.41 (a) y = 5x + cos x

Sum Rule

dy d d = (5 x ) + (cos x ) = 5 – sin x dx dx dx

(b) y = sin x cos x dy d d = sin x (cos x ) + cos x (sin x ) dx dx dx

Product Rule

= sin x(– sin x) + cos x (cos x) = cos2 x – sin2 x = cos 2x

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VECTOR & CALCULUS

Page # 27

Rule No. 9 Derivatives Of Other Trigonometric Functions Because sin x and cos x are differentiable functions of x, the related functions sin x 1 tan x = sec x = ; cos x cos x cot x =

cos x ; sin x

cos ec x =

1 sin x

are differentiable at every value of x at which they are defined. There derivatives, Calculated from the Quotient Rule, are given by the following formulas. d (tan x ) = sec 2 x ; dx

d (sec x ) = sec x tan x dx

d (cot x ) = − cos ec 2 x ; dx

d (cos ec x ) = − cos ec x cot x dx

Ex.42 Find dy / dx if y = tan x. Sol.

d d  sin x  (tan x ) =  = dx dx  cos x 

=

Ex.43 (a)

cos x

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

cos x cos x − sin x( − sin x ) 2

cos x

=

cos 2 x + sin 2 x 2

cos x

=

1 cos 2 x

= sec 2 x

d d (3x + cot x) = 3 + (cot x) = 3 – cosec2 x dx dx

d  2  d d (b) dx  sin x  = dx (2 cosec x ) = 2 dx (cosec x )  

= 2(– cosec x cot x) = – 2 cosec x cot x

Rule No. 10 Derivative Of Logrithm And Exponential Functions d 1 (loge x ) = , dx x

d x (e ) = e x dx

Ex.44 y = ex . loge (x) dy d x d = (e ). log( x ) + [log e ( x )] e x dx dx dx

⇒

dy ex = e x . loge ( x ) + dx x

Rule No. 11 Chain Rule Or ‘Outside Inside’ Rule dy dy du = . dx du dx It sometime helps to think about the Chain Rule the following way. If y = f (g(x)), dy = f’[g(x)] . g’(x) dx In words : To find dy/dx, differentiate the “outside” function f and leave the “inside” g(x) alone; then multiply by the derivative of the inside. We now know how to differntiate sin x and x2 – 4, but how do we differentiate a composite like sin(x2 – 4)? The answer is, with the Chain Rule, which says that the derivative of the composite of two differentiable functions is the product of their derivatives evaluated at appropriate points. The Chain Rule is probably the most widely used differentiation rule in mathematics. This section describes the rule and how to use it. We begin with examples.

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Ex.45 The function y = 6x – 10 = 2(3x – 5) is the composite of the functions y = 2u and u = 3x – 5. How are the derivatives of these three functions related ? Sol.

dy = 6 , dy = 2 , du = 3 dx du dx

We have

Since 6 = 2 × 3

dy dy du = . dx du dx

Is it an accident that

dy dy du = . dx du dx

?

If we think of the derivative as a rate of change, our intution allows us to see that this relationship is reasonable. For y = f(u) and u = g(x), if y changes twice as fast as u and u changes three times as fast as x, then we expect y to change six times as fast as x. Ex.46 Let us try this again on another function. y = 9x4 + 6x2 + 1 = (3x2 + 1)2 is the composite y = u2 and u = 3x2 + 1. Calculating derivatives. We see that dy du . = 2u.6 x du dx

and

= 2 (3x2 + 1). 6x = 36x3 + 12 x

dy d = (9 x 4 + 6 x 2 + 1) = 36 x3 + 12 x dx dx dy du dy . = du dx dx

Once again,

The derivative of the composite function f(g(x)) at x is the derivative of f at g(x) times the derivative of g at x. Ex.47 Find the derivation of y = x 2 + 1 Sol.

Here y = f(g(x)), where f(u) = f ′ (u) =

1 2 u

2 u and u = g(x) = x + 1. Since the derivatives of f and g are

and g′(x) = 2x,

the Chain Rule gives

1 1 dy d = f (g( x )) = f′ (g(x)).g′(x) = .g′(x) = . (2x) = 2 g( x ) dx dx 2 x2 + 1

d sin( x 2 + x ) = cos( x 2 + x ).(2x + 1) dx Inside

Ex.49 (a)

x +1

derivative of the outside

outside

Ex.48

x 2

Inside derivative left along of the inside

d 1 (1 – x 2 )1/ 4 = (1 – x 2 ) – 3 / 4 (–2x ) dx 4

u = 1 – x2 and n = 1/4

(Function defined) on [–1, 1] =

–x 2(1 – x 2 )3 / 4

(derivative defined only on (–1, 1))

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VECTOR & CALCULUS

Page # 29

(b)

d d sin 2x = cos 2x 2x = cos 2x .2 = 2 cos 2x dx dx

(c)

d d ( A sin( ωt + φ)) = A cos (ω t + φ) (ω t + φ ) = A cos (ω t + φ). ω dt dt

Rull No. 12

= A ω cos (ω t + φ)

Power Chain Rule

d n du u = nun –1 dx dx

*

If

Ex.50

d  1  d d   = (3 x – 2) –1 = – 1 (3x – 2)–2 ( 3 x – 2) dx  3 x – 2  dx dx

= – 1 (3x – 2)–2 (3) = –

3 ( 3 x – 2) 2

In part (d) we could also have found the derivation with the Quotient Rule. Ex.51 (a) Sol.

d ( Ax + B)n dx

Here u = Ax + B,

du =A dx

d ( Ax + B )n = n( Ax + B)n –1.A dx d d 1 sin( Ax + B) = cos( Ax + B).A (c) (b) log(Ax + B) = .A dx dx Ax + B ∴

(d)

d d ( Ax +B ) e = e( Ax +B ) .A tan (Ax+B) = sec2 (Ax + B).A (e) dx dx

Note : These results are important

19.

DOUBLE DIFFERENTIATION If f is differentiable function, then its derivative f' is also a function, so f' may have a derivative of its own, denoted by ( f ' )' = f ' ' . This new function f'' is called the second derivative of because it is the derivative of the derivative of f. Using Leibniz notation, we write the second derivative of y = f(x) as

d  dy  d2 y  = dx  dx  dx 2 Another notation is f''(x) = D2 f (x). Ex.52 If f(x) = x cos x, find f" (x) Sol.

Using the Product Rule, we have f '(x) = x

d d ( x ) = – x sin x + cos x (cos x) + cos x dx dx

To find f" (x) we differentiate f'(x) : d d d d (– x sin x + cos x ) = – x (sin x ) + sin x (–x) + (cos x) dx dx dx dx = – x cos x – sinx – sinx = – x cos x – 2 sin x

f"(x) =

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Page # 30

20.

APPLICATION OF DERIVATIVE DIFFERENTIATION AS A RATE OF CHANGE dy is rate of change of 'y' with respect to 'x' : dx For examples : (i) v =

dx this means velocity 'v' is rate of change of displacement 'x' with respect to time 't' dt

(ii) a =

dv this means acceleration 'a' is rate of change of velocity 'v' with respect to time 't'. dt

(iii) F =

dp this means force 'F' is rate of change of monentum 'p' with respect to time 't'. dt

(iv) τ =

dL this means torque 'τ' is rate of change of angular momentum 'L' with respect to time 't' dt

(v) Power =

dW this means power 'P' is rate of change of work 'W' with respect to time 't' dt

π 2 D . 4 How fast is the area changing with respect to the diameter when the diameter is 10 m ? The (instantaneous) rate of change of the area with respect to the diameter is

Ex.53 The area A of a circle is related to its diameter by the equation A =

Sol.

dA π πD = 2D = dD 4 2

When D =10m, the area is changing at rate (π/2) = 5π m2/m. This mean that a small change ∆D m in the diameter would result in a changed of about 5p ∆D m2 in the area of the circle. Physical Example : Ex.54 Boyle's Law state that when a sample of gas is compressed at a constant temperature, the product of the pressure and the volume remains constant : PV = C. Find the rate of change of volume with respect to pressure. Sol.

dV C =– 2 dP P

Ex.55 (a) Find the average rate of change of the area of a circle with respect to its radius r as r changed from (i) 2 to 3

(ii) 2 to 2.5

(iii) 2 to 2.1

(b) Find the instantaneous rate of change when r = 2. (c) Show that thre rate of change of the area of a circle with respect to its radius (at any r) is equal to the circumference of the circle. Try to explain geometrically when this is true by drawing a circle whose radius is increased by an amount ∆r. How can you approximate the resulting change in area ∆A if ∆r is small ? Sol.

(a) (i) 5π (ii) 4.5 π (iii) 4.1 π (b) 4 π (c) ∆A ≈ 2 π r ∆r

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VECTOR & CALCULUS

21.

Page # 31

MAXIMA & MINIMA y Suppose a quantity y depends on another quantity x in a manner shown in figure. It becomes maximum at x1 and minimum at x2. At these points the tangent to the curve is parallel to the x-axis and hence its slope is tan θ = 0. Thus, at a maxima or a minima slope ⇒

dy =0 dx

x1

x

x2

Maxima Just before the maximum the slope is positive, at the maximum it dy is zero and just after the maximum it is negative. Thus, decrease dx dy at a maximum and hence the rate of change of is negative at dx d  dy  d  dy    < 0 at maximum. The quantity   is a maximum i.e., dx  dx  dx  dx  the rate of change of the slope. It is written d2 y

dy =0 as 2 . Conditions for maxima are : (a) dx dx

(b)

d2 y dx 2

y 2

3

4

θ4 5 θ5 slope = m1 = tan θ1 m1 > m 2 >(m3 = 0) > m4 > m5 x

θ2

θ11

O For maxima, as x increases the slope decreases

0 dx  dx 

Conditions for minima are :

dy =0 (a) dx

(b)

d2 y dx 2

O

>0

slope = m1 = tan θ1 m1 < m2 90° m > 0 ⇒ θ < 90° 0° ≤ θ < 180° m

m=tanθ

m=tanθ θ

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θ

Page # 16

KINEMATICS

Ex.23 Draw the graph for the equation : 2y = 3x + 2 3 Sol. 2y = 3x + 2 ⇒ y = x + 1 2 3 m= > 0 ⇒ θ < 90° 2 c = +1 > 0 ⇒ The line will pass through (0, 1)

(0,1)

tan θ =

θ

Ex.24 Draw the graph for the equation : 2y + 4x + 2 = 0 Sol. 2y + 4x + 2 = 0 ⇒ y = – 2x – 1 m = – 2 < 0 i.e., θ > 90° c = – 1 i.e., line will pass through (0, –1)

3 2

tanθ = –2

θ

(0,–1)

: (i) If c = 0 line will pass through origin. (0,c) (ii) y = c will be a line parallel to x axis.

(ii)

(c,0)

(0,0)

(iii) x = c will be a line perpendicular to y axis

(0,0)

Parabola A general quadratic equation represents a parabola. y = ax2 + bx + c a≠ 0 if a > 0 ; It will be a opening upwards parabola. if a < 0 ; It will be a opening downwards parabola. if c = 0 ; It will pass through origin. 2

y=4x +3x e.g.

y = 4 x2 + 3x

2

y=–4x +3x

Average velocity & instantaneous velocity from Position vs time graph Average velocity from t1 to t2 =

x2 – x1 displacement = t –t time taken 2 1

B

x2

x2–x1

= tan θ = slope of the chord AB vinstantaneous

x 2 – x1 = as lim t2 → t1 t 2 – t1

x1

A t1

θ t2–t1

t2

when t2 approaches t1 point B approaches Point A and the chord AB becomes tangent to the curve. Therefore vinstantaneous = Slope of the tangent x – t curve

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Page # 17

KINEMATICS (iii)

Reading of Graph

(A)

Reading x v/s t graphs x

(1)

x0

Explanation

Body is at rest at x0.

t x

(2)

Body starts from origin and is moving with speed θ

tan θ away from origin.

t

x (3)

Body starts from rest from origin and moves away from origin with increasing speed velocity and positive acceleration.

t

x (4)

Body starts from rest from x = x0 and moves away from origin with increasing velocity or +ve acceleration.

x0 t

(5)

x0

Body starts from x = x0 and is moving toward the origin with constant velocity passes throw origin after same time and continues to move away from origin.

t x (6)

x0

Body starts from rest at x = x0 and then moves with increasing speed towards origin ∴ acceleration is –ve

t

(7)

x

Body starts moving away from origin with some initial speed. Speed of body is decreasing till t1 and it becomes 0 momentarily of t = t1 and At this instant. Its reverses its direction and move towards the origin with increasing speed.

t2

O

t1

t

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Page # 18

(8)

KINEMATICS

x

Body starts from origin moves away from origin in the –ve x-axis at t = t1 with decreasing speed and at t= t1 it comes at rest momentarily, Reverses its direction t moves towards the origin the increasing speed. Crosses the origin at t = t2.

t1 t2

x (9)

t

Body starts from origin from rest and moves away from origin with increasing speed.

(B) V-T GRAPHS v

(1)

Body is always at rest.

t

v (2)

Body is moving with constant velocity v0

v0

t

v (3)

Body is at rest initially then it starts moving with its velocity increasing at a constant rate i.e. body is moving with constant acceleration.

t

v (4)

(5)

Body starts its motion with initial velocity v0 and continues to move with its velocity increasing at a constant rate i.e. acceleration of the body is constant.

v0 t

v

v0 t0 t v (6)

t

Body starts its motion with initial velocity v0. Then it continues to move with its velocity decreasing at a constant rate i.e. acceleration of the body is negative and constant. At t = t0 the body comes to rest instantaneously and reverses its direction of motion and then continues to move with decreasing velocity or increasing speed. For 0 < t < t0 motion of the body is deaccelerated (∴ speed is decreassing) t > t0 motion of the body is accelerated (∴ speed is increasing) Body is at rest initially. Then it starts moving with increasing velocity. As time increases its velocity is increasing more rapidly. i.e. the moving with increasing acceleration.

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Page # 19

KINEMATICS

v (7)

v0

t0

t

Body starts its motion with initial velocity v0. Its velocity is decreasing with time and at t = t0 . It becomes zero after body reverse its direction of motion and continues to move with decreasing velocity or increasing speed. Since velocity of the body is decreasing for whole motion. Therefore, its acceleration is negative.For 0 < t < t0 motion of the body is deaccelerated (speed is decreassing) t > t0 motion of the body is accelerated (∵ speed is increasing)

(C) READING OF a - t GRAPHS a

(1)

acceleration of the body is zero that means the body is moving constant velocity. t

a

(2)

Acceleration of the body is constant and positive. t

a t

(3)

Acceleration of the body is constant and negative

a

Initially the acceleration of the body is zero. Then its acceleration is increasing at a constant rate.

(4) t

a

(5)

t

The body starts accelerating(initial acceleration zero) at t = 0. Its acceleration is negative for whole of its motion and is decreasing at a constant rate.

a

(6) t

Initially acceleration of the body is zero. Its acceleration is positive for whole of its motion. Its acceleration is increasing for whole of its motion.

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Page # 20

KINEMATICS

(IV)

Drawing of graphs on the basis of given information.

(a) (i)

If acceleration of the body is zero. If the velocity of the body is v0 and it starts from origin. x v x= u

0

t

v0 t

(ii)

(iii)

t

If at t = 0, x = x0 then x t 0 +v 0 x x0 x=

v

v0

t If at t = 0, x = – x0 then x 0 –x x=

t v

t t 0 +v

v0

t

–x0

t

(b)

If a body has constant acceleration : For this section (i) u0, x0 & a0 are positive constants. (ii) u ≡ initial velocity (iii) v ≡ velocity at any time t. (iv) x ≡ Position at any time t. xi ≡ initial position

(i)

if u = 0, a = a0 1 if xi = 0, x = at 2 2 x

if xi = x0, x = x0 + (1/2)at2 x

x This is wrong because it suggest the body don't have some initial velocity

x0 t

t

t v

a

a0

slope = tanθ = a0 θ

t

t

v = a0 t (ii)

If u = u0

, a = a0

x = xi + u0t +

1 a0 t 2 2

x

v = u0 + a0t

x

v

a

a0

if xi = 0

t

if xi = x0

t

t

t

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Page # 21

KINEMATICS

(iii)

if u = u0, a = – a0 1 2 x = xi + u0t – a 0 t 2

x

x

x0

if xi = 0 t

t0

v

if xi = x0

t

a

u0 t

t0 t (iv)

–a0

if u = – u0 , a = + a0 x = xi – u0t +

x

1 a0 t 2 2

x0

t if xi = x0

if xi = 0

v

a

a0

t

–u0 (v)

t

If u = u0, a = – a0 x = xi – u0t –

x

1 a0 t 2 2

x

x0

t

v

–v0

t

if xi = 0

if xi = x0

a

t

t

–a0

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Page # 22

KINEMATICS

Ex.25 Draw the (a) position vs time graph (b) velocity vs time graph (c) acceleration vs time graph for the following cases (i) If a body is projected vertically upwards with initial velocity u. Take the projection point to be origin and upward direction as positive. x = ut –

1 2 gt 2

x

v

2

u

u 2g

u g

(ii)

a 2u g

u g

t

t

t

2u g

–g

–u

If a body is dropped from a height h above the ground. Take dropping point to be origin and upward direction as +ve. x= –

1 2 gt 2

x

v 2h g

a=–g a

2h g

t

t

t

v = – gt

–h

(iii)

If a body is projected vertically upwards from a tower of height h with initial velocity u. Take the projection point to be origin and upward direction as +ve.

x u2 2g

v

u g –h

(iv)

–g

– 2gh

2u g

t v = u – gt

a u/2g

t

t

–g

A car starting from rest accelerates uniformly at 2 ms–2 for 5 seconds and then moves with constant speed acquired for the next 5 seconds and then comes to rest retarding at 2 ms–2. Draw its (a) Position vs time graph (b) Velocity vs time graph (c) acceleration vs time graph

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Page # 23

KINEMATICS acceleration vs time graph

x (in m)

x

100 a +2

v 10

15

5

t

75

–1

10ms

15

–2

5

acceleration vs time graph

(v)

10

25

t

15 5 10 Position vs time graph

velocity vs time graph

t (in sec)

A particle starts from x = 0 and initial speed 10 ms–1 and moves with constant speed 10ms–1 for 20 sec. and then retarding uniformly comes to rest in next 10 seconds. acceleration vs time graph a v –1

20

30

10ms

t (sec)

–2

–1ms

30 20 velocity vs time graph

Acceleration vs time graph

t (sec)

x

250m

200m 30sec

20

t (sec)

Position vs time graph

(V)

Conversion of velocity v/s time graph to speed v/s time graph. As we know that magnitude of velocity represent speed therefore whenever velocity goes –ve take its mirror image about time axis.

velocity

Ex-26

speed

t (sec)

velocity

Ex-27

i m

t (sec)

speed

t

e ag m i r rro

e ag m i or irr m t

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Page # 24

Conversion of displacement vs time graph to distance vs time graph For distance time graph just make the mirror image of the displacement time graph from point of zero velocity onwards.

(VII) Conversion of v - t graphs in to x-t and a-t graphs v x

A

Time

0

t

t

v

x

a

ta nθ

=

a

0

a0 ⇒

a - t graph

t

t

t

v

x

a

tan θ = – a0 (iii)

Disp-time

B

x= v

⇒

(i)

(ii)

Dist.-time

C

t

v0

D

Dist./Displacement

(VI)

KINEMATICS

t ⇒

θ

t

t0

t

t0

–a0 at t = t0 velocity reverses its direction.

v (iv)

x – t graph From t = 0 to t = t1 acceleration = 0 therefore from t = 0 to t = t1, x - t graph will be a straight line. From t = t1 to t2 acceleration is negative ∴ It will be an opening downward parabola

v0

t1

x

t1

(v)

t2

t2

t

t

v

upto t = t1 acceleration is +ve t1 < t < t2 acceleration is zero. t > t2 acceleration is –ve x

t1

t2

t

x - t graph t1

t2

t

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Page # 25

KINEMATICS Some important points : dv ⇒ dv = adt dt ⇒ ∆v = area under the a - t curve

a=

•

∫

∫

dx ⇒ dx = vdt dt ⇒ ∆x = area under the v - t curve ⇒ displacement = area under the v - t curve

•

v=

∫

∫

a

Ex-28 If at t = 0 u = 5 ms–1 then velocity at t = 10 sec = u + change in velocity = 5 + area of the shaded part = 5 + 10 × 5 = 55 ms–1

–2

5ms

a

10 sec t

Ex-29 if at t = 0, u = 2 ms–2 find out it maximum velocity 5ms–2 Since whole motion is accelerating. Therefore velocity will be max at the end of the motion which will be 1 =2+ × 5 × 10 = 27 ms–1 t 2 10sec Ex-30 if at t = 0, u = 4 ms–1 a Find out v at t = 10 sec, t = 20 sec & t = 30 sec. –2 10ms Since for whole motion acceleration of the body is positive 1 vt= 10 sec = 4 + × 10 × 10 = 54 ms–1 t 2 10sec 20sec 30sec 1 vt = 20 sec = 4 + × 10 × 10 + 10 × 10 2 –1 = 154 ms 1 vt = 30 sec = 154 + × 10 × 10 = 204 ms–1 2 (VIII) Reading of graphs if the motion of two bodies are sketched on the same axes. (a)

Reading of x - t graphs

x

x3

B

x2 x1 O

(i) (ii) (iii) (iv) (v) (vi)

A t1

t2

t3

t

Conclusions : Body A Start its motion at t = 0 from origin and is moving away from the origin with constant velocity. Finally it ends its motion at a distance of x2m from origin at t = t3. Body B starts its motion at t = t1 from origin and is moving away from origin with constant velocity. Finally it ends its motion at a distance of x3m from origin at t = t3 Since slope of B is greater than slope of A. Therefore velocity of B is greater than velocity of A. A t = t2, Both A & B are at the same distance from starting point that means B overtakes A at t = t2 ∵ velocity of both A & B are constant ∵ acceleration of both the bodies are zero. ∴ x3 > x2 ∴ At the end of the motion B is at a greater distance from the starting point.

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Page # 26

KINEMATICS x A

x2

B

x1

Ex-31

(i) (ii) (iii) (iv) (v) (vi)

x0 t1 t0 t Conclusion : Body A starts its motion at t = 0 from origin and is moving away from the origin with constant velocity. Finally its motion ends at t = t1 at x = x2 m. Body B starts its motion at t = 0 from x = x0 and then moves with constant velocity away from the origin. Finally it ends its motion at t = t1. Velocity of A is greater than that of B. At t = t0 A overtakes B acceleration of both A & B is zero. ∵ x2 > x1 ∴ At the end of the motion A is at a greater distance from the starting point then B

x

B

A

Ex-32

(i) (ii) (iii) (iv) (v) (vi)

t1 t2 t Conclusions : Both A & B starts their motion at same time t = 0 and from same point x = 0. Both are moving away from the starting point. A is moving with constant velocity while B starts its motion from rest and its velocity is increasing with time i.e. it has some positive acceleration. ∵ At t = t1 the tangent on B's graph becomes parallel to the A's graphs ∴ At t = t1 velocity of both A & B is same. For t < t1 velocity of A is greater than velocity of B. Therefore up to t = t1, separation between A & B increases with time. For t > t1 velocity of B is greater than velocity of A. Therefore after t = t1 separation between A & B starts decreasing and it becomes zero at t = t2 where B overtakes A. Now you can try Questions 14 to 38 in Exercise I and Ques. 7 to 11 in Ex.II

4.

TWO DIMENSIONAL MOTION OR MOTION IN A PLANE Motion in a plane can be described by vector sum of two independent 1D motions along two mutual perpendicular directions (as motions along two mutual directions don’t affect each other). Consider a particle moving in X-Y plane, then its equations of motions for X and Y axes are vx = ux + axt, vy = uy + ayt 1 1 x = uxt + axt2, and ; y = uyt + ayt2, and 2 2 v 2y = u 2y + 2a y y v 2v = u2x + 2a x x where symbols have their usual meanings. Thus resultant motion would be described by the equations

and v = v i + v j r = x i + y j x

y

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Page # 27

KINEMATICS 4.1

PROJECTILE MOTION It is the best example to understand motion in a plane. If we project a particle obliquely from the surface of earth, as shown in the figure below, then it can be considered as two perpendicular 1D motions - one along the horizontal and other along the vertical.

Y

⇒

O

θ

usinθ

u +

u cos θ

x

Assume that effect of air friction and wind resistance are negligible and value of ‘acceleration due to → gravity g is constant. Take point of projection as origin and horizontal and vertical direction as +ve X and Y-axes, respectively. For X-axis For Y - axis ux = u cosθ, uy = u sinθ ax = 0, ay = – g, vx = u cosθ, and vy = u sinθ – gt, and 1 x = u cosθ × t y = u sinθ t – gt2 2 It is clear from above equations that horizontal component of velocity of the particle remains constant while vertical component of velocity is first decreasing, gets zero at the highest point of trajectory and then increases in the opposite direction. At the highest point, speed of the particle is minimum. The time, which projectile takes to come back to same (initial) level is called the time of flight (T). At initial and final points, y = 0, 1 So u sinθ t – gt2 = 0 2

⇒ t = 0 and t =

2u sin θ g

So,

T=

2u sin θ g

Range (R) The horizontal distance covered by the projectile during its motion is said to be range of the projectile u 2 sin 2θ g For a given projection speed, the range would be maximum for θ = 45°. Maximum height attained by the projectile is

R = u cosθ × T =

u 2 sin 2 θ 2g at maximum height the vertical component of velocity is 0. u sin θ T Time of ascent = Time of descent = = g 2

H=

Speed, kinetic energy, momentum of the particle initialy decreases in a projectile motion and attains a minimum value (not equal to zero) and then again increases.
θ is the angle between v and horizontal which decreases to zero. (at top most point) and again

increases in the negative direction

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Page # 28

KINEMATICS

Ex.33 A body is projected with a velocity of 30 ms–1 at an angle of 30° with the vertical. Find the maximum height, time of flight and the horizontal range. Sol. Here u = 30 ms–1, Angle of projection, θ = 90 – 30 = 60° Maximum height, u 2 sin 2 θ 30 2 sin 2 60° = = 34.44 m 2g 2 × 9.8 Time fo flight, 2u sin θ 2 × 30 sin 60° T= = = 5.3 s g 9.8 Horizontal range,

H=

R=

u 2 sin 2θ 30° sin 120° 30 2 sin 60° = = = 79.53 m. g 9.8 9.8

Ex.34 Find out the relation between uA, uB, uC (where uA, uB, uC are the initial velocities of particles A, B, C, respectively)

A

Sol.

B C

∵ Hmax is same for all three particle A, B, C

⇒ Hmax =

u2y

2g ⇒ uy is same for all

∴ uyA = uyB = uyC

 2u y  ⇒ TA = TB = TC  g    from figure

RC > RB > RA

⇒ uxC > uxB > uxA

(C)

⇒

∵R =

2u xu y g

uA < uB< uC

Coordinate of a particle after a given time t : Particle reach at a point P after time t then

Y

vy

x = ucosθ .t y = usinθ.t –

1 2 gt 2

usinθ

Position vector

P(x,y) u

vx

y θ

O ucosθ


1   r = (u cos θ.t )ˆi +  (u sin θ)t – gt 2 ˆj 2  

(D)

v α

x

X

Velocity and direction of motion after a given time : After time 't' vx = ucosθ and vy = usinθ – gt Hence resultant velocity v =

tan α =

vy vx

=

u sin θ – gt u cos θ

2

vx + vy

2

=

u2 cos 2 θ + (u sin θ – gt) 2

–1  u sin θ – gt   ⇒ α = tan  u cos θ 

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Page # 29

KINEMATICS (E)

Velocity and direction of motion at a given height : At a height 'h', vx = ucosθ

u2 sin2 θ – 2gh

And

vy =

∴

Resultant velocity v=

v x2 + v y 2 =

v=

u2 – 2gh

(u cos θ) 2 + u2 sin2 θ – 2gh

Note that this is the velocity that a particle would have at height h if it is projected vertically from ground with u.

Ex.35 A body is projected with a velocity of 20 ms–1 in a direction making an angle of 60° with the horizontal. Calculate its (i) position after 0.5 s and (ii) velocity after 0.5 s. Sol. Here u = 20 ms–1, θ = 60° , t = 0.5 s (i) x = (u cosθ)t = (20 cos60°) × 0.5 = 5 m y = (u sin θ) t – –

1 2 gt = (20 × sin 60°) × 0.5 2

1 × 9.8 × (0.5)2 = 7.43 m 2

(ii) vx = u cos θ = 20 cos 60° = 10 ms–1 vy = u sin θ – gt = 20 sin 60° – 9.8 × 0.5 = 12.42 ms–1 ∴ v=

v 2x + v 2y =

–1 (10) 2 + (12.42) 2 =15.95 ms

tan β =

vy vx

=

12.42 = 1.242 10

∴ β = tan–1 1.242 = 51.16°.

Equation of trajectory of a projectile. Suppose the body reaches the point P(x, y) after time t. Y

vy x

usinθ

u θ

v A α v x P(x,y) Max. y height=h

Path of projectile

m

vx=u cosθ

O ucosθ

B

R uy

θ

X

v

∵ The horizontal distance covered by the body in time t, x = Horizontal velocity × time = u cos θ. t x ucos θ For vertical motion : u = u sinθ, a = –g, so the vertical distance covered in time t is given by

or t =

s = ut +

1 2 at 2

or y = x tanθ –

or

y = u sin θ.

1 x2 g 2 2 u cos 2 θ

x 1 x2 – g. 2 u cos θ 2 u cos 2 θ

...(1)

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Page # 30

KINEMATICS

or y = px – qx2, where p and q are constants. Thus y is a quadratic function of x. Hence the trajectory of a projectile is a parabola. From equation (1)



y = x tan θ 1 –



gx cos θ

 gx    ⇒ y = x tan θ 1 – 2u 2 cos θ sin θ  2u cos θ sin θ  2

2

x  y = x tan θ 1 –  R 

...(2)

Equation (2) is another form of trajectory equation of projectile

Ex.36 A ball is thrown from ground level so as to just clear a wall 4 m high at a distance of 4 m and falls at a distance of 14 m from the wall. Find the magnitude and direction of the velocity. Sol. The ball passes through the point P(4, 4). So its range = 4 + 14 = 18m. The trajectory of the ball is, Now x = 4m, y = 4m and R = 18 m y 4   7 ∴ 4 = 4 tan θ 1 –  = 4 tanθ . P(4,4)  18  9 u or tan θ =

9 , sin θ = 7

9 130

, cosθ =

7

4m

130

θ

18 × 9.8 × 130 or u2 = = 182 2×9×7

or u =

4m

x

14m

–1 182 = 13.5 ms

Also θ = tan–1(9/7) = 52.1°

Ex.37 A particle is projected over a triangle from one end of a horizontal base and grazing the vertex falls on the other end of the base. If α and β be the base angles and θ the angle of projection, prove that tan θ = tan α + tan β. Sol. If R is the range of the particle, then from the figure we have tan α + tan β =

y(R – x) + xy y y + = x(R – x) x R– x

or tanα + tan β =

y R × x (R – x)

Y P(x,y)

...(1)

Also, the trajectrory of the particle is

θ

x  y = x tan θ 1–   R

or tanθ =

O

y β

α x

B

R–x

A

y R × x (R – x)

From equations (1) and (2), we get tan θ = tan α + tan β .

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x

Page # 31

KINEMATICS 4.2

Projectile fired parallel to horizontal. As shown in shown figure suppose a body is projected horizontally with velocity u from a point O at a certain height h above the ground level. The body is under the influence of two simultaneous independent motions: (i) Uniform horizontal velocity u. (ii) Vertically downward accelerated motion with constant acceleration g. Under the combined effect of the above two motions, the body moves along the path OPA.

u

O

x

y x

vx

P β

h

v

vy R Y

A Ground

Trajectory of the projectile. After the time t, suppose the body reaches the point P(x, y). The horizontal distance covered by the body in time t is x x = ut ∴ t= u The vertical distance travelled by the body in time t is given by 1 2 s = ut + at 2 1 2 1 2 gt = gt 2 2 [For vertical motion, u = 0]

or

y=0×1+

or

y=

or

y = kx2 [Here k =

2

1 x  g g  =  2 2 u  2u

x  ∵ t = u   

 2 x 

g

= a constant] 2u2 As y is a quadratic function of x, so the trajectory of the projectile is a parabola. Time of flight. It is the total time for which the projectile remains in its flight (from 0 to A). Let T be its time of flight. For the vertical downward motion of the body, we use 1 2 s = ut + at 2 or h = 0 × T +

1 gT2 2

or

T=

2h g

Horizontal range. It is the horizontal distance covered by the projectile during its time of flight. It is equal to OA = R. Thus R = Horizontal velocity × time of flight = u × T 2h g Velocity of the projectile at any instant. At the instant t (when the body is at point P), let the velocity of the projectile be v. The velocity v has two rectangular components: Horizontal component of velocity, vx = u Vertical component of velocity, vy = 0 + gt = gt ∴ The resultant velocity at point P is

or

R=u

v = v2x + v2y =

u2 + g2 t2

If the velocity v makes an angle β with the horizontal, then vy  gt  gt tan β = = or β = tan–1   vx u u

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Page # 32

KINEMATICS

Ex.38 A body is thrown horizontally from the top of a tower and strikes the ground after three seconds at an angle of 45° with the horizontal. Find the height of the tower and the speed with which the body was projected. Take g = 9.8 ms–2. Sol. As shown in figure, suppose the body is thrown horizontally from the top O of a tower of height y with velocity u. The body hits the ground after 3s. Considering verticlly downward motion of the body, 1 2 1 y = uyt + gt = 0 × 3 + ×9.8 × (3)2 = 44.1 m [∴ Initial vertical velocity, uy = 0] 2 2 Final vertical velocity, vy = uy + gt = 0 + 9.8 × 3 = 29.4 ms–1 Final horizontal velocity, vx = u As the resultant velocity u makes an angle of 45° with the horizontal, so tan 45° =

vy vx

or 1 =

29.4 x

or u = 29.4 ms–1.

Ex.39 A particle is projected horizontally with a speed u from the top of plane inclined at an angle θ with the horizontal. How far from the point of projection will the particle strike the plane? Sol. The horizontal distance covered in time t, x x = ut or t = u u The vertical distance covered in time t, θ y=0+ Also

1 x2 1 2 gt = g × 2 2 2 u

[using (1)]

y

y gx 2 = tan θ or y = x tan θ ∴ = x tan θ x 2u 2

D θ x=ut

 gx  or x  2 – tan θ  = 0  2u  2u 2 tan θ g The distance of the point of strike from the point of projection is

As x = 0 is not possible, so x =

D=

x 2 + y2 =

=x

x2 + (x tan θ)2

1 + tan2 θ = x sec θ or D =

2u2 tan θ sec θ g

Ex.40 A ball rolls off the top of a stairway with a constant horizontal velocity u. If the steps are h metre high and w meter wide, show that the ball will just hit the edge of nth step if n = Sol.

Refer to figure. For n th step, net vertical displacement = nh net horizontal displacement = nω Let t be the time taken by the ball to reach the nth step. Then R = ut nω or nω = ut or t= u 1 2 Also, y = uy t + gt 2 2

or nh = 0 +

1  nω  1 2 gt = g  2  u  2

or n =

2hu 2 gw 2

u

1st 2nd h w

nth R

2hu2

gω2

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Page # 33

KINEMATICS 4.3

Projectile at an angle θ from height h Consider the projectile as shown in the adjacent figure. Take the point of projection as the origin the X and Y-axes as shown in figure.

u

Y θ

For X-axis, = u cosθ x ax = 0 vx = u cosθ, and x = u cos θ × t u

h

For Y-axis, x

uy = u sin θ, ay = –g,

gt 2 2 Ex.41 From the top of a tower 156.8 m high a projectile is projected with a velocity of 39.2 ms–1 in a direction making an angle 30° with horizontal. Find the distance from the foot of tower where it strikes the ground and time taken to do so. Sol. The situation is shown –1 u=39.2 ms Here height of tower OA = 156.8 m H u = 39.2 ms–1 uy = usinθ θ = 30° time for which projectile remain is air = t = ? θ=30° Horizontal distance covered R = OD = ? A Now ux = u cos θ and ux = u cosθ B
uy = u sin θ be the components of velocity u . Motion of projectile from O to H to D

vy = u sin θ – gt, and y = u sin θ t –

Using equation y = uyt +

1 ay t2 2

Here : y = 156.8 m ; uy = – u sinθ = 39.2 sin 30° ay = 9.8 m/s2 ; t = ? 156.8 = – 39.2 × 0.5 t + 4.9 t2 156.8 = – 19.6 t + 4.9 t2 or 4.9 t2 – 19.6 t – 156.8 = 0 (t – 8) (t + 4) = 0 or t2 – 4t – 32 = 0 ⇒ We get t = 8 s; t = – 4s t = – 4 s is not possible, thus we take t = 8s. Now horizontal distance covered in this time R = ux × t = u cos θ × t = 39.2 × cos 30° × t R = 271.57 m

4.4

156.8 m

O

C

D

Projectile Motion in Inclined Plane Here, two cases arise. One is up the plane and the other is down the plane. Let us discuss both the cases separately. (i) Up the Plane : In this case direction x is chosen up the plane and direction y is chosen perpendicular to the plane. Hence, ax = – g sin β ux = u cos α , uy = u sin α and ay = – g cos β

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Page # 34

KINEMATICS

y

x

B

u

gsinβ β

α

β

β

O

gcosβ

g

O

C

Now, let us derive the expressions for time of flight (T) and range (R) along the plane.

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

1 ay t2 , 2

we get 0

=

u

t

s i n

1 (– g cos β ) t2 2

α+

t = 0, corresponds to point O and t =

T=

∴

t = 0 and

2u sin α g cos β

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

2u sin α g cos β

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

(ii)

1 axT2 2

⇒

R = u cos α T –

Down the inclined plane : along x - axis y-axis (1) ux = ucos α (1) uy = usinα (2) ax = g sin β (2) ay = g cos β velocity at P vy = uy + ay T vx = ux + axT Time of flight T =

1 sin β T2 2

y

u sin α (0,0)

2u y ay

2u sin α = g cos β

1 2 Range Sx = ux T + a x T 2 1 2 = u cos α T + g sin β. T 2

u α

uc os α

β os c g

β

gs P in β

g

β

x

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Page # 35

KINEMATICS

Ex.42 A particle is projected at an angle α with horizontal from the foot of a plane whose inclination to horizontal is β . Show that it will strike the plane at right angles if cotβ = 2 tan (α – β) Let u be the velocity of projection so that u cos (α – β ) and u sin (α – β ) are the initial velocities Sol. respectively parallel and perpendicular to the inclined plane. The acceleration in these two directions are (–g sin β ) and (–g cos β ). The initial component of velocity perpendicular to PQ is u sin (α – β ) and the acceleration in this direction is (–g cosβ ). If T is the time the particle takes to go from P to Q then in time T the space described in a direction perpendicular to PQ is zero. 0 = u sin (α – β ).T – T=

u

1 g cos β .T2 2

Q

2u sin(α – β) g cos β

If the direction of motion at the instant when the particle hits the plane be perpendicular to the plane, then the velocity at that instant parallel to the plane must be zero. ∴ u cos (α – β ) – g sin β T = 0

α β N

P

u cos(α – β) 2u sin(α – β) =T= g sin β g cos β

∴ cosβ = 2 tan (α – β )

Ex.43 Two inclined planes OA and OB having inclinations 30° and 60° with the horizontal respectively intersect each other at O, as shwon in u

figure. a particle is projected from point P with velocity u = 10 3 m / s along a direction perpendicular to plane OA. If the particle strikes A plane OB perpendicular of flight, then calculate.

h

(a) time of flight

(d) distance PQ. (Take g = 10 m/s2) Let us choose the x and y directions along OB and OA respectively. Then, ux = u = 10 3 m/s, uy = 0 ax = – g sin 60° = – 5 3 m/s2 and ay = – g cos 60° = – 5 m/s2 (a) At point Q, x-component of velocity is zero. Hence, substituting in vx= ux + axt 10 3

= 2s Ans. 5 3 (b) At point Q, v = vy = uy + ayt ∴ v = 0 – (5) (2) = –10 m/s Ans. Here, negative sign implies that velocity of particle at Q is along negative y direction. (c) Distance PO = |displacement of particle along y-direction| = |sy| Here, sy = uyt +

∴

t=

1 1 ay t2 = 0 – (5)(2)2 = – 10 m 2 2

PO = 10 m

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B Q 60°

30°

(c) height h of point P from point O

0 = 10 3 – 5 3t ⇒

v

P

(b) velocity with which the particle strikes the plane OB,

Sol.

x

y

O

Page # 36

KINEMATICS

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

or

h = 5m Ans.

(d) Distance OQ = displaement of particle along x-direction = sx Here, sx = uxt + or

1 1 ax t2 = (10 3)(2) – (5 3)(2)2 = 10 3 m 2 2

OQ = 10 3 m PQ =

(PO)2 + (OQ)2

PQ = 20 m

=

(10)2 + (10 3)2 =

100 + 300 = 400

Ans.

Now you can try Questions 45 to 68 in Exercise I and Ques. 12 to 20 in Ex.II

5.

RELATIVE MOTION The word 'relative' is a very general term, which can be applied to physical, nonphysical, scalar or vector quantities. For example, my height is five feet and six inches while my wife's height is five feet and four inches. If I ask you how high I am relative to my wife, your answer will be two inches. What you did? You simply subtracted my wife's height from my height. The same concept is applied everywhere, whether it is a relative velocity, relative acceleration or anything else. So, from the above discussion → we may now conclude that relative velocity of A with respect of B (written as v AB ) is → → → v AB = v A – v B

Similarly, relative acceleration of A with respect of B is → → → a AB = a A – a B

If it is a one dimensional motion we can treat the vectors as scalars just by assigning the positive sign to one direction and negative to the other. So, in case of a one dimensional motion the above equations can be written as vAB = vA – vB and aAB = aA – aB Further, we can see that → → → → v AB = – v BA or a BA = – a AB

Ex.44 Seeta is moving due east with a velocity of 1 m/s and Geeta is moving the due west with a velocity of 2 m/s. What is the velocity of Seeta with respect to Geeta? Sol. It is a one dimensional motion. So, let us choose the east direction as positive and the west as negative. Now, given that vs = velocity of Seeta = 1 m/s and vG = velocity of Geeta = – 2m/s Thus, vSG = velocity of Seeta with respect to Geeta = vS – vG = 1 – (–2) = 3 m/s Hence, velocity of Seeta with respect to Geeta is 3 m/s due east.

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Page # 37

KINEMATICS

IMPORTANT NOTE : PROCEDURE TO SOLVE THE VECTOR EQUATION.


...(1) A =B+C (a)

Their are 6 variables in this equation which are following :
(1) Magnitude of A and its direction
(2) Magnitude of B and its direction
(3) Magnitude of C and its direction.

(b)

We can solve this equation if we know the value of 4 varibales [Note : two of them must be directions]

(c)

If we know the two direction of any two vectors then we will put them on the same side and other on the different side.

For example


If we know the directions of A and B and C' s direction is unknown then we make equation as follows :


C = A –B

(d)

Then we make vector diagram according to the equation and resolve the vectors to know the unknown values.

Ex.45 Car A has an acceleration of 2 m/s2 due east and car B, 4 m/s2 due north. What is the acceleration of car B with respect to car A? N Sol. It is a two dimensional motion. Therefore, → a BA = acceleration of car B with respect to car A E W → → = aB = – a A → Here, a B = acceleration of car S B = 4 m/s2 (due north) → and a A = acceleration of car A = 2 m/s2 (due east) → → → a B = 4m / s 2 a BA | a BA |= (4)2 + (2)2 = 2 5m / s2 4 α = tan–1   = tan–1(2) 2 → Thus, a BA is 2 5 m/s2 at an angle of α = tan–1(2) from west towards north.

and

α → – a A = 2m / s 2

Ex.46 Three particle A, B and C situated at the vertices of an equilateral triangle starts moving simultaneously at a constant speed "v" in the direction of adjacent particle, which falls ahead in the anti-clockwise direction. If "a" be the side of the triangle, then find the time when they meet. Sol.

Here, particle "A" follows "B", "B" follows "C" and "C" follows "A". The direction of motion of each particle keeps changing as motion of each particle is always directed towards other particle. The situation after a time "t" is shown in the figure with a possible outline of path followed by the particles before they meet.

A

O

B

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C

Page # 38

KINEMATICS

This problem appears to be complex as the path of motion is difficult to be defined. But, it has a simple solution in component analysis. Let us consider the pair "A" and "B". The initial component of velocities in the direction of line v joining the initial position of the two particles is "v" and "vcosθ" as shown in the figure here : The component velocities are directed towards eachother. v cos θ Now, considering the linear (one dimensional) motion in the 60° B direction of AB, the relative velocity of "A" with respect to v "B" is : vAB = vA – vB vAB = v – (– v cos θ) = v + vcosθ In equilateral triangle, θ = 60° v 3v vAB = v + vcos60° = v + = 2 2 The time taken to cover the displacement "a" i.e. the side of the triangle 2a t= 3v

A

O

v C

QUESTIONS BASED ON RELATIVE MOTION ARE USUALLY OF FOLLOWING FOUR TYPES : (a) Minimum distance between two bodies in motion (b) River-boat problems (c) Aircraft-wind problems (d) Rain problems

(a)

Minimum distance between two bodies in motion When two bodies are in motion, the questions like, the minimum distance between them or the time when one body overtakes the other can be solved easily by the principle of relative motion. In these type of problems one body is assumed to be at rest and the relative motion of the other body is considered. By assuming so two body problem is converted into one body problem and the solution becomes easy. Following example will illustrate the statement.

Ex.47 Car A and car B start moving simultaneously in the same direction along the line joining them. Car A with a constant acceleration a = 4 m/s2, while car B moves with a constant velocity v = 1 m/s. At time t = 0, car A is 10 m behind car B. Find the time when car A overtakes car B. Sol. Given : uA = 0, uB = 1 m/s, aA = 4m/s2 and aB = 0 Assuming car B to be at rest, we have uAB = uA – uB = 0 – 1 = – 1 m/s aAB = aA – aB = 4 – 0 = 4 m/s2 Now, the problem can be assumed in simplified form as follow : 2 2 a=4m/s v=1m/s A

10m

B

+ve

Substituting the proper values in equation 2

uAB= –1m/s, aAB= 4m/s A

10m

B

At rest

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Page # 39

KINEMATICS

s = ut + we get 10 = – t +

1 2 at 2

1 (4)(t2 ) 2

or

2t2 – t – 10 = 0

1±9 1 ± 1 + 80 1 ± 81 = = or t = 2.5 s 4 4 4 Ignoring the negative value, the desired time is 2.5s. Ans.

or

t=

and – 2 s

Note : The above problem can also be solved without using the concept of relative motion as under. At the time when A overtakes B, sA = sB + 10 1 × 4 × t 2 = 1 × t + 10 ∴ 2 or 2t2 – t – 10 = 0 Which on solving gives t = 2.5 s and – 2 s, the same as we found above. As per my opinion, this approach (by taking absolute values) is more suitable in case of two body problem in one dimensional motion. Let us see one more example in support of it. Ex.48 An open lift is moving upwards with velocity 10m/s. It has an upward acceleration of 2m/s2. A ball is projected upwards with velocity 20 m/s relative to ground. Find : (a) time when ball again meets the lift. (b) displacement of lift and ball at that instant. (c) distance travelled by the ball upto that instant. Take g = 10 m/s2 Sol. (a) At the time when ball again meets the lift, sL = sB 1 1 × 2 × t2 = 20 t – × 10t2 2 2 Solving this equation, we get

∴

10t +

t=0

∴

t=

and

2m/s2

20m/s

Ball

+ve

2

10m/s

5 s 3

Ball will again meet the lift after

10m/s

L

Lift

B

Ball

5 s. 3

(b) At this instant 2

sL = sB = 10 ×

5 1 175 5 + ×2×  = m = 19.4 m 3 2 3 9  

(c) For the ball u ↑ ↓a . Therefore, we will first find t0, the time when its velocity becomes zero. t0 =

u 20 = = 2s a 10

 5  As t  = s  < t0 , distance and displacement are equal  3 

or d = 19.4 m Ans. Concept of relative motion is more useful in two body problem in two (or three) dimensional motion. This can be understood by the following example.

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KINEMATICS

Ex.49 Two ships A and B are 10 km apart on a line running south to north. Ship A farther north is streaming west at 20 km/h and ship B is streaming north at 20km/h. What is their distance of closest approach and how long do they take to reach it ? Sol. Ships A and B are moving with same speed 20 km/h in N the directions shown in figure. It is a two dimensional, vA A two body problem with zero acceleration. Let us find E
vBA vB


vBA = vB − v A B
AB=10km Here, | vBA |= (20)2 + (20)2 = 20 2 km / h
i.e., vBA is 20 2 km / h at an angle of 45º from east towards north. Thus, the given problem can be simplified as :

45º


A is at rest and B is moving with vBA in the direction shown in figure. Therefore, the minimum distance between the two is

C

vBA

45º

smin = AC = AB sin 45º

 1  = 10   km  2

A

B = 5 2 km

Ans.

and the desired time is BC 5 2 t=
= | vBA | 20 2

(BC = AC = 5 2 km )

=

(B)

1 h = 15 min 4

Ans.

River - Boat Problems In river-boat problems we come across the following three terms : B B

→ v br

W

θ

θ

A → vr

and •

v br cos θ

y x

vbr sinθ A

vr → v r = absolute velocity of river → v br = velocity of boatman with respect to river or velocity of boatman is still water → = absolute velocity of boatman. vb → → Here, it is important to note that v br is the velocity of boatman with which he steers and v b is the actual velocity of boatman relative to ground. → → → Further, v b = v br + v r Now, let us derive some standard results and their special cases.

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Page # 41

KINEMATICS

→ A boatman starts from point A on one bank of a river with velocity v br in the direction shown in fig. → River is flowing along positive x-direction with velocity v r . Width of the river is w, then → → → v b = v br + v r Therefore, vbx = vrx + vbrx = vr – vbr sinθ and vby = vry + vbry = 0 + vbr cosθ = vbr cosθ Now, time taken by the boatman to cross the river is : w w = v by v br cos θ w or t = v cos θ ...(i) br Further, displacement along x-axis when he reaches on the other bank (also called drift) is : w x = vbx t = (vr – vbr sin θ) v cos θ br w or x = (vr – vbr sinθ) v cos θ ...(ii) br Three special are :

t=

(i)

Condition when the boatman crosses the river in shortest interval of time B From Eq.(i) we can see that time (t) will be minimum when θ = 0°, i.e., the boatman should steer his boat perpendicular to the river → vbr current. w Also, tmin = v as cos θ = 1 A → br vr

(ii)

Condition when the boatman wants to reach point B, i.e., at a point just opposite from where he started In this case, the drift (x) should be zero. B ∴ x=0 or or or

w (vr – vbr sinθ) v cos θ = 0 br

vr = vbr sin θ vr sinθ = v br

–1  v  or θ = sin  r   v br 

→ v br

θ A

→ vr

–1  v  Hence, to reach point B the boatman should row at an angle θ = sin  r  upstream from AB.  v br 

Further, since sinθ not greater than 1. So, if vr ≥ vbr, the boatman can never reach at point B. Because if vr = vbr, sin θ = 1 or θ = 90° and it is just impossible to reach at B if θ = 90°. Moreover it can be seen that vb = 0 if vr = vbr and θ = 90°. Similarly, if vr > vbr, sinθ > 1, i.e., no such angle exists. Practically it can be realized in this manner that it is not possible to reach at B if river velocity (vr) is too high.

(iii)

Shortest path Path length travelled by the boatman when he reaches the opposite shore is s=

w 2 + x2 Here, w = width of river is constant. So for s to be minimum modulus of x (drift) should be minimum. Now two cases are possible.

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Page # 42

KINEMATICS

When vr < vbr : In this case x = 0, –1  v r  when θ = sin  v   br 

or

–1  v r   at θ = sin   v br 

smin = w

When vr > vbr : In this case x is minimum, where or

d dθ

dx =0 dθ

  w (vr – vbr sin θ) = 0   vbr cos θ 

or or

–vbr cos2θ – (vr – vbr sinθ) (– sinθ) = 0 – vbr + vr sinθ = 0  vbr   or θ = sin–1   vr  Now, at this angle we can find xmin and then smin which comes out to be

 vr smin = w   vbr

  at 

–1  vbr   θ = sin   vr 

Ex.50 A man can row a boat with 4 km/h in still water. If he is crossing a river where the current is 2 km/h. (a) In what direction will his boat be headed, if he wants to reach a point on the other bank, directly opposite to starting point? (b) If width of the river is 4 km, how long will the man take to cross the river, with the condition in part (a)? (c) In what direction should he head the boat if he wants to cross the river in shortest time and what is this minimum time? (d) How long will it take him to row 2 km up the stream and then back to his starting point ? Sol. (a) Given, that vbr = 4 km/h and vr = 2 km/h  vr ∴ θ = sin–1  v  br

 2 1  = sin–1   = sin–1   = 30° 4   2 

Hence, to reach the point directly opposite to starting point he should head the boat at an angle of 30° with AB or 90° + 30° = 120° with the river flow. (b) Time taken by the boatman to cross the river w = width of river = 4 km vbr = 4 km/h and θ = 30°

∴

t=

2 4 h = 4 cos 30° 3

Ans.

(c) For shortest time θ = 0° w 4 tmin = v cos 0° = = 1h 4 br Hence, he should head his boat perpendicular to the river current for crossing the river in shortest time and this shortest time is 1 h.

and

vbr–vr D (d)

t = tCD + tDC

or

t=

vbr+vr C

CD DC + v db – v r v br + v r

D

=

C

2 2 1 4 + = 1+ = h 4–2 4+2 3 3

Ans.

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Page # 43

KINEMATICS

30 °

(a) We have


vm,g = v m,r + v r,g

vm

,g

Ex.51 A man can swim at a speed of 3 km/h in still water. He wants to cross a 500 m wide river flowing at 2 kh/h. He keeps himself always at an angle of 120° with the river flow while swimming. (a) Find the time he takes to cross the river. (b) At what point on the opposite bank will he arrive ? Sol. The situation is shown in figure
vr,g = velocity of the river with respect to the ground Here Y B C
vm,r = velocity of the man with respect to the river
vm,g = velocity of the man with respect to the ground.

vm,r = 3km/h

...(i)

θ

Hence, the velocity with respect to the ground is along AC. Taking y-components in equation (i),

A

vr,g = 2km/h


3 3 vm,g sinθ = 3 km/h cos 30° + 2 km/h cos 90° = km/h 2 Time taken to cross the river =

1/ 2km 1 displacement along the Y - axis = h = velocity along the Y - axis 3 3 / 2 km / h 3 3

(b) Taking x-components in equation (i),


1 vm,g cos θ = –3km/h sin 30° + 2 km/h = km / h 2 Displacement along the X-axis as the man crosses the river = (velocity along the X-axis) (time)  1  1  1km  h = km  ×  =  2h  3 3  6 3

Ex.52 A boat moves relative to water with a velocity v and river is flowing with 2v. At what angle the boat shall move with the stream to have minimum drift? (A) 30° (B) 60° (C) 90° (D) 120° Sol. (D) Let boat move at angle θ to the normal as shown in figure then time to cross the river =

1 v cos θ

1 drift x = (2v – v sin θ) for x to be minimum v cos θ dx = 0 = 1 (2 sec θ tan θ – sec2θ) or sin θ = 1/2 dθ or θ = 30° and φ = 90 + 30 = 120°

(C)

ub = u

u sinθ

ucosθ I = width of river

ur=2v

Aircraft Wind Problems → → This is similar to river boat problem. The only difference is that v br is replaced by v aw (velocity of → → aircraft with respect to wind or velocity of aircraft in still air), v r is replaced by v w (velocity of wind) → → → → → and v b is replaced by v a (absolute velocity of aircraft). Further,, v a = v aw + v . The following w example will illustrate the theory.

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Page # 44

KINEMATICS
A

NOTE : SHORT - TRICK



If their are two vectors A and B and their resultent

α make an anlge α with A and β with B . β A sinα then A sin α = β sin β
B sinβ Means component of A perpendicular to resultant is equal in
magnitude to the compopent of B also perpendicular to resultant.



Ex.53 If two vectors A and B make angle 30° and 60°
with their resultent and B has magnitude equal to
10, then find magnitude of A . So B sin 60° = A sin 30° ⇒ 10 sin 60° = A sin 30°

⇒




C = A +B


B


B 60°

Bsin60°

30°


A

A sin 30°

A = 10 3

Ex.54 An aircraft flies at 400 km/h in still air. A wind of 200 2 km/h is blowing from the south. The pilot wishes to travel from A to a point B north east of A. Find the direction he must steer and time of his journey if AB = 1000 km. Sol.

Given that vw = 200 2 km/h → → vaw = 400 km/h and v a should be along AB or in north-east direction. Thus, the direction of v aw → → should be such as the resultant of v w and v aw is along AB or in north - east direction. → N Let v aw makes an angle α with AB as shown in figure. B Applying sine law in triangle ABC, we get → AC BC v a 45° → = v w = 200 2km / h sin 45° sin α 45°  200 2  1 → C 1  BC  α v aw = 400 km / h   A =  sin 45° =  or sin α =   AC   400  2 2 E

∴ α = 30° Therefore, the pilot should steer in a direction at an angle of (45° + α) or 75° from north towards east. → 400 | v a| Further, = or sin 45° sin(180°–45°–30° )

sin 105° km → | v a | = sin 45° × (400) h

 cos 15°   0.9659  km km  (400)  (400) =  =   sin 45°  0 . 707 h h

= 546.47 km/h ∴ The time of journey from A to B is

(D)

AB 1000 h ⇒ t = 1.83 h t= → = 546.47 | v a| Rain Problems → → → In these type of problems we again come across three terms v r , vm and vrm , Here, → v r = velocity of rain

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Page # 45

KINEMATICS → vm = velocity of man (it may be velocity of cyclist or velocity of motorist also) and

→ vrm = velocity of rain with respect to man.

→ Here, v is the velocity of rain which appears to the man. Now, let us take one example of this. rm

Ex.55 A man standing on a road has to hold his umbrella at 30° with the vertical to keep the rain away. He throws the umbrella and starts running at 10 km/h. He finds that raindrops are hitting his head vertically. Find the speed of raindrops with respect to (a) the road, (b) the moving man. When the man is at rest with respect to the ground, the rain comes to him at an angle 30° with the Sol. vertical. This is the direction of the velocity of raindrops with respect to the ground. The situation when the man runs is shown in the figure

30°

vm,g

30°

v r,m

(b)

(a)

vr,g


Here vr,g = velocity of the rain with respect to the ground

vm,g = velocity of the man with respect to the ground and vr,m = velocity of the rain with respect to the man. We have,




vr,g = vr,m + vm,g

...(i)

Taking horizontal components, equation (i) gives vr,g sin30° = um,g = 10 km/h or, v,g =

10 km / h = 20km / h sin 30°

Taking vertical components, equation (i) gives vr,g cos30° = vr,m or, vr,m = (20 km/h)

3 = 10 √ 3 km/h. 2

Ex.56 To a man walking at the rate of 3 km/h the rain appears to fall vertically. When the increases his speed to 6 km/h it appears to meet him at an angle of 45° with vertical. Find the speed of rain. Sol.

Let i and j be the unit vectors in horizontal and vertical directions respectively.. Let velocity of rain Vertical ( j )

→ ˆ v r = aiˆ + bj Then speed of rain will be → | v r |=

a2 + b2

...(i)

Horizontal ( i )

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Page # 46

KINEMATICS

→ In the first case v m = velocity of man = 3 i → → → ∴ ˆ v rm = v r – v m = (a – 3)iˆ + bj It seems to be in vertical direction. Hence, a – 3 = 0 or a = 3 → In the second case v m = 6 i

∴

→  ˆ = – 3 i + b j v rm = (a – 6)iˆ + bj

This seems to be at 45° with vertical. Hence, |b| = 3 Therefore, from Eq. (ii) speed of rain is

→ | v r |= (3)2 + (3)2 = 3 2 km / h Ans.

Relative Motion between Two Projectiles Let us now discuss the relative motion between two projectiles or the path observed by one projectile of the other. Suppose that two particles are projected from the ground with speeds u1 and u2 at angles α1 and α2 as shown in Fig.A and B. Acceleration of both the particles is g downwards. So, relative acceleration between them is zero because a12 = a1 – a2 = g – g = 0

Y

Y

u1

u2

α1

X

(A)

α2

X

(B)

i.e., the relative motion between the two particles is uniform. Now u2x = u2 cos α2 u1x = u1 cos α1, u1y = u1 sin α1 and u2y = u2 sin α2 Therefore, u12x = u1x – u2x = u1 cos α1– u2cos α2 and u12y = u1y – u2y = u1 sin α1– u2 sin α2 u12x and u12y are the x and y components of relative velocity of 1 with respect to 2. Hence, relative motion of 1 with respect to 2 is a straight

u  line at an angle θ = tan −1 12 y  with positive x-axis.  u12 x 

y

u12y u12 θ

u12x

a12=0

x

Now, if u12x = 0 or u1 cos α1 = u2 cos α2, the relative motion is along y-axis or in vertical direction (as θ = 90º). Similarly, if u12y = 0 or u1 sin α1 = u2 sin α2, the relative motion is along x-axis or in horizontal direction (as θ = 0º).

Note : Relative acceleration between two projectiles is zero. Relative motion between them is uniform. Therefore, condition of collision of two particles in air is that relative velocity of one with respect to
the other should be along line joining them, i.e., if two projecticles A and B collide in mid air, then VAB
should be along AB or VBA along BA.

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Page # 47

KINEMATICS

Condition for collision of two projectiles : Consider the situation shown in the figure. For projectiles to collide, direction of velocity of A with respect to B has to be along line AB. Here, vABx = u1 cos α1 + u2 cos α2

u2

vABy = u1 sin α1 – u2 sin α2

Y

Let, direction of velocity vector of A(wrt B) is making an angle β with +ve X-axis, which is given by tan β =

v ABy v ABx

=

h1

u1 sin α1 − u2 sin α2 u1 cos α1 + u2 cos α2

B

u1

h2

A

X

x

For collision to take place, h2 − h1 x Ex.57 A particle A is projected with an initial velocity of 60 m/s. at an angle 30º to the horizontal. At the same time a second particle B is projected in opposite direction with initial speed of 50 m/s from a point at a distance of 100 m from A. If the particles collide in air, find (a) the angle of projection α of particle B, (b) time when the collision takes place and (c) the distance of P from A, where collision occurs. (g = 10 m/s2) 60m/s 50m/s

tan β = tan θ =

A

30º 100m

Sol. (a) Taking x and y directions as shown in figure.

Here, a = −gˆj , a = −gˆj A

B

Y

B

uAx = 60 cos 30º

= 30 3 m / s

X uAy = 60 sin 30º = 30 m/s u AB uBx = – 50 cos α and uBy = 50 sin α

Relative acceleration between the two is zero as a A = aB . Hence, the relative motion between the two
is uniform. It can be assumed that B is at rest and A is moving with u AB . Hence, the two particles will
collide, if u AB is along AB. This is possible only when

uAy = uBy i.e., component of relative velocity along y-axis should be zero. or 30 = 50 sin α ∴ = sin–1 (3/5) Ans. a

(b) Now,


| u AB |= u Ax – uBx = (30 3 + 50 cosα)m/s =

4   30 3 + 50 ×  m / s = (30 3 + 40) m/s  5

Therefore, time of collision is

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

or

t = 1.09 s Ans.

(c)

Distance of point P from A where collision takes place is

s=

1   (u Ax t) 2 +  u Ay t – gt 2    2

2

=

1   ( 30 3 × 109 . ) 2 +  30 × 109 . – × 10 × 109 . × 109 .    2

2

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or s = 62.64 m Ans.

Page # 48

KINEMATICS y

Ex.58 Two projectile are projected simultaneously from a point on the A ground "O" and an elevated position "A" respectively as shonw in the figure. If collision occurs at the point of return of two projectiles on H the horizontal surface, then find the height of "A" above the ground and the angle at which the projectile "O" at the ground should be projected.

5m/s

10m/s

θ

O

C

x

Sol. There is no initial separation between two projectile is x-direction. For collision to occur, the relative motion in x-direction should be zero. In other words, the component velocities in x-direction should be equal to that two projetiles cover equal horizontal distance at any given time. Hence, uOx = uAx ⇒

u0cosθ = uA ⇒

cosθ =

uA 5 1 = = = cos60° uO 10 2

⇒

θ = 60°

We should ensure that collision does occur at the point of return. It means that by the time projectiles travel horizontal distances required, they should also cover vertical distances so that both projectile are at "C" at the same time. In the nutshell, their times of flight should be equal. For projectile from "O". 2uO sin θ T= g For projectile from "A",

 2H     g 

T= For projectile from "A",

2uo sin θ  2H  =   g  g  Squaring both sides and putting values, T=

⇒ H=

4u2O sin2 θ 4 × 102 sin2 60° ⇒ H= 2g 2 × 10 2

 3 = 15m H = 20   2    We have deliberately worked out this problem taking advantage of the fact that projectiles are colliding at the end of their flights and hence their times of flight should be equal. We can, however, proceed to analyze in typical manner, using concept of relative velocity. The initial separation between two projectiles in the vertical direction is "H". This separation is covered with the component of relative in vertical direction.

⇒ vOAy = uOy – uAy = u0 sin60° – 0 = 10 ×

3 = 5 3m/s 2

Now, time of flight of projectile from ground is : 2uO sin θ 2x10x sin 60° = = 3 T= g 10 Hence, the vertical displacement of projectile from "A" before collision is :

⇒

H = vOAy X T = 5 3 x 3 = 15 m/s

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Page # 49

KINEMATICS

Ex.59 Two projectiles are projected simultaneously from two towers as shwon in figure. If the projectiles collide in the air, then find the distance "s" between the towers. 10 m/s

B

10 2 m / s A

30m

45°

10m

Sol. We see here that projectiles are approaching both horizontally and vertically. Their movement in two component directions should be synchronized so that they are at the same position at a particular given time. For collision, the necessary requirement is that relative velocity and displacement should be in the same direction. It is given that collision does occur. It means that two projectiles should cover the displacement with relative velocity in each of the component directions. 10 m/s Y B In x-direction, vABx = uAx – uBx = 10 2 cos 45° – (–10) = 10 2

1 2

If "t" is time after which collision occurs, then ⇒ s = vAy – uBy

⇒

vABy = ucos45° – 0 = 10 2 ×

1 2

10 2 m / s

+ 10 = 20 m/s

= 10m / s

A

30m

45°

10m O

S

x

The initial vertical distance between points of projection is 30 – 10 = 20 m. This vertical distance is covered with component of relative velocity in vertical direction. Hence, time taken to collide, "t", is :

⇒t=

20 =2 10

Putting this value in the earlier equation for "s", we have : ⇒ s = 20t = 20x2 = 40 m

Now you can try all the questions related to relative motion.

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Page # 50

KINEMATICS

QUESTIONS FOR SHORT ANSWER
1. A vector a is turned without a change in its length
through a small angle dθ. What are | a| and ∆a?

Sol.

2. Does the speedometer of a car measure speed or velocity ? Explain Sol.

3. When a particle moves with constant velocity, its average velocity and its instantaneous velocity & speed are equal. Comment on this statement. Sol.

4. In a given time interval, is the total displacement of a particle equal to the product of the average velocity and the time interval, even when the velocity is not constant? Explain. Sol.

5. Can you have zero displacement and a non zero average velocity? Can you have a zero displacement and a non zero velocity? Illustrate your answer on a x-t graph. Sol.

6. At which point on its path a projectile has the smallest speed? Sol.

7. A person standing on the edge of a cliff at some height above the ground below throws one ball straight up with initial speed u and then throws another ball straight down with the same initial speed. Which ball, if either, has the larger speed when it hits the ground? Neglect air resistance. Sol.

8. An airplane on floor relief mission has to drop a sack of rice exactly in the center of a circle on the ground while flying at a predetermined height and speed. What is so difficult about that? Why doesn’t it just drop the sack when it is directly above the circle. Sol.

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Page # 51

KINEMATICS

9. Which of the following graphs cannot possibily represent one dimensional motion of a particle? x
l |v| t

12. Give an example from your own experience in which the velocity of an object is zero for just an instant of time, but its acceleration is not zero. Sol.

t

t

l - length of path

x - displacement Sol.

10. Can you suggest a suitable situation from observation around you for each of the following ? x x

t

t

x - displacement x

13. A ball is dropped from rest from the top of a building and strikes the ground with a speed vf. From ground level, a second ball is thrown straight upward at the same instant that the first ball is dropped. The initial speed of the second ball is v0 = vf, the same speed with which the first ball will eventually strike the ground. Ignoring air resistance, decide whether the balls cross paths at half the height of the building above the halfway point, or below the halfway point. Give your reasoning. Sol.

t

Sol.

11. One of the following statements is incorrect. (a) The car traveled around the track at a constant velocity (b) The car traveled around the track at a constant speed. Which statement is incorrect and why ? Sol.

14. The muzzle velocity of a gun is the velocity of the bullet when it leaves the barrel. The muzzle velocity of one rifle with a short barrel is greater than the muzzle velocity of another rifle that has a longer barrel. In which rifle is the acceleration of the bullet larger? Explain your reasoning. Sol.

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Page # 52

KINEMATICS

15. On a riverboat cruise, a plastic bottle is accidentally dropped overboard. A passenger on the boat estimate that the boat pulls ahead of the bottle by 5 meters each second. Is it possible to conclude that the boat is moving at 5 m/s with respect to the shore? Account for your answer. Sol.

18. A child is playing on the floor of a recreational vehicle (RV) as it moves along the highway at a constant velocity. He has a toy cannon, which shoots a marble at a fixed angle and speed with respect to the floor. The connon can be aimed toward the front or the rear of the RV. Is the range towards the front the same as, less than, or greater than the range towards the rear? Answer this question (a) from the child’s point of view and (b) from the point of view of an observer standing still on the ground. Justify your answers. Sol.

16. A wrench is accidentally dropped from the top of the mast on a sailboat. Will the wrench hit at the same place on the deck whether the sailboat is at rest or moving with a constant velocity? Justify your answer. Sol. 19. Three swimmers can swim equally fast relative to the water. They have a race to see who can swim across a river in the least time. Swimmer A swims perpendicular to the current and lands on the far shore downstream, because the current has swept him in that direction. Swimmer B swims upstream at an angle to the current and lands on the far shore directly opposite the starting point. Swimmer C swims downstream at an angle to the current in an attempt to take advantage of the current. Who crosses the river in the least time? Account for your answer. Sol. 17. Is the acceleration of a projectile equal to zero when it reaches the top of its trajectory? If not, why not? Sol.

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Page # 53

KINEMATICS

(Objective Problems)

Exercise - I 1. A hall has the dimensions 10m × 10m × 10 m. A fly starting at one corner ends up at a diagonally opposite corner. The magnitude of its displacement is nearly (A) 5 3 m

(B) 10 3 m

(C) 20 3 m

(D) 30 3 m

Sol.

3. A body covers first 1/3 part of its journey with a velocity of 2 m/s, next 1/3 part with a velocity of 3 m/s and rest of the journey with a velocity 6m/s. The average velocity of the body will be (A) 3 m/s

(B)

11 m/s 3

(C)

8 m/s 3

(D)

4 m/s 3

Sol.

2. A car travels from A to B at a speed of 20 km h–1m and returns at a speed of 30 km h–1. The average speed of the car for the whole journey is (A) 5 km h–1 (B) 24 km h–1 (C) 25 km h–1(D) 50 km h–1 Sol.

4. A car runs at constant speed on a circular track of radius 100 m taking 62.8 s on each lap. What is the average speed and average velocity on each complete lap ? (A) velocity 10 m/s speed 10 m/s (B) velocity zero, speed 10 m/s (C) velocity zero, speed zero (D) velocity 10 m/s, speed zero Sol.

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Page # 54

KINEMATICS

5. The displacement of a body is given by 2s = gt2 where g is a constant. The velocity of the body at any time t is (D) gt3/3 (A) gt (B) gt/2 (C) gt2/2 Sol.

Sol.

6. A particle has a velocity u towards east at t = 0. Its acceleration is towards west and is constant, Let xA and xB be the magnitude of displacements in the first 10 seconds and the next 10 seconds. (A) xA < xB (B) xA = xB (C) xA > xB (D) the information is insufficient to decide the relation of xA with xB. Sol.

8. A stone is dropped into a well in which the level of water is h below the top of the well. If v is velocity of sound, the time T after which the splash is heard is given by (A) T = 2h/v (C) T =

2h h + g 2v

(B) T =

2h h + g v

(D) T =

h 2h + 2g v

Sol.

7. A body starts from rest and is uniformly accelerated for 30 s. The distance travelled in the first 10s is x1, next 10 s is x2 and the last 10 s is x3. Then x1 : x2 : x3 is the same as (A) 1 : 2 : 4 (B) 1 : 2 : 5 (C) 1 : 3 : 5 (D) 1 : 3 : 9

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Page # 55

KINEMATICS

9. The co-ordinates of a moving particle at a time t, are given by, x = 5 sin 10 t, y = 5 cos 10 t. The speed of the particle is (A) 25 (B) 50 (C) 10 (D) None Sol.

11. A body of mass 1 kg is acted upon by a force
F = 2 sin 3 πt i + 3 cos 3 πt j find its position at t = 1 sec if at t = 0 it is at rest at origin. 3   3 (A)  2 ,   3 π 9π 2 

2   2 (B)  2 ,   3π 3π 2 

2   2 (C)  ,   3π 3π 2 

(D) none of these

Sol.

10. A body moves with velocity v = lnx m/s where x is its position. The net force acting on body is zero at . (C) x = em (D) x = 1 m (A) 0 m (B) x = e2m Sol.

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Page # 56

KINEMATICS

12. A force F = Be–Ct acts on a particle whose mass is m and whose velocity is 0 at t = 0. It’s terminal velocity (velocity after a long time) is : (A)

C mB

(B)

B mC

(C)

BC m

(D) –

Sol.

B mC

Sol.

13. A particle starts moving rectilinearly at time t = 0 such that its velocity ‘v’ changes with time ‘t’ according to the equation v = t2 – t where t is in seconds and v is in m/s. The time interval for which the particle retards is (A) t < 1/2 (B) 1/2 < t < 1 (C) t > 1 (D) t < 1/2 and t > 1 Sol. 15. A particle is projected vertically upwards from a point A on the ground. It takes t1 time to reach a point B but it still continues to move up. If it takes further t2 time to reach the ground from point B then height of point B from the ground is (A)

1 g( t1 + t 2 ) 2 2

(B) g t1 t2

(C)

1 g( t1 + t 2 ) 2 8

(D)

1 gt 1t 2 2

Sol.

14. A ball is thrown vertically down with velocity of 5m/s. With what velocity should another ball be thrown down after 2 seconds so that it can hit the 1st ball in 2 seconds (A) 40 m/s (B) 55 m/s (C) 15 m/s (D) 25 m/s

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Page # 57

KINEMATICS 16. Balls are thrown vertically upward in such a way that the next ball is thrown when the previous one is at the maximum height. If the maximum height is 5m, the number of balls thrown per minute will be (A) 40 (B) 50 (C) 60 (D) 120 Sol.

18. The displacement-time graph of a moving particle is shown below. The instantaneous velocity of the particle is negative at the point x D E

F

C t

(A) C Sol.

17. A disc arranged in a vertical plane has two groves of same length directed along the vertical chord AB and CD as shown in the fig. The same particles slide down along AB and CD. The ratio of the time tAB/tCD is A

Sol.

(D) F

v(m/s)

C

60º

20 10

B

(B) 1: 2

(C) E

19. The variation of velocity of a particle moving along straight line is shown in the figure. The distance travelled by the particle in 4 s is

D

(A) 1 : 2

(B) D

(C) 2 : 1

(D)

2 :1

1 2 3 4

(A) 25m Sol.

(B) 30m

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t(s)

(C) 55m

(D) 60m

Page # 58

KINEMATICS

20. The displacement time graphs of two particles A and B are straight lines making angles of respectively 30º and 60º with the time axis. If the velocity of A is vA vA and that of B is vB then the value of v is B (A) 1/2

(B) 1 / 3

3

(C)

22. Acceleration versus velocity graph of a particle moving in a straight line starting form rest is as shown in figure. The corresponding velocity-time graph would be a

(D) 1/3

Sol. v v

v

(A)

(B) t

t

v

21. If position time graph of a particle is sine curve as shown, what will be its velocity-time graph

v

(C)

(D) t

t

Sol.

x t

v

v (A)

(B)

t

t v

v (C)

(D)

t

t

Sol.

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Page # 59

KINEMATICS 23. A man moves in x - y plane along the path shown. At what point is his average velocity vector in the same direction as his instantaneous velocity vector. The man starts from point P. y

PB

C

Question No. 25 to 27 (3 questions) The x-t graph of a particle moving along a straight line is shown in figure x

D

parabola

A x

(A) A Sol.

(B) B

(C) C

0

(D) D

T

2T

25. The v-t graph of the particle is correctly shown by

v (A)

v

0

T

2T t

v

24. The acceleration of a particle which moves along the positive x-axis varies with its position as shown. If the velocity of the particle is 0.8 m/s at x = 0, the

(C)

0

T

0

(B)

2T t

v T

2T

t

(D)

0

T

2T

t

Sol.

velocity of the particle at x = 1.4 is (in m/s) 2

a (in m/s ) 0.4 0.2

O

(A) 1.6 Sol.

(B) 1.2

0.4 0.8

1.4 x (in m)

(C) 1.4

(D) none

26. The a-t graph of the particle is correctly shown by

a (A)

0

a 2T T

t

(B)

0

t

v

a (C)

0

t

(D)

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0

t

Page # 60

KINEMATICS

28. Choose the incorrect statement. The particle comes to rest at (A) t = 0 s (B) t = 5 s (C) t = 8 s (D) none of these Sol.

Sol.

27. The speed-time graph of the particle is correctly shown by speed

(A)

0

speed 2T t

T

speed

(C)

0

0

(B)

T

2T t

29. Identify the region in which the rate of change of
∆v of the particle is maximum velocity ∆t (A) 0 to 2s Sol.

(B) 2 to 4s

(C) 4 to 6s (D) 6 to 8 s

speed 0

2T t (D)

T

T

2T t

Sol.

30. If the particle starts from the position x0 = –15 m, then its position at t = 2s will be (A) – 5m (B) 5m (C) 10 m (D) 15 m Sol.

Question No. 28 to 33 (6 questions) The figure shows a velocity-time graph of a particle moving along a straight line v(ms–1) 10 0

2

4

6

8 t(s)

–20

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Page # 61

KINEMATICS

31. The maximum of displacement of the particle is (A) 33.3 m (B) 23.3 m (C) 18.3 (D) zero Sol.

33. The correct displacement-time graph of the particle is shown as x (m)

x (m)

(A)

(B) 0 2 x (m)

4

6

0 2 x (m)

8 t(s)

(C)

4

6

8 t(s)

(D) 0 2 4

6

0 2 4 6 8 t(s)

8 t(s)

Sol.

34. The velocity-time graph of a body falling from rest under gravity and rebounding from a solid surface is represented by which of the following graphs ? V

(A)

V

t

(B)

V

(C) 32. The total distance travelled by the particle is (A) 66.6 m (B) 51.6 m (C) zero (D) 36.6 m Sol.

t V

t

(D)

Sol.

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t

Page # 62

KINEMATICS

35. Shown in the figure are the displacement time graph for two children going home from the school. Which of the following statements about their relative motion is true after both of them started moving ? Their relative velocity: X

37. A body A is thrown vertically upwards with such a velocity that it reaches a maximum height of h. Simultaneously another body B is dropped from height h. It strikes the ground and does not rebound. The velocity of A relative to B v/s time graph is best represented by : (upward direction is positive)

C1

VAB

VAB

(A)

C2

(B) t

O

t

VAB

VAB

T

(A) first increases and then decreases (B) first decreases and then increases (C) is zero (D) is non zero constant Sol.

36. Shown in the figure are the velocity time graphs of the two particles P1 and P2. Which of the following statements about their relative motion is true ? Theire relative velocity (A) is zero (B) is non-zero but constant (C) continuously decreases (D) continuously increases Sol.

t

v

(C)

t

(D)

t Sol.

P1 P2

O

t

T

38. An object A is moving with 10 m/s and B is moving with 5 m/s in the same direction of positive x-axis. A is 100 m behind B as shown. Find time taken by A to Meet B

10m/s

A

(A) 18 sec. Sol.

5m/s

B

100m (B) 16 sec. (C) 20 sec. (D) 17 sec.

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Page # 63

KINEMATICS 39. It takes one minute for a passenger standing on an escalator to reach the top. If the escalator does not move it takes him 3 minute to walk up. How long will it take for the passenger to arrive at the top if he walks up the moving escalator? (A) 30 sec (B) 45 sec (C) 40 sec (D) 35 sec Sol.

40. A body is thrown up in a lift with a velocity u relative to the lift and the time of flight is found to be t. The acceleration with which the lift is moving up is u – gt (A) t

2u – gt (B) t

u + gt (C) t

2u + gt (D) t

42. A point mass is projected, making an acute angle
with the horizontal. If angle between velocity v and
acceleration g is θ, then θ is given by (A) 0º < θ < 90º (C) θ = 90º Sol.

(B) θ = 90º (D) 0º < θ < 180º

43. The velocity at the maximum height of a projectile is half of its initial velocity u. Its range on the horizontal plane is 2u 2 (A) 3g

(B)

3 u2 2g

u2 (C) 3g

u2 (D) 2g

Sol.

Sol.

Question No. 44 to 46 A projectile is thrown with a velocity of 50 ms–1 at an angle of 53º with the horizontal 41. A ball is thrown upwards. It returns to ground describing a parabolic path. Which of the following remains constant ? (A) speed of the ball (B) kinetic energy of the ball (C) vertical component of velocity (D) horizontal component of velocity. Sol.

44. Choose the incorrect statement (A) It travels vertically with a velocity of 40 ms–1 (B) It travels horizontally with a velocity of 30 ms–1 (C) The minimum velocity of the projectile is 30 ms–1 (D) None of these Sol.

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Page # 64

KINEMATICS

45. Determine the instants at which the projectile is at the same height (A) t = 1s and t = 7s (B) t = 3s and t = 5s (C) t = 2s and t = 6s (D) all the above Sol.

48. A ball is thrown from a point on ground at some angle of projection. At the same time a bird starts from a point directly above this point of projection at a height h horizontally with speed u. Given that in its flight ball just touches the bird at one point. Find the distance on ground where ball strikes (A) 2u

h g

(B) u

2h g

(C) 2u

2h g

(D) u

h g

Sol.

46. The equation of the trajectory is given by (B) 180 y = x2 – 240x (A) 180y = 240 x – x2 2 (C) 180y = 135x – x (D) 180y = x2 – 135x Sol.

49.A ball is hit by a batsman at an angle of 37º as shown in figure. The man standing at P should run at what minimum velocity so that he catches the ball before it strikes the ground. Assume that height of man is negligible in comparison to maximum height of projectile.

47. A particle is projected from a horizontal plane (xz plane) such that its velocity vector at time t is
given by V = aˆi + (b – ct )ˆj . Its range on the horizontal plane is given by ba 2ba (A) (B) c c Sol.

3 ba (C) c

(D) None

(A) 3 ms–1 Sol.

(B) 5 ms–1

(C) 9 ms–1

(D) 12 ms–1

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Page # 65

KINEMATICS

Question No. 53 & 54 (2 questions) At t = 0 a projectile is fired from a point O (taken as origin) on the ground with a speed of 50 m/s at an angle of 53° with the horizontal. It just passes two points A & B each at height 75 m above horizontal as shown.

50m/s

50. A projectile is fired with a speed u at an angle θ with the horizontal. Its speed when its direction of motion makes an angle ‘α’ with the horizontal is (A) u secθ cosα (B) u secθ sinα (C) u cosθ secα (D) u sinθ secα Sol.

53°

51. Two projectiles A and B are thrown with the same speed such that A makes angle θ with the horizontal and B makes angle θ with the vertical, then (A) Both must have same time of flight (B) Both must achieve same maximum height (C) A must have more horizontal range than B (D) Both may have same time of flight Sol.

52. Suppose a player hits several baseballs. Which baseball will be in the air for the longest time? (A) The one with the farthest range. (B) The one which reaches maximum height (C) The one with the greatest initial velocity (D) The one leaving the bat at 45° with respect to the ground. Sol.

B

A 75m

O 53. The horizontal separation between the points A and B is (A) 30 m (B) 60 m (C) 90 m (D) None Sol.

54 The distance (in metres) of the particle from origin at t = 2 sec. (A) 60 2 Sol.

(B) 100

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(C) 60

(D) 120

Page # 66

KINEMATICS

55. Particle is dropped from the height of 20 m from horizontal ground. There is wind blowing due to which horizontal acceleration of the particle becomes 6 ms–2. Find the horizontal displacement of the particle till it reaches ground. (A) 6 m (B) 10 m (C) 12 m (D) 24 m Sol.

56. A ball is projected from top of a tower with a velocity of 5 m/s at an angle of 53º to horizontal. Its speed when it is at a height of 0.45 m from the point of projection is (A) 2 m/s (B) 3 m/s (C) 4 m/s (D) data insufficient Sol.

58. One stone is projected horizontally from a 20 m high cliff with an initial speed of 10 ms–1. A second stone is simultaneously dropped from that cliff. Which of the following is true ? (A) Both strike the ground with the same velocity (B) The ball with initial speed 10ms–1 reaches the ground first (C) Both the balls hit the ground at the same time (D) One cannot say without knowing the height of the building Sol.

59. An aeroplane flying at a constant velocity releases a bomb. As the bomb drops down from the aeroplane. (A) it will always be vertically below the aeroplane (B) it will always be vertically below the aeroplane only if the aeroplane is flying horizontally (C) it will always be vertically below the aeroplane only if the aeroplane is flying at an angle of 45° to the horizontal. (D) it will gradually fall behind the aeroplane if the aeroplane is flying horizontally Sol.

57. Find time of flight of projectile thrown horizontally with speed 10 ms–1 from a long inclined plane which makes an angle of θ = 45º from horizontal. (A)

2 sec

(B) 2 2 sec

(C) 2 sec

(D) none

Sol.

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Page # 67

KINEMATICS 60. A particle is projected at angle 37º with the incline plane in upward direction with speed 10 m/s. The angle of incline plane is given 53º. Then the maximum height above the incline plane attained by the particle will be (A) 3m (B) 4m (C) 5m (D) zero Sol.

Sol.

63. If time taken by the projectile to reach Q is T, than PQ = v 90° P

θ

61. On an inclined plane of inclination 30°, a ball is thrown at an angle of 60° with the horizontal from the foot of the incline with velocity of 10 3 ms–1. If g = 10 ms–2, then the time in which ball with hit the inclined plane is (A) 1.15 sec. (B) 6 sec (C) 2 sec (D) 0.92 sec Sol.

62. A projectile is fired with a velocity at right angle to the slope which is inclined at an angle θ with the horizontal. The expression for the range R along the incline is (A)

2v 2 sec θ g

(B)

2v 2 tan θ g

(C)

2v 2 tan θ sec θ g

(D)

v2 tan2 θ g

Q

(A) Tvsinθ Sol.

(B) Tvcosθ

(C) Tv secθ (D) Tv tanθ

Question No. 64 to 67 (4 questions) Two projectiles are thrown simultaneously in the same plane from the same point. If their velocities are v1 and v2 at angles θ1 and θ2 respectively from the horizontal, then answer the following questions. 64. The trajectory of particle 1 with respect to particle 2 will be (A) a parabola (B) a straight line (C) a vertical straight line (D) a horizontal straight line

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Page # 68

KINEMATICS Sol.

Sol.

65. If v1 cosθ1 = v2 cosθ2, then choose the incorrect statement (A) one particle will remain exactly below or above the other particle (B) the trajectory of one with respect to other will be a vertical straight line (C) both will have the same range (D) none of these Sol. 68. A helicopter is flying south with a speed of 50 kmh–1. A train is moving with the same speed towards east. The relative velocity of the helicopter as seen by the passengers in the train will be towards. (A) north east (B) south east (C) north west (D) south west Sol.

66. If v1sinθ1 = v2sinθ2, then choose the incorrect statement (A) the time of flight of both the particles will be same (B) the maximum height attained by the particles will be same (C) the trajectory of one with respect to another will be a horizontal straight line (D) none of these Sol.

69. Two particles are moving with velocities v1 and v2. Their relative velocity is the maximum, when the angle between their velocities is (A) zero (B) π/4 (C) π/2 (D) π Sol.

67. If v1 = v2 and θ1 > θ2, then choose the incorrect statement (A) Particle 2 moves under the particle 1 (B) The slope of the trajectory of particle 2 with respect to 1 is always positive (C) Both the particle will have the same range if θ1 > 45° and θ2 < 45° and θ1 + θ2 = 90° (D) none of these

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Page # 69

KINEMATICS 70. A ship is travelling due east at 10 km/h. A ship heading 30º east of north is always due north from the first ship. The speed of the second ship in km/h is (A) 20 2 (B) 20 3 / 2 (C) 20 (D) 20 / 2 Sol.

73. A swimmer’s speed in the direction of flow of river is 16 km h–1. Against the direction of flow of river, the swimmer’s speed is 8 km h–1. Calculate the swimmer’s speed in still water and the velocity of flow of the river. (A) 12 km/h, 4 km/h (B) 10 km/h, 3 km/h (C) 10 km/h, 4 km/h (D) 12 km/h, 2 km/h Sol.

71. A particle is kept at rest at origin. Another particle starts from (5, 0) with a velocity of – 4 ˆi + 3 ˆj . Find their closest distance of approach. (A) 3 m (B) 4 m (C) 5 m (D) 2 m Sol.

74. A pipe which can rotate in a vertical plane is mounted on a cart. The cart moves uniformly along a horizontal path with a speed v1 = 2 m/s. At what angle α to the horizontal should the pipe be placed so that drops of rain falling with a velocity v2 6 m/s move parallel to the walls of the pipe without touching them ? consider the velocity of the drops as constant due to the resistance of the air.

v1

72. Four particles situated at the corners of a square of side ‘a’ move at a constant speed v. Each particle maintains a direction towards the next particle in succession. Calculate the time particles will take to meet each other. (A) Sol.

a v

(B)

a 2v

(C)

a 3v

(D)

2a 3v

(A) tan –1( 3)

–1  1  (B) tan   3

–1  1  (C) tan   2

(D) None of these

Sol.

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Page # 70

KINEMATICS 77. A flag is mounted on a car moving due North with velocity of 20 km/hr. Strong winds are blowing due East with velocity of 20 km/hr. The flag will point it direction (A) East (B) North-East (C) South-East (D) South-West Sol.

75. A swimmer swims in still water at a speed = 5 km/ hr. He enters a 200 m wide river, having river flow speed = 4 km/hr at point A and proceeds to swim at an angle of 127° (sin37° = 0.6) with the river flow direction. Another point B is located directly across A on the other side. The swimmer lands on the other bank at a point C, from which he walks the distance CB with a speed = 3 km/hr. The total time in which he reachrs from A to B is (A) 5 minutes (B) 4 minutes (C) 3 minutes (D) None Sol.

76. A boat having a speed of 5 km/hr. in still water, crosses a river of width 1 km along the shortest possible path in 15 minutes. The speed of the river in Km/hr. (A) 1

(B) 3

(C) 4

(D)

41

78. A man is crossing a river flowing with velocity of 5 m/s. He reaches a point directly across at a distance of 60 m in 5 sec. His velocity in still water should be (A) 12 m/s (B) 13 m/s (C) 5 m/s (D) 10 m/s Sol.

79. Wind is blowing in the north direction at speed of 2 m/s which causes the rain to fall at some angle with the vertical. With what velocity should a cyclist drive so that the rain appears vertical to him (A) 2 m/s south (B) 2 m/s north (C) 4 m/s west (D) 4 m/s south

Sol.

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Page # 71

KINEMATICS

Exercise - II

(One or more than one option is correct)

1. The displacement x of a particle depend on time t as x = αt2 – β t3 (A) particle will return to its starting point after time α/β. 2α (B) the particle will come to rest after time 3β (C) the initial velocity of the particle was zero but its initial acceleration was not zero. α (D) no net force act on the particle at time 3β Sol.

Sol.

3. Mark the correct statements for a particle going on a straight line (A) if the veloci ty is zero at any instant, the acceleration should also be zero at that instant (B) if the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval (C) if the velocity and acceleration have opposite sign, the object is slowing down (D) if the position and velocity have opposite sign, the particle is moving towards the origin. Sol.

2. A particle has intial velocity 10 m/s. It moves due to constant retarding force along the line of velocity which produces a retardation of 5 m/s2. Then (A) the maximum displacement in the direction of initial velocity is 10 m (B) the distance travelled in first 3 seconds is 7.5 m (C) the distance travelled in first 3 seconds is 12.5 m (D) the distance travelled in first 3 seconds is 17.5 m

4. A particle initially at rest is subjected to two forces. One is constant, the other is a retarding force proportion at to the particle velocity. In the subsequent motion of the particle. (A) the acceleration will increase from zero to a constant value (B) the acceleration will decrease from its initial value to zero (C) the velocity will increase from zero to maximum & then decrease (D) the velocity will increase from zero to a constant value.

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Page # 72

KINEMATICS

Sol.

Sol.



5. Let v and a denote the velocity and acceleration respectively of a body in one-dimensional motion

(A) | v| must decrease when a < 0
(B) Speed must increase when a > 0

(C) Speed will increase when both v and a are < 0

(D) Speed will decrease when v < 0 and a > 0 Sol.

6. Which of the following statements are true for a moving body? (A) If its speed changes, its velocity must change and it must have some acceleration (B) If its velocity changes, its speed must change and it must have some acceleration (C) If its velocity changes, its speed may or may not change, and it must have some acceleration (D) If its speed changes but direction of motion does not changes, its velocity may remain constant

7. Let v and a denote the velocity and acceleration respectively of a body (A) a can be non zero when v = 0 (B) a must be zero when v = 0 (C) a may be zero when v ≠ 0 (D) The direction of a must have some correlation with the direction of v Sol.

8. A bead is free to slide down a A sm ooth wi re ti g ht l y st ret ched θ between points A and B on a vertical R B circle. If the bead starts from rest at A, the highest point on the circle (A) its velocity v on arriving at B is proportional to cosθ (B) its velocity v on arriving B is proportional to tanθ (C) time to arrive at B is proportional to cosθ (D) time to arrive at B is independent of θ

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Page # 73

KINEMATICS

10. The figure shows the velocity (v) of a particle plotted against time (t)

Sol.

+v0

O –v0

v T

t

2T

(A) The particle changes its direction of motion at some point (B) The acceleration of the particle remains constant (C) The displacement of the particle is zero (D) The initial and final speeds of the particle are the same Sol.

9. Velocity-time graph for a car is semicircle as shown here. Which of the following is correct : v 1m/s 2 sec (A) Car must move in circular path (B) Acceleration of car is never zero (C) Mean speed of the particle is π/4 m/s. (D) The car makes a turn once during its motion Sol.

11. A block is thrown with a velocity of 2 ms–1 (relative to ground) on a belt, which is moving with velocity 4 ms–1 in opposite direction of the initial velocity of block. If the block stops slipping on the belt after 4 sec of the throwing then choose the correct statements(s) (A) Displacement with respect to ground is zero after 2.66 sec and magnitude of displacement with respect to ground is 12 m after 4 sec. (B) Magnitude of displacement with respect to ground in 4 sec is 4 m. (C) Magnitude of displacement with respect to belt in 4 sec is 12 m. (D) Displacement with respect to ground is zero in 8/ 3 sec. Sol.

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Page # 74

KINEMATICS 13. An observer moves with a constant speed along the line joining two stationary objects. He will observe that the two objects (A) have the same speed (B) have the same velocity (C) move in the same direction (D) move in opposite directions Sol.

12. A particle moves with constant speed v along a regular hexagon ABCDEF in the same order. Then the magnitude of the average velocity for its motion from A to (A) F is v/5 (B) D is v/3 (D) B is v (C) C is v √3/2 Sol.

14. A man on a rectilinearly moving cart, facing the direction of motion, throws a ball straight up with respect to himself (A) The ball will always return to him (B) The ball will never return to him (C) The ball will return to him if the cart moves with constant velocity (D) The ball will fall behind him if the cart moves with some acceleration Sol.

15. A projectile of mass 1 kg is projected with a velocity of

20 m/s such that it strikes on the same level as

the point of projection at a distance of 3 m. Which of the following options are incorrect. (A) the maximum height reached by the projectile can be 0.25 m (B) the minimum velocity during its motion can be 15 m/s

3 sec. 5 (D) maximum potential energy during its motion can be 6J.

(C) the time taken for the flight can be

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Page # 75

KINEMATICS

Sol.

17. If T is the total time of flight, h is the maximum height & R is the range for horizontal motion, the x & y co-ordinates of projectile motion and time t are related as : t  t (A) y = 4h   1 −   T  T

X  X  (B) y = 4h   1 −   R  R

T  T  (C) y = 4h   1 −   t t

R  R  (D) y = 4h   1 −   X  X

Sol.

16. Choose the correct alternative (s) (A) If the greatest height to which a man can throw a stone is h, then the greatest horizontal distance upto which he can throw the stone is 2h. (B) The angle of projection for a projectile motion whose range R is n times the maximum height is tan–1(4/n) (C) The time of flight T and the horizontal range R of a projectile are connected by the equation gT2 = 2Rtanθ where θ is the angle of projection. (D) A ball is thrown vertically up. Another ball is thrown at an angle θ with the vertical. Both of them remain in air for the same period of time. Then the ratio of heights attained by the two ball 1 : 1. Sol.

18. A particle moves in the xy plane with a constant acceleration ‘g’ in the negative y-direction. Its equation of motion is y = ax – bx2, where a and b constants. Which of the following are correct? (A) The x-component of its velocity is constant. (B) At the origin, the y-component of its velocity is g 2b (C) At the origin, its velocity makes an angle tan–1(a) with the x-axis (D) The particle moves exactly like a projectile. Sol. a

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Page # 76

KINEMATICS

20. A particle is projected from the ground with velocity u at angle θ with horizontal. The horizontal range, maximum height and time of flight are R, H and T respectively. They are given by, R=

2u sin θ u 2 sin 2θ u 2 sin2 θ ,H= and T = g g 2g

Now keeping u as fixed, θ is varied from 30° to 60°. Then, (A) R will first increase then decrease, H will increase and T will decrease (B) R will first increase then decrease while H and T both will increase (C) R will decrease while H and T will increase (D) R will increase while H and T will increase Sol. 19. A ball is rolled off along the edge of a horizontal table with velocity 4 m/s. It hits the ground after time 0.4s. Which of the following are correct? (A) The height of the table is 0.8 m (B) It hits the ground at an angle of 60° with the vertical (C) It covers a horizontal distance 1.6 m from the table (D) It hits the ground with vertical velocity 4 m/s Sol.

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Page # 77

KINEMATICS

Exercise - III

(Subjective Problems)

1. The position vector of a particle moving in x-y
plane is given by r = ( t 2 − 4)i + ( t − 4)j . Find (a) Equation of trajectory of the particle Sol.

3. At time t the position vector of a particle of mass
m = 3kg is given by r = 6 t i − t 3 j + cos tk . Find the re
sultant force F ( t) , magnitude of its acceleration when t=

(b) Time when it crosses x-axis and y-axis Sol.

π , & speed when t = π. 2

Sol.

2. A p arti cl e move s al ong the sp ac e curv e
r = ( t 2 + t) i + ( 3 t − 2) j + (2t 3 − 4 t 2 ) k . (t in sec, r in m) Find at time t = 2 the (a) velocity, (b) acceleration, (c) speed or magnitude of velocity and (d) magnitude of acceleration. Sol.

4. The velocity time graph of a body moving in a straight line is shown. Find its

velocity in m/sec

y

60°

30° time in sec 2

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x 2.5 sec

Page # 78

KINEMATICS

(a) instantaneous velocity at t = 1.5 sec Sol.

(b) average acceleration from t = 1.5 sec. to t = 2.5 sec Sol.

6. Velocity of car v is given by v = at – bt2, where a and b are positive constants & t is time elapsed. Find value of time for which velocity is maximum & also corresponding value of velocity. Sol.

(c) draw its acceleration time graph from t = 0 to t = 2.5 sec Sol.

5. The curvilinear motion of a particle is defined by vx = 50 – 16t and y = 100 – 4t2 , where vx is in metres per second, y is in metres and t is in seconds. It is also known that x = 0 at t = 0. Determine the velocity (v) and acceleration (a) when the position y = 0 is reached. Sol.

7. The force acting on a body moving in a straight line is given by F = (3t2 – 4t + 1) Newton where t is in sec. If mass of the body is 1kg and initially it was at rest at origin. Find (a) displacement between time t = 0 and t = 2 sec Sol.

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Page # 79

KINEMATICS

9. A particle is moving along x-axis. Initially it is located 5 m left of origin and it is moving away from the origin and slowing down. In this coordinate system, the signs of the initial velocity and acceleration, are + y

+ x

– (0, 0)

(b) distance travelled between time t = 0 and t = 2 sec Sol.

v0

a

–

Sol.

8. A particle goes from A to B with a speed of 40 km/ h and B to C with a speed of 60 km/h. If AB = 6BC the ave rage speed i n k m/h betwe en A and C i s ____________ [Hint : Average speed =

10. Find the change in velocity of the tip of the minute hand (radius = 10 cm) of a clock in 45 minutes. Sol.

total dis tan ce travelled ] time taken

Sol.

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Page # 80

KINEMATICS

11. At a distance L = 400 m from the traffic light brakes are applied to a locomotive moving at a velocity v= 54 km/hr. Determine the position of the locomotive relative to the traffic light 1 min after the application of the breaks if its acceleration is –0.3 m/sec2. Sol.

12. A train starts from rest and moves with a constant acceleration of 2.0 m/s2 for half a minute. The brakes are then applied and the train comes to rest in one minute. Find (a) the total distance moved by the train, (b) the maximum speed attained by the train and (c) the position (s) of the train at half the maximum speed. Sol.

13. A car is moving along a straight line. It is taken from rest to a velocity of 20 ms–1 by a constant acceleration of 5ms–2. It maintains a constant velocity of 20 ms–1 for 5 seconds and then is brought to rest again by a constant acceleration of –2 ms–2. Draw a velocity-time graph and find the distance covered by the car. Sol.

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Page # 81

KINEMATICS

14. A stone is dropped from a height h. Simultaneously another stone is thrown up from the ground with such a velocity that it can reach a height of 4 h. Find the time when two stones cross each other. Sol.

15. A bal loon is ascending vertical ly with an acceleration of 0.2 m/s2 Two stones are dropped from it at an interval of 2 sec. Find the distance between them 1.5 sec after the second stone is released (use g = 9.8 m/s2). Sol.

16. From the velocity-time plot shown in figure, find the distance travelled by the particle during the first 40 seconds. Also find the average velocity during this period. V 5m/s t(s) 0 20 40 –5m/s Sol.

17. The velocity-time graph of the particle moving along a straight line is shown. The rate of acceleration and deceleration is constant and it is equal to 5 ms–2. If the average velocity during the motion is 20ms–1, then find the value of t.

o

t

Sol.

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25 sec

Page # 82

KINEMATICS

18. The fig. shows the v-t graph of a particle moving in straight line. Find the time when particle returns to the starting point. v

20. A particle is projected upwards with a velocity of 100 m/sec at an angle of 60º with the vertical. Find the time when the particle will move perpendicular to its initial direction, taking g = 10 m/sec2. Sol.

20 10

10

20

25 t

Sol.

gx2 . The 2 angle of projectile is ________ and initial velocity is _______. Sol.

21. The equation of a projectile is y = 3 x −

19. A particle is projected in the X-Y plane. 2 sec after projection the velocity of the particle makes an angle 45º with the X-axis. 4 sec after projection, it moves horizontally. Find the velocity of projection (use g = 10 ms–2). Sol.

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Page # 83

KINEMATICS

22. A ball is projected at an angle of 30º above with the horizontal from the top of a tower and strikes the ground in 5 sec at an angle of 45º with the horizontal. Find the height of the tower and the speed with which it was projected. [g =10 m/s2] Sol.

23. A rocket is launched at an angle 53º to the horizontal with an initial speed of 100 ms–1. It moves along its initial line of motion with an acceleration of 30 ms–2 for 3 seconds. At this time its engine falls & the rocket proceeds like a free body. Find : (i) the maximum altitude reached by the rocket (ii) total time of flight (iii) the horizontal range. [sin 53º = 4/5] Sol.

24. A ball is thrown horizontally from a cliff such that it strikes ground after 5 sec. The line of sight from the point of projection to the point of hitting makes an angle of 37º with the horizontal. What is the initial velocity of projection. 37º

Sol.

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Page # 84

KINEMATICS

25. A ball is projected on smooth inclined plane in direction perpendicular to line of greatest slope with velocity of 8m/s. Find it’s speed after 1 sec. 8 m/s 37º

27. The horizontal range of a projectiles is R and the maximum height attained by it is H. A strong wind now begins to blow in the direction of motion of the projectile, giving it a constant horizontal acceleration = g/2. Under the same conditions of projection, find the horizontal range of the projectile. Sol.

Sol.

26. Find range of projectile on the inclined plane which is projected perpendicular to the incline plane with velocity 20m/s as shown in figure. -1

u = 20 ms

28. A butterfly is flying with velocity 10 i + 12j m / s and wind is blowing along x axis with velocity u. If butterfly starts motion from A and after some time reaches point B, find the value of u. y

37º

B

Sol. A

37° x

Sol.

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Page # 85

KINEMATICS

29. In the figure shown, the two projectiles are fired simultaneously. What should be the initial speed of the left side projectile for the two projectile to hit in mid-air ? u

Sol.

20m/s 60º 45º \\\\\\\\\\\\\\\\\\\\\\\\\\ 10m

30. In the figure shown, the two projectiles are fired simultaneously. Find the minimum distance between them during their flight?

31. Two particles are moving along two long straight lines, in the same plane, with the same speed = 20 cm/s. The angle between the two lines is 60°, and their intersection point is O. At a certain moment, the two particles are located at distance 3m and 4m from O, and are moving towards O. Find the shortest distance between them subsequently? Sol.

20 3 m / s 20 m/s

60°

30° 20 m

Sol.

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Page # 86

KINEMATICS

32. Rain is falling vertically with a speed of 20 ms–1 relative to air. A person is running in the rain with a velocity of 5 ms–1 and a wind is also blowing with a speed of 15 ms–1 (both towards east). Find the angle with the vertical at which the person should hold his umbrella so that he may not get drenched. Sol.

33. A glass wind screen whose inclination with the vertical can be changed is mounted on a car. The car moves horizontally with a speed of 2 m/s. At what angle α with the vertical should the wind screen be placed so that the rain drops falling vertically downwards with velocity 6 m/s strike the wind screen perpendicularly? Sol.

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

35. A man crosses a river in a boat. If he crosses the river in minimum time he takes 10 minutes with a drift 120 m. If he crosses the river taking shortest path, he takes 12.5 minute, find (i) width of the river (ii) velocity of the boat with respect to water (iii) speed of the current. Assume vb/r > vr Sol.

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Page # 87

KINEMATICS

Exercise - IV

(Tough Subjective Problems)

1. A speeder in an automobile passes a stationary policeman who is hiding behind a bill board with a motorcycle. After a 2.0 sec delay (reaction time) the policeman acceleraties to his maximum speed of 150 km/hr in 12 sec and catches the speeder 1.5 km beyond the billboard. Find the speed of speeder in km/hr. Sol.

2. A large number of bullets are fired in all direction with the same speed v. What is the maximum area on ground on which these bullets can spread? Sol.

3. The speed of a particle when it is at its greatest height is

2 / 5 times of its speed when it is at its half the maximum height. The angle of projection is _________ and the velocity vector angle at half the maximum height is _________. Sol.

4. A projectile is to be thrown horizontally from the top of a wall of height 1.7m. Calculate the initial velocity of projection if it hits perpendicularly an incline of angle 37° which starts from the ground at the bottom of the wall. The line of greatest slope of incline lies in the plane of motion of projectile. Sol.

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Page # 88

KINEMATICS

5. Two inclined planes OA and OB having inclination (with horizontal) 30° and 60° respectively, intersect each other at O as shown in figure. A particle is projected from point P with velocity u = 10 3 ms –1 along

6. A particle is thrown horizontally with relative velocity 10 m/s from an inclined plane, which is also moving with acceleration 10 m/s2 vertically upward. Find the time after which it lands on the plane (g = 10 m/s2)

a direction perpendicular to plane OA. If the particle strikes plane OB perpendicularly at Q, calculate A

h

u

2

10 m/s

B Q

P

30°

Sol.

60°

30° O

(a) velocity with which particle strikes the plane OB, (b) time of flight, (c) vertical height h of P from O, (d) maximum height from O attained by the particle and (e) distance PQ Sol.

7. A, B & C are three objects each moving with constant → velocity. A’s speed is 10 m/sec in a direction PQ . The velocity of B relative to A is 6 m/sec at an angle of, cos–1(15/24) to PQ. The velocity of C relative to B is → 12 m/sec in a direction QP , then find the magnitude of the velocity of C. Sol.

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Page # 89

KINEMATICS 8. A particle is projected from point P with velocity 5 2 m/s perpendicular to the surface of a hollow right angle cone whose axis is vertical. It collides at Q normally. Find the time of the flight of the particle. y

P

Q 45°

x

Sol.

10. A hunter is riding an elephant of height 4m moving in straight line with uniform speed of 2m/sec. A deer running with a speed V in front at a distance of 4 5 m moving perpendicular to the direction of motion of the elephant. If hunter can throw his spear with a speed of 10 m/sec. relative to the elephant, then at what angle θ to it’s direction of motion must he throw his spear horizontally for a successful hit. Find also the speed ‘V’ of the deer. Sol.

9. A glass wind screen whose inclination with the vertical can be changed, is mounted on a cart as shown in figure. The cart moves uniformly along the horizontal path with a speed of 6 m/s. At what maximum angle α to the vertical can the wind screen be placed so that the rain drops falling vertically downwards with velocity 2 m/s, do not enter the cart? α v=6m/s

Sol.

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Page # 90

KINEMATICS

Exercise - V

JEE-Problems

1. Two guns, situated at the top of a hill of height 10 m, fire one shot each with the same speed 5 3 m/s at some interval of time. One gun fires horizontally and other fires upwards at an angle of 60° with the horizontal. The shots collide in air at a point P. Find (a) the time interval between the firings, and (b) the coordinates of the point P. Take origin of the coordinates system at the foot of the hill right below the muzzle and trajectories in X-Y plane.[JEE’ 1996] Sol.

3. A large heavy box is sliding without friction down a smooth plane of inclination θ. From a point P on the bottom of a box, a particle is projected inside the box. The initial speed of the particle with respect to box is u and the direction of projection makes an angle α with the bottom as shown in figure.

P

α

Q

θ

2. The trajectory of a projectile in a vertical plane is y = ax – bx2, where a, b are constants & x and y are respectively the horizontal & vertical distances of the projectile from the point of projection. The maximum height attained is ___________ & the angle of projection from the horizontal is ______, [JEE’ 1997] Sol.

(a) Find the distance along the bottom of the box between the point of projection P and the point Q where the particle lands. (Assume that the particle does not hit any other surface of the box. Neglect air resistance). (b) If the horizontal displacement of the particle as seen by an observer on the ground is zero, find the speed of the box with respect to the ground at the instant when the particle was projected. [JEE’ 1998] Sol.

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Page # 91

KINEMATICS

5. A ball is dropped vertically from a height d above the ground it hits the ground and bounces up vertically to a height d/2. Neglecting subsequent motion and air resistances, its velocity v varies with the height h above the ground as [JEE’ 2000 (Scr)] v

v

d

(A)

h

v

v d

(C)

h

d

(B)

d

h

(D)

h

Sol.

4. In 1.0 sec. a particle goes from point A to point B moving in a semicircle of radius 1.0 m. The magnitude of average velocity is [JEE ‘99] A 1m

(A) 3.14 m/sec (C) 1.0 m/sec Sol.

B

(B) 2.0 m/sec (D) zero

6. An object A is kept fixed at the point x = 3 m and y = 1.25 m on a plank P raised above the ground. At time t = 0 the plank starts moving along the +x direction with an acceleration 1.5 m/s2. At the same instant a stone is projected from the origin with a velocity u as shown. A stationary person on the ground observes the stone hitting the object during its downward motion at an angle of 45° to the horizontal. All the motions are in x-y plane. Find u and the time after which the stone hits the object. Take g = 10 m/s2 [JEE 2000] y A 1.25m P

u O Sol.

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3.0 m x

Page # 92

KINEMATICS

7. On a frictionless horizontal surface, assumed to be the x-y plane, a small trolley A is moving along a straight line parallel to the y-axis (see figure) with a constant velocity of ( 3 – 1) m/s. At a particular instant, when the line OA makes an angle of 45° with the x-axis, a ball is thrown along the surface from the origin O. Its velocity makes an angle φ with the x-axis and it hits the trolley. y A

45° x O (a) The motion of the ball is observed from the frame of trolley. Calculate the angle θ made by the velocity vector of the ball with the x-axis in this frame. (b) Find the speed of the ball with respect to the 4θ surface, if φ = . [JEE 2002] 3 Sol.

8. A particle starts from rest. Its acceleration (a) varsus time (t) is as shown in the figure. The maximum speed of the particle will be - [JEE’ 2004 (Scr)] a 2

10m/s

11 t(s)

(A) 110 m/s Sol.

(B) 55 m/s (C) 550 m/s (D) 660 m/s

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Page # 93

KINEMATICS

9. A small block slides without friction down an inclined plane starting from rest. Let Sn be the distance Sn travelled from time t = n – 1 to t = n. The S is n+1 [JEE’ 2004 (Scr)] (A)

2n – 1 2n

(B)

2n + 1 2n – 1

2n – 1 2n (D) 2n + 1 2n + 1

(C)

Sol.

10. The velocity displacement graph of a particle moving along a straight line is shown. The most suitable acceleration-displacement graph will be [JEE’ 2005 (Scr)]

11. STATEMENT-1 For an observer looking out through the window of a fast moving train, the nearby objects appear to move in the opposite direction to the train, while the distant objects appear to be stationary. STATEMENT-2 If the observer and the object are moving at velocities

V1 and V2 respectively with reference to a laboratory frame, the velocity of the object with respect to the

observer is V2 – V1 (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 [JEE’ 2008] Sol.

v

v0

x0

a

a

(A)

x

x

(B) x

a (C)

Sol.

a

x (D)

x

12. A train is moving along a straight line with a constant acceleration 'a'. A boy standing in the train throws a ball forward with a speed of 10 m/s, at an angle of 60° to the horizontal. The boy has to move forward by 1.15 m inside the train to catch the ball back at the initial height. The acceleration of the train [JEE’ 2011] in m/s2 is Sol.

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Page # 94

KINEMATICS

ANSWER KEY

QUESTIONS FOR SHORT ANSWER 1.



Magnitude | a| will remain uncharged. B = a



∆a = B − A

= a 2 + a 2 + 2a 2 cos( π − dθ)

= 2a2 (1 − cos θ)

⇒ 2a2 (1 − 1 + 2 sin 2dθ / 2) = 2a sin d θ/2

–a

2

Speedometer measure speed of car as it only gives the magnitude.

3

When particle is moving with constant velocity its average velocity and instantaneous velocity will be

4 5

same and magnitude of instantaneous velocity will also be same.


∆S VAvg = , ∆S = VAvg × ∆t ∆t


∆S
VAvg = with zero displacement non zero VAvg is not ∆ S = 0 ∆t

x


possible zero displacement and non zero V is possible if particle

is reversing and coming to starting point. Show on x-t graph by an example.

t

6

Speed of projectile is smallest at the highest point.

7

Both the ball will hit the ground with same speed.

8

If sack of rice is dropped when it is just above the centre it will fall ahead of circle because sack will have velocity same as plane in horizontal direction.

9

Ist Curve : at particular time x has more than one value hence not a 1-D motion.
IInd Curve : | V| cannot be negative IIIrd Curve : Length of a moving body can not decrease with time

10

Ist Curve : A ball moving forward collides with surface rebounds and stops after IInd collision IInd Curve : A ball repeatedly making inelastic collisions with floor. IIIrd Curve : Collision of a ball with surface. {Surface has large velocity for short time}

11

(a) is incorrect car can not travel around track with constant velocity as direction is continuously changing. (b) correct

12 13

Ball at maximum height V = 0 for just an instant but acceleration due to gravity. 1 . Let balls meet after t sec. h1 Vf = 2gH 1 2 1 2 H X h1 = gt and h2 = Vf t = gt 2 2 h2 H V0=Vf H = 2 gH t t = 2 h1 + h2 = H = Vf t 2g 1 H H ∴ h1 = g = hence they will meet above half height of building. 2 2g 4

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Page # 95

KINEMATICS

14

Initially bullet is at rest u = 0 V2 = 0 + 2as ∴ a=

V2 muzzle velocity is more for short barrl and S is also less hence acceleration will be more in that case. 2S

Hence we can not conclude that velocity of boat is 5 m/sec w.r.t. shore

15

VBottle = Vriver 16

;

VB – VR = 5

Yes wrench will hit at the same place on the deck irrespective of that boat is at rest or moving because when boat is at rest wrench will have zero horizontally velocity and when boat is moving both will have same horizontal velocity.

17

Acceleration of the projectile remains constant throughout the journey = g

18

(a) In child point of view range will be same in both the cases. (b) In ground frame of reference VCT = VC – VT VC = VCT + VT

19

For front range Vcannon = VC cos θ + VT

Range will be more

For Rear range Vcannon = VC cos θ – VT

Range will be less

d t=

Vbr

d for tmin cos θ = 1 maximum Hence A will reach opposite end in least time Vbr cos θ

EXERCISE - I

ANSWER KEY 1. B 2. B 3. A 4. B 5. A 6. D 7. C 8. B 9. B 10.

D 11.

C 12.

B 13.

B 14.

A

15.

D 16. C 17.

B 18.

C 19.

C 20

D 21.

C 22.

D 23.

C 24.

B 25.

B

26.

D 27. C 28.

B 29.

C 30.

A 31.

A 32.

A 33.

C 34.

A 35.

D 36.

D

37.

C 38. C 39.

B 40.

B 41.

D 42.

D 43.

B 44.

A 45.

D 46.

A 47.

B

48.

C 49. B 50.

C 51.

D 52.

B 53.

B 54.

A 55.

C 56.

C 57.

C 58.

C

59.

A 60. A 61.

C 62.

C 63.

D 64.

B 65.

C 66.

D 67.

B 68.

D 69.

D

70.

C 71. A 72.

A 73.

A 74.

A 75.

B 76.

B 77.

C 78.

B 79.

B

EXERCISE - II

ANSWER KEY 1. A,B,C,D 2. A,C 10. A,B,C,D

3. B,C,D 4. B,D

11. B,C,D

5. C,D

6. A,C

7. A,C

8. A,D

12. A,C,D 13. A,B,C 14. C,D 15. D,C 16. A,B,C,D

19. A,C,D 20. B

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9. C

17. A,B 18. A,B,C,D

Page # 96

KINEMATICS

EXERCISE - III

ANSWER KEY

1. (a) y2 + 8y + 12 = x ; (b) crosses x axis when t = 4 sec, crosses y axis when t = ± 2 sec. 3. –18 tj – 3 cos t k ; 3π ; 3 4 + π 4

2. (a) 5i + 3j + 8k, (b) 2i + 16k, (c) 7 2 , (d) 2 65

4. (a)

7. (a)

1 3

m / s , (b)

3 m / s 2 , (c) 2



5. v = –30 i – 40 j, a = –16 i – 8 j 6. a/2b, a2/4b

38 2 m , (b) m 8. 42 km/hr 3 3

9.

v0

a

–

+

π 2 10.  3  cm/min 11. 25 m  

vel

12. (a) 2.7 km; (b) 60 m/s; (c) 225 m and 2.25 km 13. 240 m 14.

16. 100 m, zero 17. 5 s 18. 36.2 sec. 19. 20 5 22. u = 50 ( 3 – 1) m/sec., H = 125 (– 24. 100/3 m/s

30. 10 m

25. 10 m/s

31. 50 3 cm

 h    8g 

20. 20 sec

15. 50 m

21. 60, 2 m/sec.

3 + 2)m 23. (i) 1503.2 m (ii) 35.54 sec (iii) 3970.56 m

26. 75 m

27. R + 2H

32. tan–1 (1/2)

28. 6 m/s 29. 20 ×

33. tan–1(3)

2/3

–1  1  4 34. θ = tan   , 2 3

35. 200 m, 20 m/min, 12 m/min

EXERCISE - IV

ANSWER KEY 1. 122.7 km/hr 2.

πv 4

3. 60°, tan

g2

16.25 m, (e) 20 m 6.

1 3

sec

–1

(

3/2

)

7. 5 m/sec

4. u = 3m/s 5. (a) 10 ms–1,(b) 2 sec, (c)5 m, (d) 8. 1 sec

9. 2 tan–1 (1/3) 10. θ = 37°, v = 6 m/s

EXERCISE - V

ANSWER KEY

1. (a) 1 sec, (b) ( (5 3 m, 5 m)

2.

a2 , tan –1 a 4b

3. (a)

u 2 sin 2α u cos(α + θ) , (b) v = g cos θ cos θ

4. B

5. A

6. u = 7.29 m/s, t = 1 sec 7. (a) 45°, (b) 2m/sec

8. B

9. C

10. B

11. B

12. 5 m/s2

394,50 - Rajeev Gandhi Nagar Kota, Ph. No. : 93141-87482, 0744-2209671 IVRS No : 0744-2439051, 52, 53, www. motioniitjee.com , [email protected]

CONSTRAINED MOTION MOTION, TION, N.L.M., FRICTION THEORY AND EXERCISE BOOKLET CONTENTS

S.NO.

TOPIC

PAGE NO.

CONSTRAINED MOTION 1. String Constraint ............................................................................. 3 – 9 2. Wedge Constraint ........................................................................... 9 – 10

NEWTON'S LAW OF MOTION 1.

Force ....................................................................................... 11– 15

2.

Newton's Ist Law of Motion ................................................................ 16

3.

Newton's 2nd Law of Motion ........................................................ 16 – 25

4.

Newton's 3rd Law of Motion ......................................................... 25 – 27

5.

Spring Force ............................................................................. 27 – 32

6.

Pseudo Force ........................................................................... 32 – 36

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www. motioniitjee.com, [email protected]

Page # 2

CONSTRAINED MOTION

FRICTION S.NO.

TOPIC

PAGE NO.

1.

Friction ................................................................................... 37 – 40

2.

Minimum Force Required to move the particle .................................. 40 – 41

3.

Friction as the component of contact force .................................... 41 – 42

4.

Motion on a Rough Inclined Plane .................................................. 42 – 43

5.

Angle of Repose ............................................................................... 43

6.

Two blocks on an inclined Plane .................................................... 44 – 45

7.

Range of force for which Acceleration of body is zero ....................... 45 – 48

8.

Pulley block system involve friction ............................................... 48 – 49

9.

Two block system ...................................................................... 49 – 53

10. Friction involve pseudo concept ................................................... 54 – 55 11. Exercise - I .............................................................................. 56 – 87 12. Exercise - II ............................................................................. 88 – 96 13. Exercise - III .......................................................................... 97 – 106 14. Exercise - IV ......................................................................... 107 – 111 15. Exercise - V .......................................................................... 112 – 117 16. Answer key ........................................................................... 118 – 120

IIT-JEE Syllabus : Newton's law of motion; inertial frame of reference; Uniformly accelerated frames of reference, Static and dynamic friction.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 3

CONSTRAINED MOTION

1.

CONSTRAINED MOTION :

1.1

String constraint : When the two object are connected through a string and if the string have the following properties : The length of the string remains constant i.e., it is inextensible string Always remains taut i.e., does not slacks. Then the parameters of the motion of the objects along the length of the string have a definite relation between them.

• •

Ist format : - (when string is fixed) A

s

B

v

The block B moves with velocity v. i.e. each particle of block B moves with velocity v. If string remain attached to block B it is necessary that velocity of each particle of string is same = v (vs = v) Now we can say that Block A also moves with velocity v. v v

A

B vA = vB = v

: If pulley is fixed then the velocity of all the particles of string is same along the string. v

B

Ex.1

Sol.

A vA =?

In the above situation block B is moving with velocity v. Then speed of each point of the string is v along the string. ∴ speed of the block A is also v v B v A vA=v

Ex.2

VA = 8 m/s

A 37°

B

vB=? Sol.

∵ Block A is moving with velocity 8 ms–1. 8 m/s ∴ velocity of every point on the string must be 8m/s along the string. The real velocity of B is vB. Then the string will not break only when the compoent of vB along string is 8 m/s. ⇒ vB cos 37° = 8 ⇒

vB =

8 = 10 m/sec cos 37°

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

A 8 m/s 8 m/s 37° vB

B

Page # 4 Ex.3

CONSTRAINED MOTION

Find out the velocity of block B in a pulley block system as shown in figure.

37°

53° 10 m/s Sol.

B

A

In a given pulley block system the velocity of all the particle of string is let us assume v then.

v 53°

10m/s

A

v 37°

B

53° 10cos53° 10 m/s is the real velocity of block A then its component along string is v. ⇒ 10 cos 53° = v ...(1) If vB is the real velocity of block B then it component along string is v then vBcos37° = v ...(2)

v

37° vB

B

from (1) & (2) vB cos37° = 10 cos53° ⇒

vB =

10 × 3 / 5 30 15 = m / sec = 4/5 4 2

50/3 m/s 53°

Ex.4

A

What is the velocity of block A in the figure as shown above. Sol.

The component of velocity of ring along string = velocity of A =

50 cos 53° = vA ⇒ vA = 10 m/s 3

: In the first format only two points of string are attached or touched to moving bodies.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 5

CONSTRAINED MOTION IInd format (when pulley is also moving) To understand this format we consider the following example in which pulley is moving with velocity vp and both block have velocity vA & vB respectively as shwon in figure. If we observe the motion of A and B with respect to pulley. Then the pulley is at rest. Then from first format. vAP = – vBP

vP

vB

vA A

(–ve sign indicate the direction of each block is opposite with respect to Pulley) v A – v p = – vB + v P ⇒ vP = :-

v A + vB 2

To solve the problem put the values of vA, vB, & vP with sign.

vP

10 m/s

Ex.5 A v=? A

B

Sol.

v A + vB 2 Putting vp = 10 ms–1, vB = 0, we get vA = 20 ms–1 (upward direction)

vP =

vP= 10m/s Ex.6

5m/s

Sol.

A

B v=? B

If we take upward direction as +ve then –5 + vB 2 vB = 25 m/sec (in upward direction)

10 =

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

B

Page # 6

CONSTRAINED MOTION

7 m/s

E 8m/s

F

Ex.7

Sol.

A

D

B C 5m/s

2m/s

Find out the velocity of Block D From 2nd format of constrained motion vE =

v A + vB 2

2–5 = –3/2 (If upward direction is taken to be +ve) 2 vE = –3/2 m/s

vE =

Now

vE + vF –3 / 2 + vF = 7 m/s ⇒ 7 = 2 2

⇒

14 + 3/2 = vF ⇒ vF =

31 2

v C + vD 8 + vD 31 = vF ⇒ = ⇒ vD = 31 – 8 2 2 2 vD = 23 m/s (upward direction)

Now

C

B

D E

Ex.8 A H

F

G m

10 m/sec

Sol.

Find the velocity of point G. In string ABCD from first format of constrain VD = 10 m/s↑ Now

+ 10 =

vH + vE 2 vH = 10 m/s ↓ if upward direction is taken to be positive

vD =

−10 + v E ⇒ vE = 30 m/s ↑ 2

vF + v G –10 + v G = vE ⇒ 30 = 2 2 60 + 10 = vG vG = 70 m/s↑

Now

: In IInd format three or four Points of the string is attached to the moving bodies.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 7

CONSTRAINED MOTION III format :

1.

SOLVING STRATEGY : First choose the longest string in the given problem which contains the point of which velocity/ acceleration to be find out.

2.

Now mark a point on the string wherever it comes in contact or leaves the contact of real bodies.

3.

If due to motion of a point, length of the part of a string with point is related, increases then its speed will be taken +ve otherwise –ve.

A D

I

E H

J C Ex.9

B 5m/s

C F A

vC=?

G B

2m/s

Sol. Step 1.

We choose a longest string ABCDEFGHIJ in which we have to find out velocity of point J (vc)

Step 2.

Mark all the point A, B ................

Step 3.

Write equation vA + vB + vC + vD + vE + vF + vG + vH + vI + vJ = 0 vA = vD = vE = vH = vI = 0

(No movement of that point because attached to fixed objects) ⇒ vB + vc + vF + vG + vJ = 0

...(1)

vB = vC = 5 m/s

(increases the length)

vF = vG = 2m/s

(It also increases the length)

Let us assume C is moving upward with velocity vc so vc negative because it decreasing the length ⇒ 5 + 5 + 2 + 2 – vc = 0 vC = 14 m/sec (upward) Ex.10

4m/s F

c ↑ 2m / sec

8m / s↑ A

B ↓ 2m / s

E

1 m / s↓ D Find out the velocity of block E as shown in figure.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 8 Sol. Step-1

CONSTRAINED MOTION We first choose the longest string in which point j (block E) lie. (abcdefghij) 4m/s c

F

d

b

e x

a k 8m / s ↑ A

h

y

c ↑ 2m / s

i j

f

B ↓ 2m / s

g

E

z 1m / s ↓ D Step 2 : Now write equation according to the velocity of each point (either increase or decrease the length) va + vb + vc + vd + ve + vf + vg + vh + vi + vj= 0 ...(1) Now find value of va, vb ..... in a following way v A + vB (from second format) 2 8–2 = = 3 m/sec. (upward) 2

vk =

va =

vK + v C (from 2nd format) 2

3+2 = 5/2 m/sec. (upward) 2 vx = 4m/s (from first format of constrain) vy + vz from 2nd format of constrain vx = ∴ vz = 0 (fixed) 2 ⇒ vy = 2 vx = 8 m/s (upward) ⇒ Now va = – 5/2 m/s (decreases the length) vb = vc = vd = ve = 0 (attached to fixed object) vf = vg = 1m/s (increases the length) vh = vi = vy = 8 m/s (increase the length) Let us assume block E move upward then vj = – vE (decrease the length) Puting the above values in eq. (1) ⇒ –5/2 + 1 + 1 + 8 + 8 – vE = 0 vE = 31/2 m/s (upward)

=

: In the following figure pulley is moving with velocity v at an angle θ with the horizontal.

v sin θ

v

A

B

θ C

D

*

v

A

B

θ C v cos θ

D

Only v cos θ is responsible to increase or decrease the length AB and v sin θ is responsible to either decrease or increase the length CD. Further solving strategy is same as 3rd format

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 9

CONSTRAINED MOTION Ex.11 Find out the relation between acceleration a and b as shown in following figure.

b B Sol. Step 1.

A

a

θ

Mark the points on the string which is attached to the real object (e.f,g,h)

f

e

b

g

f

b cos θ

b θ

a

θ

A

B

os bc

h

b

g

b

θ

h

θ

a Step 2. Acceleration of each point which are responsible to effect the length of string ae = 0 (because it is attached to fixed object) af = –b (attach to pulley which is moving with wedge's acceleration & –ve because it decreases the length) ag = b cos θ (only this component is responsible to effect the length of string) ah = (a – b cos θ) (resultant velocity at point h along the string) So now from 3rd format ae + af + ag + ah = 0 ⇒ 0 + (–b) + b cos θ + (a – b cos θ) = 0 a–b=0 ⇒ a=b

2.

(i) (ii)

WEDGE CONSTRAINT : Conditions : Contact must not be lost between two bodies. Bodies are rigid. The relative velocity / acceleration perpendicular to the contact surface of the two rigid object is always zero. Wedge constraint is applicable for each contact.

v3

v2

v3

v 3 = v1 sin θ

v1

v 1 sin θ Contact Plane

θ

In other words, Components of velocity and acceleration perpendicular to the contact surface of the two objects is always equal if there is no deformation and they remain in contact.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 10

CONSTRAINED MOTION

Ex.12 Find the relation between velocity of rod and that of the wedge at any instant in the figure shown.

v θ Sol.

u

Using wedge constraint. Component of velocity of rod along perpendicular to inclined surface is equal to velocity of wedge along that direction. u cos θ = v sin θ u = tan θ v u = v tan θ

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 11

NEWTON’S LAW OF MOTION

1.

FORCE A pull or push which changes or tends to change the state of rest or of uniform motion or direction of motion of any object is called force. Force is the interaction between the object and the source (providing the pull or push). It is a vector quantity. Effect of resultant force : may change only speed may change only direction of motion. may change both the speed and direction of motion. may change size and shape of a body

• • • •

unit of force : newton and

kg.m (MKS System) s2

g.cm (CGS System) s2 1 newton = 105 dyne Kilogram force (kgf) The force with which earth attracts a 1 kg body towards its centre is called kilogram force, thus

dyne and

Force in newton g Dimensional Formula of force : [MLT–2] For full information of force we require → Magnitude of force → direction of force → point of application of the force

kgf =

•

Force

Electromagnetic force

Gravitational force

Contact force

Nuclear force

Normal reaction

Tension

friction

1.1

Electromagnetic Force

• • • •

Force exerted by one particle on the other because of the electric charge on the particles is called electromagnetic force. Following are the main characteristics of electromagnetic force These can be attractive or repulsive These are long range forces These depend on the nature of medium between the charged particles. All macroscopic force (except gravitational) which we experience as push or pull or by contact are electromagnetic, i.e., tension in a rope, the force of friction, normal reaction, muscular force, and force experienced by a deformed spring are electromagnetic forces. These are manifestations of the electromagnetic attractions are repulsions between atoms/molecules.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 12

1.2

NEWTON’S LAW OF MOTION

Gravitational force : It acts between any two masses kept anywhere in the universe. It follows inverse square rule (F ∝ 1 ) and is attractive in nature. dis tan ce 2 GM1M2 F= R2 The force mg, which Earth applies on the bodies, is gravitational force.

1.3

Nuclear force : It is the strongest force. It keeps nucleons (neutrons and protons) together inside the nucleus inspite of large electric repulsion between protons. Radioactivity, fission, and fusion, etc. result because of unbalancing of nuclear forces. It acts within the nucleus that too upto a very small distance.

1.4

Contact force : Forces which are transmitted between bodies by short range atomic molecular interactions are called contact forces. When two objects come in contact they exert contact forces on each other.

1.4.1 Normal force (N) : It is the component of contact force perpendicular to the surface. It measures how strongly the surfaces in contact are pressed against each other. It is the electromagnetic force. A table is placed on Earth as shown in figure. •

3

1

4

Here table presses the earth so normal force exerted by four legs of table on earth are as shown in figure. N2

N1 N3

•

2

N4

ground

Now a boy pushes a block kept on a frictionless surface. Block

Here, force exerted by boy on block is electromagnetic interaction which arises due to similar charges appearing on finger and contact surface of block, it is normal force. (by boy) N

•

Block

A block is kept on inclined surface. Component of its weight presses the surface perpendicularly due to which contact force acts between surface and block.

θ

Normal force exerted by block on the surface of inclined plane is shown in figure.

N θ

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 13

NEWTON’S LAW OF MOTION Force acts perpendicular to the surface : • •

Normal force acts in such a fashion that it tries to compress the body Normal is a dependent force, it comes in role when one surface presses the other.

Ex.1

Two blocks are kept in contact on a smooth surface as shown in figure. Draw normal force exerted by A on B. A

B

Sol.

In above problem, block A does not push block B, so there is no molecular interaction between A and B. Hence normal force exerted by A on B is zero.

Ex.2

Draw normal forces on the massive rod at point 1 and 2 as shown in figure. 2

1

Sol.

Normal force acts perpendicular to extended surface at point of contact. N2

N1

50N 30°

Ex.3

Two blocks are kept in contact as shown in figure. Find (a) forces exerted by surfaces (floor and wall) on blocks (b) contact force between two blcoks. N1

Sol.

F.B.D. of 10 kg block N1 = 10 g = 100 N N2 = 100 N

...(1) ...(2)

100 N

N2 10 g

F.B.D. of 20 kg block N2 = 50 sin 30° + N3 ∴ N3 = 100 – 25 = 75 N and N4 = 50 cos30° + 20 g N4 = 243.30 N

N4

50 N

30°

...(3)

N2

N3 20 g

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

100N

10kg 20kg

Page # 14

NEWTON’S LAW OF MOTION

R=5m B

A

3m 1m

Ex.4

Find out the normal reaction at point A and B if the mass of sphere is 10 kg. N1

N2

y O

Sol.

A

5m3m 53°

N1

N2

37° 1m 4m

53°

37° O x'

B

Now F.B.D.

x

10 g y'

Now resolve the forces along x & y direction 3N2 5 y N1sin53° = 4N1/5 N1 N2sin37°=

N2

37° 4N2 O N2cos37°= 5

∵

53° N1cos53°=

3N1 5

100

The body is in equilibrium so equate the force in x & y direction In x-direction

3N1 4N2 = 5 5

...(1)

3N2 4N1 + = 100 5 5 after solving above equation N1 = 80 N, N2 = 60 N

In y-direction

...(2)

1.4.2 Tension : Tension in a string is an electromagnetic force. It arises when a string is pulled. If a massless string is not pulled, tension in it is zero. A string suspended by rigid support is pulled by a force ‘F’ as shown in figure, for calculating the tension at point ‘A’ we draw F.B.D. of marked portion of the string; Here string is massless. F.B.D of marked portion T A

A

F

F

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 15

NEWTON’S LAW OF MOTION

⇒ T=F String is considered to be made of a number of small segments which attracts each other due to electromagnetic nature. The attraction force between two segments is equal and opposite due to newton’s third law. Conclusion : T = mg (i) Tension always acts along the string and in such a direction that it tries to reduce the length of string (ii) If the string is massless then the tension will be same along the string but if the string have some mass then the tension will continuously change along the string.

Ex.5

m mg

The system shown in figure is in equilibrium. Find the magnitude of tension in each string ; T1, T2, T3 and T4. (g = 10 m/s–2)

60°

T3 T1

Sol.

T

30°

B

T4

T A 2 10 kg

F.B.D. of block 10 kg

F.B.D. of point ‘A’

y

T0

30° T0=10 g T0=100N

T1

10g

A

T2 x

T0

∑ Fy = 0 T2 cos30° = T0 = 100 N T2 =

200 3

N

∑ Fx = 0 T1 = T2 sin 30° =

200 1 100 . = N 3 2 3

F.B.D of point of ‘B’ ∑ Fy = 0 ⇒ T4 cos60° = T2 cos 30° and ∑ Fx = 0 ⇒ T3 + T2 sin 30° = T4 sin 60° ∴

T3 =

200 3

N , T = 200 N 4

y 60°

T3

B T2

T4 x

° 30

1.4.3 Frictional force : It is the component of contact force tangential to the surface. It opposes the relative motion (or attempted relative motion) of the two surfaces in contact. (which is explained later) 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 16

2.

NEWTON’S LAW OF MOTION

NEWTON’S FIRST LAW OF MOTION : According to this law “A system will remain in its state of rest or of uniform motion unless a net external force act on it. 1st law can also be stated as “If the net external force acting on a body is zero, only then the body remains at rest.” The word external means external to the system (object under observation), interactions within the system has not to be considered. The word net means the resultant of all the forces acting on the system. Newton’s first law is nothing but Galileo’s law of inertia. Inertia means inability of a body to change its state of motion or rest by itself. The property of a body that determines its resistance to a change in its motion is its mass (inertia). Greater the mass, greater the inertia. An external force is needed to set the system into motion, but no external force is needed to keep a body moving with constant velocity in its uniform motion. Newton’s laws of motion are valid only in a set of frame of references, these frames of reference are known as inertial frames of reference. Generally, we take earth as an inertial frame of reference, but strictly speaking it is not an inertial frame. All frames moving uniformly with respect to an inertial frame are themselves inertial. We take all frames at rest or moving uniformly with respect to earth, as inertial frames.

3.

NEWTON’S SECOND LAW OF MOTION : Newton’s second law states, “The rate of change of a momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts”    dp dp or F = k i.e., F ∝ dt dt where k is a constant of proportionality.     p = mv, So F = k (dmv) dt For a body having constant mass,    dv ⇒ F = km = k ma dt From experiments, the value of k is found to be 1.   So, Fnet = ma Force can’t change the momentum along a direction normal to it, i.e., the component of velocity normal to the force doesn’t change. Newton’s 2nd law is strictly applicable to a single point particle. In case of rigid bodies or system of   particles or system of rigid bodies, F refers to total external force acting on system and a refers to acceleration of centre of mass of the system. The internal forces, if any, in the system are not to be  included in F . Acceleration of a particle at any instant and at a particular location is determined by the force (net) acting on the particle at the same instant and at same location and is not in any way depending on the history of the motion of the particle.

PROBLEM SOLVING STRATEGY : Newton’s laws refer to a particle and relate the forces acting on the particle to its mass and to its acceleration. But before writing any equation from Newton’s law, you should be careful about which particle you are considering. The laws are applicable to an extended body too which is nothing but collection of a large number of particles. Follow the steps given below in writing the equations : 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 17

NEWTON’S LAW OF MOTION

Step 1 : Select the body The first step is to decide the body on which the laws of motion are to be applied. The body may be a single particle, an extended body like a block, a combination of two blocks-one kept over another or connected by a string. The only condition is that all the parts of the body or system must have the same acceleration. Step 2 : Identify the forces Once the system is decided, list down all the force acting on the system due to all the objects in the environment such as inclined planes, strings, springs etc. However, any force applied by the system shouldn’t be included in the list. You should also be clear about the nature and direction of these forces. Step 3 : Make a Free-body diagram (FBD) Make a separate diagram representing the body by a point and draw vectors representing the forces acting on the body with this point as the common origin. This is called a free-body diagram of the body. A

Ts

Tb R B m/sec2

5

R

C

Wp Wm platform man F.B.D of Diagram

100kg

50 kg

Look at the adjoining free-body diagrams for the platform and the man. Note that the force applied by the man on the rope hasn’t been included in the FBD. Once you get enough practice, you’d be able to identify and draw forces in the main diagram itself instead of making a separate one Step 4 : Select axes and Write equations When the body is in equillibrium then choose the axis in such a fashion that maximum number of force lie along the axis. If the body is moving with some acceleration then first find out the direction of real acceleration and choose the axis one is along the real acceleration direction and other perpendicular to it. Write the equations according to the newton’s second law (Fnet = ma) in the corresponding axis.

4.

APPLICATIONS :

4.1

Motion of a Block on a Horizontal Smooth Surface. Case (i) : When subjected to a horizontal pull : The distribution of forces on the body are shown. As there is no motion along vertical direction, hence, R = mg For horizontal motion F = ma or a = F/m R a

m

F

mg

Case (ii) : When subjected to a pull acting at an angle (θ) to the horizontal : Now F has to be resolved into two components, F cosθ along the horizontal and F sin θ along the vertical direction. Fsinθ R F θ m Fcosθ mg

For no motion along the vertical direction. 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 18

NEWTON’S LAW OF MOTION

we have R + F sin θ = mg or R = mg – F sin θ :

Hence R ≠ mg. R < mg For horizontal motion F cos θ m

F cos θ = ma, a =

Case (iii) : When the block is subjected to a push acting at an angle θ to the horizontal : (down ward) The force equation in this case θ F R = mg + F sin θ

R

F cos θ

θ

:

R ≠ mg, R > mg

mg

For horizontal motion F cosθ = ma, a =

4.2

F cos θ m

F F sin θ

Motion of bodies in contact. Case (i) : Two body system : Let a force F be applied on mass m1 B

F

A f m1

f m2

Free body diagrams : (vertical force do not cause motion, hence they have not been shown in diagram) F ⇒ a = m +m 1 2

m 2F and f = m + m 1 2

(i) Here f is known as force of contact. (ii) Acceleration of system can be found simply by a=

F

m1

f

f

m2

force total mass

: If force F be applied on m2, the acceleration will remain the same, but the force of contact will be different m1F i.e., f’ = m + m 1 2

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 19

NEWTON’S LAW OF MOTION Ex.6

Find the contact force between the 3 kg and 2kg block as shown in figure. F1 = 100N

Sol.

B

A 3kg

F2 = 25N

2kg

Considering both blocks as a system to find the common acceleration Fnet = F1 – F2 = 100 – 25 = 75 N common acceleration Fnet = 75 N 5kg Fnet 75 2 a= = = 15 m / s 5kg 5 R To find the contact force between A & B we draw a N F.B.D of 2 kg block 25 N 2kg from (∑Fnet)x = max ⇒ N – 25 = (2) (15) ⇒ N = 55 N 2g

a

Case (ii) : Three body system : C B A m2 m3 m F 1

Free body diagrams : F ⇒ a = m +m +m 1 2 3

For A

(m 2 + m 3 ) F and f1 = (m + m + m ) 1 2 3

For B

m1

F

m2

f1 f1

For C

f2 f2

m3

F – f1 = m1a f1 – f2 = m2a f2 = m3a m 3F f2 = (m + m + m ) 1 2 3 f1 = contact force between masses m1 and m2 f2 = contact force between masses m2 and m3 Remember : Contact forces will be different if force F will be applied on mass C

Ex.7

Find the contact force between the block and acceleration of the blocks as shown in figure.

F1 = 50N

Sol.

C

B

A

5kg

2kg

3kg

F2 = 30N

Considering all the three block as a system to find the common acceleration Fnet = 50 – 30 = 20 N 20 = 2m / s2 10 To find the contact force between B & C we draw F.B.D. of 3 kg block. a=

(∑ F ) net

x

= ma

Fnet=20N

10kg

a R

N1

⇒ N1 – 30 = 3(2) ⇒ N1 = 36 N To find contact force between A & B we draw F.B.D. of 5 kg block ⇒ N2 – N1 = 5a N2 = 5 × 2 + 36 ⇒ N2 = 46 N

3kg

30N a

mg

a

N2

5kg

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N1

Page # 20

4.3

NEWTON’S LAW OF MOTION

Motion of connected Bodies Case (i) For Two Bodies : F is the pull on body A of mass m1. The pull of A on B is exercised as tension through the string connecting A and B. The value of tension throughout the string is T only. B

T

m2

A

T

m1

F

Free body diagrams : For body A

For body B

R1

R2

a

A

T

a

B

F

m1g

m2g

R1 = m1g F – T = m1a

R2 = m2g T = m2a

T

F a = m +m 1 2

⇒

Case (ii) : For Three bodies : A

T1

m1

B m2

a

C

T2

m3

→

F

Free body diagrams :

For A

For B

R1

For C R3

R2

A

m1g R1 = m1g T1 = m1a

T1

T1

B

T2

m2g

T2

C

F

m3g

R3 = m3g R2 = m2g F – T2 = m3a T2 – T1 = m2a ⇒ F = m3a + T2 ⇒ T2 = m2a + T1 =m a+(m 3 1+m2 )a T2 = (m2 + m1)a F=(m1+m2+m3)a F ⇒ a = m +m +m 1 2 3

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Page # 21

NEWTON’S LAW OF MOTION Ex.8

A 5 kg block has a rope of mass 2 kg attached to its underside and a 3 kg block is suspended from the other end of the rope. The whole system is accelerated upward is 2 m/s2 by an external force F0. F0 5 kg

2 kg 3 kg

Sol.

(a) What is F0? (b) What is the force on rope? (c) What is the tension at middle point of the rope? (g = 10 m/s2) For calculating the value of F0, consider two blocks with the rope as a system. F.B.D. of whole system

F0 (a)

2m/s2

10 g = 100N

F0 – 100 = 10 × 2 F = 120 N ...(1) (b) According to Newton’s second law, net force on rope. F = ma = (2) (2) = 4 N ...(2) (c) For calculating tension at the middle point we draw F.B.D. of 3 kg block with half of the rope (mass 1 kg) as shown. T – 4g = 4.(2) = 48 N

4.4

T 4g

Motion of a body on a smooth inclined plane : Natural acceleration down the plane = g sin θ Driving force for acceleration a up the plane, F=m(a+ gsinθ) and for an acceleration a down the plane, F=m(a – gsinθ) N mgcos mg

mg sin

Find out the contact force between the 2kg & 4kg block as shown in figure.

2k g

4k g

Ex.9

37º

Sol.

On an incline plane acceleration of the block is independent of mass. So both the blocks will move with the same acceleration (gsin 37º) so the contact force between them is zero. 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564

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Page # 22

NEWTON’S LAW OF MOTION

20 N

2k g

3k g

Ex.10 Find out the contact force between 2kg & 3kg block placed on the incline plane as shown in figure.

37º

Considering both the block as a 5kg system because both will move the same acceleration.

20 N

5k g

Sol.

37º

20 N

5k g

N

Now show forces on the 5 kg block

4.5

c t

f o

r c e

g 5k

2 3 7 0N º

s in

( N

1

t a

N

n

5gcos 37º 37º 5g

3k g

c o

2

r

2m /s 3g si n3 7º

F o

a

5g

∵ Acceleration of 5kg block is down the incline. So choose one axis down the incline and other perpendicular to it From Newton’s second Law N = 5g cos 37º ...(i) 5gsin 37º – 20 = 5a ..(ii) 30 – 20 = 5a a = 2m/s2 (down the incline) ) between 2kg & 3kg block 1 we draw F.B.D. of 3kg block From Fnet = ma ⇒ 3gsin 37º – N1 = 3 × 2 18 – N1 = 6 N1= 12 N

N

37º 5g

Pulley block system :

Ex.11 One end of string which passes through pulley and connected to 10 kg mass at other end is pulled by 100 N force. Find out the acceleration of 10 kg mass. (g = 9.8 m/s2)

Sol.

Since string is pulled by 100N force.

100 N

So tension in the string is 100 N. F.B.D. of 10 kg block 100 – 10 g = 10 a

100 N

100 – 10 × 9.8 = 10 a a = 0.2 m/s2

10 kg

10 g

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 23

NEWTON’S LAW OF MOTION Ex.12 In the figure shown, find out acceleration of each block.

10kg

2kg 4kg

Sol.

Now F.B.D. of each block and apply Newton’s second law on each F.B.D

(1)

10 kg

10kg

a1

2T

a1

⇒

10g –2 T = 10a1

...(i)

a2

2kg 4kg

a3

10g T

(2)

2kg

a2

⇒

T – 2g = 2a2

...(2)

a3

⇒

T – 4g = 4a3

...(3)

2g T

(3)

4kg

4g

...(4) from constrain relation 2a1 = a2 + a3 Solving equations (1), (2), (3) and (4) we get 800 N 23 a1 = 70/23 m/s2 (downward), a2 = 170/23 m/s2 (upward),

T=

a3 = 30/23 m/s2 (downward)

Ex.13 Find the acceleration of each block in the figure shown below; in terms of their masses m1, m2 and g. Neglect any friction. m1

m2

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Page # 24 Sol.

NEWTON’S LAW OF MOTION

Let T be the tension in the string that is assumed to be massless. For mass m1, the FBD shows that N1 = m1g Where N1 is the force applied upward by plane on the mass m1. If acceleration of m1 along horizontal is a1. then T = m1a1 ...(i) For mass m2, the FBD shows that m2g – 2T = m2a2 ...(ii) Where a2 is vertical acceleration of mass m2. Note that upward tension on m2 is 2T applied by both sides of the string. from constrain relation a a2 = 1 2 Thus, the acceleration of m1 its twice that of m2. with this input, solving (i) and (ii) we find 2m 2 g a1 = 4m + m 1 2 m2g a2 = 4m + m 1 2

N1 T

m1g

T

T 2T

m2g

a2 m2g

Ex.14 Two blocks A and B each having a mass of 20 kg, rest on frictionless surfaces as shown in the figure below. Assuming the pulleys to be light and frictionless, compute : (a) the time required for block A, to move down by 2m on the plane, starting from rest, (b) tension in the string, connecting the blocks. A B

37º

Sol. Step 1.

Draw the FBDs for both the blocks. If tension in the string is T, then we have T

NA

mAg

NB

and

T mB g

Note that mAg, should better be resolved along and perpendicular to the plane, as the block A is moving along the plane. T NA mAg sin

mAg cos

Step 2. From FBDs, we write the force equations ‘ for block A where NA = mA g cos θ = 20 × 10 × 0.8 = 160 N and mAg sin θ – T = mA a ... (i) Where ‘a’ is acceleration of masses of blocks A and B. Similarly, force equations for block B are NB = mBg = 20 × 10 = 200 N and T = mBa ...(ii) From (i) and (ii), we obtain

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Page # 25

NEWTON’S LAW OF MOTION m A g sin θ 20 × 10 × 0.6 a = m +m = = 3 ms–2 40 A B

T = mBa = 20 × 3 = 60 N Step 3. With constant acceleration a = 3 ms–2, the block A moves down the inclined plane a distance S = 2 m in time t given by S=

1 2 at or t = 2

2S 2 = sec onds. a 3

Ex.15 Two blocks m1 and m2 are placed on a smooth inclined plane as shown in figure. If they are released from rest. Find : (i) acceleration of mass m1 and m2 (ii) tension in the string (iii) net force on pulley exerted by string N1 T Sol. F.B.D of m1 : m1g sin θ – T = m1a a m1 3 g – T = 3a ...(i) θ=30° 2

m1

m2 1kg

√3kg 30°

60°

m1g F.B.D. of m2 :

T – 1.

3 g = 1.a 2

a

m2

...(ii) θ

Adding eq. (i) and (ii) we get a = 0 Putting this value in eq. (i) we get T=

N2

T

T – m2g sin θ = m2a

m2g

3g , 2

F.B.D. of pulley

5.

FR =

2T

FR =

3 g 2

T

T

FR

NEWTONS’ 3RD LAW OF MOTION : Statement : “To every action there is equal and opposite reaction”. But what is the meaning of action and reaction and which force is action and which force is reaction? Every force that acts on body is due to the other bodies in environment. Suppose that a body A   experiences a force FAB due to other body B. Also body B will experience a force FBA due to A. According to Newton third law two forces are equal in magnitude and opposite in direction Mathematically we write it as   FAB = –FBA   Here we can take either FAB or FBA as action force and other will be the reaction force.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 26

NEWTON’S LAW OF MOTION

: (i) Action-Reaction pair acts on two different bodies. (ii) Magnitude of force is same. (iii) Direction of forces are in opposite direction. (iv) For action-reaction pair there is no need of contact

Ex.16 A block of mass ‘m’ is kept on the ground as shown in figure. (i) Draw F.B.D. of block (ii) Are forces acting on block action - reaction pair (iii) If answer is no, draw action reaction pair. Sol. (i) F.B.D. of block

m

N (Normal)

m

mg (field force)

(ii) ‘N’ and Mg are not action - reaction pair. Since pair act on different bodies, and they are of same nature. (iii) Pair of ‘mg’ of block acts on earth in opposite direction.

m

mg

earth

and pair of ‘N’ acts on surface as shown in figure.

N

5.1

Climbing on the Rope : Rope

F.B.D of man T

a

a mg

Now three condition arises. if T > mg ⇒ man accelerates in upward direction T < mg ⇒ man accelerates in downward direction T = mg ⇒ man’s acceleration is zero

*

Either climbing or decending on the rope man exerts force downward

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Page # 27

NEWTON’S LAW OF MOTION Ex.17 If the breaking strength of string is 600N then find out the maximum acceleration of the man with which he can climb up the road

600N (50 kg)

Sol.

Maximum force that can be exerted on the man by the rope is 600 N. F.B.D of man ⇒ 600 – 50 g = 50 a amax = 2 m/s2

a

50 g

Ex.18 A 60 kg painter on a 15 kg platform. A rope attached to the platform and passing over an overhead pulley allows the painter to raise himself along with the platform.

400 N

Sol.

(i) To get started, he pulls the rope down with a force of 400 N. Find the acceleration of the platform as well as that of the painter. (ii) What force must he exert on the rope so as to attain an upward speed of 1 m/s in 1 s ? (iii) What force should apply now to maintain the constant speed of 1 m/s? The free body diagram of the painter and the platform as a system can be drawn as shown in the figure. Note that the tension in the string is equal to the force by which he pulles the rope. (i) Applying Newton’s Second Law TT 2T – (M + m)g = (M + m)a 2T – (M + m)g M+m Here M = 60 kg; m = 15 kg ; T = 400 N g = 10 m/s2

or

a=

a

2( 400) – ( 60 + 15)(10 ) (M+m) g = 0.67 m/s2 60 + 15 (ii) To attain a speed of 1 m/s in one second the acceleration a must be 1 m/s2 Thus, the applied force is

a=

1 (M + m) (g + a) = (60 + 15) (10 + 1) = 412.5 N 2 (iii) When the painter and the platform move (upward) together with a constant speed, it is in a state of dynamic equilibrium Thus, 2F – (M + m) g = 0

F=

or F =

6.

(M + m)g (60 + 15)(10) = = 375 N 2 2

SPRING FORCE : Every spring resists any attempt to change its length; when it is compressed or extended, it exerts force at its ends. The force exerted by a spring is given by F = –kx, where x is the change in length and k is the stiffness constant or spring constant (unit Nm–1) When spring is in its natural length, spring force is zero.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 28

NEWTON’S LAW OF MOTION

 0+x

 0 F=0

Fext F = –kx

F F

x Graph between spring force v/s x Ex.19 Two blocks are connected by a spring of natural length 2 m. The force constant of spring is 200 N/m. Find spring force in following situations. 2m B

A

Sol.

(a) If block ‘A’ and ‘B’ both are displaced by 0.5 m in same direction. (b) If block ‘A’ and ‘B’ both are displaced by 0.5 m in opposite direction. (a) Since both blocks are displaced by 0.5 m in same direcetion, so change in length of spring is zero. Hence, spring force is zero. (b) In this case, change in length of spring is 1 m. So spring force is F = –Kx = – (200). (1) F = –200 N 2m Natural length

B

A 3m

1m B

A F

F

When spring is extended

B

A F

F

When spring is compressed

Ex.20 Force constant of a spring is 100 N/m. If a 10 kg block attached with the spring is at rest, then find extension in the spring (g = 10 m/s2) Sol.

6.1

In this situation, spring is in extended state so spring force acts in upward direction. Let x be the extension in the spring. F.B.D. of 10 kg block : Fs Fs = 10 g

⇒

Kx = 100

⇒

(100)x = (100)

⇒

x=1m

10 kg

10g

SPRING FORCE SYSTEM : Initially the spring is in natural length at A with block m. But when the block displaced towards right then the spring is elongated and now block is released at B then the block move towards left due to spring force (kx).

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 29

NEWTON’S LAW OF MOTION Analysis of motion of block : Natural Length

v a v=0

v a m

m

C v a Av a (i)

Initial position

B

From B to A speed of block increase and acceleration decreases. (due to decrease in spring force kx)

kx (ii)

a v

m

Due to inertia block crosses natural length at A. From A to C speed of the block decreases and acceleration increases.(due to increase in spring force kx)

m

(iii)

kx

a v

At C the block stops momentarily at this instant and since the spring is compressed spring force is towards right and the block starts to move towards right. From C to A speed of block increases and acceleration decreases.(due to decrease in spring force kx) kx a v

m

(iv)

Again block crosses point A due to inertia then from A to B speed decreases and acceleration increases. kx

m

a v

In this way block does SHM (to be expalined later) if no resistive force is acting on the block. Note : N.L.

A

(i)

Release

B when the block A is released then it take some finite time to reach at B. i.e., spring force doesn’t change instantaneously. N.L.

A

Release

m

(2) B

When point A of the spring is released in the above situation then the spring forces changes instantaneously and becomes zero because one end of the spring is free. (3)

In string tension may change instantaneously. 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564

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Page # 30

NEWTON’S LAW OF MOTION

Ex.21 Find out the acceleration of 2 kg block in the figures shown at the instant 1 kg block falls from 2 kg block. (at t = 0)

1kg

A Sol.

1kg

2kg

2kg B

F.B.D.s before fall of 1kg block

30N (kx) 2kg

1kg

1kg

30N (T)

2kg 30(mg)

30(mg)

after the fall of the 1 kg block tension will change instantaneously but spring force (kx) doesn’t change instantaneously. F.B.D.s just after the fall of 1 kg block

(A)

30N (kx) 2kg

20 B 2kg

20 aA =

20

30 – 20 = 5 m/s2 (upward) 2

aB = 0 m/s2

Ex.22 Two blocks ‘A’ and ‘B’ of same mass ‘m’ attached with a light spring are suspended by a string as shown in figure. Find the acceleration of block ‘A’ and ‘B’ just after the string is cut.

Sol.

When block A and B are in equilibrium position F.B.D of ‘B’ T0

A m

B m

T0=mg

...(i) mg

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Page # 31

NEWTON’S LAW OF MOTION F.B.D of ‘A’ T T = mg + T0 T = 2 mg

.....(ii)

mg T0

when string is cut, tension T becomes zero. But spring does not change its shape just after cutting. So spring force acts on mass B, again draw F.B.D. of block A and B as shown in figure F.B.D of ‘B’

T0=mg T0 – mg = m.aB aB = 0

mg F.B.D. of ‘A’ mg + T0 = m. aA 2 mg = m. aA aA = 2g (downwards)

mg T0=mg Ex.23 Find out the acceleration of 1kg, 2kg and 3kg block and tension in the string between 1 kg & 2 kg block just after cutting the string as shown in figure. Sol.

A 1kg B 2kg

F.B.D before cutting of string

6gN

A 1kg

C 3kg

1gN (mg) 5gN

B 2kg 3gN

2gN

3gN(spring force)

C 3kg 3gN(mg)

Let us assume the Tension in the string connecting blocks A & B becomes zero just after cutting the string then.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 32

NEWTON’S LAW OF MOTION 2 kg

1 kg a1 =

1g

1g = g ms–2 1

a2 =

2g + 3g (weight)

5g = 2.5 g ms–2 2

(spring force)

∵ a2 > a1 i.e., ∴ T≠0 If T ≠ 0 that means string is tight and Both block A & B will have same acceleration. So it will take as a system of 3 kg mass.

1 kg 10N

T

3kg

2 kg

a

≡

60N

30N 20N System

T 2

20m/s

Total force down ward = 10 + 30 + 20 = 60 N 60 Total mass = 3 kg ⇒ a = = 20 m/s2 3 Now apply Fnet = ma at block B. ∵

2kg 50

50 – T = 2 × 20 T = 10 N

the spring force does not change instantaneously the F.B.D of ‘C’ 3g 3kg

2

ac= 0 m/s

3g Reference Frame : A frame of reference is basically a coordinate system in which motion of object is analyzed. There are two types of reference frames. (a) Inertial reference frame : Frame of reference moving with constant velocity or stationary (b) Non-inertial reference frame : A frame of reference moving with non-zero acceleration

: (i) Although earth is a non inertial frame (due to rotation) but we always consider it as an inertial frame. (ii) A body moving in circular path with constant speed is a non intertial frame (direction change cause acceleration)

7.

PSEUDO FORCE : Consider the following example to understand the pseudo force concept support

B m

a A

The block m in the bus is moving with constant acceleration a with respect to man A at ground. Force 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 33

NEWTON’S LAW OF MOTION required for this acceleration is the normal reaction exerted by the support So,

N = ma

..(i)

This block m is at rest with respect to man B who is in the bus (a non intertial frame). So the acceleration of the block with respect to man B is zero. N = m(0) = 0 ..(ii) But the normal force is exerted in a non-inertial frame also. So the equation (ii) is wrong therefore we conclude that Newton’s law is not valid in non-inertial frame. If we want to apply Newton’s law in non-inertial frame, then we can do so by using of the cencept pseudo force. Pseudo force is an imaginary force, which in actual is not acting on the body. But after applying it on the body we can use Newton’s laws in non-inertial frames. This imaginary force is acting on the body only when we are solving the problem in a non-inertial frame of reference. In the above example. The net force on the block m is zero with respect to man B after applying the pesudo force. ma

m

N

N = ma : 1.

Direction of pseudo force is opposite to the acceleration of frame

2.

Magnitude of pseudo force is equal to mass of the body which we are analyzing multiplied by acceleration of frame

3.

Point of application of pseudo force is the centre of mass of the body which we are analysing

Ex.24 A box is moving upward with retardation ‘a’ < g, find the direction and magnitude of “pseudo force” acting on block of mass ‘m’ placed inside the box. Also calculate normal force exerted by surface on block

'm' 'Ma'

Sol.

N

Pseudo force acts opposite to the direction of acceleration of reference frame.

N + ma = mg N = mg – ma

(Pseudo force)

pseudo force = ma in upward direction F.B.D of ‘m’ w.r.t box (non-inertial)

mg Ex.25 Figure shows a pendulum suspended from the roof of a car that has a constant acceleration a relative to the ground. Find the deflection of the pendulum from the vertical as observed from the ground frame and from the frame attached with the car. a

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Page # 34

NEWTON’S LAW OF MOTION T

θ

Sol.

ax=a

a

θ mg

ay= 0

mg

Figure represents free Body diagram of the bob w.r.t ground. In an inertial frame the suspended bob has an acceleration a caused by the horizontal component of tension T. ...(i) T sin θ = ma ...(ii) T cosθ = mg From equation (i) and (ii) tan θ =

a g

 a θ = tan–1    g

⇒

In a non-inertial frame

T

θ ma

θ

a

a=0 x

ma

mg

a= y 0 mg Figure represents free Body diagram of bob w.r.t car. In the non-intertial frame of the car, the bob is in static equilibrium under the action of three froces, T, mg and ma (pseudo force) T sin θ = ma ...(iii) ....(iv) T cos θ = mg From equation (iii) and (iv) tan θ =

 a a ⇒ θ = tan–1    g g

Ex.26 A pulley with two blocks system is attached to the ceiling of a lift moving upward with an acceleration a0. Find the deformation in the spring. K m2

Sol.

a0

m1

Non-Inertial Frame T

K

2T m2T

T m1

a0

a

T a

m2g m2a0 m g m a (pseudo) 1 1 0 (pseudo)

Let relative to the centre of pulley, m1 accelerates downward with a and m2 accelerates upwards with a. Applying Newton’s 2nd law. m1a + m1a0 – T = m1a ...(i) ...(ii) T – m2g – m2a0 = m2a

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Page # 35

NEWTON’S LAW OF MOTION On adding (iv) and (v) we get  m1 – m 2  a =  m + m  (g + a0)  1 2

...(iii)

Substituting a in equation (i) We get T =

2m1m 2 (g + a 0 ) m1 + m 2

∴ x=

4m1m 2 ( g + a 0 ) F 2T = = (m + m ) k k k 1 2

Ex.27 All the surfaces shown in figure are assumed to be frictionless. The block of mass m slides on the prism which in turn slides backward on the horizontal surface. Find the acceleration of the smaller block with respect to the prism. A

m a0

Sol.

α B C Let the acceleration of the prism be a0 in the backward direction. Consider the motion of the smaller block from the frame of the prism The forces on the block are (figure) N'

N

a0 α N Mg

α

a

α

ma0

α mg

α

(i) N normal force (ii) mg downward (gravity), (iii) ma0 forward (Psuedo Force) The block slides down the plane. Components of the forces parallel to the incline give ma0 cosα + mg sin α = ma or, a = a0 cos α + g sin α ...(i) Components of the forces perpendicular to the incline give ...(ii) N + ma0 sin α = mg cos α Now consider the motion of the prism from the ground frame. No pseudo force is needed as the frame used is inertial. The forces are (figure) (i) Mg downward (ii) N normal to the incline (by the block) (iii) N’ upward (by the horizontal surface) Horizontal components give, N sin α = Ma0 or N = Ma0 / sin α, ...(iii) Putting in (ii) Ma 0 + ma0 sin α = mg cos α sin α mg sin α cos α or, a0 = M + m sin 2 α (M + m) g sin α mg sin α cos 2 α From (i) a = + g sin α = 2 M + m sin 2 α M + m sin α

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Page # 36

8.

NEWTON’S LAW OF MOTION

WEIGHING MACHING : A weighing machine does not measure the weight but measures the force. exerted by object on its upper surface or we can say weighing machine measure normal force on the man.

8.1

Motion in a lift :

(A)

If the lift is unaccelerated (v = 0 or constant) In this case no pseudo force act on the man N In this case the F.B.D. of the man N = mg In this case machine read the actual weight

(B)

If the lift is accelerated upward. (where a = constant)

weighing machine

mg

a

N

weighing machine F.B.D of man with respect to lift So weighing machine read N = m(g + a) Apparent weight N > Actual weight (mg) (c)

mg ma(pseudo)

If the lift is accelerated down ward.

a N ma (pseudo)

weighing machine F.B.D of man with respect to lift So weighing machine read

mg

N = m(g - a) Apparent weight N < Actual weight (mg) Note : (i) If a = g ⇒ N = 0 Thus in a freely falling lift, the man will experience a state of weightlessness (ii) If the lift is accelerated downwards such that a > g : So the man will be accelerated upward and will stay at the ceiling of the lift. (iii) Apparent weight is greater than or less than actual weight only depends on the direction and magnitude of acceleration. Magnitude and direction of velocity doesn’t play any roll in apparent weight. 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 37

FRICTION

1.

FRICTION : Friction is a contact force that opposes the relative motion or tendency of relative motion of two bodies. Mg

F

F f

N Consider a block on a horizontal table as shown in the figure. If we apply a force, acting to the right, the block remains stationary if F is not too large. The force that counteracts F and keeps the block in rest from moving is called frictional force. If we keep on increasing the force, the block will remain at rest and for a particular value of applied force, the body comes to state of about to move. Now if we slightly increase the force from this value, block starts its motion with a jerk and we observe that to keep the block moving we need less effort than to start its motion. So from this observation, we see that we have three states of block, first, block does not move, second, block is about to move and third, block starts moving. The friction force acting in three states are called static frictional force, limiting frictional force and kinetic frictional force respectively. If we draw the graph between applied force and frictional force for this observation its nature is as shown in figure.

1.1

Static frictional force f

b

flim fkin

a

Static region

c

Kinetic region

d

F

When there is no relative motion between the contact surfaces, frictional force is called static frictional force. It is a self-adjusting force, it adjusts its value according to requirement (of no relative motion). In the taken example static frictional force is equal to applied force. Hence one can say that the portion of graph ab will have a slope of 45º. The Direction of Static Friction The direction of static friction on a body is such that the total force acting on it keeps it at rest with respect to the body in contact. The direction of static friction is as follows. For a moment consider the surfaces to be frictionless. In absence of friction the bodies will start slipping against each other. One should then find the direction of friction as opposite to the velocity with respect to the body applying the friction.

1.2

Limiting Frictional Force This frictional force acts when body is about to move. This is the maximum frictional force that can exist at the contact surface. (i) The magnitude of limiting frictional force is proportional to the normal force at the contact surface. flim ∝ N ⇒ flim = µsN Here µs is a constant the value of which depends on nature of surfaces in contact and is called as ‘coefficient of static firction’.

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Page # 38

1.3

FRICTION

Kinetic Frictional Force Once relative motion starts between the surface in contact, the frictional force is called as kinetic frictional force. The magnitude of kinetic frictional force is also proportional to normal force. fk = µkN From the previous observation we can say that µk < µs Although the coefficient of kinetic friction varies with speed, we shall neglect any variation i.e., when relative motion starts a constant frictional force starts opposing its motion. Direction of Kinetic Friction The kinetic friction on a body A slipping against another body B is opposite to the velocity of A with respect to B. It should be carefully noted that the velocity coming into picture is with respect to the body applying the force of friction. f

f

v

Suppose we have a long truck moving on a horizontal road. A small block is placed on the truck which slips on the truck to fall from the rear end. As seen from the road, both the truck and the block are moving towards right, of course the velocity of the block is smaller than that of the truck. What is the direction of the kinetic friction acting on the block due to the truck ? The velocity of the block as seen from the truck is towards left. Thus, the friction on the block is towards right. The friction acting on the truck due to the block is towards left. Ex.1

Find the direction of kinetic friction force (a) on the block, exerted by the ground. (b) on the ground, exerted by the block. F=1N

Sol.

(a)

f1

1 kg 5 m/s w.r.t to ground

F=1N

1 kg V=5m/s ///////////////////////////////////

5m/s w.r.t to block

(b)

f2

where f1 and f2 are the friction forces on the block and ground respectively. Ex.2

Sol.

The correct relation between magnitude of f1 and f2 in above problem is : (A) f1 > f2 (B) f2 > f1 (C) f1 = f2 (D) not possible to decide due to insufficient data. By Newton’s third law the above friction forces are action-reaction pair and equal but opposite to each other in direction. Hence (C) Also note that the direction of kinetic friction has nothing to do with applied force F. 10 m/s

A

Ex.3

B

20m/s

All surfaces are rough. Draw the friction force on A & B fkBA Sol.

A

fkAB

B

fkBG Kinetic friction acts to reduce relative motion. Summary We can summarise the laws of friction between two bodies in contact as follows: 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 39

FRICTION (i)

(ii) (iii)

(iv) Ex.4

If the bodies slip over each other, the force of friction is given by fk = µk N where N is the normal contact force and µk is the coefficient of kinetic friction between the surfaces. The direction of kinetic friction on a body is opposite to the velocity of this body with respect to the body applying the force of friction. If the bodies do not slip over each other, the force of friction is given by fs ≤ µs N where µs is the coefficient of static friction between the bodies and N is the normal force between them. The direction and magnitude of static friction are such that the condition of no slipping between the bodies is ensured. The frictional force fk or fs does not depend on the area of contact as long as the normal force N is same. A block of mass 5 kg is resting on a rough surface as shown in the figure. It is acted upon by a force of F towards right. Find frictional force acting on block when (a) F = 5N (b) 25 N (c) 50 N (µs = 0.6, µk = 0.5) [g = 10 ms–2]

F

Sol.

Ex.5

Maximum value of frictional force that the surface can offer is Mg fmax = flim = µsN F = 0.6 × 5 × 10 = 30 newton Therefore, it F ≤ fmax body will be at rest and f = F or F > fmax body will more and f = fk N (a) F = 5N < Fmax So body will not move hence static frictional force will act and , fs = f = 5N (b) F = 25 N < Fmax ∴ fs = 25 N (c) F = 50 N > Fmax So body will move and kinetic frictional force will act, its value will be fk = µk N = 0.5 × 5 × 10 = 25 newton

f

A block having a mass 3 kg is initially at rest on a horizontal surface. The coefficient of static friction µs = 0.3 between the block and the surface and µk is 0.25. A constant force F of 50 N, acts on the body at the angle θ = 37º. What is the acceleration of the block ? θ

F

x

Sol.

We have two possibilities here, the block may remain at rest, or it may accelerate towards the right. The decision hinges on whether or not the x-component of the force F has magnitude, less than or greaer than the maximum static friction force. The x-component of F is Fx = Fcos θ = (50 N ) (0.8) = 40 N To find fs, max, we first calculate the normal force N, whether or not the block accelerates horizontally, the sum of the y-component of all the forces on the block is zero. N N - F sin θ – mg = 0 x f θ or N = F sin θ + mg=(50 N) (0.6) + (3 kg)(9.8ms–2) = 59.4 N The maximum static frictional force mg F fs,max = µsN = (0.3) (59.4 N) = 17.8 N This value is smaller than the x-component of F, hence the block moves. We now interpret the force f in the figure as a kinetic frictional force. This value is obtained as fK = µk N = (0.25) (59.4 N) = 14.8 N 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564

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Page # 40

FRICTION

Therefore resultant force in the x-direction is

∑F

x

= F cos θ − f = 40 N –14.8 N = 25.2 N

Then the acceleration ‘a’ of the block is a=

25.2 N = 8.4 ms −2 3 kg

Think : What would happen if the magnitude of Fx happened to be less than fs.max but larger than fk ? Ex.6

In the previous example, suppose we move the block by pulling it with the help of a massless string tied to the block as shown here. What is the force F required to produce the same acceleration in the block as obtained in the last example ? F θ m

Sol.

a

We are given that, m = 3kg, µs = 0.3, µk = 0.25, θ = 37º, and a = 8.4 ms–2 In order to determine the force F, we first draw the FBD as shown below f The equations of motion therefore, are N + Fsin θ = mg N = mg – Fsin θ F cos θ – f = ma and where f = µs N before the start of the motion, once motion is set, f = µkN. Hence, force F which produces a = 8.4 m/s2 is given by Fcosθ – µk (mg – F sin θ) = ma or F =

N

F sin F cos

mg

3(0.4 + 0.25 × 9.8) ma + µkmg = = 34.26N 0.8 + 0.25 × 0.6 cos θ + µk sin θ

: Fsin θ works out to be less than mg. Otherwise we would lift the block up in the above analysis COMMENT It is easier to pull then to push. Only about 34 N force is required to pull than 50 N required during pushing why ? Because, when we pull at an angle, the effective normal force N by which block is pressing down on surface is reduced and consequently friction is reduced. Just the contrary happens when you are pushing.

2.

MINIMUM FORCE REQUIRED TO MOVE THE PARTICLE : A body of mass m rests on a horizontal floor with which it has a coefficient of static friction µ. It is desired to make the body slide by applying the minimum possible force F. F φ m Fig. A Let the applied force F be at angle φ with the horizontal

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Page # 41

FRICTION F

R

φ

R m

R = Normal force mg

Fig. B For vertical equilibrium, R = (mg – F sinφ) ...(i) R + F sin φ = mg or, For horizontal equilibrium i.e. when the block is just about to slide, F cos φ = µR ...(ii) Substituting for R, or F = µmg / (cos φ + µ sinφ) F cosφ = µ (mg – F sinφ) for minimum F (cosφ + µsinφ) is maximum, ⇒ Let x = cos φ + µ sinφ dx = − sin φ + µ cos φ dφ for maximum of x, dx = 0 dφ tan φ = µ and at this value of φ Fmin =

3.

µmg 1 + µ2

FRICTION AS THE COMPONENT OF CONTACT FORCE : When two bodies are kept in contact, electromagnetic forces act between the charged particles at the surfaces of the bodies. As a result, each body exerts a contact force on other The magnitudes of the contact forces acting on the two bodies are equal but their directions are opposite and hence the contact forces obey Newton’s third law. N=normal force Fc=contact force

f=friction

The direction of the contact force acting on a particular body is not necessarily perpendicular to the contact surface. We can resolve this contact force into two components, one perpendicular to the contact surface and the other parallel to it. The perpendicular component is called the normal contact force or normal force and parallel component is called friction.

f 2 + N2 = N {when fmin = 0}

Contact force = Fc min

Fc max =

µ 2N2 + N2

N ≤ Fc ≤

(µ 2 + 1) N

{when fmax = µN}

0 ≤ λ ≤ tan–1µ

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Page # 42 Ex.7

Sol.

FRICTION

A body of mass 400 g slides on a rough horizontal surface. If the frictional force is 3.0 N, find (a) the angle made by the contact force on the body with the vertical and (b) the magnitude of the contact force. Take g = 10 m/s2. Let the contact force on the block by the surface be Fc which makes an angle λ with the vertical (shown figure) Fc N f The component of Fc perpendicular to the contact surface is the normal force N and the component of F parallel to the surface is the firction f. As the surface is horizontal, N is vertically upward. For vertical equilirbrium, N = Mg = (0.400 kg) (10 m/s2) = 4.0 N The frictional force is f = 3.0 N f 3 = or, λ = tan–1 (3/4) = 37º N 4 (b) The magnituded of the contact force is

(a)

tan λ =

F = N2 + f 2 =

4.

(4.0 N) 2 + (3.0N) 2 = 5.0 N

MOTION ON A ROUGH INCLINED PLANE Suppose a motion up the plane takes place under the action of pull P acting parallel to the plane N = mg cos α P N Frictional force acting down the plane F = µN = µ mg cos α Appling Newton’s second law for motion up the plane mg sin α mg cos α P – (mg sin α + f) = ma f α mg P – mg sin α – µ mg cos α = ma If P = 0 the block may slide downwards with an acceleration a. The frictional force would then act up the plane mg sin α – F = ma or, mg sin α – µ mg cos α = ma

Ex.8

A 20 kg box is gently placed on a rough inclined plane of inclination 30° with horizontal. The coefficient of sliding friction between the box and the plane is 0.4. Find the acceleration of the box down the incline. N Y X F = µN mgsinα

Sol.

mg

O

mgcosα

Y' X' In solving inclined plane problems, the X and Y directions along which the forces are to be considered, may be taken as shown. The components of weight of the box are (i) mg sin α acting down the plane and (ii) mg cos α acting perpendicular to the plane. N = mg cos α mg sin α – µ N = ma ⇒ mg sin α – µ mg cos α = ma a = g sin α – µg cos α = g (sin α – µ cos α)

1 3 = 9.8 2 – 0.4 × 2  = 4.9 × 0.3072 = 1.505 m/s2   The box accelerates down the plane at 1.505 m/s2. 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 43

FRICTION Ex.9

A force of 400 N acting horizontal pushes up a 20 kg block placed on a rough inclined plane which makes an anlge of 45° with the horizontal. The acceleration experienced by the block is 0.6 m/ s2. Find the coefficient of sliding friction between the box and incline.

Sol.

The horizontally directed force 400 N and weight 20 kg of the block are resolved into two mutually perpendicular components, parallel and perpendicular to the plane as shown. N = 20 g cos 45° + 400 sin 45° = 421.4 N The frictional force experienced by the block

a=0.6 m/s

F = µN = µ × 421.4 = 421.4 µN.

2

400 cos45°

R

As the accelerated motion is taking placed up the plane.

400 N

400 cos 45° – 20 g sin 45° – f = 20a 400 2

–

20 × 9.8 2

400 sin45°

20g sin45°

– 421.4 µ = 20a = 20 × 0.6 = 12

45°20 g 20 cos 45°

 400  196 1 282.8 – 138.6 – 12 µ= =– – 12 × = 0.3137 . =  2  4214 2 4214 . The coefficient of sliding friction between the block and the incline = 0.3137

5.

ANGLE OF REPOSE : Consider a rough inclined plane whose angle of inclination θ with ground can be changed. A block of mass m is resting on the plane. Coefficient of (static) friction between the block and plane is µ. For a given angle θ, the FBD (Free body diagram) of the block is

N

f

mg sin

mg cos

Where f is force of static friction on the block. For normal direction to the plane, we have N=mg cosθ As θ increases, the force of gravity down the plane, mg sin θ, increases. Friction force resists the slide till it attains its maximum value. fmax = µN = µ mg cos θ Which decreases with θ (because cos θ decreases as θ increases) Hence, beyond a critical value θ = θc, the blocks starts to slide down the plane. The critical angle is the one when mg sin θ is just equal of fmax, i.e., when mg sin θc = µ mg cosθc or tan θc = µ where θC is called angle of repose If θ > θc, block will slide down.For θ < θc the block stays at rest on the incline.

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Page # 44

6.

FRICTION

TWO BLOCKS ON AN INCLINED PLANE :

2

N

m

1

N

m

Consider two blocks having masses m1 & m2 placed on a rough inclined plane. µ1 & µ2 are the friction coefficient for m1 & m2 respectively. If N is the normal force between the contact surface of m1 & m2

θ

Now three condition arises. (i) If µ1 = µ2 = µ then N = 0 because, Both the blocks are in contact but does not press each other. a1 = a2 = g sin θ – µ mg cos θ (a1,a2 are acceleration of block µ1 & µ2 respectively) (ii) If µ1 < µ2 then N = 0 because, there is no contact between the blocks. a1 = g sin θ – µ1 g cos θ a2 = g sin θ – µ2 g cos θ ⇒ a1 > a2 (iii) If µ1 > µ2 then N≠0 a1 = a2

Ex.10 Mass m1 & m2 are placed on a rough inclined plane as shown in figure. Find out the acceleration of the blocks and contact force in between these surface. m2 2kg

m1 1kg

37°

As we know if µ1 > µ2 both will travel together so a1 = a2 = a F.B.D =3 .2

os gc

m2 ° 37 in s g

m1

f2

=µ

° 37

m2

2

f1 = µ1m1gcos37°=4

° 37 n i gs 37° 1

m2

m

3k g

Sol.

µ2=0.2 µ1=0.5

which is equivalent to a=

3g sin 37°–(f1 + f2 ) 18 – 7.2 a= = 3.6 m/sec2 3 2

Now F.B.D of 1 kg block is

3

f1 + f2

7° n3 37° i gs

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Page # 45

FRICTION

N 3. a=

7° in3 s g

4N 2 c /se m 6

gsin37° + N – 4 = (1) a N = 3.6 + 4 – 6 = 1.6 Newton

RANGE OF FORCE F FOR WHICH ACCELERATION OF BODY IS ZERO.

2k g

F

7.

Ex.11 37º

=8N x

2kg

12 N

m

37º

Now take different value of F Force (F) F.B.D.

fma

F =( 0. 5) m gc o ax

fm

gs in 37 º

2k g

s3 7º

F

Sol.

Find out the range of force in the above situation for which 2kg block does not move on the incline. F.B.D of 2 kg block

Acceleration

Friction Type

Fnet m

8N

a=

Direction & magnitude

N 12

2m/s2

8N

Kinetic

0m/s2

8N

Static

8N

4N

(i) F = 0N

(ii) F = 4N

4N

8N

N 12

0m/s2

(iii) F = 8N

4N

N 12

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Static

Page # 46

f= 0

12 N

FRICTION

0m/s2

(iv) F = 12N

0

Static

16 N

N 12

0m/s2

(v) F = 16N

4N

Static

8N

Static

8N

Kinetic

f= 4N

N 12

20

N

(in this condition friction change its direction to stop relative slipping)

0m/s2

(vi) F = 20N

24

N

8N

N 12

2m/s2

(vii) F = 24N 8N

N 12

From the above table block doesn't move from F = 4N (mgsinθ – µmgcosθ) to F = 20N (mgsinθ + µmgcosθ). So friction develope a range of force for which block doesn't move

: If Friction is not present then only for F = 12N the block will not move but friction develop a range of force 4N to 20N to prevent slipping. So we can write the range of force F for which acceleration of the body is zero. mg sin θ – µmg cos θ ≤ F ≤ mg sin θ + µ mg cosθ .

Ex.12 In the following figure force F is gradually increased from zero. Draw the graph between applied force F and tension T in the string. The coefficient of static friction between the block and the ground is µs.

F

Sol.

M

µs As the external force F is gradually increased from zero it is compensated by the friction and the string beares no tension. When limiting friction is achieved by increasing force F to a value till µs mg, the further increase in F is transferred to the string.

T 45° F µsmg

Ex.13 Fig. shows two blocks tied by a string. A variable force F = 5t is applied on the block. The coefficient of friction for the blocks are 0.6 and 0.5 respectively. Find the frictional force between blocks and ground as well as tension in the string at

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Page # 47

FRICTION (A) t = 1 s

(B) t = 2s B

(C) t = 3s A

1kg

2kg

µ=0.6

µ=0.5

10N

Sol.

F=5t

20N

T fA

T

fB 10N

5t

20N

(a) At t = 1s, F = 5 × 1 = 5 N Maximum value of friction force fA = µN = 0.5 × 20 = 10 N To keep the block stationary the magnitude of frictional force should be 5N. So fA = 5 N Now from the figure it becomes clear that if fA = 5N & F = 5 N, Tension T = 0 Since tension is not in application so frictional force on block B is 0 i.e., fB = 0 (b) At t = 2s, F = 5 × 2 = 10 N Maximum value of friction force f = µN = 0.5 × 20 = 10 N To keep the block stationary the magnitude of friction force should be 10 N. So fA = 10 N From the figure it is clear that if fA = 10 N and F = 10 N Tension T = 0 Hence friction force on block B is fB = 0 (c) At t = 3s, F = 5 × 3 = 15 N Maximum value of friction force f = µN = 0.5 × 20 = 10 Newton Again applying the same analogy fA = 10 N From the figure it is clear that if fA = 10 N and F = 15 N Tension T = 5 N So frictional force on block B is fB = 5 Newton

Ex.14 Find the tension in the string in situation as shown in the figure below. Forces 120 N and 100 N start acting when the system is at rest.

120 N

10 fsmax=90 N

Sol.

20

100N

fsmax=60 N

(i) Let us assume that system moves towards left then as it is clear from FBD, net force in horizontal direction is towards right. Therefore the assumption is not valid.

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Page # 48

FRICTION

120 N

10

100N

20 60 N

90 N

Above assumption is not possible as net force on system comes towards right. Hence system is not moving towards left. (ii) Similarly let us assume that system moves towards right.

120 N

10

100N

20 60 N

90 N

Above assumption is also not possible as net force on the system is towards left in this situation. Hence assumption is again not valid. Therefore it can be concluded that the system is stationary T

10

120 N

20

100N

fmax=60 N

Fmax=90 N

Assuming that the 10 kg block reaches limiting friction first then using FBD’s

10

120 N 120 = T + 90 ⇒ Also

T + f = 100

∴

30 + f = 100 ⇒

T T 90N f

20

100N

T = 30 N f = 70 N

which is not possible as the limiting value is 60 N for this surface of block.

∴ Our assumption is wrong and now taking the 20 kg surface to be limiting we have 10

120 N

Also

T T f 60 N

T + 60 = 100 N ⇒

T = 40 N

f + T = 120 N ⇒

f = 80 N

20

100N

This is acceptable as static friction at this surface should be less than 90 N. Hence the tension in the string is

T = 40 N

8.

PULLEY BLOCK SYSTEM INVOLVING FRICTION :

•

If friction force is acting and value of acceleration of a particle is negative, then it means direction of friction force is opposite to that what we assumed and acceleration would be having a different numerical value.

Ex.15 Two blocks of masses 5 kg and 10 kg are attached with the help of light string and placed on a rough incline as shown in the figure. Coefficients of friction are as marked in the figure. The system is released from rest. Determine the acceleration of the two blocks.

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Page # 49

FRICTION

10kg

5kg

50°

37°

Fixed

Sol.

Let 10 kg block is sliding down, then acceleration of both the blocks are given by,

10g sin 37°– µ1 × 10g cos 37°–5g sin 53° – µ 2 × 5g cos 53°

a=

15

= – ve

It means our assumed direction of motion is wrong and 5 kg block is going to slide down, if this would be the case, the direction of friction force will reverse and acceleration of blocks would be given by :

5g sin 53°– µ 2 × 5g cos 53°– µ1 × 10g cos 37° – 10g sin 37°

a1 =

= –ve

15

It means in this direction also there is no motion. So we can conclude that the system remains at rest and friction force is static in nature.

9.

TWO BLOCK SYSTEM :

B 2kg

Ex.16

A

F

4kg frictionless

Find out the maximum value of F for which both the blocks will move together Sol.

In the given situation 2kg block will move only due to friction force exerted by the 4 kg block

F.B.D. B 2kg f

A

4kg

f F

The maximum friction force exerted on the block B is fmax = µN fmax = (0.5) (20) = 10 N So the maximum acceleration of 2 kg block is a 2 max

10 = = 5m / s 2 2

2kg

fmax = 10N

amax is the maximum acceleration for which both the block will move together. i.e., for a ≤ 5 ms–2 acceleration of both blocks will be same and we can take both the blocks as a system.

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Page # 50

FRICTION

F.B.D a

5m/s

6kg

2

Fmax

Fmax = 6 × 5 = 30 N for 0 < F < 30 Both the block move together.

Ex.17 In the above question find the acceleration of both the block when (i) F = 18 N (ii) F = 36 N (i) Since F < 30 both the blocks will move together Sol. F.B.D

6kg a=

F = 18N

18 = 3 m / s2 6

(ii) When F = 36 N When F > 30 both the blocks will move separately so we treat each block independently F.B.D of 2 kg block

B 2kg

f = 10N (Friction force)

aB = 5 m/s2 F.B.D of 4 kg block

f = 10N

aA =

A

4kg

F = 36N

36 − 10 26 = m / s2 4 4

B 2kg Ex.18

Sol.

A

4kg

F

Find out the range of force in which both the blocks move together If f1 is friction force between block A & lower surface and f2 is friction force between both the block’s surface. F.B.D f2=10N B 2kg f2=10N F 6N = f1 A 4kg f1 max = µ1N1 = (0.1) (60) = 6 N f2 max = µ2N2 = (0.5) (20) = 10 N Upper 2kg block is move only due to friction force so maximum acceleration of that block is 2kg

f2 =10N

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Page # 51

FRICTION amax =

10 = 5 m / s2 2

This is the maximum acceleration for which both the blocks will move together. Therefore for a ≤ 5ms–2 we can take both the blocks as one system. F.B.D. 2 5m/s 6kg

f1=6N

F

For F < 6 N. Blocks will not move at all. Now the value of Fmax for which both the blocks will move together. Fmax – 6 = 6 × 5 Fmax = 36 N Conclusion if 0N < F < 6N No blocks will move 6N < F < 36 N Both blocks will move together F > 36 N Both move separately.

Ex.19

B 2kg

F

A 4kg frictionless

The lower block A will move only due to friction force F.B.D.

2kg

f

4kg

F f(frictional force)

fmax = µN = (0.5) (20) = 10 N

F.B.D. of 4 kg blocks

amax

4kg

f = 10N

The maximum acceleration of 4 kg block is ⇒

amax =

10 = 2.5 m / s 2 4

This is the maximum acceleration for which both the blocks move together 2.5 m/s2

6kg

F

Fmax for which both the blocks will move together Fmax = 2.5 × 6 = 15 N

B 2kg Ex.20

F

A 4kg

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Page # 52

FRICTION

If f2 is the friction force between A & B and f1 is the friction force between A & floor f1 max = 6 N f2 max = 10 N Lower block A will move only due to friction force So amax for 4 kg block

f1 =6N

4kg

f2 = 10N

10 − 6 = 1 m / s2 4 This is the maximum acceleration for which both the blocks will move together amax =

1m/s f1 =6N

2

6kg

F

F–6=6×1 F = 12 N If F is less than 6N both the blocks will be stationary Conclusion : 0 < F < 6 N = Both blocks are stationary 6 N < F < 12 N = Both move together F > 12 N = Both move separately

Ex.21 Find the accelerations of blocks A and B for the following cases. (A) µ1 = 0 and µ2 = 0.1 (P) aA = aB = 9.5 m/s2 (B) µ2 = 0 and µ1 = 0.1 (Q) aA = 9 m/s2, aB = 10 m/s2 (C) µ1 = 0.1 and (R) aA = aB = g = 10 m/s2 µ2 = 1.0 (D) µ1 = 1.0 and (S) aA = 1, aB = 9 m/s2 µ2 = 0.1 Sol. (a) R, (b) Q, (c) P, (d) S (i) FBD in (case (i)) {µ1 = 0, µ2 = 0.1} µ2N O

1kg

N=10

A

1kg B

µ1

1 kg A

µ2

1kg

10 N

B

N=10

µ2N mg

mg

While friction’s work is to oppose the relative motion and here if relative motion will start then friction comes and without relative motion there is no friction so both the block move together with same acceleration and friction will not come.

A

B

mg

mg

2

⇒ aA = aB = 10 m/s

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Page # 53

FRICTION

0

1 (ii)

1

B 10 1kg

A 10 1kg 10 0

10

Friction between wall and block A oppose relative motion since wall is stationary so friction wants to slop block A also and maximum friction will act between wall and block while there is no friction between block.

: Friction between wall and block will oppose relative motion between wall and block only it will not do anything for two block motion.

1 A

B

10

10 2

aA = 9 m/s ; aB = 10 m/s2

f

1 A

(iii)

10

B f

10

10

Friction between wall and block will be applied maximum equal to 1N but maximum friction available between block A and B is 10 N but if this will be there then relative motion will increase while friction is to oppose relative motion. So friction will come less than 10 so friction will be f that will be static. 1 f

A

B

f 10

10

by system (20–1) = 2 × a ⇒ a = (iv)

19 =9.5 m/s2 2

aB =

10 – 1 = 9 m / s2 1

1

10

11 – 10 = 1m / s 2 aA = 1

A 10

10

B 1 10

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Page # 54

10.

FRICTION

FRICTION INVOLVING PSEUDO CONCEPT :

Ex.22 What is the minimum acceleration with which bar A should be shifted horizontally to keep the bodies 1 and 2 stationary relative to the bar ? The masses of the bodies are equal and the coefficient of friction between the bar and the bodies equal to µ. The masses of the pulley and the threads are negligible

1

A

2

while the friction in the pulley is absent. see in fig. Sol.

Let us place the observer on A. Since we have non-inertial frame we have pseudo forces. For body ‘1’ we have, T = ma + µmg

ma

....(1)

T

1 mg

For body ‘2’ we have,

µN

N = ma

From (1) and (2) amin

ma

A

mg – T – µma = 0 ∴ mg = T + µmg

....(2)

T

2

N

mg

 1− µ  = g   1+ µ 

a

µ=0.5

F

Ex.23

M=4kg m= 1kg

Find out the range of force for which smaller block is at rest with respect to bigger block. Sol.

Smaller block is at rest w.r.t. the bigger block. Let both the block travel together with acceleration a F.B.D of smaller block w.r. to the bigger block.

f

fmax = µ × N N = ma f = µ ma ⇒ f = mg

ma (Pseudo)

...(1) ...(2)

N

from (1) & (2)

mg

a = g/µ = 20 m/s2 So F = 20 (M + m) = 20 (5) = 100 N If F ≥ 100 N Both will travel together

Ex.24 The rear side of a truck is open and a box of 40 kg mass is placed 5m away from the open end as shown. The coefficient of friction between the box & the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 ms–2 . At what distance from the starting point does the box fall off the truck (i.e. distance travelled by the truck) ? [Ignore the size of the box]

/////////////////////////////////////////

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Page # 55

FRICTION

Sol.

In the reference frame of the truck FBD of 40 kg block

µN

40 kg

ma (psuedo force)

Net force ⇒ ma – µN ⇒ 40 × 2 – mablock ⇒ 80 – 60 ⇒ ablock =

15 × 40 × 10 100

20 1 = m/s2 40 2

This acceleration of the block in reference frame of truck so time taken by box to fall down from truck Srel = urelt +

1 1 1 a t2 ⇒ 5 = 0 + × × t2 ⇒ t2 = 20 2 rel 2 2

So distance moved by the truck ⇒

1 × atruck × t2 2

⇒

1 × 2 × (20) = 20 meter.. 2

Ex.25 Mass m2 placed on a plank of mass m1 lying on a smooth horizontal plane. A horizontal force F = α0t (α0 is a constant) is applied to a bar. If acceleration of the plank and bar are a1 and a2 respectively and the coefficient of friction between m1 and m2 is µ. Then find acceleration a with time t.

m2

F

m1 Sol.

If F < µm2g then both blocks move with common acceleration, i.e., a1 = a2 When F > µm2g, then Equation for block of mass m

and

F – µm2g = m2a2

...(1)

µm2g = m1a1

...(2)

a2

From equation (1)

its slope positive and intercept negative. From equation (2) a1 is independent of time.

=a

2

a1

1

i.e., acceleration a2 varies with time linearly,

a

a

α0t – µm2g = m2a2

0

t0

t

So, the graph between a & t is as follow.

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Page # 56

N.L.M.

(Objective Problems)

Exercise - I (A) NEWTON'S LAW OF MOTION 1. At a given instant, A is moving with velocity of 5 m/s upwards. What is velocity of B at the time

3. The pulleys in the diagram are all smooth and light. The acceleration of A is a upwards and the acceleration of C is f downwards. The acceleration of B is

A

C

A B

(A) 15 m/s↓ Sol.

(B) 15 m/s↑

(C) 5 m/s ↓ (D) 5 m/s ↑

(A) 1/2 (f – a) up (C) 1/2 (a + f) up Sol.

B (B) 1/2 (a + f) down (D) 1/2 (a – f) up

4. If acceleration of A is 2 m/s2 to left and acceleration of B is 1 m/s2 to left, then acceleration of C is A

2. Find the velocity of the hanging block if the velocities of the free ends of the rope are as indicated in the figure. 2m/s

C 1m/s

(A) 3/2 m/s ↑ (C) 1/2 m/s ↑ Sol.

B

2

(A) 1 m/s upwards (C) 2 m/s2 downwards Sol.

(B) 1 m/s2 downwards (D) 2 m/s2 upwards

(B) 3/2 m/s ↓ (D) 1/2 m/s ↓

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Page # 57

N.L.M.

4 m/s

6 m/s 6m/s

5. In the figure shown the velocity of different blocks is shown. The velocity of C is

7. Find velocity of block 'B' at the instant shown in figure.

A B C D

(A) 6 m/s (C) 0 m/s Sol.

(B) 4 m/s (D) none of these

37°

A

(A) 25 m/s (C) 22 m/s Ans.

10 m/s

(B) 20 m/s (D) 30 m/s

6. If block A has a velocity of 0.6 m/s to the right, determine the velocity of block B.

A B (A) 1.8 m/s in downward direction (B) 1.8 m/s in upward direction (C) 0.6 m/s in downward direction (D) 0.6 m/s in upward direction Sol.

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B

Page # 58

N.L.M.

8.In the arrangement shown in fig. the ends P and Q of an unstretchable string move downwards with uniform speed U. Pulleys A and B are fixed. Mass M moves upwards with a speed. B

A

θ

θ

Q

P

M

(A) 2 U cos θ (C)

(B) U cos θ

2U cos θ

(D)

U cos θ

Sol.

10. The velocity of end ‘A’ of rigid rod placed between two smooth vertical walls moves with velocity ‘u’ along vertical direction. Find out the velocity of end ‘B’ of that rod, rod always remains in constant with the vertical walls.

A 'u' 9. Block B moves to the right with a constant velocity v0. The velocity of body A relative to B is :

v0

B (A) u tan 2θ (C)u tan θ Sol.

(B) u cot θ (D) 2u tan θ

B

A

(A)

v0 , towards left 2

(B)

v0 , towards right 2

(C)

3v0 , towards left 2

(D)

3v0 , towards right 2

Sol.

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Page # 59

N.L.M. 11. Find the acceleration of C w.r.t. ground.

A

C

a

Sol.

B b

(A) ai
– (2a + 2b)
j

(B) a ˆi – (2a + b)ˆj

(C) a ˆi – (a + 2b)ˆj Sol.

(D) b ˆi – (2a + 2b)ˆj

13. The 50 kg homogeneous smooth sphere rests on the 30° incline A and bears against the smooth vertical wall B. Calculate the contact forces at A and B. A

B

30°

(A) NB = (B) NA = (C) NA = (D) NA =

N, N = A

3 1000

N, N = B

3 100 3

N, N = B

1000 3

N, N = B

Sol.

12. Find the acceleration of B.

A

1000

B

a

acos α 1 (A) cos α 2

(B)

a sin α1 a cos α 2 (C) cos α 2 cos α1

(D)

cos α1 cos α 2

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500 3 500 3 500 3 50 3

N N N N

Page # 60

N.L.M. 15. A sperical ball of mass m = 5 kg rests between two planes which make angles of 30° and 45° respectively with the horizontal. The system is in equilibrium. Find the normal forces exerted on the ball by each of the planes. The planes are smooth.

30 °

45°

14. Find out the reading of the weighing machine in the following cases.

g 2k

M

W 30º

(A) 10 3 Sol.

(B) 10 2

W

2k g

(A) N45 = 96.59 N, N30 = 136.6 N (B) N30 = 96.59 N, N45 = 136.6 N (C) N45 = 136.6 N, N30 = 96.56 N (D) none of these Sol.

M

30º

(C) 20 3

(D) 30 3

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Page # 61

N.L.M. Question No. 16 to 17 (2 questions) In the figure the tension in the diagonal string is 60 N.

18. A mass M is suspended by a rope from a rigid support at A as shown in figure. Another rope is tied at the end B, and it is pulled horizontally with a force θ with the vertical in equilibrium, then the tension in the string AB is : F .

F1

45°

I f

t h

e

r o

p

e

A

B

m

a

k

e

 16. Find the magnitude of the horizontal force F1 and  F2 that must be applied to hold the system in the

position shown. 3

N

(B)

20 2

N

(C)

n

θ

W

60

a

40 2

a

n

g

l e

A

F3

(A)

s

N

(D)

60 2

B

F

M

(A) F sin θ Sol.

(B) F/sin θ

(C) F cos θ (D) F/cos θ

N

Sol.

17. In the above questions what is the weight of the suspended block ? (A) Sol.

60 2

N

(B)

40 2

N

(C)

60 3

N

(D)

50 2

N 19. Objects A and B each of mass m are connected by light inextensible cord. They are constrained to move on a frictionless ring in a vertical plane as shown in figure. The objects are released from rest at the positions shown. The tension in the cord just after release will be A T T

B

mg C

mg

(A) mg 2

(B)

mg 2

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(C)

mg 2

(D)

mg 4

Page # 62

N.L.M. 21. Two masses m and M are attached to the strings as shown in the figure. If the system is in equilibrium, then

Sol.

45°

45°

M

m

2M m 2M (C) cot θ = 1 + m Sol. (A) tan θ = 1 +

2m M 2m (D) cot θ = 1 + M (B) tan θ = 1 +

20. Three blocks A, B and C are suspended as shown in the figure. Mass of each blocks A and C is m. If system is in equilibrium and mass of B is M, then :

A

(A) M = 2m Sol.

B

(B) M < 2 m

C

(C) M > 2m (D) M = m

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Page # 63

N.L.M. 22. A flexible chain of weight W hangs between two fixed points A & B which are at the same horizontal level. The inclination of the chain with the horizontal at both the points of support is θ. What is the tension of the chain at the mid point ? A

Sol.

B θ

θ

W

W .cos ec θ 2 W .cot θ (C) 2 Sol. (A)

(B)

W .tan θ 2

(D) none

24.A stunt man jumps his car over a crater as shown (neglect air resistance) (A) during the whole flight the driver experiences weightlessness (B) during the whole flight the driver never experiences weightlessness (C) during the whole flight the driver experiences weightlessness only at the highest point

(D) the apparent weight increases during upward journey Sol. 23.A weight can be hung in any of the following four ways by string of same type. In which case is the string most likely to break ?

(A)

W

(B)

(C)

W

W (D)

W (A) A

(B) B

(C) C

(D) D

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Page # 64 Question No. 25 to 27 (3 questions) A particle of mass m is constrained to move on x-axis. A force F acts on the particle. F always points toward the position labeled E. For example, when the particle is to the left of E, F points to the right. The magnitude of F is constant except at point E where it is zero.

N.L.M. Sol.

A

+ve x E m The system is horizontal. F is the net force acting on the particle. The particle is displaced a distance A towards left from the equilibrium position E and released from rest at t = 0

25. What is the period of the motion ?  2Am   (A) 4  F  

 2Am   (B) 2  F   27. Find minimum time it will take to reach from x = −

 2 Am   (C)  F  

(D) None

A 2

to 0. mA ( 2 − 1) F mA ( 2 − 1) (C) 2 F Sol.

Sol.

(A)

3 2

(B)

mA ( 2 − 1) F

(D) none

26. Velocity-time graph of the particle is v

v t

(A)

v

v

(C)

t

(B)

t

(D)

t

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Page # 65

N.L.M. 28. A particle of mass 50 gram moves on a straight line. The variation of speed with time is shown in figure. find the force acting on the particle at t = 2, 4 and 6 seconds.

Sol.

v(m/s)

15 10 5 0

2

4

6

8 t(s)

(A) 0.25 N along motion, zero, 0.25 opposite to motion (B) 0.25 N along motion, zero, 0.25 along to motion (C) 0.25 N opposite motion, zero, 0.25 along to motion (D) 0.25 N opposite motion, zero, 0.25 opposite to motion Sol.

30. A constant force F is applied in horizontal direction as shown. Contact force between M and m is N and between m and M’ is N’ then

F

M'>M M m M'

(A) N or N’ equal (C) N’ > N Sol.

smooth (B) N > N’ (D) cannot be determined

29. Two blocks are in contact on a frictionless table. One has mass m and the other 2m. A force F is applied on 2m as shown in the figure. Now the same force F is applied from the right on m. In the two cases respectively, the ratio force of contact between the two block will be :

2m m (A) same

(B) 1 : 2

F

F

2m m

(C) 2 : 1

(D) 1 : 3

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Page # 66

N.L.M. 33. Force exerted by support on string. (A) 10 N (B) 15 N (C) 20 N Sol.

Question No. 31 to 33 (3 questions) A block of mass 1kg is suspended by a string of mass 1 kg, length 1m as shown in figure. (g = 10 m/ s2) Calculate :

(D) 25 N

34. A body of mass 8 kg is hanging another body of mass 12 kg. The combination is being pulled by a string –2 . The tension T1 and T2 will be respectively : (use g = 9.8 m/s2) w

i t h

a

n

a

c c e

l e

r a

t i o

n

o

f

2

. 2

m

s

T1

12kg 1m

T2 1 kg

a

8kg

31. The tension in string at its lowest point. (A) 10 N (B) 15 N (C) 20 N (D) 25 N Sol.

32. The tension in string at its mid-point (A) 10 N (B) 15 N (C) 20 N Sol.

(A) 200 N, 80 N (C) 240 N, 96 N Sol.

(B) 220 N, 90 N (D) 260 N, 96 N

(D) 25 N

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Page # 67

N.L.M. 35. A rope of mass 5 kg is moving vertically in vertical position with an upwards force of 100 N acting at the upper end and a downwards force of 70 N acting at the lower end. The tension at midpoint of the rope is (A) 100 N (B) 85 N (C) 75 N (D) 105 N Sol.

37. Two blocks, each having mass M , rest on frictionless surfaces as shown in the figure. If the pulleys are light and frictionless, and M on the incline is allowed to move down, then the tension in the string will be

M

fixed

M

(A)

2 Mgsin θ 3

(B)

(C)

Mgsin θ 2

(D) 2 Mg sin θ

Sol.

36. A particle of small mass m is joined to a very heavy body by a light string passing over a light pulley. Both bodies are free to move. The total downward force in the pulley is (A) mg (B) 2 mg (C) 4 mg (D) can not be determined Sol.

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3 Mgsin θ 2

Page # 68 38. The pulley arrangements shown in figure are identical the mass of the rope being negligible. In case I, the mass m is lifted by attaching a mass 2m to the other end of the rope. In case II, the mass m is lifted by pulling the other end of the rope with cosntant downward force F = 2mg, where g is acceleration due to gravity. The acceleration of mass in case I is

N.L.M. 39. Two masses M1 and M2 are attached to the ends of a light string which passes over a massless pulley attached to the top of a double inclined smooth plane of angles of inclination α and β . The tension in the string is :

F=2mg

m 2m

m

(I)

(II)

(A) zero (B) more than that in case II (C) less than that in case II (D) equal to that in case II Sol.

α

fixed

M2

M1

β

(A)

M2 (sin β) g M1 + M2

(B)

M1(sin α ) g M1 + M2

(C)

M1 M2 (sin β + sin α ) g M1 + M2

(D) zero

Sol.

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N.L.M. 40. Calculate the acceleration of the block B in the above figure, assuming the surfaces and the pulleys P1 and P2 are all smooth and pulleys and string and light 2m

F

A

Page # 69 41. In previous Question surface is replaced by block C of mass m as shown in figure. Find the acceleration of block B. 2m

4m

4m P1

F

B

P1

A

c

P2

B P2

(A) a =

3F m/s2 17m

(B) a =

2F m/s2 17m

3F (A) a = 20m m/s2

(B) a =

(C) a =

3F m/s2 15m

(D) a =

3F m/s2 12m

2F (C) a = 21m m/s2

3F (D) a = 18m m/s2

Sol.

Sol.

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3F m/s2 21m

Page # 70 42. In the arrangement shown in the fig, the block of mass m = 2 kg lies on the wedge on mass M = 8 kg. Find the initial acceleration of the wedge if the surfaces are smooth and pulley & strings are massless.

N.L.M. 43. In the arrangement shown in figure, pulleys are massless and frictionless and threads are inextensible. The Block of mass m1 will remain at rest, if

P1

m

60°

M

P2

60°

(A) a =

30 3 m/s2 23

30 2 (C) a = m/s2 23 Sol.

(B) a =

20 3 m/s2 23

(D) none of these

m1 m2

1 1 1 (A) m = m + m 1 2 3 4 1 1 (C) m = m + m 1 2 3

m3

(B) m1 = m2 + m3 (D)

1 2 3 = + m3 m2 m1

Sol.

44. Both the blocks shown here are of mass m and are moving with constant velocity in direction shown in a resistive medium which exerts equal constant force on both blocks in direction opposite to the velocity. The tension in the string connecting both of them will be (Neglect friction) 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 71

N.L.M. Sol.

A B (A) mg Sol.

(B) mg/2

(C) mg/3

(D) mg/4

47. Block of 3 kg is initially in equilibrium and is hanging by two identical springs A and B as shown in figures. If spring A is cut from lower point at t = 0 then, find acceleration of block in ms–2 at t = 0. 45. A monkey of mass 20 kg is holding a vertical rope. The rope can break when a mass of 25 kg is suspended from it. What is the maximum acceleration with which the monkey can climb up along the rope? (A) 7 ms–2 (B) 10 ms–2 (C) 5 ms–2 (D) 2.5 ms–2 Sol.

46. Fi nd the accel eration of 3 kg mass when acceleration of 2 kg mass is 2 ms–2 as shown in figure. 3 kg

2 kg

10N

–2

(A) 3 ms–2

(B) 2 ms–2

2ms (C) 0.5 ms–2

(D) zero

A

B 3 kg

(A) 5 Sol.

(B) 10

(C) 15

(D) 0

48. Two masses of 10 kg and 20 kg respectively are connected by a massless spring as shown in figure. A force of 200 N acts on the 20 kg mass at the instant when the 10 kg mass has an acceleration of 12 ms–2 towards right, the aceleration of the 20 kg mass is : 10kg

20kg 200 N

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Page # 72 (A) 2 ms Sol.

–2

N.L.M. (B) 4 ms

–2

(C) 10 ms

–2

–2

(D) 20 ms

50. A man of mass 60 kg is standing on a weighing machine placed in a lift moving with velocity 'v' and acceleration 'a' as shown in figure. Calculate the reading of weighing machine in following situation: (g = 10 m/s2) (i) a = 0, v=0 (A) 600 N (B) 500 N (C) 450 N (D) 700 N (ii) a = 0, v = 2m/s (A) 600 N (B) 500 N (C) 450 N (D) 700 N (iii) a = 0, v = –2m/s (A) 450 N (B) 500 N (C) 600 N (D) 700 N 49. Two blocks are connected by a spring. The combination is suspended, at rest, from a string attatched to the ceiling, as shown in the figure. The string breaks suddenly. Immediately after the string breaks, what is the initial downward acceleration of the upper block of mass 2m ?

2m

m

(A) 0 Sol.

(B) 3g/2

(C) g

(D) 2g

a

v

W.M.

(iv) a = 2m/s2, v=0 (A) 600 N (B) 500 N (C) v=0 (v) a = –2m/s2 (A) 600 N (B) 480 N (C) (vi) a = 2m/s2, v = 2m/s (A) 600 N (B) 480 N (C) (vii) a = 2 m/s2, v = –2m/s (A) 600 N (B) 720 N (C) (viii) a = –2m/s2 v = –2 m/s (A) 600 N (B) 480 N (C) Sol.

450 N

(D) 720 N

450 N

(D) 700 N

450 N

(D) 720 N

450 N

(D) 700 N

450 N

(D) 700 N

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Page # 73

N.L.M.

Question No. 52 to 54 (3 questions) An object of mass 2 kg is placed at rest in a frame (S1) moving with velocity 10
i + 5
j m/s and having 51. What will be the reading of spring balance in the figure shown in following situations. (g = 10 m/s2) (i) a = 0, v = 0 (A) 100 N (B) 80 N (C) 120 N (D) 150 N (ii) a = 0, v = 2 m/s (A) 100 N (B) 80 N (C) 120 N (D) 150 N (iii) a = 0, v = – 2m/s (A) 100 N (B) 80 N (C) 120 N (D) 150 N a

acceleration 5
i + 10
j m / s 2 . The object is also seen by an observer standing in a frame (S2) moving with velocity 5
i + 10
j m / s 52. Calculate 'Pseudo force' acting on object. Which frame is responsible for this force. (A) F = – 10
i – 20
j due to acceleration of frame S1 (B) F = – 20
i – 20
j due to acceleration of frame S1 (C) F = – 10
i – 30
j due to acceleration of frame S1 (D) none of these Sol.

v

M = 10 kg

(iv) a = 2 m/s2, v =0 (A) 100 N (B) 80 N (C) 120 N (v) a = – 2m/s2, v = 0 (A) 100 N (B) 80 N (C) 120 N (vi) a = 2 m/s2, v = 2 m/s (A) 100 N (B) 80 N (C) 120 N (vii) a = 2 m/s2, v = –2m/s (A) 100 N (B) 80 N (C) 120 N (viii) a = – 2 m/s2, v = – 2 m/s (A) 100 N (B) 80 N (C) 120 N Sol.

(D) 150 N (D) 150 N (D) 150 N (D) 150 N (D) 150 N

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Page # 74 53. Calculate net force acting on object with respect to S2 frame. (B) F = 10
i + 20
j (A) F = 20ˆi + 20ˆj (C) F = 5ˆi + 20 ˆj Sol.

N.L.M.

(D) F = 10 ˆi + 5 ˆj

54. Calculate net force acting on object with respect of S1 frame. (A) 0 (B) 1 (C) 2 (D) none of these Sol.

56. A block of mass m resting on a wedge of angle θ as shown in the figure. The wedge is given an acceleration a. What is the minimum value of a so that the mass m falls freely ?

A m

a θ

B (A) g Sol.

(B) g cos θ

(C) g cot θ

C (D) g tan θ

gs in θ

55. A trolley is accelerating down an incline of angle θ with acceleration gsinθ. Which of the following is correct. (α is the constant angle made by the string with vertical)

α

m

θ

(A) α = θ (B) α = 0º (C) Tension in the string, T = mg (D) Tension in the string, T = mg sec θ Sol.

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Page # 75

N.L.M. 57. In the figure the reading of the spring balanced will be : [g = 10 m/s2] 2m/s

10kg

Sol.

2

5kg

30°

(A) 50 N Sol.

(B) 40 N

(C) 60 N

(D) 70 N

(B) FRICTION 59. Find the direction of friction forces on each block and the ground (Assume all surfaces are rough and all velocities are with respect to ground). E D C

58. A pendulum of mass m hangs from a support fixed to a trolley. The direction of the string when the trolley rolls up of plane of inclination α with acceleration a0 is (String and bob remain fixed with respect to trolley)

a0

B

5 m/s

Sol.

θ

α

(A) θ = tan–1α

 a0  (B) θ = tan–1    g

 g (C) θ = tan–1  a   0

–1  a 0 + g sin α   (D) θ = tan   g cos α 

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A

2 m/s 1 m/s 3 m/s 5 m/s

Page # 76

N.L.M. 61. A monkey of mass m is climbing a rope hanging from the roof with acceleration a. The coefficient of static friction between the body of the monkey and the rope is µ. Find the direction and value of friction force on the monkey.

(A) Upward, F = m(g + a) (B) downward, F = m(g + a) (C) Upward, F = mg (D) downward, F = mg Sol.

60. In the following figure, find the direction of friction on the blocks and ground

VA=3m/s 7N VB=6m/s

A

F = 5N

B

Sol.

62. A body is placed on a rough inclined plane of inclination θ. As the angle θ is increased from 0º to 90º the contact force between the block and the plane (A) remains constant (B) first remains constant then decreases (C) first decreases then increases (D) first increases then decreases

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Page # 77 64. If force F is increasing with time and at t=0, F=0 where will slipping first start ?

N.L.M. Sol.

F

3 2 1

(A) between 3 kg and 2 kg (B) between 2 kg and 1 kg (C) between 1 kg and ground (D) both (A) and (B) Sol.

 63. A force F =
i + 4
j acts on block shown. The force of friction acting on the block is F

y 1 Kg x (A) −
i Sol.

(B) − 18 .
i

(C) − 2.4
i

(D) − 3
i

65. A wooden block of mass m resting on a rough µ) is pulled by a force F as shown in figure. The acceleration of the block moving horizontally is : h

o

r i z o

n

t a

l

t a

b

l e

( c o

e

f f i c i e

n

t

o

f

f r i c t i o

n

=

F θ

m

(A)

F cos θ m

(B)

(C)

F (cos θ + µ sin θ) – µg m

(D) none

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µF sin θ M

Page # 78

N.L.M.

Sol.

Sol.

66. In the arrangement shown in the figure, mass of the block B and A is 2m and m respectively. Surface between B and floor is smooth. The block B is connected to the block C by means of a string pulley system. If the whole system is released, then find the minimum value of mass of block C so that block A remains stationary w.r.t. B. Coefficient of friction between A and B is µ.

67. Two masses A and B of 10 kg and 5 kg respectively are connected with a string passing over a fricionless pulley fixed at the corner of a table as shown in figure. The coefficient of friction of A with the table is 0.2. The minimum mass of C that may be placed on A to prevent if from moving is equal to :

A

C 10kg A

B 5 kg B

C (A)

m µ

(B)

2m + 1 µ +1

(C)

3m µ −1

(D)

6m µ +1

(A) 15 kg

(B) 10 kg

(C) 5 kg

(D) zero

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Page # 79

N.L.M.

69. A body of mass m moves with a velocity v on a surface whose friction coefficient is µ. If the body covers a distance s then v will be :

Sol.

(A)

2µgs (B) µgs

(C)

µgs / 2

(D)

3µgs

Sol.

68. If the coefficient of friction between an insect and bowl is µ and the radius of the bowl, is r, the maximum height to which the insect can crawl in the bowl is : 1+ µ2

 1  r (B) 1 – 1+ µ2 

(C) r 1 + µ 2

(D) r 1 + µ 2 – 1

r

(A)

Sol.

70. With what minimum velocity should block be projected from left end A towards end B such that it reaches the other end B of conveyer belt moving with constant velocity v. Friction coefficient between block and belt is µ.

A

    (A)

µgL (B)

m

v0

B

v

2µgL (C)

Sol.

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L 3µgL

(D) 2 µgL

Page # 80

N.L.M.

71. A box 'A' is lying on the horizontal floor of the compartment of a train running along horizontal rails from left to right. At time 't', it decelerates. Then the reaction R by the floor on the box is given best by A

(A)

R

R

(B)

floor

A floor

R

R

(C)

A

A

(D)

floor

floor

Sol.

73. A small mass slides down an inclined plane of θ with the horizontal. The co-efficient of friction is µ = µ0x where x is the distance through which the mass slides down and µ0, a constant. Then the distance covered by the mass before it stops is : i n

c l i n

a

t i o

n

2 4 1 1 (A) µ tan θ (B) µ tan θ (C) 2µ tan θ (D) µ tan θ 0 0 0 0

Sol. 72. A block of mass m lying on a rough horizontal plane is acted upon by a horizontal force P and another force Q inclined an at an angle θ to the vertical. The minimum value of coefficient of friction between the block and the surface for which the block will remain in equilibrium is :

Q θ

P /////////////////////////////////////// (A)

P + Q sin θ mg + Q cos θ

(B)

P cos θ + Q mg – Q sin θ

(C)

P + Q cos θ mg + Q sin θ

(D)

P sin θ – Q mg – Q cos θ

Sol.

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Page # 81

N.L.M. 74. In the above question the speed of the mass when travelled half the maximum distance is (A)

g tan θ sin θ µ0

(B)

g tan θ sin θ 2µ 0

(C)

g tan θ sin θ 8µ 0

(D) none of these

Sol.

76. A body is moving down a long inclined plane of slope 37º. The coefficient of friction between the body and plane varies as µ = 0.3 x, where x is distance travelled down the plane. The body will have maximum 3 and g = 10 m/s2) speed. (sin 37º = 5 (A) at x = 1.16 m (B) at x = 2m (C) at bottom of plane (D) at x = 2.5 m Sol.

75. For the equilibrium of a body on an inclined plane of inclination 45º. The coefficient of static friction will be (A) greater than one (B) less than one (C) zero (D) less than zero Sol.

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Page # 82 77. Block B of mass 100 kg rests on a rough surface of friction coefficient µ = 1/3. A rope is tied to block B as shown in figure. The maximum acceleration with which boy A of 25 kg can climbs on rope without making block move is

(A) Sol.

4g 3

(B)

g 3

(C)

g 2

(D)

N.L.M. 78. Starting from rest a body slides down a 45° inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is : (A) 0.75 (B) 0.33 (C) 0.25 (D) 0.80 Sol.

3g 4

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N.L.M. 79. A block of mass 5 kg and surface area 2 m2 just begins to slide down an inclined plane when the angle of inclination is 30°. Keeping mass same, the surface area of the block is doubled. The angle at which this starts sliding down is : (A) 30° (B) 60° (C) 15° (D) none Sol.

Page # 83 81. A block placed on a rough inclined plane of inclination (θ = 30º) can just be pushed upwards by applying a force “F” as shown. If the angle of inclination of the inclined plane is increased to (θ = 60º), the same block can just be prevented from sliding down by application of a force of same magnitude. The coefficient of friction between the block and the inclined plane is

F

(A) (C)

3 +1 3 −1 3 −1 3 +1

Sol. 80. Two blocks of masses m1 and m2 are connected with a massless unstretched spring and placed over a plank moving with an acceleration 'a' as shown in figure. the coefficient of friction between the blocks and platform is µ. m1

m2 a

(A) spring will be stretched if a > µg (B) spring will be compressed if a ≤ µg (C) spring will neither be compressed nor be stretched for a ≤ µg (D) spring will be in its natural length under all conditions. Sol.

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(B)

2 3 −1 3 +1

(D) none of these

oo

Q 3m

h

Sm

N.L.M.

ug

P

Ro

m

th

Page # 84 82. A fixed wedge with both surface inclined at 45° to the horizontal as shown in the figure. A particle P of mass m is held on the smooth plane by a light string which passes over a smooth pulley A and attached to a particle Q of mass 3m which rests on the rough plane. The system is released from rest. Given that g the acceleration of each particle is of magnitude 5 2 then A

45° fixed 45°

(a) the tension in the string is : 6mg mg (A) mg (B) (C) 5 2 2 Sol.

(D)

mg 4

(c) In the above question the magnitude and direction of the force exerted by the string on the pulley is : (A)

6mg downward 5

(B)

6mg upward 5

(C)

mg downward 5

(D)

mg downward 4

Sol.

(b) In the above question the coefficient of friction between Q and the rough plane is : (A)

4 5

(B)

1 5

(C)

3 5

(D)

2 5

Sol.

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N.L.M. 83. A force F = t is applied to block A as shown in figure. The force is applied at t = 0 seconds when the system was at rest and string is just straight without tension. Which of the following graphs gives the friction force between B and horizontal surface as a function a time 't'. B A m

m

f

For Q.84. to Q.88 refer given figure (5 questions) 84. When F = 2N, the frictional force between 5 kg block and ground is 10kg

F

5kg

F

µs = µk

µ s > µk

Page # 85

(A) 2N Sol.

(B) 0

(C) 8 N

(D) 10 N

f

(A)

(B) t

t

f

f

(C)

(D) t

t

Sol.

85. When F = 2N, the frictional force between 10 kg block and 5 kg block is (A) 2N (B) 15N (C) 10N (D) None Sol.

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Page # 86 86. The maximum “F” which will not cause motion of any of the blocks (A) 10N (B) 15N (C) data insufficient (D) None Sol.

N.L.M. 89. A truck starting from rest moves with an acceleration of 5 m/s2 for 1 sec and then moves with constant velocity. The velocity w.r.t. ground v/s time graph for block in truck is (Assume that block does not fall off the truck) µ = 0.2

5 m/s

3 m/s

(A)

(B)

1 sec

87. The maximum acceleration of 5 kg block (B) 3 m/s2 (C) 0 (D) None (A) 1 m/s2 Sol.

1 sec

5 m/s (C)

(D) None of these

2.5 sec Sol.

88. The acceleration of 10 kg block when F = 30 N (B) 3 m/s2 (C) 1 m/s2 (D) None (A) 2 m/s2 Sol.

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N.L.M. 90. A board is balanced on a rough horizontal semicircular log. Equilibrium is obtained with the help of addition of a weight to one of the ends of the board when the board makes an angle θ with the horizontal. Coefficient of friction between the log and the board is

Page # 87 91. A stationary body of mass m is slowly lowered onto a massive plateform of mass M (M >> m) moving at a speed V0 = 4 m/s as shown in fig. How far will the body slide along the platform (µ = 0.2 and g = 10 m/ s2 ) ?

(A) 4 m Sol. (A) tan θ Sol.

(B) cos θ

(C) cot θ

(B) 6 m

(D) sin θ

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(C) 12 m

(D) 8 m

Page # 88

N.L.M.

(One or more than one option is correct)

Exercise - II (A) N. L. M

1. A student calculates the acceleration of m1 in figure (m1 – m 2 )g shown as a1 = m + m . Which assumption is not 1 2 required to do this calculation.

3. Two men of unequal masses hold on to the two sections of a light rope passing over a smooth light pulley. Which of the following are possible?

m1 m2 (A) pulley is frictionless (C) pulley is massless Sol.

(B) string is massless (D) string is inextensible

2. Which graph shows best the velocity-time graph for an object launched vertically into the air when air resistance is given by |D| = bv? The dashed line shows the velocity graph if there were no air resistance.

v

v t

(A)

t

(B)

v (C)

(A) The lighter man is stationary while the heavier man slides with some acceleration (B) The heavier man is stationary while the lighter man climbs with some acceleration (C) The two men slide with the same acceleration in the same direction (D) The two men move with accelerations of the same magnitude in opposite directions Sol.

v t

(D)

t

Sol.

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Page # 89

N.L.M. 4. Adjoining figure shows a force of 40 N acting at 30° to the horizontal on a body of mass 5 kg resting on a smooth horizontal surface. Assuming that the acceleration of free-fall is 10 ms–2, which of the following statements A, B, C, D, E is (are) correct?

5 kg

6. In the system shown in the figure m1 > m2. System is held at rest by thread BC. Just after the thread BC is burnt :

40 N 30°

[1] The horizontal force acting on the body is 20 N [2] The weight of the 5 kg mass acts vertically downwards [3] The net vertical force acting on the body is 30 N (A) 1, 2, 3 (B) 1, 2 (C) 2 only (D) 1 only Sol.

spring k

B m2

m1 A C

(A) acceleration of m2 will be upwards (B) magnitude of acceleration of both blocks will be  m1 – m 2  equal to  m + m  g  1 2 (C) acceleration of m1 will be equal to zero (D) magnitude of acceleration of two blocks will be non-zero and unequal. Sol.

5. For ordinary terrestrial experiments, which of the following observers below are inertial. (A) a child revolving in a “giant wheel”. (B) a driver in a sports car moving with a constant high speed of 200 km/h on a straight road. (C) the pilot of an aeroplane which is taking off. (D) a cyclist negotiating a sharp turn. Sol. 7. A particle is resting on a smooth horizontal floor. At t = 0, a horizontal force starts acting on it. Magnitude of the force increases with time according to law F = α.t, where α is a constant. For the figure shown which of the following statements is/are correct ? y

O

2 1

x

(A) Curve 1 shows acceleration against time (B) Curve 2 shows velocity against time (C) Curve 2 shows velocity against acceleration (D) none of these

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Page # 90

N.L.M.

Sol.

9. Figure shows the displacement of a particle going along the x-axis as a funtion of time :

x

E

A

8. Two blocks A and B of equal mass m are connected through a massless string and arranged as shown in figure. Friction is absent everywhere. When the system is released from rest.

B

C

D t

(A) the force acting on the particle is zero in the region AB (B) the force acting on the particle is zero in the region BC (C) the force acting o the particle is zero in the region CD (D) the force is zero no where. Sol.

A fixed

30°

mg 2 mg (B) tension in string is 4 (C) acceleation of A is g/2 3 (D) acceleration of A is g 4 Sol.

B

(A) tension in string is

10. A force of magnitude F1 acts on a particle so as to accelerate it from rest to a velocity v. The force F1 is then replaced by another force of magnitude F2 which decelerates it to rest. (A) F1 must be the equal to F2 (B) F1 may be equal to F2 (C) F1 must be unequal to F2 (D) None of these Sol.

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Page # 91

N.L.M. 11. In the figure, the blocks A, B and c of mass m each have acceleration a1. a2 and a3 respectively. F1 and F2 are external forces of magnitudes 2 mg and mg respectively.

m F1=2mg A

m B

2m

m C

Sol.

F2=mg

(A) a1 = a2 = a3 (B) a1 > a2 > a3 (C) a1 = a2, a2 > a3 (D) a1 > a2 , a2 = a3 Sol.

12. A rope is stretched between two boats at rest. A sailor in the first boat pulls the rope with a constant force of 100 N. First boat with the sailor has a mas of 250 kg whereas mass of second boat is double of that mass. If the initial distance between the boats was 100 m, the time taken for two boats to meet each other is -

(A) 13.8 s

(B) 18.3 s

(C) 3.18 s

(D) 31.8 s

13. A chain of length l is placed on a smooth spherical surface of radius r with one of its ends fixed at the top of the surface. Length of chain is assumed to be πr l< . Acceleration of each element of chain when 2 upper end is released is -

(A)

g  r  1– cos  r  

(B)

rg    1– cos  r 

(C)

g    1– sin  r  r

(D)

rg  r  1– sin   

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Page # 92

N.L.M.

Sol.

(B) FRICTION 15. A block of mass 2.5 kg is kept on a rough horizontal surface. It is found that the block does not slide if a horizontal force less than 15 N is applied to it. Also it is if found that it takes 5 second to slide throughout the first 10 m if a horizontal force of 15 N is applied and the block is gently pushed to start the motion. Taking g = 10 m/s2, then (A) µs = 0.60 (B) µk = 0.52 (C) µk = 0.60 (D) µs = 0.52 Sol.

14. Five persons A, B, C, D & E are pulling a cart of mass 100 kg on a smooth surface and cart is moving with acceleration 3 m/s2 in east direction. When person ‘A’ stops pulling, it moves with acceleration 1 m/s2 in the west direction. When person ‘B’ stops pulling, it moves with acceleration 24 m/s2 in the north direction. The magnitude of acceleration of the cart when only A & B pull the cart keeping their directions same as the old directions, is : (A) 26 m/s2 (C) 25 m/s2 Sol.

(B) 3 71 m / s 2 (D) 30 m/s2

16. The coefficient of friction between 4 kg and 5 kg blocks is 0.2 and between 5 kg block and ground is 0.1 respectively. Choose the correct statements P

4 kg

Q

5 kg

F

(A) Minimum force needed to cause system to move is 17N (B) When force is 4N static friction at all surfaces is 4 N to keep system at rest. (C) Maximum acceleration of 4 kg block is 2 m/s2 (D) Slipping between 4 kg and 5 kg blocks start when F is 17 N 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 93

N.L.M. Sol.

Sol.

17. In a tug-of-war contest, two men pull on a horizontal rope from opposite sides. The winner will be the man who (A) exerts greater force on the rope (B) exerts greater force on the ground (C) exerts a force on the rope which is greater than the tension in the rope (D) makes a smaller angle with the vertical Sol.

Q

u

e

s

t i o

n

N

o

.

1

9

t o

2

0

(

2

q

u

e

s

t i o

n

s

)

In figure, two blocks M and m are tied together with an inextensible and light string. The mass M is placed on a rough horizontal surface with coefficient of friction µ and the mass m is hanging vertically against a smooth vertical wall. The pulley is frictionless. M Rough Smooth

18. A man pulls a block heavier than himself with a light horizontal rope. The coefficient of friction is the same between the man and the ground, and between the block and the ground.

m

19. Choose the correct statement(s) (A) The system will accelerate for any value of m (B) The system will accelerate only when m > M (C) The system will accelerate only when m > µM (D) Nothing can be said Sol.

(A) The block will not move unless the man also moves (B) The man can move even when the block is stationary (C) If both move, the acceleration of the man is greater than the acceleration of the block (D) None of the above assertions is correct

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Page # 94 20. Choose the correct statement(s) related to the tension T in the string (A) When m < µM, T = mg (B) When m < µM, T = Mg (C) When m > µM, µMg < T < mg (D) When m > µM, mg < T < µMg Sol.

Question No. 21 to 23 (3 questions) Imagine a situation in which the horizontal surface of block M0 is smooth and its vertical surface is rough with a coefficient of friction µ

N.L.M. Sol.

22. In above problem, choose the correct value(s) of F which the blocks M and m remain stationary with respect to M0 g µ

(B)

(C) (M0 + M + m)

mg M

(D) none of these

Smooth M F

Mo

m

m(M0 + M + m)g M – µm

(A) (M0 + M + m)

Sol.

Rough

21. Identify the correct statement(s) (A) If F=0, the blocks cannot remain stationary (B) For one unique value of F, the blocks M and m remain stationary with respect to M0 (C) The limiting friction between m and M 0 is independent of F (D) There exist a value of F at which friction force is equal to zero.

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Page # 95

N.L.M.

24. The value(s) of mass m for which the 100 kg block remains is static equilibrium is

100

23. Consider a special situation in which both the faces of the block M0 are smooth, as shown in adjoining figure. Mark out the correct statement(s) (A) 35 kg Sol.

Smooth M

F

M0

µ = 0.3

37°

(B) 37 kg

m Smooth

(A) If F = 0, the blocks cannot remain stationary (B) For one unique value of F, the blocks M and m remain stationary with respect to block M0 (C) There exists a range of F for which blocks M and m remain stationary with respect to block M0 (D) Since there is no friction, therefore, blocks M and m cannot be in equilibrium with respect to M0 Sol.

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m

(C) 83 kg

(D) 85 kg

Page # 96

N.L.M. 26. Car is accelerating with acceleration = 20 m/s2. A box that is placed inside the car, of mass m = 10 kg is put in contact with the vertical wall as shown. The friction coefficient between the box and the wall is µ = 0.6. µ = 0.6 10kg

20m/s2

(A) The acceleration of the box will be 20 m/sec2 (B) The friction force acting on the box will be 100 N (C) The contact force between the vertical wall and

25. The contact force exerted by one body on another body is equal to the normal force between the bodies. It can be said that : (A) the surface must be frictionless (B) the force of friction between the bodies is zero (C) the magnitude of normal force equals that of friction (D) It is possible that the bodies are rough and they do not slip on each other. Sol.

the box will be 100 5 N (D) The net contact force between the vertical wall and the box is only of electromagnetic in nature. Sol.

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Page # 97

N.L.M.

(SUBJECTIVE

Exercise - III

PROBLEMS)

1. Two masses A and B, lie on a frictionless table. They are attached to either end of a light rope which passes around a horizontal movable pulley of negligible mass. Find the acceleration of each mass MA = 1 kg, MB = 2 kg, MC = 4 kg. The pulley P2 is vertical.

P1

B

P2

A

C

Sol. 3. To Paint the side of a building, painter normally hoists himself up by pulling on the rope A as in figure. The painter and platform together weigh 200 N . The rope B can withstand 300 N. Find

2. Block A of mass m/2 is connected to one end of light rope which passes over a pulley as shown in the fig. Man of mass m climbs the other end of rope with a relative acceleration of g/6 with respect to rope find acceleration of block A and tension in the rope.

(a) the maximum acceleration of the painter. (b) tension in rope A (i) when painter is at rest (ii) when painter moves up with an acceleration 2 m/s2. Sol.

g/6 m

m/2 Sol.

A

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Page # 98 4. An inclined plane makes an angle 30º with the horizontal. A groove OA = 5 m cut in the plane makes an angle 30º with OX. A short smooth cylinder is free to slide down the influence of gravity. Find the time taken by the cylinder to due to reach from A to O. (g = 10 m/s2)

Sol.

37º

30°

30°

Sol.

x

5. Same spring is attached with 2 kg, 3 kg and 1 kg blocks in three different cases as shown in figure. If x1, x2 and x3 be the extensions in the spring in these three cases then find the ratio of their extensions.

2 kg

2 kg

(a)

2.5 kg

Spring balance M

er ind A cyl

O

N.L.M. 6. Find the reading of spring balance as shown in figure. Assume that mass M is in equilibrium. (All surfaces are smooth)

3 kg

2 kg

(b)

7. At what value of m1 will 8 kg mass be at rest.

8kg

1 kg

2 kg

(c) 5kg

Sol.

m1

Sol.

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Page # 99

N.L.M. 8. What force must man exert on rope to keep platform in equilibrium ?

10. Find force in newton which mass A exerts on mass B if B is moving towards right with 3 ms–2. Also find mass of A. (All surfaces are smooth) A

2

3m/s

1kg B 37º

Sol.

Sol.

9. Inclined plane is moved towards right with an acceleration of 5ms–2 as shown in figure. Find force in newton which block of mass 5 kg exerts on the incline plane. (All surfaces are smooth) 5kg

37º 5 m/s2

Sol.

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Page # 100 11. Force F is applied on upper pulley. If F = 30t where t is time in seconds. Find the time when m1 loses contact with floor.

N.L.M.

F 30t N

m1 m2 m1 = 4kg m2 = 1kg

13. The vertical displacement of block A in meter is given by y = t2/4 where t is in second. Calculate the downward acceleration aB of block B.

Sol.

A

y

B

Sol.

12. The 40 kg block is moving to the right with a speed of 1.5 m/s when it is acted upon by forces F1 & F2. These forces vary in the manner shown in the graph. Find the velocity of the block after t = 12 s Neglect friction and masses of the pulleys and cords. F(N)

F2

40

F2

F1

30

F1

20 10

t(s) 0

2

4

6

12

Sol.

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Page # 101

N.L.M. 14. An object of mass m is suspended in equilibrium using a string of length l and a spring having spring constant K (< 2 mg/l) and unstreched length l/2.

Sol.



m

(a) Find the tension in the string (b) What happens if K > 2 mg / l ?

16. A person of mass m is standing on a platform of mass M and wants to raise this platform. Massless pulleys are configured in two different ways as shown. We would like to know which configuration makes it easier to raise the platform. Answer the following questions in terms of m, M, a and constant as appropriate. [Note : Assume the rope is also massless and does not stretch.]

m

M

M

Fig(1)

15. Three monkeys A, B, and C with masses of 10, 15 & 8 kg respectively are climbing up & down the rope suspended from D. at the instant represented, A is descending the rope with an acceleration of 2 m/s2 & C is pulling himself up with an acceleration of 1.5 m/s2. Monkeys B is climbing up with a constant speed of 0.8 m/s. Treat the rope and monkeys as a complete system & calculate the tension T in the rope at D. (g = 10 m/s–2)

m Fig(2)

(a) For configuration (1) find the force, F, the person must exert straight up in order to accelerate the platform + person system with an acceleration a. Include a freebody diagram in your solution. (b) What force does the platform exert on the person when the acceleration of the system is a? Include a freebody diagram in your solution. (c) If platform is massless, M = 0, and he wants to raise it with a constant velocity find F. Does this configuration offer a mechanical advantage ? (That is, is F < mg ?) (d) Now repeat the above for configuration (2). First, find the force, F, the person must exert straight down in order to accelerate the platform+ person system with an upward acceleration a. Include a freebody diagram in your solution. (e) Now, what force does the platform exert on the

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Page # 102

N.L.M.

person when the acceleration of the system is a? Include a freebody diagram in your solution. (f) Again, if the platform is massless, M = 0, and he wants to raise it with a constant velocity find F. Does this configuration offer a mechanical advantage ? (That is, is F < mg?) Sol.

(B) FRICTION 17. Give the acceleration of blocks : 50N

µ

= 0.5

(A) µ ks = 0.4 5kg

40N

(B)

µ s = 0.5 µ k = 0.4 10kg

37°

µ s = µ k = 0.6 10kg

(C) 5kg

Sol.

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N.L.M. 18. Determine the coefficient of friction (µ), so that rope of mass m and length l does not slide down.

l/3

Page # 103 20. A rope so lies on a table that part of it lays over. The rope begins to slide when the length of hanging part is 25 % of entire length. The co-efficient of friction between rope and table is : Sol.

Sol.

21. A worker wishes to pile a cone of sand into a circular area in his yard. The radius of the circle is r, and no sand is to spill onto the surrounding area. If µ is the static coefficient of friction between each layer of sand along the slope and the sand, the greatest volume of sand that can be stored in this manner is : Sol.

19. A wooden block A of mass M is placed on a frictionless horizontal surface. On top of A, another lead block B also of mass M is placed. A horizontal force of magnitude F is applied to B. Force F is increased continuously from zero. Then draw the graph between A and F. [µk < µs] B A

F

Sol.

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Page # 104

N.L.M.

22. A block of mass 15 kg is resting on a rough inclined plane as shown in figure. The block is tied up by a horizontal string which has a tension of 50 N. The coefficient of friction between the surfaces of contact is (g = 10 m/s2) T

24. A block of mass 1 kg is horizontally thrown with a velocity of 10 m/s on a stationary long plank of mass 2 kg whose surface has µ = 0.5. Plank rests on frictionless surface. Find the time when m1 comes to rest w.r.t. plank. Sol.

m horizontal

45°

Sol.

25. Block M slides down on frictionless incline as shown. Find the minimum friction coefficient so that m does not slide with respect to M.

m M

23. In the figure, what should be mass m so that block A slide up with a constant velocity.

37º Sol.

A g 1k 37º

m

=0.5

Sol.

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Page # 105

N.L.M. 26. The coefficient of static and kinetic friction between the two blocks and also between the lower block and the ground are µs = 0.6 and µk = 0.4. Find the value of tension T applied on the lower block at which the upper block begins to slip relative to lower block.

M = 2kg M = 2kg

T

Sol.

28. A body of mass 2kg rests on a horizontal plane having coefficient of friction µ = 0.5. At t = 0 a horizontal  force F is applied that varies with time F = 2t. The time constant t0 at which motion starts and distance moved in t = 2t0 second will be _______ and ________ respectively. Sol.

27. A thin rod of length 1 m is fixed in a vertical position inside a train, which is moving horizontally with constant acceleration 4 m/s2. A bead can slide on the rod, and friction coefficient between them is 1/2. If the bead is released from rest at the top of the rod, find the time when it will reach at the bottom. Sol.

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Page # 106

N.L.M.

29. Find the acceleration of the blocks and magnitude & direction of frictional force between block A and table, if block A is pulled towards left with a force of 50N.

F = 50N

A

x

5Kg

g = 10m/s2

B 4Kg

Sol.

31. Coefficient of friction between 5 kg and 10 kg block is 0.5. If friction between them is 20N. What is the value of force being applied on 5 kg. The floor is frictionless.

5kg

F

10kg Sol.

30. A block A of mass 2kg rests on another block B of mass 8kg which rests on a horizontal floor. The coefficient of friction between A and B is 0.2 while that between B and floor is 0.5. When a horizontal force F of 25N is applied on the block B, the force of friction between A and B is _________.

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Page # 107

N.L.M.

(TOUGH

Exercise - IV

1. The diagram shows particles A and B, of masses 0.2 kg and m kg respectively, connected by a light inextensible string which passes over a fixed smooth peg. The system is released from rest, with B at a height of 0.25m above the floor. B descends, hitting the floor 0.5s later. All resistances to motion may be ignored.

SUBJECTIVE PROBLEMS)

2. An ornament for a courtyard at a word’s fair is to be made up of four identical, frictionless metal sphere, each weighing 2 6 Newton. The spheres are to be arranged as shown, with three resting on a horizontal surface and touching each other; the fourth is to rest freely on the other three. The bottom three are kept from separating by spot welds at the points of contact with each other. Allowing for a factor of satety of 3N, how much tension must the spot welds with stand.

A 0.2 kg

B

m kg 0.25m

(a) Find the acceleration of B as it descends. (b) Find the tension in the string while B is descending and find also the value of m. (c) When B hits the floor it comes to rest immediately, and the string becomes slack. Find the length of time for which B remains at rest on the ground before being jerked into motion again. Sol.

Top View

Sol.

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Horizontal View

Page # 108

N.L.M.

3. A 1kg block ‘B’ rests as shown on a bracket ‘A’ of same mass. Constant foroes F1 = 20 N and F2 = 8N start to act at time t = 0 when the distance of block B from pulley is 50 cm. Time when block B reaches the pulley is _______.

Sol.

50cm F2

F1

B

A

Sol.

5. In figure shown, pulleys are ideal m1 > 2 m2. Initially the system is in equilibrium and string connecting m2 to rigid support below is cut. Find the initial acceleration of m2?

m2 m1

4. Two men of masses m1 and m2 hold on the opposite ends of a rope passing over a frictionless pulley. The mass m1 climbs up the rope with an acceleration of 1.2 m/s2 relative to the rope. The man m2 climbs up the rope with an acceleration of 2.0 m/s2 relative to the rope. Find the tension in the rope if m1 = 40 kg and m2 = 60 kg. Also find the time after which they will be at same horizontal level if they start from rest and are initially separated by 5m.

Sol.

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Page # 109

N.L.M. 6. The system shown adjacent is in equilibrium. Find the acceleration of the blocks A, B & C all of equal masses m at the instant when (Assume springs to be ideal) (a) The spring between ceiling & A is cut.

7. In the system shown. Find the initial acceleration of the wedge of mass 5M. The pulleys are ideal and the cords are inextensible. (there is no friction anywhere). M 5M

K A

2M

Sol.

B K C

(b) The string (inextensible) between A & B is cut. (c) The spring between B & C is cut. Also find the tension in the string when the system is at rest and in the above 3 cases. Sol.

8. A smooth right circular cone of semi vertical angle α = tan–1(5/12) is at rest on a horizontal plane. A rubber ring of mass 2.5 kg which requires a force of 15N for an extension of 10cm is placed on the cone. Find the increase in the radius of the ring in equilibrium. Sol.

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Page # 110

N.L.M. 10. A car begins to move at time t = 0 and then accelerates along a straight track with a speed given 2 ms–1 for 0 ≤ t ≤ 2 After the end of acceleration, the car continues to move at a constant speed. A small block initially at rest on the floor of the car begins to slip at t = 1sec. and stops slipping at t = 3sec. Find the coefficient of static and kinetic friction between the block and the floor. Sol. b

y

V

( t )

=

2

t

9. A block of mass m lies on wedge of mass M as shown in figure. Answer following parts separately. (a) With what minimum acceleration must the wedge be moved towards right horizontally so that block m falls freely. m

M

(b) Find the minimum friction coefficient required between wedge M and ground so that it does not move while block m slips down on it. Sol.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 111

N.L.M. 11. In the figure shown, (i) For what maximum value of force F can all these blocks move together. (ii) Find the value of force F at which sliding starts at other rough surfaces F

1kg 2kg 3kg

12. A particle having a mass m and velocity Vm in the y-direction is projected on to a horizontal belt that is moving with uniform velocity Vb in the x-direction as shown in figure. µ is the coefficient of friction between particle and belt. Assuming that the particle first touches the belt at the origin of the fixed xy coordinate system and remains on the belt, find the coordinates (x, y) of the point where the sliding stops. y

(iii) Find acceleration of all blocks, nature and value of friction force of force F = 18N. Sol.

vb

belt

x m

Sol.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 112

N.L.M.

Exercise - V

(JEE-PROBLEMS)

1. A spring of force constant k is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of (A) (2/3) k (B) (3/2) k (C) 3k (D) 6K [JEE 1999] Sol.

2. In the figure masses m1, m2 and M are 20 kg, 5 kg and 50 kg respectively. The co-efficient of friction between M and ground is zero. The co-efficient of friction between m1 and M and that between m2 and ground is 0.3. The pulleys and the string are massless. The string is perfectly horizontal between P1 and m1 and also between P2 and m2. The string is perfectly vertical between P1 and P2. An external horizontal force F is applied to the mass M. Take g = 10 m/s2. P1 m2

P2

m1 M

F

(a) Draw a free - body diagram for mass M, clearly showing all the forces. (b) Let the magnitude of the force of friction between m1 and M be f1 and that between m2 and ground be f2. For a particular F it is found that f1 =2f2. Find f1 and f2. Write down equations of motion of all the masses. Find F, tension in the string and accelerations of the masses. [JEE 2000] Sol.

3. The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle θ should be [JEE (Scr) 2001]

B

A

P m (A) 0°

(B) 30°

1/2

2 m (C) 45°

Q m (D) 60°

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 113

N.L.M. 5. A block of mass

Sol.

3 kg is placed on a rough horizontal

surface whose coefficient of friction is 1 / 2 3 minimum value of force F (shown in figure) for which the block starts to slide on the surface. (g = 10m/s2)

60°

(A) 20 N

(B) 20 3 N

(C) 10 3 N Sol.

(D) None of these

4. A string of negligible mass going over a clamped pulley of mass m supports a block of mass M as shown in the figure. The force on the pulley by the clamp is given [JEE (Scr) 2001]

M

(A)

2 Mg

(B)

2 mg

(C)

(M + m)2 + m2 g

(D)

(M + m) 2 + M2 g

Sol.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 114

N.L.M.

6. Two blocks A and B of equal masses are released from an inclined plane of inclination 45° at t = 0. Both the blocks are initially at rest. The coefficient of kientic friction between the block A and the inclined plane is 0.2 while it is 0.3 for block B. Initially, the block A is

Sol.

2 m behind the block B. When and where their front faces will come in line. [Take g = 10m/s2].

2m

A

B

45°

A

B

Sol.

7. Two blocks A and B masses 2m and m, respectively, are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in the figure. The magnitudes of acceleration of A and B, immediately after the string is cut, are respectively. [JEE 2006]

8. A circular disc with a groove along its diameter is placed horizontally. A block of mass 1 kg is placed as shown. The co-efficient of friction between the block and all surfaces of groove in contact is µ = 2/5. The disc has an acceleration of 25 m/s 2 . Find the acceleration of the block with respect to disc. [JEE 2006]

a=25m/s2 2m A m B

(A) g, g

(B) g, g/2

(C) g/2, g

(D) g/2, g/2

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 115

N.L.M. Sol.

Sol.

9. Two particles of mass m each are tied at the ends of a light string of length 2a. The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance ‘a’ from the center P (as shown in the figure). Now, the midpoint of the string is pulled vertically upwards with a small but constant force F. As a result, the particles move towards each other on the surfaces. The magnitude of acceleration, when the separation between them becomes 2x, is [JEE 2007] F

m

a F a (A) 2m 2 a − x2

(C)

F x 2m a

m

P

10. STATEMENT-1 A cloth Covers a table. Some dishes are kept on it. The cloth can be pulled out without dislodging the dishes from the table because STATEMENT-2 For every action there is an equal and opposite reaction (A) Statement-1 is True, Statement-2 is True; State me nt -2 i s a correc t ex pl anat i on 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 [JEE 2007] Sol.

a F x (B) 2m 2 a − x2

(D)

F a2 − x2 2m x

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Page # 116

N.L.M.

11. STATEMENT-1 It is easier to pull a heavy object than to push it on a level ground. and STATEMENT-2 The magnitude of frictional force depends on the nature of the two surfaces in contact. (A) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 (B) Statement-1 i s true, statement-2 is true’ statement-2 is NOT a correct explanati on for statement-1 (C) Statement-1 is true, statement-2 is false (D) Statement-1 is false, statement-2 is true [ JEE 2008] Sol.

13. A block of mass m is on an inclined plane of angle θ. The coefficient of friction betwen the block and the plane is µ and tan θ > µ. The block is held stationary by applying a force P parallel to the plane. The direction of force pointing up the plane is taken to the positive. As P is varied from P = mg (sin θ – µ cos θ ) to Pz = mg (sin θ + µ cos θ), the frictional force f versus P graph will look like

P

θ

f

f

P2

(A)

P1

P

(B)

f

12. A block of base 10 cm × 10 cm and height 15 cm is kept on an inclined plane. The coefficient of friction between them is 3 . The inclination θ of this inclined plane from the horizontal plane is gradually increased from 0º. Then (A) at θ = 30º, the block will start sliding down the plane (B) the block will remain at rest on the plane up to certain θ and then it will topple (C) at θ = 60º, the block will start sliding down the plane and continue to do so at higher angles (D) at θ = 60º, the block will start sliding down the plane and on further increasing θ, it will topple at certain θ [JEE 2009] Sol.

P1

P2

P1

P2

f

P1

(C)

P

P

P2

P

(D)

[JEE 2010] Sol.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

N.L.M. 14. A block is moving on an inclined plane making an angle 45º with horizontal and the coefficient of friciton is µ. the force required to just push it up the inclined plane is 3 times the force requried to just prevent it from sliding down. If we define N = 10µ, then N is [JEE 2011]

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 117

Page # 118

N.L.M.

EXERCISE - I

ANSWER KEY

(A) NEWTONS'S LAW OF MOTION 1.

A

2.

A

3.

A

4.

A

5.

B

6.

A

7.

A

8.

D

9.

B

10.

C

11.

A

12.

A

13.

B

14.

A

15.

A

16.

D

17.

A

18.

B

19.

B

20.

B

21.

A

22.

C

23.

C

24.

A

25.

A

26.

A

27.

B

28.

A

29.

B

30.

B

31.

A

32.

B

33.

C

34.

C

35.

B

36.

C

37.

C

38.

C

39.

C

40.

A

41.

B

42.

A

43.

C

44.

B

45.

D

46.

B

47.

A

48.

B

49.

B

50.

(i)

A

(ii)

A

(iii)

C

(iv)

D

(v)

B

(vi)

D

(vii)

B

(viii)

B

(i)

A

(ii)

A

(iii)

A

(iv)

C

(v)

B

(vi)

C

(vii)

C

(viii)

B

A

53.

B

54.

A

55.

A

56.

C

57.

C

58.

D

51.

52.

(B) FRICTION

fDC

fED 59.

E

fED

2 m/s

1 m/s

D

3 m/s

C

fDC

fCS fBA

fCB A

5 m/s

5 m/s

B

fAg

fAg

fBA fkAB

fkAB

60.

fkAB

61.

A

62.

B

63.

A

64.

C

65.

C

66.

C

67.

A

68.

B

69.

A

70.

B

71.

C

72.

A

73.

A

74.

A

75.

A

76.

D

77.

B

78.

A

79.

A

80.

D

81.

C

82. (a) B (b) D (c) A

83.

A

84.

A

85.

A

86.

A

87.

C

88.

90.

A

91.

A

A

A

89.

C

B

fkBG

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 119

N.L.M.

EXERCISE - II

ANSWER KEY

(A) NEWTONS'S LAW OF MOTION

1.

C

7. 13.

2.

B

3.

A,B,D

4.

C

5.

B

6.

A,C

A,B,C 8.

B,D

9.

A,B,C

10.

B

11.

B

12.

B

B

C

14.

(B) FRICTION

15.

A,B

16.

C

17.

B

18.

A,B,C

19.

C

20.

A,C

21.

A,D

22.

B,C

23.

A,B

24.

B,C

25.

B,D

26.

A,B,C,D

EXERCISE - III

ANSWER KEY

(A) NEWTON'S LAW OF MOTION 1.

4 g 2g 3 g , , 5 5 5

2. a =

4g 13mg ,T= 9 18

3. (a) 5m/s2, (b) (i) 100N, (ii) 120N

4. 2 sec

5. x2 > x1 > x3 x1 : x2 : x3 : 15 : 18 : 10

6. 12 N

7. 10/3 kg

8. 300 N

9. 55

11. 2sec

12. 12 m/s

13. aB = 4m/s2 ( ↑ )

10. 5N, 16/31 kg 14. (a) T = mg –

kl , (b) length of spring will less than ‘l’ and T = 0 in the string. 2

15. 322 N 16. (a) T =

(m + M)( a + g) (m + M)(a + g) mg , (b) N = m(a + g) + T, (c) T = , (d) T = 2 3 2

(e) N = m (a + g) – T,

(f) T =

mg 3

(B) FRICTION

17. (A)4m/s2, (B) 1.2 m/s2, (C) 0 a

18. µ = 2

aB aA

1 µ π r3 3

20. 0.33

21.

24. 4/3 sec

25. 3/4

26. 40 N

29. 10
i

30. 0

31. 30 N

19.

22. 1/2

23. 1kg

27. 1/2 sec

28. 5 sec, 125/6 m

F

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 120

N.L.M.

EXERCISE - IV

ANSWER KEY 1. (a) 2 ms–2, (b) 2.4 N 0.3 (c) 0.2 s  m1 − 2m 2  g 5.  2m   2

6. (a) a A =

(c) a A = aB = g / 2 ↑ ; T =

7. 2g/23

4. 556.8 N , 1.47 sec

8. ∆ r =

mg cot α 4 π 2k

, 1cm

m sin θ cos θ

10. µs = 0.4 , µk = 0.3

m cos 2 θ + M

12. x =

11. 12 N, 21 N, 4 m/s2, 2 m/s2, 4 N, 6 N

Vb V Vm2 + Vb2 y = m Vm2 + Vb2 2µg 2µg

EXERCISE - V

ANSWER KEY

1. B

3. 0.5 sec

3g ↓ = aB ; a C = 0 ; T = mg / 2 (b) a A = 2g ↑ ; aB = 2g ↓ ; aC = 0, T = 0 2

3mg ; T = 2mg 2

9. (a) a = g cotθ, (b) µ min =

2. 2 N

2. (b) a = 3/5 m/s2, T = 18 N, F = 60 N

6. 11.313 m

7. B

8. 10 m/s

12. B

13. A

14. 5N

2

3. C

4. D

9. B

10. B

5. A 11. B

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR MOTION & W.P.E THEORY AND EXERCISE BOOKLET CONTENTS

TOPIC

S. NO.

PAGE NO.

1.

Circular Motion ........................................................................ 3

2.

Kinematics of Circular Motion .............................................. 3 – 10

3.

Dynamics of circular Motion ................................................ 11 – 12

4.

Simple Pendulam ................................................................... 13

5.

Circular Motion in Horizontal Plane ......................................... 13

6.

Motion of motorcyclist on a curved ...................................... 14 – 16 path

7.

Circular Turning on roads .................................................... 16 – 17

8.

Death Well .............................................................................. 18

9.

Motion of a cyclist on a circular path .................................. 18 – 19

10.

Effect of earth rotation on Apparent .................................... 19 – 20 weight

11.

Some Solved Examples ..................................................... 20 – 23

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CIRCULAR MOTION

Page # 2

WORK POWER ENERGY 1. Work ................................................................................ 24–25 2. Units of Work ...................................................................... 25 3 Work Done by multiple forces ......................................... 25–27 4. Work Done by a variable force ....................................... 27 – 28 5. Area under force Displacement curve ................................. 28 6. Internal work .................................................................... 28 – 29 7. Conservative force .......................................................... 29 – 31 8. Non-conservative forces .................................................. 31–32 9. Energy ............................................................................ 32 – 33 10. Conservative force and Potential .................................. 34 – 35 energy 11. Work energy Theorem ................................................... 35–41 12. Power ........................................................................... 41 – 43 13. Vertical Circular Motion ................................................ 43 – 48 14. Exercise - I.................................................................... 49 – 68 15. Exercise - II ................................................................... 69 – 73 16. Exercise - III .................................................................. 74 – 84 17. Exercise - IV ................................................................. 85 – 89 18. Exercise- V ................................................................... 90 – 97 19. Answer key .................................................................. 98 – 100

IIT - JEE Syllbabus : Circular Motion (uniform and non-uniform), Work, Power, Kinetic Energy, Potential Energy, Conservation of Mechanical Energy.

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CIRCULAR MOTION

1.

Page # 3

CIRCULAR MOTION When a particle moves in a plane such that its distance from a fixed (or moving) point remains constant then its motion is called as the circular motion with respect to that fixed (or moving) point. That fixed point is called centre and the distance between fixed point and particle is called radius. v v v

A θ1 θ2

B The car is moving in a straight line with respect to the man A. But the man B continuously rotate dθ =0 his face to see the car. So with respect to man A dt dθ ≠0 But with respect to man B dt Therefore we conclude that with respect to A the motion of car is straight line but for man B it has some angular velocity

2.

KINEMATICS OF CIRCULAR MOTION :

2.1

Variables of Motion :

(a)

Angular Position : The angle made by the position vector with given line (reference line) is called angular position Circular motion is a two dimensional motion or motion in a plane. Suppose a particle P is moving in a circle of radius r and centre O. The position of the particle P at a given instant may be described by the angle θ between OP and OX. This angle θ is called the angular position of the particle. As the particle moves on the circle its angular position θ change. Suppose the point rotates an angle ∆θ in

Y P' ∆θ P θ r O

x

time ∆t. (b)

Angular Displacement : Definition : Angle rotated by a position vector of the moving particle in a given time interval with some reference line is called its angular displacement.

______________________________________________________________________________________________ Important point : •

It is dimensionless and has proper unit SI unit radian while other units are degree or revolution 2π rad = 360° = 1 rev

•

Infinitely small angular displacement is a vector quantity but finite angular displacement is not because the addition of the small angular displacement is cummutative while for large is not.



dθ1 + dθ 2 = dθ 2 + dθ1 but θ1 + θ2 ≠ θ2 + θ1

•

Direction of small angular displacement is decided by right hand thumb rule. When the fingers are directed along the motion of the point then thumb will represents the direction of angular displacement.

•

Angular displacement can be different for different observers

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CIRCULAR MOTION

Page # 4

______________________________________________________________________________________________ (c)

Angular Velocity ω (i) Average Angular Velocity ωav =

Total Angle of Rotation ; Total time taken

θ 2 – θ1 ∆θ ωav = t – t = ∆t 2 1

where θ1 and θ2 are angular position of the particle at time t1 and t2 respectively. (ii) Instantaneous Angular Velocity The rate at which the position vector of a particle with respect to the centre rotates, is called as instantaneous angular velocity with respect to the centre. dθ ∆θ lim ω = ∆t→0 = dt ∆t

_______________________________________________________________________________________________ Important points : •

It is an axial vector with dimensions [T–1] and SI unit rad/s.

•

For a rigid body as all points will rotate through same angle in same time, angular velocity is a characteristic of the body as a whole, e.g., angular velocity of all points of earth about its own axis is (2π/24) rad/hr.

•

If a body makes ‘n’ rotations in ‘t’ seconds then angular velocity in radian per second will be 2πn t If T is the period and ‘f’ the frequency of uniform circular motion ωav =

ωav =

•

2π × 1 = 2πf T

If θ = a – bt + ct2 then ω =

dθ = – b + 2ct dt

Relation between speed and angular velocity : ω=

dθ ∆θ lim = dt ∆t→θ ∆t

The rate of change of angular velocity is called the angular acceleration (α). Thus, α=

dω d2 θ = 2 dt dt

Y

P'

The linear distance PP’ travelled by the particle in time ∆t is ∆s = r∆θ

∆S or ∆lim t→0 ∆t

∆θ = r ∆lim t→ 0 ∆ t

or

P

∆θ O

r

X

∆s dθ =r or v = rω ∆t dt

Here, v is the linear speed of the particle It is only valid for circular motion

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CIRCULAR MOTION

Page # 5

Ex.1

If θ depends on time t in following way θ = 2t2 + 3 then (b) ω at 3 sec (a) Find out ω average upto 3 sec.

Sol.

ωavg =

Total angular displacement θ f – θi = total time t 2 – t1

θf = 2 (3)2 + 3 = 21 rad θi = 2 (0) + 3 = 3 rad.

So,

ωavg =

21 – 3 = 6 rad/sec 3

dθ = 4t dt = 4 × 3 = 12 rad/sec

ωinstantaneous = ωat t = 3 sec

(d)

Relative Angular Velocity Angular velocity is defined with respect to the point from which the position vector of the moving particle is drawn Here angular velocity of the particle w.r.t. ‘O’ and ‘A’ will be different P'

A

ω PO =

O

P Ref lin

dα dβ ; ω PA = dt dt

Definition : Relative angular velocity of a particle ‘A’ with respect to the other moving particle ‘B’ is the angular velocity of the position vector of ‘A’ with respect to ‘B’. That means it is the rate at which position vector of ‘A’ with respect to ‘B’ rotates at that instant

A

VA

VB r

B

ω AB =

=

( VAB ) ⊥ rAB

here VAB ⊥ = Relative velocity ⊥ to position vector AB

Re lative velocity of A w.r.t. B perpendicu lar to line AB Seperation between A and B ( VAB ) ⊥ = VA sin θ1 + VB sin θ 2 rAB = r ω AB =

VA sin θ1 + VB sin θ 2 r

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CIRCULAR MOTION

Page # 6

_______________________________________________________________________________________ Important points : • If two particles are moving on the same circle or different coplanar concentric circles in same direction with different uniform angular speed ωA and ωB respectively, the rate of change of →

→

angle between OA and OB is B

B

A

dθ = ωB − ω A dt

O

A Initial line

O

Initial line

So the time taken by one to complete one revolution around O w.r.t. the other T=

•

If two particles are moving on two different concentric circles with different velocities then angular velocity of B relative to A as observed by A will depend on their positions and velocities. consider the case when A and B are closest to each other moving in same direction as shown in figure. In this situation

B vB v rel = | vB − v A | = vB − v A rrel

so,

TT 2π 2π = = 1 2 ω rel ω 2 − ω 1 T1 − T2



= | rB − rA | = rB − rA

ω BA =

( v rel ) ⊥ vB − v A = rrel rB − rA

vA

A rA

r

rB

O

( v rel ) ⊥ = Relative velocity perpendicular to position vector

_______________________________________________________________________________________ Ex.2

Sol.

Two particles move on a circular path (one just inside and the other just outside) with angular velocities ω and 5 ω starting from the same point. Then, which is incorrect. 2π when their angular velocities are (a) they cross each other at regular intervals of time 4ω oppositely directed (b) they cross each other at points on the path subtending an angle of 60° at the centre if their angular velocities are oppositely directed π (c) they cross at intervals of time if their angular velocities are oppositely directed 3ω (d) they cross each other at points on the path subtending 90° at the centre if their angular velocities are in the same sense If the angular velocities are oppositely directed, they meet at intervals of 2π 2π π time t = ω = = 6ω 3ω rel Angle subtended at the centre by the crossing points π θ = ωt = = 60° 3 When their angular velocities are in the same direction,

Ans. Ex.3

2π 2π π π π ×ω = t’ = ω = = and θ’ = 4ω 2ω 2ω 2 rel (a)

Two moving particles P and Q are 10 m apart at a certain instant. The velocity of P is 8 m/s making 30° with the line joining P and Q and that of Q is 6 m/s making 30° with PQ in the figure. Then the angular velocity of Q with respect to P in rad/s at that instant is 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR MOTION

Page # 7 6 m/s 30°

P 30°

10 m

Q

8 m/s

(A) 0

(B) 0.1

(C) 0.4

(D) 0.7

6 m/s 30°

P

Sol.

30°

10 m

Q

8 m/s Angular velocity of Q relative to P =

Projection of VQP perpendicular to the line PQ Separation between P and Q

VQ sin θ 2 – VP sin θ1 6 sin 30°–(–8 sin 30° ) = = 0.7 rad/s PQ 10 ∴ (D)

(e)

Angular Acceleration α : (i) Average Angular Acceleration : Let ω1 and ω2 be the instantaneous angular speeds at times t1 and t2 respectively, then the average angular acceleration αav is defined as ω 2 – ω1 ∆ω αav = t – t = ∆t 2 1

(ii) Instantaneous Angular Acceleration : It is the limit of average angular acceleration as ∆t approaches zero, i.e., lim ∆ω = dω = ω dω α = ∆t→0 dt dθ ∆t

_______________________________________________________________________________________________ Important points : •

It is also an axial vector with dimension [T–2] and unit rad/s2

•

If α = 0, circular motion is said to be uniform.

•

As ω =

dθ dω d2θ , α= = , dt dt dt 2

i.e., second derivative of angular displacement w.r.t time gives angular acceleration. •

α is a axial vector and direction of α is along ω

if ω increases and opposite to ω if ω decreases

_______________________________________________________________________________________ (f)

Radial and tangential acceleration Acceleration of a particle moving in a circle has two components one is along e t (along tangent) and the other along − eˆ r (or towards centre). Of these the first one is the called the tangential acceleration. (at) and the other is called radial or centripetal acceleration (ar). Thus.

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CIRCULAR MOTION

Page # 8

at =

dv = rate of change of speed dt 2

2 v a r = ω2 r = r   = v and r r Here, at is the component which is responsible for changing the magnitude of speed of the particle in circular motion. ar is the component which is responsible for changing the direction of particle in circular motion. the two component are mutually perpendicular. Therefore, net acceleration of the particle will be :

 dv  a = ar2 + a 2t = (rω 2 ) 2 +    dt 

2

2

2  v2   dv  =   +    dt   r 

Following three points are important regarding the above discussion : 1. In uniform circular motion, speed (v) of the particle is constant, i.e.,

dv = 0 . Thus, dt

at = 0 and a = ar = rω2 dv = positive, i.e., at is along e t or tangential acceleration of dt
dv 

et particle is parallel to velocity v because v = rω e t and ar = dt dv 3. In decelerated circular motion, = negative and hence, tangential acceleration is anti-parallel dt
to velocity v .

2. In accelerated circular motion,

(g)

Relation between angular acceleration and tangential acceleration we know that v = rω Here, v is the linear speed of the particle Differentiating again with respect to time, we have dv dω =r or at = rα dt dt dv Here, at = is the rate of change of speed (not the rate of change of velocity). dt

at =

Ex.4 Sol.

A particle travels in a circle of radius 20 cm at a speed that uniformly increases. If the speed changes from 5.0 m/s to 6.0 m/s in 2.0s, find the angular acceleration. The tangential acceleration is given by at =

v 2 – v1 dv = t –t dt 2 1

( ∵ Here speed increases uniformly at =

∆v dv = ) ∆t dt

6.0 – 5.0 m/s2 = 0.5 m/s2 2.0 The angular acceleration is α = at/r

=

= Ex-5

Sol.

0.5 m / s 2 = 2.5 rad/s2 20 cm

A particle moves in a circle of radius 20 cm. Its linear speed at any time is given by v = 2t where v is in m/s and t is in seconds. Find the radial and tangential acceleration at t = 3 seconds and hence calculate the total acceleration at this time. The linear speed at 3 seconds is 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR MOTION

Page # 9

v = 2 × 3 = 6 m/s The radial acceleration at 3 seconds 6×6 v2 = = 180 m/s2 0.2 r The tangential acceleration is given by dv = 2 , because v = 2t. dt ∴ tangential acceleration is 2 m/s2.

=

Net Acceleration =

ar 2 + a t 2 =

(180)2 + (2)2 = 180.01 m/s2

T-1

Is it possible for a car to move in a circular path in such a way that it has a tangential acceleration but no centripetal acceleration ?

Ex.6

A particle moves in a circle of radius 2.0 cm at a speed given by v = 4t, where v is in cm/s and t in seconds. (a) Find the tangential acceleration at t = 1 s. (b) Find total accleration at t = 1 s. (a) Tangential acceleration

Sol.

at =

dv dt

or

at =

ac =

v2 (4) 2 = =8 R 2

⇒

a=

d (4 t) = 4 cm/s2 dt a 2t + a2c =

2 (4) 2 + (8) 2 = 4 5 m / s

Ex.7

A boy whirls a stone in a horizontal circle of radius 1.5 m and at height 2.0 m above level ground. The string breaks, and the stone files off horizontally and strikes the ground after traveling a horizontal distance of 10 m. What is the magnitude of the cetripetal acceleration of the stone while in circular motion ?

Sol.

t=

Ex.8 Sol.

2h = g

2×2 = 0.64 s 9.8

v=

10 = 15.63 m/s t

a=

vB2 = 0.45 m/s2 R

Find the magnitude of the acceleration of a particle moving in a circle of radius 10 cm with uniform speed completing the circle in 4 s. The distance covered in completing the circle is 2 π r = 2π × 10 cm. The linear speed is v = 2 π r/t =

2π × 10cm = 5 π cm/s. 4s 2

The acceleration is ar = Ex.9 Sol.

v 2 ( 5 πcm / s) = =2.5 π2 cm/s2 10 cm r

A particle moves in a circle of radius 20 cm. Its linear speed is given by v = 2t where t is in second and v in meter/second. Find the radial and tangential acceleration at t = 3s. The linear speed at t = 3s is v = 2t = 6 m/s The radial acceleration at t = 3s is

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CIRCULAR MOTION

Page # 10

ar = v2/r =

36m 2 / s 2 = 180 m/s2 0.20 m

The tangential acceleration is at =

dv d( 2t) = = 2 m/s2 dt dt

Y

Ex.10 Two particles A and B start at the origin O and travel in opposite directions along the circular path at constant speeds vA = 0.7 m/s and vB = 1.5 m/s, respectively. Determine the time when they collide and the magnitude of the acceleration of B just before this happens. Sol.

1.5 t + 0.7 t = 2πR = 10 π a=

∴ t=

10 π = 14.3 s 2.2

5. 0

cm

A

B

vB =1.5 m/s O x vA=0.7m/s

vB2 = 0.45 m/s2 R Non-uniform circular motion

Uniform circular motion

speed of the particle is not constant i.e. ωis not constant
d| v| ≠0 at = dt

(1) Speed of the particle is constant i.e., ω is constant
d| v| (ii) a t = =0 dt

v2 ≠0 r

ar =

ar ≠ 0


anet = ar + a t

∴ anet = ar

anet

at

ar= anet ar

(h)

Relations among Angular Variables These relations are also referred as equations of rotational motion and are ω = ω0 + αt ...(1) 1 2 θ = ω0t + αt ...(2) 2 ω2 = ω02 + 2αθ ...(3) These are valid only if angular acceleration is constant and are analogous to equations of translatory motion, i.e.,

v = u + at ; s = ut +

1 2 at and 2

v2 = u2 + 2as

dθ, ω or α

ar vo

ra

O

ds r

t

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CIRCULAR MOTION

3.

Page # 11

DYNAMICS OF CIRCULAR MOTION : In circular motion or motion along any curved path Newton’s law is applied in two perpendicular directions one along the tangent and other perpendicular to it. i.e., towards centre. The compnent of net force towards the centre is called centripetal force. The component of net force along the tangent is called tangential force. tangential force (Ft) = Mat = M

dv = M α r ; where α is the angular acceleration dt

mv 2 r Ex.11 A small block of mass 100 g moves with uniform speed in a horizontal circular groove, with vertical side walls, of radius 25 cm. If the block takes 2.0s to complete one round, find the normal contact force by the slide wall of the groove. Sol. The speed of the block is

centripetal force (Fc) = m ω2 r =

v=

2π × ( 25 cm) = 0.785 m/s 2.0 s

The acceleration of the block is (0.785m / s) 2 v2 = = 2.46 m/s2 0.25 r towards the center. The only force in this direction is the normal contact force due to the side walls. Thus from Newton’s second law, this force is N = ma = (0.100 kg) (2.46 m/s2) = 0.246 N

a=

3.1

Centripetal Force : Concepts : This is necessary resultant force towards the centre called the centripetal force. mv 2 = mω2r r A body moving with constant speed in a circle is not in equilibrium. It should be remembered that in the absence of the centripetal force the body will move in a straight line with constant speed. It is not a new kind of force which acts on bodies. In fact, any force which is directed towards the centre may provide the necessary centripetal force.

F=

(i) (ii) (iii)

Ex.12 A small block of mass 100 g moves with uniform speed in a horizontal circular groove, with vertical side walls, of radius 25 cm. If the block takes 2.0s to complete one round, find the normal contact force by the slide wall of the groove. Sol. The speed of the block is v=

2π × ( 25 cm) = 0.785 m/s 2.0 s

The acceleration of the block is (0.785m / s) 2 v2 = = 2.5 m/s2 0.25 r towards the center. The only force in this direction is the normal contact force due to the slide walls. Thus from Newton’s second law, this force is N = ma = (0.100 kg) (2.5 m/s2) = 0.25 N

a=

3.2

Centrifugal Force : When a body is rotating in a circular path and the centripetal force vanishes, the body would leave the circular path. To an observer A who is not sharing the motion along the circular path, the body appears to fly off tangentially at the point of release. To another observer B, who is sharing the motion along the circular path (i.e., the observer B is also rotating with the body which is released, 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564

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CIRCULAR MOTION

Page # 12

it appears to B, as if it has been thrown off along the radius away from the centre by some force. This inertial force is called centrifugal force.) mv 2 . Centrifugal force is a fictitious force r which has to be applied as a concept only in a rotating frame of reference to apply Newton’s law of motion in that frame) FBD of ball w.r.t non inertial frame rotating with the ball.

Its magnitude is equal to that of the centripetal force =

ω T

mω 2r mg

Suppose we are working from a frame of reference that is rotating at a constant, angular velocity ω with respect to an inertial frame. If we analyse the dynamics of a particle of mass m kept at a distance r from the axis of rotation, we have to assume that a force mrω2 act radially outward on the particle. Only then we can apply Newton’s laws of motion in the rotating frame. This radially outward pseudo force is called the centrifugal force. T-2

A particle of mass m rotates in a circle of radius r with a uniform angular speed ω. It is viewed from a frame rotating about same axis with a uniform angular speed ω0. The centrifugal force on the particle is 2

2

 ω + ω0  (C) m  r  2 

2 0

(A) mω r

(B) mω r

(D) mω0ωr

B

: A rod move with ω angular velocity then we conclude following for point A & B in a rod. αA = αB s B > sA θA = θB vB > vA ωA = ωB atB > atA

A

ω

Ex.13 Find out the tension T1, T2 is the string as shown in figure 2 rad/sec. T1

1kg

1m

T2 1m

2kg

ω rad/sec.

T1

m1

T2

m2

We know that ωm1 = ωm 2 ⇒

So

T1 = m1ω2R1 + T2 T2 = m ω2 R2 T2 = 2 × 4 × 2 = 16 N T1 = (1) (2)2 (1) + 16 N = 4 + 16 N T1 = 20 N 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR MOTION

4.

Page # 13

SIMPLE PENDULUM

Ex.14 A simple pendulum is constructed by attaching a bob of mass m to a string of length L fixed at its upper end. The bob oscillates in a vertical circle. It is found that the speed of the bob is v when the string makes an angle θ with the vertical. Find the tension in the string at this instant. Sol. The force acting on the bob are (figure) (a) the tension T (b) the weight mg. As the bob moves in a vertical circle with centre at O, the radial acceleration is v2/L towards O. Taking the components along this radius and applying Newton’s second law, we get mgcos T – mgcos θ = mv2/L or, T = m(gcos θ + v2/L) mg mgsin
|Fnet | =

5.

 mv 2   (mg sin θ) +   L   

2

2

2 2 = m g sin θ +

v4 L2

CIRCULAR MOTION IN HORIZONTAL PLANE ω A ball of mass m attached to a light and inextensible string rotates in a horizontal circle of radius r with an angular speed

θ

ω about the vertical. If we draw the force diagram of the ball. We can easily see that the component of tension force along

the centre gives the centripetal force and component of tension along vertical balances the gravitation force. Such a system is called a conical pendulum.

T T cos θ θ

T sinθ

mω 2r

mg FBD of ball w.r.t ground

Ex. 15 A particle of mass m is suspended from a ceiling through a string of length L. The particle moves in a horizontal circle of radius r. Find (a) the speed of the particle and (b) the tension in the string. Sol.

The situation is shown in figure. The angle θ made by the string with the vertical is given by sinθ = r/L

... (i)

T

θ

The forces on the particle are (a) the tension T along the string and

r

(b) the weight mg vertically downward. The particle is moving in a circle with a constant speed v. Thus, the radial acceleration towards the centre has magnitude v2/r. Resolving the forces along the radial direction and applying. Newton’s second law, Tsin θ = m(v2/r)

mg

...(ii)

As there is no acceleration in vertical directions, we have from Newton’s law, Tcosθ = mg

...(iii)

Dividing (ii) by (iii),

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L

CIRCULAR MOTION

Page # 14

tanθ =

v2 rg

or v =

And from (iii), T = Using (i), v =

6.

rgtan θ mg cos θ

r g 2

and

2 1/ 4

(L – r )

T=

mgL 2

(L – r 2 )1/ 2

MOTION OF A MOTORCYCLIST ON A CURVED PATH. A cylist having mass m move with constant speed v on a curved path as shown in figure. B D

A

C

E

We divide the motion of cyclist in four parts : (1) from A to B (2) from B to C (3) from C to D (4) from D to E (1 and 3 are same type of motion)

(A)

Motion of cyclist from A to B N+

mv 2 = mg cos θ R

mv N = mg cosθ – R f = mg sin θ

⇒

N+

mv 2 R

f

B

2

...(1) ..(2)

mgsinθ

θ

θ os gc m θ

(1)

As cyclist move upward A mg ∵ θ decreases & cos θ increases ∴ N increases and ∵ θ decreases sin θ decreases ∴ friction force required to balance mg sin θ (As cyclist is moving with constant speed) also decreases

(B)

Motion of cyclist from B to C B

2

N+

mv = mg cos θ R

⇒ N = mg cos θ –

mv 2 R

...(1)

f = mg sinθ ...(2) Therefore from B to C Normal force decrease but friction force increase becuse θ increases. (C)

N+

f mgcos θ

mv 2 R

mg sin θ

mg

C

Motion of cyclist from D to E N=

mv 2 + mg cos θ R

f = mg sin θ

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CIRCULAR MOTION

Page # 15

from D to E θ decreases therefore mg cos θ increase So N increase but f decreases

D

f mv 2 R

E

mg Ex.16 A hemispherical bowl of radius R is rotating about its axis of symmetry which is kept vertical. A small ball kept in the bowl rotates with the bowl without slipping on its surface. If the surface of the bowl is smooth and the angle made by the radius through the ball with the vertical is α. Find the angular speed at which the bowl is rotating. Sol. Let ω be the angular speed of rotation of the bowl. ω Two force are acting on the ball. 1. normal reaction N 2. weight mg The ball is rotating in a circle of radius r (= R sin α) with R α centre at A at an angular speed ω. Thus, N N sin α = mrω2 = mRω2 sin α A r N = mRω2 ...(i) ...(ii) and N cos α = mg mg Dividing Eqs. (i) by (ii),

we get

1 ω 2R = cos α g

∴

ω=

g Rcos α

Ex.17 If friction is present between the surface of ball and bowl then find out the range of ω for which ball does not slip (µ is the friction coefficient) Friction develop a range of ω for which the particle will be at rest. Sol. (a) When ω > ω0 In this situation ball has a tendency to slip upwards α so the friction force will act downwards. So F.B.D of ball N = mω2r sin α + mg cos α. ... f + mg sin α = mω2r cos α ...(2) ∴ fmax = µN = µ(mω2r sin α + mg cosα) 2 r = R sin α mω r Substituting the values of fmax & r in eq. (2) we get

N α

m α

⇒ µ (mω2r sin α + mg cosα) ≥ mω2r cosα –mg sin α ∴ µ(mω2R sin 2α + mg cos α) ≥ mω2 R sinα cosα – mg sin α ω≤

(b)

f

mg

µg cos α + g sin α R sin α(cos α – µ sin α )

N 2

mω r

f + mω2r cos α = mg sin α f = m (g sinα – ω2r cosα ) ...(1) N = mg cos α + mω2r sin α ...(2) fmax = µN = µ(mg cos α + mω2r sinα) for equillibrium

α

f

when ω < ω0 In this situation ball has a tendency to slip downwards so the friction force will act upwards. So F.B.D of ball ⇒

r

α

m r

α

mg

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CIRCULAR MOTION

Page # 16 ⇒ µ(g cos α + ω2r sinα ) ≥ (gsinα – ω2 r cos α) ⇒ Substituting r = Rsinα then ω≥

7.

g(sin α – µ cos α ) R sin α(µ sin α + cos α )

CIRCULAR TURNING ON ROADS : When vehicles go through turnings, they travel along a nearly circular arc. There must be some force which will produce the required centripetal acceleration. If the vehicles travel in a horizontal circular path, this resultant force is also horizontal. The necessary centripetal force is being provided to the vehicles by following three ways. 1. By Friction only 2. By Banking of Roads only 3. By Friction and Banking of Roads both. In real life the necessary centripetal force is provided by friction and banking of roads both. Now let us write equations of motion in each of the three cases separately and see what are the constant in each case.

7.1

By Friction Only Suppose a car of mass m is moving at a speed v in a horizontal circular arc of radius r. In this case, the necessary centripetal force to the car will be provided by force of friction f acting towards center mv 2 r Further, limiting value of f is µN or fL = µN = µmg (N = mg) Thus, f =

Therefore, for a safe turn without sliding

mv 2 ≤ fL r

v2 mv 2 µmg or µ ≥ or v ≤ µrg ≤ rg r Here, two situations may arise. If µ and r are known to us, the speed of the vehicle should not

or

exceed

7.2

µrg and if v and r are known to us, the coefficient of friction should be greater than

v2 . rg

By Banking of Roads Only Friction is not always reliable at circular turns if high speeds and sharp turns are involved to avoid dependence on friction, the roads are banked at the turn so that the outer part of the road is some what lifted compared to the inner part. Applying Newton’s second law along the

N

radius and the first law in the vertical direction. Nsinθ =

mv 2 r

or

N cosθ = mg θ

from these two equations, we get v2 tanθ = rg

or

v=

W

rgtanθ

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CIRCULAR MOTION 7.3

Page # 17

By Friction and Banking of Road Both If a vehicle is moving on a circular road which is rough and banked also, then three forces may act on the vehicle, of these force, the first force, i.e., weight (mg) is fixed both in magnitude and direction. N θ θ f

y mg

x

θ

Figure (ii)

The direction of second force i.e., normal reaction N is also fixed (perpendicular or road) while the direction of the third i.e., friction f can be either inwards or outwards while its magnitude can be vari ed upt o a maxi mum l i mi t (fL = µN). So the magnitude of normal reaction N and directions plus magnitude of friction f are so mv 2 towards the center. Of r these m and r are also constant. Therefore, magnitude of N and directions plus magnitude of friction mainly depends on the speed of the vehicle v. Thus, situation varies from problem to problem. Even though we can see that : (i) Friction f will be outwards if the vehicle is at rest v = 0. Because in that case the component weight mg sinθ is balanced by f. (ii) Friction f will be inwards if adjusted that the resultant of the three forces mentioned above is

v > rgtanθ (iii) Friction f will be outwards if v
90°), we say that the work done by the force is negative.

Example : When a body is lifted, the work done by the gravitational force is negative. This is because the gravitational force acts vertically downwards while the displacement is in the vertically upwards direction.

__________________________________________________________________________________ Important points about work : 1. Work is said to be done by a force when its point of application moves by some distance.Force does no work if point of application of force does not move (S = 0) Example : A person carrying a load on his head and standing at a given place does no work. 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

WORK, POWER & ENERGY 2.

Page # 25

Work is defined for an interval or displacement. There is no term like instantaneous work similar to instantaneous velocity. µ=0.2

2kg frictionless

3. 4. 5. 6.

2kg

10 N 2m

10 N 2m

Work done by 10 N force in both the cases are same = 20 N For a particular displacement, work done by a force is independent of type of motion i.e. whether it moves with constant velocity, constant acceleration or retardation etc. If a body is in dynamic equilibrium under the action of certain forces, then total work done on the body is zero but work done by individual forces may not be zero. When several forces act, work done by a force for a particular displacement is independent of other forces. A force is independent of reference frame. Its displacement depends on frame so work done by a force is frame dependent therefore work done by a force can be different in different reference frame.

________________________________________________________________________________________ 2.

UNITS OF WORK : In cgs system, the unit of work is erg. One erg of work is said to be done when a force of one dyne displaces a body through one centimetre in its own direction. ∴ 1 erg = 1 dyne × 1 cm = 1 g cm s–2 × 1 cm = 1 g cm2 s–2 Note : Another name for joule is newton metre. Relation between joule and erg 1 joule = 1 newton × 1 metre 1 joule = 105 dyne × 102 cm = 107 dyne cm 1 joule = 107 erg 1 erg = 10–7 joule Dimensions of Work : [Work] = [Force] [Distance] = [MLT–2] [L] = [ML2T–2] Work has one dimension in mass, two dimensions in length and ‘–2’ dimensions in time, On the basis of dimensional formula, the unit of work is kg m2 s–2. Note that 1 kg m2 s–2 = (1 kg m s–2) m = 1 N m = 1 J.

3.

WORK DONE BY MULTIPLE FORCES :


If several forces act on a particle, then we can replace F in equation W = F . S by the net force
F where



F = F1 + F2 + F3 +......

F . S ∴ W= ...(i)
This gives the work done by the net force during a displacement S of the particle. We can rewrite equation (i) as :





W = F1. S + F2 . S + F3 . S+..... or W = W1 + W2 + W3 + ......... So, the work done on the particle is the sum of the individual work done by all the forces acting on the particle.

∑ ∑

[∑ ]

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WORK, POWER & ENERGY

Page # 26

Ex.1

Sol.

A block of mass M is pulled along a horizontal surface by applying a force at an angle θ with horizontal. Coefficient of friction between block and surface is µ. If the block travels with uniform velocity, find the work done by this applied force during a displacement d of the block. The forces acting on the block are shown in Figure. As the block moves with uniform velocity the resultant force on it is zero. ∴ F cos θ = µN ...(i) F sin θ + N = Mg ...(ii) Eliminating N from equations (i) and (ii), N F cos θ = µ(Mg – Fsin θ) F F=

µMg cos θ + µ sin θ

M

Work done by this force during a displacement d W = F . d cos θ =

Ex.2

Sol.

µMgdcos θ cos θ + µ sin θ

Mg


A particle moving in the xy plane undergoes a displacement S = (2.0 ˆi + 3.0ˆj)m while a constant
force F = (5.0 ˆi + 2.0ˆj)N acts on the particle.

(a) Calculate the magnitude of the displacement and that of the force. (b) Calculate the work done by the force.

(a) s = (2.0 i + 3.0 j ) F = (5.0 i + 2.0 j )
| s | = x2 + y 2 = (2.0)2 + (3.0)2 = 13 m
|F|=

Fx2 + Fy2 =

(5.0 )2 + ( 2.0) 2 = 5.4N



(b) Work done by force, W = F . s

= (5.0 ˆi + 2.0 ˆj) . (2.0ˆi + 3.0ˆj) N. m Ex.3

= 10 + 0 + 0 + 6 = 16 N.m = 16 J

A block of mass m is placed on an inclined plane which is moving with constant velocity v in horizontal direction as shown in figure. Then find out work done by the friction in time t if the block is at rest with respect to the incline plane. v=const.

m µ θ

Sol.

F.B.D of block with respect to ground.

f f

m

v

π−θ

v θ θ m mg g sin θ B

A

θ

mg

mgsin θ

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WORK, POWER & ENERGY

Page # 27

Block is at rest with respect to wedge ⇒ f = mg sin θ In time t the displacement of block with respect to ground d = vt Work done by friction for man A Wf = (component of friction force along displacement) × displacement Wf = mgsinθ.vt cos(180°–θ) Wf = – mg vt cosθ sin θ Wf for man B = 0 (displacement is zero with respect to man B)

4.

WORK DONE BY A VARIABLE FORCE :

(A)

When F as a function of x, y, z When the magnitude and direction of a force vary in three dimensions, it can be expressed as a function of the position. For a variable force work is calculated for infinitely small displacement and for this displacement force is assumed to be constant.

dW = F. ds The total work done will be sum of infinitely small work B

WA → B =

∫



F. ds =

A

B







∫ (F cos θ)ds A

It terms of rectangular components,
F = Fx i + Fy j + Fzk
ds = dx i + dy j + dzk xB

WA → B =

∫

yB

Fx dx +

xA

∫

zB

∫ F dz

Fy dy +

z

yA

zA

Ex.4

A force F = (4.0 x i + 3.0 y j ) N acts on a particle which moves in the x-direction from the

Sol.

or ig in to x = 5.0 m. Find the work done on the object by the force. Here the work done is only due to x component of force because displacement is along x-axis. x2

i.e., W =

∫

5

Fx dx =

x1

Ex.5

Sol.

∫ 4x dx = [2x ] 2

0

5 0

= 50 J

A force F = 0.5x + 10 acts on a particle. Here F is in newton and x is in metre. Calculate the work done by the force during the displacement of the particle from x = 0 to x = 2 metre. → Small amount of work done dW in giving a small displacement dx is given by →→ dW = F . dx or dW = Fdx cos 0° or dW = Fdx [∴ cos 0° = 1] x =2

Total work done, W =

∫

x =2

Fdx =

x =0 x=2

=

x =2

∫ 0.5xdx + ∫ 10dx

x =0

x =0

F

∫ (0.5x + 10)dx

x = 0.5 2

+ve

0

x =0

2

10

2

x= 2 x= 2

+ 10 x x= 0 x= 0

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x

WORK, POWER & ENERGY

Page # 28 0.5 2 [2 – 02] + 10[2 – 0] = (1 + 20) = 21 J 2

= (B)

When F is given as a function of Time(t) :

Ex.6

The force F = 2t2 is applied on the 2 kg block. Then find out the work done by this force in 2sec. Initially at time t = 0, block is at rest.

at t = 0, v = 0 F=2t2

2kg

Sol.

F = ma ⇒ 2t2 = 2a ⇒

a =t2 v

dv = t2 ⇒ dt

∫

⇒

t

dv =

0

∫ t dt 2

(At t = 0 it is at rest)

0

t3 3 Let the displacement of the block be dx from t = t to t = t +dt then, work done by the force F in this time interval dt is. dw = F.dx = 2t2.dx

⇒ v=

2 dw = 2t . w

∫

2

0

5.

∫

dw = 2t 2 . 0

dx . dt ⇒ dt

t3 dt ⇒ 3

dw = 2t 2 ( v)dt 2

2  t6  W= 36  

2 5 W = 3 t dt ⇒

∫ 0

2

= 0

64 Joule 9

AREA UNDER FORCE DISPLACEMENT CURVE : Graphically area under the force-displacement is the work done +ve work

Fx

Fy

+ve work

+ve work

Fz

y

x

Ex.7 Sol.

6.

–ve work

z

The work done can be positive or negative as per the area above the x-axis or below the x-axis respectively. Force acting on a particle varies with x as shown in figure. Calculate the work done by the force as the particle moves from x = 0 to x = 6.0 m. The work done by the force is equal to the area under the curve from x = 0 to x = 6.0 m. Fx(N) This area is equal to the area of the rectangular section from x = 0 to x = 4.0 m plus the area of the triangular section from 5 x = 4.0 m to x = 6.0 m. The area of the rectangle is (4.0) (5.0) N.m = 20 1 J, and the area of the triangle is (2.0), (5.0) N.m = 5.0 J. 0 1 2 3 4 5 6 x(m) 2 Therefore, the total work done is 25 J.

INTERNAL WORK : Suppose that a man sets himself in motion backward by pushing against a wall. The forces acting on the man are his weight 'W' the upward force N exerted by the ground and the horizontal force 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

WORK, POWER & ENERGY

Page # 29

N′ exerted by the wall. The works of 'W' and of N are zero because they are perpendicular to the motion. The force N' is the unbalanced horizontal force that imparts to the system a horizontal acceleration. The work of N', however, is zero because there is no motion of its point of application. We are therefore confronted with a curious situation in which a force is responsible for acceleration, but its work, being zero, is not equal to the increase in kinetic energy of the system. N'

N

The new feature in this situation is that the man is a composite system with several parts that can move in relation to each other and thus can do work on each other, even in the absence of any interaction with externally applied forces. Such work is called internal work. Although internal forces play no role in acceleration of the composite system, their points of application can move so that work is done; thus the man's kinetic energy can change even though the external forces do no work. "Basic concept of work lies in following lines Draw the force at proper point where it acts that give proper importance of the point of application of force. Think independently for displacement of point of application of force, Instead of relation the displacement of applicant point with force relate it with the observer or reference frame in which work is calculated.  displacement vector of po int of  W = (Force vector ) ×  application of force as seen by  observer 

7.

    

CONSERVATIVE FORCE : 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 does not depend on the nature of path followed between the initial and final positions. m

m

m

m

m

m

(a)

(b)

(c)

Consider a body of mass m being raised to a height h vertically upwards as shown in above figure. The work done is mgh. Suppose we take the body along the path as in (b). The work done during horizontal motion is zero. Adding up the works done in the two vertical parts of the paths, we get the result mgh once again. Any arbitrary path like the one shown in (c) can be broken into elementary horizontal and vertical portions. Work done along the horizontal path is zero. The work done along the vertical parts add up to mgh. Thus we conclude that the work done in raising a body against gravity is independent of the path taken. It only depends upon the intial and final positions of the body. We conclude from this discussion that the force of gravity is a conservative force.

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WORK, POWER & ENERGY

Page # 30

(i) (ii) (iii) (iv)

• • •

•

Examples of Conservative forces. Gravitational force, not only due to Earth due in its general form as given by the universal law of gravitation, is a conservative force. Elastic force in a stretched or compressed spring is a conservative force. Electrostatic force between two electric charges is a conservative force. Magnetic force between two magnetic poles is a conservative force. Forces acting along the line joining the centres of two bodies are called central forces. Gravitational force and Electrosatic forces are two important examples of central forces. Central forces are conservative forces. Properties of Conservative forces Work done by or against a conservative force depends only on the initial and final position of the body. Work done by or against a conservative force does not depend upon the nature of the path between initial and final position of the body. Work done by or against a conservative force in a round trip is zero. If a body moves under the action of a force that does no total work during any round trip, then the force is conservative; otherwise it is non-conservative. The concept of potential energy exists only in the case of conservative forces. The work done by a conservative force is completely recoverable. Complete recoverability is an important aspect of the work done by a conservative force.

Work done by conservative forces Ist format : (When constant force is given)


Ex.8 Sol.

Calculate the work done to displace the particle from (1, 2) to (4, 5). if F = 4 ˆi + 3ˆj


dw = F.d r ( dr = dxi + dyj + dzk ) dw = (4 i + 3 j ).(dxi + dyj) 4

w

∫

dw =

∫

5

4dx +

1

0

⇒ dw = 4dx + 3dy

∫ 3dy 2

4

5

⇒ w = [4 x]1 + [3 y]2

w = (16 – 4) + (15 – 6) ⇒ w = 12 + 9 = 21 Joule II format : (When F is given as a function of x, y, z)
If F = Fx i + Fy j + Fzk then dw = (Fx i + Fy j + Fzk ).( dx i + dyj + dzk ) ⇒ dw = Fxdx + Fydy + FZdz

Ex.9



An object is displaced from position vector r1 = (2 ˆi + 3ˆj)m to r2 = (4 ˆi + 6ˆj)m under a force
F = (3x 2 ˆi + 2yˆj)N . Find the work done by this force.
rf

Sol.


r2



W = F.dr = (3 x 2 ˆi + 2yˆj ) • (dx ˆi + dyˆj + dzkˆ ) =

∫ ri

∫
r1


r2

∫ (3x dx + 2ydy) = [x 2


r1

3

+ y 2 ](( 42,,36)) = 83 J Ans.

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WORK, POWER & ENERGY

Page # 31

IIIrd format (perfect differential format)
Ex.10 If F = yiˆ + xjˆ then find out the work done in moving the particle from position (2, 3) to (5, 6)

Sol. dw = F. ds

Now

dw = ( y i + xj ).( dxi + dyj ) dw = ydx + xdy ydx + xdy = d(xy) (perfect differential equation) ⇒ dw = d(xy) for total work done we integrate both side

∫ dw = ∫ d( xy) Put xy = k then at (2, 3) ki = 2 × 3 = 6 at (5, 6) kf = 5 × 6 = 30 30

then

w=

∫ dk = [k]

30 6

⇒ w = (30 – 6) = 24 Joule

6

8.

NON-CONSERVATIVE FORCES :

A force is said to be non-conservative if work done by or against the force in moving a body depends upon the path between the initial and final positions. The frictional forces are non-conservative forces. This is because the work done against friction depends on the length of the path along which a body is moved. It does not depend only on the initial and final positions. Note that the work done by fricitional force in a round trip is not zero. The velocity-dependent forces such as air resistance, viscous force, magnetic force etc., are non conservative forces.
Ex.11 Calculate the work done by the force F = y i to move the particle from (0, 0) to (1, 1) in the following condition (a) y = x (b) y = x2 Sol. We know that

dw = F.ds ⇒ dw = ( y i ) .(dx i ) dw = ydx ...(1) In equation (1) we can calculate work done only when we know the path taken by the particle. either y = x or y = x2 so now (a) when y = x 1 1 dw = xdx ⇒ w = Joule 0 2 (b) when y = x2

∫

∫

∫

1

dw =

∫ x dx 2

0

⇒

w=

1 Joule 3

Difference between conservative and Non-conservative forces

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WORK, POWER & ENERGY

Page # 32

S. No. 1 2 3

4

5

9.

Non-Conse rva tive force s W ork done does not W ork done depends on depend upon path path. W ork done in a round W ork done in a round trip trip is zero. is not zero. Forc es are veloc ityCentral in nature. dependent and retarding in nature. W hen only a c ons ervative forc e ac ts within a sy s tem , the W ork done agains t a nonk inetic energy and c ons ervative forc e m ay potential energy can be disipiated as heat c hange. However their energy. s um , the m ec hanical energy of the s y stem , does not c hange. Conse rva tive force s

W ork done is W ork done in not c om pletely rec overable. c om pletely recoverable.

ENERGY A body is said to possess energy if it has the capacity to do work. When a body possessing energy does some work, part of its energy is used up. Conversely if some work is done upon an object, the object will be given some energy. Energy and work are mutually convertiable. There are various forms of energy. Heat, electricity, light, sound and chemical energy are all familiar forms. In studying mechanics, we are however concerned chiefly with mechanical energy. This type of energy is a property of movement or position.

9.1

• 9.2

1. 2. 3. 4. 5.

Kinetic Energy Kinetic energy (K.E.), is the capacity of a body to do work by virtue of its motion. If a body of mass m has velocity v its kinetic energy is equivalent to the work, which an external force would have to do to bring the body from rest up to its velocity v. The numerical value of the kinetic energy can be calculated from the formula 1 2 ...(8) K.E. = mv 2 2 Since both m and v are always positive, K.E. is always positive and does not depend upon the direction of motion of the body. Potential Energy Potential energy is energy of the body by virtue of its position. A body is capable to do work by virtue of its position, configuration or state of strain. Now relation between Potential energy and work done is W.D = – ∆U where ∆U is change in potential energy There are two common forms of potential energy, gravitational and elastic. Important points related to Potential energy : Potential energy is a straight function (defined only for position) Potential energy of a point depends on a reference point Potential energy difference between two position doesn't depend on the frame of reference. Potential energy is defined only for conservative force because work done by conservative force is path independent. If we define Potential energy for non conservative force then we have to define P.E. of a single point through different path which gives different value of P.E. at single point that doesn't make any sense.

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WORK, POWER & ENERGY

Page # 33

9.2.1 (a) Gravitational Potential Energy : It is possessed by virtue of height. When an object is allowed to fall from one level to a lower level it gains speed due to gravitational pull, i.e., it gains kinetic energy. Therefore, in possessing height, a body has the ability to convert its gravitational potential energy into kinetic energy. The gravitational potential energy is equivalent to the negative of the amount of work done by the weight of the body in causing the descent. If a mass m is at a height h above a lower level the P.E. possessed by the mass is (mg) (h). Since h is the height of an object above a specified level, an object below the specified level has negative potential energy. Therefore GPE = ± mgh ...(9) mg

mg

h1 Specific level where P.E. is zero

h P.E.=mgh h2

fig(a)

fig(b)

P.E. of m1 is m1gh1 P.E of m2 is –m2gh2 m2g

•

The chosen level from which height is measured has no absolute position. It is important therefore to indicate clearly the zero P.E. level in any problem in which P.E. is to be calculated. • GPE = ± mgh is applicable only when h is very small in comparison to the radius of the earth. We have discussed GPE in detail in 'GRAVITATION'. 9.2.2 (b) Elastic Potential Energy : It is a property of stretched or compressed springs. The end of a stretched elastic spring will begin to move if it is released. The spring. therefore possesses potential energy due to its elasticity. (i.e., due to change in its configuration) The amount of elastic potential energy stored in a spring of natural length a and spring constant k when it is extended by a length x (from the natural length) is equivalent to the amount of work necessary to produce the extension. 1 2 kx ...(10) 2 It is never negative whether the spring is extended or compressed.

Elastic Potential Energy =

Proof :

N.L.

K

N.L.

K

M

M

x0 Consider a spring block system as shown in the figure and let us calculate work done by spring when the block is displaceed by x0 from the natural length. At any moment if the elongation in spring is x, then the force on the block by the spring is kx towards left. Therefore, the work done by the spring when block further displaces by dx dW = – kx dx x0

∴ Total work done by the spring, W = –

∫ kxdx = – 0

1 2 kx 0 2

Similarly, work done by the spring when it is given a compression x0 is – : We assume zero potential energy at natural length of the spring : 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

1 2 kx 0 . 2

WORK, POWER & ENERGY

Page # 34

10.

CONSERVATIVE FORCE AND POTENTIAL ENERGY : ∂U ∂s i.e. the projection of the force field , the vector F, at a given point in the direction of the displacement r equals the derivative of the potential energy U with respect to a given direction, taken with the opposite sign. The designation of a partial derivative ∂/∂s emphasizes the fact of deriving with respect to a definite direction.

Fs = –

So, having reversed the sign of the partial derivatives of the function U with respect to x, y, z, we obtain the projection Fx, Fy and Fz of the vector F on the unit vectors i, j and k. Hence, one can readily find the vector itself :  ∂U ∂U ∂U  F = Fxi + Fy j + Fzk, or F = –  ∂ x i + ∂ y j + ∂z k    The quantity in parentheses is referred to as the scalar gradient of the function U and is denoted

by grad U or ∇ U. We shall use the second, more convenient, designation where ∇ (“nabla”) signifies the symbolic vector or operator ∂ ∂ ∂ ∇ = i ∂x + j ∂y + k ∂z

Potential Energy curve : •

A graph plotted between the PE of a particle and its displacement from the centre of force field is called PE curve.

•

Using graph, we can predict the rate of motion of a particle at various positions.

•

Force on the particle is F(x) = –

dU dx

Q

S

U

B P1

A

P2 R

P O

D

C

x

Case : I

On increasing x, if U increases, force is in (–) ve x direction i.e. attraction force.

Case : II

On increasing x, if U decreases, force is in (+) ve x-direction i.e. repulsion force.

Different positions of a particle : Position of equilibrium If net force acting on a body is zero, it is said to be in equilibrium. For equilibrium

dU = 0. Points P,, dx

Q, R and S are the states of equilbrium positions. Types of equilirbium : •

Stable equilibrium : When a particle is displaced slightly from a position and a force acting on it brings it back to the initial position, it is said to be in stable equilibrium position. dU d2U = 0,and 2 =+ ve dx dx In figure P and R point shows stable equilibrium point.

Necessary conditions:–

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WORK, POWER & ENERGY

•

Page # 35

Unstable Equilibrium : When a particle is displaced slightly from a position and force acting on it tries to displace the particle further away from the equilibrium position, it is said to be in unstable equilibrium. dU d2U = 0 potential energy is maximum i.e. = = – ve dx dx2 Q point in figure shows unstable equilibrium point Neutral equilibrium : In the neutral equilibrium potential energy is constant. When a particle is displaced from its position it does not experience any force acting on it and continues to be in equilibrium in the displaced position. This is said to be neutral equilibrium. In figure S is the neutral point

Condition :

•

Condition :

dU d2U =0 , =0 dx dx2

a b – , where a x 12 x 6 and b are positive constants and x is the distance between the atoms. The system is in stable equilibrium when -

Ex.12 The potential energy between two atoms in a molecule is given by, U(x) =

(A) x = 0 Sol.

 2a  (C) x =    b 

a (B) x = 2b

1/6

 11a   (D) x =  5b 

(C) Given that, U(x) =

6b x7

12

x

–

b x6

du dx = (–12) a x–13 – (–6 b) x–7= 0

We, know

or

a

F=–

=

12a 4 x13

or

x6 = 12a/6b = 2a/b or

 2a  x=    b

1/ 6

Ex.13 The potential energy of a conservative system is given by U = ax2 – bx where a and b are positive constants. Find the equilibrium position and discuss whether the equilibrium is stable, unstable or neutral. Sol.

In a conservative field F = –

dU dx

∴F=–

For equilibrium F = 0 or b – 2ax = 0 ∴ x =

From the given equation we can see that

Therefore, x =

11.

d ( ax2 – bx) = b – 2ax dx

b 2a

d2U dx2

= 2a (positive), i.e., U is minimum.

b is the stable equilibrium positon. 2a

WORK ENERGY THEOREM : If the resultant or net force acting on a body is Fnet then Newton's second law states that ...(1) Fnet = ma If the resultant force varies with x, the acceleration and speed also depend on x. 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564

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WORK, POWER & ENERGY

Page # 36

then

a= v

dv dx

...(2)

from eq. (1) Fnet

dv = mv dx

m

⇒ Fnet.dx = m v dv

vi

Fnet

m vf

vf

∫F

net . dx

=

∫ mvdv vi

1 1 mv 2f – mv i2 2 2 Wnet = kf – ki Wnet = ∆K ...(3)

Wnet =

Work done by net force Fnet in displacing a particle equals to the change in kinetic energy of the particle i.e. we can write eq. (3) in following way (W.D)c + (W.D)N.C + (W.D)ext. + (W.D)pseudo = ∆K

...(4)

where (W.D)c = work done by conservative force (W.D)N.C = work done by non conservative force. (W.D)ext = work done by external force (W.D)pseudo = work done by pseudo force. we know that (W.D)c = – ∆U ⇒ – ∆U + (W.D)N.C + (W.D)ext + (W.D)pseudo = ∆K ⇒ (W.D)N.C + (W.D)ext. + (W.D)pseudo = (kf + uf) – (ki + ui) ∵ k + u = Mechanical energy. ⇒ work done by forces (except conservative forces) = change is mechanical energy. If (W.D)N.C = (W.D)ext = (W.D)pseudo = 0 Kf + Uf = Ki + Ui Initial mechanical energy = final mechanical energy This is called mechanical energy conservation law. Questions Based on work Energy Theorem : (A)

When only one conservative force is acting

Ex.14 The block shown in figure is released from rest. Find out the speed of the block when the spring is compressed by 1 m. N.L.

2kg 2m

A

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WORK, POWER & ENERGY

Sol.

Page # 37

In the above problem only one conservative force (spring force) is working on the block so from mechanical energy conservation kf + uf = ki + ui

...(i)

at A block is at rest so 1 1 ui = kx12 = k( 2) 2 2 2

ki = 0

N.L.

= 2k Joule

At position B if speed of the block is v then kf =

1 1 mv 2 = × 2 × v 2 = v 2 2 2

uf =

1 2 1 k kx2 = × k × 1 = 2 2 2

B

A

2kg

2kg

x2=1m

x1=2m

Putting the above values in equation (i), we get ⇒

v2 +

k = 2k 2

2 ⇒ v =

3k ⇒ v= 2

3k m / sec 2

Ex.15 A block of mass m is dropped from height h above the ground. Find out the speed of the block when it reaches the ground. Sol. Initial situation

Ugi = mgh , ki = 0

h

Final situation v

Ugf = 0 , K f =

1 mv 2 2

Figure shows the complete description of the problem only one conservative force is working on the block. So from mechanical energy conservation kf + uf = ki + ui

⇒

1 mv 2 + 0 = 0 + mgh 2

v = 2gh m / sec

(B)

When two conservative force are acting in problem.

Ex.16 One end of a light spring of natural length d and spring constant k is fixed on a rigid wall and the other is attached to a smooth ring of mass m which can slide without friction on a vertical rod fixed at a distance d from the wall. Initially the spring makes an angle of 37º with the horizontal as shown in fig. When the system is released from rest, find the speed of the ring when the spring becomes horizontal. [sin 37º = 3/5]

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WORK, POWER & ENERGY

Page # 38

A

Ring v=0

h

I

37°

d

B

v Rod

Sol.

If l is the stretched length of the spring, then from figure d 4 = cos 37º = , i.e., 5 l

So, the stretch

l=

y=l −d=

5 d 4

5 d d− d = 4 4

5 3 3 d× = d 4 5 4 Now, taking point B as reference level and applying law of conservation of mechanical energy between A and B, EA = EB

and

h = l sin 37º =

1 2 1 ky = mv 2 2 2 [as for, B, h = 0 and y = 0]

or

mgh +

2

3 1  d 1 mgd + k   = mv 2 4 2  4 2

or

v=d

or

3g k + 2d 16m

[as for A, h =

3 1 d and y = d ] 4 4

Ans.

Ex.17 The block shown in figure is released from rest and initially the spring is at its natural length. Write down the energy conservation equation. When the spring is compressed b

y

l1 ? m

Here two conservative forces are included in the problem. (i) Gravitational force (ii) spring force N.L.

at A as shown in figure. from mechanical energy conservation kf + uf = ki + ui

...(i)

1 1 mv 2 + k 21 = mg(  1 +  ) sin θ 2 2

final position

m

m

( + 1 )sin θ

initial position

We assume zero gravitational potential energy

1

Sol.

B

Ug = mg( +  1) sin θ

Us=0, K = 0 1 Ug = 0, Us = k 12 2 1 K = mv 2 2

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WORK, POWER & ENERGY

(C)

Page # 39

When only one non conservative force is included in problem.

Ex.18 Find out the distance travelled by the block as shown in figure. If the initial speed of the block is v and µ is the friction coefficient between the surface of block and ground. m

Sol.

(D)

v

Applying work energy theorem, we get ⇒

1  2 (–µ mg ) = (0 + 0) –  mv + 0 2

⇒

1 2 v µg  = 2

⇒

v=0

v m

m

Initial

final

mg

v2 = 2µ g

When both conservative and non-conservative force in the problem

Ex.19 A particle slides along a track with elevated ends and a flat central part as shown in figure. The flat portion BC has a length l = 3.0 m. The curved portions of the track are frictionless. For the flat part the coefficient of kinetic friction is µk = 0.20, the particle is released at point A which is at height h = 1.5 m above the flat part of the track. Where does the particle finally comes to rest? Sol.

As initial mechanical energy of the particle is mgh and final is zero, so loss in mechanical energy = mgh. This mechanical energy is lost in doing work against friction in the flat part, So, loss in mechanical energy = work done against friction or mgh = µ mgs i.e.,

s=

h 15 . = = 7.5 m µ 0.2

After starting from B the particle will reach C and then will rise up till the remaining KE at C is converted into potential energy. It will then again descend and at C will have the same value as it had when ascending, but now it will move from C to B. The same will be repeated and finally the particle will come to rest at E such that

h

BC + CB + BE = 7.5 or i.e.,

D

A

B

E

C

3 + 3 + BE = 7.5 BE = 1.5

So, the particle comes to rest at the centre of the flat part. Ex.20 A 0.5 kg block slides from the point A on a horizontal track with an initial speed 3 m/s towards a weightless horizontal spring of length 1 m and force constant 2 N/m. The part AB of the track is frictionless and the part BC has the coefficient of static and kinetic friction as 0.22 and 0.20 respectively. If the distance AB and BD are 2 m and 2.14 m respectively, find the total distance through which the block moves before it comes to rest completely. [g = 10 m/s2] Sol. As the track AB is frictionless, the block moves this distance without loss in its initial 1 1 mv 2 = × 0.5 × 32 = 2.25 J. In the path BD as friction is present, so work done against 2 2 friction = µk mgs = 0.2 × 0.5 × 10 × 2.14 = 2.14 J So, at D the KE of the block is = 2.25 – 2.14 = 0.11 J. Now, if the spring is compressed by x

KE =

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WORK, POWER & ENERGY

Page # 40

0.11 =

1 × k × x2 + µk mgx 2

i.e.,

0.11 =

1 × 2 × x2 + 0.2 × 0.5 × 10x 2

or

x2 + x – 0.11 = 0

A

B

D

C

which on solving gives positive value of x = 0.1 m After moving the distance x = 0.1 m the block comes to rest. Now the compressed spring exerts a force : F = kx = 2 × 0.1 = 0.2 N on the block while limiting frictional force between block and track is fL = µs mg = 0.22 × 0.5 × 10 = 1.1 N. Since, F < fL. The block will not move back. So, the total distance moved by block = AB + BD + 0.1 = 2 + 2.14 + 0.1 = 4.24 m (E) Important Examples : Ex.21 A smooth sphere of radius R is made to translate in a straight line with a constant acceleration a. A particle kept on the top of the sphere is released from there at zero velocity with respect to the sphere. Find the speed of the paritcle with respect to the sphere as a function of the angle θ it slides. Sol. We solve the above problem with respect to the sphere. So apply a pseudo force on the particle m ma R

a with respect to sphere

⇒

Now from work energy theorem. work done by ma = change in mechanical energy ma R sin θ = (kf + uf) – (ki + ui) 1 mv 2 − mgR (1 − cos θ) ⇒ 2 ⇒ v2 = 2R(a sin θ + g – g cos θ) ⇒

maR sin θ =

1 mv 2 = maR sin θ + mgR (1 – cos θ) 2 v = [2R (a sin θ + g – g cos θ)]1/2 m/sec

Ex.22 In the arrangement shown in figure mA = 4.0 kg and mB = 4.0 kg. The system is released from rest and block B is found to have a speed 0.3 m/s after it has descended through a distance of 1m. Find the coefficient of friction between the block and the table. Neglect friction elsewhere. (Take g = 10 m/s2) Sol.

A

B

From constraint relations, we can see that v A = 2 vB Therefore,

vA = 2(0.3) = 0.6 m/s

as

vB = 0.3 m/s (given)

Applying

Wnc = ∆U + ∆K

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WORK, POWER & ENERGY

Page # 41 1 1 mAvA2 + mBvB2 2 2

we get – µ mA g SA = – mB g SB +

Here, SA = 2SB = 2m as SB = 1 m (given) ∴

– µ(4.0) (10) (2) = – (1) (10) (1) +

or – 80 µ = – 10 + 0.72 + 0.045

or

1 1 (4) (0.6)2 + (1) (0.3)2 2 2

80µ = 9.235 or µ = 0.115

Ex.23 A body of mass ‘m’ was slowly hauled up the hill as shown in the figure by a force F which at each point was directed along a tangent to the trajectory. Find the work performed by this force, if the height of the hill is h, the length of its base is l and the coefficient of

Ans.

m

F

h

friction is µ. Sol.

l

Four forces are acting on the body : 1. weight (mg) 2. normal reaction (N) 3. friction (f) and 4. the applied force (F) Using work-energy theorem Wnet = ∆KE or Wmg + WN + Wf + WF = 0 Here, ∆KE = 0, because Ki = 0 = Kf

ds

Wmg = – mgh ⇒ WN = 0 (as normal reaction is perpendicular to displacement at all points) Wf can be calculated as under :

A

f = µ mg cos θ

∴

F

dl

(dWAB)f = – f ds = – (µ mg cos θ) ds = – µ mg (dl)

∴

B

f = – µ mg ∑ dl

(as ds cos θ = dl)

= – µ mgl

Substituting these values in Eq. (i), we get WF = mgh + µmgl

: Here again, if we want to solve this problem without using work-energy theorem we will first find


magnitude of applied force F at different locations and then integrate dW ( = F. dr ) with proper limits.

12.

POWER Power is defined as the time rate of doing work. When the time taken to complete a given amount of work is important, we measure the power of the agent doing work. The average power (P or Pav ) delivered by an agent is given by ∆W

P or Pav = ∆ t =

Total work done Total time

where ∆W is the amount of work done in time ∆ t. Power is the ratio of two scalars-work and time. So, power is a scalar quantity. If time taken to complete a given amount of work is more, then power is less. 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

WORK, POWER & ENERGY

Page # 42

• •

dW
The instantaneous power is, P = where dW is the work done by a force F in a small time dt. dt
dr


dW
F = F. v where v is the velocity of the body.. = . P= dt dt By definition of dot product,

P = Fvcosθ

where θ is the smaller angle between F and v

This P is called as instantaneous power if dt is very small. 12.1

Unit of Power : A unit power is the power of an agent which does unit work in unit time. The power of an agent is said to be one watt if it does one joule of work in one second. 1 watt = 1 joule/second = 107 erg/second Also,

1 watt =

1 newton × 1 metre = 1 N ms–1. 1 sec ond

Dimensional formula of power [Power ] =

[ Work ] [ML2 T –2 ] = = [ML2T–3] [ Time] [ T]

Ex.24 A one kilowatt motor pumps out water from a well 10 metre deep. Calculate the quantity of water pumped out per second. Sol.

Power, P = 1 kilowatt = 103 watt S = 10 m ; Time, t = 1 second ; Mass of water, m = ? ∴

103 =

or

m=

Power =

mg × S t

m × 9.8 × 10 1

10 3 kg 9.8 × 10

= 10.204 kg

Ex.25 The blades of a windmill sweep out a circle of area A. (a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through in time t? (b) What is the kinetic energy or the air? (c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30m2, v = 36 km h–1 and the density of air is 1.2 kg m–3. What is the electrical power produced? Sol.

(a) Volume of wind flowing per second = Av Mass of wind flowing per second = Avρ Mass of air passing in t second = Avρt (b) Kinetic energy of air

=

1 1 1 mv 2 = ( Avρt)v 2 = Av 3 ρt 2 2 2

(c) Electrical energy produced = Electrical power

=

25 1 3 Av 3 ρt × Av ρt = 100 2 8

Av 3 ρt Av 3 ρ = 8t 8

Now, A = 30 m2, v = 36 km h–1 = 36 ×

5 m s–1 18

= 10 m s–1, ρ = 1.2 kg ms–1

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WORK, POWER & ENERGY

∴ Electrical power =

Page # 43 30 × 10 × 10 × 12 . W = 4500 W = 4.5 kW 8

Ex.26 One coolie takes one minute to raise a box through a height of 2 metre. Another one takes 30 second for the same job and does the same amount of work. Which one of the two has greater power and which one uses greater energy? Sol.

Power of first coolie =

Work M × g × S = Time t

Power of second coolie =

=

M × 9.8 × 2 –1 Js 60

 M × 9.8 × 2  M × 9.8 × 2 –1  J s–1 = 2 × Power of first coolie Js = 2    60 30

So, the power of the second coolie is double that of the first. Both the coolies spend the same amount of energy. We know that W = Pt For the same work, W = p1t1 = P2t2 or

P2 t1 = P1 t 2

=

1minute =2 30 s

or

P2 = 2P1

13.

VERTICAL CIRCULAR MOTION

A) B)

To understand this consider the motion of a small body (say stone) tied to a string and whirled in a vertical circle. Now we study the circular motion of the body in two parts. Motion of a body from A to B. Motion of a body from B to C.

A.

C

B

R T

T1

v1

2 m A u mg mv1 R

Motion of a body from A to B.

mv12 ...(1) R During the motion of the body from A to B. θ will increase so cos θ will decrease. Due to which mg cos θ will decrease. From A to B speed of the body also decreases due to which

T1 = mg cos θ +

mv 2 decreases. Therefore tenstion in the string decreases from A to B. R But due to mg cos θ tension can never be zero.

B.

Motion of a body from B to C. T2 =

mv 22

R From B → C

− mg cos θ

speed decreases due to which

C

...(2)

T2

mg mv 22

decreases.

R θ decreases due to which mg cos θ increases. Therefore from B →C. Tension in the string decreases.

String slacks at a point where

v2

mv 22 = mg cos θ i.e., T = 0 R

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mv2 R

B

2

WORK, POWER & ENERGY

Page # 44

13.1

Minimum velocity at point A for which body can complete the vertical circle The condition for the body to complete the vertical circle is that the string should be taut all the time i.e. the tension is greater than zero. So the body can complete the vertical circle if the tension is not zero in between the region B to C. Initially. mv 2 = TC + mg ...(1) R Apply energy conservation from A to C then Kf + Uf = Ki + Ui 1 1 mv 2 + 2mgR = mu 2 + 0 ...(2) 2 2 body can complete vertical circle, when TC ≥ 0

finally

from figure (b)

mv 2 – mg ≥ 0 R 2 ⇒ v ≥ gR ...(3) Put the value from (3) to (2) and u = umin 1 1 2 ⇒ m(Rg) + 2mgR = mumin 2 2

v

mv2 C R

Tc+mg

TA u

A mg fig(a)

fig(b)

2 ⇒ umin = 5gR ⇒ umin = 5gR

It the velocity is greater than equal to

5gR then the body will complete the vertical circle.

Tension at A TA = mg +

mu2 R

If u = umin = then

TA

5gR

5mgR TA = mg + R

u mu2 mg + R

⇒ TA = 6mg

Tension at B mv 2 R energy conservation from A to B

v

TB =

1 1 2 2 mumin = mgR + mv 2 2 ⇒ v2 = 3gR ⇒ TB = 3mg

13.2

A

Condition for the body to reach B : Let us calculate the umin such that the body just reaches B. Work done by tension = 0 Only gravitational force is working on the body which is a conservative force. Therefore Applying conservation of energy, we get mgR = ∴ if u ≤

1 2 mumin ⇒ umin = 2

TB

2gR

B

mv2 R

mg

O R A u min

R

v=0, Ug=mgR K= 0

Ug=0 1 2 K = mumin 2

2gR then the body will oscillate about A.

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WORK, POWER & ENERGY

13.3

When

Page # 45

2gR < u < 5gR

If the velocity of projection is greater than 2gR but less than

5gR , the particle rises above the

horizontal diameter and the tension vanishes before reaching the highest point. We have seen that the tension in the string at the highest point is lower than the tension at the lowest point. At the point D, the string OD makes an angle φ with the vertical. The radial component of the weight is mg cos φ towards the centre O. T + mg cos φ =

mv 2 R

⇒

 v2  – g cos φ  T = m  R 

...(i)

1 mv 2 2 Potential energy at D = mg(AN) = mg (AO + ON) ⇒ mg(R + R cosφ) = mgR(1 + cosφ) From conservation of energy

Kinetic energy at D =

1 1 mu 2 = mv 2 + mgR(1+ cosφ) 2 2 v2 = u2 – 2gR(1 + cos φ) Substituting in equation (i),

B

N

φT

D

O θ mg

A

 u2  T = m  – 2g(1 + cos φ) – g cos φ R    u2 2   T = m – 3g cos φ +  R 3    This equation shows that the tension becomes zero. if u2 2  = 3g cos φ +  R 3 

...(ii)

If the tension is not to become zero. 2  u2 > 3Rg  cos φ +   3

Equation (ii) gives the values of φ at which the string becomes slack. cosφ +

u2 2 = 3Rg 3

cosφ =

2 u2 – 3Rg 3

cos φ =

u 2 – 2gR 3gR

It is the angle from the vertical at which tension in the string vanishes to zero. And after that its motion is projectile. 13.4

Tension in the string versus θ We may find an expression for the tension in the string when it makes an angle θ with the vertical. At C, the weight of the body acts vertically downwards, and the tension in the string is towards the centre O.

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WORK, POWER & ENERGY

Page # 46

The weight mg is resolved radially and tangentially. The radial component is mg cos θ and the tangential component is mg sin θ. T – mg cos θ =

mv 2 , where v is the velocity at C. R

 v2  + g cos θ  ...(i) i.e., T = m  R  The velocity v can be expressed in terms of velocity u at A.

B

O

v θ T M C A u θ mg cos θ mg

1 2 The total energy at A = mu 2 1 mv 2 2 The potential energy at C = mg (AM) = mg (AO – MO) = mg (R – R cosθ) = mgR (1 – cos θ) The kinetic energy at C =

1 mv 2 + mgR(1 – cos θ) 2 ∴ From conservation of energy

The total energy at C =

1 1 mu 2 = mv 2 + mgR (1 – cosθ) 2 2 u2 = v2 + 2gR (1 – cos θ) or Substituting in equation (v),

2

v2 = u – 2gR(1 – cos θ)

  mu 2 2  u2 + 3mg cos θ –  – 2g(1 – cos θ) = T = mg cos θ + R 3 R    

...(ii)

This expression gives the value of the tension in the string in terms of the velocity at the lowest point and the angle θ. Equation (i) shows that tension in the string decreases as θ increases, since the term 'g cos θ' decreases as θ increases. when u =

5gR

⇒ T = 3mg (1 + cos θ) Now θ = 0 ⇒ cos θ = 1 ⇒ if, θ = 90° ⇒ TB = 3 mg θ = 180°, cosθ = – 1 Tc = 0

T 6mg

3mg –1

1

TA = 6 mg

13.5

Different situations :

(A)

A BODY MOVING INSIDE A HOLLOW TUBE OR SPHERE The previous discussion holds good for this case, but instead of tension in the string we have the normal reaction of the

v2

N' mg R

surface. If N is the normal reaction at the lowest point, then the condition u ≥

cosθ

5Rg for the body to complete the

circle holds for this case also. All other equations (can be) similarly obtained by replacing tension T by normal reaction N.

N

u

mg

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WORK, POWER & ENERGY (B)

Page # 47

WHEN BODY IS ATTACHED TO A ROD OF LENGTH R In this case since the body is attached to a rigid rod. The body can not leave the circular path. Therefore, if the speed of the body becomes zero before the highest point C. It's motion will be oscillatory about the centre of the rod. Condition for completing the circle : If the body just reaches the highest point then it will completes the vertical circle Applying energy conservation between the lowest and highest point of circle, we get v=0

Uf = mg2R Kf = 0

u

2mgR =

Ui = 0, Ki =

1 mu 2 ⇒ u = 2

1 mu2 2

4gR

So, If the velocity at point A is greater than equal to

4gR then

body will compete the vertical circle. (C)

VERTICAL MOTION IN A DUAL RING

R u

This system will behave as the preivious system. So umin to complete vertical circle umin =

4gR

Angle at which the normal reaction on the body will change its direction from inward to outward the ring is given by cosφ = (D)

u 2 – 2gR 3 gR

BODY MOVING ON A SPHERICAL SURFACE The small body of mass m is placed on the top of a smooth sphere of radius R and the body slides down the surface. At any instant, i.e., at point C the forces are the normal reaction N and the weight mg. The radial component of the weight is mgcos φ acting towards the centre. The centripetal force is

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WORK, POWER & ENERGY

Page # 48

mv 2 , R where v is the velocity of the body at O.

m

mg cos φ – N =

B C

D

 v 2  N = m  g cos φ – ...(i) R   The body flies off the surface at the point where N becomes zero. v2 v2 ; cos φ = ...(ii) Rg R To find v, we use conservation of energy

φ O

N

mg

A

i.e., g cos φ =

1 mv 2 = mg (BD) 2 = mg (OB – OD) = mgR (1 – cos φ) v2 = 2Rg (1 – cos φ)

i.e.,

2(1 – cos φ) =

v2 Rg

...(iii)

From equation (ii) and (iii) we get cos φ = 2 – 2 cos φ ; 3 cos φ = 2 2  2 ; φ = cos–1   ...(iv) 3  3 This gives the angle at which the body goes of the surface. The height from the ground of that point = AD = R(1 + cos φ) cos φ =

 =R  1 + 

5 2  = R 3  3

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CIRCULAR & W.P.E

Page # 49

Exercise - I

(Objective Problems)

(A) CIRCULAR MOTION 1. A wheel is at rest. Its angular velocity increases uniformly and becomes 80 radian per second after 5 second. The total angular displacement is : (A) 800 rad (B) 400 rad (C) 200 rad (D) 100 rad Sol.

3. A spot light S rotates in a horizontal plane with a constant angular velocity of 0.1 rad/s. The spot of light p moves along the wall at a distance 3 m. What is the velocity of the spot P when θ = 45° ? Wall

P

θ

3m

(Top view) S(Spot light)

(A) 0.6 m/s (C) 0.4 m/s Sol.

2. The second’s hand of a watch has length 6 cm. Speed of end point and magnitude of difference of velocities at two perpendicular positions will be : (A) 2π & 0 mm/s

(B) 2 2 π & 4.44 mm/s

(C) 2 2 π & 2π mm/s

(D) 2π & 2 2 π mm/s

(B) 0.5 m/s (D) 0.3 m/s

4. Two moving particle P and Q are 10 m apart at a certain instant. The velocity of P is 8m/s making an angle 30° with the line joining P and Q and that of Q is 6m/s making an angle 30° with PQ as shown in the figure. Then angular velocity of P with respect to Q is 6m/s

Sol.

P

30°

10m 30°

(A) Zero (C) 0.4 rad/sec Sol.

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8m/s

Q (B) 0.1 rad/sec (D) 0.7 rad sec

CIRCULAR & W.P.E 5. The magnitude of displacement of a particle moving in a circle of radius a with constant angular speed ω varies with time t as ωt (A) 2 a sin ωt (B) 2a sin 2 ωt (C) 2a cos ωt (D) 2a cos 2 Sol.

6. Two bodies A & B rotate about an axis, such that angle θ A (in radians) covered by first body is proportional to square of time, & θB (in radians) covered by second body varies linearly. At t = 0, θA = θB = 0. If A completes its first revolution in π sec. & B needs 4π sec. to complete half revolution then; angular velocity ωA : ωB at t = 5 sec. are in the ratio (A) 4 : 1 (B) 20 : 1 (C) 80 : 1 (D) 20 : 4 Sol.

Page # 50  20  7. A particle moves along a circle of radius   m π with constant tangential acceleration. If the velocity of the particle is 80 m/s at the end of the second revolution after motion has begun, the tangential acceleration is : (B) 40 π m/s2 (A) 160 π m/s2 2 (C) 40 m/s (D) 640 π m/s2 Sol.

8. The graphs below show angular velocity as a function of time. In which one is the magnitude of the angular acceleration constantly decreasing?

(A)

t

(B)

(C)

t

(D)

Sol.

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t

t

CIRCULAR & W.P.E

Page # 51

9. A particle moves with deaceleration along the circle of radius R so that at any moment of time its tangential and normal accelerations are equal in moduli. At the initial moment t = 0 the speed of the particle equals v0, then : (i) the speed of the particle as a function of the distance covered s will be (B) v = v0es/R (A) v = v0 e–s/R –R/s (C) v = v0e (D) v = v0eR/s (ii) the total acceleration of the particle as function of velocity and distance covered (A) a = (C) a =

v2 R R 2 v 2

(B) a =

2

(D) a =

2R v

Sol.

v R

Sol.

11. A particle moves along an arc of a circle of radius R. Its velocity depends on the distance covered as v = a s , where a is a constant then the angle α between the vector of the total acceleration and the vector of velocity as a function of s will be R (A) tanα = (B) tanα = 2s / R 2s 2R s (C) tan α = (D) tanα = s 2R Sol.

10. If angular velocity of a disc depends an angle rotated θ as ω = θ2 + 2θ, then its angular acceleration α at θ = 1 rad is (B) 10 rad/sec2 (A) 8 rad/sec2 2 (C) 12 rad/sec (D) None 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR & W.P.E

Page # 52

Sol.

13.Tangential acceleration of a particle moving in a circle of radius 1 m varies with time t as (initial velocity of particle is zero). Time after which total cceleration of particle makes and angle of 30º with radial acceleration is

60º time(sec)

(A) 4 sec 2/3

(C) 2

(B) 4/3 sec

sec

(D)

2 sec

Sol. 12. A particle A moves along a circle of radius R = 50 cm so that its radius vector r relative to the point O (Fig.) rotates with the constant angular velocity ω = 0.40 rad/s. Then modulus of the velocity of the particle, and the modulus of its total Aacceleration will be r

R

0

(A) v = 0.4 m/s, a = 0.4 m/s2 (B) v = 0.32 m/s, a = 0.32 m/s2 (C) v = 0.32 m/s, a = 0.4 m/s2 (D) v = 0.4 m/s, a = 0.32 m/s2 Sol.

14. A particle is going in a uniform helical and spiral path separately as shown in figure with constant speed.

(A)

(B)

(A) The velocity of the particle is constant in both cases (B) The acceleration of the particle is constant in both cases (C) The magnitude of acceleration is constant in (A) and decreasing in (B) (D) The magnitude of acceleration is decreasing continuously in both the cases Sol.

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CIRCULAR & W.P.E

Page # 53 17. Three identical particles are joined together by a thread as shown in figure. All the three particles are moving on a smooth horizontal plane about point O. If the speed of the outermost particle is v0, then the ratio of the tensions in the three sections of the string is : (Assume that the string remains straight) O

15. If the radii of circular paths of two particles of same masses are in the ratio of 1 : 2, then in order to have same centripetal force, their speeds should be in the ratio of : (A) 1 : 4 (B) 4 : 1 (C) 1 :

2

(D)

(A) 3 : 5 : 7 (C) 7 : 11 : 6 Sol.

B C

(B) 3 : 4 : 5 (D) 3 : 5 : 6 A




2 :1

Sol.

16. A stone of mass of 16 kg is attached to a string 144 m long and is whirled in a horizontal smooth surface. The maximum tension the string can withstand is 16 newton. The maximum speed of revolution of the stone without breaking it, will be : (A) 20 ms–1 (B) 16 ms–1 (C) 14 ms–1 (D) 12 ms–1 Sol.

18. A particle is kept fixed on a turntable rotating uniformly. As seen from the ground, the particle goes in a circle, its speed is 20 cm/s and acceleration is 20 cm/s2. The particle is now shifted to a new position to make the radius half of the original value. The new values of the speed and acceleration will be (A) 10 cm/s, 10 cm/s2 (B) 10 cm/s, 80 cm/s2 (C) 40 cm/s, 10 cm/s2 (D) 40 cm/s, 40 cm/s2 Sol.

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CIRCULAR & W.P.E

Page # 54 Sol.

19. A particle moving along a circular path due to a centripetal force having constant magnitude is an example of motion with : (A) constant speed and velocity (B) variable speed and velocity (C) variable speed and constant velocity (D) constant speed and variable velocity. Sol.

20. A curved section of a road is banked for a speed v. If there is no friction between road and tyres of the car, then : (A) car is more likely to slip at speeds higher than v than speeds lower than v (B) car cannot remain in static equilibrium on the curved section (C) car will not slip when moving with speed v (D) none of the above Sol.

22. A particle of mass m is fixed to one end of a light spring of force constant k and unstretched length
. The system is rotated about the other end of the spring with an angular velocity ω, in gravity free space. The increase in length of the spring will be

k

(A) (C)

mω 2
k mω 2
k + mω 2

Sol.

21. The kinetic energy k of a particle moving along a circle of radius R depends on the distance covered s as k = as2 where a is a positive constant. The total force acting on the particle is :

s2 (A) 2a R

 s2  (B) 2as 1 + 2   R 

(C) 2 as

(D) 2a

1/ 2

R2 S

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(B)

m

mω 2
k − mω 2

(D) None of these

CIRCULAR & W.P.E

23. A unifrom circular ring of mass per unit length λ and radius R is rotating with angular velocity ω about its own axis in a gravity free space. Tension in the ring is 1 (A) zero (B) λ R2 ω2 2 (C) λ R2 ω2 (D) λ R ω2 Sol.

24. A uniform rod of mass m and length
rotates in a horizontal plane with an angular velocity ω about a vertical axis passing through one end. The tension in the rod at distance x from the axis is : 1 1 x2 2 (A) mω x (B) mω 2 2 2
2 x 1   2 1 mω [
2 – x2 ] (C) mω
 1 –  (D) 2
2


Page # 55 25. Water in a bucket is whirled in a vertical circle with a string attached to it. The water does not fall down even when the bucket is inverted at the top of its path. We conclude that : mv 2 R mv 2 (C) mg < R Sol.

(A) mg =

(B) mg >

mv 2 R

(D) none of these

26. A man is standing on a rough (µ = 0.5) horizontal disc rotating with constant angular velocity of 5 rad/ sec. At what distance from centre should he stand so that he does not slip on the disc ? (B) R > 0.2 m (A) R ≤ 0.2 m (C) R > 0.5 m (D) R > 0.3 m Sol.

Sol.

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CIRCULAR & W.P.E

Page # 56

27. A car travelling on a smooth road passes through a curved portion of the road in form of an arc, of circle of radius 10m. If the mass of car is 500 kg, the reaction on car at lowest point P where its speed is 20 m/s is

29. A conical pendulum is moving in a circle with angular velocity ω as shown. If tension in the string is T, which of following equations are correct ? l m

P

(A) 35 kN (C) 25 kN Sol.

(B) 30 kN (D) 20 kN

28. A pendulum bob is swinging in a vertical plane such that its angular amplitude is less than 90°. At its highest point, the string is cut. Which trajectory is possible for the bob afterwards.

(A)

(B)

(C)

(D)

(A) T = mω2l (C) T = mg cosθ Sol.

(B) T sinθ = mω2l (D) T = mω2l sinθ

30. A road is banked at an angle of 30° to the horizontal for negotiating a curve of radius 10 3 m. At what velocity will a car experience no friction while negotiating the curve? (A) 54 km/hr (B) 72 km/hr (C) 36 km/hr (D) 18 km/hr Sol.

Sol.

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CIRCULAR & W.P.E

Page # 57

31. The ratio of period of oscillation of the conical pendulum to that of the simple pendulum is : (Assume the strings are of the same length in the two cases and θ is the angle made by the string with the vertical in case of conical pendulum) (A) cosθ (C) 1 Sol.

(B) cosθ (D) none of these

33. Which vector in the figures best represents the acceleration of pendulum mass of the intermediate point in its swing?

(A)

(B)

(C)

(D)

Sol.

32. A particle is moving in a circle: (A) The resultant force on the particle must be towards the centre. (B) The cross product of the tangential acceleration and the angular velocity will be zero. (C) The direction of the angular acceleration and the angular velocity must be the same. (D) The resultant force may be towards the centre. Sol.

34. The dumbell is placed on a frictionless horizontal table. Sphere A is attached to a frictionless pivot so that B can be made to rotate about A with constant angular velocity. If B makes one revolution in period P, the tension in the rod is 2M

d

B

A

(A)

4 π 2Md P2

(B)

8 π 2Md P2

Sol.

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(C)

2Md 4π 2Md (D) P P

CIRCULAR & W.P.E

Page # 58

35. Two racing cars of masses m1 and m2 are moving in circles of radii r1 and r2 respectively. Their speeds are such that each makes a complete circle in the same time t. The ratio of the angular speeds of the first to the second car is (A) 1 : 1 (B) m1 : m2 (C) r1 : r2 (D) m1m2 : r1r2 Sol.

37. A block of mass m is suspended by a light thread from an elevator. The elevator is accelerating upward with uniform acceleration a. The work done by tension on the block during t seconds is :

T a

m

(A)

m (g + a)at 2 2

(B)

m ( g – a)at 2 2

(C)

m gat 2 2

(D) 0

Sol.

(B) WORK, POWER AND ENERGY

36. A rigid body of mass m is moving in a circle of radius r with a constant speed v. The force on the mv 2 and is directed towards the centre. What r is the work done by this force in moving the body over half the cirumference of the circle.

body is

mv 2

(A)

πr 2

mv 2

(C)

r

2

(B) Zero

(D)

πr 2 mv 2

Sol.

38. Equal force F(> mg) is applied to string in all the three cases. Starting from rest, the point of application of force moves a distance of 2 m down in all cases. In which case the block has maximum kinetic energy?

F

m (1)

(A) 1 (C) 3

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F

F m

m (3)

(2) (B) 2 (D) equal in all 3 cases

CIRCULAR & W.P.E

Page # 59

Sol.

41. A body of mass m accelerates uniformly from rest to a speed v0 in time t0. The work done on the body till any time t is

39. Two springs have their force constant as k1 and k2(k1 > k2). When they are stretched by the same force (A) No work is done by this force in case of both the springs (B) Equal work is done by this force in case of both the springs (C) More work is done by this force in case of second spring (D) More work is done by this force in case of first spring Sol.

 2  1 2 t (A) 2 mv 0  2   t0 

(B)

2 t  (C) mv 0  t   0

2 t  (D) mv 0    t0 

1 t  mv 20  0   t 2 3

Sol.

40. The work done by the frictional force on a surface in drawing a circle of radius r on the surface by a pencil of negligible mass with a normal pressing force N (coefficient of friction µk) is : (A) 4 πr 2 µ kN

(B) –2πr 2 µ kN

(C) –3πr 2 µ kN Sol.

(D) –2πrµ kN

 42. A force F = k[ yi + xj] where k is a positive constant acts on a particle moving in x-y plane starting from the point (3, 5), the particle is taken along a straight line to (5, 7). The work done by the force is : (A) zero (B) 35 K (C) 20 K (D) 15 K

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CIRCULAR & W.P.E

Sol.

Page # 60 45.When a conservative force does positive work on a body (A) the potential energy increases (B) the potential energy decreases (C) total energy increases (D) total energy decreases Sol.

43. A light spring of length 20 cm and force constant 2 kg/cm is placed vertically on a table. A small block of mass 1 kg falls on it. The length h from the surface of the table at which the ball will have the maximum velocity is (A) 20 cm (B) 15 cm (C) 10 cm (D) 5 cm Sol.

44. The work done is joules in increasing the extension of a spring of stiffness 10 N/cm from 4 cm to 6 cm is : (A) 1 (B) 10 (C) 50 (D) 100 Sol.

46. The P.E. of a certain spring when stretched from natural length through a distance 0.3 m is 10 J. The amount of work in joule that must be done on this spring to stretch it through an additional distance 0.15 m will be (A) 10 J (B) 20 J (C) 7.5 J (D) 12.5 J Sol.

47. A 10 kg block is pulled in the vertical plane along a frictionless surface in the form of an arc of a circle of radius 10 m. The applied force is 200 N as shown in the figure. If the block started from rest at A, the velocity at B would be : 60° B

F

A

(A) 1.732 m/s (C) 173.2 m/s Sol.

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(B) 17.32 m/s (D) None of these

CIRCULAR & W.P.E

Page # 61

50. In the figure shown all the surfaces are frictionless, and mass of the block, m = 1kg. The block and wedge are held initially at rest. Now wedge is given a horizontal acceleration of 10 m/s2 by applying a force on the wedge, so that the block does not slip on the wedge. Then work done by the normal force in ground frame on the block in 3 seconds is

10m/s2 m

M 48. A man who is running has half the kinetic energy of the boy of half his mass. The man speeds up by 1 m/s and then has the same kinetic energy as the boy. The original speed of the man was 2m/ s

(A)

(C) 2 m/s

(A) 30 J

(B) 60 J

(C) 150 J

(D) 100

3J

Sol.

(B) ( 2 – 1)m / s (D) ( 2 + 1)m / s

Sol.

49. A particle is released from rest at origin. It moves under influence of potential field U = x2 – 3x, kinetic energy at x = 2 is (A) 2 J (B) 1 J (C) 1.5 J (D) 0 J Sol.

51. A 1.0 kg block collides with a horizontal weightless spring of force constant 2.75 Nm–1 as shown in figure. The block compresses the spring 4.0 m from the rest position. If the coefficient of kinetic friction between the block and horizontal surface is 0.25, the speed of the block at the instant of collision is

(A) 0.4 ms–1 (B) 4 ms–1 (C) 0.8 ms–1 (D) 8 ms–1 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR & W.P.E

Sol.

Page # 62 53. The correct statement is (A) The block will cross the mean position (B) The block come to rest when the forces acting on it are exactly balanced (C) The block will come to rest when the work done by friction becomes equal to the change in energy stored in spring. (D) None Sol.

Question No. 52 to 53 (2 questions) A spring block system is placed on a rough horizontal floor. The block is pulled towards right to give spring 2µmg µmg but more than an elongation less than K K and released. 52. Which of the following laws/principles of physics can be applied on the spring block system

54. A toy car of mass 5 kg moves up a ramp under the influence of force F plotted against displacement x. The maximum height attained is given by F ymax x=0

x=11m

100 80 60 40 20 0 2 4 6 8 10 12 x

(A) conservation of mechanical energy (B) conservation of momentum (C) work energy principle (D) None Sol.

(A) ymax = 20 m (C) ymax = 11 m Sol.

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(B) ymax = 15 m (D) ymax = 5 m

CIRCULAR & W.P.E

Page # 63

55. A wedge of mass M fitted with a spring of stiffness 'k' is kept on a smooth horizontal surface. A rod of mass m is kept on the wedge as shown in the figure. System is in equilibrium. Assuming that all surfaces are smooth, the potential energy stored in the spring is :

Question No. 57 to 62 (6 questions) A block of mass m moving with a velocity v0 on a smooth horizontal surface strikes and compresses a spring of stiffness k till mass comes to rest as shown in the figure. This phenomenon is observed by two observers:

V0

m

k M

m

θ

(A)

mg2 tan 2 θ 2K

(B)

m 2 g tan2 θ 2K

(C)

m 2 g2 tan 2 θ 2K

(D)

m 2 g2 tan 2 θ K

Sol.

K

A : standing on the horizontal surface B : standing on the block 57. To an observer A, the work done by spring force is (A) negative but nothing can be said about its magnitude 1 2 (B) – mv 0 2 (C) positive but nothing can be said about its magnitude 1 2 (D) + mv 0 2 Sol.

56. A block of mass m is hung vertically from an elastic thread of force constant mg/a. Initially the thread was at its natural length and the block is allowed to fall freely. The kinetic energy of the block when it passes through the equilibrium position will be : (A) mga (B) mga/2 (C) zero (D) 2mga Sol.

58. To an observer A, the work done by the normal reaction N between the block and the spring on the block is 1 2 (A) zero (B) – mv 0 2 1 2 (C) + mv 0 (D) none of these 2 Sol.

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CIRCULAR & W.P.E

Page # 64

59. To an observer A, the net work done on the block is 2 (A) –mv 0

1 2 (C) – mv 0 2 Sol.

(B) +mv 20 (D) zero

60. According to the observer A (A) the kinetic energy of the block is converted into the potential energy of the spring (B) the mechanical energy of the spring-mass system is conserved (C) the block loses its kinetic energy because of the negative work done the conservative force of spring (D) all the above Sol.

61. To an observer B, when the block is compressing the spring (A) velocity of the block is decreasing (B) retardation of the block is increasing (C) kinetic energy of the block is zero (D) all the above Sol.

62. According to observer B, the potential energy of the spring increases (A) due to the positive work done by pseudo force (B) due to the positive work done by normal reaction between spring & wall (C) due to the decrease in the kinetic energy of the block (D) all the above Sol.

63. A car of mass 'm' is driven with acceleration 'a' along a straight level road against a constant external resistive force 'R'. When the velocity of the car is 'V', the rate at which the engine of the car is doing work will be : (A) RV (B) maV (C) (R + ma) V (D) (ma – R)V Sol.

64. A truck of mass 30,000 kg moves up an inclined plane of slope 1 in 100 at a speed of 30 kmph. The power of the truck is (given g = 10 ms–2) (A) 25 kW (B) 10 kW (C) 5 kW (D) 2.5 kW Sol.

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CIRCULAR & W.P.E

Page # 65 Sol.

65. A part i c l e m ov es wi th a v el oc i t y  v = (5 i – 3 j + 6k ) m / s under the influence of a constant  force F = (10 i + 10 j + 20k )N. The instantaneous power applied to the particle is : (A) 200 J/s (B) 40 J/s (C) 140 J/s (D) 170 J/s Sol.

66. Assume the aerodynamic drag force on a car is proportional to its speed. If the power output from the engine is doubled, then the maximum speed of the car. (A) is unchanged (B) increases by a factor of 2 (C) is also doubled (D) increases by a factor of four. Sol.

68. The diagrams represent the potential energy U of a function of the inter-atomic distance r. Which diagram corresponds to stable molecules found in nature. U U

(A)

(B)

O U

r

(C)

Sol.

O U

r

(D)

O

67. A body of mass 1 kg starts moving from rest at t = 0, in a circular path of radius 8 m. Its kinetic energy varies as a function of time as : K.E. = 2t2 Joules, where t is in seconds. Then (A) tangential acceleration = 4m/s2 (B) power of all forces at t = 2 sec is 8 watt (C) first round is completed in 2 sec. (D) tangential force at t = 2 sec is 4 newton. 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

r

O

r

CIRCULAR & W.P.E

Page # 66 71. A particle originally at rest the highest point of a smooth vertical circle is slightly displaced. It will leave the circle at a vertical distance h below the highest point, such that (A) h = R (B) h = R/3 (C) h = R/2 (D) h = 2R Sol.

 69. The potential energy for a force field F is given by U(x, y) = sin (x + y). The force acting on the  π particle of mass m at  0,  is 4

(A) 1

(B)

2

(C)

1 2

(D) 0

Sol.

72. A ball whose size is slightly smaller than width of the tube of radius 2.5 m is projected from bottommost point of a smooth tube fixed in a vertical plane with velocity of 10 m/s. If N1 and N2 are the normal reactions exerted by inner side and outer side of the tube on the ball

D

A

C B O

10 m/s (A) N1 > 0 for motion in ABC, N2 > 0 for motion in CDA (B) N1 > 0 for motion in CDA, N2 > 0 for motion in ABC (C) N2 > 0 for motion in ABC & part of CDA (D) N1 is always zero. Sol. 70. F = 2x2 – 3x –2. Choose correct option (A) x = –1/2 is position of stable equilibrium (B) x = 2 is position of stable equilibrium (C) x = –1/2 is position of unstable equilibrium (D) x = 2 is position of neutral equilibrium Sol.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR & W.P.E

Page # 67

73. A bob attached to a string is held horizontal and released. The tension and vertical distance from point of suspension can be represented by.

T

Sol.

T

(A)

(B) h

h

T

T

(C)

(D) h

h

Sol.

74. A small cube with mass M starts at rest point 1 at a height 4R, where R is the radius of the circular part of the track. The cube slides down the frictionless track and around the loop. The force that the track exerts on the cube at point 2 is nearly _________ times the cube’s weight Mg.

1

1R

2 R

(A) 1

(B) 2

(C) 3

(D) 4

75. The tube AC forms a quarter circle in a vertical plane. The ball B has an area of cross-section slightly smaller than that of the tube, and can move without friction through it. B is placed at A and displaced slightly. It will A B

C (A) always be in contact with the inner wall of the tube (B) always be in contact with the outer wall of the tube (C) initially be in contact with the inner wall and later with the outer wall (D) initially be in contact with the outer wall and later with the inner wall

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR & W.P.E

Page # 68

Sol.

76. A particle is rotated in a vertical circle by connecting it to a light rod of length l and keeping the other end of the rod fixed. The minimum speed of particle when the light rod is horizontal for which the particle will complete the circle is (A)

gl

(C) 3gl Sol.

(B)

2gl

(D) none

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR & W.P.E

Page # 69

(Multiple Correct Problems)

Exercise - II

(A) CIRCULAR MOTION  1. A person applies a constant force F on a particle of mass m and finds that the particle moves in a circle of radius r with a uniform speed v as seen (in the plane of motion) from an inertial frame of reference. (A) This is not possible. (B) There are other forces on the particle. (C) The resultant of the other forces is

Sol.

mv 2 towards r

the centre. (D) The resultant of the other forces varies in magnitude as well as in direction. Sol.

3. A simple pendulum of length l and mass (bob) M is oscillating in a plane about a vertical line between angular limits –φ and φ. For an angular displacement θ, [|θ| < φ] the tension in the string and velocity of the bob are T and v respectively. The following relations hold good under the above conditions : (A) T cos θ = Mg Mv 2 L (C) Tangential acc. = g sin θ (D) T = Mg cos θ Sol.

(B) T – Mg cos θ =

2.A machine, in an amusement park, consists of a cage at the end of one arm, hinged at O. The cage revolves along a vertical circle of radius r (ABCDEFGH) ab out i ts hi nge O, at c onst ant l i near spe ed v = gr . The cage is so attached that the man of weight ‘w’ standing on a weighing machine, inside the cage, is always vertical. Then which of the following is correct E × F D ×

×

r ×C

G× × A

B

×

×

H

(A) the reading of his weight on the machine is the same at all positions (B) the weight reading at A is greater than the weight reading at E by 2 w. (C) the weight reading at G = w (D) the ratio of the weight reading at E to that at A = 0 (E) the ratio of the weight reading at A to that at C = 2

(B) W.P.E 4. No work is done by a force on an object if (A) the force is always perpendicular to its velocity (B) the force is always perpendicular to its acceleration (C) the object is stationary but the point of application of the force moves on the object. (D) the object moves in such a way that the point of application of the force remains fixed.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR & W.P.E

Page # 70 Sol.

Sol.

7. When total work done on a particle is positive (A) KE remains constant (B) Momentum increases (C) KE decreases (D) KE increases Sol.

5. One end of a light spring of spring constant k is fixed to a wall and the other end is tied to a block pl aced on a smooth hori zontal surface. In a displacement, the work done by the spring is

1 2 kx . 2

The possible cases are : (A) the spring was initially compressed by a distance x and was finally in its natural length (B) it was initially stretched by a distance x and finally was in its natural length (C) it was initially in its natural length and finally in a compressed position. (D) it was initially in its natural length and finally in a stretched position. Sol.

8. A particle with constant total energy E moves in one dimension in a region where the potential energy is U(x). The speed of the particle is zero where (A) U(x) = E (C)

dU( x) =0 dx

Sol.

6. Work done by force of friction (A) can be zero (B) can be positive (C) can be negative (D) information insufficient

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

(B) U(x) = 0 (D)

d2U( x) dx2

=0

CIRCULAR & W.P.E

Page # 71

9. A block of mass m slides down a plane inclined at an angle θ. Which of the following will NOT increase the energy lost by the block due to friction ? (A) Increasing the angle of inclination (B) Increasing the distance that the block travels (C) Increasing the acceleration due to gravity (D) Increasing the mass of the block Sol.

11. A ball of mass m is attached to the lower end of light vertical spring of force constant k. The upper end of the spring is fixed. The ball is released from rest with the spring at its normal (unstreched) length, comes to rest again after descending through a distance x. (A) x = mg/k (B) x = 2 mg/k (C) The ball will have no acceleration at the position where it has descended through x/2. (D) The ball will have an upward acceleration equal to g at its lowermost position. Sol.

10. A box of mass m is released from rest at position on the frictionless curved track shown. It slides a distance d along the track in time t to reach position 2, dropping a vertical distance h. Let v and a be the instantaneous speed and instantaneous acceleration, respectively, of the box at position 2. Which of the following equations is valid for this situation? 1 m h 2 d (A) h = vt

(B) h = (1/2)gt2

(C) d = (1/2)at2 Sol.

(D) mgh = (1/2)mv2

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR & W.P.E

Page # 72

12. A cart moves with a constant speed along a horizontal circular path. From the cart, a particle is thrown up vertically with respect to the cart (A) The particle will land somewhere on the circular path (B) The particle will land outside the circular path (C) The particle will follow an elliptical path (D) The particle will follow a parabolic path Sol.

Question No. 14 to 16 (3 questions) A particle of mass m is released from a height H on a smooth curved surface which ends into a vertical loop of radius R, as shown m 13. The potential energy in joules of a particle of mass 1 kg moving in a plane is given by U = 3x + 4y, the position coordinates of the point being x and y, measured in meters. If the particle is initially at rest at (6, 4), then (A) its acceleration is of magnitude 5 m/s2 (B) its speed when it crosses the y-axis is 10 m/s (C) it crosses the y-axis (x = 0) at y = –4 (D) it moves in a straight line passing through the origin (0, 0) Sol.

R H

C θ

14. Choose the correct alternative(s) if H = 2R (A) The particle reaches the top of the loop with zero velocity (B) The particle cannot reach the top of the loop (C) The particle breaks off at a height H = R from the base of the loop (D) The particle break off at a height R < H < 2R

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR & W.P.E

Page # 73

Sol.

16. The minimum value of H required so that the particle makes a complete vertical circle is given by (A) 5 R (B) 4 R (C) 2.5 R (D) 2 R Sol.

15. If θ is instantaneous angle which the line joining the particle and the centre of the loop makes with the vertical, then identify the correct statement(s) related to the normal reaction N between the block and the surface (A) The maximum value N occurs at θ = 0 (B) The minimum value of N occurs at N = π for H > 5R/2 (C) The value of N becomes negative for π/2 < θ < 3π/2 (D) The value of N becomes zero only when θ ≥ π/2 Sol.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR & W.P.E

Page # 74

(Subjective Problems)

Exercise - III (A) CIRCULAR MOTION 1. The 10 kg block is in equilibrium. (A)

10 kg

(B)

(i) Find the tension in string A. (ii) Find the tension in string A just after the string B is cut? Sol.

2. A particle moves in the x-y plane with the velocity  v = ai + bt j . At the instant t = a 3 / b the magnitude

3. A particle moves clockwise in a circle of radius 1 m with centre at (x, y) = (1m, 0). It starts at rest at the origin at time t = 0. Its speed increases at the  π constant rate of   m/s2. (a) How long does it take 2 to travel halfway around the circle ? (b) What is the speed at that time ? Sol.

4. A point moves along a circle having a radius 20 cm with a constant tangential acceleration 5 cm/s2. How much time is needed after motion begins for the normal acceleration of the point to be equal to tangential acceleration? Sol.

of tangential, normal and total acceleration are _________________, _______ & __________. Sol.

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CIRCULAR & W.P.E

Page # 75

5. A ring rotates about z axis as shown in figure. The plane of rotation is xy. At a certain instant the acceleration of a particle P (shown in figure) on the ring is  (6 i – 8 j ) m/s2. find the angular acceleration of the ring & the angular velocity at that instant. Radius of the ring is 2m. y P

O

x

Sol.

7. Figure shows the total acceleration and velocity of a particle moving clockwise in a circle of radius 2.5 m at a given instant of time. At this instant, find :

°

30

2. 5m

a=25 m/s

6. A particle is revolving in a circle of radius 1m with an angular speed of 12 rad/s. At t = 0, it was subjected to a constant angular acceleration α and its angular speed increased to (480/π) rpm in 2 sec. Particle then continues to move with attained speed. Calculate (a) angular acceleration of the particle, (b) tangential velocity of the particle as a function of time. (c) acceleration of the particle at t = 0.5 second and at t = 3 second (d) angular displacement at t = 3 second. Sol.

a

v

(a) the radial acceleration, (b) the speed of the particle and (c) its tangential acceleration Sol.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

2

CIRCULAR & W.P.E

Page # 76

8. A stone is launched upward at 45° with speed v0. A bee follows the trajectory of the stone at a constant speed equal to the initial speed of the stone. (a) Find the radius of curvature at the top point of the trajectory. (b) What is the acceleration of the bee at the top point of the trajectory? For the stone, neglect the air resistance. Sol.

Sol.

9. A particle moves in circle of radius R with a constant speed v. Then, find the magnitude of average acceleration during a time interval

πR . 2v

Sol.

11. A rod of length 1 m is being rotated about its end in a gravity free space with a constant angular acceleration of 5 rad/s2 starting from rest. A sleeve is fitted on the rod at a distance of 0.5 m from the centre. The coefficient of friction between the rod and the sleeve is 0.05. Find the time after which sleeve will start slipping on the rod. Sol.

10. A 4 kg block is attached to a vertical rod by means of two strings of equal length. When the system rotaes about the axis of the rod, the strings are extended as shown in figure. (a) How many revolutions per minute must the system make in order for the tension in the upper chord to be 20 kgf? (b) What is the tension in the lower chord? 5m 8m

θ

A

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR & W.P.E

Page # 77

12. A mass m rotating freely in a horizontal circle of radius 1 m on a frictionless smooth table supports a stationary mass 2m, attached to the other end of the string passing through smooth hole O in table, hanging vertically. Find the angular velocity of rotation.

O

14. The blocks are of mass 2 kg shown is in equilibrium. At t = 0 right spring in fig. (i) and right string in fig. (ii) breaks. Find the ratio of instantaneous acceleration of blocks ? 37° 37°

m

2m

2 kg

Sol.

37°

37°

2 kg figure (ii)

Sol.

13. A beam of mass m is attached to one end of a spring of natural length

3 R and spring constant

( 3 + 1)mg . The other end of the spring is fixed at R point A on a smooth fixed vertical ring of radius R as shown in the figure. What is the normal reaction at B just after the bead is released? B k=

A

Sol.

60°

(B) WORK, POWER AND ENERGY 15. A block of mass m is pulled on a rough horizontal surface which has a friction coefficient µ. A force F isapplied which is capable of moving the body uniformly with speed v. Find the work done on the block in time t by (a) weight of the block, (b) Normal reaction by surface on the block, (c) friction, (d) F. Sol.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR & W.P.E

Page # 78

(Force dyne)

18. The relationship between force and position is shown in the figure given (in one dimensional case). What will be the work done by the force in displacing a body from x = 1 cm to x = 5 cm. 20 10 0

1

10

2

3

4

5 6 x(cm)

20

Sol.

16. Calculate the work done against gravity by a coolie in carrying a load of mass 10 kg on his head when he walks uniformly a distance of 5 m in the (i) horizontal direction (ii) vertical direction. (Take g = 10 m/s2) Sol.

17. A body is constrained to move in the y-direction. It is subjected to a force (–2i + 15 j + 6k ) newton. What is the work done by this force in moving the body through a distance of 10 m ? Sol.

19. It is well known that a raindrop or a small pebble falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to be proportional to the speed of the drop but is otherwise undetermined. Consider a drop or small pebble of 1 g falling from a cliff of height 1.00 km. It hits –1 . What is the work done by the unknown resistive force ? Sol. t h

e

g

r o u

n

d

w

i t h

a

s p

e

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

e

d

o f

5

0

. 0

m

s

CIRCULAR & W.P.E

Page # 79

20. A rigid body of mass 2 kg initially at rest moves under the action of an applied horizontal force 7 N on a table with coefficient of kinetic friction = 0.1. Calculate the (a) work done by the applied force on the body in 10 s. (b) work done by friction on the body in 10 s. (c) work done by the net force on the body in 10 s. (d) change in kinetic energy of the body is 10 s. Sol.

21. A rigid body of mass 0.3 kg is taken slowly up an inclined plane of length 10 m and height 5 m, and then allowed to slide down to the bottom again. The coefficient of friction between the body and the plane is 0.15. Using g = 9.8 m/s2 find the

(a) work done by the gravitational force over the round trip. (b) work done by the applied force (assuming it to be parallel to the inclined plane) over the upward journey (c) work done by frictional force over the round trip. (d) kinetic energy of the body at the end of the trip? Sol.

22. A block of mass m sits at rest on a frictionless table in a rail car that is moving with speed vc along a straight horizontal track (fig.) A person riding in the car pushes on the block with a net horizontal force F for a time t in the direction of the car's motion.

Ground

Train

m

s

F

s1

(a) What is the final speed of the block according to a person in the car ? 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR & W.P.E

Page # 80

Sol. (g) How much work does each say the force did ? Sol.

(b) According to a person standing on the ground outside the train? Sol.

(h) Compare the work done to the K gain according to each person. Sol.

(c) How much did K of the block change according to the person in the car ? Sol.

(i) What can your conclude from this computation? Sol. (d) According to the person on the ground ? Sol.

(e) In terms of F, m, & t, how far did the force displace the object according to the person in car ? Sol.

23. In the figure shown, pulley and spring are ideal. Find the potential energy stored in the spring (m1 > m2) Sol.

(f) According to the person on the ground ? Sol.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

k

m1 m2

CIRCULAR & W.P.E

Page # 81

24. A block of mass m placed on a smooth horizontal surface is attached to a spring and is held at rest by a force P as shown. Suddenly the force P changes its direction opposite to the previous one. How many times is the maximum extension l2 of the spring longer compared to its initial compression l2 ?

26. A labourer lifts 100 stones to a height of 6 metre in two minute. If mass of each stone be one kilogram, calculate the average power. Given : g = 10 ms–2. Sol.

l1 P

Sol.

27. An engine develops 10 kW of power. How much time will it take to lift a mass of 200 kg through a height of 40 m? Given : g = 10 ms–2 Sol.

25. (a) Power applied to a particle varies with time as P = (3t2 – 2t + 1) watt, where t is in second. Find the change in its kinetic energy between time t = 2 s and t = 4 s. Sol.

28. Two trains of equal masses are drawn along smooth level lines by engines; one of then X exerts a constant force while the other Y works at a constant rate. Both start from rest & after a time t both again have the same velocity v. Find the ratio of travelled distance during the interval. Sol.

(b) The potential function for a conservative force is given by U = k(x + y). Find the work done by the conservative force in moving a particle from the point A(1, 1) to point B (2, 3). Sol.

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CIRCULAR & W.P.E

Page # 82 30. A particle moves along a straight line. A force acts on the particle which produces a constant power. It starts with initial velocity 3 m/s and after moving a distance 252 m its velocity is 6 m/s. Find the time taken. Sol.

29. Water is pumped from a depth of 10m and delivered through a pipe of cross section 10–2m2 upto a height of 10m. If it is needed to deliver a volume 0.2 m3 per second, find the power required. [Use g = 10 m/s2] Sol.

31. A force F = x2y2i + x2y2j (N) acts on a particle which moves in the XY plane.

Y

a

C

D a X A B Find the work done by F as it moves the particle from A to C (fig.) along each of the paths ABC, ADC, and AC. 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR & W.P.E

Sol.

Page # 83 32. Calculate the forces F(y) associated with the following one-dimensional potential energies : (a) U = –ωy (b) U = ay3 – by2 (c) U = U0 sin β y Sol.

33. Consider the shown arrangement when a is bob of mass ‘m’ is suspended by means of a string connected  to peg P. If the bob is given a horizontal velocity u having magnitude 3gl , find the minimum speed of the bob in subsequent motion. P l

Sol.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

u

CIRCULAR & W.P.E

34. A person rolls a small ball with speed u along the floor from point A. If x = 3R, determine the required speed u so that the ball returns to A after rolling on the circular surface in the vertical plane from B to C and becoming a projectile at C. What is the minimum value of x for which the game could be played if contact must be maintained to point C ? Neglect friction.

Page # 84 35. A toy rocket of mass 1 kg has a small fuel of mass 0.02 kg which it burns out in 3 s. Starting from rest on a horizontal smooth track, it gets a speed of 20 ms–1 after the fuel is burnt out. What is the average thrust of the rocket? What is the energy content per unit mass of the fuel? (Ignore the small mass variation of the rocket during fuel burning). Sol.

Sol.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR & W.P.E

Page # 85

(Tough Subjective Problems)

Exercise - IV Q.1 A particle which moves along the curved path shown passes point O with a speed of 12 m/s and slows down to 5m/s at point A in a distance of 18 m measured along the curve from O. The deceleration measured along the curve it proportional to distance from O. If the total acceleration of the particle is 10 m/s2 on it passes A. Find the radius of curvature of A.

A

O

Sol.

Q.3 A small is block can move in a straight horizontal line a along AB. Flash lights from one side projects its shadow on a vertical wall which has horizontal cross section as a circle. Find tangential & normal acceleration of shadow of the block on the wall as a function of time if the velocity of the block is constant (v). B

R

v=const

A Top View

Q.2 A ball of mass 1 kg is released from position A inside a wedge with a hemispherical cut of radius 0.5 m as shown in the figure. Find the force exerted by the vertical wall OM on wedge, when the ball is in position B. (neglect friction everywhere) Take (g = 10m/s2) M A C

Sol.

60° O

N

Sol.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR & W.P.E

Page # 86

Q.4 A particle is confined to move along the +x axis under the ac tion of a force F(x) that is derivable from thePotential U(x) = ax3 – bx. U

Sol.

x x1 x0 (a) Find the expression for F(x) (b) When the total energy of the particle is zero, the particle can be trapped with in the interval x = 0 to x = x1. For this case find the values of x1. (c) Determine the maximum kinetic energy that the trapped particle has in its motion. Express all answers in terms a and b. Sol.

O

Q.5 A particle of mass 5 kg is free to slide on a smooth ring of radius r = 20 cm fixed in a vertical plane. The particle is attached to one end of a spring whose other end is fixed to the top point O of the ring. Initially the particle is at rest at a point A of the ring such that ∠OCA = 60º, C being the centre of the ring. The natural length of the spring is also equal to r = 20 cm. After the particle is released and slides down the ring the contact force between the particle & the ring becomes zero when it reaches the lowest position B. Determine the force constant of the spring.

Q.6 Two blocks of mass m1 = 10kg and m2 = 5kg connected to each other by a massless inextensible string of length 0.3 m are placed along a diameter of a turn table. The coefficient of friction between the table and m1 is 0.5 while there is no friction between m2 and the table. The table is rotating with an angular velocity of 10 rad/sec about a vertical axis passing through its centre. The masses are placed along the diameter of the table on either side of the centre O such that m1 is at a distance of 0.124 m from O. The masses are observed to be at rest with respect to an observer on the turn table. (i) Calculate the frictional force on m1 (ii) What should be the minimum angular speed of the turn table so that the masses will slip from this position (iii) How should the masses be placed with the string remaining taut, so that there is no frictional force acting on the mass m1. Sol.

O

A 60°

C

B

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CIRCULAR & W.P.E

Page # 87 Q.8 Two identical beads of mass 1 kg each are connected by an inextensible massless string & they can slide along the two arms AB and BC of a rigid smooth wire frame in vertical plane. If the system is released from rest, find the speeds of the particles when they have moved by a dis tance of 0.1 m. Also find tension in the string. 0.4m B A

0.3m

C Q.7 A ring of mass m can slide over a smooth vertical rod. The ring is connected to a spring of force con4mg stant K = where 2R is the natural length of the R spring. The other end of the spring is fixed to the ground at a horizontal distance 2R from the base of the rod. The mass is released at a height of 1.5 R from ground

Sol.

3R/2

A 2R (a) calculate the work done by the spring (b) calculate the velocity of the ring as it reaches the ground. Sol.

Q.9 The ends of spring are attached to blocks of mass 3 kg and 3 kg. The 3 kg block rests on a horizontal surface and the 2 kg block which is vertically above it is in equilibrium producing a compression of 1cm of the spring. The 2kg mass must be compressed further by at least ________, so that when it is released, the 3 kg block may be lifted off the ground.

2kg

3 kg Sol.

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Page # 88

Q.10 A uniform rod of mass m length L is sliding along its length on a horizontal table whose top is partly smooth & rest rough with friction coefficient µ. If the rod after moving through smooth part, enters the rough with velocity v0. (a) What will be the magnitude of the friction force when its x length (< L) lies in the rough part during sliding. (b) Determine the minimum velocity v0 with which it must enter so that it lies completely in rough region before coming to rest. (c) If the velocity is double the minimum velocity as calculated in part (a) then what distance does its front end A would have travelled in rough region before rod comes to rest.

Sol.

L

v0 B

M

A

m

Sol. Q.12 A small bead of mass m is free to slide on a fixed smooth vertical wire, as indicated in the diagram. One end of a light elastic string, of unstretched length a and force constant 2 mg/a is attached to B. The string passes through a smooth fixed ring R and the other end of the string is attached to the fixed point A, AR being horizontal. The point O on the wire is at same horizontal level as R, and AR = RO = a. (i) In the equilibrium, find OB (ii) The bead B is raised to a point C of the wire above O, where OC = a, and is released from rest. Find the speed of the bead as it passes O, and find the greatest depth below O of the bead in the subsequent motion.

O B Sol.

Q.11 Find the velocity with which a block of mass 1 kg must be horizontally projected on a conveyer belt moving uniformly at a velocity of 3 m/s so that maximum heat is liberated. Take coefficient of friction of 0.1. Also find the corresponding amount of heat liberated. What happens when belt velocity is 5 m/s ? 1kg v=3m/s 8m

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a R

A

CIRCULAR & W.P.E

Page # 89

Q.13 A small block of mass m is projected horizontally from the top the smooth hemisphere of radius r with speed u as shown. For values of u ≥ u0, it does not slide on the hemisphere (i.e. leaves the surface at the top itself) (a) For u = 2u0 it lands at point P on ground Find OP. (b) For u = u0/3, Find the height from the ground at which it leaves the hemisphere. (c) Find its net acceleration at the instant it leaves the hemisphere. r

Sol.

u

o

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Page # 90

JEE-Problems

Exercise - V  Q.1 A force F = −K( y i + x j ) where K is a positive con-

stant, acts on a particle moving in the x-y plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work  done by the force F on the particle is [JEE-98] (A) –2Ka2 (B) 2Ka2 (C) –Ka2 (D) Ka2 Sol.

Q.3 A particle is suspended vertically from a point O by an inextensible massless string of length L. A vertical line AB is at a distance L/8 from O as shown. The object given a horizontal velocity u. At some point, its motion ceases to be circular and eventually the object passes through the line AB. At the instant of crossing AB, its velocity is horizontal. Find u. [JEE-99] A O

u

Sol.

Q.2 A stone is 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 at it reaches a position where the string is horizontal is [JEE-98] (A)

(u 2 − 2gl )

(B)

2gl

(C)

(u2 − gl )

(D)

2(u2 − gl )

Sol.

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B

CIRCULAR & W.P.E

Page # 91

Q.4 A long horizontal rod has bead which can slide along its length, and initially placed at a distance L from one end of A of the rod. The rod is set in angular motion about A with constant angular acceleration α. If the coefficient of friction between the rod and the bead is µ and gravity is neglected, then the time after which is bead starts slipping is [JEE-2000] µ α

(A)

(B)

µ α

1

(C)

µα

(D) infinitesimal

(A)

v

(B)

v

(C)

v

(D)

v

Sol.

Sol.

Q.6 An insect crawls up a hemispherical surface very slowly (see the 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 α with the vertical, the maximum possible value of α is given by [JEE(Scr.)-2001]

(A) cot α = 3 (C) sec α = 3 Sol.

Q.5 A small block is shot into each of the four tracks as shown below. Each of the tracks risks to the same height. The speed with which the block enters the track is the same in all cases. At the highest point of the track, the normal reaction is maximum in [JEE-2001] 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

(B) tan α = 3 (D) cosec α = 3

CIRCULAR & W.P.E

Q.7 A small ball of mass 2 × 10–3 Kg having a charge of 1µc is suspended by a string of length 0.8m. Another identical ball having the same charge is kept at the point of suspension. Determine the minimum horizontal velocity which should be imparted to the lower ball so that it can make complete revolution. [JEE-2001] Sol.

Page # 92

(A)

(B)

(C)

(D)

Sol.

Q.9 A particle, which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance x of the particle x of the particle from the origin as F(x) = –kx + ax2. Here k and a are positive constants. For x ≥ 0, the functional form of the potential energy U(x) of the particle is [JEE(Scr.)-2002] U(x)

U(x) x

(A) Q.8 A simple pendulum is oscillating without damping. When the displacement of the bob is less that maxi mum, its acceleration vector a is correctly shown in [JEE(Scr.)-2002]

(B)

U(x)

(C)

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x

U(x)

x

(D)

x

CIRCULAR & W.P.E

Sol.

Page # 93 (a) Express the total normal reaction force exerted by the spheres on the ball as a function of angle θ. Sphere B

d

Q.10 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 [JEE(Scr.)-2002] (A) 4 Mg/k (B) 2 Mg/k (C) Mg/k (D) Mg/2k Sol.

R

O

Sphere A (b) Let NA and NB denote the magnitudes of the normal reaction force on the ball exerted by the spheres A and B, respectively. Sketch the variations of NA and NB as functions of cosθ in the range 0 ≤ θ ≤ π by drawing two separate graphs in your answer book, taking cosθ on the horizontal axes. Sol.

Q.12 In a region of only gravitational field of mass ‘M’ a particle is shifted from A to B via three different paths in the figure. The work done in different paths are W1, W2, W3 respectively then [JEE(Scr.)-2003] Q.11 A spherical ball of mass m is kept at the highest point in the space between two fixed, concentric spheres A and B (see figure.) The smaller sphere A has a radius R and the space between the two spheres has a width d. The ball has a diameter very slightly less than d. All surfaces are frictionless. The ball is given a gentle push (towards the right in the figure). The angle made by the radius vector of the ball with the upward vertical is denoted by θ (shown in the figure) [JEE-2002]

(3)

B

C

M (2) (1) A

(A) W1 = W2 = W3 (C) W1 > W2 > W3

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(B) W1 = W2 > W3 (D) W1 < W2 < W3

CIRCULAR & W.P.E

Page # 94 U(x)

Sol. U(x) x

(A)

U(x)

U(x)

(C)

x

(B)

x

(D)

x

Sol. Q.13 A particle of mass m, moving in a circular path of radius R with a constant speed v2 is located at point (2R, 0) at time t = 0 and a man starts moving with velocity v1 along the +ve y-axis from origin at time t = 0. Calculate the linear momentum of the particle w.r.t. the man as a function of time. [JEE-2003] y v2 v1 R (0,0)

m

x

Sol.

Q.14 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 (where U is the potential energy function) [JEE(Scr.)-2004]

Q.15 STATEMENT-1 A block of mass m starts moving on a rough horizontal surface with a velocity v. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of 30º with the horizontal and the same block is made to go up on the surface with the same initial velocity v. The decrease in the mechanical energy in the second situation is smaller than that in the first situation. because STATEMENT-2 The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination. [JEE-2007] (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

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Page # 95

Sol.

[JEE 2009] Sol. 16. A bob of mass M is suspended by a massless string of length L. The horizontal velocity V at position A is just sufficient to make it reach the point B. The angle θ at which the speed of the bob is half of that at A, satisfies

[JEE 2008]

B

L A (A) θ =

π 4

π 3π (C) < θ < 2 4 Sol.

V (B)

π π v1

v 2 – v1 e = u –u 1 2 Note : Coefficient of restitution is a factor between two colliding bodies which is depends on the material of the body but independent of shape. We can say e is a factor which relates deformation and reformation of the body. 0  e1

Ex.35 If a body falls normally on a surface from height h, what will be the height regained after collision if coefficient of restitution is e?

h

Sol.

If a body falls from height h, from equations of motion we know that it will hit the ground with a velocity say u =

2gh which is also the velocity of approach here.

Now if after collision it regains a height h1 then again by equations of motion v =

2gh1 which is also

the velocity of separation. So, by definition of e, e=

2gh1 2gh

or h1 = e2h

Ex.36 A block of mass 2 kg is pushed towards a very heavy object moving with 2 m/s closer to the block (as 2m/s very shown). Assuming elastic collision and frictionless heavy 10m/s surface, find 2kg object the final velocities of the blocks. Sol. Let v1 and v2 be the final velocities of 2kg block 2m/s very and heavy object respectively then, heavy 14m/s v1 = u1 + 1 (u1 – u2) = 2u1 – u2 2kg object = – 14 m/s v2 = – 2m/s Ex.37 A ball is moving with velocity 2 m/s towards a heavy wall moving towards the ball with speed 1 m/s as shown in fig. Assuming collision to be elastic, find the velocity of the ball immediately after the collision.

2m/s

Sol.

1m/s

The speed of wall will not change after the collision. So, let v be the velocity of the ball after collision in the direction shown in figure. Since collision is elastic (e = 1).

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CENTRE OF MASS

2m/s

Page # 31

1m/s

Before Collision

1m/s

v

After Collision

separation speed = approach speed or v–1=2+1 or v = 4 m/s Ans. Ex.38 A ball is dropped from a height h on to a floor. If in each collision its speed becomes e times of its striking value (a) find the time taken by ball to stop rebounding (b) find the total change in momentum in this time (c) find the average force exerted by the ball on the floor using results of part (a) and (b). Sol. (a) When the ball is dropped from a height h, time taken by it to reach the ground will be

t0 =

2h and its speed v0 = g

2gh

h

v0

v1 v2

t0

t1

t0

Now after collision its speed will becomes e times, i.e., v1 = ev0 = e 2gh and so, it will take time to go up till its speed becomes zero = (v1/g). The same time it will take to come down. So total time between I and II collision will be t1 = 2v1/g. Similarly, total time between II and III collision t2 = 2v2/g. So total time of motion T = t0 + t1 + t2 +......... or

T = t0 +

2 v1 2v 2 + ....... g g

or

T = t0 +

2ev 0 2e 2 v 0 + ....... g g [as v2 = ev1 = e2v0]

or

T =



2h [1  2e(1  e  e2 ....)] g

2h   1    1  2e g  1 e 

2h  1  e  g  1 – e 

(b) Change in momentum in I collision = mv1 – (–mv0) = m (v1 + v0) Change in momentum in II collision = m(v2 + v1) Change in momentum in nth collision = m(vn + vn–1) Adding these all total change in momentum p = m[v0 + 2v1 + ....+ 2vn–1 + vn] or

p = mv0[1 + 2e + e2 + .....]

or

 1 e   1  p = mv0 1  2e 1 – e    m 2gh  1 – e     

...(2)

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CENTRE OF MASS

Page # 32   dp p (C) Now as F  so, Fav =  T dt

Substituting the value of T and p from Eqns. (1) and (2) 1 e  Fav = m 2gh   × 1 – e 

7.1

g 1 – e  = mg 2h  1  e 

...(3)

Line of Motion The line passing through the centre of the body along the direction of resultant velocity.

7.2

Line of Impact The line passing through the common normal to the surfaces in contact during impact is called line of impact. The force during collision acts along this line on both the bodies. Direction of Line of impact can be determined by : (a) Geometry of colliding objects like spheres, discs, wedge etc. (b) Direction of change of momentum. If one particle is stationary before the collision then the line of impact will be along its motion after collision. Examples of line of impact (i) Two balls A and B are approaching each other such that their centres are moving along line CD. Line of impact and line of motion

C

D

A

B

(ii) Two balls A and B are approaching each other such that their centre are moving along dotted lines as shown in figure.

B Line of motion of ball B

Line of motion of ball A D

A Line of impact (iii)

Ball is falling on a stationary wedge.

Line of motion of ball

Line of impact

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CENTRE OF MASS

Page # 33

Note : In previous discussed examples line of motion is same as line of impact. But in problems in which line of impact and line of motion is different then e will be

velocity of seperation along line of impact velocity of approach along line of impact

e=

Ex.40 A ball of mass m hits a floor with a speed v making an angle of incident  with the normal. The coefficient of restitution is e. Find the speed of the reflected ball and the angle of reflection of the ball. Sol.

Suppose the angle of reflection is  and the speed after the collision is v  (shown figure) The floor exerts a force on the ball along the normal during the collision. There is no force parallel to the surface. Thus, the parallel component of the velocity of the ball remains unchanged. This gives v  sin   = v sin 

...(i)

For the components normal to the floor, the velocity of separation is v cos  and the velocity of approach is v cos .

v

 '

v'

Hence, v  cos   = ev cos  sin 2   e 2 cos 2 

From (i) and (ii), v  = v

tan e For elastic collision, e = 1, so that  =  and v = v.

v cos 

tan   =

Hence,

v'cos '

v sin

Initial velocity

v' sin  ' Final velocity

Ex.41 A ball is projected from the ground at some angle with horizontal. Coefficient of restitution between the ball and the ground is e. Let a, b and c be the ratio of times of flight, horizontal range and maximum height in two successive paths. Find a, b and c in terms of e?

1 2

Sol.

Let us assume that ball is projected with speed u at an angle  with the horizontal. Then Before first collision with the ground. Time fo flight T 

2u y

euy

g

Horizontal range R 

u I

u y  u sin 

2u x u y g

u x  u cos 

u2y Maximum Height Hmax =

ux



...(1)

2g

uy

After striking the ground the component uy is change into e uy, so 2eu y

Time of flight T =

'  Hmax

g

, R' 

2u x ( eu y ) g

(eu y )2 2g

...(2)

from eq (1) & (2) Now

T 1 a T' e

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II

CENTRE OF MASS

Page # 34 Hmax 1 R 1  2 =c b ; ' Hmax e R' e

Ex.42 A ball is projected from the ground with speed u at an angle  with horizontal. It collides with a wall at a distance a from the point of projection and returns to its original position. Find the coefficient of restitution between the ball and the wall. Sol.

A ball is a projected with speed u at an angle  with horizontal. It collides at a distance a with a wall parallel to y-axis as shown in figure. Let vx and vy be the components of its velocity along x and y-directions at the time of impact with wall. Coefficient of restitution between the ball and the wall is e. Component of its velocity along y-direction (common tangent) vy will remain unchanged while component of its velocity along x-direction (common normal) vx will becomes evx is opposite direction. *Further, since vy does not change due to collision, the time of flight (time taken by the ball to return to the same level) and maximum height attained by the ball remain same as it would had been in the absence of collision with the wall. Thus, vy

v B

A

u

vx

evx

C

vy

B

y

O



O

x

a

a

From O A B, R = a = u cos  . tOAB from BCO, R = a = eucos. tBCO T = tOAB + tBCO

or

2u sin a a = + g u cos  eu cos 

or

a 2u 2 sin  cos  – ag  eu cos  gu cos 

2u sin a a – = g eu cos  u cos 

or

ag 

e=

2

2u sin  cos  – ag

1 e =  u 2 sin 2  Ans.  – 1 ag   Ex.43 To test the manufactured properties of 10 N steel balls, each ball is released from rest as shown and strikes a 45° inclined surface. If the coefficient of restitution is to be e = 0.8. determine the distance s to where the ball must strike the horizontal plane at A. At what speed does the ball strike at A? (g = 9.8 m/s2) or

B 1.5m

1.0m

45° A s

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CENTRE OF MASS Sol.

v0 =

2gh =

Page # 35 2  9.8  15 . = 5.42 m/s

Component of velocity parallel and perpendicular to plane at the time of collision. v0

v1 = v2 =

= 3.83 m/sec.

2 ev2=0.8v2

C

C

v1

v2

v1

1.0 m 45°

v0 45°

D

x

A

E s x

y

Component parallel to plane (v1) remains unchanged, while component perpendicular to plane becomes ev2, where ev2 = 0.8 × 3.83 = 3.0 m/s  Component of velocity in horizontal direction after collision ( v1  ev 2 ) vx =

(3.83  3.0) =

2

2

= 4.83 m/s

While component of velocity in vertical direction after collision. v1 – ev 2 vy =

2

3.83 – 3.0 =

2

= 0.59 m/s

Let t be the time, the particle takes from point C to A, then 1.0 = 0.59 t +

1 × 9.8 × t2 ; 2

t = 0.4 sec

Solving this we get, 

DA = vxt = (4.83)(0.4) = 1.93 m



S = DA – DE = 1.93 – 1.0 S = 0.93 m vyA = vyc + gt = (0.59) + (9.8) (0.4) = 4.51 m/s vxA = vxC = 4.83 m/s



vA =

( v xA ) 2  ( v yA ) 2 = 6.6 m/s

Ex.44 A ball of mass m = 1 kg falling vertically with a velocity v0 = 2m/s strikes a wedge of mass M = 2kg kept on a smooth, horizontal surface as shown in figure. The coefficient of restitution 1 between the ball and the wedge is e = . Find the velocity of the wedge and the ball immediately 2 after collision.

m v0 M

Sol.

30°

Given M = 2kg and m = 1kg

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CENTRE OF MASS

Page # 36

v3

v1

Jcos30°

M

m

v2

m

Jsin30° J

30°

J Jsin30°

30°

Jcos30°

Let, J be the impulse between ball and wedge during collision and v1, v2 and v3 be the components of velocity of the wedge and the ball in horizontal and vertical directions respectively. Applying

impulse = change in momentum

we get

J sin 30° = Mv1 = mv2 J = 2v1 = v2 2

or

J cos 30° = m(v3 + v0)

3 J = (v3 + 2) 2

or

...(i)

...(ii)

Applying, relative speed of separation = e (relative speed of approach) in common normal direction, we get (v1 + v2) sin 30° + v3 cos30° =

1 ( v 0 cos 30 ) 2 Common normal direction

3 ...(iii) 2 Solving Eqs. (i), (ii) and (iii), we get or

v1 + v2 +

30°

1 v1 =

3v3 =

3

m/s

2

m / s and v = 0 3 3 Thus, velocities of wedge v2 =

1 and ball are v1 = 1

v1 

3

m/s

v2 

m/s

3

2

m/s

3

30°

2 and v2 =

3

m / s in horizontal direction as shown in figure.

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8.

Page # 37

COLLISION OR IMPACT Collision is an event in which an impulsive force acts between two or more bodies for a short time, which results in change of their velocities. Note :

•

In a collision, particles may or may not come in physical contact.

•

The duration of collision, t is negligible as compared to the usual time intervals of observation of motion.

•

In a collision the effect of external non impulsive forces such as gravity are not taken into account as due to small duration of collision (t) average impulsive force responsible for collision is much larger than external forces acting on the system.

The collision is in fact a redistribution of total momentum of the particle : Thus law of conservation of linear momentum is indepensible in dealing with the phenomenon of collision between particles. Consider a situation shown in figure. Two balls of masses m1 and m2 are moving with velocities v1 and v2 (< v1) along the same straight line in a smooth horizontal surface. Now let us see what happens during the collision between two particles. v1

v2

m1

m2

figure (a) v1 '

N

N

v2 '

N

N

figure(b)

figure(c) figure (a) : Balls of mass m1 is behind m2. Since v1 > v2, the balls will collide after some time. figure (b) : During collision both the balls are a little bit deformed. Due to deformation two equal and opposite normal forces act on both the balls. These forces decreases the velocity of m1 and increase the velocity of m2 figure (c): Now velocity of ball m1 is decrease from v1 to v1 and velocity of ball m2 is increase from v2 to v2. But still v1 > v2 so both the ball are continuously deformed. figure(d) : Contact surface of both the balls are deformed till the velocity of both the balls become equal. So at maximum deformation velocities of both the blocks are equal v1 ''

v 2 ''

figure(d)

at maximum deformation v1 ''  v 2 '' figure(e) : Normal force is still in the direction shown in figure i.e. velocity of m1 is further decreased

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Page # 38

and that of m2 increased. Now both the balls starts to regain their original shape and size. v1 ''

v 2 '' N

N

v 2 ''  v1 ''

figure(e) figure (f) : These two forces redistributes their linear momentum in such a manner that both the blocks are separated from one another, Velocity of ball m2 becomes more than the velocity of block m1 i . e . , v2 > v1 v1

m1

v2

m2

v2>v1

figure(f)

The collision is said to be elastic if both the blocks regain their original form, The collision is said to be inelastic. If the deformation is permanent, and the blocks move together with same velocity after the collision, the collision is said to be perfectly inelastic.

8.1

Classification of collisions

(a)

On the basis of line of impact (i) Head-on collision : If the velocities of the colliding particles are along the same line before and after the collision. (ii) Oblique collision : If the velocities of the colliding particles are along different lines before and after the collision.

(b)

On the basis of energy : (i) Elastic collision : (a) In an elastic collision, the colliding particles regain their shape and size completely after collision. i.e., no fraction of mechanical energy remains stored as deformation potential energy in the bodies. (b) Thus, kinetic energy of system after collision is equal to kinetic energy of system before collision. (c)

e=1

(d) Due to Fnet on the system is zero linear momentum remains conserved. (ii) Inelastic collision : (a) In an inelastic collision, the colliding particles do not regain their shape and size completely after collision. (b) Some fraction of mechanical energy is retained by the colliding particles in the form of deformation potential energy. Thus, the kinetic energy of the particles no longer remains conserved. (c) However, in the absence of external forces, law of conservation of linear momentum still holds good. (d) (Energy loss)Perfectly Inelastic > (Energy loss)Partial Inelastic (e) 0 < e < 1

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(iii) Perfectly Inelastic collision : (i) In this the colliding bodies do not return to their original shape and size after collision i.e. both the particles stick together after collision and moving with same velocity (ii) But due to Fnet of the system is zero linear momentum remains conserved. (iii) Total energy is conserved. (iv) Initial kinetic energy > Final K.E. Energy (v) Loss in kinetic energy goes to the deformation potential energy (vi) e = 0

8.2

Value of Velocities after collision : Let us now find the velocities of two particles after collision if they collide directly and the coefficient of restitution between them is given as e. m2

m1

m2

m1

u1

u2

v1

v2

(a) Before Collision

(b) After Collision

u1 > u2

v2 > v1

v 2 – v1 e = u –u 1 2  (u1 – u2)e = (v2 – v1)

...(i)

By momentum conservation m1u1 + m2u2 = m1v1 + m2v2

...(ii)

v2 = v1 + e(u1 – u2)

...(iii)

from above equation

v1 =

m 1u1  m 2u 2  m 2 e(u 2 – u1 ) m1  m 2

...(iii)

v2 =

m 1u1  m 2 u 2  m 1e(u 1 – u 2 ) m1  m 2

...(iv)

Special cases : u

1.

If m1 >> m2 and u2 = 0 and u1 = u and

m1

e=1 m1 = m2

from eq. (iii) & (iv) v1 =

m1u – m 2u u(m1 – m 2 ) = m1  m 2 m1  m 2

v1 ~– u

v2 =

m1u  m 2u 2m1u = m1  m 2 m1  m 2

; v2 = 2u

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m2

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Page # 40 2.

u1

If m1 = m2 = m and e = 1 then

m

from eq. (iii) & (iv)

v1 =

u2 m

m(u1  u 2 )  m(u2 – u1 ) 2m

v 1 = u2 In this way v2 = u1

8.3

i.e when two particles of equal mass collide elastically and the collision is head on, they exchange their velocities. Collision in two dimension (oblique) :

1.

A pair of equal and opposite impulses act along common normal direction. Hence, linear momentum of individual particles change along common normal direction. If mass of the colliding particles remain constant during collision, then we can say that linear velocity of the individual particles change during collision in this direction.

2.

No component of impulse act along common tangent direction. Hence, linear momentum or linear velocity of individual particles (if mass is constant) remain unchanged along this direction.

3.

Net impulse on both the particles is zero during collision. Hence, net momentum of both the particles remain conserved before and after collision in any direction.

4.

Definition of coefficient of restitution can be applied along common normal direction, i.e., along common normal direction we can apply Relative speed of separation = e (relative speed of approach)

Ex.45 A ball of mass m makes an elastic collision with another identical ball at rest. Show that if the collision is oblique, the bodies go at right angles to each other after collision. Sol.

In head on elastic collision between two particles, they exchange their velocities. In this case, the component of ball 1 along common normal direction, v cos  becomes zero after collision, while v sin v sin

1 v

1 v cos

2 2 v cos After collision

Before collision

that of 2 becomes v cos . While the components along common tangent direction of both the particles remain unchanged. Thus, the components along common tangent and common normal direction of both the balls in tabular form are given a head :

Ball

Component along common tangent direction Before collision After collision

Component along common normal direction Before collision

After collision

1

v sin 

v sin 

v cos 

0

2

0

0

0

v cos 

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From the above table and figure, we see that both the balls move at right angles after collision with velocities v sin  and v cos . Note : When two identical bodies have an oblique elastic collision, with one body at rest before collision, then the two bodies will go in  directions. Ex.46 Two spheres are moving towards each other. Both have same radius but their masses are 2kg and 4kg. If the velocities are 4m/s and 2m/s respectively and coefficient of restitution is e = 1/3, find. (a) The common velocity along the line of impact. (b) Final velocities along line of impact. 2kg A R

4m/s 2m/s

R B

4kg

(c) Impulse of deformation. (d) impulse of reformation (e) Maximum potential energy of deformation (f) Loss in kinetic energy due to collision.

Sol.

In ABC sin =

BC R 1 = = AB 2R 2

or  = 30°

A 4m/s 2kg  R

C

Line of motion

R R 4kg Line of motion 2m/s B

Line of impact

(a) By conservation of momentum along line of impact. LOI 2kg

4sin30°

4m/s

30°

4sin30°

4cos30° 2cos30°

2m/s

30°

B 4kg

2sin30°

v 2cos30° Maximum Deformed State

Just Before Collision Along LOI

2(4 cos 30°) – 4(2cos30°) = (2 + 4)v or

v = 0 (common velocity along LOI)

(b) Let v1 and v2 be the final velocity of A and B respectively then, by conservation of momentum along line of impact, 2(4 cos 30°) – 4(2cos30°) = 2(v1) + 4(v2)

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Page # 42 4sin30°

A 2kg

v1

4kg

B

v2

2sin30° Just After Collision Along LOI

or

0 = v1 + 2v2 ........(1)

By coefficient of restitution, e=

velocity of separation along LOI velcoity of approach along LOI v 2 – v1 1 = 4 cos 302 cos 30 3

or

or

v2 – v1 =

3

...(2)

from the above two equations, –2 v1 =

3

m / s and v = 2

1 3

m/s

(c) J0 = m1(v – u1) = 2 (0 – 4 cos 30°) = – 4 3 N-s (d) JR = eJ0 =

1 4 N s (–4 3 ) = – 3 3

(e) Maximum potential energy of deformation is equal to loss in kinetic energy during deformation upto maximum deformed state,

U=

1 1 1 1 1 1 m1(u1 cos ) 2  m 2 (u 2 cos ) 2 – (m1  m 2 )v 2 = 2( 4 cos 30 ) 2  4(–2 cos 30 ) 2 – ( 2  4)(0) 2 2 2 2 2 2

or U = 18 Joule (f)

Loss in kinetic energy

KE =

1 1 1 1 2 2 m1(u1 cos ) 2 + m 2 (u 2 cos ) 2 –  m1v1  m 2 v 2    2 2 2 2

 1  2  2 1  1  2 1 1    4   = 2 (4 cos 30°) + 4 (–2 cos 30°) –  2 2 2  3   3 2 2 

KE = 16 Joule

9.

VARIABLE MASS In our discussion of the conservation linear momentum, we have so far dealt with systems whose system whose mass remains constant. We now consider those mass is variable, i.e., those in which mass enters or leaves the system. A typical case is that of the rocket from which hot gases keep on escaping thereby continuously decreasing its mass.  In such problem you have nothing to do but apply a thrust force (Ft ) to the main mass in addition to the all other force acting on it. This thrust force is given by,    dm  Ft  v rel    dt 

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Page # 43

 Here v rel is the velocity of the mass gained or mass ejected relative to the main mass. In case of

rocket this is sometimes called the exhaust velocity of the gases.

dm is the rate at which mass is dt

increasing or decreasing.

vr

v

dm

m

v + dv m–dm system

The expression for the thrust force can be derived from the conservation of linear momentum in the absence of any external forces on a system as follows :  Suppose at some moment t = t mass of a body is m and its velocity is v . After some time at t = t + dt   its mass becomes (m – dm) and velocity becomes v  dv . The mass dm is ejected with relative velocity     v r . Absolute velocity of mass ‘dm’ is therefore ( v r  v  dv) . If no external forces are acting on the system, the linear momentum of the system will remain conserved,

or

  Pi  Pf

or

        mv  mv  mdv  dmv  (dm)(dv)  dm v  v r dm  (dm)((dv)



  mdv   v r dm

or

  dv    dm  m    vr     dt   dt   

Here,

    dv  m   = thrust force (F1 )  dt   

or

      mv  (m  dm)( v  dv)  dm ( v r  v  dv)

and



dm = rate at which mass is ejecting dt

Problems related to variable mass can be solved in following three steps 1.

2.

Make a list of all the forces acting on the main mass and apply them on it.   dm   Apply an additional thrust force Ft on the mass, the magnitude of which is v r   dt  and direction is   given by the direction of v r in case the mass is increasing and otherwise the direction of  v r if it is decreasing.

3.

Find net force on the mass and apply 

 dv Fnet  m dt

(m = mass at that particular instant)

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Page # 44 9.1

Rocket Propulsion Let m0 be the mass of the rocket at time t = 0. m its mass at any time t and v its velocity at that moment. Initially let us suppose that the velocity of the rocket is u. u

u At

t=0 v=u m = m0

At

t=t m=m v=v

Exhaust velocity = vr   dm   be the mass of the gas ejected per unit time and vr the exhaust velocity of the Further, let  dt    dm   and vr are kept constant throughout the journey of the rocket. Now, let us gases. Usually  dt  write few equations which can be used in the problems of rocket propulsion. At time t = t

1. Thrust force on the rocket  dm  Ft  v r     dt 

(upwards)

2. Weight of the rocket W = mg

(downwards)

3. Net force on the rocket Fnet = Ft – W or

(upwards)

 – dm  Fnet  v r   – mg  dt 

4. Net acceleration of the rocket a 

or

dv v r   dm     g dt m  dt 

or

  dm  dv  v r    g dt  m 

or

 dv  v 

or

m  v – u = vr In  0   gt  m

Thus,

 m0   v = u – gt + vr In  m

v u

m r

m0

F m

t dm  g dt 0 m



...(i)

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Page # 45

dm  dm  Note : 1. Ft  v r   is negative.  is upwards, as vr is downwards and dt  dt   m0  . 2. If gravity is ignored and initial velocity of the rocket u = 0, Eq. (i) reduces to v = vr In  m

Ex.47 A uniform chain of mass per unit length  begins to fall with a velocity v on the table. Find the thrust force exerted by the chain on the table. Sol.

Let us assume that the mass of the chain is m and length . We assume that after time t, x length of the chain has fallen on the table. Then the speed of the upper part of the chain is

2gx as shown in figure.

m

x

2gx  v  v r



at t =0

at time t = t

Now its time t + dt, length of chain has fallen on the table is v dt. Then the mass of chain has fallen on the table is dm 

m .vdt 

Now the rate of increase of mass

x

t t + dt

vdt

dm m m  v 2gx   dt

Here v is downward and mass is increasing so thrust force act in down ward direction and is given by ft  v r

=

dm dt

2gx

at time t + dt

m ( 2gx ) 

ft =  v2

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Page # 46

Objective Problems

Exercise - I 1. A semicircular portion of radius 'r' is cut from a uniform rectangular plate as shown in figure. The distance of centre of mass 'C' of remaining plate, from point 'O' is r

2r

C

O

(A)

2r (3 – )

(B)

3r 2(4 – )

(C)

2r (4   )

(D)

2r 3(4 – )

2. From a circle of radius a, an isosceles right angled triangle with the hypotenuse as the diameter of the circle is removed. The distance of the centre of gravity of the remaining position from the centre of the circle is (  – 1)a (A) 3( – 1)a (B) 6 a a (C) (D) 3(  – 1) 3(   1) 3. In the figure shown a hole of radius 2 cm is made in semicircular disc of radius 6 at a distance 8 cm from the centre C of the disc. The distance of the centre of mass of this system from point C is 2cm

8cm (A) 4 cm (C) 6 cm

(B) 8 cm (D) 12 cm

4. Centre of mass of two thin uniform rods of same length but made up of different materials & kept as shown, can be, if the meeting point is the origin of co-ordinates

5. A man of mass M stands at one end of a plank of length L which lies at rest on a frictionless surface. The man walks to other end of the plank. If the mass M of the plank is , then the distance that the man 3 moves relative to ground is : 3L L (A) (B) 4 4 4L L (C) (D) 5 3 6. A particle of mass 3m is projected from the ground at some angle with horizontal. The horizontal range is R. At the highest point of its path it breaks into two pieces m and 2m. The smaller mass comes to rest and larger mass finally falls at a distance x from the point of projection where x is equal to (A)

3R 4

(B)

3R 2

(C)

5R 4

(D) 3R

7. A man weighing 80 kg is standing at the centre of a flat boat and he is 20 m from the shore. He walks 8 m on the boat towards the shore and then halts. The boat weight 200 kg. How far is he from the shore at the end of this time? (A) 11.2 m (B) 13.8 m (C) 14.3 m (D) 15.4 m 8. Two particles having mass ratio n : 1 are interconnected by a light inextensible string that passes over a smooth pulley. If the system is released, then the acceleration of the centre of mass of the system is : 2

2

(A) (n –1) g 2

 n – 1  g (C)   n  1

 n  1  g (B)   n – 1  n  1 g (D)  n – 1

Question No. 9 to 10 (2 questions) A uniform chain of length 2L is hanging in equilibrium position, if end B is given a slightly downward displacement the imbalance causes an acceleration. Here pulley is small and smooth & string is inextensible

y

L

x L

(A) (L/2, L/2)

(B) (2L/3, L/2)

(C) (L/3, L/3)

(D) (L/3, L/6) A

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B

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Page # 47

9. The acceleration of end B when it has been displaced by distance x, is x 2x x g (A) g (B) (C) g (D) g L L 2 10. The velocity v of the string when it slips out of the pulley (height of pulley from floor > 2L) (A)

gL 2

(B)

(C)

gL

(D) none of these

2gL

11. Internal forces can change (A) the linear momentum but not the kinetic energy of the system. (B) the kinetic energy but not the linear momentum of the system. (C) linear momentum as well as kinetic energy of the system. (D) neither the linear momentum nor the kinetic energy of the system. 12. A small sphere is moving at a constant speed in a vertical circle. Below is a list of quantities that could be used to describe some aspect of the motion of the sphere I - kinetic energy II - gravitational potential energy III - momentum Which of these quantities will change as this sphere moves around the circle ? (A) I and II only (B) I and III only (C) III only (D) II and III only 13. Which of the following graphs represents the graphical relation between momentum (p) and kinetic energy (K) for a body in motion ?

ln p

16. A system of N particles is free from any external forces (a) Which of the following is true for the magnitude of the total momentum of the system ? (A) It must be zero (B) It could be non-zero, but it must be constant (C) It could be non-zero, and it might not be constant (D) It could be zero, even if the magnitude of the total momentum is not zero. (b) Which of the following must be true for the sum of the magnitudes of the momenta of the individual particles in the system ?

ln p

(D) none

17. An isolated rail car of mass M is moving along a straight, frictionless track at an initial speed v0. The car is passing under a bridge when a crate filled with N bowling balls, each of mass m, is dropped from the bridge into the bed of the rail car. The crate splits open and the bowling balls bounce around inside the rail car, but none of them fall out. (a) Is the momentum of the rail car + bowling balls system conserved in this collision ?

ln K

ln K

ln p (C)

15. There are some passengers inside a stationary railway compartment. The track is frictionless. The centre of mass of the compartment itself (without the passengers) is C1, while the centre of mass of the compartment plus passengers system is C2. If the passengers move about inside the compartment along the track. (A) both C1 and C2 will move with respect to the ground (B) neither C1 nor C2 will move with respect to the ground (C) C1 will move but C2 will be stationary with respect to the ground (D) C2 will move but C1 will be stationary with respect to the ground

(A) It must be zero (B) It could be non-zero, but it must be constant (C) It could be non-zero, and it might not be constant (D) The answer depends on the nature of the internal forces in the system

(B)

(A)

(A) depends on the direction of the motion of the balls (B) depends on the masses of the two balls (C) depends on the speeds of the two balls (D) is equal to g

ln K 14. Two balls are thrown in air. The acceleration of the centre of mass of the two balls while in air (neglect air resistance)

(A) Yes, the momentum is completely conserved (B) Only the momentum component in the vertical direction is conserved (C) Only the momentum component parallel to the track is conserved (D) No components are conserved

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(b) What is the average speed of the rail car + bowling balls system some time after the collision ? (A) (M + Nm)v0/M (B) Mv0/(Nm + M) (C) Nmv0/M (D) The speed cannot be determined because there is not enough information Question No. 18 to 21 A small ball B of mass m is suspended with light inelastic string of length L from a block A of same mass m which can move on smooth horizontal surface as shown in the figure. The ball is displaced by angle  from equilibrium position & then released. A L



L u=0

B

18. The displacement of block when ball reaches the equilibrium position is L sin (B) L sin  (A) 2 (C) L (D) none of these 19. Tension in string when it is vertical, is (A) mg (B) mg(2 – cos ) (C) mg (3 – 2 cos) (D) none of these 20. Maximum velocity of block during subsequent motion of the system after release of ball is (A) [gl(1 – cos )]1/2 (B) [2gl(1 – cos )]1/2 (C) [glcos]1/2 (D) informations are insufficient to decide 21. The displacement of centre of mass of A + B system till the string becomes vertical is L (A) zero (B) (1 – cos ) 2 L (C) (1 – sin ) (D) none of these 2 Question No. 22 to 26 (5 questions) Two persons of mass m1 and m2 are standing at the two ends A and B respectively, of a trolley of mass M as shown. m2

m1

22. When the person standing at A jumps from the trolley towards left with urel with respect to the trolley, then (A) the trolley moves towards right m1urel (B) the trolley rebounds with velocity m  m  M 1 2 (C) the centre of mass of the system remains stationary (D) all the above 23. When only the person standing at B jumps from the trolley towards right while the person at A keeps standing, then (A) the trolley moves towards left m 2urel (B) the trolley mones with velocity m  m  M 1 2 (C) the centre of mass of the system remains stationary (D) all the above 24. When both the persons jump simultaneously with same speed then (A) the centre of fmass of the systyem remains stationary (B) the trolley remains stationary (C) the trolley moves toward the end where the person with heavier mass is standing (D) None of these 25. When both the persons jump simultaneously with urel with respect to the trolley, then the velocity of the trolley is | m1  m 2 |urel | m1  m 2 |urel (B) (A) m  m  M M 1 2 m1urel m 2urel (C) m  M  m  M 2 1

(D) none of these

26. Choose the incorrect statement, if m1 = m2 = m and both the persons jump one by one, then (A) the centre of mass of the system remains stationary (B) the final velocity of the trolley is in the direction of the person who jumps first murel   murel   (C) the final velocity of the trolley is  M  m M  2m  (D) none of these 27. In the diagram shown, no friction at any contact surface. Initially, the spring has no deformation. What will be the maximum deformation in the spring ? Consider all the strings to be sufficiency large. Consider the spring constant to be K

M A

B

L

F

2M

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M

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(A) 4F / 3K (C) F / 3K

(B) 8F / 3K (D) none

28. A super-ball is to bounce elastically back and forth between two rigid walls at a distance d from each other. Neglecting gravity and assuming the velocity of super-ball to be v0 horizontally, the average force being exerted by the super-ball on each wall is :

1 mv 20 (A) 2 d

mv 20 (B) d

2mv 20 (C) d

4mv 20 (D) d

is masselss and frictionless. Initially the system is at rest when, a bullet of mass 'm' moving with a velocity 'u' as shown hits the block 'B' and gets embedded into it. The impulse imparted by tension force to the block of mass 3m is :

m u m B

A 3m 5mu 4 2mu (C) 5

4mu 5 3mu (D) 5

(A) 29. In the figure (i), (ii) & (iii) shown the objects A, B & C are of same mass. String, spring & pulley are massless. C strikes B with velocity ‘u’ in each case and sticks to it. The ratio of velocity of B in case (i) to (ii) to (iii) is

(i)

(iii)

(ii) C B

A

A

B

C A

B

(B)

34. A ball strikes a smooth horizontal ground at an angle of 45° with the vertical. What cannot be the possible angle of its velocity with the vertical after the collision. (Assume e  1). (A) 45° (B) 30° (C) 53° (D) 60°

C

(A) 1 : 1 : 1 (C) 3 : 2 : 2

(B) 3 : 3 : 2 (D) none of these

30. A force exerts an impulse I on a particle changing its speed from u to 2u. The applied force and the initial velocity are oppositely directed along the same line. The work done by the force is 3 1 (A) Iu (B) Iu 2 2 (C) Iu (D) 2 Iu

35. As shown in the figure a body of mass m moving vertically with speed 3 m/s hits a smooth fixed inclined plane and rebounds with a velocity vf in the horizontal direction. If  of inclined is 30°, the velocity vf will be vf

(A) 3 m/s 31. A parallel beam of particles of mass m moving with velocity v impinges on a wall at an angle  to its normal. The number of particles per unit volume in the beam is n. If the collision of particles with the wall is elastic, then the pressure exerted by this beam on the wall is (A) 2 mn v2 cos  (B) 2 mn v2 cos2  (C) 2 mn v cos  (D) 2 mn v cos2  32. A boy hits a baseball with a bat and imparts an impulse J to the ball. The boy hits the ball again with the same force, except that the ball and the bat are in contact for twice the amount of time as in the first hit. The new impulse equals. (A) half the original impulse (B) the original impulse (C) twice the original impulse (D) four times the original impulse

(B)

m

3 m/s

(C) 1 / 3 m/s (D) this is not possible 36. Two massless string of length 5 m hang from the ceiling very near to each other as shown in the figure. Two balls A and B of masses 0.25 kg and 0.5 kg are attached to the string. The ball A is released from rest at a height 0.45 m as shown in the figure. The collision between two balls is completely elastic Immediately after the collision. the kinetic energy of ball B is 1 J. The velocity of ball A just after the collision is ,

A

33. A system of two blocks A and B are connected by an inextensible massless strings as shown. The pulley

0.45m

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(A) 5 ms–1 to the right (C) 1 ms–1 to the right

(B) 5 ms–1 to the left (D) 1 ms–1 to the left

37. Two balls A and B having masses 1 kg and 2 kg, moving with speeds 21 m/s and 4 m/s respectively in opposite direction, collide head on. After collision A moves with a speed of 1 m/s in the same direction, then the coefficient of restitution is (A) 0.1 (B) 0.2 (C) 0.4 (D) None 38. A truck moving on horizontal road east with velocity 20ms–1 collides elastically with a light ball moving with velocity 25 ms–1 along west. The velocity of the ball just after collision (A) 65 ms–1 towards east (B) 25 ms–1 towards west (C) 65 ms–1 towards west (D) 20 ms–1 towards east 39. Two perfectly elastic balls of same mass m are moving with velocities u1 and u2. They collide elastically n times. The kinetic energy of the system finally is : (A)

1m 2 u1 2 u

(B)

1m 2 (u1  u 22 ) 2 u

(C)

1 m(u12  u22 ) 2

(D)

1 mn(u12  u 22 ) 2

40. In the figure shown, the two identical balls of mass M and radius R each, are placed in contact with each other on the frictionless horizontal surface. The third ball of mass M and radius R/2, is coming down vertically and has a velocity = v0 when it simultaneously hits the two balls and itself comes to rest. The each of the two bigger balls will move after collision with a speed equal to

(A) 4 v 0 / 5

(B) 2 v 0 / 5

(C) v 0 / 5

(D) none

41. In the above, suppose that the smaller ball does not stop after collision, but continues to move downwards with a speed = v0/2, after the collision. Then, the speed of each bigger ball after collision is (A) 4 v 0 / 5

(B) 2 v 0 / 5

(C) v 0 / 2 5

(D) none

42. A body of mass ‘m’ is dropped from a height of ‘h’. Simultaneously another body of mass 2m is thrown up vertically with such a velocity v that they collide at the height h/2. If the collision is perfectly inelastic, the velocity at the time of collision with the ground will be (A)

5gh 4

(B)

(C)

gh 4

(D)

gh 10 gh 3

43. A sphere of mass m moving with a constant velocity hits another stationary sphere of the same mass, if e is the coefficient of restitution, then ratio of speed of the first sphere to the speed of the second sphere after collision will be  1– e  (A)  1 e

 1 e  (B)  1– e

 e  1  (C)  e – 1

 e – 1  (D)  e  1

44. In a smooth stationary cart of length d, a small block is projected along it’s length with velocity v towards front. Coefficient of restitution for each collision is e. The cart rests on a smooth ground and can move freely. The time taken by block to come to rest w.r.t. cart is d

v

(A)

ed (1 e) v

(B)

ed (1 e) v

(C)

d e

(D) inifinite

45. Three blocks are initially placed as shown in the figure. Block A has mass m and initial velocity v to the right. Block B with mass m and block C with mass 4 m are both initially at rest. Neglect friction. All collisions are elastic. The final velocity of block A is V B A C m m 4m (A) 0.60 v to the left (C) v to the left

(B) 1.4 v to the left (D) 0.4 v to the left

46. A block of mass m starts from rest and slides down a frictionless semi-circular track from a height h as shown. When it reaches the lowest point of the

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track, it collides with a stationary piece of putty also having mass m. If the block and the putty stick together and continue to slide, the maximum height that the block-putty system could reach is

h (A) h/4 (C) h

(B) h/2 (D) independent of h

47. Two billiard balls undergo a head-on collision. Ball 1 is twice as heavy as ball 2. Initially, ball 1 moves with a speed v towards ball 2 which is at rest. Immediately after the collision, ball 1 travels at a speed of v/3 in the same direction. What type of collision has occured ? (A) inelastic (B) elastic (C) completely inelastic (D) Cannot be determined from the information given 48. The diagram shows the velocity - time graph for two masses R and S that collided elastically. Which of the following statements is true ?

V(ms–1) 1.2 0.8 0.4

R

S

1 2 3 4 t(s) I. R and S moved in the same direction after the collision. II. The velocities of R and S were equal at the mid time of the collision. III. The mass of R was greater than mass of S. (A) I only (B) II only (C) I and II only (D) I, II and III 49. A ball is dropped from a height h. As is bounces off the floor, its speed is 80 percent of what it was just before it hit the floor. The ball will then rise to a height of most nearly (A) 0.80 h (B) 0.75 h (C) 0.64 h (D) 0.50 h 50. A ball is thrown vertically downwards with velocity 2gh from a height h. After colliding with the ground it just reaches the starting point. Coefficient of restitution is (A) 1 / 2

(B) 1/2

(C) 1

(D)

2

51. A ball is dropped from height 5m. The time after which ball stops rebounding if coefficient of restitution between ball and ground e = 1/2, is

(A) 1 sec (C) 3 sec

(B) 2 sec (D) infinite

52. A ball is projected from ground with a velocity V at an angle  to the vertical. On its path it makes an elastic collision with a vertical wall and returns to ground. The total time of flight of the ball is (A)

2v sin g

(B)

2v cos  g

(C)

v sin2 g

(D)

v cos  g

53. The Gardener water the plants by a pipe of diameter 1 mm. The water comes out at the rate of 10 cm3/ sec. The reactionary force exerted on the hand of the Gardener is : (density of water is 103 kg/m3) (A) zero (B) 1.27 × 10–2 N –4 (C) 1.27 × 10 N (D) 0.127 N 54. An open water tight railway wagon of mass 5 × 103 kg coasts at an initial velocity 1.2 m/s without friction on a railway track. Rain drops fall vertically downwards into the wagon. The velocity of the wagon after it has collected 103 kg of water will be (A) 0.5 m/s (B) 2 m/s (C) 1 m/s (D) 1.5 m/s 55. If the force on a rocket which is ejecting gases with a relative velocity of 300 m/s, is 210 N. Then the rate of combustion of the fuel will be (A) 10.7 kg/sec (B) 0.07 kg/sec (C) 1.4 kg/sec (D) 0.7 kg/sec 56. A rocket of mass 4000 kg is set for vertical firing. How much gas must be ejected per second so that the rocket may have initial upwards acceleration of magnitude 19.6 m/s2. [Exhaust speed of fuel = 980 m/ s] (A) 240 kg s–1 (B) 60 kg s–1 –1 (C) 120 kg s (D) none 57. A wagon filled with sand has a hole so that sand leaks through the bottom at a constant rate . An  external force F acts on the wagon in the direction of motion. Assuming instantaneous velocity of the wagon to be v and initial mass of system to be m0, the force equation governing the motion of the wagon is :       dv dv  v – v (B) F  m 0 (A) F  m 0 dt dt      dv dv F  ( m –  t ) F  ( m –  t )  v (C) (D) 0 0 dt dt

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Multiple Choice Problem

Exercise - II

1. A body has its centre of mass at the origin. The xcoordinates of the particles (A) may be all positive (B) may be all negative (C) may be all non-negative (D) may be positive for some cases and negative in other cases 2. An object comprises of a uniform ring of radius R and its uniform chord AB (not necessarily made of the same material) as shown. Which of the following can not be the centre of mass of the object. y

(A) must not move (C) may move

(B) must not accelerate (D) may accelerate

Question No. 6 to 12 (7 Questions) A particle of mass m moving horizontal with v0 strikes a smooth wedge of mass M, as shown in figure. After collision, the ball starts moving up the inclined face of the wedge and rises to a height h.

h

B

m v0 A

x

6. The final velocity of the wedge v2 is (A)

(A) (R/3, R/3)

(B) (R3, R/2)

(C) (R/4, R/4)

(D) (R / 2, R / 2 )

3. In which of the following cases the centre of mass of a an rod is certainly not at its centre ? (A) the density continuously increases from left to right (B) the density continuously decreases from left to right (C) the density decreases from left to right upto the centre and then increase (D) the density increases from left to right upto the centre and then decreases 4. Consider following statements [1] CM of a uniform semicircular disc of radius R = 2R/ from the centre [2] CM of a uniform semicircular ring of radius R = 4R/3 from the centre [3] CM of a solid hemisphere of radius R = 4R/3 from the centre [4] CM of a hemisphere shell of radius R = R/2 from the centre Which statements are correct? (A) 1, 2, 4 (B) 1, 3, 4 (C) 4 only (D) 1, 2 only 5. If the external forces acting on a system have zero resultant, the centre of mass

M

mv 0 M

(C) v0

(B)

mv 0 Mm

(D) insufficient data

7. When the particle has risen to a height h on the wedge, then choose the correct alternative(s) (A) The particle is stationary with respect to ground (B) Both are stationary with respect to the centre of mass (C) The kinetic energy of the centre of mass remains constant (D) The kinetic energy with respect to centre of mass is converted into potential energy 8. The maximum height h attained by the particle is 2  m  v0  (A)  m  M 2g 2  M  v0  (C)  m  M 2g

2  m v0 (B)   M 2g

(D) none of these.

9. Identify the correct statement(s) related to the situation when the particle starts moving downward. (A) The centre of mass of the system remains stationary (B) Both the particle and the wedge remain stationary with respect to centre of mass (C) When the particle reaches the horizontal surface its velocity relative to the wedge is v0 (D) None of these

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10. Suppose the particle when reaches the horizontal surfaces, its velocity with respect to ground is v1 and that of wedge is v2. Choose the correct statement (s)

v2

v1

m

(A) mv1 = Mv2 (C) v1 + v2 = v0

M (B) Mv2 – mv1 = mv0 (D) v1 + v2 < v0

11. Choose the correct statement(s) related to particle m  mM  (A) Its kinetic energy is K f    gh  m  M  M  m  (B) v1  v 0  M  m

(C) The ratio of its final kinetic energy to its initial kinetic energy is

Kf  M    K i  m  M

2

(D) It moves opposite to its initial direction of motion 12. Choose the correct statement related to the wedge M

 4 m2   gh (A) Its kinetic energy is K f    m  M  2m   v0 (B) v 2   m  M  4 mM   1  mv 20   (C) Its gain in kinetic energy is K   2    (m  M)   2 (D) Its velocity is more that the velocity of centre of mass 13. Two blocks A (5kg) and B(2kg) attached to the ends of a spring constant 1120 N/m are placed on a smooth horizontal plane with the spring undeformed. Simultaneously velocities of 3m/s and 10m/s along the line of the spring in the same direction are imparted to A and B then

A

3m/s

10m/s

5

2

B

(A) when the extension of the spring is maximum the velocities of A and B are zero. (B) the maximum extension of the spring is 25 cm

(C) maximum extension and maximum compression occur alternately. (D) the maximum compression occur for the first time  after sec. 56 14. Two identical balls are interconnected with a massless and inextensible thread. The system is in gravity free space with the thread just taut. Each ball is imparted a velocity v, one towards the other ball and the other perpendicular to the first, at t = 0. Then, (A) the thread will become taut at t = (L/v) (B) the thread will become taut at some time t < (L/ v). (C) the thread will always remain taut for t > (L/v) (D) the kinetic energy of the system will always remain mv2. 15. A ball moving with a velocity v hits a massive wall moving towards the ball with a velocity u. An elastic impact lasts for a time t. (A) The average elastic force acting on the ball is m(u  v) t (B) The average elastic force acting on the ball is 2m(u  v) t (C) The kinetic energy of the ball increases by 2mu (u + v) (D) The kinetic energy of the ball remains the same after the collision. 16. A particle moving with kinetic energy = 3 joule makes an elastic head on collision with a stationary particle when has twice its mass during the impact. (A) The minimum kinetic energy of the system is 1 joule (B) The maximum elastic potential energy of the system is 2 joule. (C) Momentum and total kinetic energy of the system are conserved at every instant. (D) The ratio of kinetic energy to potential energy of the system first decreases and then increases. 17. Two balls A and B having masses 1 kg and 2 kg, moving with speeds 21 m/s and 4 m/s respectively in opposite direction, collide head on. After collision A moves with a speed of 1 m/s in the same direction, then correct statements is :

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(A) The velocity of B after collision is 6 m/s opposite to its direction of motion before collision. (B) The coefficient of restitution is 0.2. (C) The loss of kinetic energy due to collision is 200 J. (D) The impulse of the force between the two balls is 40 Ns. 18. The diagram to the right shown the velocity-time graph for two masses R and S that collided elastically. Which of the following statements is true? –1 v(ms )

1.2 0.8

S R

0.4 1

2 3 4 t (milli sec)

(I) R and S moved in the same direction after the collision. (II) Kinetic energy of the system (R & S) is minimum at t = 2 milli sec. (III) The mass of R was greater than mass of S. (A) I only (B) II only (C) I and II only (D) I, II and III 19. In an inelastic collision, (A) the velocity of both the particles may be same after collision. (B) kinetic energy is not conserved (C) linear momentum of the system is conserved. (D) velocity of separation will be less than velocity of approach.

L 2V L (B) The last block starts moving at t = (n – 1) V (C) The centre of mass of the system will have a final speed v/n (D) The centre of mass of the system will have a final speed v (A) The last block starts moving at t = n(n–1)

22. An isolated rail car original moving with speed v0 on a straight, frictionless, level track contains a large amount of sand. A release value on the bottom of the car malfunctions, and sand begins to pour out straight down relative to the rail car. (a) Is momentum conserved in this process? (A) The momentum of the rail car along is conserved (B) The momentum of the rail car + sand remaining within the car is conserved (C) The momentum of the rail car + all of the sand, both inside and outside the rail car, is conserved (D) None of the three previous systems have momentum conservation (b) What happens to the speed of the rail car as the sand pours out? (A) The car begins to roll faster (B) The car maintains the same speed (C) The car begins to slow down (D) The problem cannot be solved since momentum is not conserved

20. In a one-dimensional collision between two par ticles, their relative velocity is v1 before the collision  and v 2 after the collision   (A) v1  v 2 if the collision is elastic   (B) v1  – v 2 if the collision is elastic   (C) | v 2 | | v1| in all cases   (D) v1  – kv 2 in all cases, where k  1 21. A set of n-identical cubical blocks lie at rest parallel to each other along a line on a smooth horizontal surface. The separation between the near surfaces of any two adjacent blocks is L. The block at one end is given a speed V towards the next one at time t = 0. All collision are completely inelastic, then

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Subjective Problem

Exercise - III 1. The mass of an uniform ladder of length l increases uniformly from one end A to the other end B, (a) Form an expression for linear mass density as a function of distance x from end A where linear mass density 0. The density at one end being twice that of the other end. (b) find the position of the centre of mass from end A. 2. Find the distance of centre of mass of a uniform plate having semicircular inner and other boundaries of radii a and b from the centre O.

6. In the arrangement, mA = 2 kg and mB = 1 kg. String is light and inextensible. Find the acceleration of centre of mass of both the blocks. Neglect friction everywhere. 7. A small cube of mass m slides down a circular path of radius R cut into a large block of mass M. M rests on a table and both blocks move without friction. The blocks initially are at rest and m starts from the top of the path. Find the velocity v of the cube as it leaves the block. m R

M

a b O 3. A thin sheet of metal of uniform thickness is cut into the shape bounded by the line x = a and y = ± k x2, as shown. Find the coordinates y of the centre of mass. y=kx2 x

a y=–kx2

4. Two balls of equal masses are projected upwards simultaneously, one from the ground with speed 50 m/s and other from a 40m high tower with initial speed 30 m/s. Find the maximum height attained by their centre of mass. 5. In the figure shown, when the persons A and B exchange their positions, then

B

A m1

M

m2

2m m1=50kg, m2 = 70kg, M = 80 kg

8. A (trolley + child) of total mass 200 kg is moving with a uniform speed of 36 km/h on a frictionless track. The child of mass 20 kg starts running on the trolley from one end to the other (10 m away) with a speed of 10 ms–1 relative to the trolley in the direction of the trolley's motion and jumps out of the trolley with the same relative velocity. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run? 9. In the figure shown the spring is compressed by ‘x0’ and released. Two blocks ‘A’ and ‘B’ of masses ‘m’ and ‘2m’ respectively are attached at the ends of the spring. Blocks are kept on a smooth horizontal surface and released. k A

B

(a) Find the speed of block A by the time compression of the spring is reduced to x0/2. (b) Find the work done by the spring on ‘A’ by the time compression of the spring reduced to x0/2. 10. The figure showns the force versus time graph for a particle. (i) Find the change in momentum p of the particle (ii) Find the average force acting on the particle

100 N (i) the distance moved by the centre of mass of the system is__________. (ii) the plank moves towards_________ (iii) the distance moved by the plank is _________. (iv) the distance moved by A with respect to ground is ____________ (v) the distance moved by B with respect to ground is ____________.

0

0.2

0.4

t(a)

11. A force F acts on an object (mass = 1kg) which is initially at rest as shown in the figure. Draw the graph showing the momentum of the object varying during the time for which the force acts.

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F(N) 5

50

100 t(ms)

12. A man hosing down his driveway hits the wall by mistake. Knowing that the velocity of the stream is 25 m/s and the crosssectional area of the stream is 300 mm2, determine the force exerted on the wall. Assume that streamstrikes wall horizontally and after striking the wall, stream comes to rest. Also find the pressure exerted on the wall by stream.

13. A bullet of mass m strikes an obstruction and deviates off at 60° to its original direction. If its speed is also changed from u to v, find the magnitude of the impulse acting on the bullet. 14. A neutro intially at rest, decays into a proton, an electron and an antineutrio. The ejected electron has a momentum of p1 = 1.4 × 10–28 kg-m/s and the antineutrino p2 = 6.5 × 10–27 kg-m/ s. Find the recoil speed of the proton (a) if the electron and the antineutrino are ejected along the the same direection and (b) if they are ejected along perpendicular direction. Mass of the proton mp = 1.67 × 10–27 kg. 15. A steel ball of mass 0.5 kg is dropped from a height of 4 m on to a horizontal heavy steel slab. The collision is elastic and the ball rebounds to its original height. (a) Calculate the impulse delivered to the ball during impact. (b) If the ball is in contact with the slab for 0.002 s, find the average reaction force on the ball during impact. 16. A particle A of mass 2 kg lies on the edge of a table of height 1m. It is connected by a light inelastic string of length 0.7 m to a second particle B of mass 3 kg which is lying on the table 0.25 m from the edge (line joining A & B is perpendicular to the edge). If A is pushed gently so that it start falling from table then, find the speed of B when it starts to move. Also find the imulsive tension in the string at that moment. 17. Two particles, each of mass m, are connected by a light inextensible string of length 2l. Initially they lie on a smooth horizontal table at points A and B distant l apart. The particle at A is projected across the table with velocity u. Find the speed

Page # 56 with which the second particle begins to move if the direction of u is, (a) along BA, (b) at an angle of 120° with AB, (c) perpendicular to AB. In each case calculate (in terms of m and u) the impulsive tension in the string. 18. Two particle P of mass 2m and Q of mass m are subjected to mutual force of attraction and no other act on them. At time t = 0, P is at rest at a fixed O and Q is directly moving away from O with a speed 5 u. At a later instant when t = T before any collision has taken place Q is moving towards O with speed u. (a) Find in terms of m and u the total work done by the forces of attraction during the time interval 0  t  T. (b) At the instantt = T, impulses of magnitude J and K and applied to P and Q bringing them to rest. Find the values of J and K 19. A block of mass m moving with a velocity v, enters a region where it starts colliding with the stationary dust particles. If the desnsity of dust particles is , & all colliding particle stick to its front surface of cross-sectional area A. The velocity of block after it has covered a distance x in this region is __________________. 20. A football approaches a player at v = 12 m/s. At what speed u and in which direction should the player’s foot move in order to stop the ball upon contact? Assume that the mass of the foot is much greater than that of the ball and that the collision is elastic. 21. Three carts move on a frictioless track with inertias and velocities as shown. The carts collide and stick together after successive collisions.

m1 = 2 kg v1 = 1 m/s

m2 = 1 kg v2 = 1 m/s

A

B

m3 = 2 kg v3 = 2 m/s C

(a) Find loss of mechanical energy when B & C stick together. (b) Find magnitude of impulse experienced by A when it sticks to combined mass (B & C). 22. A sphere of mass m1 in motion hits directly another sphere of mass m2 at rest and sticks to it, the total kinetic energy after collision is 2/3 of their total K.E. before collision. Find the ratio of m1 : m2. 23. A body is thrown vertically upwards from ground with a speed of 10 m/s. If coefficient of restitution of ground, e = 1/2. Find (a) the total distance travelled by the time it almost stops. (b) time elapsed (after the ball has been thrown) when it is at its subsequent maximum height for the third time.

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CENTRE OF MASS

Page # 57

24. A ball of mass 'm' is suspended by a massless string of length '' from a fixed point. A ball of mass 2m strikes in the direction of  = 45° from horizontal and sticks to it. (a) What should be the initial velocity of 2m so that system  deflects by  = 2 (b) If at  = 60° the stirng is cut then what will be the velocity at highest point of trajectory.



2m 45°

m

A

2kg

90° 1.5 m 4 kg B

(b) the maximum displacement of B after the impact. 28. A small block of mass 2m initially rests at the bottom of a fixed circular, vertical track, which has a radius of R. The contact surface between the mass and the loop is frictionless. A bullet of mass m strikes the block horizontally with initial speed v0 and remain embedded in the block as the block and the bullet circle the loop. Determine each of the following in terms m, v0, R and g.

25. A ball is thrown horizontally from a cliff 10 m high with a velocity of 10 m/s. It strikes the smooth ground and rebounds as shown. The coefficient of restitution e for collision with the ground e = 1/ 2 . Find

R

10 m/s

m, v0 (a) The speed of the masses immediately after the impact.

(a) velocity of ball just before striking ground. (b) angle of velocity vector with horizontal before striking. (c) angle of velocity vector with horizotal after striking. (d) range of ball after first collision. 26. A wedge free to move of mass 'M' has one face making an angle  with horizotnal and is resting on a smooth rigid floor. A particle of mass 'm' hits the inclined face of the wedge with a horizontal velocity v0. It is observed that the particle rebounds in vertical direction after impact. Neglect friction between particle and the wedge & take M = 2m, v0 = 10 m/s, tan  = 2, g = 10 m/s2



(b) The minimum initial speed of the bullet if the block and the bullet are to successfully execute a complete ride on the loop. 29. A Cart of total mass M0 is at rest on a rough horizontal road. It ejects bullets at rate of  kg/s at an angle  with the horizontal and at velocity 'u' (constant) relative to the cart. The coefficient of friction between the cart and the ground is . Find the velocity of the cart in terms of time 't'. The cart moves with sliding.

M

(a) Determine the coefficient of restitution for the impact. (b) Assume that the inclined face of the wedge is sufficiently long so that the particle hits the same face once more during its downward motion. Calculate the time elapsed between the two impacts. 27. A sphere A is released from rest in the position shown and strikes the block B which is at rest. If e = 0.75 between A and B and k = 0.5 between B and the support, determine (a) the velocity of A just after the impact

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

CENTRE OF MASS

Page # 58

Tough Subjective Problem

Exercise - IV

1. In a game of Carom Board, the Queen (a wooden disc of radius 2 cm and mass 50 gm) is placed at the exact center of the horizontal board. The striker is a smooth plastic disc of radius 3 cm and mass 100 gm. The board is frictionless. Th striker is given an initial velocity ‘u’ parallel to the sides BC or AD so that is hits the Queen inelastically with same coefficient of restitution = 2/3. The impact parameter for the collision is ‘d’ (shown in the figure). The Queen rebounds from the edge AB of the board inelastically with same coefficient of restitution = 2/3. and enters the hole D following the dotted path shown. The side of the board is L. Find the value of impact parameter ‘d’ and the time which the Queen takes to enter hole D after collision with the striker. A

L

L

B

5. A massive vertical wall is approaching a man at a speed u. When it is at a distance of 10m, the man throws a ball with speed 10 m/s an at angle of 37° which after completely elastic rebound reaches back directly into his bands. Find the velocity u of the wall. 6. Mass m1 hits & sticks with m2 while sliding horizontally with velocity v along the common line of centres of the three equal masses (m1 = m2 = m3 =m). Initially masses m2 and m3 are stationary and the spring is unstretched. Find

m1

k

v

m2

m3

Frictionless (a) the velocities of m1, m2 and m3 immediately after impact.

(b) the maximum kinetic energy of m3.

u d

(c) the minimum kinetic energy of m2. (d) the maximum compression of the spring.

D

C

2. A flexible chain has a length l and mass m. It is lowered on the table top with constant velocity v. Find the force that the chain exerts on the table as a function of time.

7. Two masses A and B connected with an inextensible string of length l lie on a smooth horizontal plane. A is giv en a velo city of v m/s alon g the gro und perpendicular to line AB as shown in figure. Find the tension in string during their sub sequent motion. B m

3. A 24-kg projectile is fired at an angle of 53° above the horizontal with an initial speed of 50 m/s. At the highest point in its trajectory, the projectile explodes into two fragments of equal mass, the first of which falls vertically with zero initial speed. (a) How far from the point of firing does the second fragment strike the ground? (Assume the ground is level.) (b) How much energy was released during the explosion?

l

A 2m

v

8. The simple pendulum A of mass mA and length l is suspended from the trolley B of mass mB. If the system is released from rest at  = 0, determine the velocity vB of the trolley and tension in the string when  = 90°. Friction is negligible.

4. A particle is projected from point O on level ground towards a smooth vertical wall 50m from O and hits the wall. The initial velocity of the particle is 30m/s at 45° to the horizontal and the coefficient of restitution between the particle and the wall is e. Find the distance from O of the point at which the particle hits the ground again if (a) e = 0, (b) e = 1, (c) e = 1/2

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B l

A

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Page # 59

9. A ball with initial speed 10m/s collides elastically with two other identical ball whose centres are on a line perpendicular to the initial velocity and which are initially in contact with each other. All the three ball are lying on a smooth horizontal table. The first ball is aimed directly at the contact point of the other two balls All the balls are smooth. Find the velocities of the three balls after the collision.

11. A cart is moving along +x direction with a velocity of 4m/s. A person in the cart throws a stone with a velocity of 6m/s relative to himself. In the frame of reference of the cart the stone is thrown in y-z plane making an angle of 30° with the vertical z-axis. At the highest point of its trajectory, the stone hits an object of equal mass hung vertically from branch of a tree by means of a string of length L. A completely inelastic collosion occurs, in which the stone gets embedded in the object. Determine (a) the speed of the combined mass immediately after collision with respect to an observer on the ground.

10 m/s . 10. A mass m1 with initial speed v0 in the positive xdirection collides with a mass m2 = 2m1 which is initially at rest at the origin, as shown in figure. After the collision m1 moves off with speed v1 = v0/2 in the negative y-direction, and m2 moves off with speed v2 at angle . (A) Find the velocity (magnitude and direction) of the center of mass before the collision, as well as its velocity after the collision. (B) Write down the x and y-components of the equation of conservation of momentum for the collision. (C) Determine tan, and find v2 in terms of v0.

(b) the length L of the string such that tension in the string becomes zero when the string becomes horizontal during the subsequent motion of the combined mass.

12. Twp equal sphere of mass ‘m’ are suspended by vertical strings so that they are in contact with their centres at same level. A third equal spheres of mass m falls vertically and strikes the other two simultaneously so that their centres at the instant of impact form an equilateral triangle in a vertical plane. If u is the velocity of m just before impact, find the velocities just after impact and the impulsive tension of the strings.

(D) Determine how much (if any) energy was gained or lost in the collision, and state whether the collision was elastic or inelastic. y

u

v2

y

m

m2

v 

x v0 m2 = 2m1

v

m

m1 After

Before

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m

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Page # 60

JEE-Problems

Exercise - V 1. Two trolleys A and B of equal masses M are moving   in opposite directions with velocities v and  v respectively on separate horizontal frictionless parallel tracks. When they start crossing each other, a ball of mass m is thrown from B to A and another of same is  thrown from A to B with velocities normal to v . The balls may be thrown in following two ways (i) balls from A to B to A are thrown simultaneously. (ii) ball is thrown from A to B after the ball thrown from B reaches A. Which procedure would lead to a larger change in the velocities of the trolleys ? [REE-2000] 2. A wind-powered generator converts wind energy into electrical energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed v, the electrical power output will be proportional to [IIT(Scr.)-2000] (A) v (B) v2 3 (C) v (D) V4 3. Two particles of masses m1 and m2 in projectile   motion have velocities v 1 and v 2 respectively at time t = 0. They collide at time t0. Their velocities become   v 1 and v 2 at time 2t0 while still moving in air. The     value of [(m1v 1  m 2 v 2 )  (m1u1  m 2u 2 )] is [IIT(Scr.)-2001)] (A) zero (B) (m1 + m2)gt0 (C) 2(m1 + m2)gt0 (D) ½(m1 + m2)gt0 4. A car P is moving with a uniform speed of 5(31/2)m/ s towards a carriage of mass, 9 Kg at rest kept on the rails at a point B as shown in fig. The height AC is 120 m. Cannon balls of 1 Kg are fired from the car with an initial velocity 100 m/s at an angle 30º with the horizontal. The first canon ball hits the stationary carriage after a time t0 ans stricks to it. Determine t0. At t0, the second cannon ball is fired. Assume that the resistive force between the rails and the carriage is constant and ignore the vertical motion of the carriage throughout. If the second ball also hits and sticks to the carriage. What will be the horizontal velocity of the carriage just after the second impact ? [IIT-2001] C P

5. Two block of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the centre of mass is [IIT(Scr.)-2002] (A) 30 m/s (B) 20 m/s (C) 10 m/s (D) 5 m/s 6. STATEMENT-1 In an elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision. because STATEMENT-2 In an elastic collision, the linear momentum of the system is conserved (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 correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True  7. The balls, having linear momenta p1  p i and  p 2  –p i , undergo a collision in free space. There is   no external force acting on the balls. Let p1 and p 2 be their final momenta. The following options(s) is (are) NOT ALLOWED for any non-zero value of p, a1, a2, b1, b2, c1 and c2. [JEE 2008]  (A) p'1  a1ˆi  b1ˆj  c 1kˆ    (B) p'1  c1kˆ ; p' 2  a 2 ˆi  b 2 ˆj ; p'2  c 2kˆ  (C) p'1  a1ˆi  b1ˆj  c 1kˆ    (D) p'1  a1ˆi  b1ˆj ; p' 2  a 2 ˆi  b 2 ˆj – c1k ; p' 2  a 2 ˆi  b1ˆj

Paragraph for Question No. 8 to 10 A small block of mass M moves on a frictionless surface of an inclined plane, as shown in figure. The angle of the incline suddenly changes from 60° to 30° at point B. The block is initially at rest at A. Assume that collisions between the block and the incline are totally inelastic (g = 10 m/s2). Figure : [JEE 2008] M A v 60° B

30° A

B

3m

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3 3m

C

CENTRE OF MASS

Page # 61

8. The speed of the block at point B immediately after it strikes the second incline is (A)

60 m / s

(B)

45 m / s

(C)

30 m / s

(D)

15 m / s

9. The speed of the block at point C, immediately before it leaves the second incline is (A)

120 m / s

(B)

105 m / s

(C)

90 m / s

(D)

75 m / s

10. If collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the block at point B, immediately after it strikes the second incline is (A)

30 m / s

(B)

15 m / s

(D) – 15 m / s

(C) 0

11. Look at the drawing given in the figure which has been drawn with ink of uniform line-thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is m. The mass of the ink unsed to draw the outer circle is 6m. The coordinates of the centres of the different parts are outer circle (0, 0), left inner circle (–a, a), right inner circle (a, a), vertical line (0, 0) and horizontal line (0, – a). The y-coordinate of the centre of mass of the ink in this drawing is [JEE 2009] m (-a, a)

m (a, a) (0, 0) 7m

13. Three objects A, B and C are kept in a straight line on a frictionless horizontal surface. These have masses m, 2m and m, respectively. The object A moves towards B with a speed 9 ms–1 and makess an elastic collision with it. There after, B makes completely inelastic collision with C. All motions occur on the same straight line. Find the final speed (in ms–1 ) of the object C. [JEE 2009]

2m

m A

B

m C

14. A point mass of 1 kg collides elastically with a stationary point mass of 5 kg. After their collision, the 1 kg mass reverse its direction and moves with a speed of 2 ms–1. Which of the following statement(s) is (are ) correct for the system of these two masses ? (A) Total momentum of the system is 3 kg ms–1 (B) Momentum of 5 kg mass after collision is 4 kg ms–1 (C) Kinetic energy of the centre of mass is 0.75 J (D) Total kinetic energy of the system is 4 J [JEE 2010] 15. A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet of mass 0.01 kg traveling with a velocity V m/s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. The initial velocity V of the bullet is v m/s

m (0, -a)

12. Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circle 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 particulars will again reach the point A ?

V

(A) 4 (C) 2

A

0

(A) 250 m/s (C) 400 m/s

20

(B) 250 2 m/s (D) 500 m/s

2V

(B) 3 (D) 1

100

[JEE 2009]

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[JEE 2011]

Page # 62

CENTRE OF MASS

Exercise-I 1.

D

2.

C

8. 15. 22. 29. 36. 43.

C C D B D A

9. 16. (a) 23. 30. 37. 44.

50. 57.

A D

51.

3.

4.

D

5.

B

6.

C

7.

C

A 10. C B, (b) C 17. (a) C, (b) B D 24. A B 31. B B 38. A D 45. A

11. 18. 25. 32. 39. 46.

B A A C C A

12. 19. 26. 33. 40. 47.

D D D D C B

13. 20. 27. 34. 41. 48.

D A B B C D

14. 21. 28. 35. 42. 49.

D B B B D C

C

53.

D

54.

C

55.

D

56.

C

B,C

6.

B

7.

B,D

52.

B

B

Exercise-II 1.

C,D

2.

B,D

3.

A,B

4.

C

5.

8.

C

9.

C

10.

B,C

11.

B

12. A,B,C,D

13.

B,C

14.

A,C

15.

B,C

16.

17.

A,B,C

18.

D

19.

20.

B,D

21.

A,C

22.

(a) A,C (b) B

4.

100 m

A,B,D

A,B,C,D

Exercise-III 4  b3  a3  2. y  3  2 2  b – a 

5 x , (b) L L 9

1.

(a) ( x)   

5.

(i) zero; (ii) right ; (iii) 20 cm ; (iv) 2.2 m ; (v) 1.8 m

8.

9 m/s, 9 m

9. (a)

Kx20 kx 20 , (b) 4 2m

3.

3 a 4

6. g/9 downwards

7. v 

2gR m 1 M

10. (i) 20Ns, (ii) 50 N

P(N-sec) 0.25

11.

0.125

12. 187.5 N, 625 kPa 50

13. m  u 2  uv  v 2

100 t(ms)

p1  p 2  12.3 m / sec , (b) mp

14.

(a)

16.

1.5 m/s, 3.6 Ns

18.

W = –3mu2; J = 6 mu, K = mu

p12  p 2 2 mp

= 9.4 m/s

15. (a) 4 5 N , (b) 2000 5 N

17. (a) u/2, mu/2; (b) u 13 / 8 , m u 13 / 8 (c) u 3 / 4 , mu 3 / 4 mv 19. (m  Ax)

20. 6 m/s in the direction of football’s velocity

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Page # 63

CENTRE OF MASS 12 Ns 5

21.

(a) 3J, (b)

22. 2 : 1

24.

(a) v  3 gl , (b) v =

26.

(a) e = cot2 +

28.

(a) v0/3, (b) 3 5gR

gl 2

23. (a)

40 m , (b) te = 3.25 s 3

25. (a) 10 3 , (b) tan–1 2 , (c) 45º, (d) 20 m

m 3 = , (g) t = 3 sec. M 4

27. v A  g / 12 m / s , Smax = 49/48 m  M0  29. v = (ucos –  u sin  ) ln  M – t  – gt  0 

Exercise-IV m v( v  gt) 

3. (a) 360 m, (b) 10800 J

4. (a) 50 m, (b) 10 m, (c) 30 m

1. ( 5 / 17 cm , 153L / 80u

2.

5. 13/3 m/s

6. (a) v/2, v/2, 0; (b) 2mv2/9, (c) mv2/72, (d) x  m / 6k v

7. 2mv2/3l

mA 8. vB  m B

2gl 2m2A g ; T = 3m g + A 1+ m A / mB mB

10. (a) v0/3, (b) mv0 = 2mv2 cos, 0 = 2mv2sin–

11. 2.5 m/sec, 0.312 m

12. v 

9. -2m/s, 6.93 m/s 30º

mv 0 mv 20 1 5 v 0 , (d) , (c) , 2 2 4 16

2 3u 5u 6 , u  , T  mv 7 7 7

Exercise-V 1. 2 in case I

2. C

3. C

4. t0 = 12 sec, v = 100 3 /11 1

5. C

6. B

7. A, D

8. B

9. B

10. C

11. a/10

12. C

13. 4

14. A,C

15. D

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IIT-JEE|AIEEE CBSE|SAT|NTSE OLYMPIADS

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ROTATIONAL MOTION THEORY AND EXERCISE BOOKLET CONTENTS S.NO.

TOPIC

PAGE NO.

1.

Rigid Body .............................................................................................. 3 – 5

2.

Moment of Inertia ................................................................................... 5 – 10

3.

Theorems of Moment of Inertia ............................................................. 11 – 14

4.

Cavity Problems ................................................................................... 14 – 15

5.

Torque ................................................................................................... 15 – 18

6.

Body In Equilibrium ............................................................................... 18 – 22

7.

Relation between Torque & ................................................................... 23 – 25 Angular acceleration

8.

Angular Momentum ............................................................................... 25 – 27

9.

Conservation of Angular Momentum ..................................................... 27 – 30

10.

Angular Impulse .................................................................................... 30 – 31

11.

Combined Translational & ..................................................................... 31 – 39 Rotational Motion

12.

Uniform Pure Rolling ............................................................................. 39 – 47

13.

Toppling ................................................................................................ 47 – 49

14.

Instantaneous Axis of Rotation ............................................................. 49 – 51

15.

Exercise - I ........................................................................................... 52 – 59

16.

Exercise - II .......................................................................................... 60 – 62

17.

Exercise - III ......................................................................................... 63 – 66

18

Exercise - IV ........................................................................................ 67 – 69

19.

Exercise - V .......................................................................................... 70 – 76

20.

Answer key ........................................................................................... 77 – 78

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ROTATIONAL DYNAMICS

Syllabus Rigid body, moment of inertia, parallel and perpendicular axes theorems, moment of inertia of uniform bodies with simple geometrical shapes; Angular momentum; Torque; Conservation of angular momentum; Dynamics of rigid bodies with fixed axis of rotation; Rolling without slipping of rings, cylinders and spheres; Equilibrium of rigid bodies; Collision of point masses with rigid bodies.

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Page # 3

ROTATIONAL DYNAMICS 1.

RIGID BODY : Rigid body is defined as a system of particles in which distance between each pair of particles remains constant (with respect to time) that means the shape and size do not change, during the motion. Eg. Fan, Pen, Table, stone and so on. Our body is not a rigid body, two blocks with a spring attached between them is also not a rigid body. For every pair of particles in a rigid body, there is no velocity of seperation or approach between the particles. In the figure shown velocities of A and B with respect to   ground are VA and VB respectively

A VA sin1

A

A

VA cos 1

 1 VA B

B

VBA

B

VB  2 VB sin 2

VB cos  2

If the above body is rigid VA cos 1 = VB cos 2 Note : With respect to any particle of rigid body the motion of any other particle of that rigid body is circular. VBA = relative velocity of B with respect to A. Types of Motion of rigid body

Pure Translational Motion

1.1.

Pure Rotational Motion

Combined Translational and Rotational Motion

Pure Translational Motion :

A body is said to be in pure translational motion if the displacement of each particle is same   during any time interval however small or large. In this motion all the particles have same s, v  & a at an instant. example. A box is being pushed on a horizontal surface.

10

6

6

10 16   Vcm  V of any particle,   Scm  S of any particle

  a cm  a of any particle

For pure translational motion :v m2

v m1 v m4

v

m7

v m5

v m3

v vm6

m8

m2

m1

m3

m4

m5

m7

m6 m8

    Fext  m1a1  m 2 a 2  m 3 a 3 ............. IIT-JEE|AIEEE CBSE|SAT|NTSE OLYMPIADS

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ROTATIONAL DYNAMICS

Where m1, m2, m3, ......... are the masses of different particles of the body having accelerations    a1, a 2 , a 3 ,............... respectively..     But acceleration of all the particles are same So, a1  a 2  a 3  .........  a   Fext  Ma Where M = Total mass of the body  a = acceleration of any particle or of centre of mass of body     P  m1v1  m 2 v 2  m 3 v 3 ............. Where m1, m2, m3 ...... are the masses of different particles of the body having velocities    v 1, v 2 , v 3 ............. respectively     But velocities of all the particles are same so v1  v 2  v 3 ..........  v   P  Mv  Where v = velocity of any particle or of centre of mass of the body.. Total Kinetic Energy of body =

1 1 1 m1v 12  m 2 v 22  .......... .  Mv 2 2 2 2

1.2. Pure Rotational Motion : A body is said to be in pure rotational motion if the perpendicular distance of each particle remains constant from a fixed line or point and do not move parallel to the line, and that line    is known as axis of rotation. In this motion all the particles have same ,  and  at an instant. Eg. : - a rotating ceiling fan, arms of a clock. For pure rotation motion :m3 m5

m2

m2

s Where  = angle rotated by the particle r s = length of arc traced by the particle. r = distance of particle from the axis of rotation. 



m1

m3 m6

m1

m5

m6

m4

m4

d  Where  = angular speed of the body.. dt

d Where  = angular acceleration of the body.. dt All the parameters ,  and  are same for all the particles. Axis of rotation is perpendicular to the plane of rotation of particles. Special case : If  = constant,  = 0 +  t Where 0 = initial angular speed 

  0t  2 = 02

1 2 t t = time interval 2 + 2

Total Kinetic Energy 



1 1 m1v12  m 2 v 22 ................. 2 2

1 [m1r12  m 2r22 ................]  2 2

1 2 I Where I = Moment of Inertia = m1r12  m 2r22 ....... 2  = angular speed of body. 

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ROTATIONAL DYNAMICS

1.3

Combined translation and rotational Motion A body is said to be in translation and rotational motion if all the particles rotates about an axis of rotation and the axis of rotation moves with respect to the ground.

2.

MOMENT OF INERTIA

Like the centre of mass, the moment of inertia is a property of an object that is related to its mass distribution. The moment of inertia (denoted by I) is an important quantity in the study of system of particles that are rotating. The role of the moment of inertia in the study of rotational motion is analogous to that of mass in the study of linear motion. Moment of inertia gives a measurement of the resistance of a body to a change in its rotaional motion. If a body is at rest, the larger the moment of inertia of a body the more difficuilt it is to put that body into rotational motion. Similarly, the larger the moment of inertia of a body, the more difficult to stop its rotational motion. The moment of inertia is calculated about some axis (usually the rotational axis). Moment of inertia depends on : (i) density of the material of body (ii) shape & size of body (iii) axis of rotation In totality we can say that it depends upon distribution of mass relative to axis of rotation. Note : Moment of inertia does not change if the mass : (i) is shifted parallel to the axis of the rotation (ii) is rotated with constant radius about axis of rotation 2.1

2.2

Moment of Inertia of a Single Particle For a very simple case the moment of inertia of a

r

single particle about an axis is given by, I = mr2 ...(i) Here, m is the mass of the particle and r its distance from the axis under consideration. Moment of Inertia of a System of Particles The moment of inertia of a system of particles about an axis is given by, I=

2 i i

m r

...(ii)

i

r1

m1

r2

m2

r3

m3

where ri is the perpendicular distance from the axis to the ith particle, which has a mass mi. Ex.1

Two heavy particles having masses m1 & m2 are situated in a plane perpendicular to line AB at a distance of r1 and r2 respectively. C A

E

(i) (ii)

r1

m1

r2 m2

F

D B What is the moment of inertia of the system about axis AB? What is the moment of inertia of the system about an axis passing through m1 and perpendicular to the line joining m1 and m2 ? IIT-JEE|AIEEE CBSE|SAT|NTSE OLYMPIADS

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Page # 6 (iii) Sol.

(i)

(ii)

(iii)

ROTATIONAL DYNAMICS What is the moment of inertia of the system about an axis passing through m1 and m2? Moment of inertia of particle on left is I1 = m1r12. Moment of Inertia of particle on right is I2 = m2r22. Moment of Inertia of the system about AB is I = I1+ I2 = m1r12 + m2r22 Moment of inertia of particle on left is I1 = 0 Moment of Inertia of the system about CD is I = I1 + I2 = 0 + m2(r1 + r2)2 Moment of inertia of particle on left is I1 = 0 Moment of inertia of particle on right is I2 = 0 Moment of Inertia of the system about EF is I = I1 + I2 = 0 + 0

Ex.2

Sol.

Three light rods, each of length 2, are joined together to form a triangle. Three particles A, B, C of masses m, 2m, 3m are fixed to the vertices of the triangle. Find the moment of inertia of the resulting body about (a) an axis through A perpendicular to the plane ABC, (b) an axis passing through A and the midpoint of BC. (a) B is at a distant 2 from the axis XY so the moment of inertia of B (IB) about XY is 2 m (2)2 Similarly Ic about XY is 3m (2)2 and IA about XY is m(0)2

X

A m Y

2l Therefore the moment of inertia of the body about XY is 2m (2)2 + 3 m(2)2 + m(0)2 = 20 m2 B (b) IA about X' Y' = m(0)2 2 IB about X' Y' = 2m () 2m 2 IC about X' Y' = 3m () Therefore the moment of inertia of the body about X' Y' is m(0)2 + 2m()2 + 3m()2 = 5 m2

2l

C 3m

X' A

m

B

C 3m

2m Y' Ex.3

Four particles each of mass m are kept at the four corners of a square of edge a. Find the moment of inertia of the system about a line perpendicular to the plane of the square and passing through the centre of the square.

Sol.

The perpendicular distance of every particle from the given line is a / 2 . The moment of inertia of

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Page # 7

ROTATIONAL DYNAMICS one particle is, therefore, m(a / 2 ) 2 =

1 ma 2 . The 2

m

m

moment of inertia of the system is, /

2

1 ma2 = 2 ma2. 2

a

therefore, 4 

m

2.3

m

Moment of Inertia of Rigid Bodies For a continuous mass distribution such as found in a rigid body, we replace the summation of I

2 i i

m r

by an integral. If the system is divided

i

r

into infinitesimal element of mass dm and if r is the distance from a mass element to the axis of rotation, the moment of inertia is, I=

r

2

dm

where the integral is taken over the system. (A)

Uniform rod about a perpendicular bisector Consider a uniform rod of mass M and length l figure and suppose the moment of inertia is to be calculated about the bisector AB. Take the origin at the middle point O of the rod. Consider the element of the rod between a distance x and x + dx from the origin. As the rod is uniform, Mass per unit length of the rod = M / l so that the mass of the element = (M/l)dx. The perpendicular distance of the element from the line AB is x. The moment of inertia of this element about AB is dI 

A

B

x

dx

0

M dx x 2 . l

When x = – l/2, the element is at the left end of the rod. As x is changed from – l/2 to l/2, the elements cover the whole rod. Thus, the moment of inertia of the entire rod about AB is l/2

I



M x3  M 2 x dx    l  l 3 

l / 2

(B)

l/2

 –l / 2

Ml 2 12

Moment of inertia of a rectangular plate about a line parallel to an edge and passing through the centre The situation is shown in figure. Draw a line parallel to AB at a distance x from it and another at a distance x + dx. We can take the strip enclosed between the two lines as the small element. x A b dx l It is “small” because the perpendiculars from different points of the strip to AB differ by not B

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Page # 8

ROTATIONAL DYNAMICS

more than dx. As the plate is uniform, its mass per unit area =

M bl

M M b dx  dx . l bl The perpendicular distance of the strip from AB = x. Mass of the strip =

The moment of inertia of the strip about AB = dI =

M dx x 2 . The moment of inertia of the given l

plate is, therefore, l/2

I



M 2 Ml 2 x dx  l 12

l / 2

The moment of inertia of the plate about the line parallel to the other edge and passing through the centre may be obtained from the above formula by replacing l by b and thus,

Mb 2 . 12 Moment of inertia of a circular ring about its axis (the line perpendicular to the plane of the ring through its centre) Suppose the radius of the ring is R and its mass is M. As all the elements of the ring are at the same perpendicular distance R from the axis, the moment of inertia of the ring is I

(C)

I  r 2 dm  R2 dm  R2 dm  MR2 .



(D)





Moment of inertia of a uniform circular plate about its axis Let the mass of the plate be M and its radius R. The centre is at O and the axis OX is perpendicular to the plane of the plate. X dx

0 x R

Draw two concentric circles of radii x and x + dx, both centred at O and consider the area of the plate in between the two circles. This part of the plate may be considered to be a circular ring of radius x. As the periphery of the ring is 2 x and its width is dx, the area of this elementary ring is 2xdx. The area of the plate is  R2. As the plate is uniform, M

Its mass per unit area = M

Mass of the ring 

2

 R2

2  x dx 

2 M x dx

R R2 Using the result obtained above for a circular ring, the moment of inertia of the elementary ring about OX is  2 Mx dx  2 dI   x .  R2  The moment of inertia of the plate about OX is R

I

2M

R 0

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2

x 3 dx 

MR 2 . 2

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Page # 9

ROTATIONAL DYNAMICS (E)

Moment of inertia of a hollow cylinder about its axis Suppose the radius of the cylinder is R and its mass is M. As every element of this cylinder is at the same perpendicular distance R from the axis, the moment of inertia of the hollow cylinder about its axis is

I  r 2 dm  R 2 dm  MR 2



(F)



Moment of inertia of a uniform solid cylinder about its axis Let the mass of the cylinder be M and its radius R. Draw two cylindrical surface of radii x and x + dx coaxial with the given cylinder. Consider the part of the cylinder in between the two surface. This part of the cylinder may be considered to be a hollow cylinder of radius x. The area of cross-section of the wall of this hollow cylinder is 2 x dx. If the length of the cylinder is l, the volume of the material of this elementary hollow cylinder is 2 x dxl. The volume of the solid cylinder is  R2 l and it is uniform, hence its mass per unit volume is M 

 R2 l

The mass of the hollow cylinder considered is M  R2 l

2 x dx l 

2M R2

x dx .

dx

As its radius is x, its moment of inertia about the given axis is

x  2M  dI   2 x dx x2 . R 

The moment of inertia of the solid cylinder is, therefore, R

I

2M

R 0

(G)

2

x 3 dx 

MR 2 2 .

Note that the formula does not depend on the length of the cylinder. Moment of inertia of a uniform hollow sphere about a diameter Let M and R be the mass and the radius of the sphere, O its centre and OX the given axis (figure). The mass is spread over the surface of the sphere and the inside is hollow. Let us consider a radius OA of the sphere at an angle  with the axis OX and rotate this radius about OX. The point A traces a circle on the sphere. Now change  to  + d and get another circle of somewhat larger radius on the sphere. The part of the sphere between these two circles, shown in the figure, forms a ring of radius R sin. The width of this ring is Rd and its periphery is 2R sin. Hence, the area of the ring = (2R sin) (Rd). x

M Mass per unit area of the sphere 

M The mass of the ring 

4 R

4 R2

R sin A Rd

.

(2R sin )(Rd)  2

M sin  d. 2

R 0

d

The moment of inertia of this elemental ring about OX is M  d I   sin  d. (R sin ) 2  M R2 sin 3  d 2  2

As  increases from 0 to , the elemental rings cover the whole spherical surface. The

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Page # 10

ROTATIONAL DYNAMICS

moment of inertia of the hollow sphere is, therefore, 

I

    M 2 MR2  MR 2   (1  cos 2 )sin  d   (1  cos 2 ) d (cos ) R sin3  d  2 2  2    0 0   0  





MR2  2

(H)





 cos 3   2 2 cos     MR 3  3  0

Moment of inertia of a uniform solid sphere about a diameter Let M and R be the mass and radius of the given solid sphere. Let O be centre and OX the given axis. Draw two spheres of radii x and x + dx concentric with the given solid sphere. The thin spherical shell trapped between these spheres may be treated as a hollow sphere of radius x.

x dx

0 x

The mass per unit volume of the solid sphere

=

M 3M  3 4 R 3 4 R 3

The thin hollow sphere considered above has a surface area 4x2 and thickness dx. Its volume is 4  x2 dx and hence its mass is  3M  3M  (4  x 2 dx) = 3 x 2 dx =  3 R  4 R  Its moment of inertia about the diameter OX is, therefore, 2  3M 2  x dx x2 3  R 3 

2M

x 4 dx R3 If x = 0, the shell is formed at the centre of the solid sphere. As x increases from 0 to R, the shells cover the whole solid sphere.

dl =

=

The moment of inertia of the solid sphere about OX is, therefore, R

I=

2M

R

3

x 4 dx =

0

Ex.4

2 MR 2 . 5

Find the moment of Inertia of a cuboid along the axis as shown in the figure.

I

b a c

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Page # 11

ROTATIONAL DYNAMICS

M(a 2  b 2 ) 12

Sol.

After compressing the cuboid parallel to the axis I =

3.

THEOREMS OF MOMENT OF INERTIA There are two important theorems on moment of inertia, which, in some cases enable the moment of inertia of a body to be determined about an axis, if its moment of inertia about some other axis is known. Let us now discuss both of them. Theorem of parallel axes A very useful theorem, called the parallel axes theorem relates the moment of inertia of a rigid body about two parallel axes, one of which passes COM through the centre of mass.

3.1

Two such axes are shown in figure for a body of mass M. If r is r the distance between the axes and ICOM and I are the respective moments of inertia about them, these moments are related by, I = ICOM + Mr2 Theorem of parallel axis is applicable for any type of rigid body whether it is a two dimensional or three dimensional

*

A

Ex 5. Three rods each of mass m and length l are joined together to form an equilateral triangle as shown in figure. Find the moment of inertia of the system about an axis passing through its centre of mass and perpendicular to the plane of triangle. Sol.

Moment of inertia of rod BC about an axis perpendicular to plane of triangle ABC and passing through the midpoint of rod BC (i.e., D) is

COM B

C

ml 2 12 From theorem of parallel axes, moment of inertia of this rod about the asked axis is I1 =

A

2

 l  ml 2 ml 2   m   I2 = I1 + mr = 12 6 2 3

COM

2

r  Moment of inertia of all the three rod is

B  ml 2  ml 2   I  3I2  3 2  6 

30° D

C

Ex.6. Find the moment of inertia of a solid sphere of mass M and radius R about an axis XX shown in figure.

x

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Page # 12

ROTATIONAL DYNAMICS

x Sol.

From theorem of parallel axis, IXX = ICOM + Mr2 =

2 MR2  MR2 5

=

7 MR 2 5

COM

x

r=R Ex.7. Consider a uniform rod of mass m and length 2l with two particles of mass m each at its ends. Let AB be a line perpendicular to the length of the rod passing through its centre. Find the moment of inertia of the system about AB. A Sol. IAB = Irod + Iboth particles





3.2

m(2l )2  2(ml 2 ) 12 7 2 ml 3

I

I

m

m Ans.

B

Theorem of perpendicular axes The theorem states that the moment of inertia of a plane lamina about an axis perpendicular to the plane of the lamina is equal to the sum of the moments of inertia of the lamina about two axes perpendicular to each other, in its own plane and intersecting each other, at the point where the perpendicular axis passes through it. Let x and y axes be chosen in the plane of the body and z-axis perpendicular, to this plane, three axes being mutually perpendicular, then the theorem states that.

z y

xi ri

P yi x

O

(i) (ii) (iii) Ex.8

Iz = Ix + Iy Important point in perpendicular axis theorem This theorem is applicable only for the plane bodies (two dimensional). In theorem of perpendicular axes, all the three axes (x, y and z) intersect each other and this point may be any point on the plane of the body (it may even lie outside the body). Intersection point may or may not be the centre of mass of the body. Find the moment of inertia of uniform ring of mass M and radius R about a diameter. B

Z

C

Sol.

0

D

A Let AB and CD be two mutually perpendicular diameters of the ring. Take them ax X and Y-

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Page # 13

ROTATIONAL DYNAMICS

axes and the line perpendicular to the plane of the ring through the centre as the Z-axis. The moment of inertia of the ring about the Z-axis is I = MR2. As the ring is uniform, all of its diameter equivalent and so Ix = Iy, From perpendicular axes theorem, Iz = Ix + Iy

Hence Ix =

Iz MR2 = 2 2

Similarly, the moment of inertia of a uniform disc about a diameter is MR2/4 Ex.9

Two uniform identical rods each of mass M and length  are joined to form a cross as shown in figure. Find the moment of inertia of the cross about a bisector as shown dotted in the figure.

Sol.

Consider the line perpendicular to the plane of the figure through the centre of the cross. The moment of inertia of each rod about this line is

M 2 and hence the moment of inertia of the 12

M 2 . The moment of inertia of the cross about the two bisector are equal by 6 symmetry and according to the theorem of perpendicular axes, the moment of inertia of the

cross is

cross about the bisector is

M 2 . 12

Ex.10 In the figure shown find moment of inertia of a plate having mass M, length  and width b about axis 1,2,3 and 4. Assume that C is centre and mass is uniformly distributed 1

2

4

C 3

b 

Sol.

Moment of inertia of the plate about axis 1 (by taking rods perpendicular to axis 1) l1 = Mb2/3 Moment of inertia of the plate about axis 2 (by taking rods perpendicular to axis 2) I 2 = M2/12 Moment of inertia of the plate about axis 3 (by taking rods perpendicular to axis 3)

Mb 2 12 Moment of inertia of the plate about axis 4(by taking rods perpendicular to axis 4) I4 = M2/3 I3 

3.3

Moment of Inertia of Compound Bodies Consider two bodies A and B, rigidly joined together. The moment of inertia of this compound body, about an axis XY, is required. If IA is the moment of inertia of body A about XY. IB is the moment of inertia of body B about XY.Then, moment of Inertia of compound body I = IA + IB Extending this argument to cover any number of bodies rigidly joined together, we see that the moment of inertia of the compound body, about a specified axis, is the sum of the moments of inertia of the separate parts of the body about the same axis. IIT-JEE|AIEEE CBSE|SAT|NTSE OLYMPIADS

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Page # 14

ROTATIONAL DYNAMICS

A Y

X B

Ex.11 Two rods each having length l and mass m joined together at point B as shown in figure.Then findout moment of inertia about axis passing thorugh A and perpendicular to the plane of page as shown in figure.  A B × 

Sol.

C We find the resultant moment of inertia I by dividing in two parts such as I = M.I of rod AB about A + M.I of rod BC about A I = I1 + I2 ... (1) first calculate I1 : B  A ×

m 2 ...(2) 3 Calculation of I2 : use parallel axis theorem I2 = ICM + md2 I1 =



/2

...(3)

m  2 m  2 5 2m   3 12 4

I= 4.

d

COM ×

 2  m 2 m 2 5 2   2  m   m = 12  4  = 12 4   Put value from eq. (2) & (3) into (1) I=

×

m 2 (4  1  15 ) 12



I=

5m 2 3

CAVITY PROBLEMS :

Ex.12 A uniform disc having radius 2R and mass density  as shown in figure. If a small disc of radius R is cut from the disc as shown. Then find out the moment of inertia of remaining disc around the axis that passes through O and is perpendicular to the plane of the page.

2R O

Sol.

R

We assume that in remaning part a disc of radius R and mass density ±  is placed. Then

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Page # 15

ROTATIONAL DYNAMICS M1  (2R) 2

M2  –R2 2R

2R O   R

× I1

 when  is taken

I2

R

+ × when – is takes

Total Moment of Inertia I = I1 + I2 I1 =

M1(2R)2 2

4R2 .4R2 = 8  R4 2 To calculate I2 we use parallel axis theorem. I2 = ICM + M2R2 I1 =

I2 =

M2R2 + M2R2 2

3 3 M2R2 = (– R 2 )R2 2 2 Now I = I1 + I2 I2 =

4 I = 8 R –

3 R4 2

I2 = –

I=

3 R4 2

13 R 4 2

R/ 3

Ex.13 A uniform disc of radius R has a round disc of radius R/3 cut as shown in Fig. The mass of the remaining (shaded) portion of the disc equals M. Find the moment of inertia of such a disc relative to the axis passing through geometrical centre of original disc and perpendicular to the plane of the disc.

Sol.

O

R

Let the mass per unit area of the material of disc be . Now the empty space can be considered as having density –  and . Now I0 = I + I–  (R2)R2/2 = M.I of  about O – (R / 3) 2 (R / 3) 2  [– (R / 3) 2 ]( 2R / 3) 2 2

I– =

= M.I of –  about 0 

5.

I0 =

4 R 4 Ans. 9

TORQUE : Torque represents the capability of a force to produce change in the rotational motion of the body

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Page # 16

ROTATIONAL DYNAMICS Line of action of force

F

P r

r sin Q

5.1 Torque about point :     Torque of force F about a point   r F  where F = force applied P = point of application of force Q = point about which we want to calculate the torque.  r = position vector of the point of application of force from the point about which we want to determine the torque.    rF sin  = rF = rF

where  = angle between the direction of force and the position vector of P wrt. Q. r = perpendicular distance of line of action of force from point Q. F = force arm

SI unit to torque is Nm Torque is a vector quantity and its direction is determined using right hand thumb rule. Ex.14 A particle of mass M is released in vertical plane from a point P at x = x0 on the x-axis it falls vertically along the y-axis. Find the torque  acting on the particle at a time t about origin? x0 P O x 

r 

Sol. 

mg Torque is produced by the force of gravity    r F sin  k or

  r F  x 0 mg

Ex.15 Calculate the total torque acting on the body shown in figure about the point O

10N

15N

O 4cm

90°

6c m

37°

3c 30° m

150°

20N

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Page # 17

ROTATIONAL DYNAMICS

15sin37°

15N

10N 90°

4cm

O

4cm

Sol.

6c m

37° 5N 20N

30°

150° 20sin30° 0 = 15sin37 × 6 + 20 sin 30° × 4 = 54 + 40 – 40 = 54 N-cm 0 = 0.54 N-m

– 10 × 4

Ex.16 A particle having mass m is projected with a velocity v0 from a point P on a horizontal ground making an angle  with horizontal. Find out the torque about the point of projection acting on the particle when

V0 

(a) it is at its maximum height ? P

Q

(b) It is just about to hit the ground back ? Sol. (a)

F = mg ; r 

p =

r

R 2

v 20 sin 2 R  P = mg = mg  2g 2

(b)

v0

Particle is at maximum height then  about point P is  p  r F

mg



P

mv 20 sin 2 2

when particle is at point Q then  about point P is  p'  rF r  R ; F = mg

 p'

v 02 sin2  mgR = mg g

Q

P mg

Ex.17 In the previous question, during the motion of particle from P to Q. Torque of gravitational force about P is : (A) increasing (B) decreasing (C) remains constant (D) first increasing then decreasing Sol.

Torque of gravitational force about P is increasing because r is increasing from O to R. (Range)

5.2

Torque about axis :      r F  where  = torque acting on the body about the axis of rotation  r = position vector of the point of application of force about the axis of rotation. IIT-JEE|AIEEE CBSE|SAT|NTSE OLYMPIADS

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Page # 18

ROTATIONAL DYNAMICS  F = force applied on the body..      net   1   2   3 .....

To understand the concept of torque about axis we take a general example which comes out in daily life. Figure shows a door ABCD. Which can rotate about axis AB. Now if we apply force. F at point.

D

A r

×

y x

B

C F

in inward direction then AB = r F and direction of this AB is along y axis from right hand thumb rule. Which is parallel to AB so gives the resultant torque. Now we apply force at point C in the direction as shown   figure. At this time r & F are perpendicular to each other which gives  AB  rF

But door can’t move when force is applied in this direction because the direction of  AB is perpendicular to AB according to right hand thumb rule. So there is no component of  along AB which gives res  0  Now conclude Torque about axis is the component of r  F parallel to axis of rotation. Note : The direction of torque is calculated using right hand thumb rule and it is always perpendicular to the plane of rotation of the body.

F2 r2

F3

r3 × r1 F1

If F1 or F2 is applied to body, body revolves in anti-clockwise direction and F3 makes body revolve in clockwise direction. If all three are applied.   resul tan t  F1r1  F2r2 – F3 r3 (in anti-clockwise direction) 6.

BODY IS IN EQUILIBRIUM : We can say rigid body is in equillibrium when it is in (a) Translational equilibrium  i.e. Fnet  0

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Page # 19

ROTATIONAL DYNAMICS Fnet x = 0 and Fnet y = 0 and (b) Rotational equillibrium   net  0 i.e., torque about any point is zero

Note : (i) If net force on the body is zero then net torque of the forces may or may not be zero. example. A pair of forces each of same magnitude and acting in opposite direction on the rod.

B

A F (2)

F C

2

 A  2F If net force on the body is zero then torque of the forces about each and every point is same  about B

 B  F  + F

 B  2F  about C

 C  2F

Ex.18 Determine the point of application of third force for which body is in equillibrium when forces of 20 N & 30 N are acting on the rod as shown in figure 20N

A 10cm C 20cm Sol.

(i)

(ii)

B

30N Let the magnitude of third force is F, is applied in upward direction then the body is in the equilibrium when  Fnet  0 (Translational Equillibrium)  20 + F = 30  F = 10 N So the body is in translational equilibrium when 10 N force act on it in upward direction. Let us assume that this 10 N force act. 10N 20N Then keep the body in rotational equilibrium x So Torque about C = 0 i.e. c = 0 B A C 20cm  30 × 20 = 10 x 30N x = 60 cm so 10 N force is applied at 70 cm from point A to keep the body in equilibrium.

Ex.19 Determine the point of application of force, when forces are acting on the rod as shown in figure.

10N

5N 5cm

5cm 3N

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Page # 20

Sol.

ROTATIONAL DYNAMICS

 Since the body is in equillibrium so we conclude F net  0 and torque about any point is zero  i.e.,  net  0 10N

6

5N F2 A

x 

F

37° 8N

F1 3N

Let us assume that we apply F force downward at A angle  from the horizontal, at x distance from B From

 F net  0



Fnet x = 0 which gives F2 = 8 N From Fnet y = 0  5 + 6 = F1 + 3  F1 = 8 N If body is in equillibrium then torque about point B is zero.  3 × 5 + F1. x – 5 × 10 = 0  15 + 8x – 50 = 0 x=

35 9



x = 4.375 cm

Ex.20 A uniform rod length , mass m is hung from two strings of equal length from a ceiling as shown in figure. Determine the tensions in the strings ?

/4

Sol.

B A Let us assume that tension in left and right string is TA and TB respectively. Then   Rod is in equilibrium then Fnet  0 & net  0  From Fnet  0

mg = TA + TB ...(1) From net = 0 about A mg



 3  TB  0 2 4

TB =

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TB  /4

 /2

B

A mg

2mg 3

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Page # 21

ROTATIONAL DYNAMICS

from eq. (1) TA 

2mg = mg  3

TA =

mg 3

Ladder Problems : Ex.21 A stationary uniform rod of mass ‘m’, length ‘’ leans against a smooth vertical wall making an angle  with rough horizontal floor. Find the normal force & frictional force that is exerted by the floor on the rod?

smooth



rough Sol.

As the rod is stationary so the linear acceleration and angular acceleration of rod is zero. i.e., acm = 0 ;  = 0. A N2 N2 = f  acm =0 N = mg Torque about any point of the rod should also be zero =0 A = 0  mg cos 

N1 cos  = sin  f +

f=

mgcos 

1

N1  mg



B f Free Body Diagram

 + f  sin  = N1 cos .  2

mgcos  2

mgcos  mgcot  = 2 sin  2

Ex.22 The ladder shown in figure has negligible mass and rests on a frictionless floor. The crossbar connects the two legs of the ladder at the middle. The angle between the two legs is 60°. The fat person sitting on the ladder has a mass of 80 kg. Find the contanct force exerted by the floor on each leg and the tension in the crossbar.

W 1m 60°

T

N 1m

Sol.

N

° 30

The forces acting on different parts are shown in figure. Consider the vertical equilibrium of “the ladder plus the person” system. The forces acting on this system are its weight (80 kg)

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Page # 22

ROTATIONAL DYNAMICS

g and the contact force N + N = 2 N due to the floor. Thus 2 N = (80 kg) g or N = (40 kg) (9.8 m/s2) = 392 N Next consider the equilibrium of the left leg of the ladder. Taking torques of the forces acting on it about the upper end, 2 N (2m) tan 30° = T (1m)

or

T=N

3

2 = (392 N) ×

3

= 450 N

Ex.23 A thin plank of mass m and length  is pivoted at one end and it is held stationary in horizontal position by means of a light thread as shown in the figure then find out the force on the pivot.

Sol.

(i)

T

N2

Free body diagram of the plank is shown in figure.  Plank is in equilibrium condition So Fnet & net on the plank is zero

N1

O

A mg

from Fnet = 0  Fnet x = 0 N1 = 0 Now Fnet  0 y 

N2 + T = mg

...(i)

from net = 0  net about point A is zero so N2 .  = mg . /2 

N2 

mg 2

Ex.24 A square plate is hinged as shown in figure and it is held stationary by means of a light thread as shown in figure. Then find out force exerted by the hinge.

square plate

T

Sol.

F.B.D.

 Body is in equilibrium and N

T and mg force passing through one line so from

net = 0,

N=0

mg

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Page # 23

ROTATIONAL DYNAMICS

7.

RELATION BETWEEN TORQUE AND ANGULAR ACCELERATION The angular acceleration of a rigid body is directly proportional to the sum of the torque components along the axis of rotation. The proportionality constant is the inverse of the moment of inertia about that axis, or 

 I

Thus, for a rigid body we have the rotational analog of Newton's second law ;

   I

...(iii)

Following two points are important regarding the above equation. (i) The above equation is valid only for rigid bodies. If the body is not rigid like a rotating tank of water, the angular acceleration  is different for different particles. (ii) The sum  in the above equation includes only the torques of the external forces, because all the internal torques add to zero. Ex.25 A uniform rod of mass m and length  can rotate in vertical plane about a smooth horizontal axis hinged at point H.  × A H (i) Find angular acceleration  of the rod just after it is released from initial horizontal position from rest? (ii) Calculate the acceleration (tangential and radial) of point A at this moment. Sol. (i) H = IH   3g m 2  = = 2 2 3

mg.

(ii) aA = a =

3g 3g . = 2 2

aCA = 2r = 0. = 0

(  = 0 just after release)

×



Ex.26 A uniform rod of mass m and length  hinged at point H can rotate in vertical plane about a smooth horizontal axis. Find force exerted by the hinge just after the rod is released from rest, from an initial position making an angle of 37° with horizontal ?

Sol.

37°

H Just After releasing at 37º from horizontal F.B.D. of plank from

net = I

 m 2 .  about point A = A = mg cos 37° = 3 2

 N1

R

37º

mg mgcos 37º

A

N2

6g = rad/sec2 5 Now Tangential acceleration of centre of mass  3g m / s2 = 2 5 just after release vcm = 0  ar = 0 Now resolving of at in horizontal and vertical direction as shwon in figure

at = .

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Page # 24

ROTATIONAL DYNAMICS

a t || 

N1

37°

R

N2

a t 

9g 25

N1

3g/5

mg

12g 25

N2

from Fnet = ma in both horizontal and vertical direction  9g  N2= m  25

Now

R=



N1 =

13mg 25

N12  N22

mg 10 5 PULLEY BLOCK SYSTEM If there is friction between pulley and string and pulley have some mass then tension is different on two sides of the pulley. R=

Reason : To understand this concept we take a pulley block system as shown in figure.

 B

a

R

C

T1

A M

D a M>m m

Let us assume that tension induced in part AB of the string is T1 and block M move downward. If friction is present between pulley and string then it opposes the relative slipping between pulley and string, take two point e and f on pulley and string respectively. If friction is there then due to this, both points wants to move together. So friction force act on e and d in the direction as shown is figure T2 This friction force f acting on point d increases the tension T1 by a small amount dT. f Then T1 = T2 + dT C d e or we can say T2 = T1 – f f In this way the tension on two side of pulley is different If there is no relative slipping between pulley and string at a then   = R R

T1

Ex.27 The pulley shown in figure has moment of inertia l about its axis and radius R. Find the acceleration of the two blocks. Assume that the string is light and does not slip on the pulley. Sol. Suppose the tension in the left string is T1 and that in the right string is T2. Suppose the block of mass M goes down with an acceleration a and the other block moves up with the same acceleration. This is also the tangential acceleration of the rim of the wheel as the string

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Page # 25

ROTATIONAL DYNAMICS does not slip over the rim. The angular acceleration of the wheel  =

a . R

The equations of motion for the mass M, the mass m and the pulley are as follows ; Mg – T1 = Ma ...(i) T2 – mg = ma ...(ii)

R

m

Ia M T1R – T2R = I = ...(iii) R Substituting for T1 and T2 from equations (i) and (ii) in equation (iii) [M(g – a) – m (g + a)]R =

Ia R

Solving, we get

(M – m)gR2 a=

I  (M  m)R2

8.

ANGULAR MOMENTUM

8.1

Angular momentum of a particle about a point.     L = r p sin  L  r P  |L|  r  P  |L|  P  r  Where P = momentum of partilcle  r = position of vector of particle with respect to point about which

Pcos    r

 P

P sin 

angular momentum is to be calculated.    = angle between vectors r & p

O r = perpendicular distance of line of motion of particle from point O.

P = perpendicular component of momentum. SI unit of angular momentum is kgm2/sec. Ex.28 A particle of mass m is moving along the line y = b, z = 0 with constant speed v. State whether the angular momentum of particle about origin is increasing, decreasing or constant. Sol.

y

 | L |  mvr sin  = mvr mvb



P  r

 v

r  b

  X | L | = constant as m, v and b all are constants. O   Direction of r  v also remains the same. Therefore, angular momentum of particle about origin remains constant with due course of time.

 Note : In this problem | r | is increasing,  is decreasing but r sin , i.e., b remains constant. Hence, the angular momentum remains constant.

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Page # 26

ROTATIONAL DYNAMICS

Ex.29 A particle of mass m is projected with velocity v at an angle  with the horizontal. Find its angular momentum about the point of projection when it is at the highest point of its trajectory. Sol. At the highest point it has only horizontal velocity y vx = v cos . Length of the perpendicular to the horizontal velocity from 'O' is the maximum height, where Hmax 

v 2 sin 2  2g

H

 Angular momentum L =

8.2

mv 3 sin 2  cos  2g

O

x

Angular Momentum of a rigid body rotating about a fixed axis Suppose a particle P of mass m is going in a circle of radius r and at some instant the speed of the particle is v. For finding the angular momentum of the particle about the axis of rotation, the origin may be chosen anywhere on the axis. We choose it at the centre of the

   circle. In this case  r and P are perpendicular to each other and r  P is along the axis.   Thus, component of r  P along the axis is mvr itself. The angular momentum of the whole rigid body about AB is the sum of components of all particles, i.e., L=

m r v

i i i

i

Here, 

vi = r i  L=

m r

i i

2

 i or L = 

i

m r

2

i i

i

or L = I Here, I is the moment of inertia of the rigid body about AB.  Note : Angular momentum about axis is the component of I along the axis. In most of the cases angular momentum about axis is I.

Ex.30 Two small balls A and B, each of mass m, are attached rigidly to the ends of a light rod of length d. The structure rotates about the perpendicular bisector of the rod at an angular speed . Calculate the angular momentum of the individual balls and of the system about the axis of rotation. d A

Sol.

O

B

Consider the situation shown in figure. The velocity of the ball A with respet to the centre O d is v = . 2 The angular momentum of the ball with respect to the axis is

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Page # 27

ROTATIONAL DYNAMICS

1  d   d  L1 = mvr = m    = md2. The same the angular momentum L2 of the second ball. The  2   2 4 angular momentum of the system is equal to sum of these two angular momenta i.e., L = 1/2 md2. 9.

CONSERVATION OF ANGULAR MOMENTUM : The time rate of change of angular momentum of a particle about some referenence point in an inertial frame of reference is equal to the net torques acting on it.   dL ....(i) or net  dt    dL Now, suppose that  net  0 , then  0 , so that L = constant. dt "When the resultant external torque acting on a system is zero, the total vector angular momentum of the system remains constant. This is the principle of the conservation of angular momentum. For a rigid body rotating about an axis (the z-axis, say) that is fixed in an inertial reference frame, we have Lz = I It is possible for the moment of inertia I of a rotating body to change by rearrangement of its parts. If no net external torque acts, then Lz must remains constant and if I does change, there must be a compensating change in . The principle of conservation of angular momentum in this case is expressed. I = constant.

Ex.31 A wheel of moment of inertia I and radius R is rotating about its axis at an angular speed 0. It picks up a stationary particle of mass m at its edge. Find the new angular speed of the wheel. Sol. Net external torque on the system is zero. Therefore, angular momentum will remain conserved. Thus, I1 1 I11 = I22 or 2 = I 2 Here, I1 = I, 1 = 0, I2 = I + mR2 I 0  2 = I  mR 2 Note : A

Hinge m

L O m

u

Case I

O m

u m

Case II

Comments on Linear Momentum : In case I : Linear momentum is not conserved just before and just after collision because during collision hinge force act as an external force. In case II : Linear momentum is conserved just before and just after collision because no external force on the string. Comments on Angular Momentum : In case I : Hinge force acts at an external force during collision but except point A all the other reference point given net  0. So angular momentum is conserved only for point A.

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Page # 28

ROTATIONAL DYNAMICS

In case II : angular momentum is conserved at all points in the world. Ex.32 A uniform rod of mass m and length  can rotate freely on a smooth horizontal plane about a vertical axis hinged at point H. A point mass having same mass m coming with an initial speed u perpendicular to the rod, strikes the rod in-elastically at its free end. Find out the angular velocity of the rod just after collision? m,  × H u

Sol.

m Angular momentum is conserved about H because no external force is present in horizontal plane which is producing torque about H.   m 2 2 3u mu =  3  m    w =   4

Ex.33 A uniform rod of mass m and length  can rotate freely on a smooth horizontal plane about a vertical axis hinged at point H. A point mass having same mass m coming with an initial speed u perpendicular to the rod, strikes the rod and sticks to it at a distance of 3/4 from hinge point. Find out the angular velocity of the rod just after collision?

m,  H× u m m,

H

Sol.

Initial position

3/4

u m

from angular momentum conservation about H initial angular momentum = final angular momentum 2

m. u



ml 3  3   m   +  4 4 3

3mu 1 9   m 2    4  3 16 

2

 m,



H m

3u  16  27    4   48 





36 u 43 

Ex.34 A uniform rod AB of mass m and length 5a is free to rotate on a smooth horizontal table about a pivot through P, a point on AB such that AP = a. A particle of mass 2m moving on the table strikes AB perpendicularly at the point 2a from P with speed v, the rod being at rest. If the coefficient of restitution between them is

Sol.

1 , find their speeds 4

immediately after impact. Let the point of impact be Q so that IIT-JEE|AIEEE CBSE|SAT|NTSE OLYMPIADS

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Page # 29

ROTATIONAL DYNAMICS PQ = 2a Let P be the point of pivot that AP = a 5a Pm

A

Q

C

a

B

v 2m

Before Collision

Let the velocities of point, Q and the particle after impact be vq and vp respectively then from momentum conservation about point P. Li = L f 2a(2mv) = Ip + (2a) (2mvp) ...(i) Vq  P 2

IP 

1  5a   3a  m   m    2 3  2

2

C

use parallel    axis theorem

13 ma 2 3 use equation (ii) in equation (i) 

...(ii)

13 ma2  3 12(v – vp) = 13a

....(iii)

Q

3a/2

Vp

After Collision

4ma(v – vp) =

coefficient of restitution e =

velocity of seperation velocity of approach

1 vq  vp  4 v



vq – vp =

v 4

...(iv)

vq = 2a ...(v) Put value of  from eq (iii) to equation (v)  12  vq  2   (v  vp )  13  So now from equation (iv) v 24 83 v ( v – vp )  v p   vp  13 4 148 So in this way we get  

15 v 37 a

Ex.35 A person of mass m stands at the edge of a circular platform of radius R and moment of inertia. A platform is at rest initially. But the platform rotate when the person jumps off from the platform tangentially with velocity u with respect to platfrom. Determine the angular velocity of the platform. Sol. Let the angular velocity of platform is . Then the velocity of person with respect to ground v. vmD = vmG – VDG u = vm + R vm = u –  R Now from angular momentum conservation Li = L f 0 = mvmR – I  

I  = m (u –  R) . R

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

R M R

u

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Page # 30

ROTATIONAL DYNAMICS muR



=

I  mR 2

Ex.36 Consider the situation of previous example. If the platform is rotating intially with angular velocity 0 and then person jumps off tangentially. Determine the new angular velocity of the platform. Sol. Let the angular velocity of platfrom after jumps off the mass is . Then velocity Of man. 0



R

R

u

Initially

vm = vmp + vp vm = u – R From Angular momentum conservation (I + mR2) 0 = I  – m (u –  R) R I0 + mR2 0 = I  – m u R + m  R2 

10.



(I  mR 2 )0  mu R (I  mR 2 )

ANGULAR IMPULSE t2 

The angular impulse of a torque in a given time interval is defined as



t1

 dt

Here,   is the resultant torque acting on the body. Further, since   dL       dt  d L dt t2     dt = angular impulse = L 2 – L 1 or



t1

Thus, the angular impulse of the resultant torque is equal to the change in angular momentum. Let us take few examples based on the angular impulse. Ex.37 Figure shows two cylinders of radii r1 and r2 having moments of inertia I1 and I2 about their respective axes. Initially, the cylinders rotate about their axes with angular speeds 1 and 2 as shown in the figure. The cylinders are moved closer to touch each other keeping the axes parallel. The cylinders first slip over each other at the contact but the slipping finally ceases due to the friction between them. Find the angular speeds of the cylinders after the slipping ceases. 2 1 I2 r I1 r 1

Sol.

2

When slipping ceases, the linear speeds of the points of contact of the two cylinders will be equal. If  1 ' and  2 ' be the respective angular speeds, we have  1 ' r1   2 ' r2

.....(i)

The change in the angular speed is brought about by the frictional force which acts as long as the slipping exists. If this force f acts for a time t, the torque on the first cylinder is fr1 and that on the second is f r2. Assuming 1 r1 > 2 r2, the corresponding angular impluses are – f r1 t and f r2 t. We, therefore, have

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Page # 31

ROTATIONAL DYNAMICS – f r1 t = I 1 (  1 '   1 ) and

f r2 t = I 2 (  2 '   2 )

or,

–

I1 I ( 1 '– 1 ) = 2 ( 2 '– 2 ) r1 r2

...(ii)

Solving (i) and (ii),  1' =

I1 1r2  I2  2r1 I2r12  I1r22

r2 and  2 ' 

I1 1r2  I2  2r1 I2r12  I1r22

r1

Kinetic Energy of a rigid body rotating about a fixed axis. Suppose a rigid body is rotating about a fixed axis with angular speed . Then, kinetic energy of the rigid body will be : K=

 i

1 2 = 2

1 mi v i2 = 2 2 i i

m r

=

i

 i

 ri

1 mi (ri ) 2 2

mi

1 2 I 2

(as

2 i i

m r

 I)

i

1 2 I 2 Sometimes it is called the rotational kinetic energy. Ex.38 A uniform rod of mass m and length  is kept vertical with the lower end clamped. It is slightly pushed to let it fall down under gravity. Find its angular speed when the rod is passing through its lowest position. Neglect any friction at the clamp. What will be the linear speed of the free end at this instant? Sol. As the rod reaches its lowest position, the centre of mass is lowered by a distance . Its gravitational potential energy is decreased by mg. As no energy is lost against friction, this should be equal to the increase in the kinetic energy. As the rotation occurs about the horizontal axis through the clamped end, the moment of inertia is I = m 2/3. Thus, Thus, KE =

1 2 I  mg 2 or



1  m 2  2   = mg   2  3 



–

6g 

=



The linear speed of the free end is v =  =

11.

6g

COMBINED TRANSLATIONAL AND ROTATIONAL MOTION OF A RIGID BODY : We have already learnt about translational motion caused by a force and rotational motion about a fixed axis caused by a torque. Now we are going to discuss a motion in which body undergoes translational as well as rotational motion. Rolling is an example of such motion. If the axis of rotation is moving then the motion is combined translational and rotational motion. To understand the concept of combined translational and rotational motion we consider a uniform disc rolling on a horizontal surface. Velocity of its centre of mass is Vcom and its angular speed is  as shown in figure.

v R

A

v

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Page # 32

ROTATIONAL DYNAMICS

Path of point A with respect to ground will be a cycloid as shown in figure. A

 v A

A

A A Motion of point A with respect to center of mass is pure rotational while center of mass itself is moving in a straight line. So for the analysis of rolling motion we deal translational motion seperately and rotational motion seperately and then we combine the result to analyses the over all motion. The velocity of any point A on the rigid body can be obtained as    VA  VCOM  VA COM  | VCOM |  V

 | V A.COM | r in the direction  to line OA   Thus, the velocity of point A is the vector sum of VCOM and VP.COM as shown in figure  VA O

r A

VCOM

Important points in combined Rotational + translation motion : 1.

Velocity of any point of the rigid body in combined R + T motion is the vector sum of v(velocity of centre of mass) and r for example A disc of radius r has linear velocity v and angular velocity  as shown in figure then find velocity of point A. B, C, D on the disc C

r B

D

v

 A We divide our problem in two parts (1) Pure Rotational + (2) Pure Translational about centre of mass. C

r

r

v

D

B r

+

v

v

r

r

A

v

Then combine the result of above both

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Page # 33

ROTATIONAL DYNAMICS r

C

D

v

2

(

r )2

v  r

v

B 2

v  (r )2

A v – r r

In combined rotational and translational motion angular velocity of any point of a rigid body with respect to other point in the rigid body is always same. For example : C 2v v 2v

v

B

D

v

v

D 2r

2.

(v = r)

r v=0

2v v

A

2v

A

2v Now for DA

DA =

2r

=

v r

2v

C

2R

For CA :

A

CA

2v v = = 2r r

For DB :

2v B

2r

D 2v

2v



3.

DB =

2r

vDB =

2v

2v 2v

v

=

r

Distance moved by the centre of mass of the rigid body in one full rotation is 2R.

v 

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ROTATIONAL DYNAMICS

This can be shown as under : In one rotation angular displacement  = 2 = t  2  s = v . T = ( R)    2R  

In forward slipping

s > 2R

and in backward slipping 4.

s < 2R

(as v > R) (as v < R)

The speed of a point on the circumference of the body at any instant t is 2R sin

Proof : vxp = v – v cos  vyp = v sin 

= v[1 – cos ]



 | v p |  v 2 sin2   v 2 (1 – cos ) 2

v= =

t 2

v

v = R

2 v 2 – 2 v 2 cos 

 v

P

2v(1 – cos )1/ 2

 ωt   t  = 2v sin    = 2 v sin  = 2R  sin    2  2   2 5.

The path of a point on circumference is a cycloid and the distance moved by this point in one full rotation is 8R. A3 A2

A4

A5

A1

In the figure, the dotted line is a cycloid and the distance A1 A2 ......A5 is 8R. This can be proved as under. According to point (3), speed of point A at any moment is,  t  vA = 2R sin   2

Distance moved by A in time dt is,

c

c ds = vA

 t  dt = 2R sin   dt 2

Therefore, total distance moved in one full rotation is,

v  A t=0

A



t=t

T  2 / 

 ds

S=

0

T  2 / 

or

S=

 0

 t  2 R sin  dt  2

On integration we get,

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Page # 35

ROTATIONAL DYNAMICS 6.

x and y coordinates of the bottommost point at any time t. At time t the bottommost point will rotate an angle  = t with respect to the centre of the disc C. The centre C will travel a distance s = vt. y In the figure, PQ = R sin  = R sin t CQ = R cos = R cost Coordinates of point P at time t are,

C P

x = OM – PQ = vt – R sin t and y = CM – CQ = R – R cos t  (x, y)  (vt – R sin t, R – R cos t) 11.1

R

C 

Q

M

O s=vt

Angular momentum of a rigid body in combined rotation and translation

COM  r0



 v0

O Let O be a fixed point in an inertial frame of reference. Angular momentum of the body about O is.     L  L cm  M( r 0  v 0 )

 The first term L cm represents the angular momentum of the body as seen from the centre of   mass frame. The second term M( r 0  v 0 ) equals the angular momentum of centre of mass

about point O.

Ex.39 A circular disc of mass m and radius R is set into motion on a horizontal floor with a v in clockwise direction R as shown in figure. Find the magnitude of the total angular momentum of the disc about bottommost point O of the disc.     ...(i) L  L cm  m( r0  v 0 )  v Here, L cm  I (perpendicular to paper inwards)  1 2  v  O   mR    2  R  r0 1  mvR 2   90º  and m( r0  v 0 )  mRv (perpendicular to paper inwards) v0 Since, both the terms of right hand side of Eq. (i) are in the linear speed v in the forward direction and an angular speed  

Sol.

same direction.  or

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Ans.

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Page # 36 11.2

ROTATIONAL DYNAMICS

Kinetic Energy of a Rolling Body If a body of mass M is rolling on a plane such that velocity of its centre of mass is V and its angular speed is , its kinetic energy is given by KE =

1 1 Mv 2  I  2 2 2

I is moment of inertia of body about axis passing through centre of mass. In case of rolling without slipping. KE =



1 1 M  2 R2 + I  2 [ v = R] 2 2

1 MR2  I  2 2





=

1 Ic  2 2

Ic is moment of inertia of the body about the axis passing through point of contact. Ex.40 A uniform rod of mass M and length a lies on a smooth horizontal plane. A particle of mass m moving at a speed v perpendicular to the length of the rod strikes it at a distance a/4 from the centre and stops after the collision. Find (a) the velocity of the centre of the rod and (b) the angular velocity of the rod about its centre just after the collision.



a/4

A

a

r0 A

v

Sol.

(a)

(b)

The situation is shown in figure. Consider the rod and the particle together as the system. As there is no external resultant force, the linear momentum of the system will remain constant. Also there is no resultant external torque on the system and so the angular momentum of the system about any line will remain constant. Suppose the velocity of the centre of the rod is V and the angular velocity about the centre is . (a) The linear momentum before the collision is mv and that after the collision is MV. Thus, mv = MV, or V =

m v M

(b) Let A be the centre of the rod when it is at rest. Let AB be the line perpendicular to the plane of the figure. Consider the angular momentum of "the rod plus the particle" system about AB. Initially the rod is at rest. The angular momentum of the particle about AB is L = mv(a/4) After the collision, the particle mass to rest. The angular momentum of the rod about A is

    L  L cm  M( r 0  V ) As

  r 0 || V ,

  r0  V  0   L  L cm

Thus,

Hence the angular momentum of the rod about AB is

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Page # 37

ROTATIONAL DYNAMICS

Ma 2  12

L  I 

Thus,

mva Ma 2   4 12

or,

=

3mv Ma

Ex.41 A uniform rod of length  lies on a smooth horizontal table A particle moving on the table has a mass m and a speed v before the collision and it sticks to the rod after the collision. The rod has a mass M then find out. (a) (b) Sol.

The moment of inertia of the system about the vertical axis passing through the centre of mass C after the collision. The velocity of the centre of mass C and the angular velocity of the system about the centre of mass after the collision. Figure shows the situation of system just before and just after collision. Initially the centre of mass of the rod is at point O. After collision when the particle sticks to the rod. Centre of mass is shifted from point O to C as shown in figure. Now the system is rotated about axis passing through C

v

A 1 

m /2

v'

C

O

M



M 2(m  M)

2 

O

m 2(M  m)

 /2

M Before collision After collision Now from linear momentum conservation mv = (M + m) v (a)



v' 

mv Mm

Let us assume that moment of inertia of the system about C is 1. Then I  I(rod)C  I(part ) C

I  I0  M 22  m 21 M 2 Mm2  2 mM2  2 M(M  4m) 2   I   2 2 12 12(m  M) 4(m  M) 4(m  M) From Angular momentum conservation about A Li = L f 0 + 0 = I  – (m + M) v 1  I = (m + M) v 1 Put the value of I, v, & 1 we get I

(b)

= 11.3 (A)

6mv (M  4m)

Acceleration of a point on the circumference of the body in R + T motion : Both  & v are constant : C  2r



r

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Page # 38 (B)

ROTATIONAL DYNAMICS

When  is constant and v is variable. 



v1, a

v2= v1+at

t=0 t=t So acceleration of different point on the body is given by following figure.

a 2

a

2

 2r

+

 r  2r a 2  r

=

 2r

(C)

a

a (Combined R + T)

(Rotational)

(Translational)

 2r

 r

 2r

When  is variable and v is constant :    i  t  i ,

v

v

t=0 t=t So acceleration of different point on the body is given by following way R



 2r

R

 2R

 2r  2r

2



2

 R

 R

 2r

R

R

(D)

a=0

 2R

When both  & v are variable :  i,   f   i  t v2=v1+at

v1, a

 time t =0 time t =t Now the net acceleration of different points on the rigid body is given by following way. R a R R R  2r  2R  2r  2R = a + a 2  2R 2  r  R R (Translational)

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R

R

(Rotational)

R

a

(combined Rotational + Translational)

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Page # 39

ROTATIONAL DYNAMICS

Ex.42 A force F acts at the centre of a thin spherical shell of mass m and radius R. Find the acceleration of the shell if the surface is smooth. N Sol.  Force F, mg & N passes through centre so net = 0, i.e., body is in rotational equilibrium R  F But F net  F so body moves with constant acceleration a = m

F

mg

Ex.43 In a previous problem if force F applied at a distance x above the centre then find out linear and angular acceleation. Sol. This force F translate the body linearly as well as rotate it. So, Net toruqe about O it 0 = Fx N From rotational motion 0 = I  F  Fx x R   3Fx a I 2MR2   O 2 MR 2 mg 3 smooth From linear motion of sphere F = ma 

a=

F m

Ex.44 A rigid body of mass m and radius r starts coming down an inclined plane of inclination . Then find out the acceleration of centre of mass if friciton is absent. Sol. Friction is absent so body is moving down the incline with out rolling so acceleration of centre of mass is g sin   in gs 

12.

UNIFORM PURE ROLLING Pure rolling means no relative motion (or no slipping at point of contact between two bodies.) For example, consider a disc of radius R moving with linear velocity v and angular velocity  on a horizontal ground. The disc is said to be moving without slipping if velocities of points P and Q (shown in figure b) are equal, i.e.,

v

COM

 (a)

 R

P

v

Q (b)

vp = vQ or v – R = 0 or v = R If vp > vQ or v > R, the motion is said to be forward slipping and if vp < vQ < R, the motion is said to be backward slipping. Now, suppose an external force is applied to the rigid body, the motion will no longer remain uniform. The condition of pure rolling on a stationary ground is, a = R Thus, v = R, a = R is the condition of pure rolling on a stationary ground. Sometime it is simply said rolling. Note : We can represent the moment of inertia of a different rigid body in a following way. I = CMR2 1 value of C = 1 for circular ring (R), C = for circular disc (D) and solid cylinder (S.C.) 2 2 2 C= for Hollow sphere (H.S) , C= for solid sphere (S.S) 3 5 IIT-JEE|AIEEE CBSE|SAT|NTSE OLYMPIADS

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Page # 40

ROTATIONAL DYNAMICS

Ex.45 A rigid body I = CMR2 is set into a motion on a rough horizontal surface with a linear speed v0 in the forward direction at time t = 0 as shown in figure. After what time slipping finally stop and pure rolling starts. Find the linear speed of the body after it starts pure rolling on the surface. v0

O R

at t = 0

Sol.

According to the given condition in problem the point P in the body move with speed v0 while the point Q on the ground is at rest. So the friciton acts on the body is in backward direction which gives the resultant torque on the body and increase the angualr speed  as shown in figure. 1 v0

O f

P

(kinetic)

Q at t = 0

O  1R f (kinetic)

v0

v1

1

v1   1R O v1

Q at t = t1

v1 R v1

R friction static

v1

Q at t = t

As shown in above figure initially v > R so forward slipping takes place. After introducing the friciton speed decreases and  increases and at time t = t the relation v = r  is satisfied. Therefore pure rolling starts. Initially the friciton is kinetic untill the motion is in slipping condition. Afterwards at v = r fricition is static. We divide the above problem in two parts. (1)

(2)

Translational Motion : Linear acceleration a = – g So after time t, v = v0 – gt ...(1) Rotational Motion : From net = I  Only friction force is responsible for providing torque. So torque about O is f. R = I  mgR = CmR2 ...(2)  is angular acceleration of the body μg

from eq. (2) from

=

CR

f = i +  t  = t   =

g .t CR

=

v at pure rolling condition. R

So,

v=

μgt C from eq. (1) & (3)

...(3)

v 0C

μgt

 v0 – gt =

C

 t=

μg(1  C)

...(4)

Equation (4) gives the time after the pure rolling starts.

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Page # 41

ROTATIONAL DYNAMICS Put the value from eq. (4) to eq. (1) v = v0 –

v0 v 0C  v= 1 C (1  C)

...(5)

Equation (5) gives the linear speed at pure rolling situation. Alternate solution : Net torque on the body about the bottom most point A is zero. Therefore angular momentum of the body will remain conserved about the bottom most point Net torque about A A = 0 v  from Angular momentum conservation Li = Lf  R mv0R = I + mvR v0

v

v mv0R = CmR2 + mvR R v0 = Cv + v



v=

A f

v0 1 C

A f

Ex.46 In the previous problem take rigid body a solid cylinder then find out the work done by friciton from time t = 0 to t = t (at v = r) Sol.

Let us suppose that in between time t = 0 to t = t cylinder displaced s. t=t t=0

v0

v  R

S Translational work done by friciton + Rotational work Done by friciton Now calculate each type of work done one by one (A)

Translational work done by friciton : for solid cylinder c =

from eq. (5) v 

v0 1

1 2

1 2



2 v0 3 2

2  2   v 0   ( v 0 ) – 2gs 3 

from eq. v 2f  ui2  2as

s=

5 v 20 18 g

Translation W.D by friciton = – f.s ( w.D) f T  – mg.

(B)

5 v 20 18 g

= –

5mv 20 18

Rotational W.D. by friciton : We known that =

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 I

=I ...(a)

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Page # 42

ROTATIONAL DYNAMICS

from  f 2 –  i2  2   2

2  2v 0  .     3R  I

Put I =

mR2 2

 =

v 02m 9

Rotation W.D by friciton W = . Wf R =

(C)

v 20 m 9

So total W.D. by friciton W = Wf

W= –

T

+ Wf

R

= –

v 2m 5 mv 20  0 18 9

mv 20 6

Alternative Method : from work – Energy Theorem work done by friciton = change in kinetic energy (W.D)f = K = kf – ki Now kf =

1 1 mv 2f + I 2 2 2

1  2v 0   kf = m  2  3 

2

1 mR 2  2v 0    + 2 2  3R 

kf =

mv 20 3

ki =

1 mv 20 2

2

2v 0    v f   3  

mv 20 1 mv 20 ( w.D)f  – – mv 20  6 3 2 To calculate work done mostly prefer alternative method. So., (w.D)f =

Ex.47 A solid sphere of radius r is gently placed on a rough horizontal ground with an initial angular speed 0 and no linear velocity. If the coefficient of friciton is , find the linear velocity v and angular velocity  at the end of slipping.

0

Sol.

m be the mass of the sphere. Since, it is a case of backward slipping, force of friction is in forward direction. Limiting friciton will act in this case. Net torque on the sphere about the bottommost IIT-JEE|AIEEE CBSE|SAT|NTSE OLYMPIADS

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Page # 43

ROTATIONAL DYNAMICS point is zero. Therefore, angular momentum of the sphere will remain conserved about the bottommost point. Li = L f  I0 = I + mrv

12.1

or

2 2 2 mr  0 = mr 2   mr (r ) 5 5



=

v

0

fmax

0

2 2  0 and v = r = r 0 7 7

Pure rolling when force F act on a body : Suppose a force F is applied at a distance x above the centre of a rigid body of radius R, mass M and moment of inertia CMR2 about an axis passing through the centre of mass. Now, the applied force F can produces by itself (i) a linear acceleration a and (ii) an angular acceleration  If a = R, then there is no need of friction and force of friction f = 0, If a < R, then to support the linear momentum the force of friciton f will act in forward direction, Similarly, if a > R, then no support the angular motion the force of friciton will act in backward direction. So, in this case force of friction will be either backward, forward or even zero also. It all depends on M, I and R. For calculation you choose any direction of friction. Let we assume it in forward direction. Let, a = linear acceleration,   = angualr acceleration F from linear motion x a F + f = Ma ...(1) C from rotational motion. Fx – f R = I  f Fx – f R

Fx – f R = CMaR from eq. (1) and (2) F(x+r) = MaR (C + 1) a=

F(R  x ) MR(C  1)

a R ...(2)

= CMR2.

...(3)

Put the value from eq. (3) to eq. (1) f=

F(x – RC) R(C  1)

f should be   s mg for pure rolling Ex.48 Consider the arrangment shown in figure. The string is wrapped around a uniform cylinder which rolls without slipping. The other end of the string is passed over a massless, frictionless pulley to a falling weight. Determine the acceleration of the falling mass m in terms of only the mass of the cylinder M, the mass m and g. M

m

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Page # 44 Sol.

ROTATIONAL DYNAMICS

Let T be the tension the string and f the force of (static) friction, between the cylinder and the surface a1 = acceleration of centre of mas of cylinder towards right a2 = downward acceleration of block m  = angular acceleration of cylinder (clockwise) Equations of motion are : For block mg – T = ma2 ...(i) For cylinder, T + f = Ma1 ...(ii) 

( T – f )R 1 MR 2 2

...(iii)

The string attaches the mass m to the highest point of the cylinder, hence vm = vCOM + R Differentiating, we get a2 = a1 + R ...(iv) We also have (for rolling without slipping) a1 = R ...(v) Solving these equations, we get

a2 

8mg 3M  8m

Note : Work done by friction in pure rolling on a stationary ground is zero as the point of application of the force is at rest. Therefore, machanical energy can be conserved if all other dissipative forces are ignored. 12.2 Pure Rolling on an Inclined Plane: A rigid body of radius R, and mass m is released at rest from height h on the incline whose inclination with horizontal is  and assume that f  friciton is sufficient for pure rolling then. a = R and v = R a From figure mg sin  – f = ma ...(1)  in {Fnet = ma} gs m 2 a cmR . f.R = ...(2) R {Fnet = I} from eq. (1) & (2) g sin 1 c So body which have low value of C have greater acceleration. value of C = 1 for circular ring (R) 1 C= for circular disc (D) and solid cylinder (S.C.) 2 2 C= for Hollow sphere (H.S) 3 a=

2 for solid sphere (S.S) 5 So, descending order of a aS.S > aD = aS.C > aH.S. > aR and order of time of descend is ts.s < tD = ts.c < tH.S < tR

C=

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Page # 45

ROTATIONAL DYNAMICS Kinetic Eneregy : Work done by friction in pure rolling is zero. Therefore, Increase in kinetic energy = change in potential energy  K.E. = mgh i.e., kinetic energy is constant for all rigid body rolling down the incline. Requirement of Friction : From eq. f = Cma f

mg sin  1  1    C

...(2)

...(3)

from eq. (3) as the value of C increase requirement of friciton is increases. Ex.49 A cylinder of mass M is suspended through two strings wrapped around it as shown in figure. Find the tension in the string and the speed of the cylinder as it falls through a distance h. Sol. The portion of the strings between ceiling and cylinder are at rest. Hence the points of the cylinder where the strings leave it are at rest also. The cylinder is thus rolling without slipping on the strings. Suppose the centre of cylinder falls with an acceleration a. The angular acceleration of cylinder about its axis given by =

a R

...(i)

as the cylinder does not slip over the strings. The equation of motion for the centre of mass of cylinder is Mg – 2T = Ma and for the motion about the centre of mass it is

 MR2  MR2  , where I = 2T.R =  2  2  Ma MR2 a  2T= 2 2 R From (i) and (ii) on adding 2TR=

T

T

mg ...(ii)

Ma 3a  Ma ; g 2 2 2g a= 3 M 2g Mg  2T= .  T= 2 3 6 As the centre of cylinder starts moving from rest, the velocity after it has fallen a height h is given by

Mg =

 2g  v2 = 2  h or v = 3

4gh 3

Ex.50 A thin massless thread is wound on a reel of mass 3kg and moment of inertia 0.6 kgm2. The hub radius is R = 10 cm and peripheral radius is 2R = 20 cm. The reel is placed on a rough table and the friction is enough to prevent slipping. Find the acceleration of the centre of reel and of hanging mass of 1 kg.

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Page # 46

ROTATIONAL DYNAMICS

2R R

A

Sol.

Let,

a1 = acceleration of centre of mass of reel a2 = acceleration of 1 kg block  = angular acceleration of reel (clockwise) T = tension in the string and f = force of friction Free body diagram of reel is as shown below : (only horizontal forces are shown). Equations of motion are : T – f = 3a1 ...(i)



a1 T f

f T  f (2R) – T.R 0.2f – 0.1T   = – ...(ii) 3 6 I I 0.6 Free body diagram of mass is, Equation of motion is, 10 – T = a2 ...(iii) For no slipping condition, a1 = 2R or a1 = 0.2 ...(iv) and a2 = a1 – R or a2 = a1 – 0.1 ...(v) 

T a2

Solving the above five equations, we get 10N a1 = 0.27 m/s2 and a2 = 0.135 m/s2 Ex.51 Determine the maximum horizontal force F that may be applied to the plank of mass m for which the solid sphere does not slip as it begins to roll on the plank. The sphere has a mass M and radius R. The coefficient of static and kinetic friction between the sphere and the plank are s and k respectively. M R

F

m

Sol.

The free body diagrams of the sphere and the plank are as shown below : Writing equations of motion : For sphere : Linear acceleration  sMg  sg a1 = M

(  sMg)R 5  s g  2 2 R MR 2 5

a1

...(i)

 s Mg  s Mg

Angular acceleration 



a2 F

..(ii)

For plank : Linear acceleration

M R

B A

m

a2

a1  R F

F –  sMg ..(iii) m For no slipping acceleration of point B and A is same, a2 

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Page # 47

ROTATIONAL DYNAMICS so : a2 = a1 + R Solving the above four equation, we get 7   F   s g M  m  2  7   Thus, maximum value of F can be  s g M  m 2

Ex.52 Find out the maximum height attained by the solid sphere on a friciton less track as shown in figure.

0  v0

R

Sol.

v0 R

Let us assume that sphere attain a maximum height H on the track. 0

Final Position v=0

0 

v0 R

H

v0

R

Initial Position

As the sphere move upward speed is decreased due to gravity but there is no force to change the 0 (friction less track). So from energy conservation 1 1 1 mv 20  I 20 = mg Hmax + I 20 2 2 2 Hmax =

13.

v 20 2g

TOPPLING You might have seen in your practical life that if a force F is applied to a block A of smaller width it is more likely to topple down, before sliding while if the same force F is applied to an another block B of broader base, chances of its sliding are more compared to its toppling. Have you ever throught why it happens so. To understand it better let us take an example. F F B A Suppose a force F is applied at a height b above the base AE of the block. Further, suppose the friction f is sufficient to prevent sliding. In this case, if the normal reaction N also passes through C, then despite the fact that the block is in translational equilibium (F = f and N = mg), an unbalanced torque (due to the couple of forces F and f) is there. This torque has a tendency to topple the block about point E. To cancel the effect of this unbalanced torque the normal reaction N is shifted towards right a distance 'a' such that, net anticlockwise torque is equal to the net clockwise torque or N D B F C b f

E

A W=mg

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Page # 48

ROTATIONAL DYNAMICS

or

a=

Fb mg N

N D

B C f

A

D

B

F

C

b

a

f

E mg

A

(a)

F b

E mg (b)

Now, as F or b (or both) are increased, distance a also increases. But it can not go beyond the right edge of the block. So, in extreme case (beyond which the block will topple down), the normal reaction passes through E as shown in figure. Now, if F or b are further increased, the block will topple down. This is why the block having the broader base has less chances of toppling in comparison to a block of smaller base. Because the block of larger base has more margin for the normal reaction to shift. Why the rolling is so easy on the ground.

N F

C f mg Because in this case the normal reaction has zero margin to shift. so even if the body is in translational equilibrium (F = f, N = mg) an unbalanced torque is left behind and the body starts rolling clockwise. As soon as the body starts rolling the force of friction is so adjusted (both in magnitude and direction) that either the pure rolling starts (if friciton is sufficient enough) or the body starts sliding. Let us take few examples related to toppling. Ex.53 A uniform block of height h and width a is placed on a rough inclined plane and the inclination of the plane to the horizontal is gradually increased. If  is the coefficient of friction then under condition the block will (A) slide before toppling : The block will slide when mg sin  > f  mg sin  >  mg cos   tan  >  i.e., block is at rest when tan . ...(1) (B)

Now suppose the friction f is sufficient to prevent sliding. Then we assume that N is shifted towards downward a distance x to prevent toppling Therefore. torque about O is zero. h =Nx 2 h mg sin . = mg cos .x 2 tan .h x= 2 Maximum value of x is a/2  f.

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N

h f xO a

 in s g m 

os gc m



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Page # 49

ROTATIONAL DYNAMICS

so to prevent toppling x  

a 2

tan .h  a/2 2

 tan  

a h

...(2)

So, the block topple before sliding from (1) & (2) s >

14.

a h

INSTANTANEOUS AXIS OF ROTATION The combined effects of translation of the centre of mass and rotation about an axis through the centre of mass are equivalent to a pure rotation with the same angular speed about an axis passing through a point of zero velocity. Such an axis is called the instantaneous axis of rotation. (IAOR). This axis is always perpendicular to the plane used to represent the motion and the intersection of the axis with this plane defines the location of instantaneous centre of zero velocity (IC).

v   IC For example consider a wheel which rolls without slipping. In this case the point of contact with the ground has zero velocity. Hence, this point represents the IC for the wheel. If it is imagined that the wheel is momentarily pinned at this point, the velocity of any point on the wheel can be found using v = r. Here r is the distance of the point from IC. Similarly, the kinetic energy of the body can be assumed to be pure rotational about IAOR or, P v P vP r  r vP  r   v r  IC



1 I IAOR 2 2 Rotation + Translation  Pure rotation about IAOR passing through IC K

KE = 14.1

1 1 mv 2COM  ICOM 2  2 2

KE 

1 I IAOR 2 2

Location of the IC If the location of the IC is unknown, it may be determined by using the fact that the relative position vector extending from the IC to a point is always perpendicular to the velocity of the point. Following three possibilities exist. (i) Given the velocity of a point (normally the centre of mass) on the body and the angular velocity of the body  If v and  are known, the IC is located along the line drawn perpendicular to v at P, such that v the distance from P to IC is, r  . Note that IC lie on that side of P which causes rotation    about the IC, which is consistent with the direction of motion caused by  and v . IIT-JEE|AIEEE CBSE|SAT|NTSE OLYMPIADS

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Page # 50

ROTATIONAL DYNAMICS



P

v

r

IC

Ex.54 A rotating disc moves in the positive direction of the x-axis. Find the equation y(x) describing the position of the instantaneous axis of rotation if at the initial moment the centre c of the disc was located at the point O after which it moved with constant velocity v while the disc started rotating counter clockwise with a constant angular acceleration . The initial angular velocity is equal to zero. y

O

Sol.

x x   t  and v v The position of IAOR will be at a distance

y

t

y

v 

or

y

x

c v

IC

y c v 

O

v x v

x

x

v2 v2  constant or xy  x  This is the desired x-y equation. This equation represents a rectangular hyperbola. or

y

(ii) Given the lines of action of two non-parallel velocities   Consider the body shown in figure where the line of action of the velocities v A and v B are known. Draw perpendiculars at A and B to these lines of action. The point of intersection of these perpendiculars as shown locates the IC at the instant considered. A

 vB

 vA IC B

(iii) Given the magnitude and direction of two parallel velocities When the velocities of points A and B are parallel and have known magnitudes vA and vB then the location of the IC is determined by proportional triangles as shown in figure.  A In both the cases, and In fig. (a)

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v rA,IC  A  v rB,IC  B  rA, I C + rB, I C = d

vA

d

IC

IC  vA

A

d  vB

B

B

(a)

(b)

 vB

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Page # 51

ROTATIONAL DYNAMICS

and in fig (b) rB, I C – rA, I C = d As a special case, if the body is translating, vA = vB and the IC would be located at infinity, in which case  = 0. Ex.55 A uniform thin rod of mass m and length l is standing on a smooth horizontal surface. A slight disturbance causes the lower end to slip on the smooth surface and the rod starts falling. Find the velocity of centre of mass of the rod at the instant when it makes an angle  with horizontal. Sol. As the floor is smooth, mechanical energy of the rod will remain conserved. Further, no horizontal force acts on the rod, hence the centre of mass moves vertically downwards in a straight line. Thus velocities of COM and the lower end B are in the direction shown in figure.   The location of IC at this instant can be found by drawing perpendiculars to v C and v B at respective points. Now, the rod may be assumed to be in pure rotational motion about IAOR passing through IC with angular speed . A

COM IC

 vC

h

l (1  sin ) 2



l sin 2 

 vB

B

Applying conservation of mechanical energy. Decrease in gravitational potential energy of the rod = increase in rotational kinetic energy about IAOR mgh 



1 I IAOR 2 2

 2 1  ml 2 ml 2 l 2 or mg 2 (1  sin )  2  12  4 cos    

Solving this equation, we get 

12g(1  sin ) l (1  3 cos 2 )

 l  | v C |   cos   2 

Now,



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3gl (1  sin ) cos 2  (1  3 cos 2 )

Ans.

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Page # 52

ROTATIONAL DYNAMICS

(Objective Problems)

Exercise - I (A) MOMENT OF INERTIA

1. The moment of inertia of a uniform semicircular wire of mass M and radius r about a line perpendicular to the plane of the wire through the centre is (A) Mr2

(C)

1 2 Mr 4

(B)

1 2 Mr 2

(D)

2 2 Mr 5

2. Let IA and IB be moments of inertia of a body about two axes A and B respectively. The axis A passes through the centre of mass of the body but B does not. (A) IA < IB (B) If IA < IB, the axes are parallel. (C) If the axes are parallel, IA < IB (D) If the axes are not parallel, IA  IB 3. Three bodies have equal masses m. Body A is solid cylinder of radius R, body B is a square lamina of side R, and body C is a solid sphere of radius R. Which body has the smallest moment of inertia about an axis passing through their centre of mass and perpendicular to the plane (in case of lamina) (A) A (B) B (C) C (D) A and C both 4. For the same total mass which of the following will have the largest moment of inertia about an axis passing through its centre of mass and perpendicular to the plane of the body (A) a disc of radius a (B) a ring of radius a (C) a square lamina of side 2a (D) four rods forming a square of side 2a 5. Two rods of equal mass m and length l lie along the x axis and y axis with their centres origin. What is the moment of inertia of both about the line x = y : ml 2 ml 2 (B) (A) 3 4 2 ml ml 2 (C) (D) 12 6 6. Moment of inertia of a rectangular plate about an axis passing through P and perpendicular to the plate is I. Then moment of PQR about an axis perpendicular to the plane of the plate :

(A) about P = I/2 (C) about P > I/2

(B) about R = I/2 (D) about R > I/2

7. A thin uniform rod of mass M and length L has its moment of inertia I1 about its perpendicular bisector. The rod is bend in the form of a semicircular arc. Now its moment of inertia through the centre of the semi circular arc and perpendicular to its plane is I2. The ratio of I1 : I2 will be _________________ (A) < 1 (B) > 1 (C) = 1 (D) can’t be said 8. Moment of inertia of a thin semicircular disc (mass = M & radius = R) about an axis through point O and perpendicular to plane of disc, is given by : O R

1 1 2 MR 2 (B) MR 4 2 1 2 (C) MR (D) MR2 8 9. A rigid body can be hinged about any point on the x-axis. When it is hinged such that the hinge is at x, the moment of inertia is given by I = 2x2 – 12x + 27 The x-coordinate of centre of mass is (A) x = 2 (B) x = 0 (C) x = 1 (D) x = 3

(A)

10. A square plate of mass M and edge L is shown in figure. The moment of inertia of the plate about the axis in the plane of plate passing through one of its vertex making an angle 15° from horizontal is. axis 15° L

L

ML2 (A) 12

(C)

7 ML2 12

(B)

11ML2 24

(D) none

Q

P

R

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11. Consider the following statements Assertion (A) : The moment of inertia of a rigid body reduces to its minimum value as compared to any other parallel axis when the axis of rotation 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No:0744-2439051,0744-2439052,0744-2439053,www.motioniitjee.com, [email protected]

Page # 53

ROTATIONAL DYNAMICS passes through its centre of mass. Reason (R) : The weight of a rigid body always acts through its centre of mass in uniform gravitational field. Of these statements : (A) both A and R are true and R is the correct explanation of A (B) both A and R are true but R is not a correct explanation of A (C) A is true but R is false (D) A is false but R is true Question No. 12 to 14 (3 questions) The figure shows an isosceles triangular plate of mass M and base L. The angle at the apex is 90°. The apex lies at the origin and the base is parallel to X - axis. Y  M X 12. The moment of inertia of the plate about the z-axis is

ML2 ML2 (B) 12 24 ML2 (C) (D) none of these 6 13. The moment of inertia of the plate about the x-axis is (A)

(A)

ML2 8

ML2 (C) 24

(B)

ML2 32

ML2 (D) 6

14. The moment of inertia of the plate about its base parallel to the x-axis is ML2 ML2 (B) 18 36 ML2 (C) (D) none of these 24 15. The moment of inertia of the plate about the y-axis is

17. One end of a uniform rod of mass m and length I is clamped. The rod lies on a smooth horizontal surface and rotates on it about the clamped end at a uniform angular velocity . The force exerted by the clamp on the rod has a horizontal component (A) m2 l (B) zero 1 2 (C) mg (D) m  2 18. A rod of length 'L' is hinged from one end. It is brought to a horizontal position and released. The angular velocity of the rod when it is in vertical position is (A)

2g L

(B)

3g L

(C)

g 2L

(D)

g L

(B) TORQUE AND PURE ROTATIONAL MOTION 19. A horizontal force F = mg/3 is applied on the upper surface of a uniform cube of mass ‘m’ and side ‘a’ which is resting on a rough horizontal surface having s = 1/2. The distance between lines of action of ‘mg’ and normal reaction ‘N’ is : (A) a/2 (B) a/3 (C) a/4 (D) None 20. A man can move on a horizontal plank supported symmetrically as shown. The variation of normal reaction on support A with distance x of the man from the end of the plank is best represented by : x=0 A B

(A)

ML2 6 ML2 (C) 24

(A)

(B)

ML2 8

1m

N

1m

N

(A)

(B) x

x

N

(D) none of these

SECTION (D) ; FIXED AXIS 16. A body is rotating uniformly about a vertical axis fixed in an inertial frame. The resultant force on a particle of the body not on the axis is (A) vertical (B) horizontal and skew with the axis (C) horizontal and intersecting the axis (D) none of these

4m

N

(C)

(D) x

x

21. A weightless rod is acted on by upward parallel forces of 2N and 4N ends A and B respectively. The total length of the rod AB = 3m. To keep the rod in equilibrium a force of 6N should act in the following manner :

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ROTATIONAL DYNAMICS

(A) Downwards at any point between A and B. (B) Downwards at mid point of AB. (C) Downwards at a point C such that AC = 1m. (D) Downwards at a point D such that BD = 1m. 22. A right triangular plate ABC of mass m is free to rotate in the vertical plane about a fixed horizontal axis through A. It is supported by a string such that the side AB is horizontal. The reaction at the support A is : l

A

B

l

(A) left half (B) right half (C) both applies equal pressure (D) the answer depend upon coefficient of friction 26. Consider the following statements Assertion (A) : A cyclist always bends inwards while negotiating a curve Reason (R) : By bending he lowers his centre of gravity Of these statements, (A) both A and R are true and R is the correct explanation of A (B) both A and R are true but R is not the correct explanation of A (C) A is true but R is false (D) A is false but R is true 27. A solid cone hangs from a frictionless pivot

C

mg 2 mg (B) 3 3 mg (C) (D) mg 2 23. In an experiment with a beam balance on unknown mass m is balanced by two known mass m is balanced by two known masses of 16 kg and 4 kg as shown in figure.

(A)

l1

l1

l2

m

16kg

at the origin O, as shown. If i , j and k are unit vectors, and a, b, and c are positive constants, which of the following forces F applied to the rim of the cone at a point P results in a torque  on the cone with a negative component Z ? z k o

i x

l2

m

y j

4kg

b

The value of the unknown mass m is (A) 10 kg (B) 6 kg (C) 8 kg (D) 12 kg

(A) F = a k , P is (0, b, –c) (B) F = –a k , P is (0, –b, –c)

24. A uniform cube of side ‘b’ and mass M rest on a rough horizontal table. A horizontal force F is applied normal to one of the face at a point, at a height 3b/4 above the base. What should be the coefficient of friction () between cube and table so that is will tip about an edge before it starts slipping?

(C) F = a j , P is (–b, 0, –c) (D) None 28. A rod is hinged at its centre and rotated by applying a constant torque starting from rest. The power developed by the external torque as a function of time is : Pext

F b

Pext

(A)

3b/4

(B) time

2 1 (A)   (B)   3 3 3 (C)   (D) none 2 25. A homogeneous cubical brick lies motionless on a rough inclined surface. The half of the brick which applies greater pressure on the plane is :

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c

time

Pext

Pext

(C)

(D) time

time

29. A pulley is hinged at the centre and a massless thread is wrapped around it. The thread is pulled with a constant force F starting from rest. As the time increases, F

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Page # 55

ROTATIONAL DYNAMICS (A) its angular velocity increases, but force on hinge remains constant (B) its angular velocity remains same, but force on hinge increases (C) its angular velocity increases and force on hinge increases (D) its angular velocity remains same and force on hinge is constant. 30. The angular momentum of a flywheel having a moment of inertia of 0.4 kg m2 decreases from 30 to 20 kg m2/s in a period of 2 second. The average torque acting on the flywheel during this period is : (A) 10 N.m (B) 2.5 N.m (C) 5 N.m (D) 1.5 N.m 31. A rod hinged at one end is released from the horizontal position as shown in the figure. When it becomes vertical its lower half separates without exerting any reaction at the breaking point. Then the maximum angle ‘’ made by the hinged upper half with the vertical is : C B A

B

(A) 30°

(B) 45°

B

C (C) 60° (D) 90°

32. A block of mass m is attached to a pulley disc of equal mass m, radius r by means of a slack string as shown. The pulley is hinged about its centre on a horizontal table and the block is projected with an initial velocity of 5 m/s. Its velocity when the string becomes taut will be

(C) ANGULAR MOMENTUM 34. A particle moves with a constant velocity parallel to the X-axis. Its angular momentum with respect to the origin. (A) is zero (B) remains constant (C) goes on increasing (D) goes on decreasing. 35. A thin circular ring of mass 'M' and radius 'R' is rotating about its axis with a constant angular velocity . Two objects each of mass m, are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velcoity. (A)

M (M  m)

(B)

M (M  2m)

(C)

M (M – 2m)

(D)

(M  3m) M

36. A person sitting firmly over a rotating stool has his arms streatched. If he folds his arms, his angular momentum about the axis of rotation (A) increases (B) decreases (C) remains unchanged (D) doubles. 37. A small bead of mass m moving with velocity v gets threaded on a stationary semicircular ring of mass m and radius R kept on a horizontal table. The ring can freely rotate about its centre. The bead comes to rest relative to the ring. What will be the final angular velocity of the system?

R

O

(A) 3 m/s (C) 5/3 m/s

(B) 2.5 m/s (D) 10/3 m/s

33. A particle starts from the point (0m, 8m) and moves with uniform velocity of 3 i m/s. After 5 seconds, the angular velocity of the particle about the origin will be : y 3m/s

(A)

8 rad / s 289

x

24 rad / s (C) 289

(B)

38. A man, sitting firmly over a rotating stool has his arms streched. If he folds his arms, the work done by the man is (A) zero (B) positive (C) negative (D) may be positive or negative. 39. A particle of mass 2 kg located at the position

8m O

(A) v/R (C) v/2R

v m (B) 2v/R (D) 3v/R

3 rad / s 8

8 rad / s (D) 17

( i  j ) m has a velocity 2(  i – j  k ) m/s. Its angular momentum about z-axis in kg-m2 /s is : (A) zero (B) +8 (C) 12 (D) – 8 40. A thin uniform straight rod of mass 2 kg and length 1 m is free to rotate about its upper end when at rest. It receives an impulsive blow of 10

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ROTATIONAL DYNAMICS

Ns at its lowest point, normal to its length as shown in figure. The kinetic energy of rod just after impact is

10 NS (A) 75 J (C) 200 J

(B) 100 J (D) none

Question No. 44 & 45 (2 questions) A uniform rod is fixed to a rotating turntable so that its lower end is on the axis of the turntable and it makes an angle of 20° to the vertical. (The rod is thus rotating with uniform angular velocity about a vertical axis passing through one end.) If the turntable is rotating clockwise as seen from above. 20°

41. A ball of mass m moving with velocity v, collide with the wall elastically as shown in the figure. After impact the change in angular momentum about P is : P

d 

(A) 2 mvd (C) 2 mvd sin

(B) 2 mvd cos (D) zero

42. A uniform rod of mass M is hinged at its upper end. A particle of mass m moving horizontally strikes the rod at its mid point elastically. If the particle comes to rest after collision find the value of M/m = ?

v m M

(A) 3/4 (C) 2/3

(B) 4/3 (D) none

43. A child with mass m is standing at the edge of a disc with moment of inertia I, radius R, and initial angular velocity . See figure given below. The child jumps off the edge of the disc with tangential velocity v with respect to the ground. The new angular velocity of the disc is

44. What is the direction of the rod’s angular momentum vector (calculated about its lower end) (A) vertically downwards (B) down at 20° to the horizontal (C) up at 20° to the horizontal (D) vertically upwards 45. Is there a torque acting on it, and if so in what direction? (A) yes, vertically (B) yes, horizontally (C) yes at 20° to the horizontal (D) no 46. One ice skater of mass m moves with speed 2v to the right, while another of the same mass m moves with speed v toward the left, as shown in figure I. Their paths are separated by a distance b. At t = 0, when they are both at x = 0, they grasp a pole of length b and negligible mass. For t > 0, consider the system as a rigid body of two masses m separated by distance b, as shown in figure II. Which of the following is the correct formula for the motion after t = 0 of the skater initially at y = b/2 ? y

y

m 2v b/2

v

b

x

x

t=0



(t FC = FD (D) none Sol.

Q.15 The maximum separation between their centres after their first collision (A) 4R (B) 6R (C) 8R (D) 12R Sol.

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GRAVITATION

Page # 32

Q.17 A satellite of the earth is revolving in circular orbit with a uniform velocity V. If the gravitational force suddenly disappears, the statellite will (A) continue to move with the same velocity in the same orbit (B) move tangentially to the original orbit with velocity V (C) fall down with increasing velocity (D) come to a stop somewhere in its original orbit Sol.

Q.20

Select the correct choice(s) :

(A) The gravitational field inside a spherical cavity, within a spherical planet must be non zero and uniform. (B) When a body is projected horizontally at an appreciable large height above the earth, with a velocity less than for a circular orbit, it will fall to the earth along a parabolic path (C) A body of zero total mechanical energy placed in a gravitational field will escape the field (D) Earth’s satellite must be in equatorial plane.

Q.18 A satellite revolves in the geostationary orbit but in a direction east to west. The time interval between its successive passing about a point on the

Sol.

equator is (A) 48 hrs

(B) 24 hrs

(C) 12 hrs

(D) never

Sol.

Q.21 A satellite of mass m, initially at rest on the earth, is launched into a circular orbit at a height

Q.19

Tw o po int ma sses o f ma ss 4m and m

respectively separated by d distance are revolving under mutual force of attraction. Ratio of their kinetic energies will be (A) 1 : 4

(B) 1 : 5

(C) 1 : 1

equal to the radius of the earth. The minimum energy required is (A)

3 mgR 4

(B)

1 mgR 2

Sol.

(D) 1 : 2

Sol.

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(C)

1 3 mgR (D) mgR 4 4

GRAVITATION

Page # 33 Sol.

Q.22 The figure shows the

correct state ment about the

A Energy

variation of energy with the orbit radius of a body in circular planetary motion. Find the

C B

curves A, B and C (A) A shows the kinetic energy, B the total energy and C the potential energy of the system (B) C shows the total energy, B the kinetic energy and A the potential energy of the system (C) C and A are kinetic and potential energies respectively and B is the total energy of the system (D) A and B are kinetic and potential energies and C is the total energy of the system Sol.

Q.23 When a satellite moves around the earth in a certain orbit, the quantity which remains constant is (A) angular velocity (B) kinetic energy (C) aerial velocity Sol.

(D) potential energy

Q.24 A satellite of mass 5M orbits the earth in a circular orbit. At one point in its orbit, the satellite explodes into two pieces, one of mass M and the other of mass 4M. After the explosion the mass M ends up travelling in the same circular orbit, but in opposite direction. After explosion the mass 4M is in (A) bound orbit (B) unbound orbit (C) partially bound orbit (D) data is insufficient to determine the nature of the orbit

Q.25 A satellite can be in a geostationary orbit around earth at a distance r from the centre. If the angular velocity of earth about its axis doubles, a satellite can now be in a geostationary orbit around earth if its distance from the center is r r r r (B) (C) (D) (A) 1/ 3 (4) (2)1/ 3 2 2 2 Sol.

Q.26 A planet of mass m is in an elliptical orbit about the sun (m t2

X

(C) t1 < t2

(D) nothing can be concluded

Y

(A) Z is total energy, Y is kinetic energy and X is potential energy (B) X is kinetic energy, Y is total energy and Z is potential energy

Sol.

(C) X is kinetic energy, Y is potential energy and Z is total energy (D) Z is kinetic energy, X is potential energy and Y is total energy Sol. Q.28 If U is the potential energy and K kinetic energy then |U| > |K| at (A) Only D (B) Only C (D) neither D nor C

(C) both D & C

Sol.

Q.31 Statement - I : Assuming zero potential at infinity, gravitational potential at a point cannot be positive.

Q.29 Satellites A and B are orbiting around the earth in orbits of ratio R and 4R respectively. The ratio of their areal velocities is (A) 1 : 2 Sol.

(B) 1 : 4

(C) 1 : 8

(D) 1 : 16

Statement - 2 : Magnitude of gravitational force between two particle has inverse square dependence on distance between two particles. (A) Statement - 1 is true, statement-2 is true and statement-2 is correct explanation for statement-1 (B) Statement -1 is true, statement-2 is true and statement - 2 is NOT the correct explanation for statement-1 (C) Statement - 1 is true, statement - 2 is false. (D) Statement - 1 is false, statement - 2 is true. Sol.

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GRAVITATION

Page # 35

Exercise - II 1. Assuming the earth to be a sphere of uniform density the acceleration due to gravity (A) at a point outside the earth is inversely proportional to the square of its distance from the center (B) at a point outside the earth is inversely proportional to its distance from the centre (C) at a point inside is zero (D) at a point inside is proportional to its distance from the centre Sol.

3. In side a hollow spherical shell (A) everywhere gravitational potential is zero (B) everywhere gravitational field is zero (C) everywhere gravitational potential is same (D) everywhere gravitational field is same Sol.

4. A geostationary satellite is at a height h above the surface of earth. If earth radius is R 2.Two masses m1 and m2 (m1 < m2) are released from rest from a finite distance. They start under their mutual gravitational attraction (A) acceleration of m1 is more than that of m2 (B) acceleration of m2 is more than that of m1 (C) centre of mass of system will remain at rest in all the references frame (D) total energy of system remains constant Sol.

R R

h

(A) The minimum colatitude on earth upto which the satellite can be used for communication is sin–1 (R/R + h) (B) The maximum colatitudes on earth upto which the satellite can be used for communication is sin–1 (R/R + h) (C) The area on earth escaped from this satellite is given as 2πR2(1 + sinθ) (D) The area on earth escaped from this satellite is given as 2πR2(1 + cosθ) Sol.

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GRAVITATION

Page # 36

7. An earth satellite is moved from one stable circular orbit to another larger and stable circular orbit. The following quantities increase for the satellite as a result of this change (A) gravitational potential energy (B) angular velocity (C) linear orbital velocity (D) centripetal acceleration Sol.

5. When a satellite in a circular orbit around the earth enters the atmospheric region, it encounters small air resistance to its motion. Then (A) its kinetic energy increases (B) its kinetic energy decreases (C) its angular momentum about the earth decreases (D) its period of revolution around the earth increases Sol.

6. A communications Earth satellite (A) goes round the earth from east to west (B) can be in the equatorial plane only (C) can be vertically above any place on the earth (D) goes round the earth from west to east Sol.

8. A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth (A) the acceleration of S is always directed towards the centre of the earth (B) the angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant (C) the total mechanical energy of S varies periodically with time (D) the linear momentum of S remains constant in magnitude Sol.

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GRAVITATION

9. If a satellite orbits as close to the earth’s surface as possible, (A) its speed is maximum (B) time period of its rotation is minimum (C) the total energy of the ‘earth plus satellite’ system is minimum (D) the total energy of the ‘earth plus satellite’ system is maximum Sol.

Page # 37 10. For a satellite to orbit around the earth, which of the following must be true ? (A) It must be above the equator at some time (B) It cannot pass over the poles at any time (C) Its height above the surface cannot exceed 36,000 km (D) Its period of rotation must be > 2π R / g where R is radius of earth Sol.

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Page # 38

(SUBJECTIVE PROBLEMS)

Exercise - III m

Q.1 Four masses (each of m) are placed at the vertices of a regular pyramid (triangular base) of side ‘a’. Find the work done by the system m while taking them apart so that they a form the pyramid of side ‘2a’. Sol.

m

m

Q.2 A small mass and a thin uniform rod each of mass ‘m’ are positioned along the same straight line as shown. Find the force of gravitational attraction exerted by the rod on the small mass. 2L L m m Sol.

Q.3 An object is projected vertically upward from the surface of the earth of mass M with a velocity such that the maximum height reached is eight times the radius R of the earth. Calculate : (i) the initial speed of projection (ii) the speed at half the maximum height. Sol.

Q.4 A satellite close to the earth is in orbit above the equator with a period of rotation of 1.5 hours. If it is above a point P on the equator at some time, it will be above P again after time ___________. Sol.

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Q.5 A satellite is moving in a circular orbit around the earth. The total energy of the satellite is E = –2 ×105 J. The amount of energy to be imparted to the satellite to transfer it to a circular orbit where its po tential energ y i s U = –2×10 5 J is equal to ___________. Sol.

Q.7 A point P lies on the axis of a fixed ring of mass M and radius a, at a distance a from its centre C. A small particle starts from P and reaches C under gravitational attraction only. Its speed a C will be ________. Sol.

Q.6 Find the gravitational field strength and potential at the centre of arc of linear mass density λ subtending an angle 2α at the centre.

2α Sol.

R

Q.8 Calculate the distance from the surface of the earth at which abo ve and below the sur face acceleration due to gravity is the same. Sol.

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GRAVITATION

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Q.9 Consider two satellites A and B of equal mass m, moving in the same circular orbit of radius r around the earth E but in opposite sense of rotation and therefore on a collision course (see figure).

A

r

Sol.

B

Me

(a)In terms of G, Me, m and r find the total mechanical energy EA + EB of the two satellite plus earth system before collision. (b) If the collision is completely inelastic so that wreckage remains as one piece of tangle d material (mass = 2m), find the total mechanical energy immediately after collision. (c) Describe the subsequent motion of the wreckage. Sol.

Q.11 A satellite of mass m is orbiting the earth in a circular orbit of radius r. It starts losing energy due to small air resistance at the rate of C J/s. Then the time taken for the satellite to reach the earth is ________. Sol.

Q.12 Find the potential energy of a system of eight particles placed at the vertices of a cube of side L. Neglect the self energy of the particles. Sol.

Q.10 A particle is fired vertically from the surface of the earth with a velocity kυe, where υe is the escape velocity and k R

O Q.14 Two small dense stars rotate about their common centre of mass as a binary system with the period 1year for each. One star is of double the mass of the other and the mass of the lighter one is 1/3 of the mass of the sun. Find the distance between the stars if distance between the earth & the sun is R. Sol.

x

(ii) Show that the gravitational field inside the hole is uniform, find its magnitude and direction. Sol.

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GRAVITATION

Page # 42 Q.17 A thin spherical shell of total mass M and radius R is held fixed. There is a small hole in the shell. A mass m is released from rest a distance R from the hole along a line that passes through the hole and also through the centre of the shell. This mass subsequently moves under the gravitational force of the shell. How long does the mass take to travel from the hole to the point diametrically opposite. Sol.

Q.16 A body moving radially away from a planet of mass M, when at distance r from planet, explodes in such a way that two of its many fragments move in mutually perpendicular circular orbits around the planet. What will be (a) then velocity in circular orbits. (b) maximum distance between the two fragments before collision and (c) magnitude of their relative velocity just before they collide. Sol.

Q.18 A remote sensing satellite is revolving in an orbit of radius x the equator of earth. Find the area on earth surface in which satellite can not send message.

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GRAVITATION

Page # 43

(TOUGH SUBJECTIVE PROBLEMS)

Exercise - IV

Q.1 A satellite P is revolving around the earth at a height h = radius of earth (R) above equator. Another satellite Q is at a height 2h revolving in opposite direction. At an instant the two are at same vertical line passing through centre of sphere. Find the least time of after which again they are in this is situation.

Earth M

PQ

Q.2 A certain triple-star system consists of two stars, each of mass m, revolving about a central star, mass M, in the same circular orbit. The two stars stay at opposite ends of a diameter of the circular orbit, see figure. Derive an expression for the period of revolution of the stars; the radius of the orbit is r.

m

r M

Q.5 A ring of radius R is made from a thin wire of radius r. If ρ is the density of the material of wire then what will be the gravitational force exerted by the ring on the material particle of mass m placed on the axis of ring at a distance x from its centre. Show that the force will be maximum when x = R / 2 and the ma xi mum va lue of for ce w il l be g iven a s

Fmax =

4 π 2 Gr 2ρm ( 3) 3 / 2 R

Q.6 A man can jump over b = 4m wide trench on earth. If mean density of an imaginary planet is twice that of the earth, calculate its maximum possible radius so that he may escape from it by jumping. Given radius of earth = 6400 km. Q.7 A launching pad with a spaceship is moving along a circular orbit of the moon, whose radius R is triple that of moon Rm. The ship leaves the launching pad with a relative velocity equal to the launching pad’s
initial orbital velocity v 0 and the launching pad then falls to the moon. Determine the angle θ with the horizontal at which the launching pad crashes into the surface if its mass is twice that of the spaceship m.

m Q.3 Find the gravitational force of interaction between the mass m and an infinite rod of varying mass density λ such that λ(x) = λ/x, where x is the distance from mass m. Given that mass m is placed at a distance d from the end of the rod on its axis as shown in figure. x d O m λ λ ( x) = x Q.4 Inside an isolated fixed sphere of radius R and uniform density r, there is a spherical cavity of radius R/2 such that the surface of the cavity passes through the centre of the sphere as in figure. A particle of mass m is released from rest at centre B of the cavity. Calculate velocity with which particle strikes the centre A of the sphere.

A R

B R/2

Q.8 A satellite of mass m is in an elliptical orbit around the earth of mass M(M >>m). The speed of the satellite 6 GM 5R where R = its closest distance to the earth. It is desired to transfer this satellite into a circular orbit around the earth of radius equal its largest distance from the earth. Find the increase in its speed to be imparted at the apogee (farthest point on the elliptical orbit). Q.9 A body is launched from the earth’s surface a an angl e α = 30º to the hor i zo ntal a t a sp eed

at its nearest point to the earth (perigee) is

15 . GM . Neglecting air resistance and earth’s R rotation, find (a) the height to which the body will rise. (ii) The radius of curvature of trajectory at its top point. Q.10 Assume that a tunnel is dug across the earth (radius = R) passing through its centre. Find the time a particle takes to reach centre of earth if it is projected into the tunnel from surface of earth with speed needed for it to escape the gravitational field of earth. v0 =

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GRAVITATION

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(JEE PROBLEMS)

Exercise - V Q.1 If the distance between the earth and the sum were half its present value, the number of days in a year would have been [JEE’ 96] (A) 64.5 (B) 129 (C) 182.5 (C) 730 Sol.

Q.2 Distance between the centres of two stars is 10a. The masses of these starts are M and 16M and their radii a and 2a respectively. A body of a mass m is fired at night from the surface of the larger star towards the maller star. What should be its minimum initial speed to reach the surface of the smaller start? Obtain the expression in terms of G, M and a. [JEE’ 96] Sol.

Q.3 An artificial satellite moving in a circular orbit around the earth has a total (K.E. + P.E.)E0. Its potential energy is [JEE’ 97] (A) –E0 (B) 1.5 E0 (C) 2E0 (D) E0 Sol.

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GRAVITATION

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Q.4 A cord of length 64 m is used to connect a 100 kg astronaut to spaceship whose mass is much larger than that of the astronaut. Estimate the value of the tension in the cord. Assume that the spaceship is orbiting near earth surafce. Assume that the spaceship and the astronaut fall on a straight line from the earth centre. The radius of the earth is 6400 km. [REE 98] Sol.

Sol.

Q.6 A body is projected vertically upwards from the bottom of a crater of moon of depth R/100 where R is the radius of moon with a velocity equal to the escape velocity on the surface of moon. Calculate maximum height attained by the body from the surface of the moon. [JEE’ 2003] Sol.

In a region of only gravitational field of mass ‘M’ a particle is shifted from A to B via three different paths in the figure. The work done in different paths are W1, W2, W3 respectively then Q

. 5

B

(1) (A) W1 = W2 = W3 (C) W1 = W2 > W3

(3) M (2)

C

A [JEE’ (Scr.) 2003] (B) W1 > W2 > W3 (D) W1 < W2 < W3

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GRAVITATION

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Q.7 A system of binary stars of masses mA and mB are moving in circular orbits of radii rA and rB respectively. If TA and TB are the time periods of masses mA and mB respectively, then [JEE 2006] (A) TA > TB (if rA > rB) (B) TA > TB (if mA > mB)  TA   (C)   TB 

2

 rA  =    rB 

3

(D) TA = TB

Sol.

Q.9 STATEMENT-1 An astronaut in an orbiting space station above the Earth experiences weightlessness. [JEE 2008] and STATEMENT-2 An object moving around the Earth under the influence of Earth’s gravitaitonal force is in a state of ‘free-fall’. (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 Sol.

Q.8 A spherically symmetric gravitational system of particles has a mass density [JEE 2008]

ρ for r ≤ R ρ= 0  0 for r > R where ρ0 is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed V as a function of distance r (0 < r < ∞) from the centre of the system is represented by

v

Q.10 A thin uniform annular disc (see figure) of mass M has outer radius 4 R and inner radius 3R. The work required to take a unit mass from point P on its axis to infinity is [JEE 2010] P 4R

v 3R

4R

(B)

(A)

R

r

v

R

r

R

r

v

(C)

(D)

R

r

2GM ( 4 2 – 5) 7R GM (C) 4 R (A)

2GM ( 4 2 – 5) 7R 2GM (D) 5 R ( 2 − 1) (B) –

Sol.

Sol.

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Q.11 A binary star consists of two stars A (mass 2.2 Ms) and B (mass 11 Ms), where Ms is the mass of the sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is. [JEE 2010] Sol.

Q.13 A satellite is moving with a constant speed 'V' in a circular orbit about the earth. An object of mass 'm' is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the object is (A)

1 mV2 2

(B) mV2

(C)

3 mV2 2

(D) 2mV2

[JEE 2011] Sol.

Q.12

Gravitational acceleration on the surface of a

6 g. where g is the gravitational accelera11 tion on the surface of the earth. The average mass planet is

density of the planet is

2 times that of the earth. If 3

the escape speed on the surface of the earth is taken to be 11 kms–1, the escape speed on the surface of [JEE 2010] the planet in kms–1 will be

Q.14 Two spherical planets P and Q have the same uniform density ρ , masses Mp and MQ, and surface areas A and 4A, respectively. A spherical planet R also has uniform density ρ and its mass is (MP + MQ). The escape velocities from the planets P, Q and R, are Vp, VQ and VR, respectively. Then [JEE 2012] (A) VQ > VR > VP

(B) VR > VQ > VP

(C) VR / VP = 3

(D) VP / VQ =

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1 2

GRAVITATION

Page # 48

ANSWER EX-I B D B A C B

1. 7. 13. 19. 25. 31.

(OBJECTIVE PROBLEMS) 2. 8. 14. 20. 26.

C A B C A

A A A D B

3. 9. 15. 21. 27.

ANSWER EX-II A,D A

1. 7.

A B B C A

5. 11. 17. 23. 29.

B B C B C

6. 12. 18. 24. 30.

(MULTIPLE CHOICE PROBLEMS) A,D A

2. 8.

B,C,D 4. A,B,C 10.

3. 9.

ANSWER EX-III 3Gm 2 a

1. –

B B A D C

4. 10. 16. 22. 28.

A,C A,D

A,C

5.

B,D

6.

(SUBJECIVE PROBLEMS) 2.

Gm 2

3. (i)

2

3L

2 2 Gm Gm , (ii) 3 5R R

4 3

4. 1.6 hours if is rotating from west to east, 24/17 hours if it is rotating from west to east. 5. 1 × 105 J

6.

2Gλ (sin α), (–Gλ 2α) R

9. (a) –GmMe/r, (b) –2GmMe/r 13. (i) –

 3Gm  m + M , (ii)  R  3 

10.

7.

2GM  1  1 −  a  2

1− k 2

   
πGρ0R  1 8 
2πGρ0R  – 2 i , g= – i 15. g = + 2   6 x 3 R   x –      2

2πR3 / 2 (6 6 )

8.

GM( 2 2 + 3 3 )

GM  2 8  –   R  3 15 

2. vmin =

6. h = 99R 13. B

7. D 14.B, D

Gm ; (b) r 2 ; (c) r

2GM r

(TOUGH SUBJECTIVE PROBLEMS) 4 πr 3 / 2

2.

G(4M + m)

3.

Gmλ 2

2d

4.

2 πGρR2 3

 7  + 1 R, (b) 1.13 R 9. (a) h =   2 

ANSWER EX-V 1. B

16. (a)

x2 – R2  4 πR 2  x 

ANSWER EX-IV 1.

 3 1  + 3 +  2 3 

14. R

3

17. 2 × R3 / GM

5 –1 R 2

–4GM2 12. L

GMm  1 1 11. t = 2C  R – r   e 

R ek 2

 G m + M  R 3 

  18.  1 – 

8. h =

6.

6.4 km

7. cos θ =

3 10

 1  Re 10. T = sin –1   3 g

(JEE PROBLEMS) 3 2

5GM a 8. C

3. C

4. T = 3 × 10–2 N

9. A

10. A

5. A 11. 6

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12. 3

IIT-JEE|AIEEE CBSE|SAT|NTSE OLYMPIADS

Nurturing potential through education

ELASTICITY & THERMAL EXPANSION THEORY AND EXERCISE BOOKLET

CONTENTS

S.NO.

TOPIC

PAGE NO.

1. Elasticity ........................................................................................ 2 2. Stress ......................................................................................... 2 – 3 3. Strain .......................................................................................... 3 – 5 4. Young Modulus ............................................................................ 5 – 6 5. Thermal Expansion .................................................................... 6 – 12 6. Exercise - I ................................................................................ 13 – 16 7. Exercise - II ............................................................................... 17 – 18 8. Exercise - III .................................................................................. 19 9. Answer key ................................................................................... 20

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Page # 2

1.

(a) (b)

2.

ELASTICITY & THERMAL EXPANSION

DEFINATION Elasticity is that property of the material of a body by virtue of which the body opposes any change in its shape or size when deforming forces are applied to it, and recovers its original state as soon as the deforming forces are removed. On the basis of defination bodies may be classified in two types : Perfectly Elastic (P.E.) : If body regains its original shape ans size completely after removal of force. Nearest approach P.E. : quartz-fibre Perfectly Plastic (P.P.) : If body does not have tendency to recover its original shape and size. Nearest Approach P.P. : Peetty Limit of Elasticity : The maximum deforming force upto which a body retains its property of elasticity is called the limit of elasticity of the material of the body. STRESS When a deforming force is applied to a body, it reacts to the applied force by developing a reaction (or restoring force which, from Newton's third law, is equal in magnitude and opposite in direction to the applied force. Thereaction force per unit area of the body which is called into play due to the action of the applied force is called stress. Stress is measured in units of force per unit area, i.e. Nm–2. Thus. F A where F is the applied force and A is the area over which it acts. A

Stress =

10 N Stress = 10/A Unit of stress : N/m2 Dimension of stress : M1L-1T-2 2.1

Types of stress : Three Types of Stress :

(A)

Tensile Stress : Pulling force per unit area.

F

A

F

It is applied parallel to the length It causes increase in length or volume (B)

Compressive Stress : Pushing force per unit area. It is applied parallel to the length

F

A

F

It causes decrease in length or volume (C)

Tangential Stress : Tangential force per unit area. It causes shearing of bodies.

Note : 1. If the stress is normal to surface called normal stress. 2. Stress is always normal to surface in case of change in length of a wire or volume of body. 3. When external force compresses the body  Nature of atomic force will be repulsive. 4. When external forces expanses the body  Nature of atomic force will be attractive. Difference between Pressure v/s Stress : S. No.

1 2 3

Pressure Pressure is always normal to the area.

Stress Stress can be normal or tangential May be compressive or Always compressive in nature tensile in nature. Scalar Tensor

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ELASTICITY & THERMAL EXPANSION Ex.1

Sol.

3.

A 4.0 m long copper wire of cross sectional area 1.2 cm2 is stretched by a force of 4.8 × 103 N stress will be (A) 4.0 × 107 N/mm2 (B) 4.0 × 107 KN/m2 (C) 4.0 × 107 N/m2 (D) None [C] 4.8  10 3 N F Stress = = = 4.0 × 107 N/m2 12 .  10 4 m 2 A STRAIN When a deforming force is applied to a body, it may suffer a change is size or shape. Strain is defined as the ratio of the change in size or shape to the original size or shape of the body. Strain is a number; it has no units or dimensions. The ratio of the change in length to the original length is called longitudinal strain. The ratio of the change in volume to the original volume is called volume strain. The strain resulting from a change in shape is called shearing strain.

Strain 

L final length – original length  =  T,, L0 original length

Note : Original and final length should be at same temperature. F

3.1

Types of strain : Three Types of Strain :

(A)

Linear Strain : Change in length per unit length is called linear strain Linear Strain =

Change in length Original length

L L Volume Strain : Change in volume per unit volume is called volume strain.

=

(B)

V  V

Volume sirain

Volume Strain Change in volume V = Original volume V Shear Strain : Angle through which a line originally normal to fixed surface is turned. =

(C)

 =

x L

x

L



Note : Strain is unitless. Ex.2

A copper rod 2m long is stretched by 1mm. Strain will be - Shear strain (A) 10-4, volumetric (B) 5 × 10-4, volumetric (C) 5 × 10-4, longitudinal (D) 5 × 10-3, volumetric

Sol.

[C] Strain =

 1 10 3 = 5 × 10–4, longitudinal =  2

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Page # 4

4.

ELASTICITY & THERMAL EXPANSION

THERMAL STRESS If the ends of a rod are rigidly fixed and its temperature is changed, then compressive stresses are set up in the rod. These developed stress are called thermal stress. Thermal Stress = Y  t Y  modulus of elasticity,   Coefficient of linear expansion t  change in temperature

5.

WORK DONE IN STRETCHING A WIRE In stretching a wire work is done against internal restoring forces. This work is stored in body as elastic potential energy or strain energy. If L = length of wire & A = Cross-sectional Area. F/A YA Y =  F = x x/L L work done to increase dx length YA dW = Fdx = xdx L L YA 1 YA Total work done = W = xdx = (L)2 0 L 2 L



2

 L  W 1  Work done per unit volume = = Y  V 2  L  W 1 = Y (strain)2 V 2 W 1 = x stress x strain V 2 W 1 ( stress)2 W 1 F L =  = × V 2 AL 2 A L Y 1 1 W = F × L = load x elongation 2 2

[ V = AL]

[ Y =

Stress ] Strain

stress

D E C If we increase the load gradually on a vertical B suspended metal wire, In Region OA : A Strain is small (< 2%) Stress  Strain  Hook's law is valid. O strain Slope of line OA gives Young's modulus Y of the material. In Region AB : Stress is not proportional to strain, but wire will still regain its original length after removing of stretching force. In region BC : Wire yields  strain increases rapidly with small change in stress. This behavior is shown up to point C known as yield point. In region CD : Point D correspondes to maximum stress, which is called point of breaking or tensile strength. In region DE : The wire literally flows. The maximum stress corresponding to D after which wire begin to flow. In this region strain increase even if wire is unloaded and rupture at E.

6.

STRESS-STAIN CURVE

7.

HOOKES' LAW Hookes' law states that, within the elastic limit, the stress developed in a bodyis proportional to the strain produced in it. Thus the ratio of stress to strain is a constant. This constant is called the modulus of elasticity. Thus stress Modulus of elasticity = strain Since strain has no unit, the unit of the modulus of elasticity is the same as that of stress, namely, Nm–2

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Page # 5

ELASTICITY & THERMAL EXPANSION 8.

YOUNG'S MODULUS Suppose that a rod of length l and a uniform crossectional area a is subjected to a logitudinal pull. In other words, two equal and opposite forces are applied at its ends. F A The stress in the present case is called linear stress, tensile stress, or extensional stress. If the direction of the force is reversed so that L is negative, we speak of compressional strain and compressional stress. If the elastic limit is not exceeded, then from Hooke's law Stress  strain or Stress = Y × strain

Stress =

Y

or

stress F L  . strain A L

...(1)

where Y, the constant of proportionality, is called the Young's modulus of the material of the rod and may be defined as the ratio of the linear stess to linear strain, provided the elastic limit is not exceeded. Since strain has no unit, the unit of Y is Nm–2. Consider a rod of length  0 which is fixed between to rigid end separated at a distance  0 now if the temperature of the rod is increased by  then the strain produced in the rod will be : length of the rod at new temperatrue – natural length of the rod at new temperature natural length of the rod at new temperature

strain =

F

 0 –  0 (1  ) –  0  = =  0 (1  )  0 (1  )

F

0

 is very small so



strain = –  (negative sign in the answer represents that the length of the rod is less than the natural length that means is compressed by the ends.) We know that  

stress then F = T A strain

Note : (A) For Loaded Wire : L =

FL 2

r Y

FL  2  Y  AL & A  r   

for rigid body L = 0 so Y =  i.e. elasticity of rigid body is infinite. (B)

If same stretching force is applied to different wire of same material. L

[As F and Y are const.] r2 Greater the value L, greater will be elongation. L 

(C)

Elongation of wire by its own weight : In this case F = Mg acts at CG of the wire so length of wire which is stretched will be L/2 (Mg)  L / 2 FL MgL gL2 = = = 2 r Y AY 2AY 2Y M = AL]

    L = [

L =

gL2 2Y

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Page # 6 Ex.3

Sol.

7.2

A wire of length 1m and area of cross section 4 x 10-8 m2 increases in length by 0.2 cm when a force of 16 N is applied. Value of Y for the material of the wire will be (A) 2 × 106 N/m2 (B) 2 × 1011 kg/m2 (C) 2 × 1011 N/mm2 (D) 2 × 1011 N/m2 [D] By Hook's law F/A FL Y = =  /L A 16  1 Y = = 2 × 10111 N/m2 ( 4  10 8 ) (0.2  10 2 ) Bulk Modulus : B =

7.3

8.

Volume stress = Volume strain

P VP  B = – V V  V

Compressibility : k =

7.4

ELASTICITY & THERMAL EXPANSION

1 1 = – B V

FG V IJ H P K

Modulus of Rigidity : tan gentialstress F/A  =   =  tan gentialstrain Only solid can have shearing as these have definite shape.

D

POISSION'S RATIO L

Lateralstrain d/D dL  = =   = Linear strain L / L LD Interatomic force constant = Young Modulus x Interatomic distance. 9.

 d

THERMAL EXPANSION

Most substances expand when they are heated. Thermal expansion is a consequence of the change in average separation between the constituent atoms of an object. Atoms of an object can be imagined to be connected to one another by stiff springs as shown in figure. At ordinary temperatures, the atoms in a solid oscillate about their equilibrium positions with an amplitude of approximately 10–11 m. The average specing between the atom is about 10–10 m. As the temperature of solid increases, the atoms oscillate with greater amplitudes, as a result the average separation between them increases, consequently the object expands.

9.1

LINEAR EXPANSION When the rod is heated, its increase in length L is proportional to its original length L0 and change in temperture T where T is in °C or K.

L0

Before heating

L  L 0  L

After heating

dL =  L0dT  L =  L0T If a T B & T  then FB  and T  then FB  v(1   B T)

9.8

v 0 g  mg

Barometer Their is a capaillary tube which have coefficient of linear expansion c and a liquid of volume v of volume expansion coefficient v of volume expansion coefficient of  at

A

temperature Ti. and given 3 c    . The Area of crosssection of capillary tube is A. Now temperature increases to Tf, So volume of liquid rises in the capillary. Let it rises to height H. So volume rises in tube = V V = V[1 +  T] – V[1+ 3 c T] = V ( – 3c ) T And Area of cross section of capillary = A = A [1 + 2CT] V VT(   – 3 C ) So height in capillary tube H'  A'  A(1  2 T) C

Ti

V

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c

Page # 12

ELASTICITY & THERMAL EXPANSION

Ex.10

What will happen to the water level if the vessel is heated ?

Sol.

(i) if     c then overflow occure and overflow



= AH(1    t ) – AH (1 + c T)

H

(ii) if     c final volume Vfc = AH (1  C T) final volume Vf = AH (1   T) Now So

Note

c

AF = A[1  2 C T] H[1  y  T ] H = final height = [1  2 T ] c

If two strips of equal length but of different metals are placed on each other and riveted, the single strip so formed is called 'bimetallic strip' [see given fig.]. 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  on outer side, i.e., convex side. This strip finds its application in auto-cut or thermostat in electric heating circuits. It has also been used as thermometer by calibrating its bending. T2 T1 Fe

Fe

Cu

Cu

(T2 > T1) (A) Ex.11

(B)

When the two rods having expansion cofficient 1, 2 (2 > 1) and width d are heated then the radius of the rod after expansion. 2

2 d

T

dI

1

(2 > 1) 

1

R

d R = (  –  ) T 2 1

Proof :  2  (1   2 t)  (R  d)   1  (1  1t )  R R  d (1   2 T )  R (1   1T )

from binomial theorem

d R = ( –  )T 2 1

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Page # 13

ELASTICITY & THERMAL EXPANSION

(OBJECTIVE QUESTIONS)

EXERCISE-I 1.

A steel scale is to be prepared such that the millimeter intervals are to be accurate within 6 × 10–5 mm. The maximum temperature variation from the temperature of calibration during the reading of the millimeter marks is ( = 12 × 10–6 k–1) (A) 4.0 ºC

(B) 4.5 ºC

(C) 5.0 ºC

(D) 5.5 ºC

2.

A steel rod 25 cm long has a cross-sectional area of 0.8 cm2. Force that would be required to stretch this rod by the same amount as the expansion produced by heating it through 10ºC is : (Coefficient of linear expansion of steel is 10–5/ºC and Young’s modulus of steel is 2 × 1010 N/m2.) (A) 160 N (B) 360 N (C) 106 N (D) 260 N

3.

Two rods of different materials having coefficients of thermal expansion 1, 2 and Young’s moduli Y1, 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 the rods. If 1 : 2 = 2 : 3, the thermal stresses developed in the two rods are equal provided Y1 : Y2 is equal to (A) 2 : 3 (B) 1 : 1 (C) 3 : 2 (D) 4 : 9

4.

If I is the moment of inertia of a solid body having -coefficient of linear expansion then the change in I corresponding to a small change in temperature T is (A)  I T

5.

1  I T 2

(C) 2  I T

(D) 3  I T

A metallic wire of length L is fixed between two rigid supports. If the wire is cooled through a temperature difference T (Y = young’s modulus,  = density,  = coefficient of linear expansion) then the frequency of transverse vibration is proportional to : (A)

6.

(B)



Y 

(B)

Y



(C)

 Y

(C)

Y

A metal wire is clamped between two vertical walls. At 20°C the unstrained length of the wire is exactly equal to the separation between walls. If the temperature of the wire is decreased the graph between elastic energy density (u) and temperature (T) of the wire is u

(A)

u

u

(B) T (in °C) 20

u

(C) T (in °C)

T (in °C)

20

20

(D)

T (in °C) 20

7.

A steel tape gives correct measurement at 20°C. A piece of wood is being measured with the steel tape at 0°C. The reading is 25 cm on the tape, the real length of the given piece of wood must be : (A) 25 cm (B) < 25 cm (C) >25 cm (D) can not say

8.

A rod of length 20 cm is made of metal. It expands by 0.075 cm when its temperature is raised from 0°C to 100°C. Another rod of a different metal B having the same length expands by 0.045 cm for the same change in temperature, a third rod of the same length is composed of two parts one of metal A and the other of metal B. Thus rod expand by 0.06 cm for the same change in temperature. The portion made of metal A has the length. (A) 20 cm (B) 10 cm (C) 15 cm (D) 18 cm

9.

A sphere of diameter 7 cm and mass 266.5 gm floats in a bath of a liquid. As the temperature is raised, the sphere just begins to sink at a temperature 35°C. If the density of a liquid at 0°C is 1.527 gm/cc, then neglecting the expansion of the sphere, the coefficient of cubical expansion of the liquid is f : (A) 8.486 × 10–4 per °C

(B) 8.486 × 10–5 per °C (C) 8.486 × 10–6 per °C

(D) 8.486 × 10–3 per °C

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Page # 14 10.

ELASTICITY & THERMAL EXPANSION

The volume of the bulb of a mercury thermometer at 0°C is V0 and cross section of the capillary is A0. The coefficient of linear expansion of glass is ag per °C and the cubical expansion of mercury m per °C. If the mercury just fills the bulb at 0°C, what is the length of mercury column in capillary at T°C. V0 T(  m  3ag ) (A) A (1  2a T) 0 g

11.

V0 T(  m – 3ag ) (B) A (1  2a T) 0 g

(B) EA t/(1 +  t)

(D) E/t

(B) W 0[1 – (s – 1) t]

(C) W 0[ (s – 1) t]

(D) W 0t / (s – 1)

A thin walled cylindrical metal vessel of linear coefficient of expansion 10–3 °C–1 contains benzenr of volume expansion coefficient 10–3 °C–1. If the vessel and its contents are now heated by 10°C, the pressure due to the liquid at the bottom. (A) increases by 2%

14.

(C) EA t/(1 – t)

The loss in weight of a solid when immersed in a liquid at 0°C is W 0 and at t°C is W. If cubical coefficient of expansion of the solid and the liquid by s and 1 respectively, then W is equal to : (A) W 0[1 + (s – 1) t]

13.

V0 T(  m – 2a g ) (D) A (1  3a T ) 0 g

A metallic rod 1 cm long with a square cross-section is heated through 1°C. If Young’s modulus of elasticity of the metal is E and the mean coefficient of linear expansion is  per degree Celsius, then the compressional force required to prevent the rod from expanding along its length is : (Neglect the change of cross-sectional area) (A) EAt

12.

V0 T(  m  2ag ) (C) A (1  3a T) 0 g

(B) decreases by 1%

(C) decreases by 2%

(D) remains unchanged

A rod of length 2m at 0°C and having expansion coefficient  = (3x + 2) × 10–6 °C–1 where x is the distance (in cm) from one end of rod. The length of rod at 20 °C is : (A) 2.124 m

(B) 3.24 m

(C) 2.0120 m

(D) 3.124 m

15.

A copper ring has a diameter of exactly 25 mm at its temperature of 0°C. An aluminium sphere has a diameter of exactly 25.05 mm at its temperature of 100°C. The sphere is placed on top of the ring and two are allowed to come to thermal equilibrium, no heat being lost to the surrounding. The sphere just passes through the ring at the equilibrium temperature. The ratio of the mass of the sphere & ring is : (given : Cu = 17 × 10–6/°C, Al = 2.3 × 10–5/°C, specific heat of Cu = 0.0923 Cal/g°C and specific heat of Al = 0.215 cal/g°C) (A) 1/5 (B) 23/108 (C) 23/54 y (D) 216/23

16.

A cuboid ABCDEFGH is anisotropic with x = 1 × 10–5/°C, y = 2 × 10–5/°C, z = 3 × 10–5/°C. Coefficient of superficial expansion of faces can be (A) ABCD = 5 × 10–5 /°C

(B) BCGH = 4 × 10–5 /°C

(C) CDEH = 3 × 10–5/°C

(D) EFGH = 2 × 10–5/°C

A

B C

D F

G

x

E H z

17.

An open vessel is filled completely with oil which has same coefficient of volume expansion as that of the vessel. On heating both oil and vessel, (A) the vessel can contain more volume and more mass of oil (B) the vessel can contain same volume and same mass of oil (C) the vessel can contain same volume but more mass of oil (D) the vessel can contain more volume but same mass of oil

18.

A metal ball immersed in Alcohol weights W 1 at 0°C and W 2 at 50°C. The coefficient of cubical expansion of the metal ()m is less than that of alcohol ()Al. Assuming that density of metal is large compared to that of alcohol, it can be shown that (A) W 1 > W 2

19.

(B) W 1 = W 2

(C) W 1 < W 2

(D) any of (A), (B) or (C)

A solid ball is completely immersed in a liquid. The coefficients of volume expansion of the ball and liquid are 3 × 10–6 and 8 × 10–6 per °C respectively. The percentage change in upthrust when the temperature is increased by 100°C is (A) 0.5 %

(B) 0.11 %

(C) 1.1%

(D) 0.05 %

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Page # 15

ELASTICITY & THERMAL EXPANSION 20.

A thin copper wire of length L increase in length by 1% when heated from temperature T1 to T2. What is the percentage change in area when a thin copper plate having dimensions 2L × L is heated from T1 to T2 ? (A) 1%

21.

(B) 2%

(C) 3%

(D) 4%

If two rods of length L and 2L having coefficients of linear expansion  and 2 respectively are connected so that total length becomes 3L, the average coefficient of linear expansion of the composition rod equals : 3 5 5  (B)  (C)  (D) none of these 2 2 3 The bulk modulus of copper is 1.4 × 1011 Pa and the coefficient of linear expansion is 1.7 × 10–5 (C°)–1. What hydrostatic pressure is necessary to prevent a copper block from expanding when its temperature is increased from 20°C to 30°C ?

(A)

22.

(A) 6.0 × 105 Pa

(B) 7.1 × 107 Pa

(C) 5.2 × 106 Pa

(D) 40 atm

23.

The coefficients of thermal expansion of steel and a metal X are respectively 12 × 10–6 and 2 × 10–6 per °C, At 40°C, the side of a cube of metal X was measured using a steel vernier callipers. The reading was 100 mm. Assuming that the calibration of the vernier was done at 0°C, then the actual length of the side of the cube at 0°C will be (A) > 100 mm (B) < 100 mm (C) = 100 mm (D) data insufficient to conclude

24.

A glass flask contains some mercury at room temperature. It is found that at different temperature the volume of air inside the flask remains the same. If the volume of mercury in the flask is 300 cm3, then volume of the flask is (given that coefficient of volume expansion of mercury and coefficient of linear expansion of glass are 1.8 × 10–4(°C)–1 and 9 × 10–6(°C)–1 respectively) (A) 4500 cm3 (B) 450 cm3 (C) 2000 cm3 (D) 6000 cm3

Question No. 25 to 29 (5 question) Solids and liquids both expand on heating. The density of substance decreases on expanding according to the relation 1 2 = 1  ( T – T ) 2 1

25.

where, 1  density at T1 2  density at T2   coeff. of volume expansion of substances when a solid is submerged in a liquid, liquid exerts an upward force on solid which is equal to the weight of liquid displaced by submerged part of solid. Solid will float or sink depends on relative densities of solid and liquid. A cubical block of solid floats in a liquid with half of its volume submerged in liquid as shown in figure (at temperature T) s  coeff. of linear expansion of solid L  coeff. of volume expansion of liquid s  density of solid at temp. T L  density of liquid at temp. T The relation between densities of solid and liquid at temperature T is (A) S = 2L (B) S = (1/2) L (C) S = L (D) S = (1/4) L

26.

If temperature of system increases, then fraction of solid submerged in liquid (A) increases (B) decreases (C) remains the same (D)inadequate information

27.

Imagine fraction submerged does not change on increasing temperature the relation between L and S is (A) L = 3S (B) L = 2S (C) L = 4S (D) L = (3/2)S

28.

Imagine the depth of the block submerged in the liquid does not change on increasing temperature then (A) L = 2 (B) L = 3 (C) L = (3/2) (D) L = (4/3)

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Page # 16

ELASTICITY & THERMAL EXPANSION

29.

Assume block does not expand on heating. The temperature at which the block just begins to sink in liquid is (A) T + 1/L (B) T + 1/(2L) (C) T + 2/L (D) T + L/2

30.

The coefficient of apparent expansion of a liquid in a copper vessel is C and in a silver vessel is S. The coefficient of volume expansion of copper is C. What is the coefficient of linear expansion of silver? (A)

C   c  S) 3

(B)

C –  c  S) 3

(C)

C   c – S) 3

(D)

C –  c – S) 3

31.

An aluminium container of mass 100 gm contains 200 gm of ice at –20°C. Heat is added to the system at the rate of 100 cal/s. The temperature of the system after 4 minutes will be (specific heat of ice = 0.5 and L = 80 cal/ gm, specific heat of Al = 0.2 cal/gm/°C) (A) 40.5°C (B) 25.5°C (C) 30.3°C (D) 35.0°C

32.

Two vertical glass tubes filled with a liquid are connected by a capillary tube as shown in the figure. The tube on the left is put in an ice bath at 0°C while the tube on the right is kept at 30° C in a water bath. The differenece in the levels of the liquid in the two tubes is 4 cm while the height of the liquid column at 0° C is 120 cm. The coefficient of volume expansion of liquid is (Ignore expansion of glass tube) (A) 22 × 10–4/°C (B) 1.1 × 10–4/°C –4 (C) 11 × 10 /°C (D) 2.2 × 10–4 /°C

33. 34.

A difference of temperature of 25ºC is equivalent to a difference of : (A) 45º F (B) 72º F (C) 32º F

4 cm

Water

120cm 30°C 0°C

(D) 25º F

Two thermometers x and y have fundamental intervals of 80º and 120º. When immersed in ice, they show the reading of 20º and 30º. If y measures the temperature of a body as 120º, the reading of x is : (A) 59º (B) 65º (C) 75º (D) 80º MULTIPLE CHOICE QUESTIONS

35.

When an enclosed perfect gas is subjected to an adiabatic process : (A) Its total internal energy does not change (B) Its temperature does not change (C) Its pressure varies inversely as a certain power of its volume (D) The product of its pressure and volume is directly proportional to its absolute temperature.

36.

Four rods A, B, C, D of same length and material but of different radii r, r 2 , r 3 and 2r respectively are held between two rigid walls. The temperature of all rods is increased by same amount. If the rods donot bend, then (A) the stress in the rods are in the ratio 1 : 2 : 3 : 4 (B) the force on the rod exerted by the wall are in the ratio 1 : 2 : 3 : 4 (C) the energy stored in the rods due to elasticity are in the ratio 1 : 2 : 3 : 4 (D) the strains produced in the rods are in the ratio 1 : 2 : 3 : 4

37.

A body of mass M is attached to the lower end of a metal wire, whose upper end is fixed. The elongation of the wire is l. (A) Loss in gravitational potential energy of M is Mgl (B) The elastic potential energy stored in the wire is Mgl (C) The elastic potential energy stored in the wire is 1/2 Mgl (D) Heat produced is 1/2 Mgl

38.

When the temperature of a copper coin is raised by 80ºC, its diameter increases by 0.2%. (A) Percentage rise in the area of a face is 0.4% (B) Percentage rise in the thickness is 0.4% (C) Percentage rise in the volume is 0.6% (D) Coefficient of linear expansion of copper is 0.25 × 10–4Cº–1.

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Page # 17

ELASTICITY & THERMAL EXPANSION

EXERCISE-II

(SUBJECTIVE QUESTIONS)

1.

We have a hollow sphere and a solid sphere of equal radii and of the same material. They are heated to raise their temperature by equal amounts. How will the change in their volumes, due to volume expansions, be related ? Consider two cases (i) hollow sphere is filled with air, (ii) there is vaccum inside the hollow sphere.

2.

The time represented by the clock hands of a pendulum clock depends on the number of oscillation performed by pendulum every time it reach to its extreme position the second hand of the clock advances by one second that means second hand move by two second when one oscillation in complete (a) How many number of oscillations completed by pendulum of clock in 15 minutes at calibrated temperature 20°C (b) How many number of oscillations are completed by a pendulum of clock in 15 minute at temperature of 40°C if  = 2 × 10–5c (c) What time is represented by the pendulum clock at 40°C after 15 minutes if the initial time shown by the clock is 12: 00 pm ? (d) If the clock gains two second in 15 minutes then find - (i) Number of extra oscillation (ii) New time period (iii) change in temperature.

3.

Consider a cylindrical container of cross section area ‘A’, length ‘h’ having coefficient of linear expansion c. The container is filled by liquid of real expansion coefficient L up to height h1. When temperature of the system h is increased by  then h1 (a) Find out new height, area and volume of cyclindrical container and new volume of liquid. (b) Find the height of liquid level when expansion of container is neglected. (c) Find the relation between L and c for which volume of container above the liquid level. (i) increases (ii) decreases (iii) remains constant. (d) If y  > 3 C and h = h1 then calculate, the volume of liquid overflow (e) What is the relation between   and  c for which volume of empty space becomes independent of (f) (i) (ii) (iii) (1)

change of temp. If the surface of a cylindrical container is marked with numbers for the measurement of liquid level of liquid filled inside it. If we increase the temperature of the system be  then Find height of liquid level as shown by the scale on the vessel. Neglect expansion of liquid Find height of liquid level as shown by the scale on the vessel. Neglect expansion of container Find relation between L and c so that height of liquid level with respect to ground increases (2) decreases (3)remains constant.

4.

A loaded glass bulb weighs 156.25 g in air. When the bulb is immersed in a liquid at temperature 15ºC, it weighs 56.25 g. On heating the liquid, for a temperature upto 52ºC the apparent weight of the bulb becomes 66.25 g. Find the coefficient of real expansion of the liquid. (Given coefficient of linear expansion of glass = 9 × 10–6/ºC).

5.

A body is completely submerged inside the liquid. It is in equilibrium and in rest condition at certain temperature. It L volumetric expansion coefficient of liquid s = linear expansion coefficient by of body. It we increases temperature by  amount than find (a) New thrust force if initial volume of body is V0 and density of liquid is d0. (b) Relation between s and L so body will (i) move upward (ii) down ward (iii) remains are rest

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Page # 18

ELASTICITY & THERMAL EXPANSION

6.

A clock pendulum made of invar has a period of 0.5 sec at 20°C. If the clock is used in a climate where average temperature is 30° C, aporoximately. How much fast or slow will the clock run in 106 sec. (invar = 1 × 10–6/°C)

7.

An iron bar (Young’s modulus = 1011 N/m2,  = 10–6/°C) 1 m long and 10–3 m2 in area is heated from 0°C to 100°C without being allowed to bend or expand. Find the compressive force developed inside the bar.

8.

Three aluminium rods of equal length form an equilateral triangle ABC. Taking O (mid point of rod BC) as the origin.Find the increase in Y-coordinate of center of mass per unit change in temperature of the system. Assume the length of the each rod is 2m, and al = 4 3 × 10–6/ °C A

B

O

C

9.

If two rods of length L and 2L having coefficients of linear expansion  and 2 respectively are connected so that total length becomes 3L, determine the average coefficient of linear expansion of the composite rod.

10.

A thermostatted chamber at small height h above earth’s surface maintained at 30°C has a clock fitted in it with an uncompensated pendulum. The clock designer correctly designs it for height h, but for temperature of 20°C. If this chamber is taken to earth’s surface, the clock in it would click correct time. Find the coefficient of linear expansion of material of pendulum.(earth’s radius is R)

11.

The coefficient of volume expansion of mercury is 20 times the coefficient of linear expansion of glass Find the volume of mercury that must be poured into a glass vessel of volume V so that the volume above mercury may remain constant at all temperature.

12.

A metal rod A of 25 cm lengths expands by 0.050 cm. When its temperature is raised from 0°C to 100°C. Another rod B of a different metal of length 40cm expands by 0.040 cm for the same rise in temperature. A third rod C of 50 cm length is made up of pieces of rods A and B placed end to end expands by 0.03 cm on heating from 0° C to 50°C. Find the lengths of each portion of the composite rod.

13.

The figure shows three temperature scales with the freezing and boiling points of water indicated. 70ºx

120ºW

90ºY

–20ºx

30ºW

0ºY

Bolling Point

Freezing Point

(a) Rank the size of a degree on these scales, greatest first. (b) Rank the following temperatures, highest first 50ºX, 50ºW and 50ºY. 14.

What is the temperature at which we get the same reading on both the centigrade and Fahrenheit scales ?

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Page # 19

ELASTICITY & THERMAL EXPANSION

EXERCISE-III 1.

(JEE PROBLEMS)

The apparatus shown in the figure consists of four glass columns connected by horizontal sections. The height of two central columns B & C are 49 cm each. The two outer columns A & D are open to the atmosphere. A & C are maintained at a temperature of 95° C while the columns B & d are maintained at 5°C. The height of the liquid in A & D measured from the base line are 52.8 cm & 51 cm respectively. Determine the coefficient of thermal expansion of the liquid. [JEE ‘97]

A 95°

B 5°

C 95°

D 5°

2.

A bimetallic strip is formed out of two identical strips one of copper and the other of brass. The coefficient of linear expansion of the two metals are C and B. On heating, the temperature of the strip goes up by T and the strip bends to form an arc of radius of curvature R. Then R is : [JEE ‘99] (A) proportional at T (B) inversely proportional to T (C) proportional to |B – C| (D) inversely proportional to |B – C|

3.

Two rods one of aluminium of length l1 having coefficient of linear expansion a, and other steel of length l2 having coefficient of linear expansion S are joined end to end. The expansion in both the rods is same on l1 [JEE’ (Scr) 2003] variation of temperature. Then the value of l  l is 1 2 s s a  s (A*)    (B)  –  (C) (D) None of these s a s a s

4.

A cube of coefficient of linear expansion s is floating in a bath containing a liquid of coefficient of volume expansion l. When the temperature is raised by T, the depth upto which the cube is submerged in the liquid remains the same. Find the relation between s and l, showing all the steps. [JEE 2004]

5.

A 0.1 kg mass is suspended from a wire of negligible mass The length of the wire is 1 m and its crosssectional area is 4.9 × 10–7 m2. If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 red s–1. If the Young's modulus of the material of the wire is n × 109 Nm–2, the value of n is [JEE 2010]

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Page # 20

ELASTICITY & THERMAL EXPANSION

: : ANSWER KEY : :

ANSWER EX-I

(OBJECTIVE PROBLEMS)

1. 7. 13. 19. 25. 31.

C B C D B B

2. 8. 14. 20. 26. 32.

37.

A,C,D 38.

A B C B A C

3. 9. 15. 21. 27. 33.

C A C C A A

4. 10. 16. 22. 28. 34.

C B C B A D

5. 11. 17. 23. 29. 35.

B B D A A C,D

6. 12. 18. 24. 30. 36.

B A C C C B,C

A,C,D

ANSWER EX-II

(SUBJECTIVE PROBLEMS)

1. (i) hollow sphere > solid sphere, (ii) hollow sphere = solid sphere 1 900 s s (iii) 2. (a) 450 (b) 449 (c) 12 : 14 : 59 (d) (i) 1 (ii) 451 450  10 –5 3. (a) h1 = h {1 + c },

A1 = A {1 + 2s },

v1 = Ah {1 + 3s }

volume of liquid Vw = Ah1(1 + L ) (b) h1 = h {1 + L }

(c) (i) L < 3c (ii) L > 3c (iii) L = 3c.

(d) V = Ah (L – 3c ) 

(e) 3hc = h1L

(f) (i) h1 (1 – 3c), (ii) h1(1 + L ),

 1  3 s   5. (a) V0d0g  1     (b) (i) L < 3s (ii) L > 3s (iii) L = 3s.   L

1 –6  4. YR =   27  37  10  /  C 9

6. 5 sec slow

7. 10000 N

8. 4 × 10–6 m/°C 9. 5 /3

13. (a) All tie (b) 50°X, 50°Y, 50°W.

2 × 10–4 C

10. h/5R 11. 3V/20 12.10 cm, 40 cm

14. –40°C or –40°F

(JEE PROBLEMS )

ANSWER EX-III 1.

(iii) (1) L > 2c (2) L < 2c (3) L = 2c .

2.

B,D

3.

A

4.

l = 2s 5.

4

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FLUID THEORY AND EXERCISE BOOKLET CONTENTS S.NO.

TOPIC

PAGE NO.

1.

Fluid .......................................................................................................... 3

2.

Pressure in a fluid ................................................................................ 3 – 9

3.

Pascal's Principle ............................................................................... 9 – 10

4.

Archimede's Principle ....................................................................... 10 – 16

5.

Equation of Continuity ....................................................................... 16 – 17

6.

Bernoullis Equation ........................................................................... 17 – 21

14. Exercise - 1 ........................................................................................ 22 – 39 15. Exercise - 2 ........................................................................................ 40 – 42 16. Exercise - 3 ........................................................................................ 43 – 48 17. Exercise - 4 ........................................................................................ 49 – 51 18. Exercise - 5 ........................................................................................ 52 – 58 19. Answer key ......................................................................................... 59 – 60

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Page # 2

FLUID

IIT-JEE Syllabus : FLUID Pressure in a fluid; Pascal's law; Byoyancy, Streamline flow, Equation of continuity Bernoulli's theorem and its applications.

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Page # 3

FLUID

1.

FLUID: Fluid mechanics deals with the behaviour of fluids at rest and in motion. A fluid is a substance that deforms continuously under the application of a shear (tangential) stress no matter how small the shear stress may be. Thus, fluids comprise the liquid and gas (or vapor) phase of the physical forms in which matter exists. Density () : Mass of unit volume, Called density Density at a point of liquid described by   Lim V  0

m dm  V dV

density is a positive scalar quantity. SI unit = Kg/m3 CGS unit = gm/cm3 Dimension = [ML–3] Relative Density : It is the ratio of density of given liquid to the density of pure water at 4°C

R.D. 

Density of given liquid Density of pure water at 4C

Relative density or specific gravity is unit less, dimensionless. It is a positive scalar physical Quantity Value of R.D. is same in SI and CGS system due to dimensionless/unitless Specific Gravity : It is the ratio of weight of given liquid to the weight of pure water at 4°C Weight of given liquid

Specific Gravity =

Weight of pure water at 4C(9.81 kN / m3 )

   g =   g   = Relative density of w w

liquid i.e. than specfic gravity of a liquid is approximately equal to the relative density. For calculation they can be interchange

2.

PRESSURE IN A FLUID When a fluid (either liquid or gas) is at rest, it exerts a force pependicular to any surface in contact with it, such as a container wall or a body immersed in the fluid. While the fluid as a whole is at rest, the molecules that makes up the fluid are in motion, the force exerted by the fluid is due to molecules colliding with their surroundings. If we think of an imaginary surface within the fluid, the fluid on the two sides of the surface exerts equal and opposite forces on the surface, otherwise the surface would acceleratate and the fluid would not remain at rest. Consider a small surface of area dA centered on a point on the fluid, the normal force exerted by the fluid on each side is dF . The pressure P is defined at that point as the normal force per unit area, i.e.,, dF dA If the pressure is the same at all points of a finite plane surface with area A, then P

P

F A

where F is the normal force on one side of the surface. The SI unit of pressure is pascal, where 1 pascal = 1 Pa = 1.0 N/m2 One unit used principally in meterology is the Bar which is equal to 105 Pa. 1 Bar = 105 Pa

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Page # 4

FLUID

2.1

Atmospheric Pressure (P0) It is pressure of the earth's atmosphere. This changes with weather and elevation. Normal atmospheric pressure at sea level (an average value) is 1.013 × 105 Pa. Thus 1 atm = 1.013 × 105 Pa Note : Fluid pressure acts perpendicular to any surface in the fluid no matter how that surface is oriented. Hence, pressure has no intrinsic direction of its own, its a scalar. By contrast, force is a vector with a definite direction.

2.2

Variation in Pressure with depth If the weight of the fluid can be neglected, the pressure in a fluid is the same throughout its volume. But often the fluid's weight is not negligible and under such condition pressure increases with increasing depth below the surface. Let us now derive a general relation between the presure P at any point in a fluid at rest and the elevation y of that point. We will assume that the density  and the acceleration due to gravity g are the same throughout the fluid. If the fluid is in equilibrium, every volume element is in equilibrium.

dW

dy

dy y PA

Consider a thin element of fluid with height dy. The bottom and top surfaces each have area A, and they are at elevations y and y + dy above some reference level where y = 0. The weight of the fluid element is dW = (volume) (density) (g) = (A dy) () (g) or dW = g A dy What are the other forces in y-direction of this fluid element ? Call the pressure at the bottom surface P, the total y component of upward force is PA. The pressure at the top surface is P + dP and the total y-component of downward force on the top surface is (P + dP) A. The fluid element is in equilibrium, so the total y-component of force including the weight and the forces at the bottom and top surfaces must be zero. Fy = 0  PA – (P + dP) A – gAdy = 0 or

dP  – g dy

...(i)

This equation shows that when y increases, P decreases, i.e., as we move upward in the fluid, pressure decreases. If P1 and P2 be the pressures at elevations y1 and y2 and if  and g are constant, then integrating Eq.(i) , we get P2



P1

or

dP  – g

y2



P2 y2

P1

y1

dy

y1

P2 – P1 = – g (y2 – y1)

...(ii)

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Page # 5

FLUID

It's often convenient to express Eq. (ii) in terms of the depth below the surface of a fluid. Take point 1 at depth h below the surface of fluid and let P represents pressure at this point. Take point 2 at the surface of the fluid, where the pressure is P0 (subscript zero for zero depth). The depth of point 1 below the surface is, h = y2 – y 1 and Eq. (ii) becomes P0 – P = – g (y2 – y1) = –  gh 

P = P0 + gh

...(iii)

Thus, pressure increases linearly with dpeth, if  and g are uniform, A graph between P and h is shown below. P

P  P0  gh

P0

P0

A

B

h

P0 h

PA  PB  P0  gh Further, the pressure is the same at any two points at the same level in the fluid. The shape of the container does not matter. 2.3.

Barometer

Vacuum (P = 0)

It is a device used to measure atmospheric pressure. In principle, any liquid can be used to fill the barometer, but

h

mercury is the substance of choice because its great density

1

makes possible an instrument of reasonable size.

2

P1 = P 2 Here,

P1 = atompsheric pressure (P0)

and

P2 = 0 + gh =  gh

Here,

 = density of mercury P0 = gh

Thus, the mercury barometer reads the atmosphereic pressure (P0) directly from the height of the mercury column. For example if the height of mercury in a barometer is 760 mm, then atmospheric pressure will be, P0 = gh = (13.6 × 103) (9.8) (0.760)= 1.01 × 105 N/m2 2.4

Force on Side Wall of Vessel Force on the side wall of the vesel can not be directly determined as at different depths pressures are different. To find this we cosider a strip of width dx at a depth x from the surface of the liquid as shown in figure, and on this strip the force due to the liquid is given as : dF = xg × bdx This force is acting in the direction normal to the side wall. Net force can be evaluated by integrating equation h

F  dF 



 xgbdx 0

F

gbh2 2

x

dx dF

h b a

...(2.4)

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Page # 6 2.5

FLUID

Average Pressure on Side Wall The absolute pressure on the side wall cannot be evaluated because at different depths on this wall pressure is different. The average pressure on the wall can be given as :  p  av 

F 1 gbh 2 1  gh = bh 2 bh 2

...(2.5)

Equation (2.5) shows that the average pressure on side vertical wall is half of the net pressure at the bottom of the vessel. 2.6

Torque on the Side Wall due to Fluid Pressure As shown in figure, due to the force dF, the side wall experiences a torque about the bottom edge of the side which is given as

d  dF  (h – x) = xgb dx (h – x) h

  d  gb(hx – x2 )dx



This net torque is

 0

 h3 h3  1  gb –   gbh 3 2 3  6  2.7

Manometer It is a device used to measure the pressure of a gas inside a container. The U-shaped tube often contains mercury. P1 = P 2 Here, P1 = pressure of the gas in the container (P) and 

P0

P2 = atmospheric pressure (P0) + gh

h

P = P0+ hg 1

This can also be written as

2

P – P0 = gauge pressure = hg Here,  is the density of the liquid used in U - tube Thus by measuring h we can find absolute (or gauge) pressure in the vessel.

Ex.1

Two liquid which do not react chemically are placed in a bent tube as shown in figure. Find out the displacement of the liquid in equillibrium position.

x 2







Sol.

 The pressure at the interface must be same, calculated via either tube. Since both tube all open to the atmosphere, we must have.

x



x



2g( – x) = g( + x)  x = /3

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Page # 7

FLUID Ex.2

Three liquid which do not react chemically are placed in a bent tube as shown in figure (initially) then fluid out the displacement of the liquid in equillibrium position.

 

Sol.

3 

2

Let us assume that level of liquid having density 3 displaced below by x as shown in figure below.

x x  – x

g + 2gx = 3( – x)g 2.8

x = 2/5 Pressure Distribution in an Accelerated Frame We've already discussed that when a liquid is filled in a container, generally its free surface remains horizontal as shown in figure (a) as for its equilibrium its free surface must be normal to gravity i.e. horizontal. Due to the same reason we said that pressure at every point of a liquid layer parallel to its free surface remains constant. Similar situation exist when liquid is in an accelerated frame as shown in figure (b). Due to acceleration of container, liquid filled in it experiences a pseudo force relative to container and due to this the free surface of liquid which normal to the gravity now is filled as  a   tan –1    g

...(2.22) a

A

a 

geff



g

(b) (a) Now from equilibrium of liquid we can state that pressure at every point in a liquid layer parallel to the free surface (which is not horizontal), remains same for example if we find pressure at a point A in the acceleratd container as shown in figure (a) is given as

PA = P0 + h

a2  g2

...(2.23)

Where h is the depth of the point A below the free surface of liquid along effective gravity and P0 is the atmopheric pressure acting on free surface of the liquid.

a

l1

h 

A

(c)

a



h



l2



(d)

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Page # 8

FLUID

The pressure at point A can also obtained in an another way as shown in figure (b). If l1 and l2 are the vertical and horizontal distances of point A from the surface of liquid then pressure at point A can also be given as PA = P0 + l1 g = P0 + l2 a

(2.24)

Here l1 g is the pressure at A due to the vertical height of liquid above A and according to Pascal's Law pressure at A is given as PA = P0 + l1 g

...(2.25) h a 2  g2 g

Here we can write l1 as

l1  h sec  

or from equation (2.25)

PA  P0  h a 2  g2

Similarly if we consider the horizotnal distance of point A from free surface of liquid, which is l2 then due to pseudo acceleration of container the pressure at point A is given as P A = P 0 + l2  a Here l2 is given as

l2  h cos ec 

...(2.26)

h g2  a 2 a

PA  P0  h g2  a 2

From equation (2.24), we have

Here students should note that while evaluating pressure at point A from vertical direction we haven't mentioned any thing about pseudo acceleration as along vertical length l1, due to pseudo acceleration at every point pressure must be constant similarly in horizontal direction at every point due to gravity pressure reamins constant. Ex.3

Figure shows a tube in which liquid is filled at the level. It is now rotated at an angular frequency w about an axis passing through arm A find out pressure difference at the liquid interfaces.  B A A

Sol.

 To solve the problem we take a small mass dm from the

B

axis at ‘a’ distance x in displaced condition. Net inward force = (P + dP) A – PA dm  Adx

= AdP

P + dP

P

x0

This force is balanced by centripetal force in equilibrium 

2

 A dP = dm 2x = Adx 2x 

 dP    xdx x0

xw

dx

x  – x0



P = 2

 xdx = x0g

x0

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Page # 9

FLUID 

Ex.4

A liquid of density  is in a bucket that spins with angular velocity  as shown in figure. Show that the pressure at a radial distance r from the axis is

2 r 2 2 where P0 is the atmospheric pressure. P  P0 

Sol.

Consider a fluid particle P of mass m at coordinates (x, y). From a non-inertial rotating frame of reference two forces are acting on it, (i) pseudo force (mx 2) (ii) weight (mg) in the directions shown in figure. Net force on it should be perpendicular to the free surface (in equilibrium). Hence. tan  

mx 2 x 2  mg g

y

x

dy 







y

0



0

or

P x

dy x2  dx g

P  ( x, y )

x2 .dx g

x 2 2 2g

P

mx2 

This is the equation of the free surface of the liquid, which is a parabola.

3.

y



mg

Fnet

r 22 2g

At x = r,

y



P(r) = P0 + gy

or

P(r) = P0 +

P0 yP(r) x=r

 2r 2 2

PASCAL'S PRINCIPLE Some times while dealing with the problems of fluid it is desirable to know the pressure at one point is pressure at any other point in a fluid is known. For such types of calculations Pascal's Law is used extensively in dealing of static fluids. It is stated as "The pressure applied at one point in an enclosed fluid is transmitted uniformly to every part of the fluid and to the walls of the container." One more example can be considered better to explain the concept of Pascal's Principle. Consider the situation shown in figure, a tube having two different cross section S1 and S2, with pistons of same cross sections fitted at the two ends. 2 1 F1

F2

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Page # 10

FLUID

If an external force F1 is applied to the piston 1, it creates a pressure p1 = F1/S1 on the liquid enclosed. As the whole liquid is at the same level, everywhere the pressure in the liquid is increased by p1. The force applied by the liquid on the piston 2 can be given as F2 =p2 × S2, and as the two pistons are at same level p2 = p1. Thus F2 = p2 × S2 F1 F2 = S  S2 1

....(2.21)

Equation (2.21) shows that by using such a system the force can be amplified by an amount equal to the ratio of the cross section of the two pistons. This is the principle of hydraulic press, we'll encounter in next few pages.

3.1

The Hydraulic Lift Figure shows how Pascal's principle can be made the basis for a hydraulic lift. In operation, let an external force of magnitude F1 be exerted downward on the left input piston, whose area is S1. It result a force F2 which will act on piston 2 by the incompressible liquid in the device. Here

F2 = p2 × S2

And

p2 = pB –  gh F1

S1

S2 F2 h

Where pB is the pressure on the bottom of the device which can be given as : pB = p1 +  gh Thus p2 = p1 and F2 = p1 S2 S2 or F2 = F1 × S 1 If S2 >> S1  F2 >> F1 4.

ARCHIMEDE'S PRINCIPLE If a heavy object is immersed in water, it seems to weight less than when it is in air. This is because the water exerts an upward force called buoyant force. It is equal to the weight of the fluid displaced by the body. A body wholly or partially submerged in a fluid is buoyed up by a force equal to the weight of the displaced fluid. This result is known as Archimedes' principle. Thus, the magnitude of buoyant force (F) is given by, F  Vi L g Here, Vi = immersed volume of solid and

L = density of liquid

g = acceleration due to gravity

Note : Point of Application of buoyant force is centre of liquied displaced

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Page # 11

FLUID Proof Consider an arbitraily shaped body of volume V placed in a container filled with a fluid of density L. The body is shown completely immersed, but complete immersion is not essential to the proof. To begin with, imagine the situation before the body was immersed. The region now occupied by the body was filled with fluid, whose weight was VL g . Because the fluid as a whole was in hydrostatic

V equilibrium, the net upwards force (due to difference in pressure at different depths) on the fluid in that region was equal to the weight of the fluid occuping that region. Now, consider what happens when the body has displaced the fluid. The pressure at every point on the surface of the body is unchanged from the value at the same location when the body was not present. This is because the pressure at any point depends only on the depth of that point below the fluid surface. Hence, the net force exerted by the surrounding fluid on the body is exactly the same as the exerted on the region before the body was present. But we know the latter to be VL g , the weight of the displaced fluid. Hence, this must also be the buoyant force exerted on the body. Archimedes' principle is thus, proved. Ex.5

Beaker cicular cross-section of radius 4 cm is filled with mercury upto a height of 10 cm. Find the force exerted by the mercury on the bottom of the beaker. The atmopheric pressure = 105 N/m2. Density of mercury = 13600 kg/m3. Take g = 10 m/s2

Sol.

The pressure at the surface = atmospheric pressure = 105 N/m2. The pressure at the bottom = 105 N/m2 + hg kg   = 105 N/m2 + (0.1 m)  13600 3  m

m   10 2   s 

= 105 N/m2 + 13600 N/m2 = 1.136 × 105 N/m2 The force exerted by the mercury on the bottom = (1.136 × 105 N/m2) × (3.14 × 0.04 m × 004 m) = 571 N Ex.6

A cubical block of iron 5 cm on each side is floating on mercury in a vessel. (i) What is the height of the block above mercury level ? (ii) What is poured in the vessel until it just covers the iron block. What is the height of water column. Density of mercury = 13.6 gm/cm3 Density of iron 7.2 gm/cm3

Sol.

Case-I : Suppose h be the height of cubical block of iron above mercury. Volume of iron block = 5 ×5 × 5 = 125 cm3 Mass of iron block = 125 × 7.2 = 900 gm Volume of mercury displaced by the block = 5 × 5 × (5 – h) cm3 Mass of mercury displaced

= 5 × 5 (5 – h) × 13.6 gm

By the law floatation, weight of mercury displaced = weight of iron block 5 × 5 (5 – h) × 13.6 = 900 or

(5 – h) =

900 = 2.65  25  13.6

h = 5 – 2.65 = 2.35 cm

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Page # 12

FLUID

Case - II : Suppose in this case height of iron block in water be x. The height of iron block in mercury will be (5 – x) cm. Water

h

Mercury

Mercury

(a)

(b)

Mass of the water displaced = 5 × 5 × (x) × 1 Mass of mercury displaced = 5 × 5 × (5 – x) × 13.6 So, weight of water displaced + weight of mercury displaced = weight of iron block or 5 × 5 × x × 1 + 5 × 5 × (5 – x) × 13.6 = 900 or x = (5 – x) × 13.6 = 36 x = 2.54 cm Ex.7 A tank contianing water is placed on spring balanced. A stone of weight w is hung and lowered into the water without touching the sides and the bottom of the tank. Explain how the reading will change. Sol. The situation is shown in figure. Make free-body diagrams of the bodies separately and consider their equilibrium. Like all other forces, buoyancy is also exerted equally on the two bodies in contact. Hence it the water exerts a buoyant force, say, B on the stone upward, the stone exerts the same force on the water downward. The forces acting on the 'water + container' system are : W, weight of the system downward, B, buoyant force of the stone downard, and the force R of the spring in the upward direction. For equilibrium R=W+B Thus the reading of the spring scale will increase by an amount equal to the weight of the liquid displaced, that is, by an amount equal to the buoyant force. Ex.8

Sol.

A cylindrical vessel containing a liquid is closed by a smooth piston of mass m as shown in figure. The area of cross-section of the piston is A. If the atmopheric pressure is P0, find the pressure of the liquid just below the prism. Let the pressure of the liquid just below the piston be P. The forces acting on the piston are (a) its weight, mg (downward) (b) force due to the air above it, P0A(downward) (c) force due to the liquid below it, PA (upward) If the piston is in equilibrium PA = P0A + mg or P = P0 +

mg A

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FLUID

Ex.9 A rubber ball of mass m and radius r is submerged in water to a depth h released. What height will the ball jump up to above the surface of the water ? Neglect the resistance of water and air. Sol. Let the ball go up by x above the level of water. Let us now consider energy conservation between the initial and final positions. In both the positions kinetic energy of the body is zero. The potential energy in the first position with reference to the water 4 3  level is – mgh plus the work done by an external agent against the buoyant force which is  r g h, 3

where  is the density of the water or

4 3  –mgh +  r g h = mgx  3

x

( 4 / 3)r 3  – m h m

Ex.10 A cube of wood supporting a 200 g mass just floats in water. When the mass is removed, the cube rises by 2 cm. What is the size of the cube ? Sol. If, l = side of cube, h = height of cube above water and  = density of wood. Mass of the cube = l3  Volume of cube in water = l2 (l – h) Volume of the displaced water

= l 2 ( l – h)

As the tube is floating weight of cube + weight of wood = weight of liquid displaced or

l3 + 200 = l2 (l – h)

...(2.10)

After the removal of 200 gm mass, the cube rises 2 cm. = l2 × {l – (h + 2)} Volume of cube in water or

l2 × {l – (h + 2)} = l3 

...(2.11)

Substituting the value of l  from equation (2.11) in equation (2.10), we get 3

l2 × {l – (h + 2)} + 200 = l2 (l – h) or

l3 – l2h – 2l2 + 200 = l3 – l2 2l2 = 200  l = 10 cm

Ex.11 A boat floating in water tank is carrying a number of large stones. If the stones were unloaded into water, what will happen to water level ? Given the reason in brief. Sol.

Suppose W and w be the weights of the boat and stones respectively. First, we consider that the boat is floating. It will displaced (W + w) × 1 cm3 of water. Thus displaced water = (W + w) cm3

[As density of water = 1 gm/cm3]

Secondly, we consider that the stones are unloaded into water. Now the boat displaces only W × 1 cm3 of water. If  be the density of stones, the volume of water displaced by stones = w/ cm3 As  > 1, hence w/ < w, thus we have Now

(W + w/) < (W + w)

This shows that the volume of water displaced in the second case is less than the volume of water displaced in the first case. Hence the level of water will come down.

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Page # 14

FLUID

Ex.12 Two solid uniform spheres each of radius 5 cm are connected by a light string and totally immersed in a tank of water. If the specific gravities of the sphere are 0.5 and 2, find the tension in the string and the contact force between the bottom of tank and the heavier sphere. Sol. The situation is shown in figure Let the volume of each sphere be V m3 and density of water be  kg/m3. Upward thrust on heavier sphere = v  g Weight of the heavier sphere = V × 2 × g For heavier sphere, T T+R+Vg=V×2×g ...(2.12) where R is the reaction at the bottom. Similarly for lighter sphere T + V × 0.5 ×  g = V  g ..(2.13) R Subtracting equation (2.13) from equation (2.12), we have R + 0.5 V  g = V  g ...(2.14) or R = 0.5 V  g ...(2.15) From equation (2.13) T = 0.5 V  g 4   0.5    3.14  5 3  10 6  × 1000 × 9.8 = 2.565 N 3  R = 2.565 N Similarly A rod of length 6 m has a mass of 12 kg. If it is hinged at one end at a distance of 3 m below a water surface, (i) What weight must be attached to other end of the rod so that 5 m of the rod is submerged ? (ii) Find the magnitude and direction of the force exerted by the hinge on the rod. The specific gravity of the material of the rod is 0.5. Sol.

Let AC be the submerged part of the rod AB hinged at A

B

FB

as shown in figure. G is the centre of gravity of the rod and G is the centre of buoyancy through which force of buoyancy FB acts vertically upwards.

Water Surface

C G

R

Since the rod is uniform,

w x

G' The weight of part AC will be

5  12  10kg 6

The buoyance force on rod at G is FB 

A [Because AB = 6 m and W AC = 5 m] Hinge

10 = 20 kg weight 0.5

(i) Let x be weight attached at the end B. Balancing torques about A, we get W × AG + x × AB = FB × AG 12 + 3 + x × 6 = 20 × (5/2) [As AG = 5/2] Solving we get x = 2.33 kg (ii) Suppose R be the upward reaction on the hinge, then in equilibrium position, we have W + x = FB + R or R = W + x – FB = 12 + 2.33 – 20 = – 5.67 kg. wt. Negative sign shows that the reaction at the hinge is acting in the downward direction. The magnitude of the reaction is 5.67 kg. wt.

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Page # 15

FLUID

Ex.13 A cylinder of area 300 cm2 and length 10 cm made of material of speicifc gravity 0.8 is floated in water with its axis vertical. It is then pushed downward, so as to be just immersed. Calculate the work done by the agent who pushes the cylinder into the water. Sol. Weight of the cylinder = (300 × 10–4) × (10 × 10–2) × 800 kgf = 2.4 kgf Let x be the length of the cylinder inside the water. Then by the law of floatation 2.4 g = (300 × 10–4 x) × 1000 g or

x = 0.08 m

When completely immersed, Fb(buoyant force) = (300 × 10–4 × 0.1) × 1000 × g = 3 g N Thus to immerse the cylinder inside the water the external agent has to push it by 0.02 m against average upward thrust. Increase in upward thrust = 3g – 2.4 g = 0.6 g N Since this increase in upthrust takes place gradually from 0 to 0.6 g, we may take the average upthrust against which work is done as 0.3 g N. work done = 0.3 g × 0.02 = 0.0588 J Ex.14 A piece of an alloy of mass 96 gm is composed of two metals whose specific gravities are 11.4 and 7.4. If the weight of the alloy is 86 gm in water, find the mass of each metal in the alloy. Sol. Suppose the mass of the metal of specific gravity 11.4 be m and the mass of the second metal of specific gravity 7.4 will be (96–m) m cm 3 114 .

Volume of first metal =

96 – m cm 3 7.4

Volume of second metal =

Total volume =

Buoyancy force in water

m 96 – m  114 . 7.4

96 – m   m    gm weight  114 . 7.4 

 m  (96 – m)    Apparent wt. in water = 96 –  114 7.4   .  According to the given problem,  m  ( 96 – m)  96 –     86 .  7.4   114

or

m ( 96 – m)   10 114 . 7.4

Solving we get,

m = 62.7 gm

Thus mass of second metal is

= 96 – 62.7 = 33.3 gm

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Page # 16

FLUID

NOTE :ASSUMPTIONS OF IDEAL FLUID (1) Fluid is incompressible : density of fluid remain constant through out the fluid. (2) Fluid is non-viscous : fluid friction is absent (3) Doesn't show rotational effect : If we release any body in the flowing section there it will not rotate about its C.O.M. (4) Stream line flow : velocity of fluid at any particular point remains constant with time It may vary with position. 5.

EQUATION OF CONTINUITY

A2 This equation defines the steady flow of fluid in a tube. It states that if flow of a fluid is a steady then the mass of fluid entering per second at one end is equal to the mass of fluid leaving per second

v2

at the other end. Figure shown a section of a tube in which at the ends, the cross

A1

sectional area are A1 and A2 and the velocity of the fluid are V1 and v2 respectivley. According to the equation of continuity, if flow is steady mass of fluid entering at end A1 per second = mass of fluid leaving the end A2 per second. dV  A1v1 dt Hence mass entering per second at A1 is = A1 v1  Similarly mass leaving per second at A2 is = A2 v2  According to the definition of steady flow A1 v1  = A2 v2  or

A1v1 = A2v2

Equation above in known as equation of continuity, which gives that in steady flow the product of cross-section and the speed of fluid everywhere remains constant.

5.1

Freely Falling Liquid When liquid falls freely under gravity, the area of cross section of the stream continuously decreases, as the velocity inreases. For example, we consider water coming out from a tap, as shown in figure. Let its speed near the mouth of tap is v0 and at a depth h it is v, then we have

v 2  v 20  2gh If cross section of tap is A then according to the equation of continuity, the cross section at point M (say a) can be given as v 0 A  a v 20  2gh

or

a

v0 A v 20

 2gh

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Page # 17

FLUID 6.

BERNOULLIS EQUATION It relates the variables describing the steady laminer of liquid. It is based on energy conservation. Assumptions The fluid is incompressible, non-viscous, non rotational and streamline flow. dm

T

T A2 V2 at time t = 0

P2

at time t + dt

P2

A1

A1 S P1

A2

dx2

h2

S

P1

V1 h1

dx1

Mass of the fluid entering from side S dm1 =  A1 dx1 =  dv1 The work done in this displacement dx1 at point S is WP = F1dx1 = P1A1dx1 1

WP = P1dV1

{ A1dx1 = dV1}

1

At the same time the amount of fluid moves out of the tube at point T is dm2 = dV2 According to equation of coutniuity

dm1 dm 2  dV1 = dV2 = dv dt dt

The work done in the displacement of dm2 mass at point T WP = P2dV2 2

Now applying work energy theorem. 1  1  2 2 WP + WP = (Kf + Uf) – (Ki + Ui)  P1dV – P2dV =  dV  2  dVgh 2  –  dV 1  dVgh 1  1 2 2  2  1 1 P1 – P 2 = V22 + gh2 – V12 + gh1 2 2

P1 + gh1 +

1 1 1 V12 = P2 + gh2 + V2  P + gh + V2 = constant 2 2 2

P 1 v2 h = constant g 2 g P where = pressure head g

h = Gravitational head 1 v2 = volume head 2 g

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Page # 18 6.1

FLUID

APPLICATION OF BERNOULLIS PRINCIPLE Magnus effect : When a spinning ball is thrown, it deviates from its usual path in flight. This effect is called Magnus effect and plays an important role in tennis, cricket and soccer, etc., as by applying appropriate spin the moving ball can be made to curve in any desired direction. If a ball is moving from left to right and also spinning about a horizontal axis perpendicular to the direction of motion as shown in figure, then relative to the ball air will be moving from right to left. V+r

V O

Vertical plane Curved path Usual path

v

V A0 A 1

(A)

(B)

(C)

The resultant velocity of air above the ball will be (V+r) while below it (V –r) (shown figure). So in accordance with Bernoulli's principle pressure above the ball will be less than below it. Due to this difference of pressure an upward force will act on the ball and hence the ball will deviate from its

usual path OA0 and will hit the ground at A1 following the path OA1 (figure shown) i.e., if a ball is thrown with back spin, the pitch will curve less sharply prolonging the flight. Similarly if the spin is clockwise, i.e., the ball is thrown with top-spin, the force due to pressure difference will act in the direction of gravity and so the pitch will curve more sharply shortening the flight. A1 A0 A0

A2

Horizontal plane

Furthermore, if the ball is spinning about a vertical axis, the curving will be sideays as shown in figure. producing the so called out swing or in swing. Action of Atomiser : The action of aspirator, carburettor, paint-gun, scent-spray or insect-sprayer is based on Bernoulli's principle. In all these by means of motion of a piston P in a cylinder C high speed air is passed over a tube T dipped in liquid L to be sprayed. High speed air creates low pressure over the tube due to which liquid (paint, scent, insecticide or petrol) rises in it and is then blown off in very small droplets with expelled air.

p

T c

L

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Page # 19

FLUID

Working of Aeroplane : This is also based on Bernouilli's principle. The wings of the aeroplane are having tapering as shown in figure. Due to this specific shape of wings when the aeroplane runs, air passes at higher speed over it as compared to its lower surface. This difference of air speeds above and below the wings, in accordance with Bernoulli's principle, creates a pressure difference, due to which an upward force called ' dynamic lift' ( = pressure difference × area of wing) acts on the plane. If this force becomes greater than the weight of the plane, the plane will rise up.

v large, p small

v small, p large

A

Ex.15 If pressure and velocity at point A is P1 and V1 respectively & at point B is P2, V2 is the figure as shown. Comment on P1 and P2. Sol.

P1

From equation of continuity A1V1 = A2V2

V1

B V2

here A1 > A2  V1 < V2

....(1)

from Bernoulli's equation. We can write 1 P1 + V12 = P2 + 1/2 V22 = 2 after using equation (1) P1 > P2 TORRICIELLI'S LAW OF EFFLUX (VELOCITY OF EFFLUX) Crossectional Area of hole at A is greater than B If water is come in tank with velocity vA and going out side with velocity vB then A1 vA = A2vB  A1 > A2  vB >> vA on applying bernoulli theorem at A and B PA  ghA 

P0

h v

H B

1 2 1 v A  PB  ghB  vB2 2 2

PA = P B = P 0 and hA – hB = h 1 2 2  gh = (vB – v A ) 2 1 gh =  v2 2 v=

A

H–h

•

[vB >> vA] [vB2 – vA2 = v2]

2 gh

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Page # 20

FLUID

Range (R) Let us find the range R on the ground. Considering the vertical motion of the liquid. (H – h) =

1 2 gt 2

t

or

2(H – h) g

Now, considering the horizontal motion,

 2(H – h)    R = vt or R  ( 2gh ) or R  2 h(H – h)  g   From the expression of R, following conclusions can be drawn, (i) Rh = R H – h as R h  2 h(H – h) and R H – h  2 (H – h)h This can be shown as in Figure

h

H–h H

H and Rmax = H. (ii) R is maximum at h  2 Proof : R2 = 4 (Hh – h2)

v

O h

H–h

dR 2 0 dh or H – 2h = 0 or h = H/2 That is, R is maximum at h=H/2 For R to be maximum.

H H H –   H Proved 2 2 Ex.16 A cylindrical dark 1 m in radius rests on a platform 5 m high. Initially the tank is filled with water up to a height of 5 m. A plug whose area is 10–4 m2 is removed from an orifice on the side of the tank at the bottom Calculate (a) initial speed with which the water flows from the orifice (b) initial speed with which the water strikes the ground and (c) time taken to empty the tank to half its original volume (d) Does the time to be emptied the tank depend upon the height of stand. Sol. The situation is shown in figure (a) As speed of flow is given by A vH = ( 2gh) R max  2

and

or

=

2  10  5

~ –

5m

10 m/s

A0

(b) As initial vertical velocity of water is zero, so its vertical velocity when it hits the ground v V  2gh =

2  10  5

~ –

10 m/s

5m

So the initial speed with which water strikes the ground. v  vH2  v 2V = 10 2 = 14.1 m/s (c) When the height of water level above the hole is y, velocity of flow will be v  2gy and so rate of flow dV  A 0 v  A 0 2gy dt

or

–Ady = ( 2gy ) A0 dt

[As dV = – A dy]

Which on integration improper limits gives 0

t

Ady

A 2   12 [ H – H'] So t  A0 g 2gy 0 10 –4 H (d) No, as expression of t is independent of height of stand.







A 0 dt  t 

2 [ 5 – (5 / 2)] = 9.2 × 103s 10

~ –

2.5 h

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Page # 21

FLUID 6.1

Venturimeter Figure shows a venturimeter used to measure flow speed in a pipe of non - uniform cross-section. We apply Bernoulli's equation to the wide (point 1) and narrow (point 2) parts of the pipe, with h1 = h2 1 1 P0 P0 P1  v 12  P2  v 22 2 2 h A 1v 1 From the continuity equation v2 = A 2 v2 Substituting and rearranging, we get A v1 H 2 A2 A  1 P1 – P2   v 12  12 – 1 2  A2  Because A1 is greater than A2, v2 is greater than v1 and hence the pressure P2 is less than P1. A net force to the right acceleration the fluid as it enters the narrow part of the tube (called throat) and a net force to the left slows as it leaves. The pressure difference is also equal to  gh, where h is the difference in liquid level in the two tubes. Substituting in Eq. (i), we get v1 

2gh 2

 A1    – 1  A2  6.2

Pitot Tube It is a device used to measure flow velocity of fluid. It is a U shaped tube which can be inserted in a tube or in the fluid flowing space as shown in vg figure shown. In the U tube a liquid which is immiscible with the fluid is filled upto a level C and the short opening M is placed in the fluid flowing space against the flow so that few of the fluid particles entered into the tube and B A h exert a pressure on the liquid in limb A of U tube. Due to this the liquid level changes as shown in figure shown. At end B fluid is freely flowing, which exert approximately negligible pressure on this liquid. The pressure difference at ends A and B can be given by measuring the liquid level difference h as It is a gas, then PA – PB = hg It if the a liquid of density , then PA – PB = h( – g)g Now if we apply Bernoulli's equation at ends A and B we'l have 1 2 1 2 v g  PA – PB  hg 0 + 0 + PA = v g + 0 + PB or 2 2 Now by using equations, we can evaluate the velocity v, with which the fluid is flowing. Note : Pitot tube is also used to measure velocity of aeroplanes with respect to wind. It can be mounted at the top surface of the plain and hence the velocity of wind can be measured with respect to plane.

6.3

SIPHON : It is a pipe used to drain liquid at a lower height but the pipe initially rises and then comes down let velocity of outflow is v & the pipe is of uniform cross-section A. Applying bernoulli's equation between P (top of tank) & R (opening of pipe) we get

Q h2

P h=0 h1

R V

1 1 1 2 (P + gh + v2)P = (P + gh + v2)R  P + 0 + 0 = P0 – gh1 + v 2 2 2 here velocity is considered zero at P since area of tank is very large compared to area of pipe  v  2gh Naturally for siphon to work h1 > 0 Now as area of pipe is constant so by equation of continuity as Av = constant so velcoity of flow inside siphon is also constant between Q & R (P + gh + 1/2 v2)Q = (P + gh + 1/2 v2)R  PQ + gh2 = P0 – gh1 (v is same)  PQ = P0 – g (h1 + h2) as PQ = 0  P0  g (h1 + h2) means (h1 + h2) should not be more than P/g for siphon to work

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Page # 22

FLUID

(ONLY ONE OPTION IS CORRECT)

Exercise - I

1. STATIC FLUID 1. A bucket contains water filled upto a height = 15 cm. The bucket is tied to a rope which is passed over a frictionless light pulley and the other end of the rope is tied to a weight of mass which is half of that of the (bucket + water). The water pressure above atmosphere pressure at the bottom is (A) 0.5 kPa (B) 1 kPa (C) 5 kPa (D) None 2. Some liquid is filled in a cylindrical vessel of radius R. Let F1 be the force applied by the liquid on the bottom of the cylinder. Now the same liquid is poured into a vessel of uniform square cross-section of side R. Let F2 be the force applied by the liquid on the bottom of this new vessel. Then : F2 (A) F1 = F2 (B) F1 = (C) F1 = F2 (D) F1 = F2  3. A liquid of mass 1 kg is filled in a flask as shown in figure. The force exerted by the flask on the liquid is (g = 10 m/s2) [Neglect atmospheric pressure] (A) 10 N (B) greater than 10 N (C) less than 10 N (D) zero 4. A U-tube having horizontal arm of length 20 cm, has uniform cross-sectional area = 1 cm2. It is filled with water of volume 60 cc. What volume of a liquid of density 4 g/cc should be poured from one side into the U-tube so that no water is left in the horizontal arm of the tube? (A) 60 cc (B) 45 cc (C) 50 cc (D) 35 cc 5 . In the figure shown, the heavy cylinder (radius R)

resting on a smooth surface separates two liquids of densities 2 and 3. The height ‘h’ for the equilibrium of cylinder must be 3 R h 2 R 3 (C) R 2 (D) None 2 6. A light semi cylindrical gate of radius R is piovted at its mid point O, of the diameter as shown in the figure holding liquid of density . The force F required to prevent the rotation of the gate is equal t

(A) 3R/2

7. The pressure at the bottom of a tank of water is 3P where P is the atmospheric pressure. If the water is drawn out till the level of water is lowered by one fifth., the pressure at the bottom of the tank will now be (A) 2P (B) (13/5) P (C) (8/5) P (D) (4/5) P 8. An open-ended U-tube of uniform cross-sectional area contains water (density 1.0 gram/centimeter3) standing initially 20 centimeters from the bottom in each arm. An immiscible liquid of density 4.0 grams/ centimeter 3 is added to one arm until a layer 5 centimeters high forms, as shown in the figure above. What is the ratio h2/h1 of the heights of the liquid in the two arms ? 5cm h2

h1

(A) 3/1

(B) 5/2

(C) 2/1

(D) 3/2

9. The vertical limbs of a U shaped tube are filled with a liquid of density  upto a height h on each side. The horizontal portion of the U tube having length 2h contains a liquid of density 2. The U tube is moved horizontally with an accelerator g/2 parallel to the horizontal arm. The difference in heights in liquid levels in the two vertical limbs, at steady state will be (A) 2h/7 (B) 8h/7 (C) 4h/7 (D) None 10. The area of cross-section of the wider tube shown in figure is 800 cm2. If a mass of 12 kg is placed on the massless piston, the difference in heights h in the level of water in the two tubes is : 12kg

h

(B) R

(A) 10 cm

(B) 6 cm

(C) 15 cm

(D) 2 cm

2. ACCELERATED FLUID 11. A fluid container is containing a liquid of density  is is accelerating upward with acceleration a along the inclined place of inclination  as shwon. Then the angle of inclination  of free surface is :

O

a 

R F

(A) 2R3g (C)

2R2 lg 3



(B) 2gR3l (D) none of these



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Page # 23

FLUID

 g  (A) tan    g cos  

–1  a  g sin   (B) tan    g cos  

–1  a – g sin   (C) tan    g(1  cos ) 

–1  a – g sin   (D) tan    g(1 – cos ) 

–1

12. Figure shows a three arm tube in which a liquid is filled upto levels of height l. It is now rotated at an angular frequency  about an axis passing through arm B. The angular frequency  at which level of liquid of arm B becomes zero.

B

A

 l

2g 3l

(B)

g l

(A) 2

abc g

(B) 2

g da

(C) 2

bc da (D) 2 dg g

l

l

(A)

17. A cuboidal piece of wood has dimensions a, b and c. Its relative density is d. It is floating in a larger body of water such that side a is vertical. It is pushed down a bit and released. The time period of SHM executed by it is :

18. A slender homogeneous rod of length 2L floats partly immersed in water, being supported by a string fastened to one of its ends, as shown. The specific gravity of the rod is 0.75. The length of rod that extends out of water is

C

l

are connected by weightless wire and placed in a large tank of water. Under equilibrium the lighter cube will project above the water surface to a height of (A) 50 cm (B) 25 cm (C) 10 cm (D) zero

3g l

(C)

(D)

3g 2l

13. An open cubical tank was initially fully filled with water. When the tank was accelerated on a horizontal plane along one of its side it was found that one third of volume of water spilled out. The acceleration was (A) g/3 (B) 2g/3 (C) 3g/2 (D) None 3. PASCAL'S LAW & ARCHIMEDE'S PRINCIPLE 14. A cone of radius R and height H, is hanging inside a liquid of density  by means of a string as shown in the figure. The force, due to the liquid acting on the slant surface of the cone is (neglect atmospheric pressure)

2L

(A) L

(B)

1 L 2

(C)

1 L 4

(D) 3 L

19. A dumbbell is placed in water of density . It is observed that by attaching a mass m to the rod, the dumbbell floats with the rod horizontal on the surface of water and each sphere exactly half submerged as shown in the figure. The volume of the mass m is negligible. The value of length l is l 2M,V

M,V m Water

H

d R

(A)gHR2

(B) HR2

4 2 (C) gHR2 (D) gHR2 3 3

15. A heavy hollow cone of radius R and height h is placed on a horizontal table surface, with its flat base on the table. The whole volume inside the cone is filled with water of density . The circular rim of the cone’s base has a watertight seal with the table’s surface and the top apex of the cone has a small hole. Neglecting atmospheric pressure find the total upward force exerted by water on the cone is (A) (2/3)R2 hg (B) (1/3)R2 hg (C) R2 hg (D) None 16. Two cubes of size 1.0 m sides, one of relative density 0.60 and another of relative density = 1.15

d(V  3M) (A) 2(V  2M) 

d(V  2M) (B) 2(V  3M) 

d(V  2M) (C) 2(V  3M) 

d(V  2M) (D) 2(V  3M) 

20. Two bodies having volumes V and 2V are suspended from the two arms of a common balance and they are found to balance each other. If larger body is immersed in oil (density d1 = 0.9 gm/cm3) and the smaller body is immersed in an unknown liquid, then the balance remain in equilibrium. The density of unknown liquid is given by : (A) 2.4 gm/cm3 (B) 1.8 gm/cm3 3 (C) 0.45 gm/cm (D) 2.7 gm/cm3

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FLUID

21. A container of large surface area is filled with liquid of density . A cubical block of side edge a and mass M is floating in it with four-fifth of its volume submerged. If a coin of mass m is placed gently on the top surface of the block is just submerged. M is (A) 4m/5 (B) m/5 (C) 4m (D) 5m 22. A boy carries a fish in one hand and a bucket (not full) of water in the other hand. If the places the fish in the bucket, the weight now carried by him (assume that water does not spill) : (A) is less than before (B) is more than before (C) is the same as before (D) depends upon his speed 23. A cork of density 0.5 gcm–3 floats on a calm swimming pool. The fraction of the cork’s volume which is under water is (A) 0% (B) 25% (C) 10% (D) 50% 24. Two cyllinders of same cross-section and length L but made of two material of densities d1 and d2 are cemented together to form a cylinder of length 2L. The combination floats in a liquid of density d with a length L/2 above the surface of the liquid. If d1 > d2 then : 3 d d  d1 (A) d1  d (B)  d1 (C) (D) d < d1 4 2 4 25. A piece of steel has a weight W in air, W1 when completely immersed in water and W2 when completely immersed in an unknown liquid. The relative density (specific gravity) of liquid is : W  W1 (A) W  W 2

W  W2 (B) W  W 1

W1  W2 W1  W2 (C) W  W (D) W  W 1 2

26. A ball of relative density 0.8 falls into water from a height of 2m. The depth to which the ball will sink is (neglect viscous forces) : (A) 8m (B) 2m (C) 6m (D) 4m 27. A small wooden ball of density  is immersed in water of density  to depth h and then released. The height H above the surface of water up to which the ball will jump out of water is (A)

h 

  (B)   1  h  

(C) h

(D) zero

28. A hollow sphere of mass M and radius r is immersed in a tank of water (density w). The sphere would float if it were set free. The sphere is tied to the bottom of the tank by two wires which makes angle 45º with the horizontal as shown in the figure. The tension T1 in the wire is :

R M

T1 45º 45º

4 R 3w g  Mg 2 3 3 (A) (B) R wg  Mg 3 2 4 3 R w g  Mg 4 3 (C) 3 (D) R w g  Mg 3 2 29. A metal ball of density 7800 kg/m3 is suspected to have a large number of cavities. It weighs 9.8 kg when weighed directly on a balance and 1.5 kg less when immersed in water. The fraction by volume of the cavities in the metal ball is approximately : (A) 20% (B) 30% (C) 16% (D) 11%

30. A sphere of radius R and made of material of relative density  has a concentric cavity of radius r. It just floats when placed in a tank full of water. The value of the ratio R/r will be 1/3

   (A)     1

   1  (B)    

1/ 3

   1  (C)    

1/ 3

   1  (D)     1

1/ 3

31. A body having volume V and density  is attached to the bottom of a container as shown. Density of the liquid is d(>). Container has a constant upward acceleration a. Tension in the string is

a

(A) V[Dg – (g + a)] (C) V (d – ) g

(B) V (g + a) (d – ) (D) none

32. A hollow cone floats with its axis vertical upto one-third of its height in a liquid of relative density 0.8 and with its vertex submerged. When another liquid of relative density  is filled in it upto one-third of its height, the cone floats upto half its vertical height. The height of the cone is 0.10 m and radius of the circular base is 0.05 m. The specific gravity  is given by (A) 1.0 (B) 1.5 (C) 2.1 (D) 1.9 33. A beaker containing water is placed on the platform of a spring balance. The balance reads 1.5 kg. A stone of mass 0.5 kg and density 500 kg/m3 is immersed in water without touching the walls of beaker. What will be the balance reading now ? (A) 2 kg (B) 2.5 kg (C) 1 kg (D) 3 kg 34. There is a metal cube inside a block of ice which is floating on the surface of water. The ice melts completely and metal falls in the water. Water level in the container (A) Rises (B) Falls (C) Remains same (D) Nothing can be concluded

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FLUID 35. A uniform solid cylinder of density 0.8 g/cm3 floats in equilibrium in a combination of two non-mixing liquid A and B with its axis vertical. The densities of liquid A and B are 0.7 g/cm3 and 1.2 gm/cm3. The height of liquid A is hA = 1.2 cm and the length of the part of cylinder immersed in liquid B is hB = 0.8 cm. Then the length of the cylinder in air is (A) 0.21 m (B) 0.25 cm (C) 0.35 cm (D) 0.4 cm 36. A cylindrical block of area of cross-section A and of material of density  is placed in a liquid of density one-third of density of block. The block compresses a spring and compression in the spring is one-third of the length of the block. If acceleration due to gravity is g, the spring constant of the spring is

P

P

(C)

(D) x

x

41. A cylindrical tank of height 1 m and cross section area A = 4000 cm2 is initially empty when it is kept under a tap of cross sectional area 1 cm2. Water starts flowing from the tap at t = 0, with a speed = 2 m/s. There is a small hole in the base of the tank of crosssectional area 0.5 cm2. The variation of height of water in tank (in meters) with time t is best depicted by h

h

1

0.8

(A)

(B) t

O

O

h

(A) Ag

(B) 2Ag

(C) 2Ag/3

(D) Ag/3

37. A body of density ’ is dropped from rest at a height h into a lake of density , where  > . Neglecting all disipative froces, calculate the maximum depth to which the body sinks before returning of float on the surface. h h' h' h (A)  – ' (B)  (C)  – ' (D)  – ' 4. FLUID FLOW & BERNOULLI'S PRINCIPLE 38. A rectangular tank is placed on a horizontal ground and is filled with water to a height H above the base. A small hole is made on one vertical side at a depth D below the level of the water in the tank. The distance x from the bottom of the tank at which the water jet from the tank will hit the ground is 1 DH (A) 2 D(H  D) (B) 2 DH (C) 2 D (H  D ) (D) 2 39.A jet of water with cross section of 6 cm2 strikes a wall at an angle of 60º to the normal and rebounds elastically from the wall without losing energy. If the velocity of the water in the jet is 12 m/s, the force acting on the wall is (A) 0.864 Nt (B) 86.4 Nt (C) 72 Nt (D) 7.2 Nt 40. The cross sectional area of a horizontal tube increases along its length linearly, as we move in the direction of flow. The variation of pressure, as we move along its length in the direction of flow (xdirection), is best depicted by which of the following graphs P

P

(A)

(B) x

x

t

h

1

0.8

(C)

(D) t

O

O

t

42. A cubical box of wine has a small spout located in one of the bottom corners. When the box is full and placed on a level surface, opening the spout results in a flow of wine with a initial speed of v0 (see figure). When the box is half empty, someone tilts it at 45º so that the spout is at the lowest point (see figure). When the spout is opened the wine will flow out with a speed of

V0

(A) v0

(B) v0/2

(C) v 0 / 2

(D) v 0 / 4 2

43. Water is flowing steadily through a horizontal tube of non uniform cross-section. If the pressure of water is 4 × 104 N/m2 at a point where cross-section is 0.02 m2 and velocity of flow is 2 m/s, what is pressure at a point where cross-section reduces to 0.01 m2 (A) 1.4 × 104 N/m2 (B) 3.4 × 104 N/m2 (C) 2.4 × 10–4 N/m2 (D) none of these 44. A vertical cylindrical container of base area A and upper cross-section area A1 making an angle 30° with the horizontal is placed in an open rainy field as shown near another cylindrical container having same base area A. The ratio of rates of collection of water in the

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Page # 26

FLUID

two containers will be. p

A1 60º

60º

30º

A

(A) 2 / 3

p

(C)

A

(B) 4 / 3

(C) 2

(D) None

45. A tube is attached as shown in closed vessel containing water. The velocity of water coming out from a small hole is :

(D)

x x 50. A cylindrical vessel filled with water upto the height H becomes empty in time t0 due to a small hole at the bottom of the vessel. If water is filled to a height 4H it will flow out in time (A) t0 (B) 4t0 (C) 8t0 (D) 2t0

(B) 2 m/s 2m / s (C) depends on pressure of air inside vessel (D) None of these

51. A cylindrical vessel open at the open at the top is 20 cm high and 10 cm in diameter. A circular hole whose cross-sectional area 1 cm2 is cut at the centre of the bottom of the vessel. Water flows from a tube above it into the vessel at the rate 100 cm3s–1. The height of water in the vessel under state is (Take g = 1000 cms–2) (A) 20 cm (B) 15 cm (C) 10 cm (D) 5 cm

46. A large tank is filled with water to a height H. A small hole is made at the base of the tank. It takes T1 time to decrease the height of water to H/, ( > 1) and it takes T2 time to take out the rest of water. If T1 = T2, then the value of  is : (A) 2 (B) 3 (C) 4 (D) 2 2

52. A fire hydrant delivers water of density  at a volume rate L. The water travels vertically upward 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 the water on the corner of the hydrant is :

20cm

(A)

v

47. In the case of a fluid, Bernoulli’s theorem expresses the application of the principle of conservation of (A) linear momentum (B) energy (C) mass (D) angular momentum

v

(A) VL 48. Fountains usually seen in gardens are generated by a wide pipe with an enclosure at one end having many small holes. Consider one such fountain which is produced by a pipe of internal diameter 2 cm in which water flows at a rate 3ms–1. The enclosure has 100 holes each of diameter 0.05 cm. The velocity of water coming out of the holes is (in ms–1) : (A) 0.48 (B) 96 (C) 24 (D) 48 49. Water flows through a frictionless duct with a cross-section varying as shown in figure. Pressure p at points along the axis is represented by

p

p

(A)

(B) x

x

(B) zero

(C) 2VL

(D)

2 VL

53. A vertical tank, open at the top, is filled with a liquid and rests on a smooth horizontal surface. A small hole is opened at the centre of one side of the tank. The area of cross-section of the tank is N times the area of the hole, where N is a large number. Neglect mass of the tank itself. The initial acceleration of the tank is g g g g (A) (B) (C) (D) 2 N N 2 N 2N 54. Two water pipes P and Q having diameters 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 of that of Q (D) 1/4 times that of Q 55. Water flows into a cylindrical vessel of large crosssectional area at a rate of 10–4m3/s. It flows out from a hole of area 10–4 m2, which has been punched through the base. How high does the water rise in the vessel ? (A) 0.075 m (B) 0.051 m (C) 0.031 m (D) 0.025 m 56. A tank is filled up to a height 2H with a liquid and is placedon a platform of height H from the ground.

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Page # 27

FLUID The distance x from the ground where a small hole is punched to get the maximum range R is : (A) H (B) 1.25 H (C) 1.5 H (D) 2 H

20 m/s. The force on the pipe bend due to the turning of water is : (A) 565.7 N (B) 400 N (C) 20 N (D) 282.8 N

57. In a cylindrical vessel containing liquid of density , there are two holes in the side walls at heights of h1 and h2 respectively such that the range of efflux at the bottom of the vessel is same. The height of a hole, for which the range of efflux would be maximum will be.

63. A jet of water having velocity = 10 m/s and stream cross-section = 2 cm2 hits a flat plate perpendicularly, with the water splashing out parallel to plate. The plate experiences a force of (A) 40 N (B) 20 N (C) 8 N (D) 10 N 64. Equal volumes of two immiscible liquids of densities  and 2 are filled in a vessel as shown in figure. Two small holes are punched at depth h/2 and 3h/2 from the surface of lighter liquid. If v1 and v2 are the velocities of a flux at these two holes, then v1/v2 is :

(A) h2 – h1

(B) h2 + h1

(C)

h2 – h1 h2  h1 (D) 2 2

58. A large tank is filled with water (density = 103 kg/ m3). A small hole is made at a depth 10m below water surface. The range of water issuing out of the hole is Ron ground. What extra pressure must be applied on the water surface so that the range becomes 2R (take 1 atm = 105 Pa and g = 10 m/s2) :

R

(B) 4 atm

(C) 5 atm

1 (A)

(B) 2 2

v2

2

h

1 2

(C)

1 4

1 (D)

2

65. A horizontal pipe line carries water in a streamline flow. At a point along the tube where the crosssectional area is 10–2m2, the water velocity is 2 ms–1 and the pressure is 8000 Pa. The pressure of water at another point where the cross-sectional area is 0.5 × 10–2 m2 is : (A) 4000 Pa (B) 1000 Pa (C) 2000 Pa (D) 3000 Pa

10m

(A) 9 atm

v1

h

(D) 3 atm

59. A water barrel stands on a table of height h. If a small hole is punched in the side of the barrel at its base, it is found that the resultant stream of water strikes the ground at a horizontal distance R from the barrel. The depth of water in the barrel is (A) R/2 (B) R2/4h (C) R2/h (D) h/2 60. A cyclindrical vessel of cross-sectional area 1000 cm2, is fitted with a frictionless piston of mass 10 kg, and filled with water completely. A small hole of crosssectional area 10 mm2 is opened at a point 50 cm deep from the lower surface of the piston. The velocity of efflux from the hole will be (A) 10.5 m/s (B) 3.4 m/s (C) 0.8 m/s (D) 0.2 m/s 61. A laminar stream is flowing vertically down from a tap of cross-section area 1 cm2. At a distane 10 cm below the tap, the cross-section area of the stream has reduced to 1/2 cm2. The volumetric flow rate of water from the tap must be about (A) 2.2 litre/min (B) 4.9 litre/min (C) 0.5 litre/min (D) 7.6 litre/min 62. A horizontal right angle pipe bend has cross-sectional area = 10 cm2 and water flows through it at speed =

66. Water is pumped from a depth of 10 m and delivered through a pipe of cross section 10–2m2. If it is needed to deliver a volume of 10–1m3 per second the power required will be : (A) 10 kW (B) 9.8 kW (C) 15 kW (D) 4.9 kW 67. The three water filled tanks shown have the same volume and height. If small identical holes are punched near this bottom, which one will be the first to get empty.

(iii) (ii) (i) (A) (i) (B) (ii) (C) (iii) (D) All will take same time

68. A cylindrical vessel filled with water upto height of H stands on a horizontal plane. The side wall of the vessel has a plugged circular hole touching the bottom. The coefficient of friction between the bottom of vessel and plane is  and total mass of water plus vessel is M. What should be minimum diameter of hole so that the vessel begins to move on the floor if plug is removed (here density of water is )

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Page # 28

(A)

2M H

FLUID

(B)

M 2H

(C)

M H

(D) none

69. Which of the following is not an assumption for an ideal fluid flow for which Bernoulli’s principle is valid (A) Steady flow (B) Incompressible (C) Viscous (D) Irrotational ASSERTION - REASON 70. Statement - 1 : A helium filled balloon does not rise indefinitely in air but halts after a certain height. Statement - 2 : Viscosity opposes the motion of balloon Choose any one of the following four responses : (A) If both (A) and (R) are true and (R) is the correct explanation of (A) (B) If both (A) and (R) are true but (R) is not correct explanation of (A) (C) If both (A) if true but (R) is false (D) If (A) is false and (R) is true 71. Statement - 1 : When a body floats such that it's parts are immersed into two immiscible liquids force exerted by liquid - 1 is of magnitude 1v1g. Statement - 2 : Total Buogyant force = 1v1g + 2 v2 g

v1 v2

(A) Statement - 1 is true, statement - 2 is true and statement - 2 is correct explanation for statement - 1 (B) Statement - 1 is true, statement - 2 is true and statement - 2 is NOT the correct explanation for statement - 1 (C) Statement - 1 is true, statement - 2 is false. (D) Statement - 1 is false, statement - 2 is true. 72. Statement - 1 : When temperature rises the coefficient of viscosity of gases decreases. Statement - 2 : Gases behave more like ideal gases at higher temperature (A) Statement - 1 is true, statement - 2 is true and statement - 2 is correct explanation for statement - 1 (B) Statement - 1 is true, statement - 2 is true and statement - 2 is NOT the correct explanation for statement - 1 (C) Statement - 1 is true, statement - 2 is false. (D) Statement - 1 is false, statement - 2 is true.

73. Statement -1 : A partly filled test tube is floating in a liquid as shown. The tube will remain as atmospheric pressure changes. Statement - 2 : The buoyant force on a submerged object is independent of atmospheric pressure

(A) Statement - 1 is true, statement - 2 is true and statement - 2 is correct explanation for statement - 1 (B) Statement - 1 is true, statement - 2 is true and statement - 2 is NOT the correct explanation for statement - 1 (C) Statement - 1 is true, statement - 2 is false. (D) Statement - 1 is false, statement - 2 is true. 74. Statement -1 : Submarine sailors are advised that they should not allow it to rest on floor of the occen Statement - 2 : The force exerted by a liquid on a submerged body may be downwards. (A) Statement - 1 is true, statement - 2 is true and statement - 2 is correct explanation for statement - 1 (B) Statement - 1 is true, statement - 2 is true and statement - 2 is NOT the correct explanation for statement - 1 (C) Statement - 1 is true, statement - 2 is false. (D) Statement - 1 is false, statement - 2 is true. 75. Statement - 1 : The free surface of a liquid at rest with respect to stationary container is always  normal to the geff Statement - 2 : Liquids at rest cannot have shear stress. (A) Statement - 1 is true, statement - 2 is true and statement - 2 is correct explanation for statement - 1 (B) Statement - 1 is true, statement - 2 is true and statement - 2 is NOT the correct explanation for statement - 1 (C) Statement - 1 is true, statement - 2 is false. (D) Statement - 1 is false, statement - 2 is true.

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Page # 29

FLUID

Exercise - II 1. STATIC FLUID

4. The spring balance A reads 2

1. The vessel shown in the figure has two sections.

kg with a block m suspended from

The lower part is a rectangular vessel with area of cross-section A and height h. The upper part is a

A

it. A balance B reads 5 kg when a beaker with liquid is put on the pan of the balance. The two

conical vessel of height h with base area ‘A’ and top balances are now so arranged that area ‘a’ and the walls of the vessel are inclined at an

m

the hanging mass is inside the angle 30° with the vertical. A liquid of density  fills

liquid in the beaker as shown in

both the sections upto a height 2h. Neglecting atmo-

the figure in this situation :

spheric pressure.

(A) the balance A will read more than 2 kg

a 30°

B

(B) the balance B will read more than 5 kg

h

(C) the balance A will read less than 2 kg and B will read more than 5 kg

h

(D) the balances A and B will read 2 kg and 5 kg respectively

A (A) The force F exerted by the liquid on the base of ( A  a) the vessel is 2hg 2

4. FLUID FLOW & BERNOULLI'S PRINCIPLE 5. Figure shows a siphon. Choose the wrong statement : h1

A 2 (C) the weight of the liquid W is greater than the

2

h=0

(B) the pressure P at the base of the vessel is 2hg

h2

h3 3

force exerted by the liquid on the base (D) the walls of the vessel exert a downward force (A) Siphon works when h3 > 0

(F–W) on the liquid.

(B) Pressure at point 2 is P2 = P0 – gh3 2. ACCELERATED FLUID

(C) Pressure at point 3 is P0

2. A beaker is filled in with water is accelerated a m/s2

(D) None of the above

in +x direction. The surface of water shall make on

(P0 = atmospheric pressure)

angle (A) tan–1(a/g) backwards

(B) tan–1(a/g) forwards

6. A tank is filled upto a height h with a liq-

(C) cot–1(g/a) backwards

(D) cot–1(g/a) forwards

uid and is placed on a

3. PASCAL'S LAW & ARCHIMEDE'S PRINCIPLE

platform of height h

3. The weight of an empty balloon on a spring bal-

from the ground. To

ance is w1. The weight becomes w2 when the balloon

get maximum range xm

is filled with air. Let the weight of the air itself be w.

a

Neglect the thickness of the balloon when it is filled

punched at a distance

with air. Also neglect the difference in the densities of air inside & outside the balloon. Then : (A) w2 = w1

(B) w2 = w1 + w

(C) w2 < w1 + w

(D) w2 > w1

small

hole

is h

y

of y from the free surface of the liquid. h Then xm

(A) x m= 2h (B) xm=1.5 h (C) y = h (D) y = 0.75 h

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Page # 30

FLUID

7. Water coming out of a horizontal tube at a speed  strikes normally a vertically wall close to the mouth of the tube and falls down vertically after impact. When the speed of water is increased to 2. (A) the thrust exerted by the water on the wall will be

9. A steady flow of water passes along a horizontal tube from a wide section X to the narrower section Y, see figure. Manometers are placed at P and Q at the sections. Which of the statements A, B, C, D, E is most correct ?

doubled (B) the thrust exerted by the water on the wall will be four times

X

(C) the energy lost per second by water strikeup the

P

wall will also be four times

Y Q

(D) the energy lost per second by water striking the wall be increased eight times

(A) water velocity at X is greater than at Y

8. A cylindrical vessel is filled with a liquid up to height

(B) the manometer at P shows lower pressure than at Q

H. A small hole is made in the vessel at a distance y

(C) kinetic energy per m3 of water at X = kinetic en-

below the liquid surface as shown in figure. The liquid

ergy per m3 at Y

emerging from the hole strike the ground at distance x

(D) the manometer at P shows greater pressure than at Y

y

H x (A) if y is increased from zero to H, x will decrease

and then increase (B) x is maximum for y = H/2 (C) the maximum value of x is H/2 (D) the maximum value of x increases with the increases in density of the liquid

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Page # 31

FLUID

(SUBJECTIVE PROBLEMS)

Exercise - III

1. STATIC FLUID 1. A piston of mass M = 3 kg and radius R = 4 cm has a hole into which a thin pipe of radius r = 1 cm is inserted. The piston can enter a cylinder tightly and without friction, and initially it is at the bottom of the cylinder. 750 gm of water is now poured into the pipe so that the piston & pipe are lifted up as shown. Find the height H of water in the cylinder and height h of water in the pipe. (Neglect width of piston)

h

H

2. Compute the work which must be performed to slowly pump the water out of a hemispherical reservoir of radius R = 0.6 m. 3. A vertical uniform U tube open at both ends contains mercury. Water is poured in one limb until the level of mercury is depressed 2cm in that limb. What is the length of water column when this happens. 2. ACCELERATED FLUID 4. A spherical tank of 1.2m radius is half filled with oil of relative density 0.8. If the tank is given a horizontal acceleration of 10 m/s2. Calculate the inclination of the oil surface to horizontal and maximum pressure on the tank. 5. An open cubical tank completely filled with water is kept on a horizontal surface. Its acceleration is then slowly increased to 2m/s2 as shown in the fig. The side of the tank is 1m. Find the mass of water that would spill out of the tank.

1m

2

2m/s 1m

6. Find the speed of rotation of 1 m diameter tank, initially full of water such that water surface makes an angle of 45° with the horizontal at a radius of 30 cm. What is the slope of the surface at the wall of the tank.

3. PASCAL'S LAW & ARCHIMEDE'S PRINCIPLE 7. A solid ball of density half that of water falls freely under gravity from a height of 19.6 m and then enter water. Upto what depth will the ball go ? How much time will it take to come again to the water surface ? Neglect air resistance & velocity effects in water. 8. Place a glass beaker, partially filled with water, in a sink. The beaker has a mass 390 gm and an interior volume of 500 cm3. You now start to fill the sink with water and you find, by experiment, that if the beaker is less than half full, it will float; but if it is more than half full, it remains on the bottom of the sink as the water rises to its rim. What is the density of the material of which the beaker is made? 9. Two spherical balls A and B made up of same material having masses 2m and m are released from rest. Ball B lies at a distance h below the water surface while A is at a height of 2h above water surface in the same vertical line at the instant they are released. (a) Obtain the position where they collide. (b) If the bodies stick together due to collision, to what maximum height above water surface does the combined mass rise? Specific gravity of the material of the balls is 2/3. Neglect viscosity and loss due to splash. 10. For the system shown in the figure, the cylinder on the left at L has a mass of 600kg and a cross sectional area of 800 cm2. The piston on the right, at S, has cross sectional area 25cm2 and negligible weight. If the apparatus is filled with oil. ( = 0.75 gm/cm3) Find the force F required to hold the system in equilibrium. s

F

8m 600kg

L

11. A test tube of thin walls floats vertically in water, sinking by a length l0 = 10 cm. A liquid of density less than that of water, is poured into the tube till the levels inside and outside the tube are even. If the tube now sinks to a length l0 = 40 cm, the specific gravity of the liquid is ______. 12. In air an object weighs 15N, when immersed completely in water the same object weighs 12N. When immersed in another liquid completely, it weighs 13N. Find (a) the specific gravity of the object and (b) the specific gravity of the other liquid.

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Page # 32

FLUID

13. Block A in figure hangs by D a cord from spring balance D and is submerged in a liquid C contained in a beaker B. The B A mass of the beaker is 1kg & the mass of the liquid is 1.5 C kg. The balance D reads 2.5 kg & balance E reads 7.5 kg. E The volume of block A is 0.003 m3. (i) What is the density of block & the liquid. (ii) What will each balance read if block is pulled out of the liquid. 14. A solid cube, with faces either vertical or horizontal, is floating in a liquid of density 6 g/cc. It has two third of its volume submerged. If enough water is added from the top so as to completely cover the cube, what fraction of its volume will remain immersed in the liquid? 15. A uniform cylindrical block of length l density d1 and area of cross section A floats in a liquid of desity d2 contained in a vessel (d2>d1). The bottom of the cylinder just rests on a spring of constant k. The other end of the spring is fixed to the bottom of the vessel. The weight that may be placed on top of the cylinder such that the cylinder is just submerged in the liquid is ____ A

d1

l

diameter 8 /  cm with

A

1.8m

its crest A 1.8 m above water level as in figure.

3.6m

Find (a) velocity of flow (b) discharge rate of the flow in m3/sec. (c) absolute pressure at the crest level A. [Use P0 = 105 N/m2 & g = 10 m/s2] 18. A large tank is filled with two liquids of specific gravities 2 and . Two holes are made on the wall of the tank as shown. Find the ratio of the distances from O of the points on the ground where the jets from holes A & B strike. h/2



h/2

2

h/4

A A

h/4

O

19. Calculate the rate of flow of glycerine of density 1.25 × 103 kg/m3 through the conical section of a pipe if the radii of its ends are 0.1 m & 0.04 m and the pressure drop across its length is 10 N/m2 20. The tank in fig discharges water at constant rate for all water levels above the air inlet R. The height above datum to which water would rise in the manometer tubes M and N respectively are _________ & ___________ .

4. FLUID FLOW & BERNOULLI'S PRINCIPLE 16. Two very large open tanks A and F both contain the same liquid. A horizontal pipe BCD, having a constriction at C leads out of the bottom of tank A, and a vertical pipe E opens into the constriction at C and dips into the liquid in tank F. Assume streamline flow and no viscosity. If the cross section at C is one half that at D and if D is at a distance h1 below the level of liquid in A, to what height h2 (in terms of h1) will liquid rise in pipe E ? (above G & upto C there is air in the pipe)

A

17. A siphon has a uniform circular base of

B

C

D

Open to atmosphere

M N 40cm 20cm R

Datum

h1

G

h2

E F

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Page # 33

FLUID

(TOUGH SUBJECTIVE PROBLEMS)

Exercise - IV

1. A solid block of volume V = 10–3 m3 and density d = 800 kg/m3 is tied to one end of a string, the other end of which is tied to the bottom of the vessel. The vessel contains 2 immiscible liquids of density 1 = 1000 kg/m3 and 2 = 1500 kg/m3. The solid block is immersed with 2/5 th of its volume in the liquid higher density & 3/5th in the liquid of lower density. The vessel is placed in an elevator which is moving up with an acceleration of a = g/2. Find the tension in the string. [g = 10 m/s2] 2. An open rectangular tank 5m a × 4m × 3m high containing 3m water water upto a height of 2m is 2m accelerated horizontally along the longer side. 5m (a) Determine the maximum acceleration that can be given without spilling the water. (b) Calculate the percentage of water split over, if this acceleration is increased by 20% (c) If initially, the tank is closed at the top and is accelerated horizontally by 9m/s2, find the gauge pressure at the bottom of the front and rear walls of the tank. 3. A level controller is shown in the figure. It consists of a thin circular plug of diameter 10cm and a cylindrical float of diameter 20 cm tied together with a light rigid rod of length 10 cm. The plug fits in snugly in a drain hole at the bottom of the tank which opens into atmosphere. As water fills up and the level reaches height h, the plug opens. Find h. Determine the level of water in the tank when the plug closes again. The float has a mass 3kg and the plug may be assumed as massless. Float

h

10cm

Plug

4. A closed tube in the form of an equilateral triangle of side l contains equal volumes of three liquids which do not mix and is placed vertically with its lowest side horizontal. Find x in the figure if the densities of the liquids are in A.P. A

x B

+ + + + + +++ + C x

5. A ship sailing from sea into a river sinks X mm and on discharging the cargo rises Y mm. On proceeding again into sea the ship rises by Z mm. Assuming ship sides to be vertical at water line, find the specific gravity of sea water. 6. A conical vessel without a bottom stands on a table. A liquid is poured with the vessel & as soon as level reaches h, the pressure of the liquid raises the vessel. The radius of the base of vessel is R and half angle of the cone is  and the weight of the vessel is W. What is the density of the liquid?

A

h 2R 7. As the arrangement shown in the fig is released the rod of mass M moves down into the water. Friction is negligible and the string is inextensible (a) Find the acceleration of m L M the system w.r.t the distance moved by each mass. (b) Find the time required to completely immerse the rod into water m  –  water if M  .   = density of rod ; water = desity of water 8. The interface of two liquids of densities  and 2 respectively lies at the point A in a U tube at rest. The height of liquid column above A is 8a/3 where AB = a. The cross sectional area of the tube is S. With what angular velocity the tube must be whirled about a vertical axis at a distance ‘a’ such that the interface of the liquids shifts towards B by 2a/3.



8a/3 2

A a

B a

9. A closed cylindrical tank 2m high & 1 m in diameter contains 1.5 m of water. When the angular velocity is constant at 20.0 rad/s, how much of the bottom of the tank is uncovered ? (The cylinder is rotated about vertical axis of symmetry passing through its length.)

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Page # 34

FLUID

10. A cylinder of height H is filled with water to a height h0 (h0 90º. The resultant (R) of P and Q in this case passes through the liquid. Let us now see why the liquid surface bends near the contact with a solid. A liquid in equilibrium can not sustain trangential stress. The resultant force on any small part of the surface layer must be perpendicular to the surface at that point. Basically three forces are acting on a small part of the liquid surface near its contact with solid. These forces are, (i) P, attraction due to the molecule of the solid surface near it i.e. adhesive force which acts outwards at right angle to the wall of tube. (ii) Q, attraction due to liquid molecules near this part and i.e. cohesive force which acts at an angle of 45º to the vertical. We have considered very small part, so weight of that part can be ignored for better understanding. As we have seen in the last figures, to make the resultant (R) of P and Q perpendicular to the liquid surface the surface becomes curved (convex or concave). Note : The angle of contact between water and clean glass is zero. 4.

CAPILLARY RISE If a tube of very narrow bore (called capillary) is dipped in a liquid, it is found that the liquid in the capillary either ascends or descends relative to the surrounding liquid. This phenomenon is called capillarily. In order to calculate the height to which a liquid will rise in a capaillary, consider a glass capillary of radius R dipped in water as shown in Fig. shown. As the meniscus is concave and nearly spherical, the pressure below the meniscus will be [p0 – (2T/r)] with p0 as atmospheric pressure and r as radius of meniscus. Now as liquid flows from higher to lower pressure and at same level in a liquid pressure must be same (this is because a liquid cannot sustain tangential stress), so the liquid will ascends in the capillary till hydrostatic pressure of the liquid compensates for the decrease in pressure. i.e.,

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SURFACE TENSION & VICOSITY

Page # 8

r 

R

p0

p0

p0 p0 

p0

h

2T t

2T   p0  p0 –  hg or r  

h

2T rg

...(1)

But from figure shown it is clear that radius of meniscus r is related to the radius of capillary through the relation (R/r) = cos , i.e., r = R/cos  ...(2) where  is the anlge of contact. *So substituting the value of from Eqn. (2) in (1), we get

h

•

2T 2T cos   rg Rg

Alternate Method As it can be seen from figure that T sin  cancels out : The force due to T cos  balances the weight of liquid (mg =  vg) vol. of the curve is negligible  vol. of liquid in r2h T cos  = 2r = r2hg  h =

(1)

...(3)

T

T 

h

2T cos  rg

This is the desired result and from this it is clear that : The capillarity depends on the nature of liquid and solid both, i.e., on T, ,  and R. If  > 90°, i.e., meniscus is convex, h will be negative, i.e., the liquid will descends in the capillary as actually happens in case of mercury in a

Hg

(A) 

90;h   ve

(B)

  90;h  0

(C)

  90;h  – ve

glass tube. However, if  = 90°, i.e., meniscus is plane, h = 0 and so no capillarity.

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Page # 9

SURFACE TENSION & VICOSITY (2)

(3)

For a given liquid and solid at a given place as , T,  and g are constant, (figure shown) hr = constant  lesser the radius of capillary greater will be the rise and vice-versa. (figure shown) Here it is important to note that in equilibrium the height h is independent of the shape of capillary if the radius of meniscus remains the same. This is why the vertical height h of a liquid column in capillaries of different shapes and sizes will be same if the radius of meniscus remains the same and also the vertical height of the liquid in a capillary does not change, when it is inclined to the vertical. (figure shown)



(a)

(b)

(c)

(d)

h

(e)

4.

Capillarity has large number of applications in our daily life, e.g., (a) The oil in the wick of a lamp rise due to capillary action of threads in the wick. (b) Action of towel in soaking up moisture from the body is due to capillary action of cotton in the towel. (c) Water is retained in a piece of sponge on account of capillarity. (d) A blotting paper soaks ink by capillary action of the pores in the blotting paper. (e) The root-hairs of plants drawn water from the soil through capillary action.

5.

In Case of glass and water  = 0 here force due to surface tension balances the weight of the liquid ( × v × g) 2 3 volume of the liquid = r2h + r3 – r 3 2 where r3 – r3 is the volume of the curve which is 3 not negligible in this case  T.2r =  (r2h + r3 –

r

h

T

If two parallel plates with the spacing 'd' are placed in water reservoir, then height or rise. 2T = hdg 

7.

T r

2 3  r )g 3

1 2T = rh g + r2 g 3

6.

T

h

2T dg



h d

If two concentric tubes of radius 'r1' and 'r2' (inner one is solid) are placed in water reservoir, then height of rise? 

T[2r1  2r2 ]  [r22h – r12h] g h

2T (r2 – r1 ) g

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SURFACE TENSION & VICOSITY

Page # 10 r1

r2

h

r2

T cos 

T sin 

8.

T cos  



r

T sin 

If weight of the liquid in the meniscus is to be consider : T cos  × 2r = [r2h +

1 r2 × r]  g 3

h

r  2T cos   h  3   rg  

9.

When capillary tube (radius, 'r') is in vertical position, the upper meniscus is concave and pressure 2T due to surface tension is directed vertically upward and is given by p1 = R 1 where R1 = radius of curvature of upper meniscus. The hydrostatic pressure p2 = h  g is always directed downwards. If p1 > p2 i.e. resulting pressure is directed upward. For equilibrium, the pressure due to lower meniscus should be downward. This makes lower meniscus concave downward (fig a). The radius of 2T lower meniscus R2 can be given by R  (p1 – p 2 ) 2 (a)

(b)

(c)

If p1 < p2 i.e. resulting pressure is directed downward for equilibrium, the pressure due to lower meniscus should be upward. This makes lower meniscus convex upward (fig. b) 2T The radius of lower meniscus can be given by R  p 2 – p1 2 2T If p1 = p2, then is no resulting pressure. then, p1 – p2 = R = 0 or, R2 =  i.e. lower surface will be FLAT T 2 (fig c) Ex.5

Sol.

A drop of water volume 0.05 cm3 is pressed between two glass-plates, as a consequence of which, it spreads and occupies an area of 40 cm2. If the surface tension of water is 70 dyne/ cm, find the normal force required to seperate out the two glass plates in newton. Pressure inside the film is less than outside by an amount, 1 1  P  T    , where r and r are the radii of curvature of 1 2  r1 r2 

t and r2 = , then the force required 2 to separate the two glass plates, between which a liquid film the meniscus. Here r1 =

is enclosed (figure) is, F = P × A =

r1= t/2

t

2AT , where t is the thickness t

of the film, A = area of film.

F

2A2 T 2A2 T 2  (40  10–4 )2  (70  10–3 )   = 45 N At V 0.05  10–6

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Page # 11

SURFACE TENSION & VICOSITY 4.1

CAPILLARY RISE IN A TUBE OF INSUFFICIENT HEIGHT

O

 C

h



R' '



h

We know, the height through which a liquid rises in the capillary tube of radius r is given by

2T 2T or h R = = constant R g g When the capillary tube is cut an its length is less then h (i.e. h'), then the liquid rises upto the top of the tube and spreads in such a way that the radius (R') of the liquid meniscus increases and it becomes more flat so that hR = h' R' = Constant. Hence the liquid does not overflow. r r  If h' < h then R' > R or cos  ' cos   cos  < cos   ' >  

5.

h

VISCOSITY AND NEWTON'S LAW OF VISCOUS FORCE In case of steady flow of a fluid when a layer of fluid slips or tends to slip on adjacent layer in contact, the two layers exert tangential force on each other which tries to destroy the relative motion between them. 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 viscous force. A briskly strirred fluid comes to rest after a short while because of viscosity. As a result of large number of experiments Newton found that viscous force F acting on any layer of a fluid is directly proportional to its area A and to the velocity gradient (dv/dy)* at the layer i.e., FA

dv dy

or

F  – A

dv dy

Y v F A

v

dy X

...(1)

when  is a constant called coefficient of viscosity or simply viscosity of the fluid. The negative sign shows that viscous force on a liquid layer acts in a direction opposite to the relative velocity of flow of fluid. The Eq. (1) is known as Newton's law of viscous force. Here y is taken from the layer of which velocity is zero. Regarding viscosity of fluid it is worth noting that : (1) It depends only on the nature of fluid and is independent of area considered or velocity gradient. (2) Its dimensions are [ML–1 T–1] and SI unit poiseuille (PI) while CGS unit dyne-s/cm2 called poise (P) with 1 Pl = 10 poise (3) Viscosity of liquids is much greater (say about 100 times more) than that of gases i.e., L > G

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SURFACE TENSION & VICOSITY

Page # 12 Ex.6

A boat of area 10 m2 floating on the surface of a river is made to move horizontally with a speed of 2 m/s by applying a tangential force. If the river is 1 m deep and the water in contact with the bed is stationary, find the tangential water in contact with the bed is stationary, find the tangential force needed to keep the boat moving with same velocity. Viscosity of water is 0.01 poise.

Sol.

As velocity changes from 2 m/s at the surface to zero at the bed which is at a depth of 1 m.

Velocity gradient =

dv 2–0 = = 2 s–1 dy 1

Now from Newton's law of viscous force,

|F| =  A

dv = (10–2 × 10–1) × 10 × 2 = 0.02 N dy

Ex.7

The velocity of water in a river is 18 km/hr at the surface. If the river is 5 m deep, find the shearing stress between the horizontal layers of water. The viscosity of water is 10–3 poiseuuille.

Sol.

As velocity at the bottom of the river will be zero, velocity gradient dv 18  10 3   1s –1 dy 60  60  5

Now as the viscous force F  A(dv / dy ) is tangential to the area,

Shear stress =

F11 dv  = 10–3 × 1 = 1 × 10–3 N/m2 A dy

Ex.8

A cylinder of mass radius r1 and length  is kept inside another cylinder of radius r2 and length . The space between them is filled with a liquid of viscosity . The inner cylinder starts rotating with angular velocity  while the other cyclinder is at rest. Find time when inner cylinder stops.

Sol.

Viscous force F = –  A

dv dy



r12 r12    – 2 = –  2 r1  = r2 – r1 r2 – r1    r13  | F  r1 | Fr1 sin 90  Fr1  –2 r2 – r1

  I 

...(1)

R1

Mr12 d r13  –2 2 dt r2 – r1

from eq. (1) 0 a4r1 t d dt  –   M(r2 – r1 ) 0 

a4r1 t   n M(r2 – r1 )

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Page # 13

SURFACE TENSION & VICOSITY 6.

STOKES LAW When a body moves through a fluid, the flui in contact with the body is dragged with it. This establishes relative motion in fluid layers near the body, due to which viscous force starts operating. The fluid exerts viscous force on the body to oppose its motion. The magnitude of the viscous force depends on the shape and size of the body, its speed and the viscosity of the fluid. Stokes established that if a sphere of radius r moves with velocity v through a fluid of viscosity , the viscous force opposing the motion of the sphere is F = 6   rv

7.

TERMINAL VELCOITY (VT) Consider a small sphere falling from rest through a large column of viscous fluid. The forces acting on the shere are, (i) Weight W of the sphere acting vertically downwards Ft + Fv (ii) Upthrust Ft acting vertically upwards (iii) Viscous force Fv acting vertically upwards, i.e., in a direction opposite to velocity of the sphere. v Initially, Fv = 0 and W > Ft and the sphere accelerates downwards. As the velocity of the sphere increases, Fv W increases, Eventually a stage is reached when W = Ft + Fv After this net force on the sphere is zero and it moves downwards with a constant velocity called terminal velocity (vT). Substituting proper values in Eq. (i) we have,

Here, and

4 3 4 r g  r 3 g  6 rv T 3 3  = density of sphere,  = density of fluid  = coefficient of viscosity of fluid

From Eq. (ii), we get

vT 

2 r 2 (  ) g  9

v

vT

O

t

Figure shows the variation of the velocity v of the sphere with time. Note : From the above expression we can see that terminal velocity of a spherical body is directly proportional to the difference in the densities of the body and the fluid ( – ). If the density of fluid is greater than that of body (i.e.,  > ), the terminal velocity is negative. This means that the body instead of falling, moves upward. This is why air bubbles rise up in water. Ex.9

Two spherical radindrops of equal size are falling vertically through air with a terminal velocity of 1 m/s. What would be the terminal speed if these two drops were to coalesce to form a large spherical drop ?

Sol.

vT  r 2 Let r be the radius of small rain drops and R the radius of large drop. Equating the volumes, we have 4 4  R 2  2 r 3  3 3  R  ( 2)1 / 3  R = (2)1/3. r or r 2



vT   R      ( 2) 2 / 3 vT  r 



v T   (2) 2 / 3 v T  ( 2) 2 / 3 (1.0) m / s

= 1.587 m/sAns.

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SURFACE TENSION & VICOSITY

Page # 14

(only one option is correct)

Exercise - I 1. There is a horizontal film of soap solution. On it a thread is placed in the form of a loop. The film is pierced inside the loop and the thread becomes a circular loop of radius R. If the surface tension of the loop be T, then what will be the tension in the thread? (A) R2/T (B) R2T (C) 2RT (D) 2RT 2. A container, whose bottom has round holes with diameter 0.1 mm is filled with water. The maximum height in cm upto which water can be filled without leakage will be what? Surface tension = 75 × 10–3 N/m and g = 10 m/s2 : (A) 20 cm (B) 40 cm (C) 30 cm (D) 60 cm 3. If two soap bubbles of different radii are connected by a tube : (A) air flows from the bigger bubble to the smaller bubble till the sizes become equal (B) air flows from bigger bubble to the smaller bubble till the sizes are interchanged (C) air flows from the smaller bubble to the bigger (D) there is no flow of air.

ratio of excess pressure in the drop to the excess pres sure inside the bubble is. d R

1

1

1

 R 3 (A)    3d 

 R 3 (B)    6d 

 R 3 (C)    24d 

(A) r =

r1  r2 2

r1r2 (C) r  r  r 1 2

r1r2 (B) r  r – r 1 2

(D) r  r1r2

5. A liquid is filled in a spherical container of radius R till a height h. At this positions the liquid surface at the edges is also horizontal. The contact angle is

h

(A) 0

1  R  h  (B) cos    R 

1  h – R  (C) cos    R 

1  R – h  (D) sin    R 

6. A soap bubble has radius R and thickness d( r2) come in contact. Their common surface has radius of curvature r.

R>>d

h

r (D) R

(C)

h

h

8. A Newtonian fluid fills the clearance between a shaft and a sleeve. When a force of 800N is applied to the shaft, parallel to the sleeve, the shaft attains a speed of 1.5 cm/sec. If a force of 2.4 kN is applied instead, the shaft would move with a speed of (A) 1.5 cm/sec (B) 13.5 cm/sec (C) 4.5 cm/sec (D) None 9. A solid metallic sphere of radius r is allowed to fall freely through air. If the frictional resistance due to air is proportional to the cross-sectional area and to the square of the velocity, then the terminal velocity of the sphere is proportional to which of the following? (A) r2 (B) r (C) r3/2 (D) r1/2 10. Two drops of same radius are falling through air with steady velocity of v cm/s. If the two drops coalesce, what would be the terminal velocity? (A) 4 v (B) (4)1/3v (C) 2 v (D) 64 v 11. A cubical block of side ‘a’ and density ‘’ slides over a fixed inclined plane with constant velocity ‘v’. There is a thin film of viscous fluid of thickness ‘t’ between the plane and the block. Then the coefficient

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SURFACE TENSION & VICOSITY

Page # 15

of viscosity of the thin film will be:

15. A ball of mass m and radius r is gently released in a viscous liquid. The mass of the liquid displaced by it is m’ such that m>m’. The terminal velocity is proportional to (A)

=37°

(A)

3 a g t 5v

(B)

4 a g t 5v

(C)

 agt v

m – m' r

(B)

(m  m ') m  m' (C) (D) (m – m)r2 r2 r

16. Which of the following is the incorrect graph for a sphere falling in a viscous liquid? (Given at t = 0, velocity v = 0 and displacement x = 0.) v

(D) none of these 12. Which of the following graphs best represents the motion of a raindrop? v v

v

(A)

(B) t a

(B)

(A) v

t

t

v

x

t

(C)

(D) t

v

(C)

(D) t

t

13. A spherical ball of density  and radius 0.003m is dropped into a tube containing a viscous fluid filled up to the 0 cm mark as shown in the figure. Viscosity of the fluid = 1.260 N.m–2 and its density L = /2 = 1260 kg.m–3. Assume the ball reaches a terminal speed by the 10 cm mark. The time taken by the ball to traverse the distance between the 10 cm and 20 cm mark is. (A) 500 s (B) 50 ms (C) 0.5 s (D) 5 s (g = acceleration due to gravity = 10 ms–2)

17. The displacement of a ball falling from rest in a viscous medium is platted against time. Choose a possible option.

0 cm 10 cm

4r2 9

(B)

9r2 4

(C)

4r 9

s (B) t

20 cm

t

s

s

(C)

(D) t

14. A sphere is dropped under gravity through a fluid of viscosity . If the average acceleration is half of the initial acceleration, the time to attain the terminal velocity is ( = density of sphere, r = radius) (A)

s (A)

(D)

9r 4

t

18. There is a 1 mm thick layer of glycerine between a flat plate of area 100 cm2 & a big fixed plate. If the coefficient of viscosity of glycerine is 1.0 kg/m-s then how much force is required to move the plate with a velocity of 7 cm/s? (A) 3.5 N (B) 0.7 N (C) 1.4 N (D) None

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SURFACE TENSION & VICOSITY

Page # 16

Exercise - II

(SUBJECTIVE PROBLEMS)

1. A film of water is formed between two straight parallel wires each 10 cm long and at separattion 0.5 cm. Calculate the work required to increase 1 mm distance between wires. Surface tension of water = 72 × 10–3 Nm–1 2. Water rises in a capillary upto a certain height such that the upward force of surface tension balances the force of 75 × 10–4 N due to weight of the liquid. If the surface tension of water is 6 × 10–2 Nm–1, what must be the internal circumfernece of the capillary ? 3. A ring cut from a platinum tube, 8.5 cm internal diameter and 8.7 cm exernal diameter, is supported horizonally from the pair of a balance so that it comes in contact with the water in a vessel. If an extra weight of 3.97 g is required to pull it away from water, calculate the surface tension of water. 4. There is a soap bubble of radius 2.4 × 10–4 m in air cylinder which is originally at the pressure of 105 Nm–2. The air in the cylinder is now compressed isothermally until the radius of the bubble is halved. Calculate now the pressure of air in the cylinder. The surface tension of the soap solution is 0.08 Nm–1 5. Two separate air bubbles (radii 0.002 m and 0.004 m) formed of the same liquid (surface tension 0.07 N/m) come together to form a double bubble . Find the radius and the sense of curvature of the internal film surface common to both the bubbles. 6. Two soap bubbles of radii a and b combine to form a single bubble of radius c. If the external pressure is P, then the surface tension of soap solution is 7. A long capillary tube of radius 2 mm open at both ends is filled with water and placed vertically. What will be the height of the column of water left in the capillary ? The thickness of the capillary walls is negligible. Surface tension of water 73.5 × 10–3 Nm–1 8. The limbs of a manometer consist of uniform capillary tubes of radii 1.44 × 10–3 m and 7.2 × 10–4 m. Find out the correct pressure difference if the level of the liquid (density 103 kgm–3, surface tension 72 × 10–3 Nm–1) in the narrower tube stands 0.2 m above that in the broader tube.

9. A glass capillary sealed at the upper end is of length 0.11 m and internal diameter 2 × 10–5 m. The tube is immersed vertically into a liquid of surface tension 5.06 × 10–2 Nm–1. To what length the capillary has to be immersed so that the liquid level inside and outside and capillary becomes the same ? What will happen to liquid level inside the capillary if the seal is now broken ? Atmopsheric pressure is 1.012 × 105 Nm–2. 10. A ball is given velocity v0 (greater than the terminal velocity vT) in downward direction inside a highly viscous liquid placed inside a large container. The height of liquid in the container is H. The ball attains the terminal velocity just before striking at the bottom of the container. Draw graph between velocity of the ball and distance moved by the ball before getting terminal velocity.

11. Two arms of a U-tube have unequal diameters d1 = 1.0 mm and d2 = 1.0 cm. If water (surface tension 7 × 10–2N/m) is poured into the tube held in the vertical position, find the difference of level of water in the Utube. Assume the angle of contact to be zero. 12. A spherical ball of radius 1 × 10–4 m and density 104 kg/m3 falls freely under gravity through a distance h before entering a tank of water. If after entering the water the velocity of the ball does not change, find h. The viscosity of water is 9.8 × 10–6 N-s/m2. 13. An expansible balloon filled with air floats on the surface of a lake with 2/3 of its volume submerged. How deep must it be sunk in the water so that it is just in equilibrium neither sinking further nor rising ? It is assumed that the temperature of the water is constant & that the height of the water barometer is 9 meters.

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SURFACE TENSION & VICOSITY

Exercise - III

Page # 17

(JEE PROBLEMS)

1. When an air bubble rises from the bottom of a deep lake to a point just below the water surface, the pressure of air inside the bubble (A) is greater than the pressure outside it (B) is less than the pressure outside it (C) increases as the bubble moves up (D) decreases as the bubble moves up 2. Assertion : A helium filled balloon does not rise indefinately in air but halts after a certain height. Reason : Viscosity opposes the motion of balloon. Choose any one of the following four responses : (A) if both (A) and (R) are true and (R) is the correct explanation of (A) (B) if both (A) and (R) are true but (R) is not correct explanation of (A) (C) if (A) is true but (R) is false (D) if (A) is false and (R) is true 3. A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric 81  10 5 Vm 1 . When the field is 7 switched off, the drop is observed to fall with terminal velocity 2 × 10 –3 ms–1 Given g = 9.8 ms–2, viscosity of the air = 1.8 × 10–5 Ns m–2 and the density of oil = 900 kg m–5, the magnitude of q is : (A) 1.6 × 10–19C (B) 3.2 × 10–19C –19 (C) 4.8 × 10 C (D) 8.0 × 10–19C [JEE 2010] field of strength

JEE PROBLEMS

Paragraph for questions 4 to 6 When liquid medicine of density  is to be put in the eye, it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop. We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension T when the radius of the drop is R. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper. [JEE 2010] 4. If the radius of the opening of the dropper is r, the vertical force due to the surface tension on the drop of radius R (assuming r 0 and along +ve x-axis for x < 0 and zero for x = 0. Thus the particle will oscillate about stable equillibrium position x = 0. The force in this case is called the restoring force. If n = 1 i.e., F = – kx the motion is said to be SHM (Simple Harmonic Motion) If the restoring force / torque acting on the body in oscillatory motion is directly proportional to the displacement of body / particle w.r.t. mean position and is always directed towards equillibrium position then the motion is called Simple Harmonic motion. It is the simplest form of oscillatory motion.

3.

TYPES OF SHM : (a) Linear SHM : When a particle moves to and fro about an equilibrium point, along a straight line here A and B are extreme positions and M is mean position so AM = MB = Amplitude. A

M

B

(b) Angular SHM : When body/particle is free to rotate about a given axis and executing angular oscillations. 4.

ANALYSIS OF MOTION IN LINEAR SHM : When the particle is moved away from the mean position or equillibrium position and released, a force (–kx) comes into play to pull it back towards mean position. By the time it gets at mean position it has picked up some kinetic energy and so it overshoots, stopping some where on the other side and it is again pulled back towards the mean position. It is necessary to study the change in speed and acceleration of particle during SHM. Let us

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Page # 4

SIMPLE HARMONIC MOTION

consider a particle whose position is x = 0 at t = 0 and v = v0. Then we divide the motion of particle in one time period in four parts. Mean Extreme extreme Position position position v=v0 A

B v=0

v=0 C

Amplitude

x=0 (A) from A to B

t=0

(B) from B to A

x

(C) from A to C

(D) from C to A

NOTE :In the figure shown, path of the particle is a straight line. (1)

Motion of a particle from A to B : Initially the particle is at A (mean position) and is moving towards +ve x direction with speed v0. As the particle is moving towards B, force acting on it towards A is increasing. Consequently its acceleration towards A is increasing in magnitude while its speed decreases and finally it comes to rest momentarily at B.

(2)

Motion of a particle from B to A : Now the particle starts moving towards A with initial speed v = 0. As the particle is moving towards A, force is acting on it towards A and decreasing as it approaches A. Consequently its acceleration towards A is decreasing in magnitude while its speed increases and finally it comes to A with same speed v = v0.

(3)

Motion of a particle from A to C : The motion of a particle from A to C is qualitatively same as motion of a particle from A to B.

(4)

Motion of a particle from C to A : It is qualitatively same as motion of a particle from B to A.

Summary : Motion from

Velocity (Direction/Magnitude)

Acceleration (Direction/ Magnitude)

AB

V

a

BA

V

a

AC

V

a

CA

V

a

5.

CHARACTERISTICS OF SHM :

(1)

Mean Position : It is the position where net force on the particle is zero.

(2)

Extreme Point : Point where speed of the particle is zero.

(3)

Displacement : It is defined as the distance of the particle from the mean position at that instant.

(4)

Amplitude : It is the maximum value of displacement of the particle from its mean position.

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Page # 5

SIMPLE HARMONIC MOTION Extreme position – Mean position = Amplitude. It depends upon the energy of the system. (5)

Frequency : The frequency of SHM is equal to the number of complete oscillations per unit time. f

(6)

Time Period : Smallest time interval after which the oscillatory motion gets repeated is called time period. T=

Ex.1

Sol.

 1  sec –1 or Hz. T 2

2 

Describe the motion of a particle acted upon by a force. (A) F = 3x + 3 = 3x – 3

(B) F = – 3x – 3

(a) Given F = 3x + 3

...(i)

(C) F = – 3x + 3

(D) F

We find the mean position at which net force on the particle is zero. 

3x + 3 = 0



x=–1

If we put x = 0 in eq. (i) then F = 3N (away from M.P.)

M.P. ...(a)

x = –1

Now put x = –2 in eq. (i) F=– 3N

(away from M.P.)

...(b)

From (a) and (b) we conclude that particle doesn't perform S.H.M. (b) Given F = – 3x – 3

...(i)

at M.P. F = 0 

x = –1 Now put x = 0 in eq. (i)



F = – 3N (towards M.P.)

If x = – 2 ; F = 3N (towards M.P.) We conclude from the above calculation that in every case (whether the particle is left from M.P. or right from M.P.) force acts towards M.P. so the particle performs S.H.M. (c)

Given F = – 3x + 3 when

F=0

x = 1 (M.P.) Now put x = 0 Then F = 3N

(towards M.P.)

If x = 2

F=–3

(towards M.P.)

i.e. particle performs S.H.M. (D)

Given F = 3x – 3 Mean position at x = 1. When x = 0 ; F = – 3N x = 2 ; F = 3N

(away from M.P.) (away from M.P.)

Particle doesn't perform S.H.M.

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Page # 6

6.

SIMPLE HARMONIC MOTION

EQUATION OF SIMPLE HARMONIC MOTION : The necessary and sufficient condition for SHM is F = – kx we can write above equation in the following way: ma = – kx m

d2 x dt 2

d2 x dt

2



 –kx k x0 m

...(1)

Equation (1) is Double Differential Equation of SHM. d2 x

Now

dt 2

 2 x  0

It's solution is x = A sin(t  ) k m

where  = angular frequency =

x = displacement from mean position k = SHM constant. The equality (t + ) is called the phase angle or simply the phase of the SHM and  is the initial phase i.e., the phase at t = 0 and depends on initial position and direction of velocity at t = 0. To understand the role of  in SHM, we take two particles performing SHM in the following condition:

v0

A

M.P.

–A

x –A

M.P.

A

x

figure II

figure I

Suppose we choose t = 0 at an instant when the particle is passing through its mean position towards right (i.e.positive direction) as shown in figure Ist then In figure I

at

t=0

x=0

i.e., x = A sin t  The particle is at its mean position. In figure II i.e.,

at

t=0

x = A and the particle is moving towards the mean position.

x = A sin (t + /2)

Here /2 is the only phase possible.

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Page # 7

SIMPLE HARMONIC MOTION

Ex.2

A particle starts from mean position and moves towards positive extreme as shown below. Find the equation of the SHM. Amplitude of SHM is A. t=0 O

–A

Sol.

A

General equation of SHM can be written as x = A sin (t + ) At t = 0, x = 0  0 = A sin    = 0,    [0, 2) Also; at t = 0, v = + ve  A  cos  = + ve or, =0 Hence, if the particle is at mean position at t = 0 and is moving towards +ve extreme, then the equation of SHM is given by x = A sin t. Similarly for particle moving towards –ve extreme then t=0 +A =  equation of SHM is x = A sin (t + ) or, x = – A sin t

–A

Ex.3

Write the equation of SHM for the situation shown below :

Sol.

t=0 A O A/2 General equation of SHM can be written as x = A sin (t + ) At t = 0, x = A/2

–A

A = A sin  2   = 30° , 150° Also at t = 0, v = – ve A cos  = – ve 

7.



 = 150°

VELOCITY : It is the rate of change of particle displacement with respect to time at that instant. Let the displacement from mean position is given by x = A sin (t + ) velocity

v

dx  A cos(t  ) dt

v = A cos (t + ) v =  A 2 – x2 At mean position (x = 0), velocity is maximum. Vmax =  A At extreme position (x = A), velocity is minimum. vmin = zero.

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Page # 8 7.1

SIMPLE HARMONIC MOTION

Graph of Velocity (v) V/S Displacement (x) : 2

2

2

v   (A – x )

v   A 2 – x2

v2

v 2   2 x 2  2 A 2

2 A 2

x2 

A2

A

1

Graph would be a half ellipse.

8.

Velocity (v)

2

–A

A

x

ACCELERATION : It is the rate of change of particle's velocity w.r.t. time at that instant. Acceleration, a 

dv d [ A cos( t  )] = dt dt

a = – 2A sin (t + ) a = – 2 x

Note : Negative sign shows that acceleration is always directed towards the mean position. At mean position (x =0), acceleration is minimum. amin = zero At extreme position (x = A), acceleration is maximum. |amax | = 2A 8.1

Graph of Acceleration (A) v/s Displacement (x):

a 2 A A

x

–A

2

a=–x

– 2 A

9.

GRAPHICAL REPRESENTATION OF DISPLACEMENT, VELOCITY & ACCELERATION IN SHM:

Displacement, x = A sin t Velocity,

v = A  cos t = A  sin ( t 

or

v   A 2 – x2

 ) 2

Acceleration, a = – 2A sin t = 2A sin (t + ) or

a = – 2 x

Note : •

v   A 2 – x2 a = – 2x

These relations are true for any equation of x.

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Page # 9

SIMPLE HARMONIC MOTION

tim e , t displa ce m e nt, x Ve locity, v

0

T/4

0

A

A

0

0

a cce le ra tion, a

x

– 2 A

T/4

T/2 3T/4 T

T/2

3T/4

0

–A

0

0

–A  0

T A

2 A

0

5T/4 3T/2

A t

–A v A t

–A a 2 A t 2

– A

Ex.4

Sol.

1.

All the three quantities displacement, velocity and acceleration vary harmonically with time, having same period.

2.

The maximum velocity is  times the amplitude (Vmax = A).

3.

The acceleration is 2 times the displacement amplitude (amax = 2A).

4.

In SHM, the velocity is ahead of displacement by a phase angle of

5.

In SHM, the acceleration is ahead of velocity by a phase angle of

 . 2  . 2

  –1 The equation of particle executing simple harmonic motion is x  (5m)sin(s )t   . 3  Write down the amplitude, time period and maximum speed. Also find the velocity at t = 1 s.

Comparing with equation x = A sin (t + ), we see that the amplitude = 5m, and time period =

2 2  2s =  s –1 

The maximum speed = A  = 5 m ×  s–1 = 5  m/s The veloity at time t  At

dx = A  cos (t + ) dt

t = 1 s, 5   m/s v = (5 m) ( s–1) cos    = – 2 3 

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Page # 10

SIMPLE HARMONIC MOTION

Ex.5

A particle executing simple harmonic motion has angular frequency 6.28 s–1 and amplitude 10 cm. Find (a) the time period, (b) the maximum speed, (c) the maximum acceleration, (d) the speed when the displacement is 6 cm from the mean position, (e) the speed at t = 1/6 s assuming that the motion starts from rest at t = 0.

Sol.

(a) Time period =

2 2 = s = 1 s.  6.28 (b) Maximum speed = A = (0.1m) (6.28 s–1) (c) Maximum acceleration = A2 = (0.1m) (6.28 s–1)2 = 4 m/s2

(d) v   A 2 – x 2 = (6.28 s–1) (10cm)2 – (6cm)2 = 50.2 cm/s. (e) At t = 0, the velocity is zero i.e., the particle is at an extreme. The equation for displacement may be written as x = A cos t. The velocity is v = – A  sin t. At t 

 6.28   v = – (0.1 m) (6.28 s–1) sin  6 

1 s, 6

 = – 54.4 cm/s. (towards mean position) 3 Note : If mean position is not at the origin, then we can replace x by x – x0 and the eqn. becomes

= (–0.628 m/s) sin

x – x0 = – A sin t, where x0 is the position co-ordinate of the mean position. Ex.6

A particle of mass 2 kg is moving on a straight line under the action force F = (8 – 2x) N. It is released at rest from x = 6m. (A) Is the particle moving simple harmonically? (B) Find the equilibrium position of the particle. (C) Write the equation of motion of the particle. (D) Find the time period of SHM.

Sol.

F = 8 – 2x or

F = –2(x – 4)

for equilibrium position F = 0 

x = 4m is equilibrium position.

Hence the motion of particle is SHM with force constant 2 and equilibrium position x =4. (a) Yes, motion is SHM. (b) Equilibrium position is x = 4m. (c) At x = 6 m, particle at rest i.e. it is one of the extreme position. Hence amplitude is A = 2 m and initially particle at the extreme position. 

Equation of SHM can be written as x – 4 = 2 cos t, where  

i.e.

k  m

2  1 (sec)–1 2

x = 4 + 2 cos t

(d) Time period, T  2  2 sec . 

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Page # 11

SIMPLE HARMONIC MOTION

10.

SHM AS A PROJECTION OF UNIFORM CIRCULAR MOTION. Consider a particle Q, moving on a circle of radius A with constant angular velocity . The projection of Q on a diameter BC is P. It is clear from the figure that as Q moves around the circle the projection P excecutes a simple harmonic motion on the x-axis between B and C. The angle that the radius OQ makes with the +ve vertical in clockwise direction in at t = 0 is equal to phase constant ().



) =0 t) t= t (a

t (at

0 A Q  t Q t

B

O

P0 Pt

C

x(t)

Let the radius OQ0 makes an angle t with the OQt at time t. Then x(t) = A sin (t + ) In the above discussion the foot of projection is x-axis so it is called horizontal phasor. Similarly the foot of perpendicular on y axis will also executes SHM of amplitude A and angular frequency  [y(t) = Acost]. This is called vertical phasor. The phaser of the two SHM differ by /2.

+A

(0,0) M.P.

–A

x(t)

Problem solving strategy in horizontal phasor: (1)

First assume circle of radius equal to amplitude of S.H.M.

(2)

Assume a particle rotating in a circular path moving with constant  same as that of S.H.M in clockwise direction.

(3)

Angle made by the particle at t = 0 with the upper vertical is equal to phase constant.

(4)

Horizontal component of velocity of particle gives you the velocity of particle performing S.H.M. for example 0) t=  t t) (a t t= 0 a ( Q Qt v(t)  t ( t   ) v  A

–A

A

from figure

Q0

v(t) = A  cos (t + ) (5)

Component of acceleration of particle in horizontal direction is equal to the acceleration of particle performing S.H.M. The acceleration of a particle in uniform circular motion is only centripetal and has a magnitude a = 2 A.

t  

a(t)



From figure a(t) = – 2A sin (t + )

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t

2 A

Qt

   – (t  ) 2 

Page # 12

SIMPLE HARMONIC MOTION

Ex.7

A particle starts from A/2 and moves towards positive extreme as shown below. Find the equation of the SHM. Given amplitude of SHM is A. T R t = 0 O A v(t) –A +A A/2

Sol.

We will solve the above problem with the help of horizontal phasor.

O' A/2 Q

Step 1. Draw a perpendicular line in upward direction from

S

point P on the circle which cuts it at point R & S v(t)

Step 2. Horizontal compoment of v(t) at R gives the direction P to A while at S gives P to O. So at t = 0 particle is at R

cos  =

A/2  60  A

A

 = 30° 

So equation of the SHM



is x = A sin (t + 30°) Ex.8

O'

A particle starts from point x  t=0

–A

Sol.

P

R

T

In  O' RQ

Step 3.

O

(0,0) A/2 A M.P.(at t=0)

–A

A/2

Q

– 3 A and move towards negative extreme as shown 2

O +A

– 3 A 2

(a) Find the equation of the SHM. (b) Find the time taken by the particle to go directly from its initial position to negative extreme. (c) Find the time taken by the particle to reach at mean position. Figure shows the solution of the problem with the help of phasor R Horizontal component of velcoity at Q gives the  required direction of velocity at t = 0.  S ( 3 / 2) A O  3 / 2A 3  cos    In  OSQ    6 A 2 Now



3  8 4 –   2 6 6 3

A Q

So equation of SHM is 4   x  A sin t   –A – 3 / 2 A (0,0) A 3   (b) Now to reach the particle at left extreme point it will travel angle  along the circle. So time taken. t

    6

 t

T sec 12

(c) To reach the particle at mean position it will travel an angle  = So, time taken =

  2   2 6 3

 T = sec  3

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Page # 13

SIMPLE HARMONIC MOTION Ex.9

Two particles undergoes SHM along parallel lines with the same time period (T) and equal amplitudes. At a particular instant, one particle is at its extreme position while the other is at its mean position. They move in the same direction. They will cross each other after a further time. B B’

Sol.

O

A

O’

A’

(A) T/8 (B) 3T/8 (C) T/6 This problem is easy to solve with the help of phasor diagram.

(D) 4T/3

First we draw the initial position of both the particle on the phasor as shown in figure.

P  Q I A'

II

B'

at t = 0

A

–A

From above figure phase difference between both the particles is /2. They will cross each other when their projection from the circle on the horizontal diameter meet at one point. Let after time t both will reach at P'Q' point having phase difference /2 as shown in figure. v(t)

Q'

A 45º

P

45º





A

v(t)

Q P' B' –A

I R II

A'

O

after t = t

+A

–A / 2

Both will meet at

– A/ 2

When they meet angular displacement of P is  = /2 + /4 = 3/4 So they will meet after time t  3  4 t

3 3T T  sec 4  2 8

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Page # 14

SIMPLE HARMONIC MOTION

Ex.10 Two particles execute SHM of same amplitude of 20 cm with same period along the same line about the same equilibrium position. If phase difference is /3 then find out the maximum distance between these two. Sol.

Let us assume that one particle starts from mean position and another starts at a distance x having  = /3. This condition is shown in figure. P

/

3

 Q

A/2 A 3 2

A /6

/3

A/2 A

Q

x

–A

A

P

A A Fig ii

40cm Fig(i)

Above figure shows the situation of maximum distance between them. So maximum distance = A = 10 cm. (as 2A = 20 cm)

Ex.11 Two particles execute SHM of same time period but different amplitudes along the same line. One starts from mean position having amplitude A and other starts from extreme position having amplitude 2A. Find out the time when they both will meet? Sol.

We solve the above problem with the help of phasor diagram. First we draw the initial position of both the particle on the phasor.

(t=0)

A



A

B (t=0) 2A 

–A –2A

O O

A 2A

From figure phase difference between both the particle is /2. They will meet each other when their projection from the circle on the horizontal diameter meet at one point.

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Page # 15

SIMPLE HARMONIC MOTION Now from figure:

EF = A cos  = 2A sin  A'(t)



1 tan  = 2

A E

 1   tan 1   2

 90  

F

2A B'(t)

So time taken by the particle to cross each other t

angle travelled by A /2   t  

P –A

Q A 2A Q'

–2A P'

Ex.12 Two particles have time periods T and 5T/4. They start SHM at the same time from the mean postion. After how many oscillations of the particle having smaller time period, they will be again in the same phase ? Sol.

They will be again at m.p. and moving in same direction when the particle having smaller time period makes n1 oscillations and the other one makes n2 oscillations. 

5T  n2 4

n1T =

n1 5  n 2 4 

n1 = 5, n2 = 4

11.

ENERGY OF SHM :

11.1

Kinetic Energy (KE):

,

K.E. =

= 

K.E = K.Emax

1 1 mv v2 = mA22 cos2( t + ) 2 2

1 m 2 (A2 – x2) 2

 2 =

k m

1 K( A 2 – x 2 ) 2 1 2 = KA (at x = 0) 2

1 kA2 4 Frequency of KE = 2 × (frequency of SHM) K.Emin = 0 (at x = A) ;

11.2

KE

0–T

=

;

KE

0– A



1 2 kA 3

Potential Energy (PE): Simple harmonic motion is defined by the equation F = – kx The work done by the force F during a displacement from x to x + dx is dW = Fdx = – kx dx The work done in a displacement from x = 0 to x is x

1 W  ( kx ) dx   kx 2 2

 0

kx M.P.

x

dx

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Page # 16

SIMPLE HARMONIC MOTION

Let U(x) be the potential energy of the system when the displacement is x. As the change in potential energy corresponding to a conservative force is the negative of the work done by that force. U(x) – UM.P. = – W =

1 2 kx 2

Let us choose the potential energy to be zero when the particle is at the mean position oscillation x = 0. Then

UM.P. = 0 and U( x ) 

1 2 kx 2

 k = m2



U( x )  U=

1 m 2 x 2 2

1 m2 A 2 sin2 ( t  ) 2

But x = A sin (t + ) Kinetic energy of the particle at any instant is K

1 1 1 mv 2  mA 2 2 cos 2 (t  )  m2 ( A 2  x 2 ) 2 2 2

So the total mechanical energy at time ‘t’ is

E=U+K

E



1 m 2 A 2 2

Note : Umin = UM.P. (which is not always = 0)

Energy

(i)

1 1 m2 A 2  kA 2  E 2 2 2 U = Umaxcos t 2

K = Kmaxsin t t

O

Potential, Kinetic and total energy plotted as function of time

Energy 2 E = 1/2m A = Constant 1 U( x)  m2 x 2 2 2

K(x) (ii)

U(x)

1 K( x )  m2 ( A 2  x 2 ) 2 x=0 x= –A x=A x Potential, Kinetic and total energy are plotted as a function of displacement from the mean position.

Ex.13 A particle of mass 0.50 kg executes a simple harmonic motion under a force F = – (50 N/m)x. If it crosses the centre of oscillation with a speed of 10 m/s, find the amplitude of the motion.

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Page # 17

SIMPLE HARMONIC MOTION Sol.

The kinetic energy of the particle when it is at the centre of oscillation is E

1 mv 2 2

=

1 (0.50kg) (10 m / s)2 2

=

2.5 J.

The potential energy is zero here. At the maximum displacement x = A, the speed is zero and hence the kinetic energy is zero. The potential energy here is

1 2 kA . As there is no loss of 2

energy, 1 2 kA  2.5 J 2 The force on the particle is given by F = – (50 N/m) x. Thus the spring constant is k = 50 N/m. Equation (i) gives 1 (50 N / m)A 2  2.5J 2

12.

1 or,

A=

m.

10

METHOD TO DETERMINE TIME PERIOD AND ANGULAR FREQUENCY IN SIMPLE HARMONIC MOTION : To understand the steps which are usually followed to find out the time period we will take one example.

Ex.14 A mass m is attached to the free end of a massless spring of spring constant k with its other end fixed to a rigid support as shown in figure. Find out the time period of the mass, if it is displaced slightly by an amount x downward. m

Sol. The following steps are usually followed in this method: Step 1. Find the stable equillibrium position which is usually known as the mean position. Net force or torque on the particle at this position is zero. Potential energy is minimum. In our example initial position is the mean position.

n.

x0 M.P.

Natural Length

x0

kx0 m Mean Position mg

k(x+x0)

x m

mg

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Page # 18

SIMPLE HARMONIC MOTION

Step 2. Write down the mean position force relation. In above figure at mean position kx0 = mg ...(1) Step 3. Now displace the particle from its mean position by a small displacement x (in linear SHM) or angle  (in case of an angular SHM) as shown in figure. Step 4. Write down the net force on the particle in the displaced position. From the above figure. Fnet = mg – k (x + x0) ...(2) Step 5. Now try to reduce this net force equation in the form of F = – kx (in linear S.H.M.) or  = – k (in angular SHM) using mean position force relation in step 2 or binomial theorem. from eq. (2) Fnet = mg – kx – kx0 Using eq (i) in above equation Fnet = – kx ...(3) Equation (3) shows that the net force acting towards mean position and is proportional to x, but in this S.H.M. constant KS.H.M. is replaced by spring constant k. So T  2

m K S.H.M.

m k

 2

Note : If we apply constant force on the string then time period T is always same T  2

m K S.H.M.

k m

F0

m

k



 m   In above both cases T =  2 k   

Ex.15 The string, the spring and the pulley shown in figure are light. Find the time period of the mass m.

Sol.

Let in equilibrium position of the block, extension in spring is x0. 

kx0 = mg

m

k

m

k

m

...(1)

x0 x

Natural length Equilibrium position

Now if we displace the block by x in the downward direction, net force on the block towards mean position is F = k (x + x0) – mg

= kx

using (1)

Hence the net force is acting towards mean position and is also proportional to x. So, the particle will perform S.H.M. and its time period would be T  2

m k

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Page # 19

SIMPLE HARMONIC MOTION

Ex.16 Figure shows a system consisting of a massless pulley, a spring of force constant k and ablock of mass m. If the block is slightly displaced vertically down from its equillibrium position and then released, find the period of its vertical oscillation in cases (a) & (b).

k

k

m

m (b)

(a)

Sol.

Let us assume that in equillibrium condition spring is x0 elongate from its natural length

n. x  x0 2

kx 0

x0

kx 0

x/2

T0

T0

m x

Case (a)

T

mg

T m mg

When equillibrium

When displaced block by 'x'

In equilibrium T0 = mg and kx0 = 2T0  kx0 = 2mg ...(1) If the mass m moves down a distance x from its equilibrium position then pulley will move down by

x kx . So the extra force in spring will be . From figure 2 2

Fnet = mg – T = mg – Fnet = mg –

k x  x0   2 2

kx 0 kx – 2 4

from eq. (1) –kx ...(3) 4 Now compare eq. (3) with F = – KS.H.M x K then KS.H.M = 4 Fnet =



T  2

m

2 K S.H.M =

4m K

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Page # 20

SIMPLE HARMONIC MOTION

Case (b) : In this situation if the mass m moves down distance x from its equilibrium position, then pulley will also move by x and so the spring will stretch by 2x. At equilibrium kx0 =

T0 mg = 2 2

When block is displaced Fnet = mg – T x0+2x

x0

= mg – 2k (x0 + 2x) = – 4 kx Now

2x

F = – KSHM x then

T0

KSHM = 4 K

m

T

x

m So time period T  2 4k

mg

m mg

Ex.17 The left block in figure collides inelastically with the right block and sticks to it. Find the amplitude of the resulting simple harmonic motion. v m

Sol.

m

k

The collision is for a small interval only, we can apply the principal of conservation of momentum. 2

The common velocity after the collision is

1 1 v v . The kinetic energy  (2m)   mv2. This is 2 2 4 2  

also the total energy of vibration as the spring is unstretched at this moment. If the amplitude is A, the total energy can also be written as 1 2 1 kA  mv 2 , giving A  2 4

1 2 kA . Thus, 2

m v 2k

Ex.18 The system is in equilibrium and at rest. Now mass m1 is removed from m2. Find the time period and amplitude of resultant motion. (Given : spring constant is K.) Sol. Initial extension in the spring (m1  m2 )g k Now, if we remove m1. equillibrium position (E.P.) x

m2 g below natural length of spring. N.L K At the initial position, since velocity is zero i.e. it is the extreme position. of m2 will be

m1g K

Hence Amplitude

=

Time period

= 2

m 2g K

(m1  m2 )g K m1 m2

m2

m1g K

E.P

m2 K

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Page # 21

SIMPLE HARMONIC MOTION

Ex.19 Block of mass m2 is in equilibrium and at rest. The mass m1 moving with velocity u vertically downwards collides with m2 and sticks to it. Find the energy of oscillation.

N.L.

Sol.

m1

x0

u

m1 +m2

m2

v

At equilibrium position m2g = kx0 

x0 =

m2g K

After collision m2 sticks to m1.  By momentum conservation. m1u = (m1 + m2) v m1u v = m m 1 2 Now both the blocks are executing S.H.M. which can be interpreted as follows:

v

N.L.

m.p.

m2 g m1g K K Now, we know that v2 = 2(A2 – x2)

...(1)

k 2 = m  m 1 2 

x=

m1g k

Put the values of v, 2 & x in eq. (1) 2  k  m1u    =   m1  m 2  m1  m 2 



 m 2 u 2 1  kA =  m  m 1 2   2

  2  m1g   A –     k  

2

 

  m 1g  2       k    

 Energy of oscillation =

1 2 1 kA = 2 2

 m 2 u 2   m 2 g2    1  1   k  m  m 2     1

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Page # 22

SIMPLE HARMONIC MOTION

Ex.20 A body of mass m falls from a height h on to the pan of a spring balance. The masses of the pan and spring are negligible. The spring constant of the spring is k. Having stuck to the pan the body starts performing harmonic oscillations in the vertical direction. Find the amplitude and energy of oscillation. Sol.

Suppose by falling down through a height h, the mass m compresses the spring balance by a length x. x=

mg ,= k

k m

velocity at Q v =

2gh

 v =  A2  x2 2

2gh 

mg 2 kh k  mg  1 A2     A k mg m  k 

Energy of oscillation 

1 2 1  mg  kA  k   2 2  k 

2

 2kh   mgh  (mg)  1  2k mg  

2

2m

Ex.21 A body of mass 2m is connected to another body of mass m as shown in figure. The mass 2m performs vertical S.H.M. Then find out the maximum amplitude of 2m such that mass m doesn't lift up from the ground. m

Sol.

In the given situation 2m mass is in equilibrium condition. Let assume spring is compressed x0 distance from its natural length. 

kx0 = 2mg

2mg k The lower block will be lift up, only in the case when the spring force on it will be greater than equal to mg and in upward direction 

N.L.

x0 

x0 2m kx0

mg  kx' = mg x'   k Above situation arises when 2m block moves upward mg/k from natural length as shown in figure A

M.P.

kx0

2m x'=mg/k N.L.

m

kx'

2mg k

M.P.

kx' m

Block m doesn't lift up if the maximum amplitude of the 2m block is 

2mg mg 3mg   k k k

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Page # 23

SIMPLE HARMONIC MOTION m m

Ex.22 A block of mass m is at rest on the another block of same mass as shown in figure. Lower block is attached to the spring then determine the maximum amplitude of motion so that

K

both the block will remain in contact. Sol. N.L of spring 2mg  x0 k

m m

eq. position (M.P.)

Kx0

The blocks will remain in contact till the blocks do not go above the natural length of the spring, because after this condition the deceleration of lower block becomes more then upper block due to spring force. So they will get seprated. So maximum possible amplitude = x0 =

12.1

2mg k

Two Block Systems:

Ex.23 Two blocks of mass m1 and m2 are connected with a spring of natural length  and spring constant k. The system is lying on a smooth horizontal surface. Initially spring is compressed by x0 as shown in figure. Show that the two blocks will perform SHM about their equilibrium position. Also (a) find the time period, (b) find amplitude of each block and (c) length of spring as a function of time.   x0

k

m1 Sol.

m2

(a) Here both the blocks will be in equilibrium at the same time when spring is in its natural length. Let EP1 and EP2 be equilibrium positions of block A and B as shown in figure.

EP2

EP1 

m1

k

m2

Let at any time during oscillations, blocks are at a distance of x1 and x2 from their equilibrium positions.

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Page # 24

SIMPLE HARMONIC MOTION

x1

EP2 x2

EP1 k

m1

m2

As no external force is acting on the spring block system 

(m1 + m2)xcm = m1x1 – m2x2 = 0

or

m1x1 = m2x2

For 1st particle, force equation can be written as k(x1 + x2) = – m1

or,

a1  –

d2 x 1 dt

k(m1  m 2 ) x1 m1m 2



m1m 2  Hence, T = 2 k(m  m ) = 2 K 1 2 (b)

  m k  x 1  1 x 1   –m1a1 m2  

or,

2

2 

k(m1  m 2 ) m1m 2

m1m 2 where   (m  m ) which is known as reduced mass 1 2

Let the amplitude of blocks be A1 and A2. m1A1 = m2A2 By energy conservation ;

(c)

1 1 k( A 1  A 2 ) 2  kx 20 2 2

or,

A1 + A2 = x0

or,

A1 + A 2 = x 0

or,

A1 

or,

A1 

m 2 x0 m1  m 2

Similarly,

m1 A 1  x0 m2 m1x 0 A2 = m  m 1 2

Let equilibrium position of 1st particle be origin, i.e., x = 0.

x1 = A1 cost and

EP2

EP1 

x co-ordinate of particles can be written as x2 =  – A2cost

m1

Hence, length of spring can be written as : length = x2 – x1

m2 x=0

=  – (A1 + A2)cost 13.

COMBINATION OF SPRINGS :

13.1 Series Combination : Total displacement x = x1 + x2

k1

Tension in both springs = k1x1 = k2x2 

k2 m

Equivalent constant in series combination Keq is given by :

1/keq = 1/k1 + 1/k2



T  2

m k eq

In series combination, tension is same in all the springs & extension will be different. (If k is same then deformation is also same)

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Page # 25

SIMPLE HARMONIC MOTION 13.2

Parallel combination :

Extension is same for both springs but force acting will be different. Force acting on the system = F 

F = – (k1x + k2x)



F = – (k1 + k2) x 

k1

F = – keqx

k2

m  T  2 k eq

 keq = k1 + k2

m

Ex.24 Find the time period of the oscilltion of mass m in figure a and b. What is the equivalent spring constant of the spring in each case. ?

m m

m

k3

(b)

(a) Sol.

k2

k1

k2

k1

In figure (a) k1

k2



k 1k 2 k1  k 2



k1k 2 + k3 k1  k 2

Which gives k 1k 2 k1  k 2

k3

k 1k 2 keq = k  k  k 3 = 1 2 Now

T  2

m k eq

k 1k 2  k 2k 3  k 1k 3 k1  k 2

 2

m(k 1  k 2 ) k 1k 2  k 2k 3  k 1k 3

Case (b) If the block is displaced slightly by an amount x then both the spring are displaced by x from their natural length so it is parallel combination of springs. which gives keq = k1 + k2 T  2

m m 2 k eq = k1  k 2

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Page # 26

SIMPLE HARMONIC MOTION

Note : •

In series combination, extension of springs will be reciprocal of its spring constant.  k  1/  k11 = k22 = k33

•

If a spring is cut in 'n' equal pieces then spring constant of one piece will be nk.

Ex.25 The friction coefficient between the two blocks shown in figure is  and the horizontal plane is smooth. (a) If the system is slightly displaced and released, find the time period. (b) Find the magnitude of the frictional force between the blocks when the displacement from the mean position is x. (c) What can be the maximum amplitude if the upper block does not slip relative to the lower block? Sol.

(a)

k

m M

For small amplitude, the two blocks oscillate together. The angular frequency is 

(b)

k Mm and so the time period T  2 Mm k

The acceleration of the blocks at displacement x from the mean position is  – kx  a  – 2 x    Mm  – mkx   The resultant force on the upper block is, therefore, ma =   Mm 

This force is provided by the friction of the lower block. Hence, the magnitude of the  mk | x |     Mm 

frictional force is (c)

Maximum force of friction required for simple harmonic motion of the upper block is mkA at the extreme positions. But the maximum frictional force can only be  mg. Mm Hence mkA  mg Mm

14.

or,

A

(M  m)g k

ENERGY METHOD : Another method of finding time period of SHM is energy method. To understand this method we will consider the following example.

Ex.26 Figure shows a system consisting of pulley having radius R, a spring of force constant k and a block of mass m. Find the period of its vertical oscillation.

K

m

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Page # 27

SIMPLE HARMONIC MOTION Sol.

The following steps are usually followed in this method:

Step 1. Find the mean position. In following figure point A shows mean position. Step 2. Write down the mean position force relation from figure.

T=mg T=kx0

mg = kx0 Step 3. Assume that particle is performing SHM with amplitude A. Then displace the particle from its mean position.

T x0 N.L.

Step 4. Find the total mechanical energy (E) in the displaced position since, mechanical energy in SHM dE 0 remains constant dt *

E=

1 1 1 mv 2  I2  k( x  x 0 )2 – mgx 2 2 2

E=

1 1 v2 1 mv 2  I 2  k( x  x 0 )2 – mgx 2 2 R 2

dE 2mv dv 2Iv dv 2k( x  x 0 ) dx dx    – mg dt 2 dt 2R 2 dt 2 dt dt

Put

T m

A

M.P.

mg

...(1)

dx dv d 2 x  v and  2 dt dt dt

in eq. (1) put dE 0  dt

mv

d2 x dt 2



I v d2 x R2 dt 2

 kxv  kx0 v  mgv  0



I  d2 x  which gives  m  2  2  kx  0 R  dt  d2 x dt 2



k x0 I   m  2  R  

x+x0

...(2)

n.

15.

k I   m  2  R  

G.P.E=0 x m

compare eq. (2) with S.H.M eq. the 2 

v R



T  2

v

(m  I / R2 ) k

ANGULAR S.H.M. : If the restoring torque acting on the body in oscillatory motion is directly proportional to the angular displacement of body from its equillibrium position i.e., =–k k = S.H.M. constant  = angular displacement from M.P. S.H.M. equation is given by d2 dt 2

 2   0

Here



K I

Here I is moment of inertia of the body/particle about a given axis.

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Page # 28 16.

SIMPLE HARMONIC MOTION

SIMPLE PENDULUM :

If a heavy point-mass is suspended by a weightless, inextensible and perfectly flexible string from a rigid support, then this arrangement is called a simple penduluml. Time period of a simple pendulum T  2  . g

O  

(some times we can take g = 2 for making calculation simple)

m n   mgcos si mg mg Proof : Now taking moment of forces acting on the bob about point O.  = T + mg T = 0 

 = –(mg sin ) if  is very small then



 = – mg  

sin  ~– 

...(1)

Now compare eq. (1) with net = – KS.H.M  which gives



T  2

KS.H.M = mg  I K S.H.M

= 2

m 2 mg

 = 2 g

Note :

17.

•

Time period of second pendulum is 2 seconds.

•

Simple pendulum performs angular S.H.M. but due to small angular displacement, it is considered as linear S.H.M.

•

If time period of clock based upon simple pendulum increases then clock will become slow but if time period decreases then clock will become fast.

TIME PERIOD OF SIMPLE PENDULUM IN ACCELERATING REFERENCE FRAME : T  2

 geff . where

  geff = Effective acceleration due to gravity in reference system = | g – a |

 a = acceleration of the point of suspension w.r.t. ground.   Condition for applying this formula : | g – a | = constant 



If the acceleration a is upwards, then | g eff | g  a and T  2

 ga

Time lost or gained in time t is given by T' 

T .t T

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Page # 29

SIMPLE HARMONIC MOTION Ex.27 If T = 2 sec

Tnew = 3 sec. then

T = 1 sec.

Since time lost by clock in 3 sec is = 1 sec then time lost by clock in 1 sec =

1 sec 3

 Time lost by the clock in an hour =

1  3600 = 1200 sec. 3

Ex.28 A simple pendulum is suspended from the ceiling of a car which is accelerating uniformly on a horizontal road. The acceleration of car is a0 and the length of the pendulum is 1. Then find the time period of small oscillations of pendulum about the mean position. Sol.

We shall work in the car frame. As it is accelerated with respect to the road, we shall have to apply a psuedo force ma0 on the bob of mass m. For mean position, the acceleration of the bob with respect to the car should be zero. If 0 be the angle made by the string with the vertical, the tension, weight and the peusdo force will add to zero in this position. Hence, resultant of mg and ma0 (say F = m g2  a 02 ) has to be along the string.



tan  0 

ma 0 a 0  mg g

Now, suppose the string is further deflected by an angle  as shown in figure. O

Now, restoring torque about point O can be given by   I 

2

(F sin )  = – m  

  0

Substituting F and using sin  = , for small .

ma0 

( m g2  a 02 )   = – m2 

or,

–

g2  a 20



so ;



mg

F

2 

g 2  a 02 

This is an equation of simple harmonic motion with time period. T

18.

 2 = 2 2 ( g  a 02 )1/ 4 

COMPOUND PENDULUM / PHYSICAL PENDULUM :

s ×

When a rigid body is suspended from an axis and

s × 

made to oscillate about that then it is called compound pendulum.





c

c

C = Position of centre of mass S = Point of suspension  = Distance between point of suspension and centre of mass

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mg sin  mg mg cos 

Page # 30

SIMPLE HARMONIC MOTION

(it remains constant during motion for small angular displacement "" from mean position The restoring torque is given by  = – mg  sin   = – mg  

 for small , sin  = 

or,

I = – mg  

where, I = Moment of inertia about point of suspension.

or,

a= –

mg   I

or,

Time period, T  2

2 

mg  I

I mg 

Ex. 29 A ring is suspended at a point on its rim and it behaves as a second's pendulum when it oscillates such that its centre move in its own plane. The radius of the ring would be ( g = 2) Sol. Time period of second pendulum T = 2 cm. × I T  2 R Mgd Moment of inertia with respect to axis O I = MR2 + MR2 = 2MR2 the distance between centre of mass and the axis O d=R 2  2

2MR 2 MgR

M

C.O.M

 R = 0.5 m

Ex.30 A circular disc has a tiny hole in it, at a distance z from its center. Its mass is M and radius R(R > 2). Horizontal shaft is passed through the hole and held fixed so that the disc can freely swing in the vertical plane. For small distrubance, the disc performs SHM whose time period is minimum for z . Find the value of z.

Sol.

I The time period w.r.t the axis T  2 Mgd where I = moment of inertia w.r.t the axis O d = distance between C.O.M and O 

I=

z

M

R

×O C.O.M

MR 2  Mz 2 2

d=z



MR 2  Mz 2 R2 z 2 T  2  2  Mgz 2gz g

the time period will be minimum when Let say f 

R2  z = minimum 2z

R2 z 2z

f will be minimum when

df 0 dz

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Page # 31

SIMPLE HARMONIC MOTION R R2 1 0  z  2 2 2z

 –

Ex.31 Find out the angular frequency of small oscillation about axis O

m

l

k

k

m

l



(1)

l

(2)

mg

Sol. k

k

(k l ) l The compression in spring (1) = l  and the extension in spring (2) = l 

(k l )

l Net torque opposite to the mean position = – (2 kl  ) l – mg sin  = net 2  is small  sin    l net = – I 2  = – (2 kl  ) l – mg sin  = net 2 I=

 19.

ml 2 3

=

3 ( 4kl  mg ) 2 ml

TORSIONAL PENDULUM : In torsional pendulum, an extended object is suspended at the centre by a light torsion wire. A torsion wire is essentially inextensible, but is free to twist about its axis. When the lower end of the wire is rotated by a slight amount, the wire applies a restoring torque causing the body to oscillate rotationally when released. The restoring torque produced is given by

C

or,

 = – C I = – C

or,

= –

C  I

A

C

X

A  X

where, C = Torsional constant where, I = Moment of inertia about the vertical axis.  Time Period,

T  2

I C

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Page # 32

SIMPLE HARMONIC MOTION

: The above concept of torsional pendulum is used in inertia table to calculate the moment of inertia of unknown body. Ex.32 A uniform disc of radius 5.0 cm and mass 200 g is fixed at its centre to a metal wire, the other end of which is fixed to a ceiling. The hanging disc is rotated about the wire through an angle and is released. If the disc makes torsional oscillations with time period 0.20 s, find the torsional constant of the wire. Sol.

The situation is shown in figure. The moment of inertia of the disc about the wire is

I

mr 2 (0.200kg)(5.0  10 –2 m)2  = 2.5 × 10–4 kg-m2. 2 2

The time period is given by T  2

I C

or,

C

4 2I

4 2 ( 2.5  10 –4 kg  m 2 ) =

20.

(0.20 s)

2

T2

= 0.25

kg  m 2 s2

VECTOR METHOD OF COMBINING TWO OR MORE SIMPLE HARMONIC MOTIONS: A simple harmonic motion is produced when a force (called restoring force) proportional to the displacement acts on a particle. If a particle is acted upon by two such forces the resultant motion of the particle is a combination of two simple harmonic motions.

20.1

In Same direction : (a) Having same Frequencies: Suppose the two individual motions are represented by, x1 = A1 sin t and

x2 = A2 sin (t + )

Both the simple harmonic motions have same angular frequency . x = x1 + x2 = A1 sin t + A2 sin (t + ) = A sin (t + ) Here,

A  A 12  A 22  2A 1A 2 cos 

and

A 2 sin  tan  = A  A cos  1 2



A2



A A 2 sin 



 A 2 cos  A1 Thus, we can see that this is similar to the vector addition. The same method of vector addition can be applied to the combination of more than two simple harmonic motions.

Important points to remember before solving the questions: 1.

Convert all the trignometric ratios into sine form and ensure that t term is with +ve sign.

2.

Make the sign between two term +ve.

3.

A1 is the amplitude of that S.H.M whose phase is small.

4.

Then resultant x = Anet sin (phase of A1 + )

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Page # 33

SIMPLE HARMONIC MOTION

Where Anet is the vector sum of A1 & A2 with angle between them is the phase difference between two S.H.M.

Ex.33 x1 = 3 sin t ; x2 = 4 cos t Find (i) amplitude of resultant SHM. (ii) equation of the resultant SHM. Sol.

First right all SHM's in terms of sine functions with positive amplitude. Keep "t" with positive sign. 

x1 = 3 sin t x2 = 4 sin (t + /2)

A  3 2  4 2  2  3  4 cos 4 sin tan  

 2



 2

=

9  16 =

25 = 5

4 3

 = 53°  2 equation x = 5 sin (t + 53°) 3  4 cos

10

Ex.34 x1 = 5 sin (t + 30°) ; x2 = 10 cos (t) Find amplitude of resultant SHM. Sol. x1 = 5 sin (t + 30°) x2 = 10 sin (t +

5 Phasor Diagram

 ) 2

A  5 2  10 2  2  5  10 cos 60  =

60°

25  100  50  175 = 5 7

Ex.35 A particle is subjected to two simple harmonic motions x1 = A1 sin t and

x2 = A2 sin (t + /3)

Find (a) the displacment at t = 0, (b) the maximum speed of the particle and (c) the maximum acceleration of the particle. Sol.

(a) At t = 0, x1 = A1 sin t = 0 and

A2 3 2 Thus, the resultant displacement at t = 0 is

x2 = A2 sin (t + /3) = A2 sin (/3) =

3 2 The resultant of the two motion is a simple harmonic motion of the same angular frequency . The amplitude of the resultant motion is x = x1 + x2 = A 2

(b)

A  A 12  A 22  2A 1A 2 cos(  / 3)

=

A 12  A 22  A 1A 2

The maximum speed is umax = A  =  A 12  A 22  A 1A 2 (c)

The maximum acceleration is amax = A 2 = 2 A 12  A 22  A 1A 2

(b) Having different frequencies x1 = A1 sin t

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Page # 34

SIMPLE HARMONIC MOTION x2 = A2 sin 2t

then resultant displacement x = x1 + x2 = A1 sin 1t + A2 sin 2t This resultant motion is not SHM. 20.2

In two perpendicular directions x = A1 sin t

...(1)

y = A2 sin (t + )

...(2)

The Amplitudes A1 and A2

may be different and Phase difference  and  is same.

So equation of the path may be obtained by eliminating t from (1) & (2) x sin  t = A 1

...(3)

1–

cos t =

x2 2

A1

...(4)

On rearranging we get x2 A

2 1



y2 A2

2xy cos   sin2  A 1A 2



2

...(5)

(general eq. of ellipse) special case : (1)

If  = 0

x2 

A

2 1



y2 A2

2



2xy 0 A 1A 2

A2  y = A .x (eq. of straight line) 1 (2)

If  = 90°

x2  (3)

A

2 1



A2

y2 1 A 22

A1

(Eq. of ellipse)

If  = 90° & A1 = A2 = A

A then x2 + y2 = A2

(Eq. of circle.)

A

The above figures are called Lissajous figures.

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Page # 35

SIMPLE HARMONIC MOTION

MIND MAP 1. Equation of S.H.M (i) Linear : a = –2x (ii) Angular :  = – 2

2. Linear SHM (i) Displacement of particle : x = A sin (t + ) (ii) Velocity

3. Angular S.H.M (i) Displacement :  = 0 sin(t + )

dx = A cos(t + ) dt

(ii) Angular velocity d = 0 cos(t + ) dt

=  A 2 – x2 (iii) Acceleration :

(iii) Acceleration

2

d x dt 2

d2 

= – A2 sin(t + )

  0  2 sin(t  )   2 

dt 2

= – 2x (iv) Phase : t +  (v) Phase constant : 

(iv) Phase : t +  (v) Phase constant : 

5. Time Period : 4. Energy in S.H.M (i) K =

(ii) U =

Pendulums : (a) Simple pendulum :

1 m 2 ( A 2 – x 2 ) 2

SHM

1 m2x2 2

(iii) E = K + U =

l g (b) Physical pendulum : I T  2 mg l (c) Torsional pendulum : T  2

1 m2 A2 2

= constant

6. Mass-spring system (a) T  2 m k (b) Two bodies system : T  2

 ; k

m1 m 2 where   m  m 1 2

T  2

Combination of springs : (a) series :

7. Composition of 2 SHMs : x1 = A1 sin t

1 1 1   K eff K 1 K 2

x2 = A2 sin (t + ) x = x1 + x2

(b) parallel : Keff = K1 + K2 (c) Spring cut into two parts m : n K1 

I C

x = A sin (t + ) where, A  A 12  A 22  2A 1A 2 cos 

(m  n)K (m  n)K ,K 2  m n

and tan  

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A 2 sin  A 1  A 2 cos 

Page # 36

SIMPLE HARMONIC MOTION

Exercise - I

(SINGLE

LINEAR S.H.M 1. For a particle executing simple harmonic motion, the acceleration is proportional to. (A) displacement from the mean position (B) distance from the mean position (C) distance travelled since t = 0 (D) speed 2. The distance moved by a particle in simple harmonic motion in one time period is (A) A (B) 2A (C) 4A (D) zero 3. The time period of a particle in simple harmonic motion is equal to the time between consecutive appearance of the particle at a particular point in its motion. This point is (A) the mean position (B) an extreme position (C) between the mean position and the positive extreme. (D) between the mean position and the negative extreme. 4. Equations y = 2 A cos2t and y = A(sin t + 3 cost) represent the motion of two particles. (A) Only one of these is S.H.M (B) Ratio of maximum speeds is 2 : 1 (C) Ratio of maximum speeds is 1 : 1 (D) Ratio of maximum accelerations is 1 : 4

5. The displacement of a body executing SHM is given by x = A sin(2t + /3). The first time from t = 0 when the velocity is maximum is (A) 0.33 sec (B) 0.16 sec (C) 0.25 sec (D) 0.5 sec

CHOICE QUESTIONS)

8. The maximum acceleration of a particle in SHM is made two times keeping the maximum speed to be constant. It is possible when (A) amplitude of oscillation is doubled while frequency remains constant (B) amplitude is doubled while frequency is halved (C) frequency is doubled while amplitude is halved (D) frequency is doubled while amplitude remains constant. 9. A small mass executes linear SHM about O with amplitude a and period T. Its displacement from O at time T/8 after passing through O is : (A) a/8

(B) a/2 2

(C) a/2

(D) a / 2

10. The time period of a particle in simple harmonic motion is equal to the smallest time between the  particle acquiring a particular velocity v . The value of v is (A) vmax (B) 0 (C) between 0 and vmax (D) between 0 and –vmax 11. The average acceleration in one time period in a simple harmonic motion is (A) A 2 (B) A 2/2 (C) A 2/ 2

(D) zero

12. A mass m is performing linear simple harmonic motion, then correct graph for acceleration a and corresponding linear velocity v is

v2 6. A simple harmonic motion having an amplitude A and time period T is represented by the equation : y = 5 sin(t + 4) m Then the values of A (in m) and T (in sec) are : (A) A = 5; T = 2 (B) A = 10 ; T = 1 (C) A = 5 ; T = 1 (D) A = 10 ; T = 2 7. Two particles are in SHM on same straight line with amplitude A and 2A and with same angular frequency . It is observed that when first particle is at a distance A / 2 from origin and going toward mean position, other particle is at extreme position on other side of mean position. Find phase difference between the two particles. (A) 45° (B) 90° (C) 135° (D) 180°

v2

(A)

(B)

a2

a2

v2

v2

(C)

(D)

a2

a2

13. The time taken by a particle performing SHM to pass from point A to B where its velocities are same is 2 seconds. After another 2 seconds it returns to B. The time period of oscillation is (in seconds) (A) 2 (B) 8 (C) 6 (D) 4

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Page # 37

SIMPLE HARMONIC MOTION 14. Two particles undergo SHM along parallel lines with the same time period (T) and equal amplitudes. At a particular instant, one particle is at its extreme position while the other is at its mean position. They move in the same direction. They will cross each other after a further time B B’

O

A

O’

A’

(A) T/8 (C) T/6

(B) 3T/8 (D) 4T/3

15. A particle performing SHM is found at its equilibrium at t = 1 sec. and it is found to have a speed of 0.25 m/s at t = 2 sec. If the period of oscillation is 6 sec. Calculate amplitude of oscillation (A)

(C)

3 m 2

(B)

6 m 

(D)

3 m 4 3 8

16. A particle executes SHM with time period T and amplitude A. The maximum possible average velocity in time

(B)

4A T

(C)

8A T

(D)

4 2A T

17. Time period of a particle executing SHM is 8 sec. At t = 0 it is at the mean position. The ratio of the distance covered by the particle in the 1 st second to the 2nd second is 1 (B)

2

(D)

2 +1

1 (C)

2

(A)

2 3

(B)

(C)

 3

(D)

 2  4

21. A particle of mass 1 kg is undergoing S.H.M., for which graph between force and displacement (from mean position) as shown. Its time period, in seconds, is. F(N) 13.5 1.5

2A T

2 1

20. Two particles execute SHM of same amplitude of 20 cm with same period along the same line about the same equilibrium position. The maximum distance between the two is 20 cm. Their phase difference in radians is

T is 4

(A)

(A)

19. Two particles are in SHM in a straight line about same equilibrium position. Amplitude A and time period T of both the particles are equal. At time t = 0, one particle is at displacement y1 = +A and the other at y2 = – A/2, and they are approaching towards each other. After what time they cross each other ? (A) T/3 (B) T/4 (C) 5T/6 (D) T/6

–1.5 –13.5

(A) /3 (C) /6

xm

(B) 2/3 (D) 3/

22. A point particle of mass 0.1 kg is executing S.H.M of amplitude of 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 when the initial phase of oscillation is 45° can be given by   (A) 0.1cos 4 t    4

  (B) 0.1sin 4 t    4

(C) 0.4 sin t     4

  (D) 0.2 sin  2 t 2 

18. The angular frequency of motion whose equad2 y tion is 4

dt 2

+ 9y = 0 is (y = displacement and t

= time) (A)

9 4

(B)

4 9

(C)

3 2

(D)

2 3

23. A particle executes SHM of period 1.2 sec. and amplitude 8 cm. Find the time it takes to travel 3 cm from the positive extremely of its oscillation. (A) 0.28 sec. (B) 0.32 sec. (C) 0.17 sec. (D) 0.42 sec.

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Page # 38

SIMPLE HARMONIC MOTION

24. Two particles P and Q describe simple harmonic motions of same period, same amplitude, along the same line about the same equilibrium position O. When P and Q are on opposite sides of O at the same distance from O they have the same speed of 1.2 m/s in the same direction, when their displacements are the same they have the same speed of 1.6 m/s in opposite directions. The maximum velocity in m/s of either particle is (A) 2.8 (B) 2.5 (C) 2.4 (D) 2

with the two spring in series is T, then (A) T = t1 + t2 (B) T 2 = t12 + t22

25. A particle performs SHM with a period T and amplitude a. The mean velocity of the particle over the time interval during which it travels a distance a/2 from the extreme position is (A) a/T (B) 2a/T (C) 3a/T (D) a/2T

Question No. 31 to 33 (3 questions) The graph in figure show that a quantity y varies with displacement d in a system undergoing simple harmonic motion.

B

C

3R

D

2

(B)

(C)

3 R

(D)

1 

t12

1 

t 22

y

y

(I)

(II) O

d

d

O

y

E

y

2 2 R

(III)

27. A toy car of mass m is having two similar rubber ribbons attached to it as shown in the figure. The force constant of each rubber ribbon is k and surface is frictionless. The car is displaced from mean position by x cm and released. At the mean position the ribbons are underformed. Vibration period is

m( 2k ) k

m k

2

(B)

1 2

(D) 2

m( 2k ) k2

m k k

28. A spring mass system oscillates with a frequency . If it is taken in an elevator slowly accelerating upward, the frequency will (A) increase (B) decrease (C) remain same (D) become zero 29. A body at the end of a spring executes S.H.M. with a period t1, while the corresponding period for another spring is t2. If the period of oscillation

(IV) O

O

d

d

Which graphs best represents the relationship obtained when y is 31. The total energy of the system (A) I (B) II (C) III (D) IV 32. The time (A) I (C) III

(C) 2

T

2

R

(A)

(A) 2 

(D)

30. A particle moves along the x-axis according to : x = A. [1 + sin t]. What distance does it travel between t = 0 and t = 2.5/? (A) 4A (B) 6A (C) 5A (D) none

26. A body performs simple harmonic oscillations along the straight line ABCDE with C as the midpoint of AE. Its kinetic energies at B and D are each one fourth of its maximum value. If AE = 2R, the distance between B and D is A

1

1 1 1 (C) T  t  t 1 2

(B) II (D) IV

33. The unbalanced force acting on the system. (A) I (B) II (C) III (D) None 34. A particle executes SHM on a straight line path. The amplitude of oscillation is 2 cm. When the displacement of the particle from the mean position is 1 cm, the numerical value of magnitude of acceleration is equal to the numerical value of magnitude of velocity. The frequency of SHM (in second–1) is 2 (A) 2 3 (C)

3 2

(B)

3 1

(D)

2 3

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Page # 39

SIMPLE HARMONIC MOTION 35. The potential energy of a simple harmonic oscillator of mass 2 kg in its mean position is 5 J. If its total energy is 9J and its amplitude is 0.01 m, its time period would be (A)

 sec 10

(B)

 sec 20

(C)

 sec 50

(D)

 sec 100

36. Find the ratio of time periods of two identical springs if they are first joined in series & then in parallel & a mass m is suspended from them (A) 4 (B) 2 (C) 1 (D) 3

ticle at the time T/12 is : (T = time period) (A) 2 : 1 (B) 3 : 1 (C) 4 : 1 (D) 1 : 4 41. In the figure, the block of mass m, attached to the spring of stiffness k is in contact with the completely elastic wall, and the compression in the spring is ‘e’. The spring is compressed further by ‘e’ by displacing the blocktowards left and is then released. If the collision between the block and the wall is completely elastic then the time period of oscillations of the block will be : Wall

37. Two bodies P & Q of equal mass are suspended from two separate massless springs of force constants k1 & k2 respectively. If the maximum velocity of them are equal during their motion, the ratio of amplitude of P to Q is :

m

k1 (A) k 2

(B)

k2 k1

(A)

2 3

k2 (C) k 1

k1 k2

(C)

(D)

 3

38. Vertical displacement of a plank with a body of mass ‘m’ on it is varying according to law y = sin t + 3 cos t. The minimum value of  for which the mass just breaks off the plank and the moment it occurs first after t = 0 are given by (y is positive vertically upwards) (A)

g , 2

(C)

g  , 2 3

2 6

(B)

g 2 , 2 3

 g

(D)

2g ,

2 3g

 g

2 g

39. Two particles A and B perform SHM along the same straight line with the same amplitude ‘a’, same frequency ‘f’ and same equilibrium position ‘O’. The greatest distance between them is found to be 3a/2. At some instant of time they have the same displacement from mean position. What is the displacement? (A) a / 2 (C)

3a/2

m k m k

m k

 6

m k

(D)

42. A spring mass system performs S.H.M. If the mass is doubled keeping amplitude same, then the total energy of S.H.M. will become : (A) double (B) half (C) unchanged (D) 4 times 43. A mass at the end of a spring executes harmonic motion about an equilibrium position with an amplitude A. Its speed as it passes through the equilibrium position is V. If extended 2A and released, the speed of the mass passing through the equilibrium position will be (A) 2V (B) 4V V V (C) (D) 2 4 44. A 2 Kg block moving with 10 m/s strikes a spring of constant 2N/m attached to 2 Kg block at rest kept on a smooth floor. The time for which rear moving block remain in contact with spring will be 10m/s 2kg

(B) a 7 / 4

2kg

(D) 3 a /4

1 (A)

40. A particle starts oscillating simple harmonically from its equilibrium position then, the ratio of kinetic energy and potential energy of the par-

(B) 2

2 sec

(C) 1 sec

(B) (D)

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2

sec

1 sec 2

Page # 40

SIMPLE HARMONIC MOTION

45. In the above question, the velocity of the rear 2 kg block after it separates from the spring will be : (A) 0 m/s (B) 5 m/s (C) 10 m/s (D) 7.5 m/s

(A) 2 m / k + 4 2E / mg 2

46. A particle is subjected to two mutually perpendicular simple harmonic motions such that its x and y coordinates are given by x = 2 sin t ; y   = 2 sin  t   4 The path of the particle will be : (A) an ellipse (B) a straight line (C) a parabola (D) a circle

(D) 2 2E / mg2

47. The amplitude of the vibrating particle due   to superposition of two SHMs, y1 = sin  t   3 and y2 = sin t is : (A) 1 (C)

2

(B) 3

(D) 2

48. Two simple harmonic motions y1 = A sin t and y2 = A cos t are superimposed on a particle of mass m. The total mechanical energy of the particle is : (A)

1 m2A2 2

(B) m2A2

1 (C) m2A2 4

(B) 2 m / k (C)  m / k + 2 2E / mg2

51. A particle of mass 4 kg moves between two points A and B on a smooth horizontal surface under the action of two forces such that when it   is at a point P, the forces are 2PA N and 2PB N. If the particle is released from rest at A, find the time it takes to travel a quarter of the way from A to B.   (A) s (B) s 2 3  s (C) s (D) 4 52. In an elevator, a spring clock of time period TS (mass attached to a spring) and a pendulum clock of time period TP are kept. If the elevator accelerates upwards (A) TS well as TP increases (B) TS remain same, TP increases (C) TS remains same, TP decreases (D) TS as well as TP decreases 53. A man is swinging on a swing made of 2 ropes of equal length L and in direction perpendicular to the plane of paper. The time period of the small oscillations about the mean position is

(D) zero

49. The springs in fig. A and B are identical but length in A is three times each of that in B. The ratio of period TA/TB is

A

L

B

L

m (A) 3 (C) 3

M m (B) 1/3 (D) 1/3

(A) 2

50. A particle of mass m moves in the potential energy U shown above. The period of the motion when the particle has total energy E is U(x)

1

U = 2 kx2, x < 0

L

U = mgx, x > 0

x

L 2g L

(C) 2

2 3g

3L 2g

(B) 2

(D) 

L g

54. A ring of diameter 2m oscillates as a compound pendulum about a horizontal axis passing through a point at its rim. It oscillates such that its centre move in a plane which is perpendicular to the plane of the ring. The equivalent length of the simple pendulum is (A) 2m (B) 4m (C) 1.5 m (D) 3m

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Page # 41

SIMPLE HARMONIC MOTION 55. Two pendulums have time periods T and 5T/ 4. They start SHM at the same time from the mean position. After how many oscillations of the smaller pendulum they will be again in the same phase (A) 5 (B) 4 (C) 11 (D) 9 56. A small bob attached to a light inextensible thread of length l has a periodic time T when allowed to vibrate as a simple pendulum. The thread is now suspended from a fixed end O of a 3 vertical rigid rod of length (as in figure). If 4 now the pendulum performs periodic oscillations in this arrangement, the periodic time will be

(A) 2

2l 3g

2l (C) 2 3g

(B) 2

2 2l 3g

(D) 3

l 3g

60. In the figure shown, the spring are connected to the rod at one end and at the midpoint. The rod is hinged at its lower end. Rotational SHM of the rod (Mass m, length L) will occur only if

k

k g

O 3l 4

(A) k > mg / 3L (C) k > 2mg/5L

l

A

3T (A) 4 (C) T

T (B) 2 (D) 2T

57. A simple pendulum is oscillating in a lift. If the lift is going down with constant velocity, the time period of the simple pendulum is T1. If the lift is going down with some retardation its time period is T2, then (A) T1 > T2 (B) T1 < T2 (C) T1 = T2 (D) depends upon the mass of the pendulum bob 58. A simple pendulum with length  and bob of mass m executes SHM of small amplitude A. The maximum tension in the string will be (A) mg (1 + A/) (B) mg (1 + A/)2 (C) mg [1 + (A/)2] (D) 2 mg 59. A system of two identical rods (L-shaped) of mass m and length l are resting on a peg P as shown in the figure. If the system is displaced in its plane by a small angle , find the period of oscillations.

l

P

(B) k > 2mg/3L (D) k > 0

61. What is the angular frequency of oscillations of the rod in the above problem if k = mg/L ? (A) (3/2).[k/m]1/2 (B) (3/4).[k/m]1/2 (C) [2k/5m]1/2 (D) None 62. A ring is suspended at a point on its rim and it behaves as a second’s pendulum when it oscillates such that its centre move in its own plane. The radius of the ring would be (g = 2) (A) 0.5 m (B) 1.0 m (C) 0.67 m (D) 1.5 m 63. A rod whose ends are A & B of length 25 cm is hanged in vertical plane. When hanged from point A and point B the time periods calculated are 3 sec & 4 sec respectively. Given the moment of inertia of rod about axis perpendicular to the rod is in ratio 9 : 4 at points A and B. Find the distance of the centre of mass from point A. (A) 9 cm (B) 5 cm (C) 25 cm (D) 20 cm

l

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Page # 42

SIMPLE HARMONIC MOTION

(Multiple Choice Problems)

Exercise - II

1. A spring has natural length 40 cm and spring constant 500 N/m. A block of mass 1 kg is attached at one end of the spring and other end of the spring is attached to ceiling. The block released from the position, where the spring has length 45 cm. (A) the block will perform SHM of amplitude 5 cm.

2 . They are in same phase. 3 Their initial phase difference is :

(B) the block will have maximum velocity 30 5 cm/sec. (C) the block will have maximum acceleration 15 m/s2. (D) the minimum potential energy of the spring will be zero.

5. Two particles are in SHM with same angular frequency and amplitudes A and 2A respectively along same straight line with same mean position. They cross each other at position A/2 distance from mean position in opposite direction. The phase between them is :

2. A particle executing a simple harmonic motion of period 2s. When it is at its extreme displacement from its mean position, it receives an additional energy equal to what it had in its mean position. Due to this, in its subsequent motion, (A) its amplitude will change and become equal to 2 times its previous amplitude (B) its periodic time will become doubled i.e. 4s (C) its potential energy will be decreased (D) it will continue to execute simple harmonic motion of the same amplitude and period as before receiving the additional energy.

(A)

 1 5 – sin–1   4 6

(B)

 1  – sin–1   4 6

(C)

 1 5 – cos–1   4 6

(D)

  1 – cos –1    4 6

3. Part of a simple harmonic motion is graphed in the figure, where y is the displacement from the mean position. The correct equation describing this S.H.M is 2 (0.6)

O

t(s)

difference. At, t =

(A)

 3

(B)

2 3

(C)

4 3

(D) 

6. The equation of motion for an oscillating particle is given by x = 3sin (4t) + 4 cos(4t), where x is in mm and t is in second (A) The motion is simple harmonic (B) The period of oscillation is 0.5 s (C) The amplitude of oscillation is 5 mm (D) The particle starts its motion from the equilibrium 7. A particle is executing SHM of amplitude A, about the mean position X = 0. Which of the following cannot be a possible phase difference between the positions of the particle at x = + A/ 2 and x = – A/ 2 . (A) 75° (B) 165° (C) 135° (D) 195°

(0.3)

–2

(A) y = 4 cos (0.6t)

  10 t–  (B) y = 2 sin  3 2

8. Speed v of a particle moving along a straight line, when it is at a distance x from a fixed point on the line is given by v2 = 108 – 9x2 (all quantities in S.I. unit). Then (A) The motion is uniformly accelerated along the straight line (B) The magnitude of the acceleration at a distance 3 cm from the fixed point is 0.27 m/s2. 12 m. (D) The maximum displacement from the fixed point is 4 cm.

(C) The motion is simple harmonic about x =   10 t  (C) y = 4 sin  3 2

  10 t  (D) y = 2 cos  3 2

4. Two particles execute SHM with amplitude A and 2A and angular frequency  and 2 respectively. At t = 0 they starts with some initial phase

9. A block is placed on a horizontal plank. The plank is performing SHM along a vertical line with amplitude of 40 cm. The block just loses contact with the plank when the plank is momentarily at rest. Then

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Page # 43

SIMPLE HARMONIC MOTION (A) the period of its oscillations is 2/5 sec. (B) the block weights on the plank double its weight, when the plank is at one of the positions of momentary rest. (C) the block weights 1.5 times its weight on the plank halfway down from the mean position. (D) the block weights its true weight on the plank, when velocity of the plank is maximum. 10. The potential energy of a particle of mass 0.1 kg, moving along x-axis, is given by U = 5x (x – 4) J where x is in metres. It can be concluded that (A) the particle is acted upon by a constant force (B) the speed of the particle is maximum at x = 2 m (C) the particle executes simple harmonic motion (D) the period of oscillation of the particle is /5 s 11. A particle is executing SHM with amplitude A, time period T, maximum acceleration a0 and maximum velocity v0. Its starts from mean position at t = 0 and at time t, it has the displacement A/2, acceleration a and velocity v then (A) t = T/12 (B) a = a0/2 (C) v = v0/2 (D) t = T/8 12. The amplitude of a particle executing SHM about O is 10 cm. Then (A) When the K.E. is 0.64 of its max. K.E. its displacement is 6cm from O. (B) When the displacement is 5 cm from O its K.E. is 0.75 of its max. P.E. (C) Its total energy at any point is equal to its maximum K.E. (D) Its velocity is half the maximum velocity when its displacement is half the maximum displacement. 13. The displacement of a particle varies according to the relation x = 3 sin 100t + 8 cos2 50t. Which of the following is/are correct about this motion. (A) the motion of the particle is not S.H.M. (B) theamplitude of the S.H.M. of the particle is 5 units (C) the amplitude of the resultant S.H.M. is 73 units (D) the maximum displacement of the particle from the origin is 9 units. 14. In SHM, acceleration versus displacement (from mean position) graph : (A) is always a straight line passing through origin and slope –m2 (B) is always a straight line passing through origin and slope +m2 (C) is a straight line not necessarily passing through origin (D) none of the above

15. A particle moves in xy plane according to the law x = a sin t and y = a(1 – cos t) where a and  are constants. The particle traces (A) a parabola (B) a straight line equallyinclined to x and y axes (C) a circle (D) a distance proportional to time 16. For a particle executing S.H.M., x = displacement from equilibrium position, v = velocity at any instant and a = acceleration at any instant, then (A) v-x graph is a circle (B) v-x graph is an ellipse (C) a-x graph is a straight line (D) a-v graph is an ellipse 17. The figure shows a graph between velocity and displacement (from mean position) of a particle performing SHM v(in cm/s)

10

2.5 x (in cm)

(A) the time period of the particle is 1.57 s (B) the maximum acceleration will be 40cm/s2 (C) the velocity of particle is 2 21 cm/s when it is at a distance 1 cm from the mean position. (D) none of these 18. Two blocks of masses 3 kg and 6 kg rest on a horizontal smooth surface. The 3 kg block is attached to a spring with a force constant k = 900 Nm–1 which is compressed 2 m from beyond the equilibrium position. The 6 kg mass is at rest at 1m from mean position. 3kg mass strikes the 6 kg mass and the two stick together. 1m

3kg

2m

6kg

equilibrium position

(A) velocity of the combined masses immediately after the collision is 10 ms–1 (B) velocity of the combined masses immediately after thecollision is 5 ms–1 (C) amplitude of the resulting oscillation is 2 m (D) amplitude of the resulting oscillation is 5/2 m 19. A particle starts from a point P at a distance of A/2 from the mean position O & travels towards left as shown in the figure. If the time period of

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Page # 44

SIMPLE HARMONIC MOTION

SHM, executed about O is T and amplitude A then the equation of motion of particle is :

A/2

O

P A

 5   2  2 t   (B) x = A sin  t  (A) x = A sin   T T 6 6     2  2 t   (D) x = A cos  t  (C) x = A cos   T T 6 3

20. The angular frequency of a spring block system is 0. This system is suspended from the ceiling of an elevator moving downwards with a constant speed v0. The block is at rest relative to the elevator. Lift is suddenly stopped. Assuming the downwards as a positive direction, choose the wrong statement. v0 (A) The amplitude of the block is  0 (B) The initial phase of the block is .

23. A particle of mass m performs SHM along a straight line with frequency f and amplitude A. (A) The average kinetic energy of the particle is zero. (B) The average potential energy is m 2 f2 A2. (C) The frequency of ocillation of kinetic energy is 2f. (D) Velocity function leads acceleration by /2. 24. A linear harmonic oscillator of force constant 2 × 106Nm–1 and amplitude 0.01 m has a total mechanical energy of 160 J. Its (A) maximum potential energy is 100 J (B) maximum kinetic energy is 100J (C) maximum potential energy is 160 J (D) minimum potential energy is zero. 25. The graph plotted between phase angle () and displacement of a particle from equilibrium position (y) is a sinusoidal curve as shown below. Then the best matching is y

3 /2 2

v0 (C) The equation of motion for the block is  0 sin 0t. (D) The maximum speed of the block is v0. 21. A disc of mass 3m and a disc of mass m are connected by a massless spring of stiffness k. The heavier disc is placed on the ground with the spring vertical and lighter disc on top. From its equilibrium position, the upper disc is pushed down by a distance  and released. Then (A) if  > 3mg/k, the lower disc will bounce up (B) if  = 2mg/k, maximum normal reaction from ground on lower disc = 6 mg. (C) if  = 2mg/k, maximum normal reaction from ground on lower disc = 4 mg. (D) if  > 4 mg/k, the lower disc will bounce up

O

/2

Column A (A)K. E. versus

phase angle curve

t

Column B

(i)

(B) P.E. versus phase

angle curve

(ii)

(C) T.e. versus phase angle curve

22. A system is oscillating with undamped simple harmonic motion. Then the (A) average total energy per cycle of the motion is its maximum kinetic energy. (B) average total energy per cycle of the motion 1 times its maximum kinetic energy.. is 2 1 (C) root mean square velocity is times its 2 maximum velocity (D) mean velocity is 1/2 of maximum velocity.

P

(iii)

(D) Velocity versus

phase angle curve

(iv)

(A) (a) - (i), (b) - (ii), (c) - (iii) & (d) - (iv) (B) (a) - (ii), (b) - (i), (c) - (iii) & (d) - (iv) (C) (a) - (ii), (b) - (i), (c) - (iv) & (d) - (iii) (D) (a) - (ii), (b) - (iii), (c) - (iv) & (d) - (i)

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Page # 45

SIMPLE HARMONIC MOTION

(Subjective Problems)

Exercise - III

8. A block of mass 0.9 kg attached to a spring of force constant k is lying on a frictionless floor.

LINEAR S.H.M 1. The equation of a particle executing SHM is   x  (5m) sin( s –1 )t   . Write down the amplitude, 6  phase constant, time period and maximum speed. 2. A particle having mass 10 g oscillates according to the equation x = (2.0 cm) sin [100 s–1] t +  ]. Find (a) the amplitude, the time period and 6 the force constant (b) the position, the velocity and the acceleration at t = 0. 3. The equation of motion of a particle started at t = 0 is given by x = 5 sin (20 t + /3) where x is in centimetre and t in second. When does the particle. (a) first come to rest ? (b) first have zero acceleration ? (c) first have maximum speed ? 4. A body is in SHM with period T when oscillated from a freely suspended spring. If this spring is cut in two parts of length ratio 1 : 3 & again oscillated from the two parts separately, then the periods are T1 & T2 then find T1/T2. 5. The system shown in the figure can move on a smooth surface. The spring is initially compressed by 6 cm and then released. Find 3 kg

k = 800 N/m

6 kg

(a) Time period (b) Amplitude of 3 kg block (c) Maximum momentum of 6 kg block 6. A body undergoing SHM about the origin has its equation is given by x = 0.2 cos 5t. Find its average speed from t = 0 to t = 0.7 sec.

2 cm and the block

The spring is compressed to

is at a distance 1/ 2 cm from the wall as shown in the figure. When the block is released, it make elastic collision with the wall and its period of motion is 0.2 sec. Find the approximate value of k. Wall

m 1/ 2 2

cm cm

9. A force f = –10x + 2 acts on a particle of mass 0.1 kg, where ‘k’ is in m and F in newton. If it is released from rest at x = –2 m, find : (A) Amplitude (b) Time period (c) Equation of motion 10. Potential energy (U) of a body of unit mass moving in a one-dimension conservative force fileld is given by, U = (x2 – 4x + 3). All units are in S.I. (i) Find the equilibrium position of the body. (ii) Show that oscillations of the body about this equilibrium position is simple harmonic motion & find its timeperiod. (iii) Find the amplitude of oscillations if speed of the body at equilibrium position is 2 6 m/s. 11. The resulting amplitude A and the phase of the vibrations  S = A cos (t) +

  A cos  t   2 2

7. The acceleration-displacement (a – x) graph of a particle executing simple harmonic motion is shown in the figure. Find the frequency of oscil-

+

3   A A  = A cos (t cos (t + ) + cos  t  2 4 8

+ ) are ....... and ......... respectively.

lation.

a

–

O –

x

12. A body is executing SHM under the action of force whose maximum magnitude is 50N. Find the magnitude of force acting on the particle at the time when its energy is half kinetic and half potential.

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Page # 46

SIMPLE HARMONIC MOTION

13. A 1kg block is executing simple harmonic motion of amplitude 0.1 m on a smooth horizontal surface under the restoring force of a spring of spring constant 100 N/m. A block of mass 3 kg is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks

x

10 mm 5 0

t(s)

move together, find the frequency and the amplitude of the motion. 17. Two particles A and B execute SHM along the same line with the same amplitude a, same frequency and same equilibrium position O. If the phase difference between them is  = 2 sin–1 (0.9), then find the maximum distance between the two.

3kg 1kg

14. The springs shown in the figure are all unstretched in the beginning when a man starts pulling the block. The man exerts a constant force F on the block. Find the amplitude and the frequency of the motion of the block.

18. Two blocks A (5kg) and B(2kg) attached to the ends of a spring constant 1120 N/m are placed on a smooth horizontal plane with the spring undeformed. Simultaneously velocities of 3m/s and 10m/s along the line of the spring in the same direction are imparted to A and B then 3m/s

M

k1 k3

k2

A 5 F

15. Two identical springs are attached to a small block P. The other ends of the springs are fixed at A and B. When P is in equilibrium the extension of top spring is 20 cm and extension of bottom spring is 10 cm. Find the period of small vertical oscillations of P about its equilibrium position. (use g = 9.8 m/s2)

A

10m/s

2 B

(a) Find the maximum extension of the spring. (b) When does the first maximum compression occurs after start. 19. The motion of a particle is described by x = 30 sin (t + /6), where x is in cm and t in sec. Potential energy of the particle is twice of kinetic energy for the first time after t = 0 when the particle is at position ............. after .......... time. 20. A particle is performing SHM with accleration a = 8 2 – 4 2 x where x is coordinate of the particle w.r.t. the origin. The parameters are in S.I. units. The particle is at rest at x = –2 at t = 0. Find coordinate of the particle w.r.t. origin at any time.

P

B 16. The figure shows the displacement - time graph of a particle executing SHM. If the time period of oscillation is 2s, then the equation of motion is given by x = ................. .

21. (a) Find the time period of oscillations of a torsional pendulum, if the torsional constant of the wire is K = 102J/rad. The moment of inertia of rigid body is 10 kg m2 about the axis of rotation. (b) A simple pendulum of length l = 0.5 m is hanging from ceiling of a car. The car is kept on a horizontal plane. The car starts accelerating on the horizontal road with acceleration of 5 m/s2. Find the time period of oscillations of the pendulum for small amplitudes about the mean position.

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Page # 47

SIMPLE HARMONIC MOTION 22. An object of mass 0.2 kg executes SHM along the x-axis with frequency of (25/) Hz. At the point x = 0.04m the object has KE 0.5 J and PE 0.4 J. The amplitude of oscillation is _____________. 23. A body of mass 1 kg is suspended from a weightless spring having force constant 600 N/ m. Another body of mass 0.5 kg moving vertically upwards hits the suspended body with a velocity of 3.0 m/s and get embedded in it. Find the frequency of oscillations and amplitude of motion. 24. A block is kept on a horizontal table. The table is undergoing simple harmonic motion of frequency 3 Hz in a horizontal plane. The coefficient of static friction between block and the table surface is 0.72. Find the maximum amplitude of the table at which the block does not slip on the surface. 25. A particle of mass m moves in a one-dimensional potential energy U(x) = –ax2 + bx4, where ‘a’ and ‘b’ are positive constants. Then what is the angular frequency of small oscillations about the minima of the potential energy.

(ii) What must be the acceleration of the lift for the period of oscillation of the pendulum to be T0 ? 2 29. A simple pendulum of length  is suspended through the ceiling of an elevator. Find the time period of small oscillations if the elevator (a) is going up with an acceleration a0 (b) is going down with an acceleration a0 and (c) is moving with a uniform velocity. 30. A simple pendulum fixed in a car has a time period of 4 seconds when the car is moving uniformly on a horizontal road. When the accelerator is pressed, the time period changes to 3.99 seconds. Making an approximate analysis, find the acceleration of the car. 31. Two identical rods each of mass m and length L, are rigidly joined and then suspended in a vertical plane so as to oscillate freely about an axis normal to the plane of paper passing through ‘S’ (point of supension). Find the time period of such small oscillations.

26. A pendulum having time period equal to two seconds is called a seconds pendulum. Those used in pendulum clocks are of this type. Find the length of a seconds pendulum at a place where g = 2 m/s2

 = sin[s–1)t]. Find the length of the pendulum 90 if g = 2 m/s2. 28. A pendulum is suspended in a lift and its period of oscillation is T0 when the lift is stationary. (i) What will the period T of oscillation of pendulum be, if the lift begins to accelerate downwards 3g ? with an acceleration equal to 4

32. A simple pendulum has a time period T = 2 sec when it swings freely. The pendulum is hung as shown in figure. so that only one-fourth of its total length is free to swing to the left of obstacle. It is displaced to position A and released. How long does it take to swing to extreme displacement B and return to A? Assume that displacement angle is always small. ob st ac le

27. The angle made by the string of a simple pendulum with the vertical depends on time as 

S

B

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A

Page # 48

SIMPLE HARMONIC MOTION

(TOUGH

Exercise - IV 1. A point particle of mass 0.1 kg is executing SHM with amplitude of 0.1 m. When the particle passes through the mean position, its K.E. is 8 × 10–3J. Obtain the equation of motion of this particle if the initial phase of oscillation is 45°. 2. The particle executing SHM in a straight line has velocities 8 m/s, 7 m/s, 4 m/s at three points distant one metre from each other. What will be the maximum velocity of the particle? 3. One end of an ideal spring is fixed to a wall at origin O and the axis of spring is parallel to a xaxis. A block of mass m = 1 kg is attached to free end of the spring and it is performing SHM. Equation of position of block in coordinate system shown is x = 10 + 3 sin10t, t is in second and x in cm. Another block of mass M = 3kg, moving towards the origin with velocity 30 cm/s collides with the block performing SHM at t = 0 and gets struck to it, calculate : (i) new amplitude of oscillations. (ii) new equation for position of the combined body. (iii) loss of energy during collision. Neglect friction. 1kg 3kg

4. A mass M is in static equilibrium on a massless vertical spring as shown in the figure. A ball of mass m dropped from certain height sticks to the mass M after colliding with it. The oscillations they perform reach to height ‘a’ above the original level of scales & depth ‘b’ below it. (a) Find the force constant of the spring.;

SUBJECTIVE PROBLEMS )

 =/6 radian with respect to diameter PQ of the circle and released from rest

(a) Calculate the frequency of oscillation of the ball B. (b) What is the total energy of the system. (c) Find the speed of the ball A when A and B are at the two ends of the diameter PQ. 6. An ideal gas is enclosed in a vertical cylinderical container and supports a freely moving piston of mass m. The piston and the cylinder have equal cross-sectional area A, atmospheric pressure is P0 and when the piston is in equilibrium position. Show that the piston executes SHM and find the frequency of oscillation (system is completely isolated from the surrounding).  = Cp/Cv. Height of the gas in equilibrium position is h. 7. A massless rod is hinged at O. A string carrying a mass m at one end is attached to point A on the rod so that OA = a. At another point B (OB = b) of the rod, a horizontal spring of force constant k is attached as shown. Find the period of small vertical oscilla tions of mass m around its equilibrium position.

k

B A O

m

a

M b

(b) Find the oscillation frequency. (c) What is the height above the initial level from which the mass m was droped ? 5. Two identical balls A and B each of mass 0.1 kg are attached to two identical massless springs. The spring mass system is constrained to move inside a rigid smooth pipe in the form of a circle as in fig. The pipe is fixed in a horizontal plane. The centres of the ball can move in a circle of radius 0.06 m. Each spring has a natural length 0.06 m and force constant 0.1 N/m. Initially both the balls are displaced by an angle of

8. Two blocks A (2kg) and B(3kg) rest up on a smooth horizontal surface are connected by a spring of stiffness 120 N/m. Initially the spring is underformed. A is imparted a velocity of 2m/s along the line of the spring away from B. Find the displacement of A t second later. 3kg 2kg 2m/s A B 9. Consider a fixed ring shaped uniform body of linear mass density  and radius R. A particle at the centre of ring is displaced along the axis by a small distance, show that the particle will execute SHM under gravitation of ring & find its time period neglecting other forces.

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Page # 49

SIMPLE HARMONIC MOTION

Exercise - V

JEE-Problems

1. A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is V(x) = k|x|3 where k is a positive constant. If the amplitude of oscillations is a, then its time period T is (A) proportional to 1/ a (B) independent of a

6. A particle executes simple harmonic motion between x = –A and x = +A. The time taken for it to go from 0 to A/2 is T1 and to go from A/2 to A is T2. Then [JEE (Scr)’ 2001] (A) T1 < T2 (B) T1 > T2 (C) T1 = T2 (D) T1 = 2T2

(C) proportional to a (D) proportional to a3/2

7. A diatomic molecule has atoms of masses m1 and m2. The potential energy of the molecule for the interatomic separation r is given by V(r) = –A + B(r –r0)2, where r0 is the equilibrium separation, and A and B are positive constants. The atoms are compressed towards each other from their equilibrium positions released. What is the vibrational frquency of the molecule? [REE’ 2001]

[JEE’ 98]

2. A particle free to move along the x-axis has potential energy given by U(x) = k[1 - exp(-x2)] for –< x < +, where k is a positive constant of appropriate dimensions. Then (A) at point away from the origin, the particle is in unstable equilibrium. (B) for any finite nonzero value of x, there is a force directed away from the origin. (C) if its total mechanical energy is k/2, it has its minimum kinetic energy at the origin. (D) for small displacements from x = 0, the motion is simple harmonic. [JEE’ 99]

8. A particle is executing SHM according to y = a cos t. Then which of the graphs represents variations of potential energy : [JEE (Scr)’ 2003] P.E.

3. Three simple harmonic motions in the same direction having the same amplitude a and same period are superposed. If each differs in phase from the next by 45°, then (A) the resultant amplitude is (1+ 2 )a (B) the phase of the resultant motion relative to the first is 90°. (C) the energy associated with the resulting motion is (3 + 2 2 ) times the energy associated with any single motion. (D) the resulting motion is not simple harmonic. [JEE’ 99]

I

II

t P.E.

IV

III

x

4. The period of oscillation of simple pendulum of length L suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination  is given by [JEE’ 2000] (A) 2

L gcos 

(B) 2

L gsin

(C) 2

L g

(D) 2

L gtan

5. A bob of mass M is attached to the lower end of a vertical string of length L and cross sectional area A. The Young’s modulus of the material of the string is Y. If the bob executes SHM in the vertical direcion, find the frequency of these oscillations. [REE’ 2000]

(A) (I) & (III) (C) (I) & (IV)

(B) (II) & (IV) (D) (II) & (III)

9. Two masses m1 and m2 connected by a light spring of natural length l0 is compressed completely and tied by a string. This system while moving with a velocity v0 along +ve x-axis pass through the origin at t = 0. At this position the string snaps. Position of mass m1 at time is given by the equation.x1(t) = v0 t – A (1 – cos t) Calculate : (a) Position of the particle m2 as a function of time. (b) l0 is terms of A. [JEE’ 2003] 10. A block P of mass m is placed on a frictionless horizontal surface. Another block Q of same

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Page # 50

SIMPLE HARMONIC MOTION

mass is kept on P and connected to the wall with the help of a spring of spring constant k as shown in the figure. s is the coefficient of friction between P and Q. The blocks move together performing SHM of amplitude A. The maximum value of the friction force between P and Q is Q

k

P kA 2 (D) smg

(A) kA

s

smooth

(B)

(C) zero

[JEE’ 2004]

11. A simple pendulum has time period T1. When the point of suspension moves vertically up according to the equation y = kt2 where k = 1 m/s2 and ‘t’ is time then the time period of the pendu T1  lum is T2 then    T2 

2

is

(A)

5 6

(B)

11 10

(C)

6 5

(D)

5 4 [JEE’ 2005(Scr)]

12. A small body attached to one end of a vertically hanging spring is performing SHM about it’s mean position with angular frequency  and amplitude a. If at a height y* from the mean position the body gets detached from the spring, calculate the value of y* so that the height H attained by the mass is maximum. The body does not interact with the spring during it’s subsequent motion after detachment. (aw2>g).

y0 m 13. Function x = A sin2 t + B cos2 t + C sin t cos t represents SHM [JEE’ 2006] (A) for any value of A, B and C (except C = 0) (B) if A = – B ; C = 2B, amplitude = |B 2 | (C) if A = B; C = 0 (D) if A = B; C = 2B, amplitude = |B|

14. A student performs an experiment for deter 4 2l    mination of g   l  1 m and he commits an T 2   error of l. For The takes the time of n oscillations with the stop watch of least count T and he commits a human error of 0.1 sec. For which of the following data, the measurement of g will be most accurate? l T nAmplitude of oscillation (A)5 mm 0.2 sec 10 5 mm (B) 5 mm 0.2 sec 20 5 mm (C) 5 mm 0.1 sec 20 1 mm (D) 1 mm 0.1 sec 50 1 mm [JEE’ 2006] 15. Column I describes some situations in which a small objact moves. Column II describes some characteristics of these motions. Match the situations in Column I with the characteristics in Column II and indicate your answer by darkening appropriate bubbles in the 4 × 4 matrix given in the ORS. Column I Column II (A) The object moves (P) The object on the x-axis under a executes a SHM conservative force in such a way that its “speed” and “po sition” satisfy v = c1 c 2  x 2 , where c1 and c2 are positive constants. (B) The object moves (Q) The object does on the x-axis in such a not change its way that its velocity and direction its displacement from the origin satisfyv = –kx, where k is a positive constant. (C) The object is attached (R) The kinetic en to one end of a massless ergy of the object spring of a given spring keeps on decreasing constant. The other end of the spring is attached to the ceiling of an elevator. Initially everything is at rest. The elevator starts going upwards with a constant acceleration a. The motion of the object is observed from the elevator during the period it maintains this acceleration.

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Page # 51

SIMPLE HARMONIC MOTION (D) The object is projected (S) The object from the earth’s surface can change its vertically upwards with direction only once

M

P

Me is the mass of the earth and Re is the radius of the earth. Neglect forces from objects other than the earth. [JEE’ 2007] 16. A block (B) is attached to two unstretched springs S1 and S2 with spring constants k and 4k, respectively (see figure I). The other ends are attached to identical supports M1 and M2 not attached to the walls. The springs and supports have negligible mass. There is no friction anywhere. The block B is displaced towards wall 1 by a small distance x (figure II) and released. The block returns and moves a maximum distance y towards wall 2. Displacement x and y are measured with respect to the equilibrium position y of the block B. The ratio in Figure x

k2A (A) k 2

k2A (B) k 2

k 1A (C) k  k 1 2

k 2A (D) k  k 1 2

19. A uniform rod of length L and mass M is pivoted at the centre. Its two ends are attached to two springs of equal spring constants k. The springs are fixed to rigid supports as shown in the figure, and the rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle  in one direction and released. The frequency of oscillation is

(A)

[JEE’2008] (A) 4 1 (C) 2

K2

K1

a speed 2 GM e / R e , where

(B) 2 1 (D) 4

17. The x–t graph of particle undergoing simple harmonic motion is shown below. The acceleration of the particle at t = 4 / 3 s is

– 2 cm / s 2 32

(A)

3 2  cm / s 2 32

(B)

(C)

2 cm / s 2 32

(D) –

3 2  cm / s 2 32

18. The mass M shown in the figure oscillates in simple harmonic motion with amplitude A. The amplitude of the point P is

1 2

2k M

(B)

1 k 2 M

1 6k 1 24k (D) 2 M 2 M 20. A metal rod of length 'L' and mass 'm' is pivoted at one end. A thin disk of mass 'M' and radius 'R' ( Angular frequency for case B (D) Angular frequency for case A < Angular frequency for case B

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SIMPLE HARMONIC MOTION

Paragraph for Question Nos. 21 to 23 Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical systems in one-dimension. For such systems, phase space is a plane in which position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space diagram is x(t) vs. p(t) curve in this plane. The arrow on the curve idicates the time flow. For example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We use the sign convention in which position of momentum upwards (or to right) is positive and downwards (or to left) is negative.

22. The phase space diagram for simple harmonic motion is a circle centered at the origin. In the figure, the two circles represent the same oscillator but for different initial conditions, and E1 and E2 are the total mechanical energies respectively. Then Momentum

E1 E2 2a

a

(A) E1 = 2 E2 (C) E1 = 4E2

position

(B) E1 =2E2 (D) E1 = 15E2

Momentum

23. Consider the spring-mass system, with the mass submerged in water, as shown in the figure. The phase space diagram for one cycle of this system is

Position

21. The phase space diagram for a ball thrown vertically up from ground is

Momentum

Momentum

Momentum Momentum

(A)

(B) Position

Position

(A)

Momentum

Position

Momentum

(C)

(D) Position

(B)

Momentum

Momentum

(C)

Position

Position

(D) Position

Position

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Page # 53

SIMPLE HARMONIC MOTION

Exercise-I LINEAR S.H.M 1.

A

2.

C

3.

B

4.

C

5.

A

6.

A

7.

C

8. 15. 22. 29. 36. 43. 50.

C A B B B A C

9. 16. 23. 30. 37. 44. 51.

D D C C B C B

10. 17. 24. 31. 38. 45.

A D D A A A

11. 18. 25. 32. 39. 46.

D C C D B A

12. 19. 26. 33. 40. 47.

B D C D B C

13. 20. 27. 34. 41. 48.

B C C C A B

14. 21. 28. 35. 42. 49.

B B C D C C

A A

57.

A

ANGULAR S.H.M 52. 58. 63.

C C D

53. 59.

B B

54. 60.

C C

55. 61.

A A

56. 62.

Exercise-II 1. 7. 12. 19. 25.

B,C,D C A,B,C B,D B

2. 8. 13. 20.

A B B,D B

3. B 9. A,B,C,D 14. A 21. B,D

4. 10. 15. 22.

B,C B,C,D C,D A,C

5. 11. 16. 23.

A 6. A,B B,C,D 17. B,C 24.

A,B,C A,B,C 18. B,C

A,C

Exercise-III LINEAR S.H.M 1.

Amplitude = 5 m Initial Phase = /6 Maximum speed = 5m/sec

2.

(a) 2.0 cm, /50 sec, 100 N/m (b) 1 cm,

3 m / sec , 100 m/sec–1

   sec. , (b) sec. (c) sec. 120 30 30

1

(a)

6.

2m/sec

7.

10.

(i) x0 = 2m

(ii) T =

13.

5 Hz , 5 cm 2

F(K 2  K 3 ) 1 K 1K 2  K 2K 3  K 3K 1 14. K K  K K  K K , 2 M (K 2  K 3 ) 1 2 2 3 3 1

1 2

  2 sec

4.

3

5. (a)

 sec , (b) 4 cm, (c) 2.40 kg m/sec 10

3.

8. 100 Nm–2

(iii) 2 3

11.

9. (a)

11 m 5

 11 sec (c) x = 0.2 – cos t 5 5

(b)

–1  1  3 5 A , tan   12. 25 2 N 2 8

15.

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 7

Page # 54

SIMPLE HARMONIC MOTION

16.

X = 10 sin (t + /6 )

19.

10 6 cm ,

22.

0.06 m

17. 1.8 a

18. (a) 25 cm, (b)

1 2 1 sin –1 – sec  3 6 23.

10 5 37 Hz , cm  6

3 seconds 56

20. 2 – 4 cos 2t 21.

(a) 2 sec.

24. 2 cm

2

25.

(b) T =

2 5 1/ 4

sec.

a m

ANGULAR S.H.M

26

1m

27.

1m

28.

30.

g 10

17L 31. 2 18 g

32.

   (i) 2T0 , (ii) 3 g up wards 29. (a) 2 a  g , (b) 2 g – a , (c) 2 g 0 0 3 sec 2

Exercise-IV 1.

y = 0.1 sin (4t +/4)

4.

(a) K =

6.

f

1 2

2.

65 m / s

 M  m  ab 2mg 1  ; (c)  , m  b – a 2 b–a

(P0  mg / A )A mh

3.

2mg (b – a)(M  m)

7. (2a/b)(m/k)1/2

3cm, x = 10 – 3 sin 5t; E = 0.135 J 5.

f=

1 ; E = 42 × 10–5 J; v =2 × 10–2 m/s 

8. 0.8t + 0.12 sin 10t

9.

2R2 Gp

6.

A

Exercise-V 1.

A

7.

1 = 2 f

10. 15. 16.

2.

D

3.

m1 m 2 2B(m1  m 2 ) 8.

A, C

4.

A

5.

A

9.

(a) x2 = v0t +

1 YA 2 ML

  m1 m1 A (1 – cos t), (b) l0 =  m  1 A   2 m2

mg g  2 T3. If the thickness of the layers x1 and x2 are the same, which of the following statements are correct. T

k1

k2

x1

x2

T1 100°C 3K

T1 T3 X

K

T2 0°C





34. The ratio of the thermal resistance of the rod is RA 1 RA (B) R  3 (A) R  3 B B RA 3 (C) R  4 B

Q

P

Question No. 34. to 36 (3 questions) Two rods A and B of same cross-sectional are A and length l connected in series between a source (T1 = 100°C) and a sink (T2 = 0°C) as shown in figure. The rod is laterally insulated

Distance

31. Three identical rods AB, CD and PQ are joined as shown. P an Q are mid points of AB and CD respectively. Ends A, B, C and D are maintained at 0°C, 100°C, 30°C and 60°C respectively. The direction of heat flow in PQ is

0°C

K2 (B) 60°C (D) 20°C

(A) 80°C (C) 40°C

Distance

T1

B

100°C K1 

(B)

O

33. Two rods A and B of different materials but same cross section are joined as in figure. The free end of A is maintained at 100°C and the free end of B is maintained at 0°C. If l2 = 2l1, K1 = 2K2 and rods are thermally insulated from sides to prevent heat losses then the temperature  of the junction of the two rods is l1 l2

T2

(A) O

(A) k1 > k2 (B) k1 < k2 (C) k1 = k2 but heat flow through material (1) is larger then through (2) (D) k1 = k2 but heat flow through material (1) is less than that through (2)

(D)

4 3

35. If TA and TB are the temperature drops across the rod A and B, then TA 3 TA 1 (A) T  1 (B) T  3 B B TA 3 TA 4 (C) T  4 (D) T  3 B B 36. If GA and GB are across the rod A and GA 3 (A) G  1 B GA 3 (C) G  4 B

the temperature gradients B, then GA 1 (B) G  3 B GA 4 (D) G  3 B

37. Two sheets of thickness d and 3d, are touching each other. The temperature just outside the thinner sheet side is A, and on the side of the thicker sheet is C. The interface temperature is B. A, B and C are in arithmetic progressing, the ratio of thermal conductivity of thinner sheet and thicker sheet is

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Page # 26

HEAT

(A) 1 : 3 (C) 2 : 3

(B) 3 : 1 (D) 1 : 9

38. A cylindrical rod with one end in a steam chamber and the outer end in ice results in melting of 0.1 gm 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 is gm/sec will be (A) 3.2 (B) 1.6 (C) 0.2 (D) 0.1 39. A composite rod made of three rods of equal length and cross-section as shown in the fig. The thermal conductivities of the materials of the rods are K/2, 5K and K respectively. The end A and end B are at constant temperatures. All heat entering the face A goes out of the end B there being no loss of heat from the sides of the bar. The effective thermal conductivity of the bar is B A K/2

(A) 15K/16 (C) 5K/16

5K K (B) 6K/13 (D) 2K/13.

40. A rod of length L with sides fully insulated is of a material whose thermal conductivity varies  with temperature as K = , where  is a T constant. The ends of the rod are kept at temperature T1 and T2. The temperature T at x, where x is the distance from the end whose temperature is T1 is x

T L (A) T1  2   T1 

(C) T e 1

T2 x T1 L

x T2 (B) L ln T 1

(D) T1 

T2 – T1 x L

41. Heat flows radially outward through a spherical shell of outside radius R2 and inner radius R1. The temperature of inner surface of shell is 1 and that of outer is 2. The radial distance from centre of shell where the temperature is just half way between 1 and 2 is : R1R 2 R  R2 (A) 1 (B) R  R 2 1 2 2 R1R 2 R2 (C) R  R (D) R1  2 1 2 42. The two ends of two similar non-uniform rods of length  each and thermal conductivity ‘K’ are maintained at different but constant temperature. The temperature gradient at any point on the T . The heat flow per unit time through the  rod is I : Given T1 > T2. Then which of the following

rod is

is true: T1

T2 T2

A1

T1

B1

A2

B2

x1

x

x2 x2 Rod (lI) Rod (l) (A) I of Rod (I) = I of Rod (II) (B) I of Rod (I) > I of Rod (II) (C) I of Rod (I) < I of Rod (II) (D) data is insufficient 43. A system S receives heat continuously from an electrical heater of power 10 W. The temperature of S becomes constant at 50°C when the surrounding temperature is 20°C. After the heater is switched off, S cools from 35.1 °C to 34.9 °C in 1 minute. The heat capacity of S is (A) 100 J/°C (B) 300 J/°C (C) 750 J/°C (D) 1500 J/°C 44. A sphere of ice at 0°C having initial radius R is placed in an environment having ambient temperature > 0°C. The ice melts uniformly, such that shape remains spherical. After a time ‘t’ the radius of the sphere has reduced to r. Assuming the rate of heat absorption is proportional to the surface area of the sphere at any moment, which graph best depicts r(t). r

r

R

R

(B)

(A) r

t

r

R

t

R

(C)

(D) t

t

45. The power radiated by a black body is P and it radiates maximum energy around the wavelength 0. If the temperature of the black body is now changed so that it radiates maximum energy around wavelength 3/40, the power radiated by it will increase by a factor of (A) 4/3 (B) 16/9 (C) 64/27 (D) 256/81 46. 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) 4P (C) 2 P (D) P

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Page # 27

HEAT 47. Star S1 emits maximum radiation of wavelength 420 nm and the star S2 emits maximum radiation of wavelength 560 nm, what is the ratio of the temperature of S1 and S2 : (A) 4/3 (B) (4/3)1/4 (C) 3/4 (D) (3/4)1/2 48. Spheres P and Q are uniformly constructed from the same material which is a good conductor of heat and the radius of Q is thrice the radius of P. The rate of fall of temperature of P is x times that of Q when both are at the same surface temperature. The value of x is : (A) 1/4 (B) 1/3 (C) 3 (D) 4 49. An ice cube at temperature –20°C is kept in a room at temperature 20°C. The variation of temperature of the body with time is given by T

(B) t

t

T

T

(C)

52. Two bodies P and Q have thermal emissivities of p and Q respectively. Surface areas of these bodies are same and the total radiant power is also emitted at the same rate. If temperature of P is P kelvin then temperature of Q i.e. Q is 1/ 4

1/ 4

 Q  (A)    P  1/ 4

 Q (C)   P

T

(A)

51. The intensity of radiation emitted by the Sun has its maximum value at a wavelength of 510 nm and that emitted by the North Star has the maximum value at 350 nm. If these stars behave like black bodies then the ratio of the surface temperature of the Sun and the North Star is (A) 1.46 (B) 0.69 (C) 1.21 (D) 0.83

  

P

 P  (B)    Q 

1  P

 Q  (D)   P  P 

P 4

53. A black body calorimeter filled with hot water cools from 60ºC to 50ºC in 4 min and 40ºC to 30ºC in 8 min. The approximate temperature of surrounding is (A) 10ºC (B) 15ºC (C) 20ºC (D) 25ºC 54. The rate of emission of radiation of a black body at 273ºC is E, then the rate of emission of radiation of this body at 0ºC will be

(D) t

t

50. The spectral emissive power E for a body at temperature T1 is plotted against the wavelength and area under the curve is found to be A. At a different temperature T2 the area is found to be 9A. Then 1/2 =

T2 T1

(A) 3

(B) 1/3

(C) 1 / 3

(D)

3

(A)

E 16

(B)

E 4

E (D) 0 8 55. A body cools from 75°C to 65°C in 5 minutes. If the room temperature is 25°, then the temperature of the body at the end of next 5 minutes is : (A) 57°C (B) 55°C (C) 54°C (D) 53°

(C)

56. The temperature of a body falls from 40°C to 36°C in 5 minutes. when placed in a surrounding of constant temperature 16°C. Then the time taken for the temperature of the body to become 32°C is (A) 5 min (B) 4.3 min (C) 6.1 min (D) 10.2 min.

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Page # 28

HEAT

Exercise - II

(Multiple Choice Problems)

1. From a black body, radiation is not : (A) emitted

(B) absorbed

(C) reflected

(D) refracted

2. In accordance with Kirchhoff’s law : (A) bad absorber is bad emitter (B) bad absorber is good reflector

7. An experiment is perfomed to measure the specific heat of copper. A lump of copper is heated in an oven, then dropped into a beaker of water. To calculate the specific heat of copper, the experimenter must know or measure the value of all of the quantities below EXCEPT the (A) heat capacity of water and beaker

(C) bad reflector is good emitter (D) bad emitter is good absorber

(B) original temperature of the copper and the water

3. The energy radiated by a body depends on :

(C) final (equilibrium) temperature of the copper and the water

(A) area of body

(B) nature of surface

(C) mass of body

(D) temperature of body

4. A hollow and a solid sphere of same material and identical outer surface are heated to the same temperature : (A) in the beginning both will emit equal amount of radiation per unit time. (B) in the beginning both will absorb equal amount of radiation per unit time (C) both spheres will have same rate of fall of temperature (dT/dt) (D) both spheres will have equal temperatures at any moment. 5.The rate of cooling of a body by radiation depends on : (A) area of body (B) mass of body (C) specific heat of body (D) temperature of body and surrounding. 6. A polished metallic piece and a black painted wooden piece are kept in open in bright sun for a long time : (A) the wooden piece will absorbs less heat than the metallic piece (B) the wooden piece will have a lower temperature than the metallic piece (C) if touched, the metallic piece will feel hotter than the wooden piece (D) when the two pieces are removed from the open to a cold room, the wooden piece will lose heat at a faster rate than the metallic piece

(D) time taken to achieve equilibrium after the copper is dropped into the water 8. One end of a conducting rod is maintained at temperature 50ºC and at the other end, ice is melting at 0ºC. The rate of melting of ice is doubled if : (A) the temperature is made 200ºC and the area of cross-section of the rod is doubled (B) the temperature is made 100ºC and length of rod is made four times (C) area of cross-section of rod is halved and length is doubled (D) the temperature is made 100ºC and the area of cross-section of rod and length both are doubled. 9. Two metallic sphere A and B are made of same material and have got identical surface finish. The mass of sphere A is four times that of B. Both the spheres are heated to the same temperature and placed in a room having lower temperature but thermally insulated from each other. (A) The ratio of heat loss of A to that of B is 24/3 (B) The ratio of heat loss of A to that of B is 22/3 (C) The ratio of the initial rate of cooling of A to that of B is 2–2/3 (D) The ratio of the initial rate of cooling of A to that of B is 2–4/3 10. 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 radiate energy at the same rate. The wavelength B, corresponding to the maximum

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Page # 29

HEAT spectral radiancy in the radiation from B, is shifted from the wavelength corresponding to the maximum spectral radiancy in the radiation from A by 1.00 m. If the temperature of A is 5802 K, (A) the temperature of B is 1934 K (B) B = 1.5 m (C) the temperature of B is 11604 K (D) the temperature of B is 2901 K 11. Three bodies A, B and C have equal surface area and thermal emissivities in the ratio 1 1 e A : eB : e C  1 : : . All the three bodies are 2 4 radiating at same rate. Their wavelengths corresponding to maximum intensity are A, B and C respectively and their temperature are TA, TB and TC on kelvin scale, then select the incorrect statement.

12. Choose the correct statement(s) (A) The radiant energy is not equally distributed among all the possible wavelengths (B) For a particular wavelength the spectral intensity is maximum (C) The area under the curve is equal to the total rate at which heat is radiated by the body at that temperature (D) None of these 13. If the temperature of the body is raised to a higher temperature T,’ then choose the correct statement(s) (A) The intensity of radiation wavelength increases

for every

(B) The maximum intensity occurs at a shorter wavelength (C) The area under the graph increases

(A) TA TC  TB (C) e A TA (D)

(B)  A  C   B

e C TC  eB TB

eA A TA . eBBTB  eCCTC

(D) The area under the graph is proportional to the fourth power of temperature 14. Identify the graph which correctly represents the spectral intensity versus wavelength graph at two temperatures T and T(T < T)

Question No. 12 to 14 (3 questions)

E

E

The figure shows a radiant energy spectrum

T'

graph for a black body at a temperature T.

T'

T

(B)

(A)

O

O

T

T

E T'

O T

(C)

(D) none of these O

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Page # 30

HEAT

1. In following equation calculate the value of H. 1 kg steam at 200°C = H + 1 Kg water at 100°C (Ssteam = Constant = .5 cal/gm°C) 2. From what height should a piece of ice (0°C) fall so that it melts completely ? Only one quarter of the heat produced is absorbed by the ice. The latent heat of ice is 3.4 × 105 J kg–1 and g is 10 N kg–1. 3. A copper cube of mass 200 g slides down on a rough inclined plane of inclination 37° at a constant speed. Assume that any loss in mechanical energy goes into the copper block as thermal energy. Find the increase in the temperature of the block as it slides down through 60 cm. Specific heat capacity of copper = 420 J/kg-K. 4. 10 gm ice at –10°C, 10 gm water at 20°C and 2g steam at 100°C are mixed with each other then final equilibrium temperature. 5. Materials A, B and C are solids that are at their melting temperatures. Material A requires 200 J to melt 4 kg, material B requires 300 J to melt 5 kg, and material C requires 300 J to melt 6 kg. Rank the materials according to their heats of fusion, greatest first. 6. In a thermally isolated container, material A of mass m is placed against material B, also of mass m but at higher temperature. When thermal equilibrium is reached, the temperature changes TA and TB of A and B are recorded. Then the experiment is repeated, using A with other materials. All of the same mass m. The results are given in the table. Rank the four materials according to their specific heats, greatest first. Experiment Temperature Changes 1. TA = + 50 C° TB = – 50 C° 2. TA = + 10 C° TC = – 20 C° 3. TA = + 2 C° TD = – 40 C° 7. Indian style of cooling drinking water is to keep it in a pitcher having porous walls. Water comes to the outer surface very slowly and avaporates. Most of the energy needed for evaporation is taken from the water itself and the water is cooled down. Assume that a pitcher contains 10 kg of water and 0.2 g of water comes out per second. Assuming no backward heat transfer from the atmosphere to the water, calculate the time in

(Subjective Problems) which the temperature decreases by 5°C. Specific heat capacity of water = 4200 J/kg–°C and latent heat of vaporization of water = 2.27 × 106 J/kg. 8. An aluminium container of mass 100 gm contains 200 gm of ice at –20°C. Heat is added to the system at the rate of 100 cal/s. Find the temperature of the system after 4 minutes (specific heat of ice = 0.5 and L = 80 cal/gm, specific heat of Al = 0.2 cal/gm/°C) 9. A volume of 120 ml of drink (half alcohol + half water by mass) originally at a temperature of 25°C is cooled by adding 20 gm ice at 0°C. If all the ice melts, find the final temperature of the drink.(density of drink = 0.833 gm/cc, specific heat of alcohol = 0.6 cal/gm/°C) 10. Two identical calorimeter A and B contain equal quantity of water at 20°C. A 5 gm piece of metal X of specific heat 0.2 cal g–1 (C°)–1 is dropped into A and a 5 gm piece of metal Y into B. The equilibrium temperature in A is 22° C and in B 23°C. The initial temperature of both the metals is 40°C. Find the specific heat of metal Y in cal g–1(C°)–1. 11. Two 50 gm ice cubes are dropped into 250 gm of water into a glass. If the water was initially at a temperature of 25°C and the temperature of ice –15°C. Find the final temperature of water. (specific heat of ice = 0.5 cal/gm/°C and L = 80 cal/gm). Find final amount of water and ice. 12. A substance is in the solid form at 0°C. The amount of heat added to this substance and its temperature are plotted in the following graph. If the relative specific heat capacity of the solid substance is 0.5, find from the graph 120 100 80 60 40 20

temp(°C)

Exercise - III

C

A

B

1000 Q

2000 (calories)

(i) the mass of the substance; (ii) the specific latent heat of the melting process, and (iii) the specific heat of the substance in the liquid state. 13. A uniform slab of dimension 10cm × 10cm × 1cm is kept between two heat reservoirs at

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Page # 31

HEAT temperatures 10°C and 90°C. The larger surface areas touch the reservoirs. The thermal conductivity of the material is 0.80 W/m–°C. Find the amount of heat flowing through the slab per second. 14. One end of a steel rod (K = 42 J/m–s–°C) of length 1.0m is kept in ice at 0°C and the other end is kept in boiling water at 100°C. The area of cross-section of the rod is 0.04 cm2. Assuming no heat loss to the atmosphere, find the mass of the ice melting per second. Latent heat of fusion of ice = 3.36 × 105 J/kg.

4 cm thick layer of insulation whose thermal conductivity is 0.2 W/m/K. The outer face of the insulation is 25°C. Find the temperature of the tank in the steady state. 20. The figure shows the face and interface temperature of a composite slab containing of four layers of two materials having identical thickness. Under steady state condition, find the value of temperature . 20°C 10°C 

k

15. A rod CD of thermal resistance 5.0 K/W is joined at the middle of an identical rod AB as shown in figure. The ends A, B and D are maintained at 100°C, 0°C and 25°C respectively. Find the heat current in CD. A 100°C

C

25°C

B 0°C

D

16. A semicircular rod is joined at its end to a straight rod of the same material and same crosssectional area. The straight rod forms a diameter of the other rod. The junctions are maintained at different temperatures. Find the ratio of the heat transferred through a cross-section of the semicircular rod to the heat transferred through a cross-section of the straight rod in a given time. 17. One end of copper rod of uniform cross-section and of length 1.45 m is in contact with ice at 0°C and the other end with water at 100°C. Find the position of point along its length where a temperature of 200°C should be maintained so that in steady state the mass of ice melting is equal to that of steam produced in the same interval of time [Assume that the whole system is insulated from surroudings]. [take Lv = 540 cal/g Lf = 80 cal/g] 18. Three slabs of same surface area but different conductivities k1, k2, k3 and different thickness t1, t2, t3 are placed in close contact. After steady state his combination behaves as a single slab. Find is effective thermal conductivity. 19. A thin walled metal tank of surface area 5m2 is filled with water tank and contains an immersion heater dissipating 1 kW. The tank is covered with

2k

–5°C –10°C

k

2k

k=thermal conductivity

21. In the square frame of side l of metallic rods, the corners A and C are maintained at T1 and T2 respectively. The rate of heat flow from A to C is . If A and D are instead maintained T1 & T2 respectively find, find the total rate of heat flow. l B C l A

l D

l

22. A hollow metallic sphere of radius 20 cm surrounds a concentric metallic sphere of radius 5 cm. The space between the two spheres is filled with a nonmetallic material. The inner and outer spheres are maintained at 50°C and 10°C respectively and it is found that 160  Joule of heat passes from the inner sphere to the outer sphere per second. Find the thermal conductivity of the material between the spheres. 23. Find the rate of heat flow through a crosssection of the rod shown in figure (2 > 1). Thermal conductivity of the material of the rod is K. r1

L

r2

1 2

 2  1

24. A metal rod of cross-sectional area 1.0 cm2 is being heated at one end. At one time, the temperature gradient is 5.0°C/cm at cross-section A and is 2.6°C/cm at cross-section B. Calculate

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Page # 32

HEAT

the rate at which the tempeature is increasing in the part AB of the rod. The heat capacity of the part AB = 0.40 J/°C, thermal conductivity of the material of the rod = 200 W/m–°C. Neglect any loss of heat to the atmosphere. 25. A rod of negligible heat capacity has length 20 cm, area of cross-section 1.0 cm2 and thermal conductivity 200 W/m–°C. The tempeature of one end is maintained at 0°C and that of the other end is slowly and linearly varied from 0°C to 60°C in 10 minutes. Assuming no loss of heat through the sides, find the total heat transmitted through the rod in these 10 minutes. 26. A pan filled with hot food cools from 50.1 °C to 49.9 °C in 5 sec. How long will it take to cool from 40.1 °C to 39.9°C if room temperature is 30°C ? 27. A solid copper cube and sphere, both of same mass & emissivity are heated to same initial temperature and kept under identical conditions. What is the ratio of their initial rate of fall of temperature ? 28. Two spheres of same radius R have their densities in the ratio 8 : 1 and the ratio of their specific heats are 1 : 4. If by radiation their rates of fall of temperature are same, then find the ratio of their rates of losing heat. 29. The maximum wavelength in the energy distribution spectrum of the sun is at 4753 Å and its temperature is 6050K. What will be the temperature of the star whose energy distribution shows a maximum at 9506 Å. 30. A black body radiates 5 watts per square cm of its surface area at 27°C. How much will it radiate per square cm at 327°C. 31. A 100 W bulb has tungsten filament of total length 1.- m and radius 4 × 10–5 m. The emissivity of the filament is 0.8 and  = 6.0 × 10–8 W/m2 – K4. Calculate the temperature of the filament when the bulb is operating at correct wattage.

32. A copper sphere is suspended in an evacuated chamber maintained at 300 K. The sphere is maintained at a constant temperature of 500 K by heating it electrically. A total of 210 W of electric power is needed to do it. When the surface of the copper sphere is completely blackened, 700 W is needed to maintain the same temperature of the sphere. Calculate the emissivity of copper. 33. During a certain duration in the day, the earth is in radiative equilibrium with the sun. Find the surface temperature of the earth during that duratian. [Given, radius of sun = 6.9 × 108 m surface temperature of sun = 6000 K and the distance of earth from the sun = 1.49 × 1011 m. Assume that the sun and earth behave as black bodies.] 34. Estimate the temperature at which a body may appear blue or red. The values of mean for these are 5000 and 7500 Å respectively. [Given Wein’s constant b = 0.3 cm K] 35. Find the quantity of energy radiated from 1 cm2 of a surface in one second by a black body if the maximum energy density corresponds to a wavelength of 5000 Å (b = 0.3 cm K and  = 5.6 × 10–8 w/m2k4) 36. The following observations have been noted for a black body spectrum, taken for T = 500 K. Calculate the value of m at T = 1000 K.  (in  m ) E (in SI units)

10

8

6

4

10

14

16

12

37. A liquid cools from 70°C to 60°C is 5 minutes. Find the time in which it will further cool down to 50 °C, if its surrounding is held at a constant temperature of 30°C 38. A body cools down from 50°C to 45°C in 5 minutes and to 40°C in another 8 minutes. Find the temperature of the surrounding.

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Page # 33

HEAT

Exercise - IV 1. A copper calorimeter of mass 100 gm contains 200 gm of a mixture of ice and water. Steam at 100°C under normal pressure is passed into the calorimeter and the temperature of the mixture is allowed to rise to 50°C. If the mass of the calorimeter and its contents is now 330 gm, what was the ratio of ice and water in beginning? Neglect heat losses. Given : Specific heat capacity of copper = 0.42 × 103 J kg–1 K–1, Specific heat capacity of water = 4.2 × 103 J kg–1K–1, Specific heat of fusion of ice = 3.36 × 105 J kg–1 Latent heat of condensation of steam = 22.5 × 105 Jkg–1 2. A solid substance of mass 10 gm at –10°C was heated to –2°C (still in the solid state). The heat required was 64 calories. Another 880 calories was required to raise the temperature of the substance (now in the liquid state) to 1°C, while 900 calories was required to raise the temperature from –2°C to 3°C. Calculate the specific heat capacities of the substance in the solid and liquid state in calories per kilogram per kelvin. Show that the latent heat of fusion L is related to the melting point temperature tm by L = 85400 + 200 tm. 3. A steel drill making 180 rpm is used to drill a hole in a block of steel. The mass of the steel block and the drill is 180 gm. If the entire mechanical work is used up in producing heat and the rate of raise in temperature of the block and the drill is 0.5 °C/s. Find (a) the rate of working of the drill in watts, and (b) the torque required to drive the drill. Specific heat of steel = 0.1 and J = 4.2 J/cal. Use : P =  4. A flow calorimeter is used to measure the specific heat of a liquid. Heat is added at a known rate to a stream of the liquid as it passes through the calorimeter at a known rate. Then a measurement of the resulting temperature difference between the inflow and the outflow points of the liquid stream enables us to compute the specific heat of the liquid. A liquid of density 0.2 g/cm3 flows through a calorimeter at the rate of 10 cm3/s. Heat is added by means of a 250-W electric heating coil, and a temperature difference of 25°C is established in steady-state conditions between the inflow and the outflow points. Find the specific heat of the liquid.

(TOUGH

SUBJECTIVE PROBLEMS)

5. Ice at –20°C is filled upto height h = 10 cm in a uniform cylindrical vessel. Water at temperature °C is filled in another identical vessel upto the same height h = 10 cm. Now, water from second vessel is poured into first vessel and it is found that level of upper surface falls through h = 0.5 cm when thermal equilibrium is reached. Neglecting thermal capacity of vessels, change in density of water due to change in temperature and loss of heat due to rediation, calculate initial temperature  of water. Given, Density of water, w = 1 gm cm–3 Density of ice, i = 0.9 gm/cm3 Specific heat of water, sw = 1 cal/gm °C Specific heat of ice si = 0.5 cal/gm ºC Specific latent heat of ice, L = 80 cal/gm 6. A composite body consists of two rectangular plates of the same dimensions but different thermal conductivities KA and KB . This body is used to transfer heat between two objects maintained at different temperatures. The composite body can be placed such that flow of heat takes place either parallel to the interface or perpendicular to it. Calculate the effective thermal conductivities K|| and K  of the composite bo dy for the parallel and perpendicular orientations. Which orientation will have more thermal conductivity ? 7. A highly conducting solid cylinder of radius a and length l is surrounded by a co-axial layer of a material having thermal conductivity K and negligible heat capacity. Temperature of surrounding space (out side the layer) is T0, which is higher than temperature of the cylinder. If heat capacity per unit volume of cylinder material is s and outer radius of the layer is b, calculate time required to increase temperature of the cylinder from T1 to T2. Assume end faces to be thermally insulated. 8. A vertical brick duct (tube) is filled with cast iron. The lower end of the duct is maintained at a temperature T1 which is greater than the melting point Tm of cast iron and the upper end at a temperature T2 which is less than the temperature of the melting point of cast iron. It is given that the conductivity of liquid cast iron is equal to k times the conductivity of solid cast iron. Determine the fraction of the duct filled with molten metal.

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Page # 34

HEAT

9. A lagged stick of cross section area 1 cm2 and length 1 m is initially at a temperature of 0°C. It is then kept between 2 reservoirs of temperature 100°C and 0°C. Specific heat capacity is 10 J/ kg°C and linear mass density is 2 kg/m. Find

100°C

0°C x

(a) temperature gradient along the rod in steady state. (b) total heat absorbed by the rod to reach steady state. 10. A cylindrical block of length 0.4 m an area of cross-section 0.04 m2 is placed coaxially on a thin metal disc of mass 0.4 kg and of the same cross-section. The upper face of the cylinder is maintained at a constant temperature of 400K and the initial temperature of the disc is 300K. If the thermal conductivity of the material of the cylinder is 10 watt/m-K and the specific heat of the material of the disc in 600 J/kg-K, how long will it take for the temperature of the disc to increase to 350K ? Assume, for purposes of calculation, the thermal conductivity of the disc to be very high and the system to be thermally insulated except for the upper face of the cylinder. 11. A solid copper sphere cools at the rate of 2.8°C per minute, when its temperature is 127°C. Find the rate at which another solid copper sphere of twice the radius lose its temperature at 327°C, if in both the cases, the room temperature is maintained at 27°C. 12. End A of a rod AB of length L = 0.5 m and of uniform cross-sectional area is maintained at some constant temperature. The heat conductivity of the rod is k = 17 J/s-m°K. The other end B of this rod is radiating energy into vaccum and the wavelength with maximum energy density emitted from this end is 0 = 75000 Å. If the emissivity of the end B is e = 1, determine the temperature of the end A. Assuming that except the ends, the rod is thermally insulated.

Blackened sphere

envelop space to section

14. A liquid takes 5 minutes to cool from 80°C to 50°C. How much time will it take to cool from 60°C to 30°C? The temperature of surrounding is 20°C. Use exact method. 15. A barometer is faulty. When the true barometer reading are 73 and 75 cm of Hg, the faulty barometer reads 69 cm and 70 cm respectively. (i) What is the total length of the barometer tube ? (ii) What is the true reading when the faulty barometer reads 69.5 cm? (iii) What is the faulty barometer reading when the true barmeter reads 74 cm? 16. A vessel of volume V = 30 l is separated into three equal parts by stationary semipermeable thin membrances as shown in the Figure. The left, middle and right parts are filled with mH2  30g of hydorgen, m O 2  160g of oxygen, and mN2  70g of nitrogen respectively. The left partition lets through only hydrogen, while the right partition lets through hydrogen and nitrogen. What will be the pressure in each part of the vessel after the equilibrium has been set in if the vessel is kept at a constant temperature T = 300K? N2

O2

H2

17. Twelve conducting rods form the riders of a uniform cube of side ‘l’. If in steady state, B and H ends of the rod are at 100°C and 0°C. Find the temperature of the junction ‘A’. G

F

0°C

E

H

13. The shell of a space station is a blackened sphere in which a temperature T = 500K is maintained due to operation of appliances of the station. Find the temperature of the shell if the station is enveloped by a thin spherical black screen of nearly the same radius as the radius of the shell.

100°C B

C A

D

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Page # 35

HEAT

Exercise - V

JEE-Problems

2. Two metal cubes A & B of same size are arranged as shown in figure. The extreme ends of the combination are maintained at the indicated temperatures. The arrangement is thermally insulated. The coefficients of thermal conductivity of A & B are 300 W/m°C and 200 W/m°C respectively. After steady state is reached the tem perature T o f th e in terf ace will be _____________. [JEE’ 96]

7. 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 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien constant b = 2.88 × 106 nm K. Then [JEE’98] (A) U1 = 0 (B) U3= 0 (C) U1 > U2 (D) U2 > U1

4. A spherical black body with a radius of 12 cm radiates 450 W power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be (A) 225 (B) 450 (C) 900 (D) 1800 5. Earth receives 1400 W/m2 of solar power. If all the solar energy falling on a lens of area 0.2 m2 is focussed on to a block of ice of mass 280 grams, the time taken to melt the ice will be __________ minutes. (Latent heat of fusion of ice = 3.3 × 105 J/kg) [JEE ‘97] 6. A solid body X of heat capacity C is kept in an atmosphere whose temperature is TA = 300K. At time t = 0, the temperature of X is T0 = 400K. It cools according to Newton’s law of cooling. At time t1 its temperature is found to be 350K. At this time t1, the body X is connected to a larger body Y at atmospheric temperature TA, through a conducting rod of length L, cross-sectional area A and thermal conductivity K. The heat capacity of Y is so large that any variation in its temperature

(A)

(B)

Temperature

3. A double pane window used for insulating a room thermally from outside consists of two glass sheets each of area 1 m2 and thickness 0.01 m separated by a 0.05 m thick stagnant air space. In the steady state, the room glass interface and the glass outdoor interface are at constant temperatures of 27°C and 0°C respectively. Calculate the rate of heat flow through the window pane. Also find the temperatures of other interfaces. Given thermal conductivities of glass and air as 0.8 and 0.08 Wm–1K–1 respectively. [JEE ‘97]

Heat supplied

(C)

Heat supplied

(D)

Temperature

T

8. A block of ice at –10°C is slowly heated and converted to steam at 100°C. Which of the following curves represents the phenomenon qualitatively? [JEE(Scr)2000] Temperature

B

Temperature

A

0°C

may be neglected. The cross-sectional area A of the connecting rod is small compared to the surface area of X. Find the temperature of X at time t = 3t1 [JEE’ 98]

100°C

1. The temperature of 100 gm of water is to be raised from 24°C to 90°C by adding steam to it. Calculate the mass of the steam required for this purpose. [JEE’ 96]

Heat supplied

Heat supplied

9. The plots of intensity versus wavelength for three black bodies at temperature T1, T2 and T3 respectively are as shown. Their temperatures are such that [JEE(Scr)2000] I T1

T3 T2



(A) T1 > T2 > T3 (C) T2 > T3 > T1

(B) T1 > T3 > T2 (D) T3 > T2 > T1

10. Three rods made of the same material and having the same cross-section have been joined as shown in the figure. Each rod is of the same length. The left and right ends are kept at 0°C and 90°C respectively. The temperature of the junction of the three rods will be [JEE (Scr) 2001] 90°C

0°C 90°C

(A) 45°C

(B) 60°C

(C) 30°C (D) 20°C

11. An ideal black body at room temperature is thrown into a furnace. It is observed that (A) initially it is the darkest body and at later

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Page # 36

HEAT

times the brightest. (B) it the darkest body at all times (C) it cannot be distinguished at all times. (D) initially it is the darkest body and at later times it cannot be distinguished.[JEE(Scr)2002] 12. An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227°C. The specific heat S of the container varies with temperature T according the empirical relations = A + BT, where A = 100 cal/kg-K and B = 2 × 10–2 cal/kg-K2. If the final temperature of the container is 27°C, determine the mass of the container. (Latent heat of fusion for water = 8 × 104 cal/kg. Specific heat of water = 103 cal/kg-K) [JEE’2001] 13. 2 kg ice at –20°C is mixed with 5kg water at 20°C. Then final amount of water in the mixture would be ; Given specific heat of ice = 0.5 cal/ g°C, specific heat of water = 1 cal/g°C, [JEE’ (Scr) 2003] Latent heat of fusion of ice = 80 cal/g. (A) 6 kg (B) 5 kg (C) 4 kg (D) 2 kg 14. If emissivity of bodies X and Y are ex and ey and absorptive power are Ax and Ay then T

(B) QB is maximum (D) QA = QB = QC

17. 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 of ice in the two cases respectively. The ratio q2/q1 is [JEE’ 2004(Scr.)] (A) 1/2 (B) 2/1 (C) 4/1 (D) 1/4 18. Liquid oxygen at 50 K is heated to 300 K at constant pressure of 1 atm. The rate of heating is constant. Which of the following graphs represents the variation of temperature with time [JEE’2004(Scr.)] Temp.

Temp.

(A)

(B) Time

Time

Temp.

Temp.

(C)

(D) Time

x y t

(A) ey > ex ; Ay > Ax (C) ey > ex ; Ay < Ax

(A) QA is maximum (C) QC is maximum

(B) ey < ex; Ay < Ax (D) ey = ex ; Ay = Ax

15. Hot oil is circulated through an insulated container with a T =127°C wooden lid at the top whose t = 5 mm, emissivity = 0.6. Temperature of the top of the lid in steady state is at Tl = T 127°. If the ambie nt T =27°C temperature Ta = 27°C. Hot oil Calculate (a) rate of heat loss per unit area due to rediation from the lid. 17 (b) temperature of the oil. (Given  = × 10–8) 3 [JEE’ 2003] l

0

a

16. Three discs A, B, and C having radii 2 m, 4m and 6m respectively are coated with carbon black on their outer surfaces. The wavelengths corresponding to maximum intensity are 300 nm, 400 nm and 500 nm respectively. The power radiated by them are QA, QB and QC respectively. [JEE’ 2004 (Scr.)]

Time

19. A cube of coefficient of linear expansion s is floating in a bath containing a liquid of coefficient of volume expansion l. When the temperature is raised by T, the depth upto which the cube is submerged in the liquid remains the same. Find the relation between s and l, showing all the steps. [JEE 2004] 20. One end of a rod of length L and crosssectional area A is kept in a furnace of temperature T1. The other end of the rod is kept at a temperature T2. The thermal conductivity of the material of the rod is K and emissivity of the rod is e. It is given that T2 = TS + T where T LA = LC

6. SA = SB > SC > SD

315  C = 28.66°C 4.  = 1050 11 sec. = 7.7 min 7. 2.27

8. 25.5°C

9. 4°C

10. 27/85

11. 0°C, 125/4 g ice, 1275/4 g water

12. (i) 0.02 kg, (ii) 40,000 cal kg–1, (iii) 750 cal kg–1 K–1

13. 64 J

15. 4.0 W

14. 5 × 10–5 g/s

16. 2 : 

17. 10 cm from end in contact with water at t1  t 2  t 3 t t t 1 18.  2  3 k1 k 2 k 3

19. 65°C

22. 15 W/m–°C

23.

20. 5°C

21. 4/3 

Kr1r2 ( 2 – 1 ) L

24. 12 °C/s

25. 1800 J

26. 10 sec

31. 1700 K

32. 0.3

33. 15°C

27. ( 6 /  )1/ 3

34. 6 × 103 K ; 4 × 103 K

35. 7.31 × 1010 erg/cm2 sec.

37. 7 minutes.

38. 34°C

28. 2 : 1 29. 3025 K

30. 80 Watt.

36. m = 3 m

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Page # 39

HEAT

Exercise-IV 1. 1 : 1.26

2. 800 cal kg–1 K–1, 1000 cal kg–1 K–1

4. 5000 J/°C kg

5. 45°C

6. K11 > K  , K|| =

 T0  T1  a2 s  b 7. 2K loge  a  loge  T  T   0 2

8.

3. (a) 37.8 J/s (Watts), (b) 2.005 N-m 2K AK B K A  KB , K = K  K 2 A B

l1 k( T1  Tm )  l k( T1  Tm )  ( Tm  T2 )

10. 166.3 sec

11. 9.72°C/min 12. TA = 423K

14. 10 minutes

15. (i) 74 cm, (ii) 73.94 cm, (iii) 69.52 cm

9. (a) –100°C/m, (b) 1000 J 13. T’’ =

4

2 × 500 = 600 K

16. (i) p1 = pH2 ~– 1.25 × 106 Pa ; p2 = pH2 + pO2 + pN2 ~– 2.8125 × 106 Pa ; p3 = pH2 + pN2 ~– 1.5625 × 106 Pa 17. 60°C

Exercise-V 1. 12 gm

6. k =

2. 60°C

3. 41.53 Watt; 26.48 °C ; 0.55 °C

  KA loge 2   loge 2    2t1  ; T = 300 + 50 exp. t1   t1   LC

9. B

10. B

11. D

15. (a) 595 watt/m2, (b) T0 420K

5. 5.5 min

7. D

12. 0.5 kg 16. B

4. D

13. A 17. D

8. A 14. A 18. C

K

19. l = 2s

20.

4eLTS3  K

21. A

22. C

23. A

24. A 25. B

28. (A) S, Q ; (B) Q ; (C) P, Q ; (D) Q, R or (A) S, (B) Q, (C) P, (D) R 29. 9 31. C

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26. A,D 27. 273K 30. 8 g

IIT-JEE|AIEEE CBSE|SAT|NTSE OLYMPIADS

Nurturing potential through education

HEAT- 2 THEORY AND EXERCISE BOOKLET

CONTENTS S.NO.

TOPIC

PAGE NO.

1. Concept of an Ideal Gas ............................................................ 2 2. Kinetic theory of gases ............................................................. 2 3. Pressure of a gas .................................................................. 2 – 3 4. Gas Law ............................................................................. 3 – 12 5. Maxwell's Distribution Law ..................................................... 12 – 13 6. Degree of Freedom .............................................................. 13 – 14 7. Internal Energy ....................................................................... 15 8. Thermodynamics ................................................................. 16 – 20 9. First Law of thermodynamics ................................................. 21 – 24 10. Specific Heat .................................................................... 24 – 29 11. Exercise -I ....................................................................... 30 – 61 12. Exercise - II ..................................................................... 62 – 68 13. Exercise - III .................................................................... 69 – 79

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Page # 2

HEAT– 2

1.

CONCEPT OF AN IDEAL GAS A gas has no shape and size and can be contained in a vessel of any size or shape. It expands indefinitely and uniformly to fill the available space. It exerts pressure on its surroundings. The gases whose molecules are point massses (mass without volume) and do not attract each other are called ideal or perfect gases. It is a hypothetical concept which can't exist in reality. The gases such as hydrogen, oxygen or helium which cannot be liquified easily are called permanent gases. An actual gas behaves as ideal gas most closely at low pressure and high temperature.

1.1

Ideal gas Equation According to this equation. PV  nRT 

m RT M

In this equation n = number of moles of the gas =

m M

m = total mass of the gas. M = molecular mass of the gas R = Universal gas constant = 8.31 J/mol-K = 2.0 cal/mol- K 2.

3.

KINETIC THEORY OF GASES Kinetic Theory of gases is based on the following basic assumptions. (a) A gas consists of very large number of molecules. These molecules are identical, perfectly elastic and hard spheres. They are so small that the volume of molecules is negligible as compared with the volume of the gas. (b) Molecules do not have any preferred direction of motion, motion is completely random. (c) These molecules travel in straight lines and in free motion most of the time. The time of the collision between any two molecules is very small. (d) The collision between molecules and the wall of the container is perfectly elastic. It means kinetic energy is conserved in each collision. (e) The path travelled by a molecule between two collisions is called free path and the mean of this distance travelled by a molecule is called mean free path. (f) The motion of molecules is governed by Newton’s law of motion (g) The effect of gravity on the motion of molecules is negligible. Note : At higher temperature and low pressure or at higher temperature and low density, a real gas behaves as an ideal gas. EXPRESSION FOR THE PRESSURE OF A GAS : Let us suppose that a gas is enclosed in a cubical box having length  . Let there are ‘N’ identical molecules, each having mass ‘m’. Since the molecules are of same mass and perfectly elastic, so their mutual collisions result in the interchange of velocities only. Only collisions with the walls of the container contribute to the pressure by the gas molecules. Let us focus on a molecule having velocity v1 and components of velocity v X1 , v y1 , v z1 along x, y and z-axis as shown in figure. H G D

v z1 v y1 Fm

Y

C v1 v x1

Z

X

E 

A



B

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HEAT– 2

Page # 3 v 12  v 2x1  v 2y1  v 2z1

The change in momentum of the molecule after one collision with wall BCHE = m v x1 – (– m v x1 ) = 2 m v x1 . The time taken between the successive impacts on the face BCHE =

Time rate of change of momentum due to collision=

2 dis tan ce = v velocity x1

2mv x1 change in momentum mv 2 x1 = 2 / v = time taken x1 

Hence the net force on the wall BCHE due to the impact of n N molecules of the gas is : mv 2x1

Fx =

+



mv 2x2 

+

mv 2x3

+ ........



mv 2xn 

=

m 2 v x1  v 2x2  v 2x3 ...... v 2xn 



v  = mN 

2 x



where < vx 2 > = mean square velocity in x-direction. Since molecules do not favour any particular direction therefore < vx 2 > = < vy 2 > = < vz 2 >. But = + +  v2  . Pressure is equal to force divided by area. 3 M M = = . Pressure is independent of x, y, z directions 3V 3 3

 = P=

Fx 2

where  3 = volume of the container = V M = total mass of the gas, = mean square velocity of molecules P=

1  3

from PV = nRT n=

M Mass = (in kg/mole) M0 Molecular Weight

RT RT M 1 2 P = M V RT = M  M = Vrms  3 0 0 0

Vrms

=

3RT M0 =

3RT mN A =

Vrms =

3RT M0

3Kt m

R K = Boltzman’s const. = N A

3.1

Co-ordinate of the gases (P, V, T) is the coordinate of the gas If initial condition of gas is given by (P1 V1 T1) and final condition of gas is given by (P2, V2 T2) such as (P1 V1 T1) (P2 V2 T2) Then (P, V, T) define situation of gas. When a gas changes from one coordinate system to another co-ordinate system, then we have to follow a process.

4.

GAS LAWS Assuming permanent gases to be ideal, through experiments, it was established that gases irrespective of their nature obey the following laws :

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Page # 4

4.1

HEAT– 2

Boyle's Law According to this law, for a given mass of a gas the volume of a gas at constant temperature (called isothermal process) is inversely proportional to its pressure, i.e., 1 (T = constant) P or PV = constant or PiVi = Pf Vf Thus, P – V graph in an isothermal process is a rectangular hyperbola. Or PV versus P or V graph is a straight line parallel to P or V axis. P PV V

T = constant T = constant

V

4.2

P or V

Charle's law According to this law, for a given mass of a gas the volume of a gas at constant pressure (called) isobaric process) is directly proportional to its absolute temperature, i.e., VT V = constant T Vi V f  Ti Tf

or or

Thus, V – T graph in an isobaric process is a straight line passing through origin. Or V/T versus V or T graph is a straight line parallel to V or T axis. V V/T P = constant

P = constant

4.3

V or T T (in K) Gay Lussac's Law or Pressure law According to this law, for a given mass of a gas the pressure of a gas at constant volume (called isochoric process) is directly proportional to its absolute temperature i.e., PT P = constant T Pi P f  Ti T f

or or

Thus, P - T graph in an isochoric process is a straight line passing through origin or P/T versus P or T graph is a straight line parallel to P or T axis. P

P/T

V = constant

T (in K)

P or T

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HEAT– 2

Page # 5

4.4

Avogadro's Law Two gases at same volume pressure and temperature contain equal amount of moles (mass of gas may be different) or we can say contain equal no. of particle. 1 mole = 6.023 × 10–23 Particles

4.5

Reading of P-V diagram P

P

P=V

A Isobaric B

45º V

V

From PV = n RT P = constant V;T

P

V

A

Isothermal

Isobaric

B V

T

V = constant P= T=

T = constant V;P

B V Isobaric

A T (K)

PV = nRT  P  const  A  B V   T  When T in ºC  PV = nR(T + 273) V

V

ric ba iso

Not isobaric

T(K)

–273

T(ºC)

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Page # 6

HEAT– 2 B

P A

P

=

A

P

V

B V

V

AB P, V, T

•

PV = Constant T=C P; V

ISOTHERMAL V

P

P

A

B

B

A

V

•

T

ISOBARIC P

P

V A

B

A

B

V

•

T

T

T

ISOCHORIC P

P

A

V

A

B

A

B B V

P 2P0

Ex.1

P0

T

?) (2P 0,V0.

B

T

C (2P0,2V0, ?)

A (P ,V ,T ) 0

V0

0

0

2V0

V

Find out the values of co-ordinates at point A, B, C in terms of pressure , volume and temperature and draw curve.

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HEAT– 2

Page # 7

,2 T 0) (2P 0,V 0

P

B

2P0

Sol.

C (2P 0,2V 0,4T0)

A (P ,V ,T )

P0

0

0

V0

0

2V0

V

A  B V = constant

B–C

2P0 P0 = TB T0

V0 2V0 = 2T0 TC

TB = 2T0

TC = 4T0 V

P

B

2P0 P0

P = constant

C

V0 A

T0 2T0

T0

T 4T0

4T0 2T0 T

P Ex.2

Sol.

B

A

P0

V0

AB (Isobaric)

2V0

P

V0 2 V0 T0 = TB

C (2P0,2V0,4T0)

2P0

TB = 2T0 B  C (Isochoric)

P0

P1 P2 = T1 T2

, ,V 0 (P 0

T 0)

B (P0,2V0,2T0)

A

V0

P0 2P0  2T = T 0 2

2V0

V

 T2 = 4T0 V

P 2P0

P0

C

2P0

Find out the values of co-ordinates at point A, B, C in terms of pressure , volume and temperature and draw curve.

C

A

V0 T0

2T0

B

2V0

B

4T0 T 

C

A T0

2T0

4T0 T

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Temp. At point A = T0

V

Page # 8

Ex.3

HEAT– 2

P 2V0

Sol.

P 2V0

Find out the values of co-ordinates at point A, B, C in terms of pressure , volume and temperature and draw curve.

B

C

P0

A (P0,V 0,T 0) V0

B (2P0,V0,2T0)

2V0 V 

C (P 0,2V 0,2T0)

P0

A (P0,V0,T0) V0

2V0 V 

A  B Volume Const (Isochoric)

P1 P2 T1 = T2

B  C Temp. Const (Isothermial)

P1V1 = P2V2

V1 V2 C  A Pressure Const (Isobaric) T = T 1 2

 V 2V0

P 2P0

B

V0 A

P0 T0

Ex.4

C

T0

2T0 T

2T0 T

 V 2V0

Find out the values of co-ordinates at point A, B, C in terms of pressure , volume and temperature and draw curve.

Sol.

B

Pressure at point A = T0

V0

AB

C

A T0

Temp. Constant (isothermal)

2T0 T

P1V1 = P2V2 2P0V0 = 2V0P2  BC

P2 =

 V 2V0

P0 2

  P0 , 2 V0, T0   B 2

C (P 0,2V 0,2T0)

volume constant (Isochoric) V0

PC PB = TB TC



P0 PC 2T0 = 2T0

,T0)

A (P0,V 0 T0



2T0 T

PC = P0

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HEAT– 2

Page # 9

Note : SOME COMMONLY USED TERM

There is heat transfer from gas to surrounding and final temp is same. Conducting wall

There is no Heat Transfer. adiabatic or nonconducting or insulator A

B

Movable Piston

If it is diathermic seprator then the finally temperature is also same on both side (finally pressure is same) movable piston B

A

finally pressure on both side is same then it doesn’t move (massless) Adiabatic walls

Ex.5

If the temperature of the increases slowly from T0 to 2T0 then how much piston will move ? A Gas

Heater ,T0) (P 0,V 0

non-conducting

Sol.

V0 Vt Pressure Same = T = 2T 0 0

Vf = 2V0 for finding distance move chauqe is volume = Ax 2V0 – V0 = Ax x=

x Gas

Heater ,T ) (P 0,V 0 0

non-conducting

V0 A

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Page # 10

HEAT– 2 A

Ex.6

Sol.

If the temperature of the gas changes slowly from T0 to 2T0. Then find out the displacement of the piston.

P0,V0,T0

Pgas A = Kx + P0 A  Kx   P0  P Pgas = Pf =  A  

P0,V0,T0

n.. P. A Kx

Pags A

P0 V0 Pf ( V0  A x ) = T0 Tf

x

PRESSURE VARIATION



H

Pressure of due to liquid

Note 1 : A

N = hg A

N    Ahg

h

h



h sin 

A

 PA  h sin .g

A

Note 2 :

PA  hg

P   Hg .76  g  1atm

vaccum 76cm of Hg

P   Hg . 76 × g

x

1 atm

P   Hg . x . g x cm of Hg pressure means if we placed a straight tube vertically in vaccum. fill the tube with Hg upto x in of height. Then the pressure exerted by Hg at the bottom of the tube is equals to pressure of the gas. Ex.7

Find the new length of gas column in tube if tube is inverted (Assume temperature is constant) atm Pgas 10cm

10cm P0

75 cm of Hg

P

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HEAT– 2 Sol.

Page # 11

Initially : Pgas + 10 = 75  Pgas = 65 Finally 75 + 10 = Pgas  Pgas = 85 cm P1V1 = P2V2 1950 85 Find the new length of gas column in tube if tube is rotated at an angle 60° as shown. (Assume constant temperature)

85 × A ×  = 65 × 30 × A   

Ex.8

,

Gas 40

20cm

60º

Hg cm 20

P0 Pgas = 75 – 20 = 55 P1V1 = P2 V2 55 × 40 × A = 65 ×  × A 

[P + 20 cos 60 = 75 2P + 20 = 150 2P = 130

55  40 65

P = 65] 40

Ex.9

P0, T0

30

Sol.

P0, T0

30

Assume constant temperature if the tube is changed to vertical position and the pallet comes down by 5 cm then find out P0. For upper part P1V1 = P0V0 30+5 P1 35 A = P0 30 A A P1 

30 P0 35

...(i)

5

For lower part P2V2 = P0V0 P0 30 A = P2 25 A P2 

30 P0 25

40 30–5 = 25

...(ii) B

Again P1 + 40 = P2 From (i) and (ii)

...(iii)

30 30 P0  40  P0 35 25

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Page # 12

HEAT– 2

PRESSURE VARIATION IN ATMOSPHERE. Assuming temp. to be const.  PM   .g -dP = dH ..g  -dP = dh.   RT  RT PV = nRT  P = M PM = RT P

–

 P

dP = P

P

h

Mg

 RT . dh 0

0

Pn

h

P+dP

dh

P Mg P0 = – RT h

Mg .h RT PRESSURE VARIATION IN ROTATING ROD.

P = P0 e–



P P+dP dx (P + dP) A – PA = dm w2x AdP = Aw2d dx dP = w2x dx P



P0

dP W 2M = P RT

x

 x dx

W2M  x 2     RT  2 

x

0

P W2M x 2  . P0 RT 2

P  P0

5.

dm  Adx

0

[ln P]PP0

ln

x

W2Mx 2  2RT

MAXWELL’S DISTRIBUTION LAW dN( v) (number of molecules per unit speed interval) against c dv is known as Maxwell’s distribution curve. The total area under the curve is given by the

Distribution Curve – A plot of



integral

 0

dN( v) dv = dv



 dN( v)  N 0

[Note : - The actual formula of

dN(v) is not in JEE syllabus.] dv

Figure shows the distribution curves for two different temperatures. At any temperature the number of molecules in a given speed interval dv is given by the area under the curve in that interval (shown shaded). This number increases, as the speed increases, upto a maximum and then decreases asymptotically toward zero. Thus, maximum number of the molecules have speed lying within a small range centered about the speed corresponding the peak (A) of the curve. This speed is called the ‘most probable speed’ vp or vmp.

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HEAT– 2

Page # 13 T1 dN( v) dv

T2 (higher)

v vp v vrms dv The distribution curve is asymmetrical about its peak (the most probable speed vp) because the lowest possible speed is zero, whereas there is no limit to the upper speed a molecule can

attain. Therefore, the average speed v is slightly larger than the most probable speed vp. The root-mean-square speed, vrms, is still larger /(vrms > v > vp). Average (or Mean) Speed : v

8 RT  M0 =

8RT M0 = 1.59

kT / m

(derivation is not in the course) RMS Speed : vrms =

(v2 ) =

3RT M0 =

3kT = 1.73 m

kT m

Most Probable Speed : The most probable speed vp or vmp is the speed possessed by the maximum number of molecules, and corresponds to the maximum (peak) of the distribution curve. Mathematically, it is obtained by the condition. dN( v) = 0 [by substitution of formula of dN(v) (which is not in the course)] dv Hence the most probable speed is

vP =

2kT = m

2 RT M0

From the above expression, we can see that vrms > v < vP. R = 8.314 J/mole k = Boltzmann counstant (k = 1.38 × 10–23JK–1) 6.

DEGREE OF FREEDOM Total number of independent co-ordinates which must be known to completely specify the position and configuration of dynamical system is known as “degree of freedom f”. Maximum 1 1 1 2 2 2 possible translational degrees of freedom are three i.e.  mVx  mVy  mVz  2 2 2 1 1 1 2 2 2 Maximum possible rotational degrees of freedom are three i.e.  Ix  x  I y  y  I z  z  2 2 2 Vibrational degrees of freedom are two i.e. (Kinetic energy. of vibration and Potential energy of vibration)

Monoatonic Eg : (all inrent gases, He, Ar, etc.) f=3 (translational) (Vx, Vy, VZ)

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Page # 14

HEAT– 2

Diatomic Eg : (gases like H2, N2, O2 etc) f=5 (3 translational + 2 rotational) y

x

(Vx, Vy, Vz ; Wx, Wy, Wz)

z If temp < 70 K for diatomic molecules, then f=3 If temp is in between 250 K to 5000 K, then f=5 If temp is very high (> 5000K) f = 7 [3 translational +2 rotational +2 vibrational] y Triatomic D.O.f. = 6 (Non - linear) V , V , V , Wx , Wy , Wz xyz    3 Trans.

x

3 Rotational

If linear (CO2)

z

Total D.O.f = 5 Vx , Vy , Vz , Wy , Wz     3 Trans.

x

2 Rotational

Maxwell’s law of equipartition of energy. Energy associated with each degree of freedom=

1 KT.. 2

1 KT 2 If degree of freedom of a molecule is f then

of one pasticle is same and =

total kinetic energy of that molecule =

t KT 2

Monoatomic Energy of one particle =

3 3 3 KT, one mde = RT, n mole = nRT T 2 2 2

Diatomic Energy of one Barticle =

5 5 5 KT, one mole = RT, n mole = nRT T 2 2 2

General degree of freedom. Energy of one particle =

t t t KT, one mole = RT, n mole = nRT T 2 2 2

Internal energy of a gas only dipends on the temperature of the gas desn’t depend on the process taken by the gas to reach the tempreature.

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HEAT– 2

Page # 15

7.

INTERNAL ENERGY The internal energy of a system is the sun of kinetic and potential energies of the molecules of the system. It is denoted by U. Internal energy (U) of the system is the function of its absolute temperature (T) and its volume (V). i.e. U = f(T , V) In case of an ideal gas, intermolecular force is zero. Hence its potential energy is also zero. In this case, the internal energy is only due to kinetic energy. Which depends on the absolute f temperature of the gas. i.e. U = f(T). For an ideal gas internal energy U = nRT. 2 Ex.10 A light container having a diatomic gas enclosed with in is moving with velocity v. Mass of the gas is M and number of moles is n. mass of gas = M tempeature T

v

(i) What is the kinetic energy of gas w.r.t centre of mass of the system? (ii) What is K.E. of gas w.r.t ground? Sol.

5 nRT 2 (ii) Kinetic energy of gas w.r.t ground = Kinetic energy of centre of mass w.r.t ground + Kinetic energy of gas w.r.t center of mass.

(i) K.E. =

K.E. =

1 5 Mv2  nRT 2 2

Ex.11 Two nonconducting containers having volume V1 and V2 contain monoatomic and dimatomic gases respectively. They are containers are P1, T1 and P2, T2 respectively. Initially stop cock is closed, if the stop cock is opened find the final pressure and temperature. P1

V1

P1 V1 n1 = RT 1

V2

T2

T1

Sol.

P2

P2 V2 n2 = RT 2

n = n1 + n2 (number of moles are conserved) Finally pressure in both parts & temperature of the both the gases will be become equal. P1 V1 P2 V2 P(V1  V2 ) = RT + RT RT 1 2

From energy conservation 3 5 3 5 n1RT1  n2RT2 = n1RT  n2RT 2 2 2 2

(3P1 V1  5P2 V2 )T1 T2  T = 3P V T  5P V T 1 1 2 2 2 1  3P1V1  5P2 V2  P =  3P V T  5P V T  2 2 1   1 1 2

 P1 V1 T2  P2 V2 T2  V1  V2 

  

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Page # 16 8.

HEAT– 2

THERMODYNAMICS Thermodynamics is mainly the study of exchange of heat energy between bodies and conversion of the same into mechanical energy and vice versa. Thermodynamic System Collection of an extremely large number of atoms or molecules confined within certain boundaries such that it has a certain value of pressure (P),volume (V) and temperature (T) is called a thermodynamic system. Anything outside the thermodynamic system to which energy or matter is exchanged is called its surroundings. Taking into consideration the interaction between a system and its surroundings thermodynamic system is divided into three classes : (a) Open system : A system is said to be an open system if it can exchange both energy and matter with its surroundings. (b) Closed system : A system is said to be closed system if it can exchange only energy (not matter with its surrounding (c) Isolated system : A system is said to be isolated if it can neither exchange energy nor matter with its surroundings. Zeroth law of Thermodynamics : If two systems (B and C) are separately in thermal equilirbrium with a third one (A), then they the mselves are in thermal equilibrium with each other. adiabatic wall diathermic wall

B

C A

Equation of State (for ideal gases) : The relation between the thermodynamic variables (P, V, T) of the system is called equation of state. The equation of state for an ideal gas of n moles is given by PV = nRT, Work done by a gas : Let P and V be the pressure and volume of the gas. If A be the area of the piston. then force exerted by gas on the piston is, F = P × A Let the piston move through a small distance dx during the expansion of the gas. Work done for a small displacement dx is dW = F dx = PA dx Since A dx = dV, increase in volume of the gas is dV  dW = P dV

dx

area enclosed

P

P,V,T

vi

vf

V

or

W

 d W   P dV

Area enclosed under P-V curve gives work done during process

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HEAT– 2

Page # 17

DIFFERENT TYPES OF PROCESSES (a)

Isothermal Process : T = constant [Boyle’s law applicable] PV = constant

T

P

T

V

P

V

There is exchange of heat between system and surroundings. System should be compressed or expanded very slowly so that there is sufficient time for exchange of heat to keep the temperature constant. Slope of P-V curve in isothermal process : PV = constant = C dP P  dV V



Work done in isothermal process : If v f  vi then W is positive    If v f  vi then W is negative 

Vf W = nRT ln V i

 V  W  2.303 nRT log10 f  Vi  

P

P Compression

Expansion

vi

vf

V

vi

vf

V

Internal energy in isothermal process : U = f (T) U = 0 (b)

Iso-Choric Process (Isometric Process) : V = constant  Change in volume is zero P is constant T P = const. (Galussac-law) T Work done in isochoric process : Since change in volume is zero therefore dW = p dV = 0 Indicator diagram of isochoric process : 

P

V

V

T T Change in internal energy in isochoric process : U  n

P

f R T 2

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Page # 18

HEAT– 2

Heat given in isochoric process : Q = U  n

f R T 2

(c) Isobaric Process : Pressure remains constant in isobaric process 

V  cons tan t T Indicator diagram of isobaric process :

P = constant 

P

P

P T

T

V

(in Kelvin)

Work done in isobaric process : W = P V = P (Vfinal – Vinitial) = nR(Tfinal – Tinitial) Change in internal energy in isobaric process : f R T 2 Heat given in isobaric process : Q = U + W

U = n

f f R T + P[Vf – Vi] = n R T + nR T 2 2 Above expression gives an idea that to increase temperature by T in isobaric process heat required is more than in isochoric process.

Q = n

(d)



Cyclic Process : In the cyclic process initial and final states are same therefore initial state = final state Work done = Area enclosed under P-V diagram. Change in internal Energy U = 0 Q = U + W Q = W If the process on P-V curve is clockwise, then net work done is (+ve) and vice-versa. The graphs shown below explains when work is positive and when it is negative (–)work

(–)work

P

P

P V

(–)work

(+)work

P V

V

V

Ex.13 The cylinder shown in the figure has conducting walls and temperature of the surrounding is T, the pistion is initially in equilibrium, the cylinder contains n moles of a gas. Now the piston is displaced slowly by an external agent to make the volume double of the initial. Find work done by external agent in terms of n, R, T. n moles

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HEAT– 2 Sol.

Page # 19

1st Method : Work done by external agent is positive, because Fext and displacement are in the same direction. Since walls are conducting therefore temperature remains constant. Applying equilibrium condition when pressure of the gas is P PatmA

PA Fext

PA + Fext = PatmA Fext = Patm A – PA d

Wext

=  Fext dx = 0

d

d

 Patm Adx –

 PA dx = Patm A  dx –

0

d

0

2v

0

nRT dV V v



= Patm Ad – nRT ln 2

2nd Method Applying work energy theorem on the piston k = 0 Wall = k Wgas + Watm + Wext = 0 Vf nRT ln V – nRT + Wext = 0 i Wext = nRT (1 –ln2) Ex.14 A nonconducting piston of mass m and area of cross section A is placed on a nonconducting cylinder as shown in figure. Temperature, spring constant, height of the piston are given by T, K, h respectively. Initially spring is relaxed and piston is at rest. Find (i) Number of moles (ii) Work done by gas to displace the piston by distance d when the gas is heated slowly. (iii) Find the final temperature

Sol.

(i) PV = nRT





Kx mg dx

d

d

 Pg Adx =

 (mg  PatmA  Kx)dx

0

T

mg   mg    Patm  A  Ah P    Ah = nRT  n =   atm A   RT

(ii) 1st method Applying newton’s law on the piston mg + Patm A + Kx = PgasA Wgas =

K

Patm

0

x

Wgas = mgd + PatmdA +

PgasA

1 Kd2 2

2nd method Applying work energy theorem on the pistion Wall = KE Since piston moves slowly therefore KE = 0 Wgravity + Wgas + Watm + Wspring = 0 – mgd + Wgas + (–PatmAd) + [–(  Wgas = mgd + PatmdA +

1 Kd2 – 0)] = 0 2

1 Kd2 2

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PatmA

mass = m Area = A

Page # 20

HEAT– 2 P

Ex.15 Find out the work done in the given graph. Also draw the corresponding T-V curve and P-T curve. Sol. Since in P-V curves area under the cycle is equal to work done therefore work done by the gas is equal to P0V0. Line AB and CD are isochoric line, line BC and DA are isobaric line.  the T-V curve and P-T curve are drawn as shown. T

C

B

2P0

C

P0

A

D

O

V0

2V0

V

P

B

B

D

C

A D

A

T

V

Ex.16 T-V curve of cyclic process is shown below, number of moles of the gas n find the total work done during the cycle. Sol. Since path AB and CD are isochoric therefore work done is zero during path AB and CD. Process BC and DA are isothermal, therefore

T

B

2T0 T0

D

A

2V0 V

V0

VC WBC = nR2T0 ln V = 2nRT0 ln 2 B VA WDA = nRT0 ln V = – nRT0 ln 2 D Total work done = WBC + WDA = 2nRT0 ln 2 – nRT0 ln 2

C

= nRT0 ln 2

Ex.17 P-T curve of a cyclic process is shown. Find out the work done by the gas in the given process if number of moles of the gas are n. Sol. Since path AB and CD are isochoric therefore work done during AB and CD is zero. Path BC and DA ar isobaric. Hence WBC = nRT = nR(T3 – T2) WDA = nR(T1 – T4) Total work done = WBC + WDA = nR(T1 + T3 – T4 – T2)

P

P1

T3 C

T2 B

P2 A

T1T4

D

T

Ex.18 Consider the cyclic process ABCA on a sample of 2.0 mol of an ideal gas as shown in figure. The temperatures of the gas at A and B are 300 K and 500 K respectively. A total of 1200 J heat is withdrawn from the sample in the process. Find the work done by the gas in part BC. Take R = 8.3 J/mol-K. Sol. The change in internal energy during the cyclic C P process is zero. Hence, the heat supplied to the gas is equal to the work done by it. Hence, WAB + WBC + WCA = –1200 J. ...(i) B A The work done during the process AB is WAB = PA(VB – VA) V = nR(TB – TA) = (2.0 mol) (8.3 J/mol-K) (200 K) = 3320 J The work done by the gas during the process CA is zero as the volume remains constant. From (i) 3320 J + WBC = – 1200 J or WBC = –4520 J = –4520 J

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HEAT– 2 9.

Page # 21

FIRST LAW OF THERMODYNAMICS The first law of thermodynamics is the law of conservation of energy. It states that if a system absorbs heat dQ and as a result the internal energy of the system changes by dU and the system does a work dW, then dQ = dU + dW But, dW = P dV dQ = dU + P dV which is the mathematical statement of first law of thermodynamics. Heat gained by a system, work done by a system and increase in internal energy are taken as positive. Heat lost by a system, work done on a system and decrease in internal energy are taken as negative.

Ex.19 1 gm water at 100°C is heated to convert into steam at 100°C at 1 atm. Find out change in internal energy of water. It is given that volume of 1 gm water at 100°C = 1 cc. volume of 1 gm steam at 100°C = 1671 cc. Latent heat of vaporization = 540 cal/g. (Mechanical equivalent of heat J = 4.2 J/cal.) Sol. From first law of thermodynamic Q = u + w Q = mL = 1 × 540 cal. = 540 cal. W = PV =

105 (1671 – 1)  10–6 4.2

105 (1670)  10–6 = 40 cal. 4.2 U = 540 – 40 = 500 cal.

=

Ex.20 Two moles of a diatomic gas at 300 K are kept in a nonconducting container enclosed by a piston. Gas is now compressed to increase the temperature from 300 K to 400 K. Find work done by the gas

Diatomic 2 moles non conducting gas 300 K container

Sol.

Q = u + W Since container is conconducting therefore Q = 0 = u + w f 5 R T = – 2 × R(400 – 300) 2 2 = – 5 × 8.314 × 100 J = – 5 × 831.4 J = – 4157 J

 W = – u = – n

Ex.21 A sample of an ideal gas is taken through the cyclic process abca (figure. It ab-sorbs 50 J of heat during the part ab, no heat during bc and reflects 70 J of heat during ca. 40 J of work is done on the gas during the part bc.(a) Find the internal energy of the gas at b and c if it is 1500 J at a. (b) Calculate the work done by the gas during the part ca. P

b

c

a v

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Page # 22 Sol.

HEAT– 2

(a) In the part ab the volume remains constant. Thus, the work done by the gas s zero. The heat absorbed by the gas is 50 J. The increase in internal energy from a to b is U = Q = 50 J. As the internal energy is 1550 J at a, it will be 1550 J at b. In the part bc, the work done by the gas is W = –40J and no heat is given to the system. The increase in internal energy from b to c is U = – W = 40 J. As the internal energy is 1550 J at b, it will be 1590 J at C. (b) The change in internal energy, from c to a is U = 1500 J – 1590 J = –90 J The heat given to the system is Q = – 70 J Using Q = U + W, Wca = Q – U = – 70 J + 90 J = 20 J.

Ex.22 The internal energy of a monatomic ideal gas is 1.5 nRT. One mole of helium is kept in a cylinder of cross-section 8.5 cm2. The cylinder is closed by a light frictionless piston. The gas is heated slowly in a process during which a total of 42 J heat is given to the gas. If the temperature rises through 2°C, find the distance moved by the piston. Atmospheric pressure = 100 kPa. Sol. The change in internal energy of the gas is U = 1.5 nR (T) = 1.5 (1 mol) (8.3 J/mol-K) (2K) = 24.9 J The heat given to the gas = 42 J The work done by the gas is W = Q – U = 42 J – 24.9 J = 17.1 J If the distance moved by the piston is x, the work done is W = (100 kPa) (8.5 cm2) x. Thus, (105 N/m2) (8.5 × 10–4 m2) x = 17.1 J or, x = 0.2 m = 20 cm. Ex.23 A sample of an ideal gas has pressure p0, volume v0 and temperature T0. It is isothermally expanded to twice its original volume. It is then compressed at constant pressure to have the original volume V0. Finally, the gas is heated at constant volume to get the original temperature. (a) Show the process in a V - T diagram (b) Calculate the heat absorbed in the process. Sol. (a) The V-T diagram for the process is shown in figure. The initial state is represented by the point a. In the first step, it is isothermally expanded to a volume 2V0. This shown by ab. Then the pressure is kept constant and the gas is compressed to the volume V0. From the ideal gas equation, V/T is constant at constant pressure. Hence, the process is shown by a line bc which passes through the origin. At point c, the volume is V0. ln the final step, the gas is heated at constant volume to a temperature T0. This is shown by ca. The final state is the same as the initial state. (b) The process is cylic so that the change in V internal energy is zero. The heat supplied is, b therefore, equal to the work done by the gas. 2V0 The work done during ab is 2V0 W1 = nRT0 ln V = nRT0 ln 2 = p0V0 ln 2. 0

Also from the ideal gas equation paVa = pbVb

V0

c

a

T0

T

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HEAT– 2

or,

Page # 23

pb =

pa Va p0 V0 p0 = = . Vb 2V0 2

In the step bc, the pressure remains constant. Hence the work done is, p0 p0 V0 (V0 – 2V0) = – 2 2 In the step ca, the volume remains constant and so the work done is zero. The net work done by the gas in the cyclic process is W = W1 + W2 = p0V0 [ln 2 – 0.5] = 0.193 p0V0. Hence, the heat supplied to the gas is 0.193 p0V0.

W2 =

Ex.24 A sample of ideal gas (f = 5) is heated at constant pressure. If an amount 140 J of heat is supplied to the gas, find (a) the change in internal energy of the gas (b) the work done by the gas. Sol. Suppose the sample contains n moles. Also suppose the volume changes from V1 to V2 and the temperature changes from T1 to T2. The heat supplied is Q = U + PV = U + nRT = U +

2U f

(a) The change is internal energy is f f f 140J R(T2 – T1) = R n(T2 – T1) = Q = = 100 J 2 2 2f 1.4 (b) The work done by the gas is W = Q – U = 140 J – 100 J = 40 J Ex.25 There are two vessels. Each of them contains one mole of a monoatomic ideal gas. I nit ial volum e of t he gas in each vessel is 8.3 × 10 –3 m3 at 27°C. Equal amount of heat is supplied to each vessel. ln one of the vessels, the volume of the gas is doubled without change in its internal energy, whereas the volume of the gas is held constant in the second vessel. The vessels are now connected to allow free mixing of the gas. Find the final temperature and pressure of the combined gas system. Sol. 369.3K, 2.462 × 105 N/m2

U = n

Efficiency of cycle () : total Mechanical work done by the area under the cycle in P-V curve gas in the whole process  = Heat injected into the system Heat absorbed by the gas (only +ve)  Q2   =  1 – Q  for Heat Engine,  1



 T2   =  1 – T  for Cannot cycle  1 P

Ex.26 n moles of a diatomic gas has undergone a cyclic process ABC as shown in figure. Temperature at A is T0. Find (i) Volume at C ?

B

2P0 P0

C

A

(ii) Maximum temperature ? (iii) Total heat given to gas ? (iv) is heat rejected by the gas, if yes how much heat is rejected ? (v) Find out the efficiency

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V0

V

V

Page # 24 Sol.

HEAT– 2

(i) Since triangle O A V0 and OC V are similar therefore 2P0 P  0  V = 2V 0 V V0 (ii) Since process AB is isochoric hence PA P  B  T = 2T B TA TB TC TB Since process BC is isobaric therefore V  V B C  TC = 2TB = 4T0 (iii) since process is cyclic therefore 1 P0 V0 2 (iv) Since u and W both are negative in process CA  Q is negative in process CA and heat is rejected in process CA QCA = wCA + uCA

Q = W = area under the cycle =

1 5 [P0 + 2P0] V0 – nR (Tc – Ta) 2 2  4P0 V0 P0 V0  1 5 –  = – [P0 + 2P0]V0 – nR  nR  2 2  nR = – 9P0V0 = Heat injected. (v)  = efficiency of the cycle P0 V0 / 2 work done by the gas = == Q × 100 heat injected injected

=–

Qinj = QAB + QBC 5  5  =  nR(2T0 – T0 )   nR(2T0 )  2P0 (2V0 – V0 ) 2 2     100 19 % P0 V0 = 19 2 SPECIFIC HEAT The specific heat capacity of a substance is defined as the heat supplied per unit mass of the substance per unit rise in the temperature. If an amount Q of heat is given to a mass m of the substance and its temperature rises by T, the specific heat capacity s is given by equation

=

10.

Q mT The molar heat capacities of a gas are defined as the heat given per mole of the gas per unit rise in the temperature. The molar heat capacity at constant volume, denoted by Cv, is :

s=

f  Q   R Cv =  nT   cons tan t volume 2

and the molar heat capacity at constant pressure, denoted by Cp is,  Q  CP =  nT   cons tan t

Pr essure

f     1 R 2 

where n is the amount of the gas in number of moles and f is degree of freedom. Quite often, the term specific heat capacity or specific heat is used for molar heat capacity. It is advised that the unit be carefully noted to determine the actual meaning. The unit of specific heat capacity is J/kg-K whereas that of molar heat capacity is J/mol-K.

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HEAT– 2

Page # 25

MOLAR HEAT CAPACITY OF IDEAL GAS IN TERMS OF R : (i) For a monoatomic gas f = 3 CV = (ii)

For a diatomic gas f = 5 CV =

(iii)

CP 5 3 5  = R , CP = R  = 1.67 C 3 2 2 v

CP 5 7  1.4 R , CP = R  = CV 2 2

For a Triatomic gas f = 6 CV = 3R, CP = 4R CP 4 = C = = 1.33 3 V

[Note for CO2 ; f = 5, it is linear] ln general if f is the degree of freedom of a molecule, then CV =

CP 2 f   f  1   R , CP =   1  R ,  = C f 2 2    V

for any general process C =

f R work done by gas  2 n T

Ex.27 Two moles of a diatomic gas at 300 K are enclosed in a cylinder as shown in figure. Piston is light. Find out the heat given if the gas is slowly heated to 400 K in the following three cases. (i) Piston is free to move (ii) If piston does not move (iii) If piston is heavy and movable. Sol. (i) Since pressure is constant

Patm

light piston

300 K 2 mole Diatomic

7 × R × (400 – 300) = 700 R 2 (ii) Since volume is constant  W = 0 and Q = u (from first law)

 Q = nCPT = 2 ×

5 × R × (400 – 300) = 500 R 2 Since pressure is constant

Q = u = nCvT = 2 × (iii)

Q = nCP T = 2 ×

7 × R × (400 – 300) = 700R 2

Ex.28 P-V curve of a diatomic gas is shown in the figure. Find the total heat given to the gas in the process AB and BC Sol. From first law of thermodynamics QABC = uABC + WABC

P

P0

VC 2V0 WABC=WAB+WBC=0+nRTBln V = nRTB ln V B 0

= nRTB ln 2 = 2P0 V0 ln 2 u = nCv T =

5 (2P0V0 – P0V0)  2

QABC =

B Diatomic Isothermal

2P0

5 P V + 2P0V0 ln 2. 2 0 0

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A

C

V0

2V0

V

Page # 26

HEAT– 2

Ex.29 Calculate the value of mechanical equivalent of heat from the following data. Specific heat capacity of air at constant volume = 170 cal/kg-K,  = Cp/Cv = 1.4. and the density of air at STP is 1.29 kg/m3. Gas constant R = 8.3 J/mol-K Sol. Using pV = nRT, the volume of 1 mole of air at STP is V=

nRT (1mol)  (8.3 J / mol – K)  (273K) = = 0.0224 m3. p 1.0  105 N / m2

The mass of 1 mole is, therefore, (1.29 kg/m3) × (0.0224 m3) = 0.029 kg. The number of moles in 1 kg is

CV =

1 . The molar heat capacity at constant volume is 0.029

170 cal = 4.93 cal/mol-K (1 / 0.029) mol  K

Hence, Cp = Cv = 1.4 × 4.93 cal/mol-K or, Cp – Cv = 0.4 × 4.93 cal/mol-K =1.97 cal/mol-K Thus, 8.3 J = 1.97 cal. The mechanical equivalent of heat is 8.3 J = 4.2 J/cal. 197 . cal Average Molar Specific Heat of Metals : [Dulong and Petit law] At room temperature average molar specific heat of all metals are same and is nearly equal to 3R (6 cal. mol–1K–1) [Note : Temp. above which the metals have constant Cv is called Debye temp.] Mayer’s equation : CP – CV = R (for ideal gases only)

6

Cv

T

ADIABATIC PROCESS When no heat is supplied or extracted from the system the process is called adiabatic. Process is sudden so that there is no time for exchange of heat. If walls of a container are thermally insulated no heat can cross the boundary of the system and Gas process is adiabatic. Equation of adiabatic process is given by PV = constant [Poission law] TP1– = constant T V – 1 = constant P Slope of P-V curve in adiabatic process : dP  P    –  Since PV is constant dV  V 

dp P  –   dV V

Slope of P – T – curve in adiabatic process : Since T P1– is a constant 

insulating wall

v

P ( ) P  dV =– = (1 – ) T ( – 1) T dT

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HEAT– 2

Page # 27

P

 P dP  dT (  – 1) T

T

P

dV  1 V  –  dT  (  – 1) T 

Slope of T-V-curve : dV 1 V – dT ( – 1) T

Work done in adiabatic Process :

W = – U = nCv (Ti – Tf ) =

T

PV i i – Pf Vf ( – 1)

nR(Ti – Tf )

=

 –1

work done by system is (+ve), if Ti > Tf (hence expansion) work done on the system is (–ve) if Ti < Tf (hence compression) Ex.30 A quantity of air is kept in a container having walls which are slightly conducting. The initial temperature and volume are 27°C (equal to the temperature of the surrounding) and 800 cm3 respectively. Find the rise in the temperature if the gas is compressed to 200 cm3 (a) in a short time (b) in a long time. Take  = 1.4. Sol.

(a)

When the gas is compressed in a short time, the process is abiabatic. Thus, T2 V2–1 = T1V1–1  –1

 V1  T2 = T1    V2 

or

0.4

 800  = (300 K) ×    200 

= 522 K.

Rise in temperature = T2 – T1 = 222K (b) When the gas is compressed in a long time, the process is isothermal. Thus, the temperature remains equal to the temperature of the surrounding that is 27°C. The rise in temperature = 0. Ex.31 A monoatomic gas is enclosed in a nonconducting cylinder having a piston which can move freely. Suddenly gas is compressed to 1/8 of its initial volume. Find the final pressure and temperature if initial pressure and temperature are P0 and T0 respectively. Sol. Since process is adiabatic therefore P0

5 V3

5/3

V = Pfinal   8

CP 5 5R 3R = C = / = 3 2 2 V Since process is adiabatic therefore. 2 /3

T1V1–1 = T2 V2–1

V   T0V02/3 = Tfinal  0   8 

 T = 4T0

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Page # 28

HEAT– 2

Ex.32 A cylindrical container having nonconducting walls is partitioned in two equal parts such that the volume of the each parts is equal to V0. A movable nonconducting piston is kept between the two parts. Gas on left is slowly heated so that the gas on right is V compressed upto volume 0 . Find pressure and temperature on both sides if initial 8 pressure and temperature, were P0 and T0 respectively. Also find heat given by the heater to the gas. (number of moles in each part is n) non conducting walls non conducting movable piston

P0, V0, T0

P0,V0,T0

monoatomic

Diatomic

Sol.

Since the process on right is adiabatic therefore PV = constant.  P0V0 = Pfinal (V0/8)  Pfinal = 32 P0 T0V0–1 = Tfinal (V0/8)–1 Let volume of the left part is V1 V0 15 V0  V1 = 8 8 Since number of moles on the left parts remains constant therefore for the left part PV/T = constant. Final pressure on both sides will be same



2V0 = V1 +



P0 V0 Pfinal V1 = T0 Tfinal 

Tfinal = 60 T0

Q = u + w 5R 3R ( 60 T0 – T0 ) + n (4T0 – T0) 2 2 5nR 3nR Q = × 59T0 + × 3T0 2 2

Q = n

Free Expansion If a system, say a gas expands in such a way that no heat enters or leaves the system and also no work is done by or on the system, then the expansion is called the “free expansion”. Q = 0, U = 0 and W = 0. Temperature in the free expansion remains constant. Ex.33 A nonconducting cylinder having volume 2V0 is partitioned by a fixed nonconducting wall in two equal part. Partition is attached with a valve. Right side of the partition is a vaccum and left part is filled with a gas having pressure and temperature P0 and T0 respectively. If valve is opened find the final pressure and temperature of the two parts. Fixed V0 P0,V0,T0

nonconductor

Vacuum

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HEAT– 2 Sol.

Page # 29

From the first law thermodynamics Q = u + W Since gas expands freely therefore W = 0, since no heat is given to gas Q = 0  u = 0 and temperature remains constant. Tfinal = T0 Since the process is isothermal therefore P0 × V0 = Pfinal × 2V0  Pfinal = P0/2 Reversible and Irreversible Process A process is said to be reversible when the various stages of an operation in which it is subjected can be traversed the back in the opposite direction in such a way that substance passes through exactly the same conditions at every step in the reverse process as in the direct process. Comparison of slopes of Iso-thermal and Adiabatic Curve Adiabatic

P isobaric

P

Isothermal

isothermal poly adiabatic di mono

Isothermal

V1

Adiabatic

 adia

V

Wmono < Wdi < Wpoly < Wisothermal < Wisobaric

V

dP dV

V2

dP dV

isothermal

In compression up to same final volume : Wadia

 Wisothermal

In Expansion up to same final volume : Wisothermal  Wadia

Limitations of Ist Law of Thermodynamics : The first law of thermodynamics tells us that heat and mechanical work are interconvertible. However, this law fails to explain the following points : (i) It does not tell us about the direction of transfer of heat. (ii) It does not tell us about the conditions under which heat energy is converted into work. (iii) It does not tell us whether some process is possible or not. Mixture of non-reacting gases : (a)

Molecular weight 

n1M1  n2M2 n1  n2

M1 & M2 are molar masses. n1C V1  n2C V2

(b)

Specific heat C V 

(c)

CPmix n1CP1  n2CP2  ......  for mixture,   C n1C V1  n2C V2  ...... Vmix

(d)

Degree of freedom for mixture f =

n1  n2

,

CP 

n1CP1  n2CP2

n1  n2

n1f1  n2 f2 n1  n2

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Page # 30

HEAT– 2

(ONLY ONE OPTION IS CORRECT)

Exercise - I SECTION (A) : KINETIC THEORY OF GASES

1. When an ideal gas is compressed isothermally then its pressure increases because : (A) its potential energy increases (B) its kinetic energy increases and molecules move apart (C) its number of colisions per unit area with walls of container increases (D) molecular energy increases 2. Which of the following is correct for the molecules of a gas in thermal equilibrium ? (A) All have the same speed (B) All have different speeds which remain constant (C) They have a certain constant average speed (D) They do not collide with one another. 3. Which of the following quantities is zero on an average for the molecules of an ideal gas in equilibrium ? (A) kinetic energy (B) momentum (C) density (D) speed 4. The average momentum of a molecules in a sample of an ideal gas depends on (A) temperature (B) number of moles (C) volume (D) none of these 5. A gas behaves more closely as an ideal gas at (A) low pressure and low temperatue (B) low pressure and high temperature (C) high pressure and low temperature (D) high pressure and high temperature. 6. The temperature at which the r.m.s velocity of oxygen molecules equal that of nitrogen molecules at 100°C is nearly : (A) 426.3 K (B) 456.3 K (C) 436.3 K (D) 446.3 K 7. Suppose a container is evacuated to leave just one molecule of a gas in it. Let va and rms represent the average speed and the rms speed of the gas. (A) a > rms (B) a < rms (C) a = rms (D) rms is undefined 8. The rms speed of oxygen molecules in a gas is . If the temperature is doubled and the O2 molecule dissociated into oxygen atoms, the rms speed will become (A)  (B) 2 (C) 2  (D) 4 9. The quantity pV/kT represents (A) mass of the gas (B) kinetic energy of the gas (C) number of moles of the gas (D) number of molecules in the gas 10. Consider a mixture of oxygen and hydrogen kept

at room temperature. As compared to a hydrogen molecule an oxygen molecule hits the wall (A) With greater average speed (B) with smaller average speed (C) with greater average kinetic energy (D) with smaller average kinetic energy. 11. Keeping the number of moles, volume and temperature the same, which of the following are the same for all ideal gas ? (A) rms speed of a molecule (B) density (C) pressure (D) average magnitude of moentum. 12. Consider the quantity MkT / pV of an ideal gas where M is the mass of the gas. It depends on the (A) temperature of the gas (B) volume of the gas (C) pressure of the gas (D) nature of the gas 13. If vrms = root mean square speed of molecules, vav = average speed of molecules. vmp = most probable speed of molecules, v = speed of sound in a gas Then, identify the correct relation between these speeds. (A) vrms > vav > vmp > vs (B) vav > vmp > vrms > vs (C) vmp > vav > vrms > vs (D) vrms > vav > vs > vmp 14. Three closed vessels A, B and C are at the same temperature T and contain gases which obey the Maxwellian distribution of velcoities. Vessel A contains only O2, B only N2 and C a mixture of equal quantities of O2 and N2. If the average speed of O2 molecules in vessel A is V1, that of the N2 molecules in vessel B is V2, the average speed of the O2 molecules in vessel C will be : (A) (V1 + V2) / 2 (B) V1 (C) (V1V2)1/2

(D)

3kT / M

15. A vessel contains a mixture of one mole of oxygen and two moles of nitrogen at 300 K. The ratio of the average rotational kinetic energy per O2 molecule to that per N2 molecule is : (A) 1 : 1 (B) 1 : 2 (C) 2 : 1 (D) depends on the moments of inertia of the two molecules 16. Three particles have speeds of 2u, 10u and 11u. Which of the following statements is correct ? (A) The r.m.s speed exceeds the mean speed by about u. (B) The mean speed exceeds the r.m.s speed by about u. (C) The r.m.s speed equals the mean speed. (D) The r.m.s. speed exceeds the mean speed by more than 2u.

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HEAT– 2 2E . 3V

Here E refers to (A) translational kinetic energy (B) rotational kinetic energy (C) vibrational kinetic energy (D) total kinetic energy.

(C)

M

N

(D)

L

Temperature/K

17. The pressure of an ideal gas is written as P =

Temperature/K

Page # 31

L

N

volume

18. Which of the following quantities is the same for all ideal gases at the same temperature ? (A) the kinetic energy of 1 mole (B) the kinetic energy of 1 g (C) the number of molecules in 1 mole (D) the number of molecules in 1 g 19. Refer to fig. Let U1 and U2 be the changes in internal energy of the system in the processes A and B then P

M

volume

23. Find the approx. number of molecules contained in a vessel of volume 7 litres at 0°C at 1.3 × 105 pascal (A) 2.4 × 1023 (B) 3 × 1023 23 (C) 6 × 10 (D) 4.8 × 1023 24. A cylindrical tube of cross-sectional area A has two air tight frictionless pistons at its two ends. The pistons are tied with a straight two ends. The pistons are tied with a straight piece of metallic wire.

A

A

B

A

V

(A) U1 > U2 (C) U1 < U2

(B) U1 = U2 (D) U1  U2

20. N(< 100) molecules of a gas have velocities 1,2,3..... N/km/s respectively. Then (A) rms speed and average speed of molecules is same. (B) ratio of rms speed to average speed is (2N + 1) (N + 1) / 6N (C) ratio of rms speed to average speed is (2N + 1) (N + 1) / 6 N (D) ratio of rms speed to average speed of a molecules is 2/6 x (2N + 1)/(N + 1) 21. Five particles have speeds 1,2,3,4,5 m/s. the average velocity of the particles is (in m/s) (A) 3 (B) 0 (C) 2.5 (D) cannot be calculated

Pressure

SECTION (B) : THERMODYNAMICS 22. A fixed mass of ideal gas undergoes changes of pressure and volume starting at L, as shown in figure.

L

M

Isothermal

M

(B) volume

Temperature/K

(A)

Temperature/K

volume O Which of the following is correct :

N

25. An ideal gas mixture filled inside a balloon expands according to the relation PV 2/3 = constant. The temperature inside the balloon is (A) increasing (B) decreasing (C) constant (D) can’t be said 26. A rigid tank contains 35 kg of nitrogen at 6 atm. Sufficient quantity of oxygen is supplied to increase the pressure to 9 atm, while the temperatute remains constant. Amount of oxygen supplid to the tank is : (A) 5 kg (B) 10 kg (C) 20 kg (D) 40 kg 27. A perfect gas of a given mass is heated first in a small vessel and then in a large vesssel, such that their volumes remain unchanged. The P-T curves are (A) parabolic with same curvature (B) parabolic with different curvature (C) linear with same slopes (D) linear with different slopes 28. At a temperature T K, the pressure of 4.0 g argon in a bulb is p. The bulb is put in a bath having temperature higher by 50 K than the first one. 0.8 g of argon gas had to be removed to maintained original pressure. The temperature T is equal to (A) 510 K (B) 200 K (C) 100 K (D) 73 K

N

L

The tube contains a gas at atmospheric pressure P0 and temperature T0. If temperature of the gas is doubled then the tension inthe wire is (A) 4 P0A (B) P0A/2 (C) P0A (D) 2P0A

M L volume

N

29. When 2 gms of a gas are introduced into an evacuated flask kept at 25°C the pressure is found to be one atmosphere. If 3 gms of another gas added to the same flask the pressure becomes 1.5 atmospheres. The ratio of the molecular weights of these gases will be

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Page # 32 (A) 1 : 3

HEAT– 2 (B) 3 : 1

(C) 2 : 3

(D) 3 : 2

30. An open and wide glass tube is immersed vertically in mercury in such a way that length 0.05 m extends above mercury level. The open end of the tube is closed and the tube is raised further by 0.43 m. The length of air column above mercury level in the tube will be : Take Patm = 76 cm of mercury (A) 0.215 m (B) 0.2 m (C) 0.1 m (D) 0.4 m 31. A vessel of volume 0.02 m3 contains a mixture of hydrogen and helium at 20°C and 2 atmospheric pressure. The mass of mixture is 5 gms. Find the ratio of mass of hydrogen to that of helium in the mixture. (A) 1 : 2 (B) 1 : 3 (C) 2 : 3 (D) 3 : 2 32. An ideal gas follows a process PT = constant. The correct graph between pressure & volume is

(A)

(B) V

V

P

P

(C)

V

33. The process AB is shown in the 2P diagram. As the gas is taken from A P to B, its temperature (A) initially increases then decreases (B) initially decreases then increases (C) remains constant (D) variation depends on type of gas

A B V 2V

34. During an experiment an ideal gas obeys an addition equation of state P 2 V = constant. The initial temperature and pressure of gas are T and V respectively. When it expands to volume 2V, then its temperature will be : (B)

2T

(C) 2 T

2T

(D) 2

35. A barometer tube, containing mercury, is lowered in a vessel containing mercury until only 50 cm of the tube is above the level of mercury in the vessel. If the atmospheric pressure is 75 cm of mercury, what is the pressure at the top of the tube ? (A) 33.3 kPa (B) 66.7 kPa (C) 3.33 MPa (D) 6.67 MPa 36. One mole of a gas expands obeying the relation as shown in the P/V diagram. The maximum temperature in this process is equal to P0 V0 (A) R

39. If a mixture of 28 g of Nitrogen, 4 g of Hydrogen and 8 gm of Helium is contained in a vessel at temperature 400 K and pressure 8.3 × 105 Pa, the density of the mixture will be : (A) 3 kg/m3 (B) 0.2 kg/m3 (C) 2 g/litre (D) 1.5 g/litre

(D) V

(A) T

38. 28 gm of N2 gas is contained in a flask at a pressure of 10 atm and at a temperature of 57°. It is found that due to leakage in the flask, the pressure is reduced to half and the temperature reduced to 27°C. The quantity of N2 gas the leaked out is (A) 11/20 gm (B) 20/11 gm (C) 5/63 gm (D) 63/5 gm

40. The temperature of a gas is doubled (i) on absolute scale (ii) on centigrade scale. The increase in root mean square velocity of gas will be (A) More in case (i) (B) More in case (ii) (C) Same in both case (D) Information not sufficient

P

P

37. A vessel with open mouth contains air at 60°C. When the vessel is heated upto temperature T, one fourth of the air goes out. The value of T is (A) 80°C (B) 171°C (C) 333°C (D) 444°C

3 P0 V0 (B) R

41. A cylinder containing gas at 27°C is divided into two parts of equal volume each 100cc and at equal pressure by a piston of cross sectional area 10.85 cm2. The gas in one part is raised in temperature to 100°C while the other maintained at original temperature. The piston and wall are perfect insulators. How far will the piston move during the change in temperature? (A) 1 cm (B) 2 cm (C) 0.5 cm (D)1.5 cm 42. 12 gms of gas occupy a volume of 4 × 10–3 m3 at a temperature of 7°C. After the gas is heated at constant pressure its density becomes 6 × 10–4 gm/cc. What is the temperature to which the gas was heated. (A) 1000 K (B) 1400 K (C) 1200 K (D) 800 K 43. The expansion of an ideal gas of mass m at a constant pressure P is given by the straight line B. Then the expansion of the same ideal gas of mass 2 m at a pressure 2P is given by the straight line Volume

( V0 ,P0 )

P (2 V0 ,P0 / 2)

V

9 P0 V0 (C) 8R

(D) None

(A) C

(B) A

A B C Temperature (C) B

(D) none

44. A vessel contains 1 mole of O2 gas (molar mass 32) at a temperature T. The pressure of the gas is P. An identical vessel containing one mole of He gas (molar mass 4) at a temperature 2T has a pressure of

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HEAT– 2 (A) P/8

Page # 33 (B) P

(C) 2P

work is done by the gas. In which pair of the changes work is done on the gas ?

(D) 8P

46. An ideal gas of Molar mass M is contained in a vertical tube of height H, closed at both ends. The tube is accelerating vertically upwards with acceleration g. Then, the ratio of pressure at the bottom and the mid point of the tube will be (A) exp [2MgH/RT] (B) exp[–2MgH/RT] (C) exp[MgH/RT] (D) MgH/RT 47. The ratio of average translational kinetic energy to rotational kinetic energy of a diatomic molecule at temperature T is (A) 3 (B) 7/5 (C) 5/3 (D) 3/2 48. One mole of an ideal gas at STP is heated in an insulated closed container until the average speed of its molecules is doubled. Its pressure would therefore increase by factor. (A) 1.5

(B)

2

(C) 2

(B)

600 R

(C)

6  104 6  105 (D) R R P

51. One mole of an ideal 2(V ,4P ) diatomic gas is taken through the cycle as shown in the 3(4V , P ) figure. 1(V , P ) 1  2 : isochoric process V 2  3 : straight line on P - V diagram 3  1 : isobaric process The average of molecular speed of the gas in the states 1, 2 and 3 are in the ratio 0

0

0

0

(A) 1 : 2 : 2 (C) 1 : 1 : 1

0

0

2 : (D) 1 : 2 : 4

(B) 1 :

S

volume

(B) PQ and QR

(C) OR and RS

(D) RS and SP

53. Consider the process on a system shown in fig. During the process, the cumulative work done by the system (A) continuously increase (B) continuously decreases (C) first increases then decreases (D) first decreases then increases

P

V

54. Consider two processes on a system as shown in fig. The volume in the initial states are the same in the two processes and the volumes in the final states are also the same. Let W1 and W2 be the work done by the system in the processes A and B respectively. P B

T

(A) W1 > W2 (B) W1 = W2 (C) W1 < W2 (D) Nothing can be said about the relation between W1 and W2 55. A mass of an ideal gas undergoes a reversible isothermal compression. Its molecules will then have compared with initial state, the same (i) root mean square velocity (ii) mean momentum (iii) mean kinetic energy (A) (i), (ii), (iii) correct (B) (i), (ii) correct (C) (ii), (iii) correct (D) (i) correct 56. When a system is taken from state ‘a’ to state ‘b’ along the path ‘acb’, it is found that a quantity of heat Q = 200 J is absorbed by the system and a work W = 80 J is done by it. Along the path ‘adb’, Q = 144J. The work done along the path ‘adb’ is P

2

c

b

a

d V

(A) 6J 52. A fixed mass of gas undergoes the cycle of changes represented by PQRSP as shown in figure. In some of the changes, work is done on the gas and in others,

R

(A) PQ and RS

A

50. A diatomic gas of moleculer weight 30 gm/mole is filled in a container at 27°C. It is moving at a velocity 100 m/s. If it is suddenly stopped, the rise in temperature of gas is :

Q

P

(D) 4

49. The ratio of specific heat of a gas is 9/7, then the number of degrees of freedom of the gas molecules for translational motion is : (A) 7 (B) 3 (C) 6 (D) none

(A) 60/R

Pressure

45. A container X has volume double that of contianer Y and both are connected by a thin tube. Both contains same ideal gas. The temperature of X is 200 K and that of Y is 400 K. If mass of gas in X is m then in Y it will be : (A) m/8 (B) m/6 (C) m/4 (D) m/2

(B) 12J

(C) 18 J

(D) 24 J

57. In the above question, if the work done on the system along the curved path ‘ba’ is 52J, heat absorbed is

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Page # 34 (A) – 140 J

HEAT– 2 (B) – 172 J

(C) 140 J

(D) 172 J

58. In above question, if Ua = 40 J, value of Ub will be (A) –50 J (B) 100 J (C) –120 J (D) 160 J 59. In above question, if Ud = 88 J, heat absorbed for the path ‘db’ is (A) –72 J (B) 72 J (C) 144 J (D) –144 J 60. Ideal gas is taken through process shown in figure : P

B

C A

dimtomic gas to be constant, its specific heat at constant volume per mole (gram mole) is (A) 5/2 R (B) 3/2 R (C) R (D) 1/2 R 66. For an ideal gas, the heat capacity at constant pressure is larger than that at constant volume because (A) work is done during expansion of the gas by the external pressure (B) work is done during expansion by the gas against external pressure (C) work is done during expansion by the gas against intermolecular forces of attraction. (D) more collisions occur per unit time when volume is kept constant.

T

67. Fig shows a process on a (A) ln process AB, work done by system is positive. (B) In process AB, heat is rejected out of the system. (C) In process AB, internal energy increases (D) In process AB internal energy decreases and in process BC internal energy increases. 61. If heat is supplied to an ideal gas in an isothermal process, (A) the internal energy of the gas will increase (B) the gas will do positive work (C) the gas will do negative work (D) the said process is not possible 62. A gas is contained in a metallic cylinder fitted with a piston. The piston is suddenly moved in to compress the gas and is maintaioned at this position. As time passes the pressure of the gas in the cylinder (A) increases (B) decreases (C) remains constant (D) increases or decreases depending on the nature of the gas. 63. A system can be taken from the initial state p1, V1 to the final state p2, V2 by two different methods. Let Q and W represent the heat given to the system and the work done by the system. Which of the following must be the same in both the methods ? (A) Q (B) W (C) Q + W (D) Q – W 64. Refer to fig. Let U1 and U 2 be change in internal P energy in process A and B A respectively, Q be the net heat given to the system in B process A + B and U be the net work done by the system in V the process A + B. (A) U1 + U2 = 0 (B) U1 – U2 = 0 (C) Q – W = 0 (D) Q + W = 0 SECTION (C) : SPECIFIC HEAT CAPACITIES OF GASES 65. Supposing the distance between the atoms of a

P

gas in which pressure and volume both change. The molar heat capacity for this process V

is C. (A) C = 0

(B) C = Cv

(C) C > Cv

(D) C < Cv

68. For a solid with a small expansion coefficient, (A) Cp – Cv = R (B) Cp – Cv = R (C) Cp is slightly greater than Cv (D) Cp is slightly less than Cv 69. The molar heat capacity for the process shown in fig. is (A) C = Cp (B) C = Cv (C) C > Cv (D) C = 0

P

p

K V

V

70. In the following P – V diagram of an ideal gas, two adiabates cut two isotherms at T1 and T2. The value of VB/VC is

A

B T1

P D VA

VD

C

VB

T2

VC

V

AB  T1, DC  T2 (A) = VA / VD (C) > VA / VD

(B) < VA / VD (D) cannot say

71. Four curves A, B, C and D are drawn in the fig. for a given amount of gas. The curves which represent adiabatic and isothermal changes are

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HEAT– 2

Page # 35 P B

A

T1  T2 (A) n  n 1 2

C

D

(C) O

(A) C and D respectively (C) A and B respectively

v

(B) D and C respectively (D) B and A respectively

72. When an ideal gas undergoes an adiabatic change causing a temperature change T (i) there is no heat gained or lost by the gas (ii) the work done is equal to change in internal energy (iii) the change in internal energy per mole of the gase is Cv T, where Cv is the molar heat capacity at constant volume. (A) (i), (ii), (iii) correct (B) (i), (ii) correct (C) (i), (iii) correct (D) (i) correct 73. Starting with the same initial conditins, 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 isothemal, W2 if isobaric and W3 if adiabatic, then : (A) W2 > W1 > W3 (B) W2 > W3 > W1 (C) W1 > W2 > W3 (D) W1 > W3 > W2 74. The internal energy of an ideal gas decreases by the same amount as the work done by the system (A) The process must be adiabatic (B) The process must be isothermal (C) The process must be isobaric (D) The temperatuer must decrease Question No. 75 to 78 (4 questions) Five moles of helium are mixed with two moles of hydrogen to form a mixture. Take molar mass of helium M1 = 4g and that of hydrogen M2 = 2g 75. The equivalent molar mass of the mixture is (A) 6 g

13g (B) 7

T1 T2 (B) n  n 1 2

18g (C) 7

(D) none

76. The equivalent degree of freedom f of the mixture is (A) 3.57 (B) 1.14 (C) 4.4 (D) none 77. The equivalent value of  is (A) 1.59 (B) 1.53 (C) 1.56

(D) none

78. If the internal energy of He sample of 100J and that of the hydrogen sample is 200 J, then the internal energy of the mixture is (A) 900 J (B) 128.5 J (C) 171.4 J (D) 300 J 79. Two monoatomic ideal gas at temperature T1 and T2 are mixed. There is no loss of energy. If the masses of molecules of the two gases are m1 and m2 and number of their molecules are n1 and n2 respectively. The temperature of the mixture will be :

n2 T1  n1 T2 n1  n2

n1 T1  n2 T2 n1  n2

(D)

80. At temperature T, N molecules of gas A each having mass m and at the same temperature 2N molecules of gas B each having mass 2m are filled in a container. The mean square velocity of molecules of gas B is v2 and mean square of x component of velocity of molecules of gas A is w2. The ratio of w2/v2 is : (A) 1 (B) 2 (C) 1/3 (D) 2/3 81. A given mass of a gas expands from a state A to the state B by three paths 1,2 and 3 as shown in T V indicator diagram. If W1, W2 and W3 respectively be the work done by the gas along the three paths, then T

B

1 2 A

3

O

(A) W1 > W2 > W3 (C) W1 = W2 = W3

V

(B) W1 < W2 < W3 (D) W1 < W2, W1 > W3

82. An ideal gas undergoes the process 1  2 as shown in the figure, the heat supplied and work done in the process is Q and W respectively. The ratio Q : W is (A)  :  – 1 (B)  (C)  – 1 (D)  – 1/

V 2

1 T P 3P0

B

83. In the above thermodynamic process, the correct statement is A 2P C (A) Heat given in the complete cycle 2V V V ABCA is zero (B) Work done in the complete cycle ABCA is zero (C) Work done in the complete cycle ABCA is (1/2 P0V0) (D) None 0

0

0

4. Pressure versus temperature graph of an ideal gas is shown in figure. P

C B D A T

(A) During the process AB work done by the gas is positive

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HEAT– 2

(B) during the process CD work done by the gas is negative (C) during the process BC internal energy of the gas is increasing (D) None 85. A reversible adiabatic path on a P-V diagram for an ideal gas passes through state A where P = 0.7 × 105 N/m–2 and v = 0.0049 m3. The ratio of specific heat of the gas is 1.4. The slope of path at A is : (A) 2.0 × 107 Nm–5 (B) 1.0 × 107 Nm–5 (C) –2.0 × 107 Nm–5 (D) –1.0 × 107 Nm–5 86. An ideal gas at pressure P and volume V is expanded to volume 2V. Column I represents the thermodynamic processes used during expansion. Column II represents the work during these processes in the random order. Column I Column II (p) isobaric

(x)

PV(1 – 21–  )  –1

(q) isothermal (y) PV (r) adiabatic (z) PV ln 2 The correct matching of column I and column II is given by : (A) p – y, q – z, r – x (B) p – y, q – x, r – z (C) p–x, q–y, r–z (D) p–z, q–y, r–x P

A

87. An ideal gas is taken from point A to point C on P–V O diagram through two process B AOC and ABC as shown in the P0 C figure. Process AOC is isothermal V0 2V0 V (A) Process AOC requires more heat than process ABC. (B) Process ABC requires more heat than process AOC. (C) Both process AOC & ABC require same amount of heat. (D) Data is insufficient for comparison of heat requirement for the two processes. 88. One mole of an ideal gas is contained piston with in a cyclinder by a frictionless piston and is initially at temperature T. The pressure of the gas is kept constant while it is heated and its volume doubles. If R is molar gas constant, the work done by the gas in increasing its volume is (A) RT ln2 (B) 1/2 RT (C) RT (D) 3/2 RT 89. The figure, shows the ln P graph of logarithmic reading of pressure and volume for two ideal gases A and B undergoing adiabatic process. From figure it can be concluded that (A) gas B is diatomic

(B) gas A and B both are diatomic (C) gas A is monoatomic (D) gas B is monoatomic & gas A is diatomic 90. A thermodynamic cycle takes in heat energy at a high temperature and rejects energy at a lower temperature. If the amount of energy rejected at the low temperature is 3 times the amount of work done by the cycle, the efficiency of the cycle is (A) 0.25 (B) 0.33 (C) 0.67 (D) 0.9 91. Monoatomic, diatomic and triatomic gases whose initial volume and pressure are same, are compressed till their volume becomes half the initial volume. (A) If the compression is adiabatic then monoatomic gas will have maximum final pressure. (B) If the compression is adiabatic then triatomic gas will have maximum final pressure. (C) If the compression is adiabatic then their final pressure will be same. (D) If the compression is isothermal then their final pressure will be different. 92. If heat is added at constant volume, 6300 J of heat are required to raise the temperature of an ideal gas by 150 K. If instead, heat is added at constant pressure, 8800 joules are required for the same temperature change. When the temperature of the gas changes by 300K, the internal energy of the gas changes by (A) 5000 J (B) 12600 J (C) 17600 J (D) 22600 J 93. Three processes from a thermodynamic cycle as shown on P-V diagram for an ideal gas. Process 1  2 takes place at constant temperature (300 K). Process 2  3 takes place at constant volume. During this process 40J of heat leaves the system. Process 3  1 is adiabatic and temperature T3 is 275K. Work done by the gas during the process 3  1 is P

2 3 V

(A) – 40 J

(B) – 20 J

A

(C) + 40 J

(D) +20 J

94. When unit mass of water boils to become steam at 100°C, it absorbs Q amount of heat. The densities of water and steam at 100°C are 1 and 2 respectively and the atmospheric pressure is p0. The increase in internal energy of the water is (A) Q

B

1

1 1 (C) Q + p0   –   1   2

1 1  (B) Q + p0   –   2   1 1 1  (D) Q – p0      2   1

ln V

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HEAT– 2

Page # 37

95. A polyatomic gas with six degrees of freedom does 25 J of work when it is expanded at constant pressure. The heat given to the gas is (A) 100 J (B) 150 J (C) 200 J (D) 250 J 96. An ideal gas expands from volume V1 to V2. This may be achieved by either of the three processes : isobaric, isothermal and adiabatic, Let U be the change in internal energy of the gas, Q be the quantity of heat added to the system and W be the work done by the system on the gas. Identify which of the following statements is false for U? (A) U is least under adiabatic process (B) U is greatest under adiabatic process. (C) U is greatest under the isobaric process (D) U in isothermal process lies in-between the values obtained under isobaric and adiabatic processes. 97. In an isobaric expansion of an ideal gas, which of the following is zero ? (A) work done (B) Q (C) U (D) d2V/dT2 98. A perfect gas is found to obey the relation PV3/2 = constant, during an adiabatic process. If such a gas, initially at a temperature T, is compressed adiabatically to half’ its initial volume, then its final temperature will be (A) 2T (B) 4T (C) 2T (D) 22T 99. A ideal monoatomic gas is carried around the cycle ABCDA as shown in the fig. The efficiency of the gas cycle is

102. Two identical vessels A & B contain equal amount of ideal monoatomic gas. The piston of A is fixed but that of B is free. Same amount of heat is absorbed by A & B. If B’s internal energy increases by 100 J the change in internal energy of A is

Q (A) 100 J

(C)

4 31

2 (B) 21

(D)

2 31

(B)

Q

500 J 3

(C) 250 J

P

1 2

P

3P0

(D) none

103. Three processes compose a thermodynamics cycle shown in the PV diagram. Process 1  2 takes place at constant temperature. Process 2  3 takes place at constant volume, and process 3  1 is adiabatic. During the complete cycle, the total amount of work done is 10 J. During process 2  3, the internal energy decrease by 20J and during process 3  1, 20 J of work is done on the system. How much heat is added to the system during process 1  2 ?

B

3

C

V

(A) 0 4 (A) 21

B

A

P0

A V0

D 2V0 V

100. In thermodynamic process pressure of a fixed mass of gas is changed in such a manner that the gas releases 30 joule of heat and 18 joule of work was done on the gas. If the initial internal energy of the gas was 60 joule, then, the final internal energy will be : (A) 32 joule (B) 48 joule (C) 72 joule (D) 96 joule 101. A cyclinder made of perfectly non conducting material closed at both ends is divided into two equal parts by a heat proof piston. Both parts of the cylinder contain the same masses of a gas at a temperature t0 = 27° and pressure P0 = 1 atm. Now if the gas in one of the parts is slowly heated to t = 57°C while the temperature of first part is maintained at t0 the distance moved by the piston from the middle of the cylinder will be (length of the cyclinder = 84 cm) (A) 3 cm (B) 5 cm (C) 2 cm (D) 1 cm

(B) 10 J

(C) 20 J

(D) 30 J

104. An ideal gas undergoes an adiabatic process obeying the relation PV4/3 = constant. If its initial temperature is 300 K and then its pressure is increased upto four times its initial value, then the final temperature (in Kelvin) : (A) 300 2

(B) 300 3 2

(C) 600

(D) 1200

105. The adiabatic Bulk modulus of a diatomic gas at atmospheric pressure is (A) 0 Nm–2 (B) 1 Nm–2 4 –2 (C) 1.4 × 10 Nm (D) 1.4 × 105 Nm–2 106. A closed container is fully insulated from outside. One half of it is filled with an ideal gas X separated by a plate P from the other half Y which contains a vacuum as shown in figure. When P is removed, X moves into Y. Which of the following statements is correct ?

Y vacuum

X gas p

(A) No work is done by X (B) X decreases in temperature

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Page # 38

HEAT– 2

(C) X increases in internal energy (D) X doubles in pressure. P

107. A cyclic process ABCA is shown in PT diagram. When presented on PV, it would P

(A)

B

P

A

A

B

T

C

V

P

A

C

B

(B)

C

B

110. A vessel contains an ideal monoatomic gas which expands at constant pressure, when heat Q is given to it. Then the work done in expansion is : 3 2 2 (A) Q (B) Q (C) Q (D) Q 5 5 3 111. One mole of an ideal monoatomic gas at temperature T0 expands slowly according to the law P/V = constant. If the final temperature is 2T0, heat supplied to the gas is :

V

P

A

(A) 2RT0

B

A

112. (C)

C

(D)

C

V

V

108. Considere the thermodynamics cycle shown on PV diagram. The process A  B is isobaric, B  C is isochoric and C  A is a straight line process. The following internal energy and heat are given : P(Nm –2) 5 3×10 2×10

A

5

B C

1

3

1.5 V(m )

UA  B = + 400 kJ and QB  C = – 500 kJ The heat flow in the process QC  A is : (A) – 20 kJ (B) + 25 kJ (C) – 25 kJ (D) Data are insufficient 109. 1 kg of a gas does 20 kJ of work and receives 16 kJ of heat when it is expanded between two states. A second kind of expansion can be found between the initial and final state which requires a heat input of 9 kJ. The work done by the gas in the second expansion is : (A) 32 kJ (B) 5 kJ (C) – 4 kJ (D) 13 kJ

(B)

3 RT0 2

(C) RT0

(D)

1 RT0 2

One mole of an ideal gas at temperature T1

P  a (constant). The V2 work done by the gas till temperature of gas becomes T2 is : 1 1 R(T2 – T1) (B) R( T2 – T1 ) (A) 2 3 1 1 (C) R( T2 – T1 ) (D) R( T2 – T1 ) 4 5

expends according to the law

113. 2 moles of a diatomic gas undergoes the process : PT2 / V = constant. Then, the molar heat capacity of the gas during the process will be equal to (A) 5R/2 (B) 9R/2 (C) 3R (D) 4R P

T1 114. Fig. shows graphs of pressure vs. density for an ideal gas at T2 two temperature T1 and T2. (A) T1 > T2 (B) T1 = T2  (C) T1 < T2 (D) any of the three is possible

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HEAT– 2

Page # 39

Exercise - II 1. Consider a collision between an oxygen molecule and a hydrogen molecules in a mixture of oxygen and hydrogen kept at room temperature. Which of the following are possible ? (A) The kinetic energies of both the molecules increase. (B) The kinetic energies of both the molecules decrease (C) The kinetic energy of the oxygen molecule increases and that of the hydrogen molecules decreases. (D) The kinetic energy of the hydrogen molecules increases and that of the oxyzen molecule decreases. 2. In a process on a system, the initial pressure and volume are equal to the final pressure and volume. (A) The initial temperature must be equal to the final temperature (B) The initial internal energy must be equal to the final internal energy. (C) The net heat given to the system in the process must be zero. (D) The net work done by the system in the process must be zero. 3. A system undergoes a cyclic process in which it absorbs Q1 heat and given out Q2 heat. The efficiency of the process in  and work done is W. Select correct statement. (A) W = Q1 – Q2 (B)  = W/Q1 (C)  = Q2/Q1 (D)  = 1 – Q2/Q1 4.A gas kept in a container of finite conductivity is suddenly compressed. The process (A) must be very nearly adiabatic (B) must be very nearly isothermal (C) may be very nearly adiabatic (D) may be very nearly isothemal 5. A rigid container of neligible heat capacity contains one mole of an ideal gas. The temperatur of the gas increases by 1°C if 3.0 cal of heat is added to it. The gas may be (A) helium (B) argon (C) oxygen (D) carbon dioxide 6. When an enclosed perfect gas is subjected to an adiabatic process : (A) Its total internal energy does not change (B) Its temperature does not change (C) Its pressure varies inversely as a certain power of its volume (D) The product of its pressure and volume is directly proportional to its absolute temperature. 7. An ideal gas expands in such a way that PV2 = constant throughout the process.

(A) The graph of the process of T - V diagram is a parabola. (B) The graph of the process of T - V diagram is a straight line. (C) Such an expansion is possible only with heating. (D) Such an expansion is possible only with cooling. 8. Figure shows the pressure P versus volume V graphs for two different gas sample at a given temperature. MA and MB are masses of two samples, nA and nB are numbers of moles. Which of the following must be incorrect. P

A

B

V

(A) MA > MB

(B) MA < MB

(C) nA > nB (D) nA < nB

9. According to kinetic theory of gases, (A) The velocity of molecules decreases for each collision (B) The pressure exerted by a diatomic gas is proportional to the mean velocity of the molecule. (C) The K.E. of the gas decreases on expansion at constant temperature. (D) The mean translational K.E. of a diatomic gas increases with increase in absolute temperature. 10. A closed vessel contains a mixture of two diatomic gases A and B. Molar mass of A is 16 times that of B and mass of gas A contained in the vessel is 2 times that of B. The following statements are given (i) Average kinetic energy per molecule of A is equal to that of B. (ii) Root mean square value of translational velocity of B is four times that of A. (iii) Pressure exerted by B is eight time of that exerted by A. (iv) Number of molecules of B in the cylinder is eight time that of A. (A) (i), (ii) and (iii) are true (B) (ii), (iii) and (iv) are true (C) (i), (ii) and (iv) are true (D) All are true. 11. What is/are the same for O2 and NH3 in gaseous state (A) ratio of specific heats (B) average velocity (C) maximum no. of vibrational degree of freedom (D) None of these assumption of the kinetic theory for an ideal gas ?

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Page # 40

HEAT– 2

(A) The duration of a collision is negligible as compared to the time between successive collisions (B) The molecules have negligible attraction for each other (C) The molecules have negligible momentum change on collision with the container walls (D) There is no total kinetic energy change of the molecules on colliding with each other or with the walls of the container. 13. Select the incorrect statement(s) (A) RMS speed of 8 gm oxygen gas in container at 27° C is approximately 484 m/s (B) RMS speed of 8 gm oxygen in container at 27°C is approximately 968 m/s (C) For number of molecules greater than one, RMS speed is greater than average speed (D) A gas behaves more closely as an ideal gas at low pressures and high temperatues 14. A gas is enclosed in a vessel at a constant temperature at a pressure of 5 atmosphere and volume 4 litre. Due to a leakage in the vessel, after some time, the pressure is reduced to 4 atmosphere. As a result, the (A) volume of the gas decreased by 20% (B) average K.E. of gas molecule decreases by 20% (C) 20% of the gas escaped due to the leakage (D) 25% of the gas escaped due to the leakage. 15. A container holds 1026 molecules/m3, each of mass 3 × 10–27 kg. Assume that 1/6 of the molecules move with velocity 2000 m/s directly towards one wall of the container while the remaining 5/6 of the molecules move either away from the wall or in perpendicular direction, and all collisions of the molecules with the wall are elastic (A) number of molecules hitting 1m2 of the wall every second is 3.33 × 1028 (B) number of molecules hitting 1m2 of the wall every second is 2 × 1029 (C) pressure exerted on the wall by molecules is 24 × 105 Pa (D) pressure exerted on the wall by molecules is 4 × 105 Pa

17. For an ideal gas (A)The change in internal energy in a constant pressure process from temperature T 1 to T 2 is equal to nCV(T2 – T1) where CV is the molar specific heat at constant volume and n is the number of the moles of the gas. (B) The change in internal energy of the gas and the work done by the gas are equal in magnitude in an adiabatic process. (C) The internal energy does not change in an isothermal process (D) A, B and C 18. Two gases have the same initial pressure, volume and temperature. They expand to the same final volume, one adiabatically and the other isothermally (A) The final temperature is greater for the isothermal process (B) The final pressure is greater for the isothermal process (C) The work done by the gas is greater for the isothermal process (D) All the above options are incorrect 19. The first law of thermodynamics can be written as U = Q + W for an ideal gas. Which of the following statements is correct ? (A) U is always zero when no heat enters or leaves the gas (B) W is the work done by the gas in this written law (C) U is zero when heat is supplied and the temperature stays constant (D) Q = – W when the temperature increases very slowly 20. For two different gases X and Y, having degrees of freedom f1 and f2 and molar heat capacities at constant vloume

C V1 and C V2 respectively, the

ln P y

x

lnP versus ln V graph is plotted of adiabatic process, as shown (A) f1 > f2 (B) f2 > f1 (C) C V2  C V1

(D) C V1  C V2

16. A student records Q, U & W for a thermodynamics cycle A  B  C  A. Certain entries are missing. Find correct entry in following options.

(A) WBC = – 70 J (C) UAB = 190 J

(B) QCA = 130 J (D) UCA = – 160 J

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ln V

HEAT– 2

Page # 41

(SUBJECTIVE PROBLEMS)

2. A cubical container having each side as  is filled with a gas having N molecules in the container. Mass of each molecule is m. If we assume that at every instant half of the molecules are moving towards the positive x-axis and half of the molecules are moving towards the negative x-axis. Two walls of the container are perpendicular to the x-axis. Find the force acting on the two walls given ? Assume that all the molecules are moving with speed v0. 3. A uniform tube closed at one end, contains a pallet of mercury 10 cm long. When the tube is kept vertically with the closed end upward, the length of the air column trapped is 20 cm. Find the length of the air column trapped when the tube is inverted so that the closed end goes down. Atmospheric pressure = 75 cm of mercury. 4. An ideal gas is trapped between a mercury column and the closed end of a narrow vertical tube of uniform base containing the column. The upper end of the tube is open to the atmosphere. The lengths of the mercury column and the trapped air column are 20 cm and 43 cm respectively. What will be the length of the air column when the tube is tied slowly in a vertical plane through an angle of 60° ? Assume the temperature to remain constant. 5. Find the temperature at which average speed of oxygen molecule be sufficient so as to escape from the earth? Escape velocity from the earth is 11.0 km/ sec and the mass of one molecule of oxygen 5.34 × 10–26 kg (Boltzmann constant k = 1.38 × 10–23 joule/K) : 6. Find the average magnitude of linear momentum of a helium molecule in a sample of helium gas at 0°C Mass of helium molecule = 6.64 × 10 –27 kg and Boltazmann constant = 1.38 × 10–23 J/K. 7. Find the ratio of the mean speed of hydrogen molecules to the mean speed of nitrogen molecules in a sample containing a mixture of the two gases. 8. 0.040 g of He is kept in a closed container initially at 100.0ºC. The container is now heated. Neglecting the expansion of the container, calculate the temperature at which the internal energy is increased by 12 J. 9. Calculate the change in internal energy of a gas kept in a rigid container when 100 J of heat is supplied to it.

P

B

2P

P

C

D

A

O

V

process shown in figure ?

30 10 10 30 k(Pa) Pressure (kPa)

12. The following graph shows two isotherms for a fixed mass of an ideal gas. Find the ratio of r.m.s. speed of the moelcules at temperatures T1 and T2 ?

4 3 P 2 5 (10 Pa) 1

0

T2 T1 1

2 3 V(m3 )

4

13. A sample of an ideal gas initially having internal energy U1 is allowed to expand adiabatically performing work W. Heat Q is then supplied to it, keeping the volume constant at its new value, until the pressure rised to its original value. The internal energy is then U2. (See Fig.) Find the increase in internal enery (U2 – U1) ?

10. An ideal monoatomic gas is taken round the cycle Volume

ABCDA as shown in the P-V diagram. Find the work done by the

V

2V

11. Find the work done by gas going through a cyclic

Volume (ltr)

1. Find the average velocity of molecules of hydrogen gas in a container at temperature 300 K.

Pressure

Exercise - III

gas during the cycle ?

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Page # 42 14. An ideal gas taken around the cycle ABCA shown in P – V diagram. Find the net work done by the gas during the cycle ?

P 4P1

P1 O

B

C

A V1

16. Find the change in the internal energy of 2 kg of water as it is heated from 0°C to 4°C. The specific heat capacity of water is 4200 J/kg-K and its densities at 0°C and 4°C 999.9 kg/m 3 and 1000 kg/m 3 respectively. Atmospheric pressure = 105 Pa. 17. A substance is taken through the process abc as shown in figure. If the internal energy of the substance increases by 5000 J and a heat of 2625 cal is given to the system, calculate the value of J.

200 kPa

C

3 B

A

1

1 2

3V1 V

15. A gas is enclosed in a cylindrical vessel fitted with a frictionless piston. The gas is slowly heated for some time. During the process, 10 J of heat is supplied and the piston is found to move out 10 cm. Find the increase in the internal energy of the gas. The area of crosssection of the cylinder = 4 cm2 and the atmospheric pressure = 100 kPa.

v(lt)

24. The average degrees of freedom per molecules for a gas is 6. The gas performs 25 J of work when it expands at constant pressure. Find the heat absorbed by the gas. 25. Pressure versus temperature g r a p h of a n id e a l g a s is shown. Density of gas at point A is 0. Find the density of gas at B.

P

B

3P P

A T0 2T0

T

26. An empty pressure cooker of volume 10 litres contains air at atmospheric pressure 10 5 Pa and temperature of 27ºC. It contains a whistle which has area of 0.1 cm2 and weight of 100 gm. What should be the temperature of air inside so that the whistle is just lifted up ? whistle

c

300 kPa

23. In the P-V diagram shown in figure, ABC is a semicircle. Find the workdone in the process ABC.

P(atm)

HEAT– 2

b

a

V(lit.) 0.05 m3

0.02 m3

18. An ideal gas is taken through a process in which the pressure and the volume are changed according to the equation p = kV. Show that the molar heat capacity of the gas for the process is given by C = Cv + R/2. 19. Two ideal gases have the same value of Cp/Cv = . What will be the value of this ratio for a mixture of the two gases in the ratio 1 : 2 ? 20. A gaseous mixture consists of 16 g of helium and Cp 16 g of oxygen. Find the ratio C of the mixture ? v

21. A gas at NTP is suddenly compressed to onefourth of its original volume. If  is supposed to be 3/ 2, then find final pressure ? 22. An ideal gas at pressure 2.5 × 10 5 Pa and temperature 300 K occupies 100 cc. It is adiabatically compressed to half its original volume. Calculate (a) the final pressure, (b) the final temperature and (c) the work done by the gas in the process. Take  = 1.5.

27. V-T curve for 2 moles of a gas is straight line as shown in the graph here. Find the pressure of gas at A.

B A 53º T(K)

28. Air at temperature of 400 K and atmospheric pressure is filled in a balloon of volume 1 m 3. If surrounding air is at temperature of 300 K, find the ratio of Buoyant force on balloon and weight of air inside 29. Ideal diatomic gas is taken through a process Q = 2U. Find the molar heat capacity for the process (where Q is the heat supplied and U is change in internal energy) T 30. Figure shows a parabolic graph between T and 1/V for a mixture of a gas undergoing an adiabatic process. What is the ratio of Vrms and speed of sound in the mixtur

2T0 T0 (1/V0)

(4/V0) (1/V)

31. A piston divides a closed gas cylinder into two parts. Initially the piston is kept pressed such that one part has a pressure P and volume 5V and th other part has pressure 8P and volume V. The piston is now

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HEAT– 2 left free. Find the new pressures and volumes for the adiabatic and isothermal processes. For this gas  = 1.5. 32. A gas undergoes a process in which the pressure and volume are related by VPn = constant. Find the buk modulus of the gas. 33. A standing wave of frequency 1000Hz in a column of methane at 27°C produces nodes which are 20.4 cm apart. Find the ratio of heat capacity of mathane at constant pressure to that at constant volume (Take gas constant, R = 8.31 J.K–1mol–1) 34. One mole of an ideal gas is compressed from 0.5 lit to 0.25 lit. During the compression, 23.04 × 102 J of work is done on the gas and heat is removed to keep the temperature of the gas constant all times. Find the temperature of the gas. (Take universal gas constant R = 8.31 J mol–1K–1) 35. The pressure of an ideal gas changes with volumes as P = aV where ‘a’ is a constant. One moles of this gas is expanded to 3 time its original volume V0. Find (i) the heat transferred in the process (ii) the heat capacity of the gas. 36. 70 calorie of heat is required to raise the temperautre of 2 mole of an ideal gas at constant pressure from 40°C to 45°C. Find the amount of heat required to raise the temperature of the same through the same range at constant volume (R = 2 cal/mol-K) 37. Find the molecular mass of a gas if the specific heats of the gas are C p = 0.2 cal/gm°C and Cv = 0.15 cal/gm°C . [Take R = 2 cal/mol°C]

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Page # 43

Page # 44

HEAT– 2

(TOUGH SUBJECTIVE PROBLEMS)

Exercise - IV

1. A freely moving piston divides a vertical cylinder, closed at both ends, into two parts each containing 1 mole of air. In equilibrium, at T = 300 K, volume of the upper part in  = 4 times greater than the lower p part. At what temperature will the ratio of these volumes be equal to ’ = 2?

k

h

2. A sample of an ideal non linear tri-atomic gas has a pressure P0 and temperature T0 taken through the cycle as shown starting from A. Pressure for process C  D is 3 times P0. Calculate the heat absorbed in the cycle and work done. 5. At a temperature of T0 = 273°K, two mols of an ideal gas undergoes a process as shown. The total amount of heat imparted to the gas equals Q = 27.7 k J . Determine the ratio of molar specific heat capacities.

V 7 V0 2

B

C

A

V0

D

T0

T

T

C

3. Figure shown three processes for an ideal gas. The tempeature at ‘a’ is 600 K, pressure 16 atm and volume 1 litre. The volume at ‘b’ is 4 litre. Out of the two process ab and ac, one is adiabatic and he other is isothermal. The ratio of specific heats of the gas is 1.5. Answer the following :

A

B

V

4V

6. A fixed mass of a gas is taken through a process A  B  C  A. Here A B is isobaric. B  C is adiabatic and C  A is isothemal. Find efficiency of the process (take  = 1.5)

p atm

a

b 1

273K

4

c P

A

B

litre

(i) Which of ab and ac processes is adiabatic. Why ? (ii) Compute the pressure of th gas at b and c. (iii) Compute the temperature and b and c. (iv) Compute the volume at c. 4. An ideal gas NTP is enclosed in a adiabatic vertical cylinder having area of cross section A = 27 cm2, between two light movable pistons as shown in th figure. Spring with force constant k = 3700 N/m is in a relaxed state initially. Now the lower piston is moved upwards a height h/2, h being the initial length of gas column. It is observed that the upper piston moves up by a distance h/16. Find h taking  for the gas to be 1.5. Also find the final temperature of the gas.

C

V

4V

V

7. A cylinder containing a gas is closed by a movable piston. The cylinder is submerged in an ice-water mixture. The piston is quickly pushed down from position 1 to position 2. The piston is held at position 2 until the gas is again at 0°C and then slowly raised back to position 1. Represent the whole process on P – V diagram. If m = 100 gm of ice are melted during the cycle, how much work is done on the gas. Latent heat of ice = 80 cal/gm.

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HEAT– 2

Page # 45 (a) Final pressurs in each compartment A, B and C 1

(b) Final temperatures in each compartment A, B and C

2

(c) Heat supplied by the heater (d) Work done by gas in A and B. (e) Heat flowing across piston I. 11. How many atoms do the molecules of gas consist

8. A parallel beam of particles of mass m moving with velocities v impings on a wall at an angle  to its normal. The number of particles per unit volume in the beam is n. If the collision of particles with the wall is elastic, then find the pressure exerted by this beam on the wall.

of if  increases 1.20 times when the vibrational degrees of freedom are “frozen”? Assume that molecules are non linear. 12.Figure shows the variation of the internal energy U with the density  of one mole of idel monoatomic

9. For the thermodynamic process shown in the figure. PA = 1 × 105 Pa ; PB = 0.3 × 105 Pa PD = 0.6 × 105 Pa ; VA = 0.20 litre VD = 1.30 litre.

gas for a thermodynamic cycle ABCA. Here process AB is a part of rectangular hyperbola.

U A

5U0

P

PA

2U0

D

PD

B 20



C

B

PB

C

(a) Draw the P-V diagram for the above process. VC

VA

VD

V

(b) Find the net amount of heat absorbed by the system for the cyclic process.

(a) Find the work performed by the system along path AD. (b) In the total work done by the system along the path ADC is 85 J find the volume a point C. (c) How much work is perfomed by the system along the path CDA ? 10. The figure shows an insulated cylinder divided into three parts A, B and C. Pistons I and II are connected by a rigid rod and can move without friction inside the cylinder. Piston I is perfectly conducting while piston II si pefectly insulating. The initial state of the gas ( = 1.5) present in each compartment A, B and C is as shown. Now, compartment A is slowly given heat through a heater H such that the final volume of C 4 V0 becomes . Assume the gas to be ideal and find. 9

(c) Find the work done in the process AB. 13. An ideal monoatomic gas undergoes a process where its pressur is inversely proportional to its tempeature. (i) Calculate the specific heat for process. (ii) Find the work done by two moles of gas if the temperature changes from T1 to T2. 14. An ideal diatomic gas undergoes a process in which its internal energy ralates to the volume as U  a V where  is a constant. (a) Find the work performed by the gas and the amount of heat to be transferred to this gas to increase its internal energy by 100 J.

I

H

A

II B

(b) Find the molar specific heat of the gas for this

C

process.

P0, V0, T0

P0, V0, T0

P0, V0, T0

15. Two rectangular boxes shown in figures has a partition which can slide without friction along the

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Page # 46

HEAT– 2

length of the box. Initially each of the two chambers

(a) Find the initial pressure of the gas.

of the box has one mole of a monoatomic ideal gas (

(b) If block m is gentily pushed down and released it

= 5/3) at a pressure p0 volume V0 and temperature T0.

oscillates harmonically, find its angula frquency of

The chamber on the left is slowly heated by an electric

oscillation.

heater. The walls of the box and the partitions are

(c) When the gas in the cylinder is heated up the

thermally insulated. Heat loss through the lead wires

piston starts moving up and the spring gets compressed

of the heater is negligible. The gas in the left chamber

so that the block M is just lifted up. Determine the

expands, pushing the partition until the final pressure

heat supplied.

in both chambers becomes 243 P0/32 . Determine

Take atmospheric pressure P0 = 105 Nm–2 , g = 10 m/s2 17. A thermally insulated vessel is divided into two parts by a heat-insulating piston which can move in the vessel without the friction. The left part of the vessel contains one mole of an ideal monatomic gas, & the right part is empty. The piston is connected to

(i) the final temperature of the gas in each chamber

the right wall of the vessel through a spring whose

and

length in free state is equal to the length of the vessel as shown in the figure. Determine the heat capacity C

(ii) the work-done by the gas in the right chamber. 16. 0.01 moles of an ideal diatomic gas is enclosed in –4

an adiabatic cylinder of cross-sectional area A = 10

of the system, neglecting the heat capacities of the vessel, piston and spring.

m2. In the arrangement shown, a block of mass M = 0.8 kg is placed on a horizontal support, and another block of mass m = 1 kg is suspended from a spring of stiffness constant k = 16 N/m. Initially, the spring is relaxed and the volume of the gas is V = 1.4 × 10–4 m3 M m

k

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HEAT– 2

Exercise - V

Page # 47

(JEE PROBLEMS)

1. The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K the root mean-square velocity of the gas molecules is v, at 480 K it becomes : (A) 4v (B) 2v (C) v / 2 (D) v/4 [JEE ‘96,2]

root mean square speed and the most probable speed of the molecule in an ideal monoatomic gas at absolute temperature T. The mass of a molecule is m then :

2. The average translational energy and the rms speed of molecules in a sample of oxygen gas at 300 K are 6.21 × 10–21 J & 484 m/s respectively. The corresponding values at 600 K are nearly (assuming ideal gas behaviour ) (A) 12.42 × 10–21 J, 968 m/s (B) 8.78 × 10–21 J, 684 m/s (C) 6.21 × 10–21 J, 968 m/s (D) 12.42 × 10–21 J, 684 m/s [JEE’ 97,1]

(B) no molecule can have speed less than v p / 2

3. The average translational kinetic energy of O2 (molar mass 32) molecules at a particular temperature is 0.048 eV. The translational kinetic energy of N2 (molar mass 28) molecules in eV at the same temperature is (A) 0.0015 (B) 0.003 (C) 0.048 (D) 0.768 [JEE ‘97,3] 4. Select the correct alternative.A vessel contains 1 mole of O2 gas (molar mass 32) at a temperature T. The pressure of the gas is P. An identical vessel containing one mole of He gas (molar mass 4) at a temperature 2T has a pressure of : (A) P/8 (B) P (C) 2P (D) 8P 5. Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300 K. The piston of A is free to move, while that of B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas A is 30K, then rise in temperature of the gas in B is (A) 30 K (B) 18 K (C) 50 K (D) 42 K [JEE ‘98] 6. Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and the same volume V. The mass of the gas in A is mA and that in B is mB. The gas in each cylinder is now allowed to expands isothermally to the same final volume 2V. The change in the pressure in A and B are found to be P and 1.5 P respectively. Then [JEE’ 98] (A) 4mA = 9mB (B) 2mA = 3mB (C) 3mA = 2mB (D) 9mA = 4mB 7. A vessel contains a mixture of one mole of oxygen and two moles of nitrogen at 300 K. The ratio of the average rotational kinetic energy per O2 molecule the that per N2 molecule is [JEE’ 98] (A) 1 : 1 (B) 1 : 2 (C) 2 : 1 (D) depend on the moment of inertia of two molecules. 8. Let vav, vrms and vp respectively denote mean speed,

(A) no molecule can have speed greater than

2 v rms

(C) vp < vav < vrms (D) the average kinetic energy of a molecule is 3/4 mvp2 [JEE’98] 9.A given quantity of an ideal gas is at pressure P and absolute temperature T. The isothermal bulk modulus of the gas is : (A) 2P/3 (B) P (C) 3P/2 (C) 2P [JEE’98] 10. During the melting of a slab of ice at 273 K at atmospheric pressutre : [JEE’98] (A) positive work is done by the ice-water system on the atmosphere. (B) positive work is done on the ice-water system by the atmosphere (C) the internal energy of the ice-water system increases (D) the internal energy of ice-water system decreases. 11. The ratio of the speed of sound in nitrogen gas to that in helium gas, at 300 K is (A) (2/7) (B) (1/7) (C) (3)/5 (D) (6/5) [JEE’ 99] 12.A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system is (A) 4RT (B) 15 RT (C) 9 RT (D) 11 RT [JEE’99] 13. A weightless piston divides a thermally insulated cylinder into two parts of volumes V and 3V.2 moles of an ideal gas at pressure P = 2 atmosphere are confined to the part with volume V = 1 litre. The remainder of the cylinder is evacuated. The piston is now released and the gas expands to fill the entire space of the cylinder. The piston is then pressed back to the initial position. Find the increase of internal energy in the process and final temperature of the gas. The ratio of the specific heat of the gas  = 1.5. 14. Two moles of an ideal monatomic gas is taken through a cycle ABCA as shown in the P-T diagram. During the process AB, pressure and temperature of the gas vary such that PT = constant. If T1 = 300 K, calculate :

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Page # 48

HEAT– 2 

P

2P1

B



C (A)

(B)

P

P1

A T1

2T1

T

(a) the work done on the gas in the process AB and (b) the heat absorbed or released by the gas in each of the processes. Give answers in terms of the gas constant R.

[JEE’ 2000]

15. P-V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should correspond respectively to P

1 2 V

(A) He and O2 (C) He and Ar

(B) O2 and He (D) O2 and N2 [JEE’ 2001]

16. In a given process on an ideal gas, dW = 0 and dQ < 0. then for the gas [JEE’ 2001] (A) the temperature will decrease (B) the volume will increase (C) the pressure will remain constant (D) the temperature will increase 17. An ideal gas is taken through the cycle A  B  C  A, as shown in the figure. If the net heat supplied to the gas in the cycle is 5J, the work done by the gas in the process C  A is [JEE(Scr) 2002] 3

V(m ) 2

C

B A

1 10 P(N/m2)

(A) – 5J (C) –15J

P



(B) –10J (D) –20J

18. Which of the following graphs correctly represents the variation of  = –(dV/dP)/V with P for an ideal gas at constant temperature ?



(C)

(D)

P P 19. A cubical box of side 1 meter contains helium gas (atomic weight 4) at a pressure of 100 N/m2. During an observation time of 1 second, an atom travelling with the root mean square speed parallel to one of the edges of the cube, was found to make 500 hits with a particular wall, without any collision with other atoms. Take R = 25/3 J/mol-K and k = 1.38 × 10–23 J/K. [JEE’ 2002] (a) Evaluate the temperature of the gas (b) Evaluate the average kinetic energy per atom (c) Evaluate the total mass of helium gas in the box. 20. In the figure AC represent Adiabatic process. The corresponding PV graph is [JEE (Scr) 2003] A P C

P

B T P

A

(A)

B

C P

A

(B)

V

B C V

P A

B

B (C)

A

(D)

C V V 21. An insulated container containing monoatomic gas of molar m is moving with a velocity v0. If the container is suddenly stopped, find the change in temperature. [JEE 2003] C

22. An ideal gas expands isothermally from a volume V1 to V2 and then compressed to original volume V1 adiabatically. Initial pressure is P1 and final pressure is P3. The total work done is W. Then (A) P3 > P1, W > 0 (B) P3 < P1, W < 0 (C) P3 > P1, W < 0 (D) P3 = P1, W = 0 [JEE’ 2004 (Scr)] 23. The piston cylinder arrangement shown contains a diatomic gas at temperature 300 K. The crosssectional area of the cylinder is 1 m2. Initially the

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HEAT– 2

Page # 49

height of the piston above the base of the cylinder is 1 m. The temperature is now raised to 400 K at constant pressure. Find the new height of the piston above the base of the cylinder. If the piston is now brought back to its original height without any heat loss, find the new equilibrium temperature of the gas. You can leave the answer is fraction. [JEE’ 2004]

Paragraph for Question Nos. 28 to 30 (3 questions) A fixed thermally conducting cylinder has radius R and length L0. The cylinder is open at its bottom and has a small hole at its top. A piston of mass M is held at a distance L from the top surface, as shown in the figure. The atmospheric pressure is P0. 2R

L

L0

1m

24. An ideal gas is filled in a closed rigid and thermally insulated container. A coil of 100  resistor carrying current 1A for 5 minutes supplies heat to the gas. The change in internal energy of the gas is (A) 10 KJ (B) 20 KJ (C) 30 KJ (D) 0 KJ [JEE’ 2004 (Scr)]

Piston

28. The piston is now pulled out slowly and held at a distnace 2L from the top. The pressure in the cylinder between its top and the piston will then be [JEE 2007] (A) P0

25. When the pressure is changed from p1 = 1.01 × 105 Pa to P2 = 1.165 × 105 Pa then the volume changes by 10% the bulk modulus is (A) 1.55 × 105 Pa (B) 0.0015 × 105 Pa 5 (C) 0.015 × 10 Pa (D) none of these [JEE’ 2004 (Scr)] 26. A cylinder of mass 1 kg is given heat of 20000 J at atmospheric pressure. If initially temperature of cylinder is 20°C, find (a) final temperature of the cylinder (b) work done by the cylinder (c) change in internal energy of the cylinder. (Given that specific heat of cylinder = 400 J kg–1°C–1, Coefficient of volume expansion = 9 × 10–5 °C–1, Atmospheric pressure = 105 N/m2 and density of cylinder = 9000 kg/m3 ) [JEE 2005]

27.

Match the following for the given process : P(atm)

J

30

M

20 10

K 10

(C)

(B) P0/2

P0 Mg  2 R2

(D)

P0 Mg – 2 R 2

29. While the piston is at a distance 2L from the top, the hole at the top is sealed. The piston is then released, to a position where it can stay in equilibrium. In this condition, the distance of the piston from the top is  2P0 R2   ( 2L) (A)  2  R P0  Mg 

 P0 R 2 – Mg   ( 2L) (B)  2  R P0 

 P0 R 2  Mg   ( 2L) (C)  2  R P0 

 P0 R 2   (2L ) (D)  2   R P0 – Mg 

[JEE 2007] 30. The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is brought below the cylinder and put in a position so that the water surface in the tank is at the same level as the top of the cylinder as shown in the figure. The density of the water is . In equilibrium, the height H of the water column in the cylinder satisfies [JEE 2007]

L 20 V(m3)

L0 H

Column I (A) Process J  K (B) Process K  L (C) Process L  M (D) Process M  J

Column II (P) w > 0 (Q) w < 0 (R) Q > 0 (S) Q < 0 [JEE 2006]

(A) g(L0 – H)2 + P0(L0 – H) + L0P0 = 0 (B) g(L0 – H)2 – P0(L0 – H) – L0P0 = 0 (C) g(L0 – H)2 + P0(L0 – H) – L0P0 = 0 (D) g(L0 – H)2 – P0(L0 – H) + L0P0 = 0

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Page # 50

HEAT– 2

31.STATEMENT - 1 The total translational kinetic energy of all the molecules of a given mass of an ideal gas is 1.5 times the product of its pressure and its volume because STATEMENT - 2 The molecules of a gas collide with each other and the velocities of the molecules change due to the collision. (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 [JEE 2007] 32. An ideal gas is expanding such that PT2 = constant. The coefficient of volume expansion of the gas is 1 2 3 4 (B) (C) (D) (A) T T T T [JEE 2008] 33.Column I Contains a list of processes involving expansion of an ideal gas. Match this with Column II describing the thermodynamic change during this process. Indicate your answer by darkening the appropriate bubbles of the 4×4 matrix given in the ORS. [JEE 2008] Column I Column II (A) An insulated container (p) The temperature of has two chambers separated the by gas decreases a valve. Chamber I contains an ideal gas the Chamber II has vacaum. The valve is opened.

ideal gas

(s) The gas gains heat

v

v1

2v1 v

34. Cv and Cp denote the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then [JEE 2009] (A) Cp – Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas (B) Cp + Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas (C) Cp / Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas (D) Cp.Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas 35. The figure shows the P–V plot of an ideal gas taken through a cycle ABCDA. The part ABCis a semicircle and CDA is half of an ellipse. Then, [JEE 2009] P

A

3 2

B

D 1 0

C 1

2

3

V

(A) the process during the path A  B is isothermal (B) heat flows out of the gas during the path B  C  D (C) work done during the path A  B  C is zero (D) positive work is done by the gas in the cycle ABCDA

opened

I

(D) An ideal monoatomic gas expands such that its pressure P and volume V follows the behaviour shown in the graph

II vacuum

36. This section contains 2 questions. Each questions (B) An ideal monoatomic (q) The temperature of the gas expands to twice its gas increase or remains original volume such that constant its pressure P  1 , where V2 V is the volume of the gas (C) An ideal monoatomic (r) The gas loses heat gas expands to twice its original volume such that

contains statements given in two columns, which have to be matched. The statements in Column I are labelled A, B, C and D, while the statements in Column II are labelled p, q, r, s and t. Any given statement in Column I can have correct matching with one or more statement(s) in Column II. The appropriate bubbles

1 ,

corresponding to the answers to these questions have

V where V is its volume

to be darkened as illustrated in the following example :

its pressure P 

4/3

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HEAT– 2

Page # 51

If the correct matches are A – p, s and t; B – q and r ; C – p and q; and D – s and t; then the correct darkening of bubbles will look like the following. Column II gives certain systems undergoing a process. Column I suggests changes in some of the parameters related to the system. Match the statements in Column I to the appropriate process(es) from Column II. [JEE 2009] Column I Column II (A) The energy of the (P) System: A capacitor system is increased Initially uncharged increased Process: It is connected to a battery (B) Mechanical energy (Q) System : A gas in an is provided to the system, adiabatic container which is converted into fitted with an adiabatic energy of random motion piston of its parts Process : The gas is compressed by pushing the piston (C) Internal energy of (R) System : a gas in a the system is converted rigid container into its mechanical energy. Process : The gas gets cooled due to colder atmosphere surrounding it (D) Mass of the system (S) System : A heavy is decreased nucleus initially at rest Process : The nucleus fissions into two fragments of nearly equal masses and some neutrons are emitted (T) System : A resistive wire loop Process : The loop is placed in a time varying magnetic field perpendicular to its plane. 37. A real gas behaves like an ideal gas if its (A) pressure and temperature are both high (B) pressure and temperature are both low (C) pressure is high and temperature is low (D) pressure is low and temperature is high [JEE 2010] 38. One mole of an ideal gas in initial sate A undergoes a cyclic process ABCA, as shown in the figure. Its pressure at A is P0. Choose the correct option (s) from the following. [JEE 2010] V

B

4v0

v0

C

A

(A) Internal energies at A and B are the same. (B) Work done by the gas in process AB is P0V0 n 4 (C) Pressure at C is P0/4 (D) Temperature at C is T0/4 39. A diatomic ideal gas is compressed adiabatically to 1/32 of its initial volume. In the initial temperature of the gas is T1 (in Kelvin) and the final temperature is aT1. the value of a is [JEE 2010] 40. 5.6 liter of helium gas at STP is adiabatically compressed to 0.7 liter. Taking the initial temperature to be T1, the work done in the process is (A) 9/8 RT1 (B) 3/2 RT1 (C) 15/8 RT1 (D)9/2 RT1 [JEE 2011] 41. One mole of a monatomic ideal gas is taken through a cycle ABCDA as shown in the P-V diagram. Column II gives the characteristics involved in the cycle. Match them with each of the processes given in Column I :

B 3P

A

1P

C

0 1V

Column I (A) Process A  B (B) Process B  C (C) Process C  D (D) Process D  A

D

3V

9V

V

Column II (P) Internal energy decreases (Q) Internal energy increases (R) Heat is lost (S) Heat is gained (T) Work is done on the gas [JEE 2011]

42. A mixture of 2 moles of helium gas (atomic mass = 4 amu) and 1 mole of argon gas (atomic mass = 40 amu) is kept at 300 K in a container. The ratio of the  vrms helium   is rms speeds  v   rms  arg on  (A) 0.32 (B) 0.45 (C) 2.24

[JEE 2012] (D) 3.16

43. Two moles of ideal helium gas are in a rubber balloon at 30°C. The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to 35°C. the amount of heat required in raising the temperature is nearly (take R = 8.31 J/mol. K) (A) 62 J (B) 104J (C) 124J (D) 208J [JEE 2012]

T T0

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Page # 52

HEAT– 2

:: ANSWER KEY :: (OBJECTIVE PROBLEMS)

ANSWER EX–I 1. A

2. C

3. B

4. A

5. C

6. C

7. B

8. A

9. B

10. C

11. A

12. D

13.D

14. C

15. A

16. D

17. D

18. B

19. D

20. C

21. C

22. A

23. D

24. D

25. A

26. C

27. A

28. B

29. A

30. C

31. B

32. B

33. A

34. C

35. A

36. C

37. C

38. A

39. D

40. B

41. D

42. A

43. C

44. A

45. A

46. A

47. B

48.D

49. D

50. D

51. D

52. B

53. D

54. B

55. B

56. B

57. A,C 58. D

59. A

60. A

61. B

62. C

63. C

64. B

65. A

66. D

67. A

68. C

69. A

70. C

71. A

72. B

73. A

74. A

75. D

76. B

77. D

78. C

79. C

80. C

81. A

82. A,D 83. D

84. B

85. D

86. A

87. D

88. A

(MULTIPLE CHOICE PROBLEMS)

ANSWER EX–II 1. D 9. B,D 17. AB

2. C 10. ACD 18. AB

3. CD 11. D 19. AD

4. B 12. BD 20. B

5. ABC 13. D 21. B

6. AB 14. D 22. CD

7. C 8. AD 15. A 16. ABD 23. CD 24. BC

(SUBJECTIVE PROBLEMS)

ANSWER EX–III 1. 28.7236 × 103 sec. 2.

20012 .428  10  25 kg  m / s 3. 1 : 2 4. 

14 5. 11 RT 6. 196ºC

3 0 2

7. PROOF

8. 3.3 × 103

13. 1.25 × 104 N/m2 15. 74.9 cm

14. (i) P1 < P2, T1 < T2 ; (ii) T1 = T2 < T3 ; (iii) V2 > V1 ; (iv) P2 > P1 16. 120 R 17. 1500 J 18. 0.0091 J 19. PV 20. –100J

21. 3P1 V1

27.

88 J / cal 21

33. – 

39.

R 2

22. /2 atm-lt

9.

2T

23. Q – W

10. 1 : 2

11. 27 : 4

24. 24 J

30. (i) 765 J (ii)

12.

25. (33600 + 0.02) J 208 1921

28. 100 J

29. 3600 R

31. 3R

34. 

35. 3R, 2R, 1.5

36. 47/29

40. 12600 J

41. 50 calorie

 3 – 2  42. RT   – 1   

37. 1.5

26.

25 J / cal 6

32. PROOF

38. CV +

R V

43. the molar mass of the gas is 40 gm, the numbe of degrees of freedom of the gas molecules is 6 44. 8 atmosphere

1   3  1 – / 3   2  45. 1 – l n2

46. 1.5

47. 5R

48. P/n

49. 300 K

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HEAT– 2

Page # 53

(TOUGH SUBJECTIVE PROBLEMS)

ANSWER EX–IV 1. 750 K

2. 31P0V0 ; – 5P0V0 3. (ii) Pb = Pc = 2 atm, (iii) Tb = 300 K, TC = 600 K, (iv) VC = 8 litre

4. 1.6 m, 364 K

5. 1.63

6.

Isochoric

P

3 – 2ln2 3

7. 8000 cal.

Adlabatic

Isotherm

V

8. 2 mnv2cos2 9. (a) W AD = 88 J, (b) VC = 1.223 litre, (c) W CDA = –85J 27 21 P0 P0 = Final pressure in C, Final pressure in B = 10. (a) Final pressure in A = 8 4 21 3 T0 , Final temperature in C = T0 , (b) Final temperature in A (and B) = 4 2 (c) 18 P0V0, (d) work done by gas in A = + P0V0, work done by gas in B = 0 P 50  0U0 C 3M

(e)

17 P0 V0 2

11. four

12. (a)

20  0U0 3M

A

B

M 20

M 5 0

13.

7R , 4R (T2 – T1) 2M

14. (a) 80 J, 180 J, (b) 4.5 R

16. (a) 2 × 105 N/m2 ; (b) 6 rad/s, (c) 75 J

V

 10  , (b) Q =  ln2.5 – 2 U0 , (c) – 2U0 3

15. T1 = (207/16)T0 ; T2 =

9 15 T0 , – PV 4 8 0 0

17 C = 2R

(JEE PROBLEMS)

ANSWER EX–V 1. 9. 14. 15.

B 2. D 3. C 4. C 5. D 6. C 7. B 10. B,C 11. C 12. D 13. 400 J, 2 T0 (a) 1200 R, (b) QAB = – 2100 R, QBC = 1500 R, QCA = 1200 R ln2 B 16. A 17. A 18. A 19. 160 K, 3.3 × 10–21 J, 0.3 gm

21.

mv 0 2 T = 3R

26. 28. 33. 34.

(a) Tfinal = 70°C, (b) 0.05 J, (c) 19999.95 J 27.(A)  S; (B)  P and R ; (C)  (R); (D)  Q and S A 29. D 30. C 31. B 32. C (A)  q, (B)  p & r, (C)  p & s, (D)  q & s B,D 35. B,D 36. (A)  (PQST), (B)  (Q), (C)  (S), (D)  (S) 37. D 38. A,B

39.

4

40.

22.

A

41.

C

23.

 4 T3 = 400   3

A

20.

8. C,D

A

0.4

K

24.

C

(A)  p,t,r ; (B)  p,r ; (C)  q, s ; (D)  r,t

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25.

42.

A

D

43. D

ELECTROST ELECTROSTA OSTATICS - 1 THEORY AND EXERCISE BOOKLET CONTENTS S.NO.

TOPIC

PAGE NO.

1. Introduction ...................................................................................................... 3 – 4 2. Coloumb's Law ............................................................................................... 5 – 14 3. Electric field ................................................................................................... 14 – 29 4. Conservative force .......................................................................................... 29 – 32 5. Electrostatic Potential Energy ....................................................................... 32 – 36 6. Electric Potential ........................................................................................... 36 – 40 7. Relation Between E and V ............................................................................. 40 – 41 8. Electric lines of force ..................................................................................... 41 – 43 9. Equipotential Surface ..................................................................................... 43 – 45 10. Electric dipole .............................................................................................. 45 – 52 11. Exercise - I................................................................................................... 53 – 67 12. Exercise - II ................................................................................................. 68 – 73 13. Exercise - III ................................................................................................. 74 – 82 14. Exercise - IV ................................................................................................ 83 – 84 15. Exercise - V ................................................................................................. 85 – 89 16. Answer key .................................................................................................. 90 – 92

1

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ELECTROSTATICS - 1

Page # 2

JEE SYLLABUS : Coulomb’s law; Electric field and potential; Electrical Potential energy of a system of point charges and of electrical dipoles in a uniform electrostatic field, Electric field lines;

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ELECTROSTATICS - 1

Page # 3

1.

INTRODUCTION :

(a)

Introduction : Electromagnetism is a science of the combinatin of electrical and magnetic phenomenon. Electromagnetism can be divided into 2 parts : (1) Electrostatics : It deals with the study of charges at rest. (2) Electrodynamics : It deals with the study of charges in motion (discusses magnetic phenomenon). In this chapter we will be dealing with charges at rest i.e. electrostatics.

(b)

Structure of Atom : An atom consists of two parts (i) nucleus (ii) extra nuclear part. Nucleus consists of neutrons and protons and extra nuclear part has electrons revolving around nucleus. In a neutral atom. number of electrons = number of protons. charge of electrons = charge of protons = 1.602 × 10–19 coulomb. Normally positive charges are positron, proton and positive ions. In nature practically free existing positive charge are positive ions and negative charges are electrons.

(c)

Electric Charge Charge of a material body or particle is the property (acquired or natural) due to which it produces and experiences electrical and magnetic effects. Some of naturally charged particles are electron, proton, α-particle etc.

(d)

Types of Charge (i) Positive charge : It is the deficiency of electrons compared to protons. (ii) Negative charge : It is the excess of electrons compared to protons.

(e)

Units of Charge Charge is a derived physical quantity. Charge is measured in coulomb in S.I. unit. In practice we use mC (10–3C), µC (10 –6 C), nC (10–9C) etc. C.G.S unit of charge = electrostatic unit = esu. 1 coulomb = 3 × 109 esu of charge Dimensional formula of charge = [M°L°T1I1]

(f)

Properties of Charge (I)

Charge is a scalar quantity : It adds algebrically and represents excess, or deficiency of electrons.

(II)

Charge is transferable : Charging a body implies transfer of charge (electrons) from one body to another. Positively charged body means loss of electrons, i.e. deficiency of electrons. Negatively charged body means excess of electrons. This also shows that mass of a negatively charged body > mass of a positively charged identical body.

(III)

Charge is conserved : In an isolated system, total charge (sum of positive and negative) remains constant whatever change takes place in that system.

(IV)

Charge is quantized : Charge on any body always exists in integral multiples of a fundamental unit of electric charge. This unit is equal to the magnitude of charge on electron (1e = 1.6 × 10–19 coulomb). So charge on anybody Q = ± ne, where n is an integer and e is the charge of the electron. Millikan's oil drop experiment proved the quantization of charge or atomicity of charge.

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ELECTROSTATICS - 1

Page # 4

•

1 2 e and ± e has been postulated. These particles 3 3 are called quarks but still this is not considered as the quantum of charge because these are unstable (They have very short span of life.)

Recently, the existence of particles of charge ±

(v)

Like point charges repel each other while unlike point charges attract each other.

(vi)

Charge is always associated with mass, i.e., charge can not exist without mass though mass can exist without charge. The particle such as photon or neutrino which have no (rest) mass can never have a charge.

(vii)

Charge is relativistically invariant : This means that charge is independent of frame of reference, i.e., charge on a body does not change whatever be its speed. This property is worth mentioning as in contrast to charge, the mass of a body depends on its speed and increases with increase in speed.

(viii) A charge at rest produces only electric field around itself; a charge having uniform motion produces electric as well as magnetic field around itself while a charge having accelerated motion emits electromagnetic radiation. (g)

Conductors and Insultators : Any object can be broadly classified in either of the following two categories : (i) Conductors (ii) Insulators

(i)

Conductors : These are the materials that allow flow of charge through them. This category generally comprises of metals but may sometimes contain non-metals too. (ex. Carbon in form of graphite.)

(ii)

Insulators : These are the materials which do not allow movement of charge through them.

(h)

Charging of Bodies : An object can be charged by addition or removed of electrons from it. In general an object can either be a conductor or insulator. Thus we are going to discuss the charging of a conductor and charging of an insultor in brife.

(i)

Charging of Conductors : Conductors can be charged by (a) Rubbing or frictional electricity (b) Conduction & Induction (will be studied in later sections) (c) Thermionic emission (will be study the topic "heat") (d) Photo electric emission (will be studied under the topic modern physics)

(ii)

Charging of Insulators : Since charge cannot flow through insulators, neither conduction nor induction can be used to charge, insultators, so in order to charge an insulator friction is used. Whenever an insulator is rubbed against a body exchange of electrons takes place between the two. This results in apperance of equal and opposite charges on the insulator and the other body. Thus the insulator is charged. For example rubbing of plastic with fur, silk with glass causes charging of these things. To charge the bodies through friction one of them has to be an insultator.

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ELECTROSTATICS - 1 2.

Page # 5

COULOMB'S LAW : Coulomb, through his experiments found out that the two charges 'q1' and 'q2' kept at distance 'r' in a medium as shown in figure-1 exert a force 'F' on each other. The value of force F is given by F=

Kq1q2 r2

F

r

F

+q1

+q2

This law gives the net force experienced by q1 and q2 taking in account the medium surrounding them. Where F gives the magnitude of electrostatic force. q1 and q2 are the magnitudes of the two interacting charges. K is electrostatic constant which depends upon the medium surrounding the two charges. This force F acts along the line joining the two charges and is repulsive if q1 and q2 are of same sign and is attractive if they are of opposite sign. Let us take some examples on application of coulomb's Law. Ex.1

FAC

Charge 5.0 × 10–7 C, –2.5 × 10–7C and 1.0 × 10–7 C are fixed at the corners A, B and C of an equilateral triangle of side 5.0 cm. Find the electric force on the charge at C due to the rest two. FAC =

9 × 109 × 5 × 10–7 × 1 × 10–7

(.05)2 9

FBC =

–7

9 × 10 × –2.5 × 10

FBC

= 0.18 Nt

× 1 × 10

(.05)

a = 5 cm

a

–7

2

Net force on C is

q3 = 1.0 × 10–7C

C

= – .09 Nt

A q1 = 5 × 10–7C

→ → → F Net = F AC + F BC

→ F Net = (FAC )2 + (F BC )2 + 2(FAC )(F BC ) cos θ

a

B q2 = –2.5 × 10–7C

[θ = 120°]

= 0.15588 Nt

Ex.2

If charge q1 is fixed and q2 is free to move then find out the velocity of q2 when it reaches distance r2 after it is release from a distance of r1 from q1 as shown in figure (Assume friction is absent).

q2

q1 fixed

r1 r2

Find v of q2 when it reaches distance r2 after it is released from rest.

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ELECTROSTATICS - 1

Page # 6

Sol.

a=

kq1q2 mx2

v

∫ 0

Ex.3

vdv =

kq1q2 m

r2

dx

∫x

2

r1

kq1q2 v2 ⇒ 2 = m

1 1   −   r1 r2 

Ten charged particles are kept fixed on the X axis at point x = 10 mm, 20 mm, 30 mm, .......... 100 mm. The first particle has a charge 10–8 C, the second 8 × 10–8 C, the third 27 × 10–8C and so on. The tenth particle has a charge 1000 × 10–8C. Find the magnitude of electric force acting on a 1 C charge placed at the origin.

Force of 1C charge =

=

q2

q1

1C

Sol.

1 1  2kq1q2  −  ⇒ v= r r m 2 1

Kq1 × 1 –3 2

(10 × 10 )

+

q3

Kq2 × 1 –3 2

(20 × 10 )

q4

+

Kq3 × 1 (30 × 10–3 )2

+ .......

K × 10–8  13 23 33 103   2 + 2 + 2 + ...... 2  = 9 × 109 × 10–4 × 55 = 4.95 × 107 Nt –4 10 2 3 10   1

[This example explains that the concept of superposition holds in the case of electric forces. Net electric force at the origin is equal to sum of the individual electric forces on the 1 C charge]

Ex.4

Sol

A block 'A' of charge q1 is fixed and second block of mass m and charge q2 is allowed to free on the floor

A,q1

findout the range of q2 for which the particle is at rest.

fixed µ

Maximum friction = µ mg

r

µmgr2 kq q µmg = 12 2 ⇒ q2 = kq1 r –

2.1

m,q2

µmgr2 µmgr2 > r) Here we have to find the electric field at point p due to the given infinite wire. Using the formula learnt in above section which E

l e

c

t r i c

f i e

l d

d

u

e

t o

E|| =

kλ ( cos θ2 − cos θ1 ) r

E⊥ =

kλ ( sin θ2 + sin θ1 ) r

∴

f i n

i t e

w

i r e

 r

θ1 = θ2 =

For above case,

3.4

i n

π 2

P

Wire

kλ 2kλ (1 + 1) = r r

Enet at P =

Electric field due to semi infinite wire For this case

θ1 =

π , θ2 = 0º 2

∴

Er =

kλ kλ ; E| | = r r

r

2 kλ s r

Enet at P =

P

Wire

E11

E1 Enet

Ex.27 Consider the system shown below If the charge is slightly displaced perpendicular to the wire from its equilibrium position then find out the time period of SHM. P q,m d

λ

Sol

: At equilbrium position weight of the particle is balanced by the electric force

⇒

mg = qE

2 kλ ....(1) d Now if the particle is slightly displaced by a distance xλ (where x 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 σ2. Find the work done by the electric field on a point charge Q that moves from S1 towards S2 along a line of length a (a < d) making an angle π/4 with the normal to the sheets. Assume that the charge Q does not affect the charge distributions of the sheets. [JEE 2004] Sol.

9. A charge +Q is fixed at the origin of the co-ordinate system while a small electric dipole of dipole-moment
p pointing away from the charge along the x-axis is set free from a point far away from the origin. (a) calculate the K.E. of the dipole when it reaches to a point (d, 0) [JEE 2003] (b) calculate the force on the charge +Q at this moment. Sol.

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ELECTROSTATICS - 1

Page # 88 12. Which of the following groups do not have same dimensions (A) Young’s modulus, pressure stress (B) work, heat, energy (C) electromotive force, potential difference, voltage (D) electric dipole, electric flux, electric field Sol.

(A) The electric field at point O is

q 8 πε 0R 2

directed

along the negative x-axis (B) The potential energy of the system is zero (C) The magnitude of the force between the charges at C and B is

q2 54 πε 0R2

q (D) the potential at point O is 12 π ε R 0

Sol.

13. Positive and negative point charges of equal a a   ma gni tude ar e ke pt at  0, 0,  a nd  0, 0,–  ,   2 2 respectively. The work done by the electric field when another positive point charge is moved from (–a, 0, 0) to (0, a, 0) is -

(A) positive (B) negative (C) zero (D) depends on the path connecting the initial and final positions [JEE 2007] Sol.

15. 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

Q1

(A) |Q1| > |Q2|

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. [JEE 2010] q q 2q 14. Consider a system of three charges , and – 3 3 3 placed at points A, B and C, respectively, as shown in the figure. Take O to be the centre of the circle of [JEE 2008] radius R and angle CAB = 60° Figure :

Sol.

B C 60°

O

x

A

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ELECTROSTATICS - 1

16. Under the influence of the Coulomb field of charge +Q, a charge –q is moving around it in an elliptical orbit. Find out the correct statement(s) (A) The angular momentum of the charge –q is constant (B) The linear momentum of the charge –q is constant (C) The angular velocity of the charge –q is constant (D) The linear speed of the charge –q is constant [JEE 2010] Sol.

17. A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric 81π × 10 5 Vm −1 . When the field is 7 switched off, the drop is observed to fall with terminal velocity 2 × 10 –3 ms–1 Given g = 9.8 ms –2 , viscosity of the air = 1.8 × 10–5 Ns m –2 and the density of oil = 900 kg m –5 , the magnitude of q is : (A) 1.6 × 10–19C (B) 3.2 × 10–19C (C) 4.8 × 10–19C (D) 8.0 × 10–19C [JEE 2010] Sol.

field of strength

Page # 89 18. Four point charges, each of +q are rigidly fixed at the four corners of a square planar soap film of side 'a'. The surface tension of the soap film is γ . the system of charges and planar film are in equilibrium,  q2  and a = k    γ 

1/N

, where 'k' is a constant. Then N is [JEE 2011]

Sol.

19. A wooden block performs SHM on a frictionless surface with frequency, v0. The block carries a charge +Q on its surface. If now a uniform electric field E is switched-on as shown, then the SHM of the block will be E +Q

(A) of the same frequency and with shifted mean position (B) of the same frequency and with the same mean position (C) of changed frequency and with shifted mean position (D) of changed frequency and with the same mean position [JEE 2011] Sol.

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ELECTROSTATICS - 1

Page # 90

EXERCISE - I

ANSWER KEY 1.

D

2.

C

3.

D

4.

B

5.

D

6.

A

7.

D

8.

B

9.

B

10.

D

11.

A

12.

A

13.

A

14.

D

15.

A

16.

B

17.

B

18.

B

19.

B

20.

B

21.

D

22.

C

23.

D

24.

B

25.

B

26.

B

27.

B

28.

B

29.

D

30.

C

31.

B

32.

B

33.

B

34.

A

35.

C

36.

B

37.

A

38.

D

39.

B

40.

B

41.

B

42.

C

43.

B

44.

D

45.

C

46.

A

47.

A

48.

D

49.

B

50.

C

51.

B

52.

A

53.

A

54.

D

55.

B

56.

B

57.

A

58.

A

59.

C

60.

D

61.

C

62.

C

EXERCISE - II

ANSWER KEY

1.

C,D

2.

B,D

3.

B

4.

C,D

A,D

6.

B

7.

A,C

8.

A

9.

A,B,D

10.

A,C

11.

A,B,C 12.

A

13.

B,D

14.

B

15.

A,D

16.

B,C

17.

A,C

18.

A

D

20.

C

21.

B

22.

D

23.

A,B,D

24.

A,D

25.

B,C,D 26.

5.

19.

B,C

EXERCISE - III

ANSWER KEY

1.

2
from charge 4e (If q is positive stable, If q is negative unstable) 3

2.

d 4 Qq , , 2 2 3 3 πε0d2

4.

a)

m π 3 ε 0 d3 Qq

(b)

3. a = l(1 +

2 ), the equilibrium will be stable

m π 3 ε 0 d3 2 Qq

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ELECTROSTATICS - 1

Page # 91

E

E

5.

(i)

6.

3 –    11

10.

3σλ 2m ∈0

11.(i)

12.

(a) (i) 0

(ii) 0

13.

1 1 W = Kqq0  r – r  = 1.2 J  B A

16.

(i)

a

17.

(ii)

3 × 10–9C

7.

a

(iii)

b r

4Kqx  a2   + x2   2 

 3 1  + 3 +  2 3 

Kq2  3 1  +  3 + 2ma  2 3

 σ q0  2 tan −1   2ε 0mg 

0

3/2

(iii)

4 2Kq a

2

,

(b) (i)

Qq 14. (a) 4 πε K 0

(ii) Wext = –

(iv)

28.

|E| =

(– i ) (ii)

KP0 Q  j r3

x

2

K m

a

9.

b r

8 π 2 ε 0r 2

2Kqa 2

a   + x2   2  (ii) 0

3/2

(iii)

2Kqa x3

15. –9.0 × 10–3 J

 3 1  2Kq2 + 3 +  , Wel = a 2 3 

(v)

 3 1  + 3 +  2 3 

3 1  Kq2  + 3 +  ma  2 3

19. 20 ln 2

Q2

K.E. =

r3

4Kq

2kQ 2 mR

18.

24.

2KP0 Q

(iv)

qQ

0

2Kq2  3 1  + 3 +  ma  2 3

23. 1.8 × 105 sec

2x3 + y3 + C 3

(b)

2Kq2 a

mπε 0 V

27. (i)

b r

, along the axis, (ii) 0 (iii)

21.

–

a

8.

kq2 (3 – 2 ) 20. – a

25.

E

3/2

4Kq2 a

(iii)

b r

E

2

2ε 0 u 2 m qσ

22.

1 Q2 2 4 πε 0 d

2qp

26.

4 πε 0 r 2

(iii)

6KP0P  3KP0P ˆ i (iv) (+ j) r4 r4

K 7K ,V= [where K = 1/4 π∈0] 4 8

kP

29.

2y

3

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(– i – 2 j )

ELECTROSTATICS - 1

Page # 92

EXERCISE - IV

ANSWER KEY

Q.1

(a) 60°(b) mg + (c)

kq1q2
2

2 3 mg, mg. q1 & q2 should have unlike charges for the beads to remain stationaly & q2q2 = – mgl /k

λq 2ε 0 m


 Q.4 H2 = h1 + h2 – g    V

2

Q.2

9.30

Q.5

8 4  Kq2 W first step =  – , W second step = 0, W total = 0  3 5 r

Q.6

Q.8

Q 1 R1 = Q 2 R2

Q.11

Q.3

Q.9

λ R E0 i

Q.10 σ =

ε r ρ0 εr – 1

Q.7

3

–

4kq  i πR2

Q.12

1.

D

4.

(a) E, (b) B, (c) v0 = 3 m/s ; K.E. at the origin = (27 – 10 6 ) × 10–4 J approx. 2.5 × 10–4 J

5.

C

8.

–

10.

D

11.

16.

A

17.

2.

A,D

3. (i) A, C, (ii) D, (iii) A, C

6. 5.86 m/s

7. B

1 q2 4 . 3 3 –3 6 – 2 4 πε 0 a 6

[

(σ1 − σ2 ) Qa 2 2 ε0 D

4 πε 0Ka

EXERCISE - V

ANSWER KEY

A

v

]

QP P Q 9. (a) K.E. = 4 πε 2 , (b) along positive x-axis 2πε 0 d3 0 d

12.

D

18.

0003

13.

C

19.

A

14.

C

15.

A,D

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ELECTROST ELECTROSTA OSTATICS - 2 THEORY AND EXERCISE BOOKLET CONTENTS S.NO.

TOPIC

PAGE NO.

1. Electric Flux .................................................................................................... 3 – 6 2. Concept of solid angle ..................................................................................... 7 – 8 3. Gauss’s Law ................................................................................................... 9 – 14 4. Conductor ...................................................................................................... 15 – 24 5. Charge Induction in Metal Cavity .................................................................... 25 – 36 6. Earthing ......................................................................................................... 37 – 38 7. Conducting Plates ......................................................................................... 39 – 43 8. Electrostatics Energy of a System ................................................................ 44 – 45 9. Exercise - I .................................................................................................... 46 – 56 10. Exercise - II.................................................................................................. 57 – 59 11. Exercise - III ................................................................................................. 60 – 63 12. Exercise - IV .................................................................................................... 64 13. Exercise - V ................................................................................................. 65 – 71 14. Answer key .................................................................................................. 71 – 72

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ELECTROSTATICS - 2

Page # 2

JEE SYLLABUS : Flux of electric field; Gauss’s law and its application in simple cases, such as, to find field due to infinitely long straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell.

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ELECTROSTATICS - 2 1.

Page # 3

ELECTRIC FLUX : Any group of electric lines of forces passing through a given surface, we call electric flux and it is denoted by φ.

•

Area as a Vector : Till now we have considered area of a surface as a scalar quantity but for further analysis we treat area
of a surface as a vector quantity whose direction is along the normal to the surface. The area vector S of a surface which has surface area S can be written as
S = Sˆ n Where nˆ is the unit vector in the direction along normal to the surface. →  S n

If a surface is three dimensional we consider a small elemental area dS on this surface and direction of this elemental area vector is along the local normal of the surface at the point where elemental area is chosen as shown. Thus →

dS = dS ˆ a

Here aˆ is the unit vector in the direction along the normal at elemental area dS. dS

•

a

Electric Field Strength in Terms of Electric Flux : Earlier we’ve defined that the density of electric lines gives the magnitude of electric field strength. Mathematically the numerical value of electric field strength at a point in the region of electric field can be give as the electric flux passing through a unit normal area at that point.

→

Flux = φ = ∫ E.dA , If E is constant, φ = E. A

E

In a uniform electric field shown in figure. If φ be the flux passing through an area S which is normal to the electric field lines, the value of electric field strength at this surface can be given as E=

φ S

area = S

or flux through the surface can be given as

A

φ = ES If in an electric field, surface is not normal as shown in figure. Here the are ABCD is inclined at an angle θ from the normal to electric field. Here we resolve the area ABCD in two perpendicular components as shown in figure. One is S cos θ, which is area ABC′D′ normal to electric field direction and other is S sin θ, which is area CDC′D′ along the direction of electric field.

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B θ

C' C

D area = S

D'

ELECTROSTATICS - 2

Page # 4 Here the total flux passing through the given area ABCD is same which is passing through its normal component S cos θ, thus here the flux φ through the area can be given as φ = ES cos θ

A

[S cos θ = area of ABC′D′] θ

If we consider the direction of area vector normal to the area
surface, as shown in figure, θ would be the angle between S and
E . Thus flux through the surface ABCD can be given as

φ = E.S (a)

B

E θ → S

D C

Electric Flux in Non-uniform Electric Field : In non-uniform electric field, we can calculate electric flux through a given surface by integrating the above expression for elemental surface area of the given surface. For this consider the situation shown in figure. If we wish to calculate electric flux through the surface M shown in figure. For this we consider an elemental area dS on the surface M as shown. At this position if electric field is E then the electric flux through this elemental area dS can be given as

E

θ

→ dS

dφ = EdS cos θ

Total flux through the surface M can be given as φ = ∫ dφ =

∫ EdS cos θ M

(b)

M

Electric Flux Through a circular Disc :

φ



Figure shows a point charge q placed at a distance  from a disc of radius R. Here we wish to find the electric flux through the disc surface due to the point charge q. We know a point charge q originates electric flux in radially outward direction. The flux of q which is originated in cone shown in figure passes through the disc surface.

R

θ

E

→ dS

x

q

θ



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ELECTROSTATICS - 2

Page # 5

To calculate this flux, we consider an elemental ring on disc surface of radius x and width dx as shown. Area of this ring (strip) is dS = 2πx dx The electric field due to q at this elemental ring is given as Kq E= 2 (x + 2 ) If dφ is the flux passing through this elemental ring, we have dφ = EdS cos θ Kq  2πkql xdx × 2πx dx × 2 2 2 = (x2 + 2 ) (l + x2 )3 / 2 x + Total flux through the disc surface can be given by integrating this expression over the whole area of disc thus total flux can be given as =

φ = ∫ dφ =

R

q

∫2∈

0

O

q = 2 ∈0

x dx (2 + x2 )3 / 2 R

 q  1 q  1 x dx ∫O (2 + x2 )3 / 2 = 2 ∈0  − 2 + x2  = 2 ∈0   −  O

R

  2 + R 2  1

The above result can be obtained in a much simpler way by using the concept of solid angle and Gauss’s Law, shortly we’ll discuss it.

(c)

Electric Flux Through the Lateral Surface of a Cylinder due to a Point Charge : Figure. shows a cylindrical surface of length L and radius R. On its axis at its centre a point charge q is placed. Here we wish to find the flux coming out from the lateral surface of this cylinder due to the point charge q. For this we consider an elemental strip of width dx on the surface of cylinder as shown. The area of this strip is dS = 2πR . dx The electric field due to the point charge on the strip can be given as E=

→ dS

Kq (x2 + R 2 )

θ

If dφ is the electric flux through the strip, we can write

q

dφ = E dS cos θ =

Kq R × 2π Rdx × 2 2 (x + R ) x + R2 2

= 2π KqR 2 ×

x

→ E

θ

dx

C R

L

dx (x2 + R 2 )3 / 2

Total flux through the lateral surface of cylinder can be given by integrating the above result for the complete lateral surface, which can be given as

φ = ∫ dφ =

or

φ=

q . ∈0

qR 2 2 ∈0

+L / 2

dx 2 3/2 −L / 2 (x + R )

∫

2

 2 + 4R 2

The solution of above integration is left for students as exercise. This situation can also be easily handled by using the concepts of Gauss’s Law, we’ll discuss in next section.

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ELECTROSTATICS - 2

Page # 6 (d)

Electric flux Produced by a Point Charge : The figure. shows a point charge placed at the centre of a spherical surface of radius R from which electric lines are originated and coming out of the surface of sphere. For clarity and convenience only lower half of sphere is drawn in the picture. As the charge q is inside the sphere, whatever flux it originates will come out from the spherical surface. To find the total flux, we consider an elemental area dS on surface. The electric field on the points on surface of sphere can be given as

→ → E dS

Kq E= 2 R

The electric flux coming out from the surface dS is
→ [As θ = 0 shown in figure] dφ = E.dS = EdS Kq dS R2 Total flux coming out from the spherical surface is Kq φ = dφ = dS R2 At every point of spherical surface, magnitude of electric field remains same hence we have

Thus

dφ =

∫

or

∫

φ=

Kq dS R2 ∫

φ=

Kq × 4πR 2 R2

φ=

q ∈0

[As

∫ dS = 4πR

2

]

q Thus total flux, the charge q originates is ∈ . Similarly a 0 q charge –q absorbs ∈ electric lines (flux) into it. 0

Figure shows a charge q enclosed in a closed surface S of random shape. Here we can say that the total electric flux emerging out from the surface S is the complete flux which charge q is originating, hence flux emerging from surface is φS =

(e)

=

S

q ∈0

The above result is independent of the shape of surface it only depends on the amount of charge enclosed by the surface. Flux Calculation in the Region of Varying Electric field : In a region electric field depends on x dierection as x2 0 In the cube of edge a shown in figure from front face electric flux goes in which can be given as φin = E0 (2a)2 . a2 = 4E0a4 From the other surface flux coming out can be given as φout = E0(3a)2 . a2 = 9 E0a4 E

q

y

E = E0x2

E

φin

a

2a

φ out x

O

a a

Here φout > φin for the cubical surface hence net flux =φout – φin = 5E0a4

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ELECTROSTATICS - 2 2.

Page # 7

CONCEPT OF SOLID ANGLE : Solid angle is the three dimensional angle enclosed by the lateral surface of a cone at its vertex as shown in figure shown. Solid angle can also be defined as the three dimensional angle subtended by a spherical section at its centre of curvature. As in the figure shown point a is the centre of curvature of a ω (omega) at point A. s p

(a)

h

e

r i c a

l

s e

c t i o

n

S

o

f

r a

d

i u

s

R

w

h

i c h

s u

b

t e

n

d

a

s o

l i d

a

n

g

S

Solid angle Ω

R

l e

A

Relation in Half Angle of cone and Solid Angle at Vertex :

Rsin θ

Consider a spherical section M of radius R, which subtend a half angle φ (radian) at the centre of curvature. To find the area of this section, we consider an elemental strop on this section of radius R sin θ and angular width dθ as shown in figure. The surface area of this strip can be given as dS = 2πR sin θ × Rdθ The total area of spherical section can be given by integrating the area of this elemental strip within limits from O to φ.

∫ 2πR

2

θ

R

φ

Ω

φ

∫ dS =

M

dθ

Total area of spherical section is S=

R dθ

O

sin θdθ

0 φ

= 2πR 2 – cos θ  0 = 2πR2 (1 – cos φ) If solid angle subtended by this section at its centre O is Ω then its area can be given as S = ΩR2 From equation (1) we have ΩR2 = 2πR2 (1 – cosφ) Ω = 2π (1 – cosφ) Equation (2) gives the relation in half angle of a cone φ and the solid angle enclosed by the lateral surface of cone at its vertex.

(b)

Electric Flux Calculation due to a Point Charge Using solid Angle : Figure shows a point charge q placed at a distance  from the centre of a circular disc of radius R. Now we wish to find the electric flux passing through the disc surface due to the charge q.  2 + R2 R

φ

q



Ω

We know from a point charge q, total flux originated is point charge q,

q in all directions or we can say that from a ∈0

q flux is originated is 4π solid angle. ∈0

Here the solid angle enclosed by cone subtended by disc at the point charge can be given as  Ω = 2π(1 – cos φ) = 2π  1 – 

   +R 



2

2

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ELECTROSTATICS - 2

Page # 8

Now we can easily calculate the flux of q which as passing through the disc surface as q / ∈0 ×Ω 4π

φdisc =

or

Ex.1 Sol.

q φdisc = 2 ∈ 0

  1 – 

=

 q / ∈0 × 2π  1 –  4π 

   +R 



2

2

   +R 



2

2

Find the electric flux coming out from one face of a cube of edge a, centre of which a point charge q is placed. Here the total solid angle subtended by cube surface at the point charge q is 4π. As q is at centre of cube, we can say the each face of cube subtend equal solid angle at the centre, thus solid angle subtended by each force at point charge is Ω face =

4π steradian 6

Thus electric flux through each face is

q

q / ∈0 × Ω face φface = 4π

=

Ex.2

q q / ∈0 4π × = 6∈ 4π 6 0

A point light source of 100 W is placed at a distance x from the centre of a hole of radius R in a sheet as shown in figure. Find the power passing through the hole in sheet.

R x 100°W bulb Sol. From figure, the solid angle of cone shown in figure can be given as  x Ω = 2π (1– cos θ) = 2π  1 – 2 R + x2 

  

Ω

θ

Power in hole = power given in solid angle Ω P=

100 ×Ω 4π

 100 x = 4π × 2π  1 – 2 R + x2 

 x = 50  1 – 2 R + x2 

  

  watt 

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ELECTROSTATICS - 2 3.

Page # 9

GAUSS’S LAW : This law is the mathematical analysis of the relation between the electric flux from a closed surface and its enclosed charge. This law states that the total flux emerging out from a closed surface is equal to the product of sum of

1 enclosed charge by the surface and the constant ∈ 0 Mathematically Gauss’s law is written as

→ →

∑ qencl ∈0

∫ E .dS = M

Here the sign

∫

represents the integration over a closed surface M which encloses a total charge

∑ qencl Let us consider a surface M shown in figure which encloses three charges q1 – q2 and q5. For the surface → → M if we find surface integral of electric field  ∫ E .dS , it gives the total electric flux coming out from the M

surface, which can be given as

→ →

∫ E .dS = M

q1 + q5 – q3 ∈0

[Gauss’s Law]

q4

q2 q1 q5

–q3 q7 M

–q6

Here electric field → E is the net electric field at the points on the surface of M. Remember that the electric field we use to find the flux must be the net electric field of the system due to all the charges but the total flux coming out from the surface is the flux originated by the charges enclosed in the closed surface. Using Gauss law we can find electric field strength due to some symmetrical distribution of charges. For appllication of Gauss’s Law, we choose a closed surface over which we apply Gauss law, called Gaussian surface. Gauss Law can be used to calculate electric field strength, for this we first choose a proper Gaussian surface on which the electric field strength is to be calculated. Some times a random Gaussion surface is chosen then the integral

→ →

∫ E .dS

involves complex calcula-

tions. To make these calculations easier, we choose a Gaussian surface keeping following points in mind. (i) The Gaussion surface should be chosen in such a way that at every point of surface the magnitude of electric field is either uniform or zero. (ii) The surface should be chosen in such a way that at every point of surface electric field strength is either parallel or perpendicular to the surface.

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ELECTROSTATICS - 2

Page # 10

Following example will illustrate the applications of Gauss’s Law in calculation of electric field in the surrounding of some charge configurations. Gauss Law is a very helpful tool in finding the electric field strength due to various distribution of charges. We start with a very simple example. Now we try to find the electric field strength due to a point charge q at a distance x, using Gauss’s law.

P

x

q

To find electric field strength at P, we first consider a Gaussian surface so that point P will be on its surface. But the question is what should be the shape of Gaussion surface. Look at the following figure shown.

dS

E

P q

q

x

x

p

q

x

If we apply Gauss’s Law to the above two cases, it will require laborious calculations to find

→ →

∫ E .dS .

The Gaussion surface should be chosen in such a way to minimize the calculations. Now consider a spherical surface shown in figure. at every point of this surface electric field due to the charge q is E=

kq x2

Here if we use Gauss Law for the spherical surface, we have → →

q

∫ E .dS = ∈

0

q E . 4πx2 = ∈ 0

q ⇒ E∫ dS = ∈ 0

or

E=

1 q Kq = 2 2 4π ∈0 x x

→ Here we can see that at every point of sphere electric field vector is parallel to dS and also the → magnitude of E is uniform at every point, thus the integral

→ →

∫ E .dS

can be easily evaluted.

Basically flux is the count of number of lines of electric field crosing an area. For open surface we choose one direction as a area vector & stick to it for the whole problem

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ELECTROSTATICS - 2 Ex.3

Page # 11

In figure (a) a charge q is pla ed just outside the centre of a closed hemisphere. In figure (b) the same charge q is placed just inside the centre of the closed hemisphere and in figure (c) the charge is plced at the centre of hemisphere open from the base. Find the electric flux passing through the hemisphere in all the three cases.

q q

q (a)

(b)

(c)

Sol.

In figure (a) φ = 0 In figure (b) φ = q/ε0 In figure (c) φ = q/2ε0

Ex.4

A charge q is placed at point D of the cube. Find the electric flux passing through the face EFGH and face AEHD.

B

F E

A

G

C H

D Sol.

φ = q/6ε0

(i)

Electric field due to a point charge The electric field due to a point charge is every where radial. We wish to find the electric field at a distance r from the charge q. We select Gaussain surface, a sphere at distance r from the charge. At every point of this sphere the electric field has the same magnitude E and it is perpendicular to the surface itself. Hence we can apply the simplified form of Gauss law, ES =

qin ε0

q Here,

S = area of sphere = 4π r2

and

qin = net charge enclosing the Gaussian surface = q ∴

E(4πr2 ) =

∴

E=

r

q ε0

1 q . 2 4πε0 r

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E

ELECTROSTATICS - 2

Page # 12 (ii)

Electric Field Strength due to a Long Charged Wire : If we wish to find electric field strength due to a long charged wire having a linear charge density λ coul/m at a point P situated at a distance x from the wire. For this application of Gauss’s Law we consider a cylindrical Gaussian surface of length  and radius x as shown in figure. If we apply Gauss’s Law on this surface, we have → →

qencl ∈0

∫ E .dS =

I II

...(1)

Here the closed Gaussian surface is made of three parts, I, II and III, two flat circular faces and one cylindrical lateral surface. Here we split the closed surfae integration in three parts as

→ → ∫ E .dS =

P



→ → → → ∫ E .dS + ∫ E .dS +

∫

I

III

II

→ → E .dS

III

λ coul / m

Here we know for part I and III, electric field strength vector is perpendicular to the area vector as shown in figure hence no flux will come out of these parts. Thus we have → → → → ∫ E .dS = ∫ E .dS = 0 I

III

Now from equation (1) we have E∫ dS = II

λ ∈0

[As enclosed charged is qencl = λ l ]

For lateral surface as E is parallel to ds is parallel E∫ ds = II

λl ε0 λ ∈0

or

E.2πx =

or

λ 2kλ E = 2π ∈ x = x 0

(iii) Electric Field Strength due to a Long Uniformly Charged Cylinder : σ coul / m2

I II



x

III R

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ELECTROSTATICS - 2

Page # 13

Case - I : Conducting Cylinder Figure shows a long cylinder of radius R which is uniformly charged on its surface with surface charge density σ coul/m2 We know at interior points of a metal body electric field strength is zero. For finding electric field strength at outer points at a distance x from the axis of the cylinder, we consider a cylindrical Gaussian surface of radius x and length  as shown in figure. Now we apply Gauss’s Law on this surface we have → 


∫ E .dS =

qencl ∈0

Here enclosed charge in the cylindrical Gaussian surface can be given qencl = σ . 2π R Here also similar to previous case the electric flux through the circular faces is zero, hence according to Gauss law, we have → → σ.2πR σ.2πR σR E∫ dS = E= or ∫II E .dS = ∈0 or ∈ ∈ 0 0 x II

Case II : Uniformly Charged Non-conducting Cylinder Figure shows a long cylinder of radius R, charged uniformly with volume charge density ρ coul/m3. To find electric field strength at a distance x from the cylinder axis we again consider a cylindrical Gaussian surface shown in figure. If we apply Gauss Law on this surface, we have → →

∫ E .dS =

qencl ∈0

or

→ → ρ.πR 2  ∫ E .dS = ∈0 II

or

E∫ dS =

or

E.2πx =

[As qencl = ρ.πR2 ]

ρπR 2  ∈0 ρπR 2  ∈0

To find electric field inside the cylinder at a distance x from the axis, we consider a small cylindrical Gaussian surface of radius x and length . If we apply Gauss Law for this surface, we have → → q encl ∫ E .dS = ∈0 or

E∫ dS = II

ρπx2  ∈0

or

E.2πx =

or

E=

I



x II III

ρπx2  ρ R2 or E = ∈0 2ε0 x

ρx 2 ∈0

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ELECTROSTATICS - 2

Page # 14 (iv)

Electric field Strength due to a Non-conducting Uniformly Charged Sheet : To find the electric field strength at a point P infront of the charged sheet we consider a cylindrical Gaussian surface as shown in figure of face area S. If we apply Gauss law for this surface, we have → → qencl ∫ E .dS = ∈0 σ coul / m2 → → → → → → σS or ∫I E .dS + II∫ E .dS + III∫ E .dS = ∈0 [As here qencl = σS] → →

In this case

∫ E .dS = 0

area S

as the lateral surface of cylinder is

II

parallel to the direction of electric field strength, no flux is coming out from the lateral surface, hence we have

E

E

II x

I

P

III

σS ∫I EdS + II∫ EdS = 2 ∈0

(v)

or

2ES =

or

E=

σS ∈0

[As electric field is uniform on both sides]

σ 2 ∈0

Electric field Stength due to a Charged Conducting Sheet : Figure show a large charged conducting sheet, charged on both the surfaces with surface charge density σ coul/m2. As we know in the metal sheet there is no charge within the volume of the sheet and also the electric field inside the metal sheet is zero. To find electric field strength at a point P in front of the sheet we consider a cylindrical Gaussian surface having one face at point P where electric field is required and other face is within the volume of sheet. If we apply Gauss’s Law on this surface, we have → → qencl ∫ E dS = ∈0 → →

or

→ →

→ → σS E .dS = ∈0 III

∫ E .dS + ∫ E .dS + ∫ I

II

[As here qencl = σS]

σ coul / m2

Here on surface I of the Gaussian surface E = 0 hence → → → → ∫ E .dS = 0 and ∫ E .dS = 0 as no electric flux is coming out I

II


from the lateral surface of cylinder ( E is perpendicular to area vector of curved surface). Hence we have total flux coming out is → → σS ∫II E .dS = ∈0 or

ES =

σS ∈0

or

E=

area S

I II

P

E

σ ∈0

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ELECTROSTATICS - 2

Page # 15

CONDUCTOR TYPE OF MATERIALS

conductors (All electrons are free)

Insulators (all electrons are bounded)

CONDUCTORS : A conductors contains free electrons, which can move freely in the material, but cannot leave it. On applying an external electric field on a conductor charges of a conductor adjust themselves in such a fashion that the net electric field inside the conductor is zero under electrostatics conditions.

–

+

–

– – – E – inside – – –

+

+

+ + + + + E outside + + + ++

→ Net E = 0 ⇒ Potential is constant ∴ Conductor behaves as an equipotential surface Being an equipotential surface, electric field lines will terminate or originate perpendicularly Let us now consider the interior of a charged conducting object. Since it is a conductor, the electric field in the interior is everywhere zero. Let we analyse a Gaussian surface inside the conductor as
close as possible to the surface of the conductor. Since the electric intensity E is zero everywhere inside the conductor, it must be zero for every point of the Gaussian surface. Hence the flux through the surface.



q (∫ E.dS = ) , the net charge inside the E.dS will be zero. Therefore, according to Gauss's law ∫ ε 0

Gaussian surface and hence inside the conductor must be zero. Since there can be no charge in the interior of the conductor charge given to the conductor will reside on the surface of the conductor.

+

+ + + +++

+ + + + +

+

All the charge given to the conductor reside on the surface of the conductor Q + + + + + +

+

++

+ + + +

4.

Semi-conductors (Some electrons are free)

Till now we have only discussed the case of uniform shaped bodies on which the charge distribute itself uniformly. But what about the charge distribution on irregular shaped bodies ? Does in this case also uniform charge distribution take place ? .............. NO

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ELECTROSTATICS - 2

Page # 16

σ∝

In this case

charge per unit area •

Q

1 rC radius of curvature

++ + ++ + + + + + ++ + + + + + + + + + + + + + + + + + + +

Let us consider a random shaped body and find Electric field due to a small portion of this body. ++ However the σ is not uniform everywhere but for a small area + + + + + + + dA, we can assume that σ is constant. Considering a cylindrical + ++ + gaussian surface, we will calculate flux pasing through the + + + cross section dA. + +

φnet



q = ∫ E.ds = in ε0

φnet = φcurved surface + φouter flat surface + φ

+ + + + +

+

+

+

+ E

+ +

++

inner flat surface

φcurved surface = 0 because no flux is passing through lateral surface (electric field lines are perpendicular to area vector.)

E.ds = 0

φinner flat surface = 0
because E inside conductor = 0 qin ∴ ε = φouter flat surface 0

σdA

= E.dA ε0


σ ⇒ E= ε 0

5.

ELECTRIC PRESSURE :

(a)

Electric pressure on a Charged Metal Surface : We know when some charge is given to a metal body it will spread on the outer surface of the body due to mutual repulsion in the charge. When on surface every charge experiences an outward repulsive force due to remaining charges, every part of body experiences an outward pressure. This pressure which acts on every part of charged metal body surface due to remaining charges on the body is called electric pressure. To calculate this we consider a small segment AB on body surface of area dS as shown. If σ be the surface charge density on AB, charge on it is

A

M

dS B

N

C

dq = σdS Now we consider two points M and N just outside and inside of section AB as shown in figure. At the two point if E1 be the electric field due to section. AB then direction of the electric fields at M and N can be given as shown in figure. If we remove section AB from the body then due to removing body ACB, if E2 be the electric field strength at point M and N, the direction of E2 can be given as shown in figure.

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ELECTROSTATICS - 2

Page # 17 AM

E1

A

E2 B

N E2

M B

N E1

C

(a)

(b) Due to complete body we know net electric fields at just outside and inside points can be given as σ ...(1) EM = E1 + E2 = ε0 and EN = E1 – E2 = 0 ...(2) Solving equations (1) and (2) we get E1 = E2 σ and E1 = E2 = 2 ε0 σ Thus electric field at the location of section AB due to remaining body ACB is , using which we can 2 ε0 find the outward force on section AB, due to the rest of the body ACB as σ Force on AB is dF = dq E2 = σ dS × 2 ε0 Thus pressure experienced by the section AB can be given as Pe =

df σ2 = ds 2ε0

As net electric field outside the surface is Enet =

Thus we have

σ ∈0

Pe =

(Enet )2 2 ∈0

Pe =

1 2 ∈0 Enet 2

ELECTRIC FIELD DUE TO SPHERICAL BODIES

Conducting hollow sphere

Conducting Solid sphere

Non Conducting Hollow sphere

Non Conducting Solid sphere

Will behave in the same fashion
because E & potential depends on charge distribution & all the above three spheres have same charge distribution

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ELECTROSTATICS - 2

Page # 18 (b)


E due to conducting hollow sphere, conducting solid sphere & non conducting hollow sphere.

For the above mentioned bodies, any excess charge given to body gets distributed uniformly over its outer surface. Since the charge lines must point radially outward & also the field strength will have the same value at all points on any imaginary spherical surface concentric with the charged conducting sphere or the shell this is the symmetry which leads us to choose the gaussian surface to be a sphere.



Any arbitrary element of area ds is parallel to the local E so E. ds = Eds at all points on the surface.

Electric field Strength due to a Conducting (solid and hollow) Sphere, Non conducting hollow sphere Case I : x > R To find electric field at an outer point at a distance x from the centre of sphere, we cosider a spherical Gaussian surface of radius x. If electric field strength at every point of this surface is E, using Gauss’s law we have E dS → → q encl ∫ E .dS = ∈0 P

+ + +

Q ∈0

1 Q E = 4π ∈ . 2 x 0

R

+

E. 4πx2 =

+ + +Q

++ +

or

+

+ + +

E∫ dS =

+

Q ∈0

Here we have

+

Simlarly for surface points we can consider a spherical Gaussian surface of radius R which gives electric field strength on the sphere surface as

ES =

1 Q 4π ∈0 x 2

C

+

+

+

P

+

x

+

∫ E .dS = Thus

E=0

+ + + +

+

R

+

→ →

+

+

+

+

+

To find electric field strength at an interior point of the sphere, we consider an inner spherical Gaussian surface of radius x(x < R). Here if we apply Gauss Law for this surface, we have + + + Q + + +

+

(c)

qencl = 0 [As all charge is on surface] ∈0 [As dS ≠ 0]

For points outside the sphere, the field is same as that of a point charge at the centre of sphere.

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ELECTROSTATICS - 2

Page # 19

Case II x < R

E 1 Q 4πε 0 R2

r=R Non-conducting Uniformly Charged Sphere For outer and surface points the electric field strength can be calculated by using Gauss Law similar to the previous case of conducting sphere. For interior points of sphere, we consider a spherical Gaussian surface of radius x as shown. If we apply Gauss Law for this surface, we have ρcoul / m3 → → q encl ∫ E .dS = ∈0 + + + + + + + + + + + + + Here enclosed charge can be given as + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + 4 qencl = ρ × πx3 + + + + + + C + + + + ++ 3 + + + + + + + + + + + + + + + + + + + + + + 4 + + + + + + + + + + ρ × πx3 + +R+ + + + + + + 3 2 Thus E. 4πx = + + + + + + + ∈0

ρx E = 3∈ 0

or

Case II : At an external point (r > R) To find the electric field outside the charged sphere, we use a spherical Gaussian surface of radius r(r > R). This surface encloses the entire charged sphere. So, from Gauss's law, we have E(4πr2 ) =

or,

E=

Q ε0

1 Q 4π ε 0 r 2

...(13F)

The field at points outside the sphere is the same as that of a point charge at the centre.

Variation of E with the distance from the centre (r) E

1 Q 4π ε 0 R2 O

r=R

r

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ELECTROSTATICS - 2

Page # 20 •

Electric Potential Inside a Metal Body : As we've already discussed whenever charge is given to a metal body, it is distributed on its outer surface in such a way that net electric field at every interior point of body is zero. Thus if inside a metal body, a charge is displaced, no work is done in the process as electric field at every point is zero. Hence we can say that the whole metal body is equipotential On the basis of above explanation we can state that a region in which at every point electric field is zero, can be regarded as equipotential region.

R

+

+

++ + + + + +

+ + + + + + + +

+ + + + +

At the points on surface of sphere, the potential can be given as

+

KQ V= x

Q + + + + + + + +

• Electric Potential due to a Charged Sphere : Case I : Conducting Sphere As we know for outer points of a charged sphere we can assume whole charge is concentrated at its centre thus electric potential at a distance x from the centre of sphere outside can be given as

KQ R At the interior points of sphere as at every point electric field is zero, we can state that this is an equipotential region thus at every interior point potential is same as that of its surface. Thus we have

Vs =

Vin = •

KQ R

Variation of Potential with Distance from centre of Sphere : V

KQ R

O

V∝

1 x x

Note : Above results are also valid for a uniformly charged hollow sphere. Case II : Non-conducting Uniformly Charged Sphere For outer and surface points here also we can say that the potnetial remains same as that of a conducting sphere as KQ Vout = (for x > R) x S + + + + + KQ VS = (for x = R) + + + + + + + + R + + + + P+ + + + + + For an interior point unlike to a conducting sphere, potential + + + + + + + + + + will not remain uniform as electric field exists inside region. + + + + + + + + + + + + + + + + + + + + + ++ We known inside a uniformly charged sphere electric field is + + + + + + + + +R+ + + Q in radially outward direction thus as we move away from + + + + + + + + + + + centre, in the direction of electric field potential decreases. + + + + + + + + + + + + + + + + + + + + As shown in figure if there is a point P at a distance x from the + + + + + + + + centre of sphere, the potential difference between points P + + + + + + and S can be given as

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ELECTROSTATICS - 2

Page # 21 R

Vp – VS =

KQx dx 3 x R

∫

or

VP –

KQ KQ (R 2 – x2 ) = 2R 3 R

or

VP =

KQ KQ (R 2 – x2 ) + 3 R 2R

KQ (3R 2 – x2 ) 2R 3 Here at x = 0, we have potential at centre of sphere is

Vp =

3 3KQ = Vs 2 2R Thus at centre, potential is maximum and is equal to 3/2 times that on the surface. Vc =

•

Variation of Potential in a Uniformly Charged Sphere with Distance : 3 kQ 2 R

V

v

KQ R

V∝

x

x=R

O

6.

1 x

FIELD ENERGY OF ELECTROSTATIC FIELD → E

qE

m,q

Consider a situation shown in figure. A small body of mass m and charge q placed in an electric field E. When the body is released it starts moving in the direction of electric field due to the electric field qE acting on it. The body will gain some kinetic energy due to its motion. Who is giving energy to this particle ? Answer is simple-electric field. This shows that electric field must posses some energy in the region where field exist due to which it can do work on any charged body placed in it. This energy we call field energy of electric field. Wherever electric field exist, field energy also exsit in space. Let us calculate the amount of energy stored in the space where electric field exist.

(a)

Field Energy Density of Electric field : As discussed in previous section in every region where electric field is present, energy must exist. This field energy we can calculate by an example given here. Consider a charged conducting body shown in figure. Its surface M is having a charge distributed on it. We know the electric field just outside the surface M at a point can be given as E=

σ ε0

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ELECTROSTATICS - 2

Page # 22

M'

+ + + + + + + + + M + + + + + + + + + + + + + We also know that on the surface of metal body experience an outward electric pressure which is given as σ2 1 = ∈0 E2 2 ∈0 2

Pe =

Now if we consider that the metal surface M is flexible and allowed to expand due to electric pressure upto a small limit to M′. Here if we check electric field associated with the body, we known inside the body there is no electric field. Initially electric field only exist from surface M to infinity. Hence the field energy also exist from the surface M to infinity. When the surface expand to M′ then in the final stage the electric field as well as field energy exist from surfaec M′ to infinity. This implies that during expansion of surface field energy in the shaded volume (say dV) vanishes as before expansion there was electric field in this region and after expansion electric field becomes zero in the region as there is no electric field inside the body. We also know that the expansion is done by electric force in the body (electric pressure) hence the work done by electric field during expansion is equal to the loss in field energy in the shaded volume dv. If Pe is the electric pressure on the body surface then in the small expansion in body volume dV, work done can be given as dW = PedV And if dU the field energy stored in this volume dV then we can use dU = dW = PedV or

dU = Pe dV u=

σ2 1 2 3 = ∈0 E joul / m 2 ∈0 2

dU is the field energy stored per unit volume in the space where electric field E exist and is dV called field energy density of electric field. If in a region electric field is uniform, the total field energy stored in a given volume V of space can be given as Here u =

U=

1 ∈0 E2 × V 2

If electric field in a region is non-uniform, the total field energy stored in a given volume of space can be calulated by integrating the field energy in an elemental volume dV of space as 1 ∈0 E2 × dV 2 And total field energy in a given volume can be given as dU =

U = ∫ dU =

1

∫2∈

0

E2 dV

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ELECTROSTATICS - 2 (b)

Page # 23

Self energy of a Hollow, conducting, solid conducting & hollow non conducting sphere. We’ve dicussed whenever a system of charges is assembled, some dq work is done and this work is stored in the form of electrical potential energy of the system. Now we consider an example of charging a q conducting sphere of radius R. In the process of charging we bring charge to the sphere from infinity in steps of elemental charges dq. The charge on sphere opposes the elemental charge being brough to it. Let us assume that at an instant sphere has charge q, due to which it has a potential given as

R

Kq R

V=

If now a charge dq is brought to its surface from infinity work done in this process can be given as dW = dqV =

Kq dq R

Total work done in charging the sphere can be given as Q

W = ∫ dW =

W=

Kq dq 0 R

∫

KQ2 2R

...(1)

Equation (1) gives the total work done in charging the sphere of radius R. We’ve discussed that in space wherever electric field exist, there must be some field energy stored which has energy density, given as 1 ∈0 E2 J / m3 2 dx

+ +

R +

with distance from centre as

+

We know electric field due to a sphere at outer points varies

E=

+

Q +

+

Here we can see that when the sphere was uncharged, there was no electric field in its surroundings. But when the sphere is fully charged, electtric field exist in its surrounding from its surface to infinity. Let us calculate the field energy associated with this charged conducting sphere.

+

u=

KQ x2

To find the total field energy due to this sphere, we consider an elemental spherical shell of radius x and width dx as shown in figure. The volume enclosed in this shell is dV = 4πx2. dx Thus the field energy stored in the volume of this elemental shell is dU =

1 ∈ E2 . dV 2 0 2

1  KQ  KQ2 2 dx = ∈0  2  × 4πx dx = 2 2x2 x 

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ELECTROSTATICS - 2

Page # 24

Thus total field energy associated with the sphere can be calculated by integrating this expression from surface of sphere to infinity as electric field inside the sphere is zero. Total field energy in the surrounding of sphere is ∞

U = ∫ dU =

KQ2

∫ 2x

2

dx

R

KQ2 = 2

∞

 1 KQ2 – x  = 2R  R

...(2)

Here we can see that this result is same as equation (1). We can conclude by this total whatever work is done in charging a body is stored in its surrounding in the form of its field energy and can be regarded as self energy of that body. Once a body is charged in a given configuration, its self energy is fixed, if the body is now displaced or moved in any manner keeping its shape and charge distribution constant, its self energy does not charge. as discussed above we can say that “Self energy of a charged body is the total field energy, associated with the electric field due to this body in its surrounding.

(c)

Self Energy of a Uniformly Charged Non-conducting Sphere : We know in outside region of a non-conducting uniformly charged sphere, every point is same as that of a conducting sphere of same radius. Thus field energy in the surrounding of this sphere from surface to infinity can be given as UR →∞ =

KQ2 2R

Unlike to the case of conducting sphere, in nonconducting sphere at interior point E ≠ 0. Thus field energy also exist in the interior region. This can be calculated by considering an elemental shell inside the sphere as shown.

+ + + + + Q + + + + + + + + + + + + + + + + + + + + + + + + + + + + dx + + + + + + + + + + + + + + + + + + + + + ++ + + + + + + + + + +R+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

Here field energy in the volume of this elemental shell can be given as 2

dU =

1  KQx  ∈0  3  × 4πx2 dx 2  R  =

[As Ein =

KQx ] R3

KQ2 4 x dx 2R 6

Total field energy inside the sphere can be given as U = ∫ dU =

U0→R =

KQ2 2R 6

R 4 ∫ x dx ⇒ U = 0

R

KQ2  x5    2R 6  5  0

KQ2 10 R

Thus total self energy of this sphere can be given as U

self

= U0

Uself =

→R

+ UR → ∞

KQ2 KQ2 3 KQ2 + ⇒ Uself = 10R 2R 5 R

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ELECTROSTATICS - 2 CHARGE INDUCTION IN METAL CAVITIES :

We've discussed that there can never be any electric field inside a conductor due to static charges. Hence no electric line of force can enter into a conducting body. Consider a point charge +q inside a spherical cavity at centre within a metal body shown in figure. q The total electric flux originated by +q is ∈ . Due to this charge at the inner surface of cavity a 0

+

charge – q is induced on which this complete flux will terminate and no electric line of force exists into the metal body. A point, A inside the metal volume we know net electric field is zero. Thus the electric field at A due to the point charge +q is nullified by the electric field due to the negative induced charges on the inner surface of cavity and the positive charge induced on outer surface is automatically distributed on the surface in such a way that it does not produce any electric field with in the metal body. + + + + + + + +

+ +

+q

+

+

– – –– –– – – – – – +q –– – – – – – –– ––

+ + +

+

+

+

A

+

+ +

+

+

+

+

+

+

From the above analysis we can conclude some points about the charge induction when a charge is placed inside the cavity of a metal body. These are. (1) Whenever a charge is placed inside a metal cavity, an equal and opposite charge is induced on the inner surface of cavity. (2) A similar charge is induced on the outer surface of body with surface charge density inversely proportional to radius of curvature of body (3) When the charge inside is displaced, the induced charge distribution on inner surface of body changes in such a way that its centre of charge can be assumed to be at the point charge so as to nullify the electric field in outer region. (4) Due to movement in the point charge inside the body. The charge distribution on outer surface of body does not change as shown in figure.

– – – –– – – –– – – +q – –

+ + +

+

+

+

+

+

+

+

+

+

+ +

+

+

+

+

+

– – – – – +q – – – –

+ + +

+

+

+

+

+

+

+

+

+

+

+

+ +

+

+

+

+

+

+

+

+

+

– – – –– – – +q ––– – ––

+

++ + +

+

+

+

+

++

+

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+

– – – –– – – – +q – – –––

+

+

+

+

+

+

+

+

+

+ + +

+ + +

+

(5) If another charge is brought to the body from outside, it will only affect the outer distribution of charges not on the charge distribution inside the cavity as shown in figure

+

7.

Page # 25

ELECTROSTATICS - 2

Page # 26

c t i n

g

s p

h

e

r i c a

l

s h

e

l l

o f

i n

n

e

r

r a

d

i u

s

R1 C

r

+

R2

+

kq r

+

Q

+ +

R1

+

+

Case I where r < R1
E = 0[because net charge within this region = 0]

R2 +

+

+

charge is given to a conductor it resides on its outer surface.
Let us find E at a point distanced r

+

+

consider the system shown in figure. As we know that when a

+

+

Cavity in a conducting Material

+

+

VP =

+

and

+

kq r2

+

Eout =

+

+

+

q

+

If we find electric field and potential at a distance r from the centre outside the shell, it will be only due to the charge on outer surface as induced charge on inner surface of cavity always nullifies the effect of point charge inside it. Thus it can be given as

q

+

kq Kq Kq VC = r – R + R 1 2

+

+

+

+

+

+

R

+

u

+

d

+

c o n

+

Now consider the situation shown in figure. Inside a and outer radius 1 R2, a point charge q is placed at a distance x from the centre as shown. The electric potential at centre due to this system can be given as

Case II When R1 < r < R2 qin
E = 0 [qnet = 0, φ = ε = 0 ⇒ E = 0 ] 0

Case III When r > R2
kQ E = 2 [It is similar to case of hollow charged sphere] r

+

+

+ +

QA

B

+

C

+

S1

+

S2

+

+

+

+

•

+

+

• •

Now we consider a case when charge is placed inside a conductor. For such case charge distribution will be as follows For simplicity in the calculation, we could bifurgate the above system as A point charge A hollow sphere S1 with charge –Q

+

A hollow sphere S2 with charge +Q electric field at A Due to Q =

KQ r2

Due to S1 = 0 [∵ Point lies inside the hollow sphere] Due to S2 = 0 [∵ Point lies inside the hollow sphere]

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ELECTROSTATICS - 2

Page # 27

Electric field at B D

u

e

t o

Q

=

Due to S1 =

KQ → r2 KQ ← r2

Due to S2 = 0
Enet at B = 0 Electric field at C Due to Q =

kq → r2

Due to S1 =

kq ← r2

Due to S2 =

kq → r2


kq Enet at C = 2 → r

At point C, net E due to S1 & q is zero. E at C is only due to outside charge (S2). If we place an external charge at point C, then effect of S1 & Q on external charge is zero or we can say that effect of external charge on S1 & Q is zero. Or we can say charge placed inside the conductor & the charge induced on the inner surface of the conductor does not get affected by any external electric field this is known as electrostatic shielding that is why, equipment sensitive towards electric field are placed inside a conductor. External electric field only affects the charge distributed on the surface of conductor. We again go back to the case when a charge was placed in the conductor Potential at A

+

+ +

B

C

+

S2

+

S1

+

+

+

⇒

QA

+

kQ due to S2 = R 2

+

+

–kQ Due to S1 = R 1

+

+

kQ Due to Q = r

+

1 1 1  + Vnet = kQ  –   r R1 R 2  Potential at B Due to Q =

kq r

Due to S1 =

–kQ r

Due to S2 =

kQ R2

⇒

Vnet =

kQ R2

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ELECTROSTATICS - 2

Page # 28 Potential at C Due to Q =

kQ r

Due to S1 = – Due to S2 =

–kQ r

kQ r

Vnet =

kQ r

While writing potential at various points is case of cavity in a conducting material first distribute charge on various surfaces & then the potential due to induced charges is also considered.

8.

CAVITY IN A NON CONDUCTING SPHERE

(a)

Electric field due to a Non-uniformly Radially Charged Solid Non-conducting Sphere : If a sphere of radius R is charged with a non-uniform charge density which varies with the distance x from centre x as ρ=

P

r

ρ0 coul / m3 x

dx

Here if we wish to find electric field strength at a point situated at a distance r from centre of sphere outside it, at point P shown in figure. This can be given as

x R

KQ (where Q is the total charge of sphere) r2 For outer points we can assume whole charge of sphere to be at its centre. Now Q can be calculated by integrating the charge of an elemental shell of radius x and width dx as shown in figure. The charge dq in this shell can be given as

Ep =

dq = ρ. 4πx2 dx =

ρ0 × 4πx2dx = 4πρ0x dx x

Total charge of sphere can be given as R

R  x2  = πρ 4 πρ dq 4 xdx Q= ∫ =  = 2πρ0 R2 0  ∫0 0  2 0

Thus electric field strength at outer points can be given as EP =

K(2πρ0R 2 )

ρ0R 2 = 2 ∈0 r 2

r2 To find electric field strength at an interior point at a distance r from the centre of sphere, we first find the charge enclosed within the inner sphere of radius r of which point P is on the surface. Thus enclosed charge can be given as r

qencl =

ρ0

∫ R .4πx dx = 2πρ r 0

2

2

0

Here electric field strength at point P can be given as Kqencl K(2πρ0r2 ) EP = = 2 r r2 ρ0 EP = 2 ∈0 Here we can see that the above expression is independent of distance from centre.

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ELECTROSTATICS - 2 Ex.5

Page # 29

Figure shows a uniformly charged sphere of radius R and total charge Q. A point charge q is situated outside the sphere at a distance r from centre of sphere. Find out the following :

Q

R q

(i) Force acting on the point charge q due to the sphere. (ii) Force acting on the sphere due to the point charge. Sol.

(i) Electric field at the position of point charge
KQ E= 2 ˆ r r
KqQ
KqQ | F |= 2 F = 2 ˆ r r r (ii) Since we know that every action has equal opposite reaction so
KqQ Fsphere = – 2 ˆ r r

so,


KqQ | Fsphere | = 2 r

Ex.6

Figure shows a uniformly charged thin sphere of total charge Q and radius R. A point charge q is also situated at the centre of the sphere. Find out the following : (i) Force on charge q

A q

B

C

(ii) Electric field intensity at A. (iii) Electric field intensity at B. Sol.

(i) Electric field at the centre of the uniformly charged hollow sphere = 0 So force on charge q = 0 (ii) Electric field at A


EA = ESphere + Eq = 0+

Kq r2

; r = CA

E due to sphere = 0, because point lies inside the charged hollow sphere.


(iii) Electric field EB at point B = E Sphere + E q

Ex.7

=

KQ Kq ˆ r+ 2 ˆ r 2 r r

=

K(Q + q) .rˆ r2

; r = CB

Here we can also assume that the total charge of sphere is concentrated at the centre, for calculation of electric field at B. Q2 Two concentric uniformly charged spherical shells of Q1 radius R and R (R > R ) have total charges Q and Q 1

2

2

1

1

2

respectively Derive an expression of electric field as a function of r for following positions. (i) r < R1

(ii) R1 ≤ r < R2

(iii) r ≥ R2

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R1 R2

ELECTROSTATICS - 2

Page # 30 Sol.

(i)

for

r < R1,

therefore point lies inside both the spheres Enet = Einner + Eouter = 0 (ii)

for R1 ≤ r < R2, therefore point lies outside inner sphere but inside outer sphere : Enet = Einner + Eouter =

(iii)

for

KQ1 ˆ r+0 r2

=

KQ1 ˆ r r2

r ≥ R2

point lies outside inner as well as outer sphere therefore. ENet = Einner + Eouter =

Ex.8

=

K(Q1 + Q2 ) ˆ r r2

A solid non conducting sphere of radius R and uniform volume charge density ρ has its centre at origin. Find out electric field intensity in vector form at following positions : (i) (

Sol.

KQ1 KQ ˆ r + 22 ˆ r 2 r r

R , 0, 0) 2

 R  R , , 0 (ii)  2  2 

(iii) (R, R, 0)

R (i) at ( , 0, 0) : Distance of point from centre = 2

2

R  R 2 2 < R, so point lies inside the 2 +0 +0 = 2  

sphere so

ρr ρ Rˆ = E= [ i] 3ε0 3ε0 2

 R  R , , 0  ; distance of point from centre = (ii) At  2 2  

2

2

 R   R  2   +  +0 =R =R  2  2

so point lies at the surface of sphere, therefore

4 3
KQ
K πR ρ  R ρ  R ˆ R ˆ 3 i+ j ˆi + R ˆj =  E= 3 r =   3 3ε0  2 2  R R 2   2

(iii) The point is outside the sphere 4 3
KQ
K πR ρ ρ 3 ˆ = ˆ so, E = 3 r = [Rˆi + Rj] [Rˆi + Rj] 3 r ( 2R) 6 2 ε0

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ELECTROSTATICS - 2 (b)

Page # 31

Electric Field Inside a Cavity of Non-conducting Charged Body :

+ + + + + + + + + + + + +
+ + + + + + + + + +E + + + + + + + +2 + + + + + + + + +x
+ + +
+ + + + + +C + + + +P+y + + + + + + + + + + + +C1+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +


E1

Consider the sphere shown in figure charged uniformly with charge density ρ coul/m3. Inside the sphere a spherical cavity is created with centre at C. Now we wish to find electric field strength inside the cavity. For this we consider a point P in the cavity

at a position vector x from the centre of sphere and at a position vector y from the centre of cavity as shown.
If E1 be the electric field strength at P due to the complete charge of the sphere (inside cavity also) then we known electric field strength inside a uniformly charged sphere is given as

ρx E1 = 3 ∈0 Similarly if we assume charge is only there in the region of cavity, this will also be a uniformly charged
small sphere. If E 2 be the electric field only due to the cavity charge, it can be given as
ρy E2 = 3ε 0

Now the electric field due to the charged sphere in the cavity at point P can be given as


[As now charge of cavity is removed] Enet = E1 – E2
ρa = 3∈ 0




[As x – y = a ]

This shows that the net electric field inside the cavity is uniform
and in the direction of a i.e. along the line joining the centre

a

ρcoul / m3

of spheres and cavity. Similarly we can find the electric field strength inside a cyclindrical cavity of a long uniformly charged cylinder. If cavity axis is displaced from axis of cylinder by a displacement vector
a , by the analysis we've done for a sphere, we can say that the electric field strength inside the cavity is also uniform and can be given as

ρa E= 2 ∈0

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+ + + + + + + + + + + +

+ + + + + + + + + + + + + + + + ++ + + + + + +

+ + + + + + + + + + + +

+ + + + + + + + + + + +

+ + + + + + + + + + + +

+ ++ ++ ++ ++ ++ ++ ++ ++ ++ ++ ++ +

+ ++ ++ ++ ++ ++ ++ ++ ++ ++ ++ ++ +

ELECTROSTATICS - 2

Page # 32 (c)

Some other important results for a closed conductor :

+q

(i) If a charge q is kept in the cavity then –q will be induced on the inner surface and +q will be induced on the outer surface of the conductor (it can be proved using

–q

q

gauss theorem) +q+Q

(ii) If a charge q is kept inside the cavity of a conductor and conductor is given a charge Q then –q charge will be induced on inner surface and total charge on the outer surface will be q + Q. (it can be proved using gauss

–q q

theorem)

(iii) Resultant field, due to q (which is inside the cavity) and induced charge on S1, at any point outside S1 (like B, C) is zero. Resultant field due to q + Q on S2 and any other charge outside S2, at any point inside of surface S2 (like A, B) is zero S2

B S1 –q

C

q

q+Q A

(iv) Resultant field in a charge free cavity in a closed conductor is zero. There can be charges outside the conductor and on the surface also. Then also this result is true. No charge will be induced on the inner most surface of the conductor.

No charge

(v) Charge distribution for different types of cavities in conductors 0 S2

S2 S1 C

q

q

S1

C

(B)

(A)

charge is at the common centre (S1 S2 → spherical)

charge is not at the common centre (S1 S 2 → spherical)

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ELECTROSTATICS - 2

Page # 33

S2

S1

q

S1

C

q

C

(C)

(D)

charge is not at the centre of S2 ( S2 → spherical)

charge is at the centre of S2 ( S 2 → spherical) S2

S2 S1 C

q S1

(E)

C q

(F)

charge is at the centre of S1 (Spherical)

charge not at the centre of S1 (spherical)

S2

S2 C

S1 c

q

(G)

S1

q

(H)

charge is not at the geometrical centre charge is at the geometrical centre
Using the result that Eres in the conducting material should be zero and using result (iii) We can show that Case S1 S2

A B C D E F G H Un iform Nonuniform Nonu niform Nonuniform U niform Nonu niform Non uniform Nonuni form Un iform U niform Uniform Uni form Nonuni form Nonu niform Non uniform Nonuni form

In all cases charge on inner surface S1 = – q and on outer surface S2 = q. The distribution of charge on 'S1' will not change even if some charges are kept outside the conductor (i.e. outside the surface S2). But the charge distribution on 'S2' may change if some charges(s) is/are kept outside the conductor. Electric field due at 'A' due to –q of S1 and +q of S2 is zero individually because they are uniformly distributed At point B : VB =

Kq K(–q) Kq Kq + + = , EB = 0 OB OB R2 R2

At point C : Kq
Kq 
, EC = OC OC OC3 (ii) Force on point charge Q : VC =

Here force on 'Q' will be only due to 'q' of S2 see result (iii)
KqQ FQ = 2 ˆ r (r = distance of 'Q' from centre 'O') r Force on point charge q :
Fq = 0 (using result (iii) & charge on S1 uniform)

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ELECTROSTATICS - 2

Page # 34 Ex.9

Sol.

An uncharged conductor of inner radius R1 and outer radius R2 contains a point charge q placed at point P (not at the centre) as shown in figure? Find out the following : (i) VC (ii) VA (iii) VB (iv) EA (v) EB (vi) force on charge Q if it is placed at B

B A

R1 S2

S1

C D

P q

R2

Kq K(−q) Kq + (i) VC = CP + R R2 1

–q on S1 in nonuniformly distributed still it produces potential

K(−q) R1 at 'C' because 'C' is at distance 'R1'

from each points of 'S1'. (ii) VA =

Kq R2

(iii) VB =

Kq CB

(iv) EA = O (point is inside metallic conductor) (v) EB =

Kq ^ CB CB2

(vi) FQ =

KQq ^ CB CB2

Combination of conducting spherical shells. Let us consider a system of concentric conducting shelk with charge q1 on inner shell & q2 on outer shell.

S2

kq1 kq2 Potential at A = R + R 1 2

S1

kq1 kq2 Potential at B = r + R 2

R1 A

R2

kq1 kq2 + Potential at C = r r

q1 q2 B C

kq1 kq2 Potential of S1 = R + R 1 2 kq1 kq2 Potential of S2 = R + R 1 2

In case of combination of concentric conducting spherical shell, we do not consider the potential due to induced charge distribution as we used to consider in the case of cavties in conducting materials.

Ex.10 Consier the following system & find VC – VA C B

A

2R R

4R

q

4q 2q

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ELECTROSTATICS - 2

Sol.

VC =

R(4q) k(2q) k(q) 7kq + + = 4R 4R 4R 4R

VA =

kq k(2q) k(4q) 3kq + + = R 2R 4R R

VC – VA =

9.

Page # 35

7kq 3kq 7kq – 12kq 5kq – = =– 4R 4r R 4R

CONNECTION OF TWO CONDUCTING MATERIAL : Two conducting hollow spherical shells of radii R1 and R2 having charges Q1 and Q2 respectively and placed separately by large distance, are joined by a conducting wire q1

q2

R1

R2

Let final charges on spheres are q1 and q2 respectively. Potential on both spherical shell become equal after joining, therefore Kq1 Kq2 = R1 R2 q1 R = 1 q2 R2

....(i)

and by charge conservation, q1 + q2 = Q1 + Q2 from (i) and (ii) (Q1 + Q2 )R1 (Q1 + Q2 )R 2 q1 = q2 = , R1 + R 2 R1 + R 2 ratio of charges

q1 R = 1 q2 R2

⇒

....(ii)

σ1 4πR12 R = 1 2 R2 σ2 4πR 2

σ1 R 2 ratio of surface charge densities σ = R 2 1 q1 R1 = Ratio of final charges q2 R2 σ1 R 2 Ratio of final surface charge densities. σ = R 2 1 If two concentric hollow sphere are connected by a wire then all the charge from inner sphere will reside to the outer sphere.

Ex.11 The two conducting spherical shells are joined by a conducting wire and cut after some time when charge stops flowing. Find out the charge on each sphere after that. Sol. After cutting the wire, the potential of both the shells is equal Kx K(–2Q – x) k(x – 2Q) + = Thus, potential of inner shell Vin = R 2R 2R Kx K(–2Q – x) –KQ + = and potential of outer shell Vout = 2R 2R R As Vout = Vin

Q

–3Q

R 2R

–KQ K(x – 2Q) = ⇒ –2Q = x – 2Q ⇒ x = 0 R 2R So charge on inner spherical shell = 0

⇒

and outer spherical shell = – 2Q.

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–2Q–x

x

ELECTROSTATICS - 2

Page # 36

5Q

Ex.12 Find charge on each spherical shell after joining the inner most shell and outer most shell by a conducting wire. Also

3 2

find charges on each surface. Sol.

–2Q Q

1

R

Let the charge on the innermost sphere be x. Finally potential of shell 1 = Potential of shell 3 Kx K(–2Q) K(6Q – x) Kx K(–2Q) K(6Q – x) + + = + + R 2R 3R 3R 3R 3R

3x – 3Q + 6Q – x = 4Q

x=

; 2x = Q ;

Charge on innermost shell =

Q 2

charge on outermost shell =

5Q 2

2R

6Q–x

Q 2

3

3R

–2Q 2

x

1

R 2R 3R

Q +3Q/2 –3Q/2 –Q/2 Q/2

middle shell = – 2Q Final charge distribution is as shown in figure.

Ex.13 Two conducting hollow spherical shells of radii R and 2R carry charges –Q and 3Q respectively. How much charge will flow into the earth if inner shell is grounded ? 3Q –Q

R

2R

Sol.

When inner shell is grounded to the Earth then the potential of inner shell will become zero because potential of the Earth is taken to be zero.

3Q

x

Kx K3Q + =0 R 2R x=

–3Q , 2

=

the charge that has increased

R

2R

–3Q –Q Q – (–Q) = hence charge flows into the Earth = 2 2 2

Ex.14 An isolated conducting sphere of charge Q and radius R is connected to a similar uncharged sphere (kept at a large distance) by using a high resistance wire. After a long time what is the amount of heat loss ? Sol.

When two conducting spheres of equal radius are connected charge is equally distributed on them (Result VI).So we can say that heat loss of system ∆H = Ui – Uf  Q2   Q2 / 4 Q2 / 4  Q2 = – 0 –  + =  8πε0R   8πε0R 8πε0R  16πε0R

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ELECTROSTATICS - 2

10.

Page # 37

EARTHING OF CHARGED OR UNCHARGED METAL BODIES : In electrical analysis, earth is assumed to be a very large conducting sphere of radius 6400 kms. If some charge Q is given to earth, its potential becomes Vi =

KQ Re

As Re is very large Ve comes out to be a negligible value. Thus for very small bodies whose dimensions are negligible compared to earth we can assume that earth is always at zero potential Keeping the above fact in mind if we connect a small body to earth, charge flow takes place between earth and the body till both will be at same potential, zero potential as potential of earth will always remain zero, no matter if charge flows into earth or from earth. This implies that if a body at some positive potential is connected to earth, earth will supply some negative charge to this body so that the final potential of body will become zero. Consider a solid uncharged conducting sphere shown in figure. A point charge q is placed in front of the sphere centre at a distance x as shown. Here due to q, the potential at sphere is V=

Kq x

x

C

+q

R

S

Here we ignore induced charges due to q because potential due to induced charges on sphere is zero. If we close the switch S, earth supplies a charge qe on to the sphere to make its final potential zero. Thus the final potential on sphere can be taken as – – qe – – – – – – – – – x – C – +q – S – – – R – – – – – – – – – – Here we ignore induced charges due to q because potential due to induced charges on sphere is zero. It we close the switch S, earth supplies a charge qe on the the sphere to make its final potential zero. Thus the final potential on sphere can be taken as V=

Kq Kqe + =0 x R

or

qe = –

qR x

Here it is obvious that earth has supplied a negative charge to devlop a negative potential on sphere to nullify the initial positive potential on it due to q. Always remember whenever a metal body is connected to earth, we consider that earth suplies a charge to it (say qe) to make its final potential zero due to all the charges including the charge on body and the charges in its surrounding.

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ELECTROSTATICS - 2

Page # 38

11.

CONDUCTOR AND IT'S PROPERTIES [FOR ELECTROSTATIC CONDITION] : (i)

Conductors are materials which contains large number of free electrons which can move freely inside the conductor.

(ii)

In electrostatics conductors are always equipotential surfaces.

(iii)

Charge always resides on outer surface of conductor.

(iv)

If there is a cavity inside the conductor having no charge then charge will always reside only on outer surface of conductor.

(v)

Electric field is always perpendicular to conducting surface.

(vi)

Electric lines of force never enter into conductors.

(vii)

Electric field intensity near the conducting surface is given by formula →

E=

σ nˆ ε0

A →

C (viii)

B

EA =


σ 
σ σA ˆ ˆ and EC = C n ˆ n ; EB = B n ε0 ε0 ε0

When a conductor is grounded its potential becomes zero.

V=0

(ix)

When an isolated conductor is grounded then its charge becomes zero.

(x)

When two conductors are connected there will be charge flow till their potential becomes equal.

(xi)

Electric pressure : Electric pressure at the surface of a conductor is given by formula P =

σ2 2ε0

where σ is the local surface charge density.

Ex.14 There are 4 concentric shells A, B, C and D of radius of a, 2a, 3a, 4a respectively. Shells B and D are given charges +q and –q respectively. Shell C in now earthed. Find the potential difference VA – VC Sol Let shell C acquires charge 'q' which will be such that final potential of C is zero. kq kq '  –kq  –q Vc = + + =0 3a 3a  4a  1 1 kq kq ' kq + = ⇒ q′ = 3q  –  3a 3a 4a 4 3 q q′ = – 4 As Vc = 0 VA – VC = VA Now calculating VA we get kq k(q / 4) kq kq VA = – – ⇒ VA = 2a 3a 4a 6a kq or VA – VC = 6a

+q A a

D

C q'

B

2a 3a 4a

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ELECTROSTATICS - 2

12.

Page # 39

Let us consider two conducting plates placed parallel to each other.

I

I plate is given a charge Q1 & II plate is given a charge Q2 which distributes itself as shown in figure above where

B

Q1 A

COMBINATION OF CONDUCTING PLATES :

E

q1 + q2 = Q1

q1

Q2 II

q2

q3

q4

E

q3 + q4 = Q2 Now we take a rectangular gaussian surface ABCD.

D C Among the four faces, two faces AD & BC of this closed surfaces lie completely inside the conductor where the electric field is zero. The flux through these faces is, therefore, zero. The other parts of the closed surface AB & CD which are outside the conductor are parallel to the electric field i.e. their area
vector is perpendicular to E & hence the flux through these parts is also zero. The total flux of the electric field through the closed surface is therefore zero. From gauss's law, the total charge inside the closed surface should be zero. The charge on the inner surface of I should be equal & opposite to that on the inner surface of II. So

q2 = – q3

Now to find further relations between the charges distributed we find electric field at point P Electric field at point P
 As E due to a single layer of charge  q1   (towards right)  q1 σ  due to q1 charge ayer = 2Aε0 is 2e = 2Aε  0 0  

q1 due to q2 charge layer = 2Aε  towards left  0 q3 due to q3 charge layer = 2Aε  towards left  0 q4 due to q4 charge layer = 2Aε  towards left  0 q3 q1 q2 q4
Enet at P = 2Aε – 2Aε – 2Aε – 2Aε [Towards right] 0 0 0 0 As the point P lies inside the conductor the field should be zero Hence q + q3 + q4 q1 = 2 2Aε0 2Aε0

q1 = q2 + q3 + q4

(But q2 = – q3)

q1 = q4 Puting in equation above we get q1 = q4 =

Q1 + Q2 2

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ELECTROSTATICS - 2

Page # 40 Ex.15 Two large parallel conducting sheets (placed at finite distance) are given charges Q and 2Q respectively. Find out charges appearing on all the surfaces. Sol.

Q

2Q

P

Q

Let there is x amount of charge on left side of first plate, so on its right side charge will be Q – x, similarly for second plate there is y charge on left side and 2Q – y charge is on right side of second plate

x

Ep = 0 (By property of conductor) ⇒

y

Q–x

P

2Q –y Q

x y 2Q − y   Q − x − + + =0 2Aε0  2Aε0 2Aε0 2Aε0 

we can also say that charge on left side of P = charge on right side of P x = Q – x + y + 2Q – y ⇒ x=

3Q , 2

Q−x =

−Q 2

+3Q 2

Similarly for point Q:

−Q 2

Q 2

+3Q 2

x + Q – x + y = 2Q – y ⇒ y = Q/2, 2Q – y = 3Q/2

So final charge distribution of plates is : –Q

Ex.16 Figure shows three large metallic plates with charges –Q, 3Q and Q respectively. Determine the final charges on all the surfaces.

Sol.

3Q

Q

We assume that charge on surface 2 is x. Following conservation of charge, we see that surfaces 1 has charge (–Q – x). The electric field inside the metal plate is zero so fields at P is zero. Resultant field at P EP = 0 ⇒ ⇒ ⇒

−Q − x x + 3Q + Q = 2Aε0 2Aε0

3

3Q

4

5

+Q 2

−Q 2

Q

6

x P

–Q –x = x + 4Q x=

2

1 –Q–x

−5Q 2

We see that charges on the facing surfaces of the plates are of equal magnitude and opposite sign. This can be in general proved by gauss theorem 3Q also. Remember this it is important result. Thus the+ 2 final charge distribution on all

−5Q 2

5Q 2

+3Q 2

the surface is as shown in figure :

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ELECTROSTATICS - 2

Page # 41

Q

Ex.17 An isolated conducting sheet of area A and carrying a charge Q is placed in a uniform electric field E, such that electric field is perpendicular to sheet and covers all the sheet. Find out charges appearing on its two surfaces. Sol.

E

Let there is x charge on left side of plate and Q – x charge on right side of plate EP = 0 x Q−x +E = 2Aε0 2Aε0 ⇒

x Q = −E A ε0 2Aε0

⇒

x=

x

Q−x 2Aε 0

x +E 2Aε 0

P

Q Q − EAε0 and Q − x = + EAε0 2 2

So charge on one side is

(a)

Q–x

Q Q − EAε0 and other side + EAε0 2 2

Solve this question for Q = 0 without using the above answer and match that answers with the answers that you will get by putting Q = 0 in the above answer. Earthing of a System of Parallel Plates : Consider a large plate shown in figure charged with a charge Q. This is connected to earth with a switch S as shown. If switch S is closed, whole charge will flow to earth and the plate will become neutral as in the surrounding of a single earthed body no electric field exist. Now consider the system of two plates A and B shown here. Plate A is given a charge Q and plate B is neutral the charge distribution on plates is as shown in figure. If the switch S is now closed the total charge on outer surface of the system of plates after earthing should become zero hence whole charge on plate A will transfer to its inner surface and hence on the inner surface of plate B an equal and opposite charge –Q is developed which is given by earth as shown in figure – – Q – – 2 – – – – – – – – –

Q + + Q + 2 + + + + + + + + + A

d

+Q

– + + Q Q – – – + 2 2 – + P – + – + – + – + – + – + – + – + B

+ + + + + + + + + + + +

S

→ E

d

–Q – – – – – P – – – – – – –

S

(b)

(a)

If area of plates is A, the electric field between the system of plates can be given as Q Ef = A ∈0 Before earthing this electric field was

Ei =

E Q = f 2A ∈0 2

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ELECTROSTATICS - 2

Page # 42

Thus just after earthing the electric field between the plates is doubled and the potential difference between the two plates will also be doubled. as plate B is earthed, its potential is zero. The potential of plate. A can be given as Q VA = A ∈ d 0 Now consider another example shown in figure. In a system of three parallel plates A, B and C the middle plate B is given a charge Q due to which charges are induced on plates A and C as shown. On the basis of discussion done in the previous section we can say that if switch S1 is closed whole charge of plate B will shift on its left surface and a charge –Q is flown through S1 toward plate A and final situation will be as shwon in figure (a) + + + Q + + 2 + + + + + + + +

S1

–Q

S1

A

– – – – – – – – – – – –

d1

+ + + + + + + + + + + + +

A

+ – + – d1 + – – Q Q+ + + – – 2+ – 2 → + – E + – + – + – + – + – +

Q

B

+ – – + – d2 – + – + Q Q – – + – + 2 – 2 + – – + – + – – + – + – – + – – + – +

Q

+ + + ++ Q + 2 + + + + + + + + C S2

Q – – – – – – – – – – – –

d2

B

(a)

C

S2

S1

(A)

d1

Q + + + + + + + + + + + + + (B)

–Q d2

(C)

S2

(b)

If instead of switch S1, S2 is closed in the beginning the distribution of charges on the system will be obviously as shown according to the figure (b) and a charge –Q now flows through switch S2 from earth to plate C. If we close both the switches simultaneously, the situation will be according to shown in figure. Now the charge on plate B is distributed on the two surface as shown and equal and oppsoite charges –q1 and –q2 are developed on the inner surfaces of plates A and C.

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ELECTROSTATICS - 2

Page # 43

Here charges q1 and q2 can be calculated by equating the potential difference of plates A and B and C and B as VB – VA = VB – VC –q1 q1 q2 –q2 + + – Here the electric field between plates can be given as – Between plates A and B

– – – – – – – – – – –

q1 E1 = A ∈0

Between plates B & C

q2 E2 = Aε 0

Now we have

VBA = VBC

or And we have

q2 q1 d1 = A ∈ d2 0 A ∈0 q1d1 = q2d2 q1 + q2 = Q

Thus on solving we get

q1 =

Qd2 d1 + d2

and

q2 =

Qd1 d1 + d2

S1

A

← E1

d1

+ + + + + + + + + + + +

B

+ + + + + + + + + + + +

→

E2

d2

– – – – – – – – – – –

C

S2

Thus if both the switches are closed simultaneously, charges –q1 and –q2 will flow through the switches S1 and S2 from each of plates A and C.

Ex.18 When a charge is given to a conducting plate, the charge distributes itself on two surface. + + + + σ + + 2ε 0 = σ + + + σ ε0 + + 2 ε 0 + +

+ + σ
2ε 0 is the E due to a single layer of charge but as in the case of conducting sheet there is generation

of two surfaces or two layers of charges.

σ ∴ electric field outside the conducting plate is ε 0

Ex.19 When a charge Q is given to a non conducting plate & conducting plate. Find the ratio of electric field produced by them ? Sol. For non conducting plate For conducting plate Q Q + + + Q Q + σ= Q + + σ = 2A A + + + 2Aε0 + =1:1 + + Ratio Q σ Q +
+ + ε= = + E= σ = Q 2Aε0 ε 0 2Aε 0 + + 2ε 0 2Aε 0 + + +

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ELECTROSTATICS - 2

Page # 44 13.

(A) TOTAL ELECTROSTATIC ENERGY OF A SYSTEM OF CHARGES : Total electrostatic potential energy of system of charges can be given as U = ∑ self energy of all charged bodies + ∑ Interaction energy of all pairs of charged bodies Let us consider some cases to understand this concept. Figure shows two uniformly charged non and R2 and charged with charges Q1 and Q2 respectively separated by 1 a distance r. If we find the total electrostatic energy of this system, we can write as

c o

n

d

u

c t i n

g

s p

h

e

r e

s

o

f

r a

d

i u

s

R

r

U = Uself + Uinteraction

+

3KQ12 3KQ22 KQ1Q2 U= + + 5R1 5R 2 r

+

+

+

+

+ +

+ +

+ + + + + + R1 + + +

+

+

+ + + + + + + + + R2 + + +

+

Q

+

Q1

(B) ELECTROSTATIC ENERGY OF A SYSTEM OF CONCENTRIC SHELLS : Figure shown two concentric shells of radii a and b charged uniformly with charges q1 and q2. Here the total energy of this system can be given as Utotal = self energy of inner shell + self energy of outer shell + interaction energy of the two shell =

a

Kq12 Kq22 Kq1q2 + + 2a 2b b

q2

q1 b

Alternative Method : We know that the total electrostatic energy of a system in stored in the form of field energy of the system hence here we can calculate the total electrostatic energy of the system by integrating the field energy density in the space surrounding the shells where electric field exist.

b

Total field energy in the electric field associated with the system shown in figure can be given as b

U=

2

1

∫2∈

0

0

 Kq1  2  2  4πx dx +  x 

∞

1

0

 K(q1 + q2 )  2   4πx dx r2  

dx

2 2 2 2 2 Kq12 Kq22 Kq1q2 1 Kq1 Kq1 Kq1 Kq2 Kq2 – + + + + + = 2 a 2b b 2b b 2b 2b b

Ex.20 Figure shows a shell of radius R having charge q1 uniformly distributed over it. A point charge q is placed at the centre of shell. Find work required to increase radius of shell from R to R1 as shown in fiugre(b) Sol.

dx

x

1 2 1 1  1 2 1  = Kq1  –  + K(q1 + q2 )   2 a b 2 b 

=

x

2

∫2∈ 0

a

R1

R

R

+q q1

+q

Work = Ui – Uf

U1 = SEq + SEq1 + IE = SEq +

Kq12 Kq1q + 2R R

q1 (a)

(b)

2 1

Kq Kq1q + Uf = SEq + 2R1 R1

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ELECTROSTATICS - 2

Page # 45 Kq12 Kq1q Kq12 Kq1q + – – = Ui – Uf = 2R R 2R1 R1

work done

(Try this problem yourself using the energy density formula)

Ex.21 A point charge q = 3 µC is located at the centre of the spherical layer of uniform isotropic dielectric with relative permittivity k = 3. The inside radius of the layer is equal to a = 250 mm and the outside radius is b = 500 mm. Find the electrostatic energy inside the dielectric layer. k=a Sol. Consider a small elemental shell of thickness dx. Volume = dV = 4πx2dx Electric field at x =

Kq x2k

1 1 K2q2 2 electric energy density = ∈0 E = ∈0 4 2 2 2 xk Thus energy content in the element shell is

a x

q

b

dx

2 2

= dE =

1 K q ∈0 4 2 4πx2dx 2 xk

b

E=

=

4 K2 πq2 2 ∈ ∫a 2 0 k2x4 x dx

q2 K 2 R2

b

1

∫x

2

Kq2  1 1  Kq2  1 1  – = – 2R 2  b a  2R 2  b a 

dx =

a

Ex.22 Find the electrostatic energy stored in a cylindircal shell of length , inner radius a and outer radius b, coaxial with a uniformly charged wire with linear charge density λ C/m. Sol.

For this we consider an elemental shell of radius x and width dx. The volume of this shell dV can be given as dV = 2πx.dx The electric field due to the wire at the shell is E=

2Kλ x

The electrostatic field energy stored in the volume of this shell is dU =

1 ∈0 E2.dV 2 2

1  2Kλ  ∈0  or  .2πxλ.dx 2  x  The total electrostatic energy stored in the above mentioned volume can be obtained by integrating the above expression within limits from a to b as dU =

a

U = ∫ dU =

or

λ2  U= 4π ∈0

or

λ2  U= 4π ∈0

2

1

∫2∈

0

b b

 2Kλ    2πx.dx  x 

1

∫ x dx a b

1

∫ x dx a

2

or

U=

λ  b  n   4π ∈0 a

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Page # 46

ELECTROSTATICS - 2

(Objective Problems)

Exercise - I 1. In a region of space, the electric field is in the x
direction and is given as E = E0 xi . Consider an imaginary cubical volume of edge a, with its edges parallel to the axes of coordinates. The charge inside this volume is 1 3 1 2 (A) zero (B) ε 0E0 a3 (C) ε E0 a (D) ε 0E0 a 6 0 Sol.

Sol.

4. The volume charge density as a function of distance X from one face inside a unit cube is varying as shown in the figure. Then the total flux (in S.I. units) through the cube if (ρ0 = 8.85 × 10–12 C/m3) is : density ρ0

2. Electric flux through a surface of area 100 m2 lying
in the xy plane is (in V-m) if E = i + 2 j + 3 k (A) 100 Sol.

(B) 141.4

(C) 173.2

3. An infinite, uniformly charged sheet with surface charge density σ cuts through a spherical Gaussian surface of radius R at a distance x from its center, as shown in the figure. The electric flux Φ through the Gaussian surface is

(D) 200

R

1/4

(A) 1/4 Sol.

(B) 1/2

x 3/4 1 (in m)

(C) 3/4

(D) 1

x

(A)

πR2 σ ε0

(B)

2π (R2 − x2 )σ ε0

(C)

π(R − x ) 2 σ ε0

(D)

π (R2 − x2 )σ ε0

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Page # 47

ELECTROSTATICS - 2 5. Figure shows two large cylindrical shells having uniform linear charge densities +λ and –λ. Radius of inner cylinder is 'a' and that of outer cylinder is 'b'. A charged particle of mass m, charge q revolves in a circle of radius r, Then its speed 'v' is : (Neglect gravity and assume the radii of both the cylinders to be very small in comparison to their length.)

–λ

7. A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 V. The potential at the centre of the sphere is (A) 0 V (B) 10 V (C) same as at point 5 cm away from the surface out side sphere (D) same as a point 25 cm away from the surface Sol.

+λ

r v a b

(A)

λq 2π ∈0 m

(B)

2λq π ∈0 m

(C)

λq π ∈0 m

(D)

λq 4 π ∈0 m

Sol.

Sol.

8. Two spherical, nonconducting, and very thin shells of uniformly distributed positive charge Q and radius d are located a distance 10d from each other. A positive point charge q is placed inside one of the shells at a distance d/2 from the center, on the line connecting the centers of the two shells, as shown in the figure. What is the net force on the charge q ? Q

Q d

o

}

d/2

6. An uncharged sphere of metal is placed in a uniform electric field produced by two large conducting parallel plates having equal and opposite charges, then lines of force look like :

+ + + + ++ +

+ + + ++

10 d

qQ 36 lπε 0 d

2

to the left

(B)

36 lπε 0 d2

(C)

36 lπε 0 d2

to the left

(D)

36 lπε 0 d2

362 qQ

Sol. (B)

(A)

– – – ––

–

– –

(D) – –

–

– –

–

–

(C)

+

+

+ +

+

+

+

+ +

+

–––––––

qQ

(A)

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360 qQ

to the right

to the right

Page # 48

ELECTROSTATICS - 2

9. Potential difference between centre & the surface of sphere of radius R and uniform volume charge density ρ within it will be (A)

ρR2 6 ∈0

(B)

ρR2 4 ∈0

(C) 0

(D)

ρR2 2 ∈0

Sol. 12. n small drops of same size are charged to V volts each. If they coalesce to form a signal large drop, then its potential will be (A) V/n (B) Vn (C) Vn1/3 (D) Vn2/3 Sol.

A solid sphere of radius R is charged uniformly. At what distance from its surface is the electrostatic potential half of the potential at the centre ? (A) R (B) R/2 (C) R/3 (D) 2R Sol. 1

0

.

13. 1000 identical drops of mercury are charged to a potential of 1 V each. They join to form a single drop. The potential of this drop will be (A) 0.01 V (B) 0.1 V (C) 10 V (D) 100 V Sol.

11. Two similar conducting spherical shells having charges 40 µC and –20µC are some distance apart. Now they are touched and kept at same distance. The ratio of the initial to the final force between them is : (A) 8 : 1

(B) 4 : 1

(C) 1 : 8

(D) 1 : 1

Sol.

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Page # 49

ELECTROSTATICS - 2 14. A bullet of mass m and charge q is fired towards a solid uniformly charged sphere of radius R and total charge +q. If it strikes the surface of sphere with speed u, find the minimum speed u so that it can penetrate through the sphere. (Neglect all resistance forces or friction acting on bullet except electrostatic forces)

Sol.

q

+q m

u R

q

(A)

2πε 0 mR

q

(B)

q

(C)

8 πε 0 mR

(D)

4 πε 0 mR 3q 4 πε 0 mR

Sol.

16. A positively charged body ‘A’ has been brought near a neutral brass sphere B mounted on a glass stand as shown in the figure. The potetial of B will be: B

+ ++ + + + ++ ++ + + A ++

(A) Zero (C) Positive Sol.

15. A unit positive point charge of mass m is projected with a velocity V inside the tunnel as shown. The tunnel has been made inside a uniformly charged nonconducting sphere. The minimum velocity with which the point charge should be projected such it can it reach the opposite end of the tunnel, is equal to R/2

(A) [ρR2/4mε0]1/2 (B) [ρR2/24mε0]1/2 (C) [ρR2/6mε0]1/2 (D) zero because the initial and the final points are at same potential.

(B) Negative (D) Infinite

17. A charge ‘q’ is placed at the centre of a conducting spherical shell of radius R, which is given a charge Q. An external charge Q′ is also present at distance R′ (R′ > R) from ‘q’. Then the resultant field will be best represented for region r < R by : [where r is the distance of the point from q] Q

q

R’

R

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Q’

Page # 50

ELECTROSTATICS - 2

E

E

(A)

(B) R

R r

r

(A) Force on Q due to E is zero (B) Net force on Q is zero (C) Net force acting on Q and conducting shell considered as a system is zero (D) Net force acting on the shell due to E is zero. Sol.

E

E

(C)

(D) R r

R r

Sol.

20. The net charge given to an isolated conducting solid sphere : (A) must be distributed uniformly on the surface (B) may be distributed uniformly on the surface (C) must be distributed uniformly in the volume (D) may be distributed uniformly in the volume. Sol.

18. In the above questins, if Q’ is removed then which option is correct :

E

E

(A)

(B) R

R r

r

21. The net charge given to a solid insulating sphere: (A) must be distributed uniformly in its volume (B) may by distributed uniformly in its volume. (C) must be distributed uniformly on its surface. (D) the distribution will depend upon whether other charges are present or not. Sol.

E

E

(C)

(D) R r

R r

Sol.

19. A positive point charge Q is kept ( as shown in the figure) inside a neutral conducting shell whose centre is at C. An external uniform electric field E is applied. Then : E

22. A dipole having dipole moment p is placed in front of a solid uncharged conducting sphere as shown in the diagram. The net potential at point. A lying on the surface of the sphere is ; A r φ

P

(A) C Q

kp cos φ r2

(C) zero

(B)

kp cos 2 φ r2

(D)

2kp cos 2 φ r2

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Page # 51

ELECTROSTATICS - 2

25. Both question (a) and (b) refer to the system of charges as shown in the figure. A spherical shell with an inner radius ‘a’ and an outer radius ‘b’ is made of conducting material. A point charge +Q is placed at the centre of the spherical shell and a total charge – q is placed on the shell.

Sol.

b Q a –q

23. Three concentric conducting spherical shells carry charges as follows +4Q 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) 0 Sol.

(B) 4 Q

(C) –Q

(D) –2Q

(i) charge –q is distributed on the surfaces as (A) –Q on the inner surface, – q on outer surface (B) –Q on the inner surface, – q + Q on the outer surface (C) +Q on the inner surface, – q – Q on the outer surface (D) The charge –q is spread uniformly between the inner and outer surface Sol.

(ii) Assume that the electrostatic potential is zero at an infinite distance from the spherical shell. The electrostatic potential at a distance R(a < R < b) from the centre of the shell is 24. Three concentric metallic spherical shell A, B and C or radii a, b and c (a < b < c) have surface charge densities –σ, +σ, and –σ respectively. The potential of shell A is (A) (σ/ε0)[a + b – c] (B) (σ/ε0)[a – b + c] (C) (σ/ε0)[b – a – c] (D) none Sol.

(A) 0 (D) K

(B)

KQ a

Q−q 1 (where K = ) b 4 πε 0

Sol.

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(C) K

Q−q R

Page # 52

ELECTROSTATICS - 2

26. A positive charge q is placed in a spherical cavity made in a positively charged sphere. The centres of sphere and cavity are displaced by a small distance
l . Force on charge q is :
(A) in the direction parallel to vector l (B) in radial direction (C) in a direction which depends on the magnitude of charge density in sphere (D) direction can not be determined Sol.

27. If the electric potential of the inner metal sphere is 10 volt & that of the outer shell is 5 volt, then the potential at the centre will be a b

(A) 10 volt Sol.

(B) 5 volt

(C) 15 volt

Sol.

29. Two uni form l y c harg ed non-c onduct i ng hemispherical sheels each having uniform charge σ and radius R form a complete sphere (not stuck together) and surround a concentric spherical conducting shell of radius R/2. If hemispherical parts are in equilibrium then minimum surface charge density of inner conducting shell is : (B) –σ/2 (C) –σ (D) 2σ (A) –2σ Sol. d

e

n

s i t y

(D) 0

28. An infinite number of concentric rings carry a charge Q each alternately positive and negative. Their radii are 1, 2, 4, 8... meters in geometric progression as shown in the figure. The potential at the centre of the rings will be

30. A point charge q is borught from infinity (slowly so that heat developed in the shell is negligible) and is placed at the centre of a conducting neutral spherical shell of inner radius a and outer radius b, then work done by external agent is:

b

4 3 Q 2 Q Q 1

(A) zero

Q (B) 12πε 0

Q (C) 8 πε 0

q

Q (D) 6 πε 0

(A) 0

(B)

k q2 2b

(C)

a

k q2 k q2 k q2 k q2 − − (D) 2b 2a 2a 2b

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Page # 53

ELECTROSTATICS - 2 Sol.

Sol.

31. A charge Q is kept at the centre of a conducting sphere of inner radius R1 and outer radius R2. A point charge q is kept at a distance r ( > R2) from the centre. If q experiences an electrostatic force 10 N then assuming that no other charges are present, electrostatic force experienced by Q will be : (A) – 10 N (B) 0 (C) 20 N (D) none of these Sol.

33. Two identical conducting spheres, having charges of opposite sign, attract each other with a force of 0.108 N when separated by 0.5 m. The spheres are connected by a conducting wire, which is then removed, and thereafter, they repel each other with a force of 0.036 N. The initial charges on the spheres are (A) ± 5 × 10–6 C and ∓ 15 × 10–6 C (B) ± 1.0 × 10–6 C and ∓ 3.0 × 10–6 C (C) ± 2.0 × 10–6 C and ∓ 6.0 × 10–6 C (D) ± 0.5 × 10–6 C and ∓ 1.5 × 10–6 C Sol.

32. 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 (>a). What is the electric field at a distance r (a < r < b) from the centre ? (A)

1 Q 4πε0 r

(B)

1 3Q 4πε0 r

(C)

1 3Q 4πε0 r2

(D)

1 Q 4πε0 r2

34. Two small conductors A and B are given charges q1 and q2 respectively. Now they are placed inside a hollow metallic conductor (C) carrying a charge Q. If all the three conductors A, B and C are connected by a conducting wire as shown, the charges on A, B and C will be respectively.

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Page # 54

ELECTROSTATICS - 2 Q

36. You are travelling in a car during a thunder storm. In order to protect yoursefl from lightening, would you prefer to : (A) Remain in the car (B) Take shelter under a tree (C) Get out and be flat on the ground (D) Touch the nearest electrical pole Sol.

C

A q1

q2

B

(A)

q1 + q2 q1 + q2 , ,Q 2 2

(B)

Q + q1 + q3 Q + q1 + q2 Q + q1 + q2 , , 3 3 3

q1 + q2 + Q q1 + q2 + Q , ,0 2 3 (D) 0, 0, Q + q1 + q2 (C)

Sol. COMPREHENSION A solid conducting sphere of radius ‘a’ is surronded by a thin uncharged concentric conducting shell of radius 2a. A point charge q is placed at a distance 4a from common centre of conducting sphere and shell. The inner sphere is then grounded.

2a

4a

a 35. There are four concentric shells A, B, C and D of radii a, 2a, 3a and 4a respectively. Shells B and D are given charges +q and –q respectively. Shell C is now earthed. The potential difference VA – VC is : (A)

Kq 2a

(B)

Kq 3a

(C)

Kq 4a

(D)

Kq 6a

q

37. The charge on solid sphere is : (A) –

q 2

(B) –

q 4

(C) –

q 8

(D) –

q 16

Sol.

Sol.

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Page # 55

ELECTROSTATICS - 2 38. Pick up the correct statement : (A) Charge on surface on inner sphere is non-uniformly distributed (B) Charge on inner surface of outer shell in nonuniformly distributed. (C) Charge on outer surface of outer shell is nonuniformly distributed. (D) All the above statement are false. Sol.

REASONING TYPE QUESTION 40. Statement - 1 : If a concentric spherical Gaussian surface is drawn inside thin spheical shell of charge, electric field (E) at each point of surface must be zero. Statement - 2 : In accordance with Gauss's law φE =



Qnet enclosed E.dA = ∫ ε0

Qnet enclosed = 0 implies φE = 0 (A) Statement - 1 is true, Statement - 2 is true and statement - 2 is correct explanation for statement - 1. (B) Statement - 1 is true, Statement - 2 is true and statement - 2 is NOT correct explanation for statement - 1. (C) Statement - 1 is true, statement - 2 is false. (D) Statement - 1 is false, statement - 2 is true. Sol.

39. The potential of outer shell is : q (A) 32 πε a 0

q (B) 16 πε a 0

q (C) 8 πε a 0

q (D) 4 πε a 0

Sol.

41. Statement - 1 : Electric field of a dipole can't be found using only Gauss law. (i.e. without using superposition principle) Statement - 2 : Gauss law is valid only for symmetrical charge distribution (A) Statement - 1 is true, Statement - 2 is true and statement - 2 is correct explanation for statement - 1. (B) Statement - 1 is true, Statement - 2 is true and statement - 2 is NOT correct explanation for statement - 1. (C) Statement - 1 is true, statement - 2 is false. (D) Statement - 1 is false, statement - 2 is true. Sol.

42. Statement - 1 : In a given situation of arrangement of charges, an extra charge is placed outside the Gaussian surface. In the Gauss Theorem

Q in ∫ E.dA = ∈0
Qin remains unchanged whereas electric field E at the site of the element is changed.
Statement - 2 : Electric field E at any point on the Gaussian surface is due to inside charge only.

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Page # 56

ELECTROSTATICS - 2

(A) Statement - 1 is true, Statement - 2 is true and statement - 2 is correct explanation for statement - 1. (B) Statement - 1 is true, Statement - 2 is true and statement - 2 is NOT correct explanation for statement - 1. (C) Statement - 1 is true, statement - 2 is false. (D) Statement - 1 is false, statement - 2 is true. Sol.

43. Statement - 1 : The flux crossing through a closed surface is independent of the location of encloses charge. Statement - 2 : Upon the displacement of charges
within a closed surface, the E at any point on surface does not change. (A) Statement - 1 is true, Statement - 2 is true and statement - 2 is correct explanation for statement - 1. (B) Statement - 1 is true, Statement - 2 is true and statement - 2 is NOT correct explanation for statement - 1. (C) Statement - 1 is true, statement - 2 is false. (D) Statement - 1 is false, statement - 2 is true. Sol.

45. When two charged concentric spherical conductors have electric potential V1 and V2 respectively Statement - 1 : The potential at centre is V1 + V2 Statement - 2 : Potential is scalar quantity. (A) Statement - 1 is true, Statement - 2 is true and statement - 2 is correct explanation for statement - 1. (B) Statement - 1 is true, Statement - 2 is true and statement - 2 is NOT correct explanation for statement - 1. (C) Statement - 1 is true, statement - 2 is false. (D) Statement - 1 is false, statement - 2 is true. Sol.

46. Statement - 1 : A point charge q is placed inside a cavity of conductor as shown. Another point charge Q is placed outside the conductor as shown. Now as the point charge Q pushed away from conductor, the potential difference (VA – VB) between two point A and B within the cavity of sphere remains constant. Statement - 2 : The electric field due to charges on outer surface of conductor and outside the conductor is zero at all points inside the conductor.

A

B q

44. The electrostatic potential on the surface of a charged solid conducting sphere is 100 volts. Two statements are made in this regard Statement - 1 : At any point inside the sphere, electrostatic potetial is 100 volt. Statement - 2 : At any point inside the sphere, electric field is zero. (A) Statement - 1 is true, Statement - 2 is true and statement - 2 is correct explanation for statement - 1. (B) Statement - 1 is true, Statement - 2 is true and statement - 2 is NOT correct explanation for statement - 1. (C) Statement - 1 is true, statement - 2 is false. (D) Statement - 1 is false, statement - 2 is true. Sol.

Q

(A) Statement - 1 is true, Statement - 2 is true and statement - 2 is correct explanation for statement 1. (B) Statement - 1 is true, Statement - 2 is true and statement - 2 is NOT correct explanation for statement - 1. (C) Statement - 1 is true, statement - 2 is false. (D) Statement - 1 is false, statement - 2 is true. Sol.

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Page # 57

ELECTROSTATICS - 2

EXERCISE - II 1. Units of electric flux are (A)

N – m2 2

Coul (C) volt-m Sol.

(B)

N

Coul2 – m 2 (D) Volt-m3

2. An electric dipole is placed at the centre of a sphere. Mark the correct answer (A) the flux of the electric field through the sphere is zero (B) the electric field is zero at every point of the sphere.Ex

4. Mark the correct options (A) Gauss’s law is valid only for uniform charge distributions. (B) Gauss’s law is valid only for charges placed in vacuum. (C) The electric field calculated by Gauss’s law is the field due to all the charges. (D) The flux of the electric field through a closed surface due to all the charges is equal to the flux due to the charges enclosed by the surface. Sol.

(C) the electric potential is zero everywhere on the sphere. (D) the electric potential is zero on a circle on the surface. Sol.

5. Charges Q 1 and Q 2 l ies i nsi de and outsi de respectively of a closed surface S. Let E be the field at any point on S and φ be the flux of E over S. (A) If Q1 changes, both E and φ will change. (B) If Q2 changes, E will change but φ will not change. (C) If Q1 = 0 and q2 ≠ 0 then E ≠ 0 but φ = 0. (D) If Q1 ≠ 0 and Q2 = 0 then E = 0 but φ ≠ 0. Sol.

3. Which of the following statements are correct? (A) Electric field calculated by Gauss law is the field due to only those charges which are enclosed inside the Gaussian surface. (B) Gauss law is applicable only when there is a symmetrical distribution of charge. (C) Electric flux through a closed surface will depends only on charges enclosed within that surface only. (D) None of these Sol.

6. An electric field converges at the origin whose magnitude is given by the expression E = 100rNt/Coul, where r is the distance measured from the origin. (A) total charge contained in any spherical volume with its centre at origin in negative. (B) total charge contained at any spherical volume, irrespective of the location of its centre, is negative. (C) total charge contained in a spherical volume of radius 3 cm with its centre at origin has magnitude 3 × 10–13C. (D) total charge contained in a spherical volume of radius 3 cm with its centre at origin has magnitude 3 × 10–9 Coul.

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Page # 58

ELECTROSTATICS - 2

Sol.

7. A conducting sphere of radius r has a charge. Then (A) The charge is uniformly distributed over its surface, if there is an external electric field. (B) Distribution of charge over its surface will be non unifrom if no external electric field exist in space. (C) Electric field strength inside the sphere will be equal to zero only when no external electric field exists. (D) Potential at every point of the sphere must be same Sol.

8. For a spherical shell (A) If potential inside it is zero then it necessarily electrically neutral (B) electric field in a charged conducting spherical shell can be zero only when the charge is uniformly distributed (C) electric potential due to induced charges at a point inside it will always be zero (D) none of these Sol.

9. At distance of 5cm and 10cm outwards from the surface of a uniformly charged solid sphere, the potentials are 100V and 75V respectively. Then (A) potential at its surface is 150V (B) the charge on the sphere is (5/3) × 10–10 C (C) the electric field on the surface is 1500 V/m (D) the electric potential at its centre is 225 V Sol.

10. A thin-walled, spherical conducting shell S of radius R is given charge Q. The same amount of charge is also placed at its centre C. Which of the following statements are correct ? (A) On the outer surface of S, the charge density is Q . 2πR2 (B) The electric field is zero at all points inside S. (C) At a point just outside S, the electric field is double the field at a point just inside S. (D) At any point inside S, the electric field is inversely proportional to the square of its distance from C.

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Page # 59

ELECTROSTATICS - 2 Sol.

12. A and B are two conducting concentric spherical shells. A is given a charge Q while B is uncharged. If now B is earthed as shown in figure. Then : B

++

++

++

++ A

(A) The charge appearing on inner surface of B is –Q (B) The field inside the outside A is zero. (C) The field between A and B is not zero. (D) The charge appearing on outer surface of B is zero. Sol.

11. A hollow closed conductor of irregular shape is given some charge. Which of the following statements are correct ? (A) The entire charge will appear on its outer surface. (B) All points on the conductor will have the same potential (C) All points on its surface will have the same charge density. (D) All points near its surface and outside it will have the same electric intensity. Sol.

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Page # 60

ELECTROSTATICS - 2

(SUBJECTIVE

Exercise - III 1. What do you predict by the given statement about the nature of charge (positive or negative) enclosed by the close surface. "In a close surface lines which are leaving the surface are double then the lines which are entering in it." Sol.

PROBLEMS)

Sol.

2. The length of each side of a cubical closed surface is l. If charge q is situated on one of the vertices of the cube, then find the flux passing through shaded face of the cube.

q

Sol.

4. A charge Q is uniformly distributed over a rod of length l. Consider a hypothetical cube of edge l with the centre of the cube at one end of the rod. Find the minimum possible flux of the electric field through the entire surface of the cube. Sol.

3. A point charge Q is located on the axis of a disc of radius R at a distance a from the plane of the disc. If one fourth (1/4th) of the flux from the charge passes through the disc, then find the relation between a & R. R a Q

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Page # 61

ELECTROSTATICS - 2 5. A very long uniformly charged thread oriented along the axis of a circle of radius R rests on its centre with one of the ends. The charge on the thread per unit length is equal to λ. Find the flux of the vector E through the circle area. Sol.

Sol.

8. There are two concentric metal shells of radii r1 and r2 (> r1). If initially the outer shell has a charge q and the inner shell is having zero charge. Now inner shell is grounded. Find : (i) Charge on the inner surface of outer shell. (ii) Final charges on each sphere. (iii) Charge flown through wire in the ground. Sol.

6. A particle of mass m and charge –q moves along a diameter of a uniformly charged sphere of radius R and carrying a total charge +Q. Find the frequency of S.H.M. of the particle if the amplitude does not exceed R. Sol.

9. A point charge ‘q’ is within an electrically neutral conducting shell whose other surface has spherical shape. Find potential V at point P lying outiside shell at a distance ‘r’ from centre O of outer sphere.

O q

Sol.

7. There are 27 drops of a conducting fluid. Each has radius r and they are charged to a potential V0. They are then combined to form a bigger drop. Find its potential.

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r

P

Page # 62

ELECTROSTATICS - 2

10. Consider two concentric conducting spheres of radii a & b (b > a). Inside sphere has a positive charge q1. What charge should be given to the outer sphere so that potential of the inner sphere becomes zero? How does the potential varies between the two spheres & outside ? 12. Consider three identical metal spheres A, B and C. Spheres A carries charge +6q and sphere B carries charge –3q. Sphere C carries no charge. Spheres A and B are touched together and then separated. Sphere C is then touched to sphere A and separated from it. Finally the sphere C is touched to sphere B and separated from it. Find the final charge on the sphere C. Sol.

Sol.

11. Two thin conducting shells of radii R and 3R are shown in figure. The outer shell carries a charge +Q and the inner shell is neutral. The inner shell is earthed with the help of switch S. Find the charge attained by the inner shell. +Q

3R R

S

A metal sphere of radius r1 charged to a potential V1 is than placed in a thin-walled uncharged conducting spherical shell of radius r2. Determine the potential acquired by the spherical shell after it has been connected for a short time to the spher by a conductor. 1

Sol.

3

.

r2

r1 φ1

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Page # 63

ELECTROSTATICS - 2 Sol.

E 14. Two thin conducting plates (very large) parallel to each other carrying tota l char ge s σ A and –2σ A respectively (where A is the area of each plate), are placed in a uniform external electric field E as shown. Find the surface charge on each surface.

σA

−2σA

Sol.

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Page # 64

ELECTROSTATICS - 2

(TOUGH

Exercise - IV

1. A positive charge Q is uniformly distributed throughout the volume of a dielectric sphere of radius R. A point mass having charge +q and mass m is fired towards the centre of the sphere with velocity v from a point at distance r (r > R) from the centre of the sphere. Find the minimum velocity v so that it can penetrate R/2 distance of the sphere. Neglect any resistance other than electric interaction. Charge on the small mass remains constant throughout the motion.

2. A cavity of radius r is present inside a solid dielectric sphere of radius R, having a volume charge density of ρ. The distance between the centres of the sphere and the cavity is a. An electron e is kept inside the cavity at an angle θ = 45° as shown. How long will it take to touch the sphere again ?

SUBJECTIVE PROBLEMS )

4. A solid non conducting sphere of radius R has a non-uniform charge distribution of volume charge r density, ρ = ρ0 , where ρ0 is a constant and r is the R distance from the centre of the sphere. Show that (a) the total charge on the sphere is Q = πρ0R3 and (b) the electric field inside the sphere has a magnitude given by, E =

KQr 2

. R4 5. An electron beam after being accelerated from rest through a potential difference of 500 V in vacuum is allowed to impinge normally on a fixed surface. If the incident current is 100 µA, determine the force exerted on the surface assuming that it brings the electrons to rest. (e = 1.6 × 10–19 C; m = 9.0 × 10–31 kg)

6. A cone made of insulating material has a total charge Q spread uniformly over its sloping surface. Calculate the energy required to take a test charge q from infinity

r

e

a

to apex A of cone. The slant length is L. A

3. Figure shows a section through two long thin concentric cylinders of radii a & b with a < b. The cylinders have equal and opposite charges per unit length λ. Find the electric field at a distance r from the axis for -

AB=L

B

7. Two concentric rings, one of radius ‘a’ and the other of radius ‘b’ have the charges +q and –(2/5)–3/2 q respectively as shown in the figure. Find the ratio b/a if a charge particle placed on the axis at z = a is in equilibrium.

b

a –3/2

b

(A) r < a

(B) a < r < b

(C) r > b

a

qB=–(2/5) q

z=a

qA=+q

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Page # 65

ELECTROSTATICS - 2

Exercise - V

(JEE-PROBLEMS)


1. The magnitude of electric field E in the annular region of charged cylindrical capacitor (A) Is same throughout [IIT-96,2] (B) Is higher near the outer cylinder than near the inner cylinder (C) Varies as (1/r) where r is the distance from the axis (D) Varies as (1/r2) where r is the distance from the axis Sol.

2. A conducting sphere S1 of radius r is attached to an insulating handle. Another conducting sphere S2 of radius R is mounted on an insulating stand. S2 is initially uncharged. S1 is given a charge Q, brought into contact with S2 & removed, S1 is recharged such that the charge on it is again Q & it is again brought into contact with S2 & removed. This procedure is repeated n times. [IIT-98] (a) Find the electrostatic energy of S2 after n such contacts with S1 (b) What is the limiting value of this energy as n → ∞ ? Sol.

3. A n ellipsoidal cavity is carved within a perfect conductor. A positive charge q is placed at the center of the cavity. The points A & B are on the cavity [IIT-99,3] surface as shown in the figure. Then

A q

B

(A) electric field near A in the cavity = electric field near B in the cavity (B) charge density at A = charge density at B (C) potential at A = potential at B (D) total electric field flux through the surface of the ε0 Sol. c a

v

i t y

i s

q

/

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Page # 66

ELECTROSTATICS - 2

(ii)A non-conducting disc of radius a and uniform positive surface charge density σ is placed on the ground, with its axis vertical. A particle of mass m & positive charge q is dropped, along the axis of the disc, from a height H with zero initial velocity. The q 4ε 0 g = particle has [IIT-99] m σ (i) Find the value of H if the particle just reaches the disc. (ii) Sketch the potential energy of the particle as a function of its height and find its equilibrium position. Sol.

4. A point charge ‘q’ is placed at a point inside a hollow conducting sphere. Which of the following electric force pattern is correct ?

(A)

(B)

(C)

(D)

Sol.

5. Three large parallel plates have uniform surface charge densities as shown in the figure. What is the [JEE’ 2005 (Scr)] electric field at P.

P

4σ  (A) – ∈ k 0

4σ  (B) ∈ k 0

2σ  (C) – ∈ k 0 Sol.

2σ  (D) ∈ k 0

z=a z=–a z=–2a

6. A conducting liquid bubble of radius a and thickness t (t C ...(i) For (r2)min, r1 should be maximum and i for (r1)max ⇒ imax = 90° r1 r2

(r1 )max = C

(r2 )max = A – C Now from eq. (i) A – C > C C A A > 2C B i.e. A > 2C, all rays are reflected back from the second surface. (b)

The relation between A & C such that ray will always A cross surface BC. For this (r2)max < C i r1 r2 (A – r1)max < C A – (r1)min < C ..(2) (r1)min = 0 when imin = 0 from eq. (2) A–0 f from the mirror. Its image will have a length (A) f2/(u – f) (B) uf/(u – f) (D) uf /(u + f) (C) f2/(u + f) Sol.

28. A concave mirror gives an image three times as large as the object placed at a distance of 20 cm from it. For the image to be real, the focal length should be : (A) 10 cm (B) 15 cm (C) 20 cm (D) 30 cm Sol. 26. A candle is kept at a distance equal to double the focal length from the pole of a convex mirror. its magnification will be : (A) –1/3 (B) 1/3 (C) 2/3 (D) –2/3 Sol.

27. A boy 2 m tall stands 40 cm in front of a mirror. He sees an erect image, 1 m high. The mirror is : (A) Concave, f = 40 cm (B) Convex, f = 40 cm (C) Plane (D) Either convex or concave Sol.

29. If an object is 30 cm away from a concave mirror of focal length 15 cm, the image will be (A) erect (B) virtual (C) diminished (D) of same size Sol.

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GEOMETRICAL OPTICS

Page # 74 30. What is the distance of a needle from a concave mirror of focal length 10 cm for which a virtual image of twice its height is formed ? (A) 2.5 cm (B) 5 cm (C) 8 cm (D) 9.1 cm Sol.

33.A real inverted image in a concave mirror is represented by (u, v, f are coordinates) v/f v/f

(A)

+1 +1

u/f

(B)

+1 +1

u/f

v/f

v/f +1

+1

(C)

u/f (D)

u/f

Sol.

31. A convex mirror has a focal length f. An object of height h is placed in front of it. If an erect image of height h/n is formed. The distance of the object from the mirror is : (A) n f (B) f /n (C) (n + 1) f (D) (n – 1) f Sol.

32. A concave mirror cannot form : (A) virtual image of virtual object (B) virtual image of a real object (C) real image of a real object (D) real image of a virutal object Sol.

34. Which one of the following statements are incorrect for spherical mirrors. (A) a concave mirror forms only virtual images for any position of real object (B) a convex mirror forms only virtual images for any position of a real object. (C) a convex mirror forms only a virtual diminished image of an object placed between its pole and the focus (D) a concave mirror forms a virtual magnified image of an object placed between its pole and the focus. Sol.

35. The distance of an object from a spherical mirror is equal to focal length of the mirror. Then the image : (A) must be at infinity (B) may be at infinity (C) may be at the focus (D) none Sol.

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GEOMETRICAL OPTICS

Page # 75

36. The largest distance of the image of a real object from a convex mirror of focal length 20 cm can be : (A) 20 cm (B) infinite (C) 10 cm (D) depends on the position of the object. Sol.

37. In the figure shown, the image of a real object is formed at point I. AB is the principal axis of the mirror. The mirror must be:

O A

d1

Sol.

39. A luminous point object is moving along the principal axis of a concave mirror of focal length 12 cm towards it. When its distance from the mirror is 20 cm its velocity is 4 cm/s. The velocity of the image in cm/s at that instant is (A) 6, towards the mirror (B) 6, away from the mirror (C) 9, away from the mirror (D) 9, towards the mirror. Sol.

B d2 > d1 I

(A) concave & placed towards right I (B) concave & placed towards left of I (C) convex and placed towards right of I (D) convex & placed towards left of I. Sol.

38. A point object at 15 cm from a concave mirror of radius of curvature 20 cm is made to oscillate along the principal axis with amplitude 2 mm. The amplitude of its image will be (A) 2 mm (B) 4mm (C) 8mm (D) none

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GEOMETRICAL OPTICS

Page # 76 40. A particle is moving towards a fixed spherical mirror. The image : (A) must move away from the mirror (B) must move towards the mirror. (C) may move towards the mirror. (D) will move towards the mirror, only if the mirror is convex. Sol.

43. The circular boundary of the concave mirror subtends a cone of half angle θ at its centre of curvature. The minimum value of θ for which ray incident on this mirror parallel to the principle axis suffers reflection more than one is θ θ

(A) 30° Sol.

(B) 45°

C

(C) 60°

(D) 75°

41. A point object on the principal axis at a distance 15 cm in front of a concave mirror of radius of curvature 20 cm has velocity 2 mm/s perpendicular to the principal axis. The velocity of image at that instant will be : (A) 2 mm/s (B) 4 mm/s (C) 8 mm/s (D) none of these Sol.

42. The origin of x and y coordinates is the pole of a concave mirror of focal length 20 cm. The x-axis is the optical axis with x > 0 being the real side of mirror. A point object at the point (25 cm, 1 cm) is moving with a velocity 10 cm/s in positive x -direction. The velocity of the image in cm/s is approximately (A) –80i + 8 j

(B) 160 i + 8 j

(C) –160i + 8 j

(D) 160 i – 4 j

44. An object is placed in front of a convex mirror at a distance of 50 cm. A plane mirror is introduced covering the lower half of the convex mirror. If the distance between the object and the plane mirror is 30 cm, it is found that there is no gap between the images formed by the two mirrors. The radius of the convex mirror is: (A) 12.5 cm (B) 25 cm (C) 50 cm (D) 100 cm Sol.

Sol.

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GEOMETRICAL OPTICS

Page # 77

45. In the figure shown find the total magnification after two successive reflections first on M1 & then on M2 f=10cm

Sol.

f=–20cm

M1

M2 10cm

(A) + 1 Sol.

30cm

(B) – 2

(C) + 2

(D) – 1 47. A ray of light is incident on a concave mirror. It is parallel to the principal axis and its height from principal axis is equal to the focal length of the mirror. The ratio of the distance of point B to the distance of the focus from the centre of curvature is (AB is the reflected by) A

f B

(A)

2 3

(B)

3 2

(C)

2 3

(D)

1 2

Sol.

46. In the figure shown if the object ‘O’ moves towards the plane mirror, then the image I (which is formed after successive reflections from M1 & M2 respectively)

O M2 (A) towards right (C) with zero velocity

M1

48. A straight line joining the object point and image point is always perpendicular to the mirror (A) if mirror is plane only (B) if mirror is concave only (C) if mirror is convex only (D) irrespective of the type of mirror. Sol.

(B) towards left (D)cannot be determined

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GEOMETRICAL OPTICS

Page # 78 SECTION (C) : REFRACTION IN GENERAL, REFRACTION AT PLANE SURFACE AND T.I.R. 49. When a wave is refracted : (A) its path must change (B) its amplitude must change (C) its velocity must change (D) its frequency must change Sol.

52. A ray of light passes through a plane glass slab of thickness t and refractive index µ = 1.5. The angle between incident ray and emergent ray will be (A) 0° (B) 30° (C) 45° (D) 60° Sol.

50. A ray incident at a point at an angle of incidence of 60° enters a glass sphere of µ = √3 and it is reflected and refracted at the farther surface of the sphere. The angle between reflected and refracted rays at this surface is (A) 50° (B) 90° (C) 60° (D) 40° Sol. 53. A ray of light moving along the unit vector (– i – 2j) undergoes refraction at an interface of two media, which is the x-z plane. The refractive index for y > 0 is 2 while for y < 0, it is

5 / 2. The unit vector along which the refracted ray moves is : (–3
i – 5
j ) (–4
i – 3
j ) (A) (B) 34 5

(–3 i – 4 j ) (C) (D) None of these 5 Sol.

51. The x-z plane separates two media A and B with refractive indices µ1 and µ2 respectively. A ray of light travels from A and B. Its directions in the two media   are given by the unit vectors, r = a
i + bj
& r = α ˆi + βˆj B

respectively where
i &
j are unit vectors in the x and y directions. Then (A) µ1a = µ2α (B) µ1α = µ2a (C) µ1b = µ2β (D) µ1B = µ2b Sol.

54. How much water should be filled in a container of 21 cm in height, so that it apears half filled (of total height of the container) when viewed from the top of the container ? (Assume near normal incidence and µw = 4/3) (A) 8.0 cm (B) 10.5 cm (C) 12.0 cm (D) 14.0 cm

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GEOMETRICAL OPTICS

Page # 79 57. An under water swimmer is at a depth of 12 m below the surface of water. A bird is at a height of 18 m from the surface of water, directly above his eyes. For the swimmer, the bird appears to be at a distance X from the surface of water. (Refractive index of water is 4/3.) The value of X is (A) 24 m (B) 12 m (C) 18 m (D) 9 m Sol.

Sol.

55. A mark at the bottom of a beaker containing liquid appears to rise by 0.1m. The depth of the liquid is 1m. the refractive index of liquid is : (A) 1.33 (B) 9/10 (C) 10/9 (D) 1.5 Sol.

56. A parallel sided block of glass of refractive index 1.5 which is 36 mm thick rests on the floor of a tank which is filled with water (refractive index = 4/3.) The difference between apparent depth of floor at A & B when seen from vertically above is equal to (A) 2 mm (B) 3mm (C) 4 mm Sol.

A

B

(D) none

58. A concave mirror is placed on a horizontal table, with its axis directed vertically upwards. Let O be the pole of the mirror and C its centre of curvature. A point object is placed at C. Its has a real image, also located at C (a condition called auto-collimation). If the mirror is now filled with water, the image will be : (A) real, and will remain at C (B) real, and located at a point between C and ∞ (C) virtual, and located at a point between C and O (D) real, and located at a point between C and O Sol.

59. A bird is flying 3 m above the surface of water. If the bird is diving vertically down with speed = 6 m/s, his apparent velocity as seen by a stationary fish underwater is (A) 8 m/s (B) 6 m/s (C) 12 m/s (D) 4 m/s Sol.

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GEOMETRICAL OPTICS

Page # 80 60. A beam of light is converging towards a point. A plane parallel plate of glass of thickness t refractive index µ is introduced in the path of the beam. The convergent point is shifted by (assume near normal incidence) :

62. A ray of light is incident on one face of a transparent slab of thickness 15 cm. The angle of incidence is 60°. If the lateral displacement of the ray

µ

O

t

 1 (A) t1 –  away µ  

 (B) t 1 + 

1  away µ

 1 (C) t1 –  nearer µ 

 1 (D) t1 +  nearer µ 

on emerging from the parallel plane is 5 3 cm, the refractive index of the material of the slab is (A) 1.414 (B) 1.532 (C) 1.732 (D) none Sol.

Sol.

61. Given that, velocity of light in quartz = 1.5 × 108 9 8 m/s and velocity of light in glycerine = × 10 m/s. 4 Now a slab made of quartz is placed in glycerine as shown. The shift of the object produced by slab is 18cm Glycerine

Glycerine Observer

63. The critical angle of light going from medium A to medium B is θ. The speed of light in medium A is v. The speed of light in medium B is : v (A) (B) v sin θ (C) v cot θ (D) v tan θ sin θ Sol.

Object 20cm

Quartz

(A) 6 cm Sol.

(B) 3.55 cm

(C) 9 cm

(D) 2 cm

64. A light ray is incident on a transparent sphere of 2 , at an angle of incidence = 45°. What is the deviation of a tiny fraction of the ray, which enters the sphere, undergoes two internal reflections, and then refracts out into air? (A) 270° (B) 240° (C) 120° (D) 180°

index =

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GEOMETRICAL OPTICS

Page # 81 67. Two transparent media A and B are separated by a plane boundary. The speed of light in medium A is 2.0 × 108 m s–1 and in medium B is 2.5 × 108 ms–1. The critical angle for which a ray of light going from A to B is totally internally reflected is

Sol.

–1  1  (A) sin   2

–1  2  –1  4  –1  1  (B) sin   (C) sin   (D) sin   5 5     3

Sol.

B E n=3/2 n=6/5

65. In the figure ABC is the cross section of a right angled prism and BCDE is the cross section of a glass slab. The value of θ so that light incident normally on the face AB does not cross the face BC is A (given sin–1(3/5) = 37°) (A) θ ≤ 37° (B) θ > 37° (C) θ ≤ 53° (D) θ < 53° Sol.

CD

68. A small source of light is 4m below the surface of a liquid of refractive index 5/3. In order to cut off all the light coming out of liquid surface, minimum diameter of the disc placed on the surface of liquid is (A) 3m (B) 4m (C) 6m (D) ∞ Sol.

66. A ray of light from a denser medium strike a rarer medium. The angle of reflection is r and that of refraction is r’. The reflected and refracted rays make an angle of 90º with each other. The critical angle will be (A) sin–1 (tan r) (B) tan–1 (sin r) (C) sin–1 (tan r’) (D) tan–1 (sin r’) Sol.

69. A cubical block of glass of refractive index n1 is in contact with the surface of water of refractive index n2. A beam of light is incident on vertical face of the block (see figure). After refraction, a total internal reflection at the base and refraction at the opposite vertical face, the ray emerges out at an angle θ. The value of θ is given by

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Page # 82

θ

n1

(C) sin θ
7, the values of λ do not lie in the visible range 4000 Å to 7500 Å. But for values of n = 4, 5, 6, 7, the following wavelengths lie in the visible region :

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WAVE OPTICS (i) λ =

Page # 21

28 × 10 –7 4

= 7.0 × 10–7 m = 7000 Å

(ii) λ =

28 × 10 –7 = 5.6 × 10–7 m = 5600 Å 5

(iii) λ =

28 × 10 –7 = 4.667 × 10–7 m = 4667 Å 6

(iv) λ =

28 × 10 –7 = 4.0 × 10–7 m = 4000 Å 7

The condition for bright fringe or strong reflection is 4µt (2n + 1)λ or λ = ( 2 n + 1) 2 Substituting the values of µ and t, we get

2µ t =

λ=

4 × 14 . × 10 –6 56 + 10 –7 = m 2n + 1 2n + 1

For values of n < 4 or > 6, the values of λ do not lie in the visible range. But for n = 4, 5, 6 the following waelengths lie in the visible range : (i) λ =

•

56 × 10 –7 = 6.222 × 10–7 m = 6222 Å 2× 4 +1

THE LLOYD'S MIRROR EXPERIMENT :

Interference Area

s a O d = 2a a I

D

Screen

In this experiment the light reflected from a long mirror & the light coming directly from the source without reflection produce interference on a screen i.e. source & Image behave as coherent sources. An important feature of this experiment lies in the fact that when the screen is placed in contact with the end of the mirror, the edge of the reflecting surface comes at the centre of dark fringe instead of a bright fringe. The direct beam does not suffer any phase change, this means that the reflected beam undergoes a phase change of π radian. Hence at any point P on the sceen the condition for minima & maxima are S2P – S1P = nλ [For minima] λ  S2P – S1P =  n +  λ [For maxima] 2

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Page # 22

Ex.19 In Lloyd's interference experiment, 10 fringes occupy a space of 1.5 mm. The distance between the source and the screen is 1.25 m. If light of wavelength 6000 Å is used, find the distance of the source from the plane minor. Sol.

Here

β=

15 . mm = 0.15 × 10–3 m 10

D = 1.25 m, λ = 6000 Å = 6 × 10–7 m As

β=

Dλ d

∴

d=

Dλ 125 . × 6 × 10 –7 m = 50 × 10–4 m = 5.0 mm = β 0.15 × 10 –3

Hence distance of source from the plane mirror =

d = 2.5 min. 2

FRESENEL'S BIPRISM : M δ

A

S1

A

P B

S

S2

C δ

N

E

Fig shows the Fresnel's biprism experiment schematicaly. The thin prism P refracts light from the slit source S into two beams AC & BE. When a screen MN is placed as shown in the figure, the interference fringes are observed only in the region BC. If the screen MN is removed, the two beam will overlap over the whole region AE. If A is the angle of refraction of thin prism & µ is the refractive index of its medium, then the angle of deviation produced by the prism is δ = A (µ – 1)

If l1 is the distance between the source & the prism, then the separation between virtual sources is d = 2δ
1 = 2A(µ – 1)
1 If
2 is the distance between the prism & the screen, then the distance between virtual sources & the screen is given by D =
1 +
2 Thus, by using the result of young's experiment, the fringe width is given by β=

λ(l1 + l2 ) λD ⇒ β= 2δ l1 d

β=

 l2  λ λ  l2  1 +  ⇒ β = 1 +  2A( µ – 1)  l1  2δ  l1 

Fringes observed in the Fresnel's biprism experiment are vertical stringht lines.

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Page # 23

Ex.20 In a biprism experiment, the slit is illuminated with light of wavelength 4800 Å. The distance between the slit and diprism is 20 cm and that between biprism and eyepiece is 80 cm. If two virtual sources are 0.3 cm apart, determine the distance between the 5th bright band on one side of the central bright band and the 4th dark bank on the other side. Sol. Here λ = 4.8 × 10–7 m, d = 0.3 × 10–2 m, D = 20 + 80 = 100 cm = 1 m Distance of 5th bright from the central bright band is x5 = 5

∆λ d

Ex.21 In a biprism experiment, fringe width is measured as 0.4 mm. When the eyepiece is moved away from the biprism through 30 cm, the fringe width increases by 50%. If the two virtual sources are 0.6 mm apart, find the wavelength of light used. Sol.

150 β = 1.5 β1 100 1 D2 = D1 + 30 cm = D1 + 0.3 m, d = 0.6 mm = 0.6 × 10–3 m

Here β1 = 0.4 mm = 0.4 × 10–3 m, β2 =

D1λ d

As

β1 =

and

∴

β 1 D1 = β 2 D 2 or

∴

Wavelenth of light used,

β2 =

D2λ d

β1 D1 = 15 . β1 D + 0.3

or

D1 = 0.6 m

β1d 0.4 × 10 –3 × 0.6 × 10 –3 λ= D = = 4 × 10–7 m = 4000 Å 1 0.6

Ex.22 Interference fringes are produced by a Fresnel's biprism in the focal plane of reading microscope which is 100 cm from the slit. A lens interposed between the biprism and the microscope gives two images of the slit in two positions. If the images of the slits are 4.05 mm apart in one case, 2.90 mm in the other and the wavelength of light used is 5893 Å, find the distance between two consecutive bands. Sol.

Here

d1 = 4.05 mm = 0.405 cm,

d2 = 2.09 mm = 0.209 cm

Distance between the two coherent sources will be d= =

d1d2

(Displacement method)

0.405 × 0.209 cm = 0.2909 cm

Also D = 100 cm, λ = 5893 × 10–8 cm ∴ Fringe width, β =

Dλ 100 × 5893 × 10 –8 = cm = 0.0203 cm. d 0.2909

HUYGEN'S PRINCIPLE : The various postulates are : 1.

Each source of light is a centre of disturbance from which waves spread in all directions. All particles equidistant from the source & vibrating in same phase lie on the surface known as wavefront.

2.

Wave propagates perpendicular to wavefront

3.

Each ray take same time to reach from one wavefront to another wavefront

4.

Every point on a wavefront is a source of new disturbance which produces secondary wavelets. These wavelets are spherical & travel with the speed of light in all directions in that medium.

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Page # 24

5.

Only forward envelope enclosing the tangents at the secondary wavelets at any instant gives the new position of wavefront. There is no backward flow of energy when a wave travels in the forward direction. F

F2

F1

F1

F2 A2

Ray

B2

C2

l

Ex.23 For the given ray diagram, draw the wavefront

b

will behave as point source Sol.

Planar wavefront

Spherical Wavefront

REFLECTION AND REFRACTION : We can use a modified form of Huygens' construction to understand reflection and refraction of light. Figure (a) shows an incident wavefront which makes an angle 'i' with the surface separating twc media, for example, air and water. The phase speeds in the two media are v1 and v2. We can see that when the point A on the incident wavefront strikes the surface, the point B still has to travel a distance BC = AC sin i, and this takes a time t = BC/v1 = AC (sin i)/v1. After a time t, a secondary wavefront of radius v2t with A as centre would have travelled into medium 2. The secondary wavefront with C as centre would have just started, i.e. would have zero radius. We also show a secondary wavelet originating from a point D in between A and C. Its radius is less than v2t. The wavefront in medium 2 is thus a line passing through C and tangent to the circle centred on A. We can see that the angle r′ made by this refracted wavefront with the surface is given by AE = v2t = AC sin r′. Hence, t = AC (sin r′)/v2. Equating the two expressions for 't' gives us the law of refraction in the form sin i/ sir r′ = v1/v2. A similar picture is drawn in shown figure (b) for the reflected wave which travels back into medium 1. In this case, we denote the angle made by the reflected wavefront with the surface by r, and we find that i = r. Notice that for both reflection and refraction, we see secondary wavelets starting at different times. Compare this with the earlier application (shown figure) where we start them at the same time.

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Page # 25

The preceding argument gives a good physical picture of how the refracted and reflected waves are built up from secondary wavelets. We can also understand the laws of reflection and refraction using the concept that the time taken by light to travel along different rays from one wavefront to another must be the same. (Fig) Shows the incident and reflected wavefronts when a parallel beam of light falls on a plane surface. One ray POQ is shown normal to both the reflected and incident wavefronts. The angle of incidence i and the angle of reflection r are defined as the angles made by the incident and reflected rays with the normal. As shows in fig shown, these are also the angles between the wavefront and the surface. Medium-1

B Air A

i

v 2t

E

C r'

v1t v,t

r

i

A

r

i O (i)

(b)

(a)

P

Q

v1t

D water

P

B

A

Medium-2

i

O r' R

C

(ii)

(c)

(Fig.) (a) Huygens' construction for the (a) refracted wave. (b) Reflected wave. (c) Calculation of propagation time between wavefronts in (i) reflection and (ii) refraction. We now calculate the total time to go from one wavefront to another along the rays. From Fig. (c), we have we have Total time for light to reach from P to Q PO OQ AO sin i OB sin r OA sin i + ( AB – OA ) sin r AB sin r + OA(sin i – sin r ) = v + v = + = = v v v v1 1 1 1 1 1

Different rays normal to the incident wavefront strike the surface at different points O and hence have different values of OA. Since the time should be the same for all the rays, the right side of equation must actually be Independent of OA. The condition, for this to happen is that the coefficient of OA in Eq. (should be zero, i.e., sin i = sin r. We, thus, have the law of reflection, i = r. Figure also shows refraction at a plane surface separating medium 1 (speed of light v1) from medium 2 (speed of light v2). The incident and refracted wavefronts are shown, making angles i and r' with the boundary. Angle r' is called the angle of refraction. Rays perpendicular to these are also drawn. As before, let us calculate the time taken to travel between the wavefronts along any ray. PO OR Time taken from P to R = v + v 1 2

=

 sin i sin r '  ΟA sin i ( AC – OA ) sin r ' AC sin r ' + OA –  = + v2  v1 v2 v2  v1

This time should again be independent of which ray we consider. The coefficient of OA in Equation is, v1 sin i therefore, zero,. That is, sin r ' = v = n 21 2

where n21 is the refractive index of medium 2 with respect to medium 1. This is Snell's law of, refraction that we have already dealt with from Eq. n21 is the ratio of speed of light in the first medium (v1) to that in the second medium (v2). Equation is, known as the Snell's law of refraction. If the first medium sin i c is vacuum, we have sin r ' = v = n 2 2 where n2 is the refrective index of medium 2 with respect to vacuum, also called the absolute refractive index of the medium. A similar equation defines absolute refractive index n1 of the first medium. From Eq. we then get v1 ( c / n1) n 2 n21 = v = (c / n ) = n 2 2 1

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Page # 26

The absolute refractive index of air about 1.0003, quite close to 1. Hence, for all practical purposes, absolute refractive index of a medium may be taken with respect to air. For water, n1 = 1.33, which c , i.e. about 0.75 times the speed of light in vacuum. The measurement of the speed of 133 . light in water by Foucault (1850) confirmed this prediction of the wave theory.

means v1 =

Once we have the laws of reflection and refraction, the behaviour of prisms. lenses, and mirrors can be understood. These topice are discussed in detial in the previous Chapter. Here was just describe the behaviour of the wavefronts in these three cases (Fig) (i)

(ii)

(iii)

(iv)

Consider a plane wave passing through a thin prism. Clearly, the portion of the incoming wavefront which travels through the greatest thickness of glass has been delayed the most. Since light travels more slowly in glass. This explains the tilt in the emerging wavefront. A concave mirror produces a similar effect. The centre of the wavefront has to travel a greater distance before and after getting reflected, when compared to the edge. This again produces a converging spherical wavefront. A concave mirror produces a similar effect. The centre of the wavefront has to travel a greater distance before and after getting reflected, when compared to the edge. This again produces a convering spherical wavefront. Concave lenses and convex mirrors can be understood from time delay arguments in a simalr manner. One interesting property which is obvious from the pictures of wavefronts is that the total time taken from a point on the object to the corresponding point on the image is the same measured along any ray (Fig.). For example, when a convex lens focuses light to form a real image, it may seem that rays going through the centre are shorter. But because to the slower speed in glass, the time taken is the same as for rays travelling near the edge of the lens.

(a)

(b)

(c)

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Page # 27

WAVE OPTICS

(ONLY ONE OPTION IS CORRECT)

Exercise - I A. YOUNG’S DOUBLE SLIT EXPERIMENT

1. Figure, shows wave fronts in still water, moving in the direction of the arrow towards the interface PQ between a shallow region and a deep (denser) region. Which of the lines shown may represent one of the wave fronts in the deep region ? P

deep

IV

shallow

Sol.

III II I Q

(A) I Sol.

(B) II

(C) III

(D) IV

2. Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beam are : (A) 5I and I (B) 5I and 3I (C) 9I and I (D) 9I and 3I Sol.

3. Figure shown plane waves refracted for air to water using Huygen’s principle a, b, c, d, e are lengths on the diagram. The refractive index of water wrt air is the ratio.

air

a

c

4. When light is refracted into a denser medium, (A) Its wavelength and frequency both increases (B) Its wavelength increase but frequency remains unchanged (C) Its wavelength decrease but frequency remains unchanged (D) Its wavelength and frequency both decrease Sol.

A

5. Two point source separated by d = 5µm emit light of wavelength D λ = 2µm in phase. A circular wire of radius 20µm is placed around the source as shown in figure. C (A) Point A and B are dark and points C and D bright (B) Points A and B are bright and point C and D dark (C) Points A and C are dark and points B and D bright (D) Points A and C are bright and points B and D dark Sol.

B

are are are are

b e

(A) a/e

(B) b/e

(C)b/d

(D) d/b

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Page # 28

WAVE OPTICS

6. Plane 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 0.13 m. If the velocity of light is 3 × 108 m/ s, find the frequency of the transmitter. (A) 1.5 × 1010 Hz (B) 1010 Hz 10 (C) 3 × 10 Hz (D) 6 × 1010 Hz Sol.

8. Two coherent narrow slits emitti ng l ight of wavelength λ in the same phase are placed parallel to each other at a small separation of 3λ. The light is collected on a screen S which is placed at a distance D (>> λ) from the slits. The smallest distance x such that the P is a maxima. P x S1

O

S2 D

(A)

3D

(B)

8D

(C)

5D

(D)

5

D 2

Sol.

7. Two monochromatic (wavelength = a/5) and coherent sources of electromagnetic waves are placed on the x-axis at the points (2a, 0) and (–a, 0). A detector moves in a circle of radius R(>>2a) whose centre is at the origin. The number of maximas detected during one circular revolution by the detector are (A) 60 (B) 15 (C) 64 (D) None Sol.

9. In YDSE how many maxima can be obtained on the screen if wavelength of light used is 200nm and d = 700 nm. (A) 12 (B) 7 (C) 18 (D) none of these Sol.

10. In a YDSE, the central bright fringe can be identified. (A) as it has greater intensity than the other bright fringes (B) as it is wider than the other bright fringes (C) as it is narrower than the other bright fringes (D) by using white light instead of single wavelength light. Sol.

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Page # 29

WAVE OPTICS 11. In Young’s double slit experiment, the wavelength of red light is 7800 Å and that of blue light is 5200 Å. The value of n for which nth bright band due to red light coincides with (n + 1)th bright band due to blue light, is (A) 1 (B) 2 (C) 3 (D) 4 Sol.

12. If the Young’s double slit experiment is performed with white light, then which of the following is not true (A) the central maximum will be white (B) there will not be a completely dark fringe (C) the fringe next to the central will be red (D) the fringe next to the central will be violet Sol.

14. In Young’s double slit experiment, the two slits act as coherent sources of equal amplitude A and wavelength λ. In another experiment with the same setup the two slits are sources of equal amplitude A and wavelength λ but are incoherent. The ratio of the average intensity of light at the midpoint of the screen in the first case to that in the second case is (A) 1 : 1 (B) 2 : 1 (C) 4 : 1 (D) none of these Sol.

15. In a Young’s double slit experiment, a small detector measures an intensity of illumination of I units at the centre of the fringe pattern. If one of the two (identical) slits is now covered, the measured intensity will be (A) 2I (B) I (C) I/4 (D) I/2 Sol.

13. Two identical narrow slits S1 and S2 are illuminated by light of wavelength λ from a point source P. If, as shown in the diagram above the light is then allowed to fall on a screen, and if n is a positive integer, the condition for destructive interference at Q is that

l1

S1 l3

P l2

S2

l4

Q

(A) (l1 – l2) = (2n + 1) λ/2 (B) (l3 – l4) = (2n + 1) λ/2 (C) (l1 + l2) – (l2 + l4) = nλ (D) (l1 + l3) – (l2 + l4) = (2n + 1) λ/2 Sol.

16. In a young double slit experiment D equals the distance of screen and d is the separation between the slit. The distance of the nearest point to the central maximum where the intensity is same as that due to a single slit, is equal to Dλ Dλ Dλ 2Dλ (A) (B) (C) (D) 2 d 3 d d d Sol.

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WAVE OPTICS

17. A beam of light consisting of two wavelength 6300 Å and λ Å is used to obtain interference fringes in a Young’s double slit experiment. If 4th bright fringe of 6300 Å coincides with 5th dark fringe of λ Å, the value of λ (in Å) is (A) 5200 (B) 4800 (C) 6200 (D) 5600 Sol.

18. A beam of light consisting of two wavelengths 6500 Å and 5200 Å is used to obtain interference fringes in Young’s double slit experiment. The distance between slits is 2mm and the distance of screen from slits is 120 cm. What is the least distance from central maximum where the bright due to both wavelength coincide ? (A) 0.156 cm (B) 0.312 cm (C) 0.078 cm (D) 0.468 cm Sol.

20. In a Young’s Double slit experiment, first maxima is observed at a fixed point P on the screen. Now the screen is continuously moved away from P the plane of slits. The ratio of O intensity at point P to the intensity at point O (centre of the screen) (A) remains constant (B) keeps on decreasing (C) first decreases and then increases (D) First decreases and then becomes constant Sol.

21. In a double slit experiment, the separation between the slits is d = 0.25 cm and the distance of the screen D = 100 cm from the slits. If the wavelength of light used is λ = 6000 Å and I0 is the intensity of the central bright fringe, the intensity at a distance x = 4 × 10–5 m from the central maximum is (B) I0/2 (C) 3I0/4 (D) I0/3 (A) I0 Sol. 19. The ratio of the intensity at the centre of a bright fringe to the intensity at a point one-quarter of the fringwidth from the centre is (A) 2 (B) 1/2 (C) 4 (D) 16 Sol.

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WAVE OPTICS 22. A monochromatic light source of wavelength λ is placed at S. Three slits S1, S2 and S3 are equidistant from the source S and the point P on the screen. S1P – S2P = λ/6 and S1P – S3P = 2λ/3. If I be the intensity at P when only one slit is open, the intensity at P when all the three slits are open is

23. In young’s double slit experiment, the value of λ = 500 nm. The value of d = 1 mm, D = 1 m. Then the minimum distance from central maximum for which the intensity is half the maximum intensity will be (A) 2.5 × 10–4 m (B) 2 × 10–4 m (C) 1.25 × 10–4 (D) 10–4m Sol.

S1 S2

D

(A) 31 Sol.

(B) 51

P

S3

S

D

(C) 81

Screen

(D) zero

24. Two slits are separated by 0.3 mm. A beam of 500 nm light strikes the slits producing an interference pattern. The number of maxima observed in the angular range –30° < θ < 30°. (A) 300 (B) 150 (C) 599 Sol.

25. In the figure shown if a parallel beam of white light is incident on the plane of the d 2d/3 slits then the distance of the white spot on the screen from O is [Assume d > d

(A) the thickness of sheet is 2 ( 2 − 1) d infront of S1 (B) the thickness of sheet is ( 2 − 1) d infront of S2 (C) the thickness of sheet is 2 2 d infront of S1 (D) the thickness of sheet is ( 2 2 − 1) d infront of S1 Sol.

8. If one of the slits of a standard YDSE apparatus is covered by a thin parallel sided glass slab so that it transmit only one half of the light intensity of the other, then (A) the fringe pattern will get shifted towards the covered slit (B) the fringe pattern will get shifted away from the covered slit (C) the bright fringes will be less bright and the dark ones will be more bright. (D) the fringe width will remain unchanged Sol.

Question No. 10 to 12 (3 questions) The figure shows a schematic di agram show i ng the arrangement of Young’s Double Slit Experiment

S1

S a

O

d

Screen

S2 D

10. Choose the correct statement(s) related to the wavelength of light used (A) Larger the wavelength of light larger the fringe width (B) The position of central maxima depends on the wavelength of light used (C) If white light is used in YDSE, then the violet colour forms its first maxima closest to the central maxima (D) The central maxima of all the wavelengths coincide Sol.

9. To make the central fringe at the centre O, a mica sheet of refractive index 1.5 is introduced. Choose the correct statements (s).

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WAVE OPTICS

11. If the distance D is varied, then choose the correct statement(s) (A) the angular fringe width does not change (B) the fringe width changes in direct proportion (C) the change in fringe width is same for all wavelengths (D) The position of central maxima remains unchanged Sol.

If the distance d is varied, then identify the correct statement (A) the angular width does not change (B) the fringe width changes in inverse proportion (C) the positions of all maxima change (D) the positions of all minima change Sol.

14. In a standard YDSE appratus a thin film (µ = 1.5, t = 2.1 µm) is placed in front of upper slit. How far above or below the centre point of the screen are two nearest maxima located ? Take D = 1 m, d = 1mm, λ = 4500 Å. (Symbols have usual meaning) (A) 1.5 mm (B) 0.6 mm (C) 0.15 mm (D) 0.3 mm Sol.

13. In an interference arrangement similar to Young’s double-slit experiment, the slits S1 & 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 a distance d = 150.0 m. The intensity I(θ) is measured as a function of θ, where θ is defined as shown. If I0 is the maximum intensity then I(θ) for 0 ≤ θ ≤ 90° is given by

15. Consider a case of thin film interference as shown. Thickness of film is equal to wavelength of light is µ2.

1

2

.

S1

µ1 µ2

d S2

I0 I for θ = 30º (B) I(θ) = 0 for θ = 90º 2 4 (C) I(θ) = I0 for θ = 0º (D) I(θ) is constant for all values of θ Sol.

(A) I(θ) =

µ3 (A) Reflected light will be maxima if µ1 < µ2 < µ3 (B) Reflected light will be maxima if µ1 < µ2 > µ3 (C) Transmitted light will be maxima if µ1 > µ2 > µ3 (D) Transmitted light will be maxima if µ1 > µ2 < µ3 Sol.

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WAVE OPTICS

Exercise - III

(SUBJECTIVE PROBLEMS)

1. In a Young’s double slit experiment for interference of light, the slits are 0.2 cm apart and are illuminated by yellow light (λ = 600 nm). What would be the fringe width on a screen placed 1 m from the plane of slits if the whole system is immersed in water of index 4/3? Sol.

4. A ray of light of intensity I is incident on a parallel glass-slab at a point A as shown in figure. It undergoes partial reflection and refraction. At each reflection 20% of incident energy is reflected. The rays AB and A’B’ undergo interference. Find the ratio Imax/Imin. 2. In Young’s double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If the wavelength of the light is changed to 400 nm, find the number of fringes observed in the same segment. Sol.

B A

B' A'

Sol.

3. On slit of double slit experiment is covered by a thin glass plate of refractive index 1.4 and the other by a thin glass plate of refractive index 1.7. The point on the screen, where central bright fringe was formed before the introduction of the glass sheets, is now occupied by the 5th bright fringe. Assuming that both the glass plates have same thickness and wavelength of light used is 4800 Å, find their thickness. Sol.

5. Light of wavelength 520 nm passing through a double slit, produces interference pattern of relative intensity versus deflection angle θ as shown in the figure. Find the separation d between the slits.

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WAVE OPTICS

Sol.

6. In Young’s double slit experiment the slits are 0.5 mm apart and the interference is observed on a screen at a distance of 100 cm from the slit. It is found that the 9th bright fringe is at a distance of 7.5 mm from the second dark fringe from the centre of the fringe pattern on same side. Find the wavelength of the light used. Sol.

8. The distance between two slits in a YDSE apparatus is 3mm. The distance of the screen from the slits is 1m. Microwaves of wavelength 1 mm are incident on the plane of the slits normally. Find the distance of the first maxima on the screen from the central maxima. Sol.

7. In a YDSE apparatus, d = 1mm, λ = 600 nm and D = 1m. The slits produce same intensity on the screen Find the minimum distance between two points on the screen having 75% intensity of the maximum intensity. Sol.

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WAVE OPTICS 9. A lens (µ = 1.5) is coated with a thin film of refractive index 1.2 in order to reduce the reflection from its surface at λ = 4800 Å. Find the minimum thickness of the film which will minimize the intensity of the reflected light. Sol.

12. In a two-slit 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. If the distance between the slits is 10–3m, calculate the wavelength of the light used. Sol.

10. A long narrow horizontal slit lies 1 mm above a plane mirror. The interference pattern produced by the slit and its image is viewed on a screen distant 1 m from the slit. The wavelength of light is 600nm. Find the distance of first maximum above the mirror. Sol.

11. A broad source of light of wavelength 680nm illuminates normally two glass plates 120 mm long that meet at one end and are separated by a wire 0.048 mm in diameter at the other end. Find the number of bright fringes formed over the 120mm distance. Sol.

13. A monochromatic light of λ = 5000 Å is incident on two slits separated by a distance of 5 × 10–4 m. The interference pattern in seen on a screen placed at a distance of 1 m from the slits. A thin glass plate of thickness 1.5 × 10–6 m & refractive index µ = 1.5 is placed between one of the slits & the screen. Find the intensity at the centre of the screen, if the intensity there is I0 in the absence of the plate. Also find the lateral shift of the central maximum. Sol.

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WAVE OPTICS

14. A double-slit apparatus is immersed in a liquid of refractive index 1.33. It has slit separation of 1mm & distance between the plane of the slits & screen is 1.33 m. The slits are illuminated by a parallel beam of light whose wavelength in air is 6300 Å. (a) Calculate the fringe width. (b) One of the slits of the apparatus is covered by a thin glass sheet of refractive index 1.53. Find the smallest thickness of the sheet to bring the adjacent minima on the axis. Sol. 16. Radio waves coming at ∠α to vertical are recieved by a radar after reflection from a nearby water surface & directly. What should be height of antenna from water surface so that it records a maximum intensity. (wavelength = λ).

h

Sol.

15. A young’s double slit experiment is performed using light of wavelength λ = 5000 Å, which emerges in phase from two slits a distance d = 3 × 10–7m apart. A transparent sheet of thickness t = 1.5 × 10–7m is placed over one of the slits. The refractive index of the material of this sheet is µ = 1.17. Where does the central maximum of the interference pattern now appear ? Sol.

17. In a biprism experiment using sodium light λ = 6000 Å an interference pattern is obtained in which 20 fringes occupy 2 cm. On replacing sodium light by another source of wavelength λ2 without making any other change 30 fringes occupy 2.7 cm on the screen. What is the value of λ2 ? Sol.

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R

S

N

o

:

0 7 4 4 - 2 4 3 9 0 5 1 ,

5 2 ,

5 3 ,

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w

w

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o t i o n i i t j e e . c o m

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o t i o n i i t j e e . c o m

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WAVE OPTICS

(TOUGH SUBJECTIVE PROBLEMS)

Exercise - IV

1. If the slits of the double slit were moved symmetrically apart with relative velocity v, calculate the rate at which fringes pass a point at a distance x from the centre of the fringe system formed on a screen y distance away from the double slits if wavelength of light is λ. Assume y >> d & d >> λ. 2. (a) A thin glass plate of thickness t and refractive index µ is inserted between screen & one of the slits in a Young’s experiment. If the intensity at the centre of the screen is I, what was the intensity at the same point prior to the introduction of the sheet. (b) One slit of a Young’s experiment is covered by a glass plate (µ1 = 1.4) and the other by another glass plate (µ2 = 1.7) of the same thickness. The point of central maxima on the screen, before the plates were introduced is now occupied by the third bright fringe. Find the thickness of the plates, the wavelength of light used is 4000 Å. 3. In a YDSE a parallel beam of light of wavelength 6000 Å is incident on slits at angle of incidence 30°. A & B are two thin transparent films each of refractive index 1.5. Thickness of A is 20.4 µm. Light coming through A & B have intensities I & 4I respectively on the screen. Intensity at point O which is symmetric relative to the slits is 3 I. The central maxima is above O. A

0.1mm

30°

O

B 1m (a) What is the maximum thickness of B to do so. Assuming thickness of B to be that found in part (a) answer the following parts. (b) Find fringe width, maximum intensity & minimum intensity on screen. (c) Distance of nearest minima from O. (d) Intensity at 5 cm on either side of O. 4. In a YDSE experiment, the distance between the slits & the screen is 100 cm. For a certain distance between the slits, an interference pattern is observed on the screen with the fringe width 0.25 mm. When the distance between the slits is increased by ∆d = 1.2 mm, the fringe width decreased to n = 2/3. of the original value. In the final position, a thin glass plate of refractive index 1.5 is kept in front of one of the slits & the shift of central maximum is observed to be 20 fringe width. Find the thickness of the plate & wavelength of the incident light.

5. A screen is at a distance D = 80 cm from a diaphragm having two narrow slits S1 and S2 which are d = 2 mm apart. Slit S1 is covered by a transparent sheet of thickness t1 = 2.5 µm and S2 by another sheet of thickness t2 = 1.25 µm as shown in figure. Both sheets are made of same material having refractive index µ = 1.40. Water is filled in space between diaphragm and screen. A monochromatic light beam of wavelength λ = 5000 Å is incident normally on the diaphragm. Assuming intensity of beam to be uniform and slits of equal width, calculate ratio of intensity at C to maximum intensity of interference pattern obtained on the screen, where C is foot of perpendicular bisector of S1S2. (Refractive index of water, µw = 4/3)

6. In Young’s experiment, the source is red light of wavelength 7 × 10–7m. When a thin glass plate of refractive index 1.5 at this wavelength is put in the path of one of the interfering beams, the central bright fringe shifts by 10–3 m to the position previously occupied by the 5th bright fringe. Find the thickness of the plate. When the source is now changed to green light of wavelength 5 × 10–7m, the central fringe shifts to a position initially occupied by the 6th bright fringe due to red light. Find the refractive index of glass for the green light. Also estimate the change in fringe width due to the change in wavelength. 7. In a Young’s experiment, the upper slit is covered by a thin glass plate of refractive index 1.4 while the lower slit is covered by another glass plate having the same thickness as the first one but having refractive index 1.7. Interference pattern is observed using light of wavelength 5400 Å. It is found that the point P on the screen where the central maximum (n = 0) fell before the glass plates were inserted now has 3/4 the original intensity. It is further observed that what used to be the 5th maximum earlier, lies below the point P while the 6th minimum lies above P. Calculate the thickness of the glass plate. (Absorption of light by glass plate may be neglected).

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WAVE OPTICS

8. A coherent parallel beam of microwaves of wavelength λ = 0.5 mm falls on a Young’s double slit apparatus. The separation between the slits is 1.0mm. The intensity of microwaves is measured on screen placed parallel to the plane of the slits at a distance of 1.0 m from it, as shown in the figure. y

30º

d=1.0mm

x

D=1.0m Screen

(a) If the incident beam falls normally on the double slit apparatus, find the y–coordinates of all the interference minima on the screen. (b) If the incident beam makes an angle of 30º with the x-axis (as in the dotted arrow shown in the figure), find the y-coordinates of the first minima on either side of the central maximum. 9. In a Young’s double slit arrangement, a source of wavelength 6000 Å is used. The screen is placed 1m from the slits. Fringes formed on the screen, are observed by a student sitting close to the slits. The student’s eye can distinguish two neighbouring fringes if they subtend an angle more than 1 minute of arc. Calculate the maximum distance between the slits so that the fringes are clearly visible. Using this information calculate the position of 3rd bright & 5th dark fringe from the centre of the screen. 10. A narrow monochromatic beam of light of intensity I is incident on a glass plate as shown in figure. Another identical glass plate is kept close to the first one & parallel to it. Each glass plaate reflects 25% of the light incident on it & transmits the remaining. Find the ratio of the minimum & the maximum intensities in the interference pattern formed by the two beams obtained after one reflection at each plate.

(i) If the third intensity maximum occurs at the point A on the screen, find the distance OA. (ii) If the gap between L1 & L2 is reduced from its original value of 0.5 mm, will the distance OA increase, decrease or remain the same? 12. Two coherent sources S1 and S2 separated by distance 2λ emit light of wavelength λ in phase as shown in figure. A circular wire of radius 100λ is placed in such a way that S1S2 lies in its plane and the midpoint of S1 S2 is at the centre of wire. Find the angular positions θ on the wire for which intensity reduces to half of its maximum value. 13. In a biprism experiment with sodium light, bands of width of 0.0195 cm are observed at 100 cm from slit. On introducing a convex lens 30 cm away from the slit between and screen, two images of the slit are seen 0.7 cm apart at 100 cm distance from the slit. Calculate the wavelength of sodium light.

S1 O

S2

1 1 2 11. In the figure shown S is a monochromatic point source emitting light of wavelength = 500 nm. A thin lens of circular shape and focal length 0.10 m is cut into two identical halves L1 and L2 by a plane passing through a diameter. The two halves are placed symmetrically about the central axis SO with a gap of 0.5 mm. The distance along the axis from S to L1 and L2 is 0.15 m, while that from L1 & L2 to O is 1.30 m. The screen at O is normal to SO.

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WAVE OPTICS

(JEE PROBLEMS)

Exercise - V

1. As a wave propagates, (A) the wave intensity remains constant for a plane wave (B) the wave intensity decreases as the inverse of the distance from the source for a spherical wave (C) the wave intensity decreases as the inverse square of the distance from the source for a spherical wave (D) total power of the spherical wave over the spherical surface centered at the source remains same at all times. Sol.

2. The Young’s double slit experiment is done in a medium of refractive index 4/3. A light of 600 nm wavelength is falling on the slits having 0.45 mm separation. The lower slit S2 is covered by a thin glass sheet of thickness 10.4 µm and refractive index 1.5. The interference pattern is observed on a screen placed 1.5 m from the slits as shown y S1 S* S2

O

(a) Find the location of the central maximum (bright fringe with zero path difference) on the y-axis. (b) Find the light intensity at point O relative to the maximum fringe intensity. (c) Now, if 600 nm light is replaced by white light of range 400 to 700 nm, find the wavelengths of the light that from maxima exactly at point O. [All wavelengths in this problem are for the given medium of refractive index 4/3. Ignore dispersion] Sol.

3. A thin slice is cut out of a glass cylinder along a plane parallel to its axis. The slice is placed on a flat glass plate as shown. The observed interference fringes from this combination shall be (A) straight (B) circular (C) equally spaced (D) having fringe spacing which increases as we go outwards. Sol.

4. In a double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then, in the interference pattern. [JEE (Scr) 2000] (A) the intensities of both the maxima and the minima increase (B) the intensity of the maxima increases and the minima has zero intensity (C) the intensity of the maxima decreases and that of the minima increases. (D) the intensity of the maxima decreases and the minima has zero intensity. Sol.

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WAVE OPTICS

5. A glass plate of refractive index 1.5 is coated with a thin layer of thickness t and refractive index 1.8. Light of wavelength λ travelling in air is incident normally on the layer. It is partly reflected at the upper and the lower surfaces of the layer and the two reflected ray s i nterfere. Wri te t he c ondi ti on for the i r constructive interference. If λ = 648 nm, obtain the l east val ue of t for whi ch the rays i nterfere constructively. [JEE 2000] Sol.

Sol.

8. In a young double slit experiment, 12 fringes are observed to be formed 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 [JEE (Scr) 2001] (A) 12 (B) 18 (C) 24 (D) 30 Sol.

6. Two coherent light sources A and B with separation 2λ are placed on the x-axis symmetrically about the origin. They emit light of wavelength λ. Obtain the positions of maxima on a circle of large radius lying in the xy-plane and with centre at the origin. [JEE 2000] Sol.

9. A vessel ABCD of 10cm width has two small slits S1 and S2 sealed with identical glass plates of equal thickness. The distance between the slits is 0.8mm. POQ is the line perpendicular to the plane AB and passing through O, the middle point of S1 and S2. A monochromatic light source is kept at S, 40 cm below P and 2 m from the vessel, to illuminate the slits as shown in the figure below. Calculate the position of the central bright fringe on the other wall CD with respect to the line OQ. Now, a liquid is poured into the vessel and filled up to OQ. The central bright fringe is found to be at Q. Calculate the refractive index of the liquid. [JEE’2001]

7. 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 π/2 at point A and π at point B. Then the difference between the resultant intensities at A and B is [JEE 2001] (A) 2I (B) 4I (C) 5I (D) 7I

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WAVE OPTICS Sol.

10. A point source S emitting light of wavelength 600 nm is placed at a very small height h ab ov e the fl at refl e ct i ng surface AB (see figure). The intensity of the reflected light is 36% of the incident intensity. Int erfe re nc e fri nge s are observed on a screen placed parallel to the reflecting surface at a very large distance D from it. [JEE’2001] (a) What is the shape of the interference fringes on the screen? (b) Calculate the ratio of the minimum to the maximum intensities in the interference fringes formed near the point P (shown in the figure). (c) If the intensities at point P corresponds to a maximum, calculate the minimum distance through which the reflecting surface AB should be shifted so that the intensity at P again becomes maximum.

Sol.

11. In the ideal souble-slit experiment, when a glass plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wavelength λ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plote is : [JEE 2002] (B) 2λ/3 (C) λ/3 (D) λ (A) 2λ Sol.

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WAVE OPTICS

12. In the adjacent diagram, CP represents a wavefront and AO and BP, the corresponding two rays. Find the condition on θ for constructive interference at P between the ray BP and reflected ray OP.

Sol.

[JEE (Scr)2003]

(A) cosθ =

3λ 2d

(C) secθ – cosθ =

(B) cosθ =

λ d

λ 4d

(D) secθ – cosθ =

4λ d

14. In a YDSE bi-chromatic light of wavelengths 400 nm and 560 nm are used. The distance between the slits is 0.1 nm and the distance between the plane of the slits and the screen is 1m. The minimum distance between two successive regions of complete darkness is [JEE’ 2004 (Scr)] (A) 4 mm (B) 5.6 mm (C) 14 mm (D) 28 mm Sol.

Sol.

13. A prism (µP = 3 ) has an angle of prism A = 30°. A thin film (µf = 2.2) is coated on face AC as shown in the figure. Light of wavelength 550 nm is incident on the face AB at 60° angle of incidence. Find

A

B

C

(i) the angle of its emergence from the face AC and (ii) the minimum thickness (in nm) of the film for which the emerging light is of maximum possible intensity.

15. In a Young’s double sl it experiment, two wavelengths of 500 nm and 700 nm were used. What is the minimum distance from the central maximum where their maximas coincide again ? Take D/d = 103. Symbols have their usual meanings. [JEE 2004]

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WAVE OPTICS

17. Light travels as a (A) parallel beam in each medium (B) convergent beam in each medium (C) divergent beam in each medium (D) divergent beam in one medium and convergent beam in the other medium [JEE 2007] Sol.

Sol.

16. In Young’s double slit experiment maximum intensity is I than the angular position where the intensity becomes 1/4 is [JEE’ 2005 (Scr)] –1  λ  –1  λ  –1  λ  –1  λ  (A) sin   (B) sin   (C) sin   (D) sin   d 3d 2d 4d

Sol.

18. The phases of the light wave at c, d, e and f are φc, φd, φe and φf respectively. It is given that φc ≠ φf . (A) φc cannot be equal to φd (B) φd can be equal to φe (C) (φd – φf) is equal to (φc – φe) [JEE 2007] (D) (φd – φc) is not equal to (φf – φe) Sol.

19. Speed of light is [JEE 2007] (A) the same in medium-1 and medium - 2 (B) larger in medium - 1than in medium - 2 (C) larger in medium-2 than in medium-1 (D) different at b and d Sol.

Paragraph for Question Nos. 17 to 19 (3 questions) 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 traveling in medium-1 and incident on XY. The lines ef and gh represent wavefronts of the light wave in medium-2 after refraction. b

d medium-1

a

c

X

f

h

Y medium-2

e

g

20. In a Young’s double slit experiment, the separation between the two slits is d and the wavelength of the light is λ. The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s). (A) If d = λ, the screen will contain only one maximum (B) If λ < d < 2λ, at least one more maximum (besides the central maximum) will be observed on the screen (C) If the intensity of light falling on slit 1 is reduced so that it becomes equal to that of slit 2, the intensities of the observed dark and bright fringes will increase. (D) If the intensity of light falling on slit 2 is increased so that it becomes equal to that of slit 1, the intensities of the observed dark the bright fringes will increase. [JEE 2008]

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Page # 50

WAVE OPTICS Column I

Sol.

Column II

P2 P1 P0

S2

(A)

(P) δ (P0) = 0

S1

S2

λ 4

(B) (µ – 1) t =

S1

S2

(C) (µ – 1) t =

λ 2

S2

(D)(µ–1)t=

3λ 4

S1

P2 P1 P0

(Q) δ(P1) = 0

P2 P1 P0

(R) I (P1) = 0

P2 P1 P0 (S) I(P0) > I(P1)

S1

(T) I(P2) > I(P1) Sol.

21. Column I shows four situations of standard Young’s double slit arrangement with the screen placed far away from the slits S1 and S2 . In each of these cases S1 P0 = S2 P0 , S1 P1 – S2 P1 = λ / 4 and S1 P2 – S2 P2 = λ/3, where λ is the wavelength of the light used. In the cases B,C and D, a transparent sheet of refractive index µ and thickness t is pasted on slit S2 . The thicknesses of the sheets are different in different cases. The phase difference between the light waves reaching a point P on the screen from the two slits is denoted by δ(P) and the intensity by I (P). Match each situation given in Column I with the statement(s) in Column II valid for that situation. [JEE 2009]

22. Young's double slit experiment is carried out by using green, red and bluelight, one color at a time. The fri nge widths recorded are β G , β R and β β , respectively. Then. (A) β G > β B > β R (B) β B > β G > βR (C) β R > β B > β G (D) β R > β G > βB Sol.

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WAVE OPTICS

Page # 51

ANSWER EXERCISE - I 1.

A

2.

C

3.

C

4.

C

5.

D

6.

A

7.

A

8. 15. 22. 29.

D C A D

9. 16. 23. 30.

B C C A

10. 17. 24. 31.

D D C C

11. 18. 25. 32.

B A D C

12. 19. 26. 33.

C A A B

13. 20. 27. 34.

D C D A

14. 21. 28. 35.

B C A B

36. 42.

C D

37. 43.

C A

38. 44.

A C

39.

A

40.

B

41.

A

7.

B

EXERCISE - II

1.

BD

2.

BCD

3.

BC

4.

B

5.

AC

6.

D

8. 14.

ACD CD

9. 15.

A AD

10.

ACD

11.

ABD

12.

BD

13.

AC

EXERCISE - III

1. 0.225 mm

2. 18

3. 8 µm

6. 5000 Å 11. 141

7. 0.2 mm 12. 6000 Å

4. 81 : 1 8. 35.35 cm 13. 0, 1.5 mm

5. 1.98 × 10–2 mm

9. 10–7 m 10. 0.15 mm 14. 0.63 mm, 1.575 µm

15. y = 0.085 D; D = distance between screen & slits λ 16. 17. 5400 Å 4 cos α

EXERCISE - IV x 1. λ v y

 π( µ – 1)t  2. (a) I0 = I sec2   , (b) 4 µm λ  

3. (a) tB = 120 µm (b) β = 6mm; Imax = 9I, Imin = I (c) β/6 = 1mm (d) I (at 5cm above 0) = 9I, I (at 5 cm below 0) = 31 4. λ = 600 nm, t = 24 µm

5. 3/4

6. 7 µm, 1.6,

400 µm (decrease) 7

7. 9.3 µm

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WAVE OPTICS

Page # 52

8. (a) ±

1 15

10. 1 : 49

, ±

3 7

(b) +

1 15

,

3

9.

7

6.48 π π mm , mm ; π 3.6 2.4

11. (i) 1 mm (ii) increase

 2n + 1 –1  2n + 1  n = 0, 1, 2, 3 13. λ = 5850 Å  , n = 0, 1, 2, 3 & π ± cos  12. ± cos–1  8   8 

EXERCISE - V 1. A, C, D 2. (a) y = –13/3 mm, (b) intensity at O = 0.75Imax (c) 650 nm, 433.33 nm 5. t =

4. A

λ 3λ λ , , ..........; tminimum = = 90 nm 7.2 7.2 7.2

R R 3   , (R, 0), 6. (0, –R),  ,– 2  2

7. B

8. B

11 .

A

12. B

13. 0, 125 nm

20. A,B

3. A

R R 3   ,  , (0, R) 2 2 

 R R 3  – ,  , (–R, 0)  2 2 

9. (i) y = 2 cm, (ii) µ = 1.0016

14. D

15. 3.5 mm

16. B

 R R 3  – , –  2   2

10. (a) circular, (b)

17. A

21. (A) → (ps), (B) →(q), (C) →(t), (D) → (rst)

1 , (c) 3000Å 16

18. C

19. B

22. D

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IIT-JEE|AIEEE CBSE|SAT|NTSE OLYMPIADS

Nurturing potential through education

MODERN PHYSICS - I THEORY AND EXERCISE BOOKLET

CONTENTS S.NO.

TOPIC

PAGE NO.

1. Nature of Light .................................................................... 3 – 15 2. Force due to Radiation ......................................................... 16 – 20 3. De-Brogli Wavelength of matter wave ..................................... 20 – 21 4. Atomic Model ....................................................................... 21 –24 5. Bohr's Model of an Atom ....................................................... 25 – 30 6. Excitation and Ionization of an Atom ...................................... 31 – 33 7. Spectral Series of H-atom .................................................... 33 – 34 8. Aplication of Nucleus Motion ................................................ 34 – 36 9. Atomic collision ................................................................... 36 – 38 10. X-Rays ............................................................................ 39 – 41 11. Exercise -I ....................................................................... 42 – 55 12. Exercise - II ..................................................................... 56 – 60 13. Exercise - III .................................................................... 61 – 68 14. Exercise - IV ..................................................................... 69 – 70 15. Exercise - V ...................................................................... 71 – 82 16. Answer key ...................................................................... 83 – 84

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MODERN PHYSICS - I

SYLLABUS Photoelectric effect; Bohr’s theory of hydrogen-like atoms; Characteristic and continuous X-rays, Moseley’s law; de Broglie wavelength of matter waves.

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MODERN PHYSICS - I 1.

1.1

NATURE OF LIGHT It was a matter of great interest for scientists of know that what exactly from the light is made up of or how the light behaves. This is briefly described over here Newton's Corpuscular theory : Newton was the first scientist who said that light is made 0 up tiny elastic particles called "Corpuscles" which travels with the velocity of light. So according to Newtons, light is a particle.

1.2

Hugen's wave theory : Huygen was a scientist working parallel to Newton who come with a drastically different idea for nature of light & said that light is not a particle but a wave.

1.3

Maxwell's electromagnetic wave theory : During the time of Hygen, his views regarding nature of light were not accepted as newton was a popular scientist of his time. but, when maxwell asserted that light is a electromagnetic wave, scientists started believing that light is a wave.

1.4

Max Planck's quantum theory of light : Once again when scientists started believing that the light is a wave max plank came with different idea & asserted that light is not a wave but a photon (i.e. a particle) which he proved through balck body radiation spectrum. At this time there was a great confusion about the nature of light which was solved by de-broglie from where origin of theory of matter wave come into picture.

1.5

Debroglie Hypothesis It supports dual nature of light (wave nature and particle nature). According to him the light consists of particles associated with definite amount of energy and momemtum. These particles were later named as photons. The photon posses momentum and is given by P

h 

...(1)

P = momentum of one photon  = wavelength of wave. h = Plank's constant = 6.62 × 10–34 Js. A photon is a packet of energy. It posses energy given by E

hc 

...(2)

where c = speed of light Debroglie relates particle property (momentum) with wave property (wavelength) i.e. he favours dual nature of light. Electron volt : It is the energy gained by an electron when it is accelerated through a potential difference of one volt. 1 eV = 1.6 × 10–19 Joule. Now from eq. (2) E

6.62  10–34  3  108 in Joule. 

E

6.62  10–34  3  108 eV   1.6  10–19

E

12400 ev 

where  is in Å

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1. 2. 3. 4. 5.

MODERN PHYSICS - I

Properties of Photon : Photon travels with speed of light. The rest mass of a photon is zero. There is no concept of photon conservation. All the photons of a particular frequency or wavelength posses the same energy irrespective of the intensity of the radiation. The increase in the intensity of the radiation imply an increase in the number of photon's crossing a given area per second. When light travels from one medium to another medium then frequency = const (because it is the property of source) but v,  changes

Ex.1

Sol.

light

A beam of light having wavelength  and intensity 1 falls normally on an area A of a clean surface then find out the number of photon incident on the surface. Total energy incident in time t = I A t Energy of one photon E 

hc 

I

A

Then number of photon incident in time t 

Total energy incident I A t  energy of one photon  hc

Electron Emission Process : light

f , ,I

e–

e– (Photoelectron)

When light is incident on a metal surface it was observed that electrons are ejected from a metal surface some times even when incredicely dim light such as that from starts and distance galaxies incident on it and some time electrons not comes out from the metal surface even high energetic or high intensity light falling on the metal surface. This shows that the electron emission from a metal surface is not depends on the intensity of incident light but it is basically depends on the energy of the incident. Photons no matters in number of photons are very less in a dim light, photo electric effect can be seen. During the phenomenon of photoelectric effect one incident photon on metal surface can eject at most only one electron. A photon is an energy packet which is fully absorbed not partially. Thus one photon can not be absorbed by more than one electron. The minimum amount of energy of photon required to eject an electron out of a metal surface is called work function It is denoted by . The work function depends on the nature of the metal. 1. The electron emission from a metal is only depends on the work function or energy of one photons. 2. But how many electrons comes out from the metal is depends on intensity of the falling light on energy of the light. 3. Energy of photon incident on metal will not necessarily cause emission of an electron even if its energy is more than work function. The electron after absorption may be involved in many other process like collision etc in which it can lose energy hence the ratio of no. of electrons emitted to the no. of photons incident on metal surface is less than unity.

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MODERN PHYSICS - I 1.6

Three Major Features of the Phtoelectric effect cannot be explained in terms of the classical of the wave theory of light.

(a)

The intensity problem : Wave theory requires that the oscillating electric field vector E of the light wave increases in amplitude as the intensity of the light beam is increased. Since the force applied to the electron is eE, this suggests that the kinetic energy of the photoelectrons should also increased the light beam is made more intense. However observation shows that maximum kinetic energy is independent of the light intensity.

(b)

The frequency problem : According to the wave theory, the photoelectric effect should occur for any frequency of the light, provided only that the light is intense enough to supply the energy needed to eject the photoelectrons. However observations shows that there exists for each surface a characterstic cutoff frequency th, for frequency less than th, the photoelectric effect does not occur, no matter how intense is light beam.

(c)

The time delay problem : If the energy acquired by a photoelectron is absorbed directly from the wave incident on the metal plate, the "effective target area" for an electron in the metal is limited and probably not much more than that of a circle of diameter roughly equal to that of an atom. In the classical theory, the light energy is uniformly distributed over the wavefront. Thus, if the light is feeble enough, there should be a measurable time lag, between the impinging of the light on the surface and the ejection of the photoelectron. During this interval the electron should be absorbing energy from the beam until it had accumulated enough to escape. However, no detectable time lag has ever been measured. Now, quantum theory solves these problems in providing the correct interpretation of the photoelectric effect.

1.7

Threshold frequency and Threshold wavelength We have discussed that to start photoelectric emission the energy of incident photon on metal surface must be more than the work function of the metal. If  is the work function of the metal then there must be a minimum frequency of the incident light photon which is just able to eject the electron from the metal surface. This minimum frequency or threshold frequency vth can be given as h th   Threshold frequency th is a characteristic property of a metal as it is the minimum frequency of the light radiation required to eject a free electron from the metal surface.

As the threshold frequency is defined, we can also define threshold wavelength  th for a metal surface. Threshold wavelength is also called cut off wavelength. For a given metal surface threshold wavelength is the longest wavelength at which photo electric effect is possible. Thus we have hc   th So for wavelength of incident light    th , the energy of incident photons will become less then the work function of the metal and hence photoelectric effect will not start. Thus for a given metal surface photoelectric emission will start at    th or    th . 1.7

EINSTEIN RELATION: Einstein suggested that the energy of photon (h  ) which is more than work function of a metal when incident on the metal surface is used by the electron after absorption in two parts. (i) A part of energy of absorbed photon is used by the free electron in work done in coming out from the metal surface as work function.

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MODERN PHYSICS - I

(ii) The remaining part of the photon energy will be gained by the electron in the form of kinetic energy after ejection from the metal surface. 





e e

–

Work function =  If a light beam of frequency  (each photon energy = h) is incident on a metal surface having work function  then for h   , we have h   

1 mv2max 2

...(1)

In equation (1) the second terms on right hand side of equation is

1 2 mv max , which is the 2

maximum kinetic energy of the ejected electron. In practical cases whenever an electron absorbs a photon from incident light, it comes out from the metal surface if h   but in process of ejection it may collide with the neighbouring electrons and before ejection it may loose some energy during collisions with the neighbouring electrons. In this case after ejection the kinetic energy of ejected electrons will be certainly less then (h – ) . If we assume there are some electrons which do not loose any energy in the process of ejection, will come out from the metal surface with the maximum kinetic energy given as 1 mv2max  h   2 Thus all the ejected electrons from the metal surface may have different kinetic energies,

distributed from 0 to

1 2 mv max . 2

Graph between Kmax and f 1

2

Kmax

f(0)1

f f(0)2

W1 W2

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MODERN PHYSICS - I

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, K max  h f – W comparing it with y = mx + c, the graph between Kmax and f is a straight line with positive slope and negative intercept. From the graph we can note the following points. (i) Kmax = 0 at f = f 0 (ii) Slope of the straight line is h, a universal constant. i.e., if graph is plotted for two different metals 1 and 2, slope of both the lines is same. (iii) The negative intercept of the line is W, the work function, which is characteristic of a metal, i.e., intercepts for two different metals will be different. Further, W2 > W1  (f0)2 > (f 0)1 Here f 0 = threshold frequency as W = h f0 Ex.2

The photoelectric threshold of the photo electric effect of a certain metal is 2750 Å. Find (i) The work function of emission of an electron from this metal, (ii) Maximum kinetic energy of these electrons, (iii) The maximum velocity of the electrons ejected from the metal by light with a wavelength 1800 Å.

Sol.

(i) Given that the threshold wavelength of a metal is  th  2750 Å . Thus work function of metal can be given as hc 12431  eV  4.52 eV  th 2750 (ii) The energy of incident photon of wavelength 1800 Å on metal in eV is 

12431 eV  6.9 eV 1800 Thus maximum kinetic energy of ejected electrons is E

K Emax  E   = 6.9 – 4.52 eV = 2.38 eV (iii) If the maximum speed of ejected electrons is vmax then we have 1 mv2max  2.38 eV 2

2  2.38  1.6  1019  9.15  10 5 m / s 9.1  1031 Light quanta with a energy 4.9 eV eject photoelectrons from metal with work function 4.5 eV. Find the maximum impulse transmitted to the surface of the metal when each electrons flies out. According to Einstein’s photoelectric equation or

Ex.3

Sol.

vmax 

1 mv2max  hv    4.9  4.5  0.4 eV 2 If E be the energy of each ejected photo electron E

momentum of electrons is

P

2mE

We know that change of momentum is impulse. Here the whole momentum of electron is gained when it is ejected out thus impulse on surface is Impulse 

2mE

Substituting the values, we get Maximum impulse 

25 2  9.1  1031  0.4  1.6  1019  3.45  10 kg m / sec

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MODERN PHYSICS - I

In a experiment tungsten cathode which has a threshold 2300 Å is irradiated by ultraviolet light of wavelength 1800 Å. Calculate (i) Maximum energy of emitted photoelectron and (ii) Work function for tungsten (Mention both the results in electron-volts) Given Plank’s constant h  6.6  10 34 joule-sec, 1 eV  1.6  10 19 joule and velocity of light c  3  10 8 m/sec

Sol.

The work function of tungsten cathode is 

hc 12431  eV = 5.4 eV  th 2300

The energy in eV of incident photons is E

hc 12431  eV  1800

The maximum kinetic energy of ejected electrons can be given as KEmax  E    6.9  5.4 eV  1.5 eV Ex.5

Sol.

Light of wavelength 1800 Å ejects photoelectrons from a plate of a metal whose work functions is 2 eV. If a uniform magnetic field of 5  10 5 tesla is applied parallel to plate, what would be the radius of the path followed by electrons ejected normally from the plate with maximum energy. Energy of incident photons in eV is given as E

12431 eV 1800

As work function of metal is 2 eV, the maximum kinetic energy of ejected electrons is KEmax  E   If vmax

= 6.9 – eV = 4.9 eV be the speed of fasted electrons then we have 1 mv2max  4.9  1.6  1019 joule 2

vmax 

or

2  4.9  1.6  1019 9.1  1031

 1.31 10 6 m / s

When an electron with this speed enters a uniform magnetic field normally it follows a circular path whose radius can be given by r

or

r

mv qB

9.1  1031  1.31  106 1.6  1019  5  105

[As qv B 

or

mv 2 ] r

r = 0.149 m

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MODERN PHYSICS - I 1.8

Experimental Study of Photo Electric Effect : Experiments with the photoelectric effect are performed in a discharge tube apparatus as illustrated in figure shown. The cathode of discharge tube is made up of a metal which shows photoelectric effect on which experiment is being carried out. Incident light p > pth

Cathode

Anode

e– e–

A V + V

– S

A high potential is applied to a discharge tube through a variable voltage source and a voltmeter and an ammeter are connected a measure the potential difference across the electrodes and to measure photoelectric current. Light with frequency more than threshold frequency of cathode metal is incident on it, due to which photoelectrons are emitted from the cathode. These electrons will reach the anode and constitute the photoelectric current which the ammeter will show. Now we start the experiment by closing the switch S. Initially the variable battery source is set at zero potential. Even at zero potential variable source, ammeter will show some current because due to the initial kinetic energy some electrons will reach the anode and cause some small current will flow. But as we know majority of ejected electrons have low values of kinetic energies which are collected outside the cathode and create a could of negative charge, we call space charge, as shown in figure shown. Incident light

Space charge

Cathode

e–

Anode e– A

V – S

+ V

(small potential difference)

If the potential difference applied across the discharge tube is gradually increased from the variable source, positive potential of anode starts pulling electrons from the space charge. As potential difference increases, space charge decrease and simultaneously the photoelectric current in circuit also increases. This we can also see in the variation graph of current with potential difference as shown in figure shown.

i

is1

VP1

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V

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Page # 10

MODERN PHYSICS - I

A shown in graph, we can see as potential difference increases, current in circuit increases. But at a higher voltage VP1 space charge vanishes and at this voltage anode is able to pull the slowest electron (zero kinetic energy) ejected by the cathode. Now as all the ejected electrons from cathode start reading anode. If further potential difference is increased, it will not make any difference in the number of electrons reaching the anode hence, further increases in potential difference will not increases the current. This we can see in figure shown that beyond VP1 current in circuit becomes constant. This current is1 is called saturation current. This potential difference VP1 at which current becomes saturated is called “pinch off voltage”. Now if the frequency of incident light is kept constant and its intensity is further increased, then the number of incident photons will increase which increases the number of ejected photo electrons so current in circuit increases and now in this case at higher intensity of incident light, current will not get saturated at potential difference VP1 as now due to more electron emission, space charge will be more and it will not vanish at VP1. To pull all the electrons emitted from cathode more potential difference is required. This we can se from figure shown, that at higher intensity I2 (I2 > I1) current becomes saturated at higher value of potential difference VP2. i is2 I2 Intensity I2 > I1 is1

1.9.

I1

V VP1 VP2 Beyond VP2, we can see that all the electrons ejected from cathode are reaching the anode are current become saturated at is2 because of more electrons. Another point we can see from figure shown that when V = 0 then also current is more at high intensity incident radiation as the number of electrons of high kinetic energy are also more in the beginning which will reach anode by penetrating the space charge. Kinetic Energies of Electrons Reaching Anode We know that when electrons are ejected from cathode then kinetic energies may vary from 1 2 0 to mv max . If V is the potential difference applied across the discharge tube then it will 2 accelerates the electron while reaching the anode. the electron which is ejected from cathode with zero kinetic energy will be the slowest one reaching the anode if its speed is v1 at anode then we have 0  ve 

1 mv12 2

1 2 mv max will be the 2 fastest one when it will reach anode. If its speed is v2 at anode then we have

Similarly the electron ejected from cathode with maximum kinetic energy

1 1 mv2max  eV  mv22 2 2

1.10

Thus we can say that all the electrons reaching anode will have their speeds distributed from v1 to v2. Reversed Potential Across Discharge Tube : Now the experiment is repeated with charging the polarity of source across the discharge tube. Now positive terminal of source is connected to the cathode of discharge tube. When a light beam incident on the cathode with (h  ) , photoelectrons are ejected and move towards anode with negative polarity.

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Page # 11

MODERN PHYSICS - I Incident light v > vth

Cathode

Anode

e–

+

–

e–

–

+ V

Now the electrons which are ejected with very low kinetic energy are attracted back to the cathode because of its positive polarity. Those electrons which have high kinetic energies will rush toward, anode and may constitute the current in circuit. In this case the fastest electron ejected from cathode will be retarded during its journey to 1 2 mv max , if potential 2 difference across the discharge tube is V then the seed vf with which electrons will reach anode can be given as 1 1 mv2max  eV  mv2f ....(1) 2 2 Thus all the electrons which are reaching anode will have speed less then or equal to vf. Remaining electrons which have relatively low kinetic energy will either be attracted to cathode just after ejection or will return during their journey from cathode to anode. Only those electrons will case current of flow in circuit which have high kinetic energies more then eV which can overcome the electric work against electric forces on electron due to opposite polarity of source.

anode. As the maximum kinetic energy just after emission at cathode is

1.11

Cut off Potential or Stopping Potential : We have seen with reverse polarity electrons are retarded in the discharge tube. If the potential difference is increased with reverse polarity, the number of electrons reaching anode will decrease hence photo electric current in circuit also decreases, this we can see from figure shown which shows variation of current with increase in voltage across discharge tube in opposite direction. Here we can see that at a particular reverse voltage V0, current in circuit becomes zero. This is the voltage at which the faster electron from cathode will be retarded and stopped just before reaching the anode. i I2 I1

V0

0

VP1 VP2

Intensity I2 > I1 Frequency v (same for both radiation)

V

Reverse voltage

This voltage V0, we can calculate from equation (1) by substituting vf = 0 hence 1 mv2max  eV0  0 2

or

eV0 

1 2 mvmax 2

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Page # 12

MODERN PHYSICS - I

1 mv2max 2 or ...(2) V0  e hv   V0  or ...(3) e We can see one more thing in figure shown that the graphs plotted for two different intensities I1 and I2, V0 is same. Current in both the cases in cut off at same reverse potential V0. The reason for this is equation-(2) and (3). It is clear that the value of V0 depends only on the maximum kinetic energy of the ejected electrons which depends only on frequency of light and not on intensity of light. Thus in above two graphs as frequency of incident light is same, the value of V0 is also same. This reverse potential difference V0 at which the fastest photoelectron is stopped and current in he circuit becomes zero is called cut off potential or stopping potential.

1.12

Effect of Change in Frequency of Light on Stopping Potential : i

If we repeat the experiment by increasing the frequency of incident light with number of incident photons constant, the variation graph of current with voltage will be plotted as shown in figure shown.

Frequency (v2 > v1) is1 v2 (II) (I) v1 V02

V01

V

VP1

This graph is plotted for two incident light beams of different frequency v1 and v2 and having same photon flux. As the number of ejected photoelectrons are same in the two cases of incident light here we can see that the pinch off voltage V01 as well as saturation current is1 are same. But as in the two cases the kinetic energy of fastest electron are different as frequencies are different, the stopping potential for the two cases will be different. In graph II as frequency of incident light is more, the maximum kinetic energy of photoelectrons will also be high and to stop it high value of stopping potential is needed. These here V01 and V02 can be given as hv1   V01  ...(4) e hv2   V02  and ...(5) e In general for a given metal with work function , if Vo is the stopping potential for an incident light of frequency v then we have eV0  hv   or eV0  hv  hvth ...(6) hv th  h V0    v  or ...(7)  e e Equation (7) shows that stopping potential V0 is linearly proportional to the frequency v of incident light. The variation of stopping potential with frequency v can be shown in figure shown. Here equation .(6) can be written as 1 2 mv max  eV0  h ( v  v th ) ...(8) 2 This equation (8) is called Einstein’s Photo Electric Effect equation which gives a direction relationship between the maximum kinetic energy stopping potential frequency of incident light and the threshold frequency.

V0

tan  

h e

 hv th e

v = vth

V

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MODERN PHYSICS - I Ex.6

Find the frequency of light which ejects electrons from a metal surface fully stopped by a retarding potential of 3 V. The photo electric effect begins in this metal at frequency of 6  10 14 sec 1 . Find the work function for this metal.

Sol.

The threshold frequency for the given metal surface is v th  6  1014 Hz Thus the work function for metal surface is   h v th  6.63  10 34  6  1014  3.978  10 19 J As stopping potential for the ejected electrons is 3V, the maximum kinetic energy of ejected electrons will be KEmax  3 eV  3  1.6  10 19 J  4.8  10 19 J According to photo electric effect equation, we have hv  hv th  KEmax or frequency of incident light is v

Ex.7

Sol.

  KEmax h



3.978  1019  4.8  1019  1.32  1015 Hz 6.63  1034

Electrons with maximum kinetic energy 3eV are ejected from a metal surface by ultraviolet radiation of wavelength 1500 Å. Determine the work function of the metal, the threshold wavelength of metal and the stopping potential difference required to stop the emission of electrons. Energy of incident photon in eV is 12431 eV 1500 According to photo electric effect equation, we have E

E    KEmax  or   E  KEmax or = 8.29 – 3 eV or = 5.29 eV Threshold wavelength for the metal surface corresponding to work function 5.29 eV is given as  th 

12431 Å = 2349.9 Å 5.29

Stopping potential for the ejected electrons can be given as V0 

Ex.8

Sol.

KEmax 3eV   3 volt e e

Calculate the velocity of a photo-electron, if the work function of the target material is 1.24 eV and the wavelength of incident light is 4360 Å. What retarding potential is necessary to stop the emission of the electrons ? Energy of incident photons in eV on metal surface is 12431 eV = 2.85 eV 4360 According to photo electric effect equation we have E

1 1 mv2max mv2max  E –  or 2 2 = 2.85 – 1.24 eV = 1.61 eV The stopping potential for these ejected electrons can be given as E

V0 

1.61 eV 1 / 2 mv2max   1.61 volts e e

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Page # 14 Ex.9

Sol.

MODERN PHYSICS - I

Determine the Planck’s constant h if photoelectrons emitted from a surface of a certain metal by light of frequency 2.2 × 1015 Hz are fully retarded by a reverse potential of 6.6 V and those ejected by light of frequency 4.6 × 1015 Hz by a reverse potential of 16.5 eV. From photo electric effect equation, we have Here

hv1    eV01

...(1)

and

hv2    2 eV02

...(2)

Subtracting equation (1) from equation (2), we get h(v2  v1 )  e (v02  v01 ) or

h

(v02  v01 )(1.6  10–19 ) (v2  v1 )

or

h

(16.5  6.6)(1.6  1019 ) or (4.6  2.2)  1015

= 6.6 × 10–34 J-s

Ex.10 When a surface is irradiated with light of wavelength 4950 Å, a photo current appears which vanishes if a retarding potential greater than 0.6 volt is applied across the photo tube. When a different source of light is used, it is found that the critical retarding potential is changed to 1.1 volt. Find the work function of the emitting surface and the wavelength of second source. If the photo electrons (after emission from the surface) are subjected to a magnetic field of 10 tesla, what changes will be observed in the above two retarding potentials. Sol. In first case the energy of incident photon in eV is 12431 eV = 2.51 eV 4950 The maximum kinetic energy of ejected electrons is E1 

KEmax 1  eV01 = 0.6 eV

Thus work function of metal surface is given as   E1  KEmax 1 = 2.51 – 0.6 eV = 1.91 eV

In second case the maximum kinetic energy of ejected electrons will become KEmax 2  eV02 = 1.1 eV

Thus the incident energy of photons can be given as E2    KEmax 2

E2 = 1.91 + 1.1 eV = 3.01 eV Thus the wavelength of incident photons in second case will be 12431 Å = 4129.9 Å 3.01 When magnetic field is present there will be no effect on the stopping potential as magnetic force can not change the kinetic energy of ejected electrons.  

Ex.11 (a) If the wavelength of the light incident on a photoelectric cell be reduced from 1 to 2 Å, then what will be the change in the cut-off potential ? (b) Light is incident on the cathode of a photocell and the stopping voltages are measured from light of two difference wavelengths. From the data given below, determine the work function of the metal of the cathode in eV and the value of the universal constant hc/e. Wavelength (Å) Stopping voltage (volt) 4000 1.3 4500 0.9

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MODERN PHYSICS - I Sol.

(a) Let the work function of the surface be . If v be the frequency of the light falling on the surface, then according to Einstein’s photoelectric equation, the maximum kinetic energy KEmax of emitted electron is given by KEmax  hv   

hc  

KEmax  eV0

We know that, Where V0= cut-off potential. or Now,

hc   or  V0  V02  V01

eV0 

V0 

hc   e e

 hc    hc         e 2 e   e1 e  

hc  1 1  hc  1   2     e   2 1  e  1 2 

...(1)

(b) From equation (1), we have hc V0 (12 )  e 1   2



(1.3  0.9)[(4000  1010 )  (4500  1010 )] = 1.44 × 10–6 V/m 500  1010 hc   e e

Now,

V0 

or

 hc 1.44  106   V0   1.3  2.3 V e e 4000  1010

or

  2.3 eV

Ex.12 A low intensity ultraviolet light of wavelength 2271 Å irradiates a photocell made of molybdenum metal. If the stopping potential is 1.3 V, find the work function of the metal. Will the photocell work if it is irradiated by a high intensity red light of wavelength 6328 Å ? Sol. The energy in eV of incident photons is 12431 eV  5.47 eV 2271 As stopping potential for ejected electrons is 1.3 V, the maximum kinetic energy of ejected electrons will be E

KEmax  eV0  1.3 eV Now from photoelectric effect equation, we have E    KEmax or

  E  KEmax

or   5.47  1.3 eV = 4.17 eV Energy in eV for photons for red light of wavelength 6328 Å is E 

12431 eV  1.96 eV 6328

As E   , photocell will not work if irradiated by this red light no matter however intense the light will be.

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Page # 16 2.

MODERN PHYSICS - I

FORCE DUE TO RADIATION (PHOTON) Each photon has a definite energy and a definite linear momentum. All photons of light of a h particular wavelength  have the same energy E  hc and the same momentum p =   When light of intensity I falls on a surface, it exerts force on that surface. Assume absorption and reflection coefficient of surface be 'a' and 'r' and assuming no transmission. Assume light beam falls on surface of surface area 'A' perpendicularly as shown in figure.

For calculating the force exerted by beam on surface, we consider following cases. Case (I) : a = 1, r = 0 initial momentum of the photon =

h 

final momentum of photon = 0 change in momentum of photon =

h 

(upward)

h  energy incident per unit time = IA P 

no. of photons incident per unit time 

IA IA  h hc

 total change in momentum per unit time = n P 

IA IA h   (upward ) c hc 

force on photons = total change in momentum per unit time 

IA (upward ) c IA (downward) c

 force on plate due to photon(F) = 

pressure =

IA I F = = cA c A

Case : (II) when r = 1, a = 0 initial momentum of the photon

=

h 

(downward)

h (upward)  h h 2h    change in momentum     energy incident per unit time = I A

final momentum of photon

=

no. of photons incident per unit time =

IA hc

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MODERN PHYSICS - I

 total change in momentum per unit time = n. P 

IA 2h 2IA .  hc  C

force = total change in momentum per unit time 2IA c F 2IA 2I P   A cA c F

pressure

Case : (III) When 0 < r < 1

(upward on photons and downward on the plate)

a+r=1

2h (upward)  h change in momentum of photon when it is absorbed = (upward)  IA no. of photons incident per unit time = hc IA .r No. of photons reflected per unit time = hc IA (1 – r ) No. of photon absorbed per unit time = hc

change in momentum of photon when it is reflected =

force due to absorbed photon (Fa)



IA IA h (1 – r ). = (1 – r ) (downward)  hc c

Force due to reflected photon (Fr)



IA 2h 2IA .r  (downward) hc  c

total force

= Fa + Fr 

2IAr IA IA (1 – r )   (1  r ) c c c

Now pressure P 

I IA 1 = (1  r ) (1  r )  c c A

Ex.13 Calculate force exerted by light beam if light is incident on surface at an angle  as shown in figure. Consider all cases.



Sol.

Case - I a = 1, r = 0 initial momentum of photon (in downward direction at an angle  with vertical) is h/ final momentum of photon = 0

change in momentum (in upward direction at an angle  with vertical) =

h 

[ 

]

energy incident per unit time = I A cos  Intensity = power per unit normal area I

P A cos 

P = I A cos 

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MODERN PHYSICS - I

No. of photons incident per unit time =

IA cos  . hc

total change in momentum per unit time (in upward direction at an angle  with vertical) IA cos  IA cos   h  .  c hc





[

]

Force (F) = total change in momentum per unit time IA cos  (direction c



F

on photon and 

on the plate)

Pressure = normal force per unit Area Pressure =

F cos  A

P=

I IA cos 2  2 = cos  c cA

Case II When r = 1, a = 0  change in momentum of one photon 

2h cos  

(upward)

No. of photons incident per unit time 

energy incident per unit time h

h sin  

IA cos .  hc







h sin  

total change in momentum per unit time 



h cos  

h cos  

IA cos . 2h  cos   hc

force on the plate 

Pressure

Case III





2IA cos 2  c

2IA cos2  cA

P

2IA cos 2  (upward) c

(downward) 2I cos2  c

0 < r < 1,

a +r=1

change in momentum of photon when it is reflected =

2h cos  (downward) 

change in momentum of photon when it is absorbed =

h (in the opposite direction of incident 

beam) energy incident per unit time = I A cos ) no. of photons incident per unit time =

no. of reflected photon (nr) =

IA cos . hc

IA cos .r hc

no. of absorbed photon (nQ) =

IA cos . (1 – r) hc

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Page # 19

MODERN PHYSICS - I force on plate due to absorbed photons Fa = na. Pa 

IA cos . h (1 – r) hc 



IA cos  (1 – r) hc

(at an angle  with vertical



)

force on plate due to reflected photons Fr = nr Pr 

IA cos . 2h  cos  (vertically downward) hc 



IA cos2  2r c

now resultant force is given by FR  Fr2  Fa2  2FaFr cos  

and, pressure

IA cos  (1 – r)2  (2r)2 cos2   4r(r – 1) cos2  c

P

Fa cos   Fr A



IA cos (1 – r) cos  IA cos2 .2r  cA cA



I cos2  I cos2  I cos2  (1 – r)  2r  (1  r) c c c

Ex.14 A perfectly reflecting solid sphere of radius r is kept in the path of a parallel beam of light of large aperture. If the beam carries an intensity I, find the force exerted by the beam on the sphere. Sol. Let O be the centre of the sphere and OZ be the line opposite to the incident beam (figure). Consider a radius about OZ to get a making an angle  with OZ. Rotate this radius about OZ to get a circle on the sphere. Change  to  +d and rotate the radius about OZ to get another circle on the sphere. The part of the sphere between these circles is a ring of area 2r2 sin d. Consider a small part A of this ring at P. Energy of light falling on this part in time t is R

p

O

Q

Z

U = It(A cos ) The momentum of this light falling on A is U/c along QP. The light is reflected by the sphere along PR. The change in momentum is p = 2

U 2 cos = t (A cos2 ) c c



(direction along OP )

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MODERN PHYSICS - I

The force on A due to the light falling on it, is p 2  t c



A cos2 

(direction along OP )

The resultant force on the ring as well as on the sphere is along ZO by symmetry. The component of the force on A along ZO p 2 cos  IA cos2  t c



(along ZO )

The force acting on the ring is  /2

The force on the entire sphere is

F=

 0

 /2

F=

 0

2 I(2r2 sin d) cos3  c

dF =

4 r 2 I cos3  d c

4 r 2 I cos3  d(cos ) = – c

 /2

  0

 /2

4r2I  cos4     c  4 0



r 2I c

Note that integration is done only for the hemisphere that faces the incident beam. 3.

DE-BROGLIE WAVELENGTH OF MATTER WAVE A photon of frequency v and wavelength has energy. E = hv =

hc 

By Einstein's energy mass relation, E = mc2 the equivalent mass m of the photon is given by.

m=

or

=

E hv h  2  2  c c c

h mc

or =

.

..(i)

h p

...(ii)

Here p is the momentum of photon. By analogy de-Broglie suggested that a particle of mass m moving with speed v behaves in some ways like waves of wavelength (called de-Broglie wavelength and the wave is called matter wave) given by, =

h h = p mv

...(iii)

where p is the momentum of the particle. Momentum is related to the kinetic energy by the equation, p=

2Km

and a charge q when accelerated by a potential difference V gains a kinetic energy K = qV. Combining all these relations Eq. (iii), can be written as, h h  = mv  p 

h

h 

2Km

2qVm

(de=Broglie wavelength)....(iv)

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MODERN PHYSICS - I 3.1

de-Broglie wavelength for an electron If an electron (charge = e) is accelerated by a potential of V volts, it acquires a kinetic energy, K = eV Substituting the value of h, m and q in Eq. (iv), we get a simple formula for calculating deBroglie wavelength of an electron.

 (in Å) = 3.2

150 V(in volts)

de-Broglie wavelength of a gas molecule : let us consider a gas molecule at absolute temperature T. Kinetic energy of gas molecule is given by 3 K.E. = kT ; k = Boltzman constant 2 h

gas

4.

molecules

= 2mkT

ATOMIC MODEL A model is simply a set of hypothesis based on logical & scientific facts. Theory : When any model satisfies majority of scientific queries by experimental verification then it is termed as theory otherwise, model is simply not accepted. In Nutshell we can say that every theory is a model but every model is not a theory. So, after more & more clarity about the substances, various new models like Dalton, Thomseon, Rutherford, Bohr etc came into the pictures.

4.1

Dalton's atomic model : (i) Every element is made up of tiny indivisible particles called atoms. (ii) Atoms of same element are identical both in physical & chemical properties while atoms of different elements are different in their properties. (iii) All elements are made up of hydrogen atom. The mass of heaviest atom is about 250 times the mass of hydrogen atom while radius of heaviest atom is about 10 times the radius of hydrogen atom. (iv) Atom is stable & electrically neutral. Reason of Failure of model : After the discovery of electron by U. Thomson (1897), it was established that atom can also be divide. Hence the model was not accepted.

4.2

Thomson's atomic model (or Plum-pudding model) (i) Atom is a positively charged solid sphere of radius of the order of 10–10 m in which electrons are embedded as seeds in a watermelon (ii) Total charge in a atom is zero & so, atom is electrically neutral. electron

uniformly distributed positively charged matter

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4.3 (a)

(b)

4.4

MODERN PHYSICS - I

Achievements of model : Explained successfully the phenomenon of thermionic emission, photoelectric emission & ionization Type of line spectrum Emission line spectrum : When an atomic gas or vapour at a pressure less than the atmospheric pressure is excited by passing electric discharge, the emitted radiation has spectrum which contains certain specific bright lines only. These emission lines constitute emission spectrum. These are obtained when electron jumps from excited states to lower states. The wavelength of emission lines of different elements are different. For one element the emission spectrum is unique. It is used for the determination of composition of an unknwon substance. Absorption line spectrum : When white light is passed through a gas, the gas is found to absorb light of certain wavelength, the bright background on the photographic plate is then crossed by dark lines that corresponds to those wavelengths which are absorbed by the gas atoms. The absorption spectrum consists of dark lines on bright background. These are obtained due to absorption of certain wavelengths, resulting into transition of atom from lower energy states to higher energy states. (The emission spectrum consists of bright lines on dark background.) Failure of the model : (i) It could not explain the line spectrum of H-atom (ii) It could not explain the Rutherford's  - particle scattering experiment Ruthorford's Atomic Model In 1911, Earnest Ruthorford performed a critical experiment that showed that Thomson's model could not be correct. In this experiment a beam of positively charged alpha particles (helium nuclei) was projected into a thin gold foil. It is observed that most of the alpha particles passed through the foil as if it were empty space. But some surprising results are also seen. Several alpha particles are deflected from their original direction by large angles. Few alpha particles are observed to be reflected back, reversing their direction of travel as shown in figure-l.2.

If Thomson model is assumed true that the positive charge is spreaded uniformly in the volume of an atom then the alpha particle can never experience such a large repulsion due to which it will be deflected by such large angles as observed in the experiment. On the basis of this experiment Ruthorford presented a new atomic model. In this new atomic model it was assumed that the positive charge in the atom was concentrated in a region that was small relative to the size of atom. He called this concentration of positive charge, the nucleus of the atom. Electrons belonging to the atom were assumed to be moving in the large volume of atom outside the nucleus. To explain why these electrons were not pulled into the nucleus, Ruthorford said that electrons revolve around the "nucleus in orbits around the positively charged nucleus in the same manner as the planets orbit the sun. The corresponding atomic model can be approximately shown in figure.

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Page # 23

MODERN PHYSICS - I

–

+

–

1.

2.

Reason of Failure of model : It could not explain the line spectrum of H-atom. Justification : According to Maxwell's electromagnetic theory every accelerated moving charged particle radiates energy in the form of electromagnetic waves & therefore duringh revolution of e– in circular orbit its frequency will contanously vary (i.e. decrease) which will result in the conitnuous emission of lines & therefore spectrum of atom must be continuous but in reality, one obtains line spectrum for atoms. It could not explain the stablity of atoms. Justification : Since revolving electron will continuously radiates energy & therfore radii of circular path will continuously decrease & in a time of about 10–8 sec revolving electron must fall down in a nucleus by adopting a spiral path

+

Path of electron spiral DETERMINATION OF DISTANCE OF CLOSEST APPROACH : When a positively charged particle approaches towards stationary nucleus then due to repulsion between the two, the kinetic energy of positively charged particle gradually decreases and a stage comes when its kinetic energy becomes zero & from where it again starts retracting its original path. Definition : The distance of closest approach is the minimum distance of a stationary nucleus with a positively charged particle making head on collision from a point where its kinetic energy becomes zero. Suppose a positively charged particle A of charge q1 = (=z1e) approaches from in finity towards a stationary nucleus of charge z2e then, z2e Suppose a positively charged particle A of charge z,e + q1 ( = z1 e) approaches from in finity towards a A B Stationary nucleus

stationry nucleus of charge z2e then, Let at point B, kinetic energy of particle A becomes zero then by the law of conservation of energy at point A & B, TEA = TEB KEA + PEA = KEB + PEB E+0=0+

k(z1e)(z2 e) (in joule) r0



r0 

r0

k(z1e)(z2 e) m E

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MODERN PHYSICS - I

Ex.15 An -particle whith kinetic energy 10 MeV is heading towards a stationary point-nucleus of atomic number 50. Calculate the distance of closest approach. Sol.  TEA = TEB B A K  (2e)(50e) 6  10 × 10 e =   particle r 0

r0

–14

r0 = 1.44 × 10 m r0 = 1.44 × 10–4 A

Ex.16 A beam of  - particles of velocity 2.1 × 107 m/s is scattered by a gold (z = 79) foil. Find out the distance of closest approach of the - particle to the gold nucleus. The value of charge/mass for  - particle is 4.8 × 107 c/kg. Sol.

1 K(2e)(ze) m v2  2 r0

r0 

 2e  2K  (79 e)  m  v2



2  (9  108 )(4.8  107 )(79  1.6  10–19 ) ; r0  2.5  10–14 m (2.1  107 )2

Ex.17 A proton moves with a speed of 7.45 × 105 m/s directing towards a free proton originally at rest. Find the distance of closest approach for the two protons. Sol.  v = 7.45 × 105 m/s u=0 Proton

Free proton

Originally

v2

v1

r0 Proton free proton after movement At the time of distance of closest approach 1 ke2 1 1 2 2 2 By the law of cons. of energy 2 mv  0  r  2 mv1  2 mv1 0

...(1)

By the cons. of momentum mv + 0 = mv1 + mv1 v1 



From equation (1)

r0 

v 2

ke2  v 1  m  mv2 =  2 r0 2

2

4 4  (9  109 )(1.6  10–19 )2  ke2  ; r0  1.0  10–12 m 2 mv (1.66  10–27 )(7.45  105 )2

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MODERN PHYSICS - I 5

BOHR'S MODEL OF AN ATOM Between 1913 and 1915 Niels Bohr developed a quantitative atomic model to the Hydrogen atom that could account for its spectrum. The model incorporated the nuclear model of the atom proposed by Rutherford on the basis of his experiments. We shall see that this model was successful in its ability to predict the gross features of the spectrum emitted by Hydrogen atom. This model was developed specifically for Hydrogenic atoms. Hydrogenic atoms are tho se which consist of a nucleus with positive charge + Ze (Z = atomic number, e = charge of electron) and a single electron. More complex electronelectron interactions in an atom are not accounted in the Bohr's Model that's why it was valid only for one electron system or hydrogenic atoms. The Bohr model is appropriate for one electron systems like H, He+, Li+2 etc. and it was successful upto some extent in explaining the features of the spectrum emitted by such hydrogenic atoms. However this model is not giving a true picture of even these simple atoms. The true picture is fully a quantum mechanical affair which is different from Bohr model in several fundamental ways. Since Bohr model incorporates aspects of some classical and some modern physics, it is now called semiclassical model Bohr has explained his atomic model in three steps called postulates of Bohr's atomic model. Lets discuss one by one.

5.1

First Postulate In this postulate Bohr incorporate and analyses features of the Rutherford nuclear model of atom. In this postulate it was taken that as the mass of nucleus is so much greater then the mass of electron, nucleus was assumed to be at rest and electron revolves around the nucleus in an orbit. The orbit of electron is assumed to be circular for simplicity. Now the statement of first postulate is "During revolution of electron around the nucleus in circular orbit, the electric coulombian force on electron is balanced by the centrifugal force acting on it in the rotating frame of reference." If electron revolves with speed v in the orbit of radius r. Then relative to rotating frame attached with electron, the centrifugal force acting on it is

v

+

Fd

Fe –e

+Ze

mv2 ...(1) r The coulombian force acting on electron due to charge of nucleus (+Ze) is Fef 

K(e)(Ze) ...(2) r2 Now according to first postulate from equation (1) & (2) we have Felectric 

mv2 KZe2  r r2 mv2 KZe2  or r r2 Equation (3) is called equation of Bohr's first postulate.

5.2

...(3)

Second Postulate In the study of atom, Bohr found that while revolving around the nucleus the orbital angular momentum of the electron was restricted to only certain values, we say that the orbital angular momentum of the electron is quantized. He therefore took this as a second postulate of the model. The statement of second Postulate is, Bohr proposed that -"During revolution around the nucleus, the orbital angular momentum of electron L could not have just any value, it can take up only those values which are integral multiples of Plank's Constant divided by 2 i.e. h/2"

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MODERN PHYSICS - I

Thus the angular momentum of electron can be written as nh ...(1) 2 Where n is a positive integer, known as quantum number. In an orbit of radius r if an electron (mass m) revolves at speed v, then its angular momentum can be given as L = mvr ...(2) Now from equation (1) and (2), we have for a revolving electron nh mvr  ...(3) 2 h Equation (3) is known as equation of second postulate of Bohr model. Here the quantity 2 occurs so frequently in modern physics that, for convenience, it is given its own designation h, pronounced as "h-bar." h ...(4)  = 2 ~– 1.055 × 10–34 J-s Third Postulate While revolution of an electron in an orbit its total energy is taken as sum of its kinetic and electric potential energy due to the interaction with nucleus. Potential energy of electron revolving in an orbit of radius r can be simply given as

L=

5.3

K(e)(Ze) Kze2 = – ...(1) r r For kinetic energy of electron, we assume that relativistic speeds are not involved so we can use the classical expression for kinetic energy. Thus kinetic energy of electron in an orbit revolving at speed v can be given as 1 K  mv2 ...(2) 2 Thus total energy of electron can be given as U–

1 KZe2 mv2 – 2 r Here we can see that while revolving in a stable orbit, the energy of electron remains constant. From the purely classical viewpoint, during circular motion, as electron is accelerated, it should steadily loose energy by emitting electromagnetic radiations and it spiraled down into the nucleus and collapse the atom.

E=K+U=

Bohr in his third postulate stated that "While revolving around the nucleus in an orbit, it is in stable state, it does not emit any energy radiation during revolution. It emits energy radiation only when it makes a transition from higher energy level (upper orbit) to a lower energy level (lower orbit) and the energy of emitted radiation is equal to the difference in energies of electron in the two corresponding orbits in transition. " If an electron makes a transition form a higher orbit n2 to a lower orbit n1 as shown in figure. Then the electron radiates a single photon of energy E = E n2 – E n1  h

n2

n2

n1 –e

–e h

n1

h

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MODERN PHYSICS - I

Here En2 and E n1 are the total energies of electron in the two orbits n2 and n1. The emitted photon energy can be expressed as h where  is the frequency of radiated energy photon. If A be the wavelength of photon emitted then the energy of emitted photon can also be given as hc ...(1)  Similarly when energy is supplied to the atom by an external source then the electron will make a transition from lower energy level to a higher energy level. This process is called excitation of electron from lower to higher energy level. In this process the way in which energy is supplied to the electron is very important because the behaviour of the electron in the excitation depends only on the process by which energy as supplied from an external source. This we'll discuss in detail in later part of this chapter. First we'll study the basic properties of an electron revolving around the nucleus of hydrogenic atoms. E  h  

5.4

Properties of Electron in Bohr's Atomic Model Now we'll discuss the basic properties of an electron revolving in stable orbits, we call Bohr energy level. We have discussed that there are some particular orbits in which electron can revolve around the nucleus for which first and second postulates of Bohr model was satisfied. Thus only those orbits are stable for which the quantum number n = 1, 2, 3, ............ Now for nth orbit if we assume its radius is denoted by rn and electron is revolving in this orbit with speed vn. We can represent all the physical parameters associated with the electron in nth orbit by using a subscript n with the symbol of the physical parameters like rn, vn etc.

(a)

Radius of nth Orbit in Bohr Model Radius of electron in nth Bohr's orbit can be calculated using the first two postulates of the Bohr's model, using previous equations, we get nh 2mrn

vn 

Substituting this value of vn in equation mvr = rn 

n2h2 4 KZe2m

or

rn 

h2 n2  42Ke2m Z

or

rn  0.529 

nh , we get 2

2

n2 A Z

1.5.2 Velocity of Electron in nth Bohr's Orbit By substituting the value of rn we can calculate the value of vn as vn 

2KZe2 nh

or

vn 

2Ke2 Z  h n

or

vn  2.18  106 

Z m/s n

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Page # 28 (b)

MODERN PHYSICS - I

Time period of Electron in nth Bohr's Orbit Time period of electron of nth orbit is given by Tn 

(c)

Tn 

or

n3h3 42K2Z2 e4m

Current in nth Bohr's Orbit Electrons revolve around the nucleus in the nth Bohr's Orbit then due to revolvution there is current in the orbit and according to the definition of current, the current in the nth orbit will be total coulombs passing through a point in one seconds, and in an orbit an electron passes through a point fn times in one second so the current in the nth orbit will be In = fn × e In 

or

(d)

1 fn

42K2Z2 e5m n3h3

Energy of Electron in nth Orbit We've discussed that in nth orbit during revolution the total energy of electron can be given as sum of kinetic and potential energy of the electron as E n = K n + Un Kinetic energy of electron in nth orbit can be given as 1 mvn2 2 From equation of first postulate of Bohr Model we have for nth orbit KZe2 mvn2  rn From equation 1 1 KZe2 Kn  mv2n  2 2 rn the potential energy of electron in nth orbit is given as KZe2 Un  – rn Thus total energy of e– in nth orbit can given as Kn 

1 KZe2 KZe2 1 KZe2 –  E n = K n + UR  2 r rn 2 rn n 1 | Un | which is a very useful relation, always followed by a 2 particle revolving under the action of a force obeying inverse square law. Now substituting the value of rn we get

Here we can see that | En || K n |

En  –

1 42KZe2m KZe2  2 n2h2 –

or

22K2Z2 e4m n2h2

22K2Z2 e4m Z2  2 h2 h Substituting the value of constants in above equation we get

or

En 

Z2 eV n2 The above equation can be used to find out energies of electron in different energy level of different hydrogen atoms.

or

En  –13.6 

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MODERN PHYSICS - I (e)

Energies of Different Energy Level in Hydrogenic Atoms By the use of above equation we can find out the energies of different energy levels. Students should remember these energies for first six level as E1 = –13.6 Z2 eV E2 = –3.40 Z2 eV E3 = –1.51 Z2 eV E4 = –0.85 Z2 eV E5 = –0.54 Z2 ev E6 = –0.36 Z2 eV The above equations clearly shows that as the value of n increase, the difference between two consecutive energy levels decreases. It can be shown with the the help of figure, which shows the energy level diagram for a hydrogen atom. E=0 n n=6 E=–0.54eV n=5

E=–0.85eV

n=4

E= –1.51eV

n=3

E= –3.4eV

n=2

E= –13.6eV n=1 Now if we multiply the numerator and denominator of above equation by ch we get En  –

or

22K2 e4m Z2  ch  ch3 n2

En = – Rch ×

Z2 eV n2

22K2e4m is defined as Rydberg Constant and the value of it is given as R = ch3 10967800 m–1, which can be taken approximately as 107 m–1. For n = 1 and Z = 1 the energy is given as E = –Rch joules and is called as One Rydberg Energy 1 Rydberg = 13.6 eV = 2.17 × 10–18 joules Lets discuss some examples on Bohr's atomic model to understand it better. Where R 

Ex.18 What is the angular mometum of an electron in Bohr's Hydrogen atom whose energy is –3.4 eV ? Sol. Energy of electron in nth Bohr orbit of hydrogen atom is given by, E–

13.6 eV n2

Hence,

–3.4 = –

or or

n2 = 4 n=2

13.6 n2

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MODERN PHYSICS - I

The angular momentum of an electron in nth orbit is given as L =

L=

2h h   2 

T  n3

Thus,

T 

or

nh . Putting n = 2, we obtain 2

n3h3 4 k 2Z2 e4m 2

T1 n3  13 T2 n2

As T1 = 8T2, the above relation gives 3

 n1     8  n2 

or

n1 = 2n2

Thus the possible values of n1 and n2 are n1 = 2, n2 = 1 ; n1= 4, n2 = 2; n1 = 6, n2 = 3

and so on .........

Ex.19 An electron in the ground state of hydrogen atom is revolving in anti-clockwise direction Bˆ in the circular orbit of radius R as shown in figure nˆ v 30° (i) Obtain an expression for the orbital magnetic dipole moment of the electron. (ii) The atom is placed in a uniform magnetic induction B such that the plane normal of the electron orbit makes an anlge 30° with the magnetic induction. Find the torque experienced by the orbiting electron. Sol. (i) According to Bohr's second postulate h h mvr  n  (As for n = 1 first only) 2 2 h v1  or 2mr1 We know that the rate of flow of charge is current. Hence current in first orbit is  v  e i  ev1  e  1    v1  2r1  2r1

Now from each equation

i

Magnetic dipole moment,

e h eh   2r1 2nr1 4 2mr 2

M1 = i × A1

eh eh  r12  2 2 4m 4 mr (ii) Torque on the orbiting electron in uniform magnetic field is      MB or  = MB sin 30° M1 

or

or



eh B ehB   4m 2 8 m

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MODERN PHYSICS - I 6.

EXCITATION AND IONIZATION OF AN ATOM According to third postulate of Bohr model we've discussed when some energy is given to an electron of atom from an external source it may make a transition to the upper energy level. This phenomenon we call excitation of electron or atom and the upper energy level to which the electron is excited is called excited state. To excite an electron to a higher state energy can be supplied to it ijn two ways. Here we'll discuss only the energy supply by an electromagnetic photon. Other method of energy supply we'll discuss later in this chapter. According to Plank's quantum theory photon is defined as a packet of electromagnetic energy, which when absorbed by a physical particle, its complete electromagnetic energy is converted into the mechanical energy of particle or the particle utilizes the energy of photon in the form of increment in its mechanical energy. When a photon is supplied to an atom and an electron absorbs this photon, then the electron gets excited to a higher energy level only if the photon energy is equal to the difference in energies of the two energy levels involved in the transition. For example say in hydrogen atom an electron is in ground state (energy E1 = –13.6 eV). Now it absorbs a photon and makes a transition to n = 3 state (Energy E3 = –1.51 eV) then the energy of incident photon must be equal to E13  E 3 – E1 = (–1.51) – (13.6) eV = 12.09 eV Now we'll see what will happen when a photon of energy equal to 11 eV incident on this atom. From the above calculation of energy differences of different energy levels we can say that if the electron in ground state absorbs this photon it will jump to a state some where between energy level n = 2 and n = 3 as shown in figure. When electron in ground state absorbs a photon of 11 eV energy, its total energy becomes E = E1 + 11 = –13.6 + 11eV = –2.6 eV E  0 ev E4 = 0.85 eV E3 = 1.51 eV Ex = –2.6 eV (Hypothetical level)

E13  12.09 ev

E2 = –3.4 eV E12  10.2 ev

E1x  11 ev E1 = –13.6 eV

As discussed in previous sections, in an atom electron can not take up all energies. It can exist only in some particular energy levels which have energy given as – 13.6/n2eV. When a photon of energy 11 eV is absorbed by an electron in ground state. The energy of electron becomes –2.6 eV of it will excite to a hypothetical energy level X some where between n = 2 and n = 3 as shown in figure, which is not permissible for an electron. Thus when in ground state electron can absorb only those photons which have energies equal to the difference in energies of the stable energy level with ground state. If a photon beam incident on H-atoms having photon energy not equal to the difference of energy levels of Hatoms such as 11 eV, the beam will just be transmitted without any absorption by the Hatoms. Thus to excite an electron from lower energy level to higher levels by photons, it is necessary that the photon must be of energy equal to the difference in energies of the two energy levels involved in the transition. As we know that for higher energy levels, energy of electrons is less. When an electron is moved away from the nucleus to th energy level or at n = , the energy becomes (zero) or the electron becomes free from the attraction of nucleus or it is removed from the atom. In

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MODERN PHYSICS - I

13.6 2   fact when an electron is in an atom, its total energy is negative  E n   2 Z  . This negative n   sign shows that electron in under the influence of attractive forces of nucleus. When energy equal in magnitude to the total energy of an electron in a particular energy level is given externally, its total energy becomes zero or we can say that electron gets excited to th energy level or the electron is removed from the atom and atom is said to be ionized.

We know that removal of electron from an atom is called ionization. In other words, ionization is the excitation of an electron to n =  level. The energy required to ionize an atom is called ionization energy of atom for the particular energy level from which the electron is removed. In hydrogen atoms, the ionization energy for nth state can be given as En  E   En or

 13.6 Z2  En  0    eV n2  

or

En  

13.6 Z2 eV n2

When an electron absorbs a monochromatic radiation from an external energy source then it makes a transition from a lower energy level to a higher level. But this state of the electron is not a stable one. Electron can remain in this excited state for a very small internal at most of the order of 10–1 second. The time period for which this excited state of the electron exists is called the life time of the excited state. After the life time of the excited state the electron must radiate energy and it will jump to the ground state. Let us assume that the electron is initially in n2 state and it will jump to a lower state n1 then it will emit a photon of energy equal to the energy difference of the two states n1 and n2 as E  En2  En1 When E is the energy of the emitted photon. Now substituting the values of En2 and E n1 in above equation, we get E  

22K2Z2 e4m 22K2Z2 e4m  n22h2 n12h2

22K2 Z2 e4m  1 1  2 2 2  h  n1 n2 

or

E 

or

 1 1 E  13.6 Z2  2  2  eV  n1 n2 

Here 13.6 Z2 can be used as ionization energy for n = 1 state for a hydrogenic atom thus the energy of emitted photon can also be written as 1 1 E  IP  2  2  n n  1 2

Equation can also be used to find the energy of emitted radiation when an electron jumos from a higher orbit n2 to a lower orbit n1. If  be the wavelength of the emitted radiation then hc  This energy can be converted to eV by dividing this energy by the electronic charge e, as if wavelength is given in A, the energy in eV can be given as E 

E 

hc (in eV) e

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Page # 33

MODERN PHYSICS - I Substituting the values of h, c and e we get E 

(6.63  1034 )  (3  108 ) eV   (1.6  1019 )  1010

12431 eV ...(a)  Here in above equation, lambda is in Å units This equation is the most important in numerical calculations, as it will be very frequency used. From equation we have E 

1 hc 1  13.6Z2  2  2  eV   n1 n2  

or

1 13.6Z2  1 1   2  eV or  2  hc  n1 n2 

 1 1 1  RZ2  2  2    n1 n2 

(As Rydberg Constant R = 13.6/hc eV) Here v is called wave number of the emitted radiation and is defined as number of waves per unit length and the above relation is used to find the wavelength of emitted radiation when an electron makes a transition from higher level n2 to lower level n1 is called Rydberg formula. But students are advised to use equation (a) is numerical calculations to find the wavelength of emitted radiation using the energy difference in electron volt. If can be rearranged as 

12431 Å E(in eV)

No. of emission spectral lines : If the electron is excited to state with principal quantum number n then from the nth state, the electron may go to (n – 1)th state, ............, 2nd state or 1st state. So there are (n – 1) possible transitions starting from the nth state. The electron reaching (n – 1)th state may make (n – 2) different transitions. Similarly for other lower states. The total no. of possible transitions is (n – 1) + (n – 2) + (n – 3) + ..............2 + 1=

n(n – 1) 2

No. of absorption spectral lines : Since at ordinary temperatures, almost all the atoms remain in their lowest energy level (n = 1) & so absorption transition can start only from n = 1 level (not from n = 2, 3, 4, ..... levels). Hence, only Lyman seires is found in the absorption spectrum of hydrogen atom (which as in the emission spectrum, all the series are found No. of absorption spectral lines = (n – 1) 7.

SPECTRAL SERIES OF HYDROGEN ATOM : The wavelength of the lines of every spectral series can be calculated using the formula given by equation (a). Five special series are observed in the Hydrogen Spectrum corresponding to the five energy levels of the Hydrogen atom and these five series are named as on the names of their inventors. These series are (1) Lyman Series (2) Balmer Series (3) Paschen Series (4) Brackett Series (5) Pfund Series These spectral series are shown in figure-1.11.

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MODERN PHYSICS - I

E=0

n=8 n=7 n=6

E = –0.54 eV

n=5

Pfund Series E = –0.85 eV

n=4

Brackett Series E = –1.51 eV

n=3

Paschen Series E = –3.4 eV

n=2

Balmer Series E = –13.6 eV

n=1 Lyman Series

(1)

(2)

(3)

(4)

(5)

8.

Lyman Series : The series consists of wavelength of the radiations which are emitted when electron jumps from a higher energy level to n = 1 orbit. The wavelength constituting this series lie in the Ultra Violet region of the electromagnetic spectrum. For Lyman Series n1 = 1 and n2 = 2, 3, 4, ......... First line of Lyman series is the line corresponding to the trnasition n2 = 2 to n1 = 1, similarly second line of the Lyman series is the line corresponding to the transition n2 = 3 to n1 = 1. Balmer Series : The series consists of wavelengths of the radiations which are emitted when electron jumps from a higher energy level to n = 2 orbit. The wavelengths consisting this series lie in the visible region of the electromagnetic spectrum. Paschen Series : The series consists of wavelengths of the radiations which are emitted when electron jumps from a higher energy level to n = 3 orbit. The wavelengths constituting this series lie in the Near Infra Red region of the electromagnetic spectrum. Brackett Series : The series consists of wavelengths of the radiations which are emitted when electron jumps from a higher energy level to n = 4 orbit. The wavelengths constituting this series lie in the Infra Red region of the electromagnetic spectrum. Pfund Series : The series consists of wavelengths of the radiations which are emitted when electron jumps from a higher energy level to n = 5 orbit. The wavelengths constituting this series lie in the Deep Infra Red region of the electromagnetic spectrum. We can find out the wavelengths corresponding to the first line and the last line for remaining four spectral series as mentioned in the case of Lyman Series. APPLICATION OF NUCLEUS MOTION ON ENERGY OF ATOM Let both the nucleus of mass M, charge Ze and electron of mass m, and charge e revolve about their centre of mass (CM) with same angular velocity () but different r1 linear speeds. Let r1 and r2 be the distance of CM from M CM nucleus and electron. Their angular velocity should be same then only their separation will remain unchanged in an energy level. Let r be the distance between the nucleus and the electron. Then Mr1 = mr2 r1 + r2 = r

r2

m

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MODERN PHYSICS - I mr Mr r2  and Mm Mm Centripetal force to the electron is provided by the electrostatic force. So,



r1 

1 Ze2 4 0 r 2

mr2 2 

or

1 Ze2  Mr  2 m   .  M  m  4 0 r 2

or

Ze2  Mr  3 2 r    M  m  4 0

or

r 3  2 

e2 4 0

Mm  Mm Moment of inertia of atom about CM,

where

 Mm  2 I  Mr12  mr22   r  r 2  M  m  nh  I 2

According to Bohr's theory,

nh 2 Solving above euqations for r, we get r 2  

or

r

0n2h2 e2Z

and

r = (0.529 A)

n2 m . Z 

Further electrical potential energy of the system. U

and kinetic energy.

K

Ze2 80r

U

–Z2 e4  4 20n2h2

1 2 1 2 2 1 I  r  and K  v 2 2 2 2 v-speed of electron with respect to nucleus. (v = r ) K

2 

here



–Ze2 4r

=

Ze2 4 0 r 3 Z2e4  820 n2 h2



Total energy of the system En = K + U 4 e En  – 8 20n2h2 this expression can also be written as Z2    En  –(13.6 eV) 2 .   n  m The expression for En without considering the motion of proton is En  –

me4 , i.e., m is 820n2h2

replaced by  while considering the motion of nucleus.

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MODERN PHYSICS - I

IMPORTANT FORMULA

(1) Time period (T) : distance = time x speed 2r = T × v T 

v

2r v

+ n2h2 Tn  4k 2 z2me4 Tn 

(2) Frequency of revolution fn 

vn 4k 2z2me4  ; 2r0 n3h3

fn 

r

n3 z2

z2 n3

(3) Momentum of electron Pn 

2mkz2 nh

Pn 

1 n

(4) Angular velocity of electron n 

83k 2 z2me4 z2   , n n3h3 n3

(5) Current (1) e ev z2  ev  ; I 3 T 2r n (6) Magnetic moment of electron (M) M=iA I

M

ev evr  r 2 , M = 2r 2

e(mvr) eJ ; = 2m 2m M = nB Mn M

e  nh   eh  M e M  ,  2   n  4m  2m J 2m B = Bohr magneton = 9.3 × 10–24 Amp. m2.

(7) Magnetic field of Magnetic induction at the centre B =

9.

0i 0 ev  2r 4r 2

ATOMIC COLLISION In such collisions assume that the loss in the kinetic energy of system is possible only if it can excite or ionise.

neutron

Ex.20

Sol.

H atom at rest in ground state and free to move

K,v head on collision What will be the type of collision, if K = 14 eV, 20.4 eV, 22 eV, 24.18 eV (elastic/inelastic/perectly inelastic) Loss in energy (E) during the collision will be used to excite the atom or electron from one level to another.

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MODERN PHYSICS - I

According to quantum Mechanics, for hydrogen atom. E = {0, 10.2 eV, 12.09 eV, .........., 13.6 eV} According to Newtonion mechanics minimum loss = 0, (elastic collision) for maximum loss collision will be perfectly inelastic if neutron collides perfectly inelastically vf m m then, Applying momentum conservation v0 mv0 = 2mvf  vf = 2 1 mv20 v20 1 k 2 final K.E. =  2m  =  2 4 2 2 K maximum loss = 2 K According to classical mechanics (E) = [0, ] 2 (a) If K = 14 eV, According to quantum mechanics (E) = {0, 10, 2eV, 12.09 eV} According to classical mechanics E = [0, 7 eV] loss = 0, hence it is elastic collision speed of particle changes (b) If K = 20.4 ev According to classical mechanics loss = [0, 10.2 eV] According to quantum mechanics loss = {0, 10.2 eV, 12.09 eV,...........} loss = 0 elastic collision. loss = 10.2 eVperfectly inelastic collision (c) If K = 22 eV Classical mechanics E = [0, 1] Quantum mechanics E = {0, 10.2 eV, 12.09 eV,.......} loss = 0 elastic collision loss = 10.2 eVinelastic collision (d) If K = 24.18 eV According to classical mechanics E = [0, 12.09 eV] According to quantum mechanics E = {0, 10.2 eV, 12.09 eV, .............13.6 eV} loss = 0 elastic collision loss = 10.2 eVinelastic collision loss = 12.09 eV perfectly inelastic collision Ex.21 A He+ ion is at rest and is in ground state. A neutron with initial kinetic energy K collides head on with the He+ ion. Find minimum value of K so that there can be an inelastic collision between these two particle. Sol. Here the loss during the collision can only be used to excite the atoms or electrons. So according to quantum mechanics loss = {0, 40.8eV, 48.3eV, ........., 54.4 eV} ...(1) Z2 eV n2 Now according to newtonion mechanics Minimum loss = 0 maximum loss will be for perfectly inelastic collision. let v0 be the initial speed of neutron and vf be the final common speed. En  –13.6

so by momentum conservation mv0 = mvf + 4mvf

vf 

m n

4m K

He+

v0 5

where m = mass of Neutron

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MODERN PHYSICS - I

 mass of He+ ion = 4m so final kinetic energy of system K.E. 

maximum loss = K –

1 1 1 1 K v2 1 2 mv2f  4mv2f = (5m) 0 = .( mv0 )  2 2 5 2 5 2 25 K 4K  5 5

4K   so loss will be 0, 5    For inelastic collision there should be at least one common value other than zero is set (1) and (2)

4K > 40.8 eV 5 K > 51 eV minimum value of K = 51 eV



Ex.22 How many different wavelengths may be observed in the spectrum from a hydrogen sample if the atoms are excited to states with principal quantum number n ? Sol. From the nth state, the atom may go to (n – 1)th state, ..........2nd state or 1st state. So there are (n – 1) possible transitions starting from the nth state. The atoms reaching (n – 1)th state may make (n – 2) different transitions. Similarly for other lower states. The total number of possible transitions is (n – 1) + (n – 2) + (n – 3) + ................ 2 + 1 

n(n – 1) 2

(Remember)

Calculation of recoil speed of atom on emission of a photon momentum of photon = mc = fixed

(a)

h 

free to move

H-atom in first excited state hc  10.2 eV (b) 

v

H-atom

h '

m - mass of atom According to momentum conservation h mv  ...(i) ' According to energy conservation 1 hc mv 2   10.2 eV ' 2 Since mass of atom is very large than photon 1 2 hence mv can be neglected 2 hc h 10 . 2  10.2 eV  eV '  c 10 .2 10.2 v mv  eV cm c 10.2 recoil speed of atom = cm

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MODERN PHYSICS - I 10.

X-RAYS It was discovered by ROENTGEN. The wavelenth of x-rays of found between 0.1 Å to 10 Å. These rays are invisble to eye. They are electromagnetic waves and have speed c = 3 × 108 m/s in vacuum. Its photons have energy around 1000 times move than the visible light. v increases Rm mw I R v uv x y When fast moving electrons having energy of order of several KeV strike the metallic target then x-rays are produced.

10.1

Produced of x-rays by coolidge tube : The melting point, specific heat capacity and atomic number of target should be high. When voltage is applied across the filament then filament on being heated emits electrons from it. Now for giving the beam shape of electrons, collimator is used. Now when electron strikes the target then x-rays are produced. When electrons strike with the target, some part of energy is lost and converted into heat. Since target should not melt or it can absorbe heat so that the melting point, specific heat of target should be high. Here copper rod is attached so that heat produced can go behind and it can absorb heat and target does not get heated very high. Evacuated tube

X-rays –

+

continuous m

Cathode

Target V

Form one energetic electron, accelerating volage is increased. Form one no. of photons voltage across filam entis increased. The x-ray were analysed by mostly taking their spectrum 10.2

Variation of Intensity of x-rays with l is plotted as shown in figure. V’>V,Z

I

V,Z

Continuous x-rays

V,Z’ 

m

1.

The minimum wavelength corresponds to the maximum energy of the x-rays which in turn is equal to the maximum kinetic energy eV of the striking electrons thus hc hc 12400  min   Å eV = hvmax =   eV V(involts) min We see that cutoff wavelength min depends only on accelerating voltage applied between target and filament. It does not depend upon material of target, it is same for two different metals (Z and Z).

2.

Charactristic X-rays The sharp peaks obtained in graph are known as characteristic x-rays because they are characteristic of target mateial 1, 2, 3, 4, ................. = charecteristic wavelength of material having atomic number Z are called characteristic x-rays and the spectrum obtained is called characteristic spectrum. If target of atomic number Z is used then peaks are shifted.

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MODERN PHYSICS - I

n=5 n=4 n=3

O N M

N

M

K L  L L 

K

n=2

M

L

x-rays

K n=1

R

Characteristic x-rays emission occurs when an energetic electron collides with target and remove an inner shell electron from atom, the vacancy created in the shell is filled when an electron from higher level drops into it. Suppose vacancy created in innermost K-shell is filled by an electron droping from next higher level L-shell then K characteristic x-ray is obtained. If vaccany in K-shell is filled by an electron from M-shell, K line is produced and so on 2 similarly L, L,................. M, M lines are produced. 1

1

Ex.23 Find w hich is K  and K

Sol.

hc hc , =  E since energy difference of K is less than K Ek < Kk 1 is K and 2 is K E 

I1

I2



1 2

I

Ex.24 Find which is K and L Sol.  EK > EL 1 is K and 2 is L

11.

MOSELEY'S LAW :

1

2



Moseley meaured the frequencies of characteristic x-rays for a large number of elements and plotted the square root of frequency against position number in periodic table. He discovered that plot is very closed to a straight line not passing through origin.  1,  1 ' ,  1 ' ' ,  1 ' ' ' v I, I2 I1' I2 '  ,  ',  ' ' ,  ' ' ' 2

2

2

2

I1' ' I2 ' ' I1' ' ' I2 ' ' '

Z

Wavelength of charactristic wavelengths.

Moseley's observations can be mathematically expressed as v  a(Z – b) a and b are positive constrants for one type of x-rays & for all elements (indepedent of Z). Moseley's Law can be derived on the basis of Bohr's theory of atom, frequency of x-rays is given by

v 

1 1 CR  2 – 2  .(Z – b)  n1 n2 

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MODERN PHYSICS - I

1 1 1 2 by using the formula   Rz  2 – 2  with modification for multi electron system. n n  1 2

b  known as screening constant or shielding effect, and (Z – b) is effective nuclear charge. for K line n1 = 1, n2 = 2 v 



Here

a

3RC (Z – b) 4 3RC , 4

v  a( Z – b)

[b = 1 for K lines]

K K

I

Z1 Z2

Ex.23

1

2



Find in Z1 and Z2 which one is greater. Sol.



v 

1 1 cR  2 – 2  n n  1 2 

. (Z – b)

If Z is greater then v will be greater,  will be less  

1   2 Z1 > Z2

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MODERN PHYSICS

Exercise - I A. PHOTO ELECTRIC EFFECT 1. Let nr and nb be respectively the number of photons emitted by a red bulb and a blue bulb of equal power in a given time. (A) nr = nb (B) nr < nb (C) nr > nb (D) data insufficient 2. 10 –3 W of 5000 Å light is directed on a photoelectric cell. If the current in the cell is 0.16 mA, the percentage of incident photons which produce photoelectrons, is (A) 0.4% (B) .04% (C) 20% (D) 10% 3. In a photo-emissive cell, with exciting wavelength , the maximum kinetic energy of electron is K. If the exciting wavelength is 3 changed to the kinetic energy of the fastest 4 emitted electron will be : (A) 3K/4 (B) 4K/3 (C) less than 4K/3 (D) greater than 4K/3 4. If the frequency of light in a photoelectric experiment is doubled, the stopping potential will (A) be doubled (B) halved (C) become more than doubled (D) become less than double 5. The stopping potential for the photo electrons emitted from a metal surface of work function 1.7eV is 10.4 V. Identify the energy levels corresponding to the transitions in hydrogen atom which will result in emission of wavelength equal to that of incident radiation for the above photoelectric effect (A) n = 3 to 1 (B) n = 3 to 2 (C) n = 2 to 1 (D) n = 4 to 1 6. When a photon of light collides with a metal surface, number of electrons, (if any) coming out is (A) only one (B) only two (C) infinite (D) depends upon factors 7. The frequency and the intensity of a beam of light falling on the surface of photoelectric material are increased by a factor of two (Treating efficiency of photoelectron generation as constant). This will : (A)inc rease the m aximum energy of the photoelectrons, as will as photoelectric current by a factor of two (B) increase the maximum kinetic energy of the pho to e lectrons and would incre ase the photoelectric current by a factor of two (C)increase the maximum kinetic energy of the photoelectrons by a factor of greater than two

(Objective Problems) and will have no effect on the magnitude of photoelectric current produced. (D)not produce any effect on the kinetic energy of the emitted electrons but will increase the photoelectric current by a factor of two 8. Light coming from a discharge tube filled with hydrogen falls on the cathode of the photoelectric cell. The work function of the surface of cathode is 4eV. Which one of the following values of the anode voltage (in Volts) with respect to the cathode will likely to make the photo current zero. (A) – 4 (B) – 6 (C) – 8 (D) – 10 9. A point source of light is used in photoelectric effect. If the source is removed farther from the emitting metal, the stopping potential : (A) will increase (B) will decrease (C) will remain constant (D) will either increase or decrease 10. A point source causes photoelectric effect from a small metal plate. Which of the following curves may represent the saturation photocurrent as a function of the distance between the source and the metal ? i

i

(A)

(B) t

t

i

i

(C)

(D) t

t

11 The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4eV. The stopping potential is Volts is : (A) 2 (B) 4 (C) 6 (D) 10 12. Let K1 be the maximum kinetic energy of photoelectrons emitted by a light of wavelength 1 and K2 corresponding to 2. If 1 = 22, then : (A) 2K1 = K2 (B) K1 = 2K2 (C) K 1 

K2 2

(D) K1 > 2K2

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MODERN PHYSICS 13. In a photoe lectric V experiment, the potential difference V that must be V1 maintained between the illuminated surface and the f o f0 f1 collector so as just to prevent any electron from reaching the collector is determined for different frequencies f of the incident illumination. The graph obtained is shown. The maximum kinetic energy of the electrons emitted at frequency f1 is (A) hf1

V1 (B) (f  f ) 1 0

(C) h(f1 – f0)

(D) eV1(f1 – f0)

14. Radiation of two photon energies twice and five times the work function of metal are incident sucessively on the metal surface. The ratio of the maximum velocity of photoelectrons emitted is the two cases will be (A) 1 : 2 (B) 2 : 1 (C) 1 : 4 (D) 4 : 1 15. Cut off potentials for a metal in photoelectric effect for light of wavelength 1, 2 and 3 is found to be V1, V2 and V3 volts if V1, V2 and V3 are in Arithmetic Progression and 1, 2 and 3 will be : (A) Arithmetic Progression (B) Geometric Progression (C) Harmonic Progression (D) None 16. Photons with energy 5eV are incident on a cathode C, on a photoelectric cell. The maximum energy of the emitted photoelectrons is 2eV. When photons of energy 6eV are incident on C, no photoelectrons will reach the anode A if the stopping potential of A relative to C is (A) 3V (B) –3V (C) –1V (D) 4V 17. In a photoelectric experiment, the collector plate is at 2.0V with respect to the emitter plate made of copper ( = 4.5eV). The emitter is illuminated by a source of monochromatic light of wavelength 200 nm. (A) the minimum kinetic ene rgy of the photoelectrons reaching the collector is 0. (B) the max imum kinetic ene rgy of the photoelectrons reaching the collector is 3.7eV. (C) if the polarity of the battery is reversed then answer to part A will be 0. (D) if the polarity of the battery is reversed then answer to part B will be 1.7eV. 18. By increasing the intensity of incident light keeping frequency (v > v0) fixed on the surface

of metal (A) kinetic energy of the photoelectrons increases (B) number of emitted electrons increases (C) kinetic energy and number of electrons increases (D) no effect 19. In a photoelectric experiment, electrons are ejected from metals X and Y by light of intensity I and frequency f. The potential difference V required to stop the electrons is measured for various frequencies. If Y has a greater work function than X ; which one of the following graphs best illustrates the expected results ? v

v X

X

(A)

(B)

Y f

o

Y

v

v Y

X

(C)

X

(D)

Y o

f

o

f

o

f

20.Monochromatic light with a frequency well above the cutoff frequency is incident on the emitter in a photoelectric effect apparatus. The frequency of the light is then doubled while the intensity is kept constant. How does this effect the photoelectric current ? (A) The photoelectric current will increase (B) The photoelectric current will decrease (C) The photoelectric current will remain the same (D) None of these 21 A image of the sun of formed by a lens of focal-length of 30 cm on the metal surface of a photo-electric cell and a photo-electrci current is produced. The lens forming the image is then replaced by another of the same diameter but of focal length 15 cm. The photo-electric current in this case is (A) I/2 (B) I (C) 2I (D) 4I 22. If a parallel beam of light having intensity I is incident normally on a perfectly reflecting surface, the force exerted on the surface, equals F (Assume that the cross section of beam remains constant). (A) 2F tan  (B) F cos  (C) F cos2  (D) 2F B. MATTER WAVES 23. A proton and an electron are accelerated by same potential difference have de-Broglie wavelength p and e. (A) e = p (B) e < p

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Page # 44 (C) e > p

MODERN PHYSICS (D) none of these

24. An electron with initial kinetic energy of 100eV is acceleration through a potential difference of 50V. Now the de-Broglie wavelength of electron becomes. (A) 1 Å (C)

3Å

(B)

1. 5 Å

(D) 12.27 Å

25. If h is Planck’s constant is SI system, the momentum of a photon of wavelength 0.01 Å is: (A) 10–2 h (B) h 2 (C) 10 h (D) 1012h 26. Imagine a Young’s double slit interference experiment performed with waves associated with fast moving electrons produced from an electron gun. The distance between successive maxima will decrease maximum if (A) the accelerating voltage in the electron gun is decreased (B) the accelerating voltage is increased and the distance of the screen from the slits is decreased (C) the distance of the screen from the slits is increased. (D) the distance between the slits is decreased.

C. ATOMIC STRUCTURE 27. In a hypothetical system of particle a mass m and charge –3q is moving around a very heavy particle having charge q. Assuming Bohr’s model to be true to this system, the orbital velocity of mass m when it is nearest to heavy particle is

(A)

3q2 2 0 h

3q (C) 2 h 0

3h . Here h is Planck’s constant. 2 The kinetic energy of this electron is : (A) 4.53 eV (B) 1.51 eV (C) 3.4 eV (D) 6.8 eV

hydrogen atom is

(B)

3 q2 4 0h

3q (D) 4  h 0

28. De-Broglie wavelength of an electron in the nth Bohr orbit is n and the angular momentum is Jn, then : (A) Jn µ n

1 (B)  n  J n

(C)  n  Jn2

(D) none of these

29. The angular momentum of an electron in the

30. Consider the following electronic energy level diagram of H-atom : Photons associated with shorte st and longe st wavele ngths wo uld be emitted from the atom by the transitions labelled. (A) D and C respectively (B) C and A respectively (C) C and D respectively (D) A and C respectively

n= A D

n=4 C

n=3 B n=2 n=1

31. In a hydrogen atom, the binding energy of the electron in the nth state is E n, then the frquency of revolution of the electron in the nth orbits is : (A) 2En/nh (B) 2Enn/h (C) En/nh (D) Enn/h 32. If the electron in a hydrogen atom were in the energy level with n = 3, how much energy in joule would be required to ionise the atom ? (Ionisation energy of H-atom is 2.18 × 10–18 J): (A) 6.54 × 10–19 (B) 1.43 × 10–19 (C) 2.42 × 10–19 (D) 3.14 × 10–20 33. In hydrogen and hydrogen like atoms, the ratio of difference of energies E4n – E2n and E2n – En varies with its atomic number z and n as : (A) z2/n2 (B) z4/n4 (C) z/n (D) z0 n0 34. In a hydrogen atom, the electron is in nth excited state. It may come down to second ex cite d state by e mitting ten dif fere nt wavelengths. What is the value of n: (A) 6 (B) 7 (C) 8 (D) 5 35. Difference between nth and (n + 1)th Bohr’s radius of ‘H’ atom is equal to it’s (n – 1)th Bohr’s radius. The value of n is : (A) 1 (B) 2 (C) 3 (D) 4 36. An electron in hydrogen atom after absorbing energy photons can jump between energy states n1 and n2(n2 > n1). Then it may return to ground state after emitting six different wavelengths in emission spectrum. The energy of emitted photons is either equal to, less than or greater than the absorbed photons. Then n1 and n2 are : (A) n2 = 4, n1 = 3 (B) n2 = 5, n1 = 3 (C) n2 = 4, n1 = 2 (D) n2 = 4, n1 = 1

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MODERN PHYSICS

38. The electron in a hydrogen atom makes a transition n1  n2 whose n1 and n2 are the principal quantum numbers of the two states. Assume the Bohr model to be valid. The frequency of orbital motion of the electron in the initial state is 1/27 of that in the final state. The possible values of n1 and n2 are (A) n1 = 4, n2 = 2 (B) n1 = 3, n2 = 1 (C) n1 = 8, n2 = 1 (D) n1 = 6, n2 = 3

46. The frequency of revolution of electron in nth Bohr orbit is vn. The graph between log n and log (vn / v1) may be

40. The ionisation potential of hydrogen atom is 13.6 volt. The energy required to remove an electron from the second orbit of hydrogen is : (A) 3.4 eV (B) 6.8 eV (C) 13.6 eV (D) 27.2 eV 41. Electron in a hydrogen atom is replaced by an identically charged particle muon with mass 207 times that of electron. Now the radius of K shell will be (A) 2.56 × 10–3 Å (B) 109.7 Å (C) 1.21 × 10–3 Å (D) 22174.4 Å 42. Monochromatic radiation of wavelength  is incident on a hydrogen sample containing in ground state. Hydrogen atoms absorb the light and subsequently emit radiations of ten different wavelengths. The value of  is (A) 95 nm (B) 103 nm (C) 73 nm (D) 88 nm 43. In a sample of hydrogen like atoms all of which are in ground state, a photon beam containing photons of various energies is passed. In absorption spectrum, five dark lines are observed. The number of bright lines in the emission spectrum will be (Assume that all transitions take place) (A) 5 (B) 10 (C) 15 (D) none of these 44. When a hydrogen atom, initially at rest emits, a photon resulting in transition n = 5  n = 1, its recoil speed is about (A) 10–4 m/s (B) 2 × 10–2 m/s (C) 4.2 m/s (D) 3.8 × 10–2 m/s 45. An electron collides with a fixed hydrogen atom in its ground state. Hydrogen atom gets excited and the colliding electron loses all its

(B)

log n

(C)

log n

(D)

log(vn/v1)

log n log(vn/v1)

39. The radius of Bohr’s first orbit is a0. The electron in nth orbit has a radius : (A) na0 (B) a0/n (C) n2a0 (D) a0/n2

(A)

log(vn/v1)

kinetic energy. Consequently the hydrogen atom may emit a photon corresponding to the largest wavelength of the Balmer series. The min. K.E. of colliding electron will be (A) 10.2 eV (B) 1.9 eV (C) 12.1 eV (D) 13.6 eV

log(vn/v1)

37. The electron in a hydrogen atom makes transition from M shell to L. The ratio of magnitudes of initial to final centripetal acceleration of the electron is (A) 9 : 4 (B) 81 : 16 (C) 4 : 9 (D) 16 : 81

log n

47. Consider the spectral line resulting from the transition n = 2  n = 1 in the atoms and ions given below. The shortest wavelength is produced by : (A) hydrogen atom (B) deuterium atom (C) singly ionized helium (D) doubly ionized lithium 48. In an atom, two electrons move around the nucleus in circular orbits of radii R and 4R. The ratio of the time taken by them to complete one revolution is : (neglect electric interaction) (A) 1 : 4 (B) 4 : 1 (C) 1 : 8 (D) 8 : 1 49. The electron in hydrogen atom in a sample is in nth excited state, then the number of different spectrum lines obtained in its emission spectrum will be : (A) 1 + 2 + 3 + ....... + (n – 1) (B) 1 + 2 + 3 + ........ + (n) (C) 1 + 2 + 3 + ........+ (n + 1) (D) 1 × 2 × 3 × ..........× (n –1) 50. The total energy of a hydrogen atom in its ground state is -13.6 eV. If the potential energy in the first excited state is taken as zero then the total energy in the ground state will be : (A) –3.4 eV (B) 3.4 eV (C) –6.8 eV (D) 6.8 eV 51. A neutron collides head on with a stationary hydrogen atom in ground state (A) If kinetic energy of the neutron is less than 13.6 eV, collision must be elastic (B) if kinetic energy of the neutron is less than 13.6 eV, collision may be inelastic. (C) inelastic collision takes place when initial

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MODERN PHYSICS

kinetic energy of neutron is greater than 13.6 eV. (D) perfectly inelastic collision cannot take place. 52. The electron in a hydrogen atom make a transition from an excited state to the ground state. Which of the following statement is true? (A) Its kinetic energy increases and its potential and total energies decrease (B) Its kinetic energy decreases, potential energy increases and its total energy remains the same. (C) Its kinetic and total energies decrease and its potential energy increases. (D) its kinetic potential and total energies decreases. 53. The magnitude of angular momentum, orbit radius and frequency of revolution of electron in hydrogen atom corresponding to quantum number n are L, r and f respectively. Then according to Bohr’s theory of hydrogen atom, (A) fr2L is constant for all orbits (B) frL is constant for all orbits (C) f2rL is constant for all orbits (D) frL2 is constant for all orbits 54. Radius of the second Bohr orbit of singly ionised helium atom is (A) 0.53 Å (B) 1.06 Å (C) 0.265 Å (D) 0.132 Å 55. An electron in Bohr’s hydrogen atom has an energy of –3.4 eV. The angular momentum of the electron is (A) h/ (B) h/2 (C) nh/2 (n is an integer) (D) 2h/

Relative intensity

D. X-RAYS 56. In a characteristic X-ray spectra of some atom superimposed on continuous X-ray spectra (A) P represents K line Q P (B) Q represents K line (C) Q and P represents K and K lines respectively (D) Relative positions of K and K depend on the particular atom 57. The “K’’ X-rays emission line of tungsten occurs at  = 0.021 nm. The energy difference between K and L levels in this atom is about (A) 0.51 MeV (B) 1.2 MeV (C) 59 ke V (D) 13.6 eV 58. Which of the following wavelength falls in a X-rays region? (A) 10,000 Å (B) 1000 Å (C) 1 Å (D) 10–2 Å 59. The penetrating power of X-ray increases with the (A) Increases of its velocity (B) Increase in its intensity (C) Decrease in its velocity

(D) Increases in its frequency. 60. The wavelength of the K line for an element of atomic number 57 is . What is the wavelength of the Kline for the element of atomic number 29? (A)  (B) 2 (C) 4 (D) 8 61. If the frequency of K, K and L X-rays for a meterial K  , K  , L  respectively, then (A) K  = K  + L 

(B) L  = K  + K 

(C) K  = K  + L 

(D) none of these

62. In X-ray tube, when the accelerating voltage V is doubled, the different between the wavelength of K line and the minimum cut off of continuous X-ray spectum : (A) remains constant (B) becomes more than two times (C) becomes half (D) becomes less than 2 times. REASONING TYPE 63. Statement-1: Figure shows graph of stopping potential and frequency of incident light in photoelectric effect. For values of frequency less than threshold frequecy (v 0 ) stopping potential is negative. Vs

(0, 0)

vs

v

Statement-2 : Lower the value of frequency of incident light (for  > 0) the lower is the maxima of kinetic energy of emitted photoelectrons. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement1 (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true. 64. Statement-1: In the process of photo electric emission, all the emitted photoelectrons have same K.E. Statement-2: According to einstein’s photo electric equation KEmax = h – . (A)Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false.

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MODERN PHYSICS (D) Statement-1 is false, statement-2 is true. 65. Statement-1: Work function of aluminum is 4.2 eV. If two photons each of energy 2.5 eV strikes on a piece of aluminum, the photo electric emission does not occur. Statement-2: In photo electric effect a single photon interacts with a single electron and electron is emitted only if energy of each incident photon is greater then the work function. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true. 66.Statement-1: An electron and a proton are accelerated through the same potential difference. The deBroglie wavelength associated with the electron is longer. Statement-2: De-Broglie wavelength associated h with a moving particle is  = where, p is the p linear momentum and both have same K.E. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.

(C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true. 67. Statement-1: Two photons having equal wavelengths have equal linear momenta. Statement-2: When light shows its photon character, each photon has a linear momentum p = h .  (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true. 68. Statement-1: If the accelerating potential of a X-ray tube is increased then the characteristic wavelength decreases. Statement-2: The cut-off wavelength for X-ray tube is given by min = hc , where V is accelerating eV potential. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true.

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MODERN PHYSICS

Exercise - II

(One or more than one option is correct)

A. PHOTO ELECTRIC EFFECT 1. In photoelectric effect, stopping potential depends on (A) frequency of the incident light (B) intensity of the incident light by varies source distance (C) emitter’s properties (D) frequency and intensity of the incident light 2. In the experiment on photoelectric effect using light having frequency greater than the threshold frequency, the photocurrent will certainly increase when (A) Anode voltage is incrased (B) Area of cathode surface is increased (C) Intensity of incident light is increased (D) Distance between anode and cathode is increased. B. MATTER WAVES 3. Two electrons are moving with the same speed v. One electron enters a region of uniform electric field while the other enters a region of uniform magnetic field, then after sometime if the de-Broglie wavelengths of the two are 1 and 2 then : (A) 1 = 2 (B) 1 > 2 (C) 1 < 2 (D) 1 > 2 or 1 < 2 C. ATOMIC STRUCTURE 4. An electron in hydrogen atom first jumps from second excited state to first excited state and then, from first excited state to ground state. Let the ratio of wavelength, momentum and energy of photons in the two cases by x, y and z, then select the wrong answers : (A) z = 1/x (B) x = 9/4 (C) y = 5/27 (D) z = 5/27 5. An electron is in an excited state in hydrogenlike atom. It has a total energy of –3.4 eV. If the kinetic energy of the electron is E and its deBroglie wavelength is , then (A) E = 6.8 eV,  = 6.6 × 10–10 m (B) E = 3.4 eV,  = 6.6 × 10–10 m (C) E = 3.4 eV,  = 6.6 × 10–11 m (D) E = 6.8 eV,  = 6.6 × 10–11 m 6. A particular hydrogen like atom has its ground state binding “energy 122.4 eV. Its is in ground state. Then : (A) Its atomic number is 3 (B) An electron of 90eV can excite it. (C) An electron of kinetic energy nearly 91.8 eV can be brought to almost rest by this atom. (D) An electron of kinetic energy 2.6 eV may emerge from the atom when electron of kinetic

energy 125 eV collides with this atom. 7. The electron in a hydrogen atom makes a transition n1  n2, where n1 & n2 are the principal quantum numbers of the two states. Assume the Bohr model to be valid. The time period of the electron in the initial state is eight times that in the final state. The possible values of n1 & n2 are : (A) n1 = 4, n2 = 2 (B) n1 = 8, n2 = 2 (C) n1 = 8, n2 = 1 (D) n1 = 6, n2 = 3 8. A beam of ultraviolet light of all wavelengths passes through hydrogen gas at room temperature, in the x-direction. Assume that all photons emitted due to electron transition inside the gas emerge in the y-direction. Let A and B denote the lights emerging from the gas in the x and y directions respectively. (A) Some of the incident wavelengths will be absent in A (B) Only those wavelengths will be present in B which are absent in A (C) B will contain some visible light. (D) B will contain some infrared light. 9. If radiation of allow wavelengths from ultraviolet to infrared is passed through hydrogen agas at room temperature, absorption lines will be observed in the : (A) Lyman series (B) Balmer series (C) Both (A) and (B) (D) neither (A) nor (B) 10.In the hydrogen atom, if the reference level of potential energy is assumed to be zero at the ground state level. Choose the incorrect statement. (A) The total energy of the shell increases with increase in the value of n (B) The total energy of the shell decrease with increase in the value of n. (C) The difference in total energy of any two shells remains the same. (D) The total energy at the ground state becomes 13.6 eV. 11.Choose the correct statement(s) for hydrogen and deuterium atoms (considering motion of nucleus) (A) The radius of first Bohr orbit of deuterium is less than that of hydrogen (B) The speed of electron in the first Bohr orbit of deuterium is more than that of hydrogen. (C) The wavelength of first Balmer line of deuterium is more than that of hydrogen (D) The angular momentum of electron in the first Bohr orbit of deuterium is more than that of

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MODERN PHYSICS hydrogen. 12.Let An be the area enclosed be the nth orbit in a hydrogen atom. The graph of ln (An/A1) agains ln (n). (A) will pass through origin (B) will be a stright line will slope 4 (C) will be a monotonically increasing nonlinear curve (D) will be a circle. 13.A free hydrogen atom in ground state is at rest. A neutron of kinetic energy ‘K’ collides with the hydrogen atom. After collision hydrogen atom emits two photons in succession one of which has energy 2.55 eV. (Assume that the hydrogen atom and neutron has same mass) (A) minimum value of ‘K’ is 25.5 eV. (B) minimum value of ‘K’ is 12.75 eV. (C) the other photon has energy 10.2 eV (D) the upper energy level is of excitation energy 12.75 eV. 14.A neutron collides head-on with a stationary hydrogen atom in ground state. Which of the following statements are correct (Assume that the hydrogen atom and neutron has same mass) (A) If kinetic energy of the neutron is less than 20.4 eV collision must be elastic (B) If kinetic energy of the neutron is less than 20.4 eV collision may be inelastic (C) Inelastic collision may be take place only when initial kinetic energy of neutron is greater than 20.4 eV. (D) Perfectly inelastic collision can not take place. D. X-RAYS 15.In a Coolidge tube experiment, the minimum

wavelength of the continuous X-ray spectrum is equal to 66.3 pm, then (A) electrons accelerate through a potential difference of 12.75 kV in the Coolidge tube (B) electrons accelerate through a potential difference of 18.75 kV in the Coolidge tube (C) de-Broglie wavelength of the electrons reaching the anticathode is of the order of 10m (D) de-Broglie wavelength of the electrons reaching the anticathode is 0.01 Å. 16.The potential difference applied to an X-ray tube is increased. As a result, in the emitted radiation : (A) the intensity increases (B) the minimum wave length increases (C) the intensity decreases (D) the minimum wave length decreases 17.A X-ray tube operates at an accelerating potential of 20 kV. Which of the following wavelengths will be absent in the continuous spectrum of X-ray. (A) 12 pm (B) 45 pm (C) 65 pm (D) 95 pm 18.X-ray are produced by accelerating electrons across a given potential difference to strike a meta target of high atomic number. If the electrons have same speed when they strike the target, the X-ray spectrum will exhibit. (A) a minimum wavelength (B) a continuous spectrum (C) some discrete comparatively prominent wavelenght (D) uniform density over the whole spectrum

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MODERN PHYSICS

A. PHOTO ELECTRIC EFFECT 1. A parallel beam of uniform, monochromatic light of wavelength 2640 A has an intensity of 200 W/ m2. The number of photons in 1 mm3 of this radiation are ............... 2. When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy Ta eV and de Broglie wavelength a. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.7 eV is Tb = (Ta – 1.5) eV. If the De Broglie wavelength of these photoelectrons is b = 2 a, then find (a) The work function of a (b) The work function of b is (c) Ta and Tb 3. When a monochromatic point source of light is at a distance of 0.2 m from a photoelectric cell, the cut off voltage and the saturation current are respectively 0.6 volt and 18.0 mA. If the same source is placed 0.6 m away from the photoelectric cell, then find (a) the stopping potential (b) the saturation current 4. An isolated metal body is illuminated with monochromatic light and is observed to become charged to a steady positive potential 1.0 V with respect to the surrounding. The work function of the metal is 3.0 eV. The frequency of the incident light is ___________________. 5. 663 mW of light from a 540 nm source is incident on the surface of a metal. If only 1 of each 5×109 incident photons in absorbed and causes an electron to be ejected from the surface, the total photocurrent in the circuit is ____________________. 6. Light of wavelength 330 nm falling on a piece of metal ejects electrons with sufficient energy which requires voltage V0 to prevent a collector. In the same setup, light of wavelength 220 nm, ejects electrons which require twice the voltage V0 to stop them in reaching a collector. Find the numerical value of voltage V0. (Take plank’s constant, h = 6.6 × 10–34 Js and 1 eV = 1.6 ×10–19 J) 7. A small 10W source of ultraviolet light of wavelength 99 nm is held at a distance 0.1 m from a metal surface. The radius of an atom of the metal is approximately 0.05 nm. Find (i) the average number of photons striking an atom per second. (ii) the number of photoelectrons emitted per unit area per second it the efficiency of liberation of photoelectrons is 1%

(SUBJECTIVE

PROBLEMS )

8. The surface of cesium is illuminated with monochromatic light of various wavelengths and the stopping potentials for the wavelengths are measured. The results of this experiment is plotted as shown in the figure. Estimate the value of work function of the cesium and Planck’s constant. 2 0.49 1 0 15 0.5 1.0 1.5 v × 10Hz –1 –2 stopping potential (volt)

Exercise - III

B. MATTER WAVES 9. An electron of mass “m” and charge “e” initially at rest gets accelerated by a constant electric field E. The rate of change of DeBroglie wavelength of this e lectron at time t is ............................ 10. Assume that a particle cannot be confined to a spherical volume of diameter less then DeBroglie wavelength of the particle. Estimate the minimum kinetic energy a proton confined to a nucleus f diameter 10–14 m may have. C. ATOMIC STRUCTURE 11. A hydrogen atom in a state having a binding energy 0.85 eV makes a transition to a state of excitation energy 10.2 eV. The wave length of emitted photon is ................ nm. 12. A hydrogen atom is in 5in excited state. When the electron jumps to ground state the velocity of recoiling hydrogen atom is ................... m/s and the energy of the photon is .......... eV. 13. The ratio of series limit wavelength of Balmer series to wavelength of first line of paschen series is ............. 14. An electron joins a helium nucleus to form a He+ ion. The wavelength of the photon emitted in this process if the electron is assumed to had has no kinetic energy when it combines with nucleus is .............. nm. 15. Three energy levels of an atom are shown in the figure. The w a v e l e n g t h corresponding to three possible transition are 1, 2 and 3. The value of 3 in terms of 1 16. Imagine an atom made up of a proton and a hypothetical particle of double the mass of an electron but having the same charge as the electron. Apply the Bohr atom model and consider

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MODERN PHYSICS a possible transitions of this hypothetical particle to the first excited level. Find the longest wavelength photon that will be emitted  (in terms of the Rydberg constant R.) 17. In a hydrogen atom, the electron moves in an orbit of radius 0.5 Å making 1016 revolution per second. The magnetic moment associated with the orbital motion of the e lectron is _____________. 18. The positron is a fundamental particle with the same mass as that of the electron and with a charge equal to that of an electron but of opposite sign. When a positron and an electron collide, they may annihilate each other. The energy corresponding to their mass appears in two photons of equal energy. Find the wavelength of the radiation emitted. [Take : mass of electron = (0.5/C2) MeV and hC = 1.2 × 10–12 MeV.m where h is the Plank’s constant and C is the velocity of light in air] 19. A hydrogen like atom has its single electron orbiting around its stationary nucleus. The energy to excite the electron from the second Bohr orbit to the third Bohr orbit is 47.2 eV. The atomic number of this nucleus is _____________. 20. A single electron orbits a stationary nucleus of charge Ze where Z is a constant and e is the electronic charge. It requires 47.2 eV to excite the electron from the 2nd Bohr orbit to 3rd Bohr orbit. Find (i) the value of Z, (ii) energy required to excite the electron from the third to the fourth orbit. (iii) the wavelength of radiation required to remove the electron from the first orbit to infinity (iv) the kinetic energy, potential energy and angular momentum in the first Bohr orbit (v) the radius of the first Bohr orbit. 21. A hydrogen like atom (atomic number Z) is in higher excited state of quantum number n. This excited atom can make a transition of the first excited state by successively emitting two photons of energy 22.95 eV and 5.15 eV respectively. Alternatively, the atom from the same excited state can make transition to the second excited state by successively emitting two photons of energies 2.4 eV and 8.7 eV respectively. Find the values of n and Z. 22. Find the binding energy of an electron in the ground state of a hydrogen like atom in whose spectrum the third of the corresponding Balmer series is equal to 108.5 nm. 23. Which level of the doubly ionized lithium has the same energy as the ground state energy of the hydrogen atom. Find the ratio of the two radii of corresponding orbits.

24. Determine the number of lines in Paschen series which have a wavelength greater than 1000 nm. 25. In a block body radiation at certain temperature T1, the wavelength having maximum intensity of radiation equals 9000 Å. When the temperature is increased from T1 to T2 the total radiation increases 16 times. The peak radiation at T 2 is found to be capable of ejecting photoelectrons. The maximum kinetic energy of the photoelectrons is the same as the energy of photon that one gets when one of electrons in the M-shell of hydrogen atom jumps to L-shell. What is the work function of the metal? Section - D X-ray 26. Obtain a relation between the frequencies of K, K and L lines for a target material. 27. A 20 KeV energy electron is brought to rest in an X-ray tube, by undergoing two successive bremsstrahling events, thus emitting two photons. The wavelength of the second photon is 130 × 10–12 m greater than the wavelength of the first emitted photon. Calculate the wavelengths of the two photons. 28. Figures shows K & K X-rays along with continuous X-ray. Find the energy of L X-ray. (Use hc = 12420 evÅ).

1Å 2Å 29. Photoelectrons are emitted when 400 nm radiation is incident on a surface of work function 1.9 eV. These photoelectrons pass through a region containing a-particles. A maximum energy electron combines with an -particle to form a He+ ion, emitting a single photon in this process. He+ ions thus formed are in their fourth excited state. Find the energies in eV of the photons, lying in the 2 to 4 eV range, that are likely to be emitted during and after the combination. [Take, h = 4.14 × 10–15 eV–s]

30. The wavelength of K x-ray of tungsten is 21.3 pm. It takes 11.3 KeV to knock out an electron from the L shell of a tungsten atom. What should be the minimum accelerating voltage across ab x-ray tube having tungsten target which allows poduction of K x-ray. 31. An electron, in a hydrogen like atom, is in an excited state. It has a total energy of –3.4 eV. Calculate: (i) The kinetic energy & (ii) The DeBroglie wave length of the electron. 32. A pontial difference of 20 KV is applied across an x-ray tube. The minimum wave length of Xrays generated is _________ .

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MODERN PHYSICS

Exercise - IV

(TOUGH

1. In a photo electric effect set-up, a point source of light of power 3.2 × 10–3 W emits mono energetic photons of energy 5.0 eV. The source is located at a distance of 0.8 m from the centre of a stationary metallic sphere of work function 3.0 eV & of radius 8.0 × 10–3 m. The efficiency of photo electrons emission is one for every 106 incident photons. Assume that the sphere is isolated and initially neutral, and that photo electrons are instantly swept away after emission. (a) Calculate the number of photo electrons emitted per second. (b) Find the ratio of the wavelength of incident light to the De-Broglie wave length of the fastest photo electrons emitted. (c) It is observed that the photo electron emission stops at a certain time t after the light source is switched on. Why ? (d) Evaluate the time t. 2. A stationary He + ion emitted a photon corresponding to the first line its Lyman series. That photon liberated a photoelectron from a stationary hydrogen atom in the ground state. Find the velocity of the photoelectron. 3. A gas of identical hydrogen like atoms has some atoms in the lowest (ground) energy level A & some atoms in a a particular upper (excited) energy level B & there are no atoms in any other energy level. The atoms of the gas make transition to a higher energy level by the absorbing monochromatic light of photon energy 2.7 eV. Subsequently, the atoms emit radiation of only six different photon energies. Some of the emitted photons have energy 2.7 eV. Some have energy more and some have less than 2.7 eV. (i) Find the principal quantum number of the initially excited level B. (ii) Find the ionisation energy for the gas atoms. (iii) Find the maximum and the minimum energies of the emitted photons. 4. An energy of 68.0 eV is required to excite a hydrogen like atom from its second Bohr orbit to the third. The nuclear charge Ze. Find the value of Z, the kinetic energy of the electron in the first Bohr orbit and the wavelength of the electro magnetic radiation required to eject the electron

SUBJECTIVE PROBLEMS)

from the first Bohr orbit to infinity. 5. Electrons in hydrogen like atoms (Z=3) make transitions from the fifth to the fourth orbit & from the fourth to the third orbit. The resulting radiations are incident normally on a metal plate & eject photo electrons. The stopping potential for the photoelectrons ejected by the shorter wavelength is 3.95 volts. Calculate the work function of the metal, & the stopping potential for the photoelectrons ejected by the longer wavelength.(Rydberg constant=1.094 × 107 m–1) 6. A beam of light has three wavelengths 4144 Å, 4972 Å & 6216Å with a total intensity of 3.6 × 10–3 W.m–2 equally distributed amongest the three wavelengths. The beam falls normally on an area 1.0 cm2 of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in two seconds. 7. Monochromatic radiation of wavelength 1 = 3000 Å falls on a photocell operating in saturating mode. The corresponding spectral sensitivity of photocell is J = 4.8 × 10–3 A/w. When another monochromatic radiation of wavelength 2 = 1650 Å and power P = 5 × 10–3 W is incident, it is found that maximum velocity of photoelectrons increases n = 2 times. Assuming efficiency of photoelectron generation per incident photon to be same for both the cases, calculate (i) threshold wavelength for the cell. (ii) saturation current in second case. 8. A monochromatic point source S radiating wavelength 6000 Å with power 2 watt, an aperture A of diameter 0.1 m & a large screen SC are placed as shown in figure. A photoemissive detector D of surface area 0.5 cm2 is placed at the centre of the screen. The efficiency of the detector for the photoelectron generation per incident photon is 0.9.

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MODERN PHYSICS A

SC

S D L 0.6m 6m (i) Calculate the photon flux density at the centre of the screen and the photocurrent in the detector. (ii) If a concave lens L of focal length 0.6 m is inserted in the aperture as shown, find the new values of photon flux density & photocurrent Assume a uniform average transmission of 80% for the lens. (iii) If the work-function of the photoemissive surface is 1 eV, calculate the values of the stopping potential in the two cases (without & with the lens in the aperture.) 9. A small 10 W source of ultraviolet light of wavelength 99 nm is held at a distance 0.1 m from a metal surface. The radius of an atom of the metal is approximaterly 0.05 nm. Find : (i) the number of photons striking an atom per second. (ii) the number of photoelectrons emitted per sec ond it the e ffic ienc y of liberation of photoelectrons is 1% 10. A monochromatic light source of frequency v illuminates a metallic surface and ejects photoelectrons. The photoelectrons having maximum energy are just able to ionize the hydrogen atoms in ground state. When the whole experiment is repeated with an incident radiation of frequency (5/6)v, the photoelectrons so emitted are able to excite the hydrogen atom beam which then emits a radiation of wavelength of 1215 Å. Find the work function of the metal and the frequency v. 11. A neutron of kinetic energy 65 eV collides inelastically with a single ionized helium atom at rest. It is scattered at an angle of 90° with respect of its original direction.

(i) Find the allowed values of the energy of the neutron & that of the atom after collision. (ii) If the atom gets de-excited subsequently by emitting radiation, find the frequencies of the emitted radiation. (Given : Mass of he atom = 4 × (mass of neutron), ionization energy of H atom = 13.6 eV) 12. A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. This excited atom can make a transition to the first excited state by successively emitting two photons of energies 10.20 eV & 17.00 eV respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV & 5.95 eV respectively. Determine the values of n & Z. (Ionisation energy of hydrogen atom = 13.6 eV) 13. Assume that the de-Broglie wave associated with an electron can form a standing wave between the atoms arranged in a one dimensional array with nodes at each of the atomic sites. It is found that one such standing wave is formed if the distance ‘d’ between the atoms of the array is 2 Å. A similar standing wave is again formed if ‘d’ is increased to 2.5 Å but not for any intermediate value of d. Find the energy of the electrons in electron volts and the least value of d for which the standing wave of the type described above can form. 14. A beam of ultraviolet light of wavelength 100 nm – 200 nm is passed through a box filled with hydrogen gas in ground state. The light coming out of the box is split into two beams ‘A’ and ‘B’. A contains unabsorbed light from the incident light and ‘B’ contains emitted light by hydrogen atoms. The beam ‘A’ is incident on the emitter in a photoelectric tube. The stopping potential in the case is 5 volts. Find the work function of the emitter. In the second case the beam ‘B’ is incident on the same emitter. Find the stopping potential in this case. You can assume that the transition to higher energy states are not permitted from the excited states. Use hc = 12400 eVÅ.

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MODERN PHYSICS

Exercise - V 1. (a) Imagine an atom made up of a proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has wavelenght  (given in terms of the Rydberg constant R for the hydrogen atom) equal to [JEE’2000(Scr)] (A) 9/(5R) (B) 36/(5R) (C) 18/(5R) (D) 4/R (b) The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following statements is true ? [JEE’2000(Scr)] (A) Its kinetic energy increases and its potential and total energies decrease (B) Its kinetic energy decreases, potential energy increases and its total energy remains the same (C) Its kinetic and total energies decrease and its potential energy increases (D) Its kinetic, potential and total energies decrease 2. (a) A hydrogen-like atom of atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204eV. If it makes a transition to quantum state n, a photon of energy 40.8 eV is emitted. Find n, Z and the ground state energy (in eV) for this atom. Also, calculate the minimum energy (in eV) that can be emitted by this atom during de-excitation. Ground state energy of hydrogen atom is –13.6 eV. [JEE’2000] (b) When a beam of 10.6 eV photon of intensity 2W/m2 falls on a platinum surface of area 1 × 104m2 and work function 5.6 eV, 0.53% of the incident photons eject photoelectrons. Find the number of photoelectron emitted per sec and their minimum and maximum energies in eV. [JEE’2000] 3. Electrons with energy 80 keV are incident on the tungsten target of an X-ray tube. K-shell electrons of tungsten have 72.5 keV energy. Xrays emitted by the tube contain only (A) a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of  0.115 Å (B) a continuous X-ray spectrum (Bremsstrahlung) with all wavelengths (C) the characteristic X-ray spectrum of tungsten (D)a continous X-ray spectrum(Bremmstrahlung) with a minimum wavelength of  0.155 Å and the characteristic X-ray spectrum of tangtsen [JEE 2000]

(JEE-PROBLEMS) 4. The transition from the state n = 4 to n = 3 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition [JEE 2001] (A) 2  1 (B) 3  2 (C) 4  2 (D) 5  4 5. The intensity of X-rays from a coolidge tube is plotted agianst wavelength  as shown in the figure. The minimum wavelength found is c and the wavelength of K line is k. As the accelerating voltage is increased. [JEE 2001] l

(A) k – c increases (B) k – c decreases (C) k increases (D) k decreases 6. The potential difference applied to an X-ray tube is 5kV and the current through it is 3.2 mA. Then the number of electrons striking the target per second is [JEE’2002(Scr)] (A) 2 × 1016 (B) 5 × 1016 (C) 1 × 1017 (D) 4 × 1015 7. A Hydrogen atom and Li++ion are both in the second excited state. If l H and l Li are their respective electronic angular momenta, and EH and ELi their respective energies, then [JEE’2002(Scr)] (A) lH > lLi and |EH| > |ELi| (B) lH = lLi and |EH| < |ELi| (C) lH = lLi and |EH| > |ELi| (D) lH < lLi and |EH| < |ELi| 8. A hydrogen like atom (described by the Bohr model) is observed to emit six wavelengths, originating from all possible transition between a group of levels. These levels have energies between –0.85 eV and –0.544 eV (including both these values) (a) Find the atomic number of the atom. (b) Calculate the smallest wavelength emitted in these transitions. [JEE’2002] 9. Two metallic plates A and B each of area 5 × 10–4 m2, are placed at a separation of 1cm. Plate B carries a positive charge of 33.7 × 10–12 C. A monochromatic beam of light, with photons of energy 5 eV each, starts falling on plate A at t = 0 so that 1016 photon fall on it per square meter per second. Assume that one photoelectron is emitted for every 106 incident photons. Also assume that all the emitted photoelectrons are collected by plate B and the work function of

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MODERN PHYSICS plate A remains constant at the value 2 eV. Determine (a) the number of photoelectrons emitted up to t = 10 sec.

(C) A and B will have different intensities while A and C will have equal frequencies. (D) A and B will have equal intensities while B and C will have different frequencies.

(b) the magnitude of the electric field between the plates A and B at t = 10 s and

15.A proton has kinetic energy E = 100 keV which is eqal to that of a photon. The wavelength of photon is 2 and that of proton is 1. The ratio of 2/1 is proportional to [JEE 2004 (Scr.)]

(c) the kinetic energy of the most energetic photoelectron emitted at t = 10 s when it reaches plate B. (Neglect the time taken by photoelectron to reach plate B) [JEE’2002] 10.If the atom 100Fm257 follows Bohr model and the radius of last orbit of 100Fm257 is n times the Bohr radius, then find n [JEE 2003] (A) 100 (B) 200 (C) 4

(D)

1 4

11.The attractive potential for an atom is given by v = v0 ln (r/r0), v0 and r0 are constant and r is the radius of the orbit. The radius r of the nth Bohr’s orbit depends upon principal quantum number n as : [JEE’2003(Scr)] (A) r  n (B) r  1/n2 (C) r  n2 (D) r  1/n 12. Frequency of a photon emitted due to transition of electron of certain elemrnt from L to K shell is found to be 4.2 × 1018 Hz. Using Moseley’s law, find the atomic number of the element, given that the Rydberg’s constant R = 1.1 × 107 m–1. [JEE’2003] 13.In a photoelctric experiment set up, photons of energy 5 eV falls on the cathode having work function 3eV. (a) If the saturation current is iA = 4A for intensity 10–5 W/m2, then plot a graph between anode potential and current. (b) Also draw a graph for intensity of incident radiation of 2 × 10–5 W/m2 ? [JEE’2003] 14.In a photoelectric experiment anode potential is plotted against plate current[JEE 2004 (Scr.)] I



1 2

(A) E2

(B) E

(C) E–1

(D) E 2

1

16 In a photoelectric setup, the radiations from the Balmer series of hydrogen atom are incident on a metal surface of work function 2eV. The wavelength of incident radiations lies between 450 nm to 700 nm. Find the maximum kinetic energy of photoelectron emitted. (Given hc/e = 1242 eV-nm). [JEE 2004] 17.The wavelength of K X – ray of an element having atomic number z = 11 is . The wavelength of K X-ray of another element of atomic number z is 4l. Then z is [JEE’ 2005 (Scr)] (A) 11 (B) 44 (C) 6 (D) 4 18.A photon of 10.2 eV energy collides with a hydrogen atom in ground state inelastically. After few microseconds one more photon of energy 15 eV collides with the same hydrogen atom. Then what can be detected by a suitable detector. (A) one photon of 10.2 eV and an electron of energy 1.4 eV (B) 2 photons of energy 10.2 eV (C) 2 photons of energy 3.4 eV (D) 1 photon of 3.4 eV and one electron of 1.4 eV [JEE’ 2005 (Scr)] 19.In Young’s double slit experiment an electron beam is used to form a fringe pattern instead of light. If speed of the electrons is increased then the fringe width will : (A) increase (B) decrease (C) remains same (D) no fringe pattern will be formed 20.The potential energy of a particle of mass m is given by

C

B

A V

(A) A and B will have different intensities while B and C will have different frequencies. (B) B and C will have different intensities while A and C will have different frequencies.

E0 0  x  1  V(x) =  x1  0 1 and 2 are the de-Broglie wavelengths of the particle, when 0 x  1 and x > 1 respectively. If the total energy of particle is 2E0, find 1/2 [JEE 2005] 21.Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. The ratio of radii of nucleus to that of helium

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MODERN PHYSICS

nucleus is (14)1/3. Find (a) atomic number of the nucleus (b) the frequency of K line of the X-ray produced. (R = 1.1 × 107 m–1 and c = 3 × 108 m/s) [JEE 2005] 22.The graph between 1/ and stopping potential (V) of three metals having work functions 1, 2 and 3 in an experiment of photoelectric effect is plotted as shown in the figure. Which of the following statement(s) is/are correct? [Here  is the wavelength of incident ray]. metal 1 metal 2 metal 3 V

 0.001 0.002 0.004 1/ nm–1 (A) Ratio of work functions 1 : 2 : 3 = 1 : 2 : 4 (B) Ratio of work functions 1 : 2 : 3= 4 : 2 : 1 (C) tan  is directly proportional to hc/e, where h is Planck’s constant and c is the speed of light (D) The vio let colo ur light can eje ct photoelectrons from metals 2 and 3.[JEE 2006]

23.In hydrogen-like atom (z = 11), nth line of Lyman series has wavelength  equal to the deBroglie’s wavelength of electron in the level from which it originated. What is the value of n? [Take : Bohr radius (r0) = 0.53 Å and Rydberg constant (R) = 1.1 × 107 m–1] [JEE 2006] 24.STATEMENT-1 If the accelerating potential in an X-ray tube is increased, the wavelengths of the characteristic X-rays do not change. [JEE 2007] because STATEMENT-2 When an electron beam strikes the target in an X-ray tube, part of the kinetic energy is converted into X-ray energy. (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 25.Electrons with de-Broglie wavelength  fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is[JEE 2007] (A)  0 

(C)  0 

2mc2 h

2m 2 c 2 3 h2

(B)  0 

2h mc

(D) 0 = 

26.The largest wavelength in the ultraviolet region of the hydrogen spectrum is 122 nm. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is : (A) 802 nm (B) 823 nm [JEE 2007] (C) 1882 nm (D) 1648 nm 27.Which one of the following statements is WRONG in the context of X-rays generated from a X-ray tube? (A) Wavele ngth of characteris tic X-rays decreases when the atomic number of the target increases. (B) Cut-off wavelength of the continuous X-rays depends on the atomic number of the target. (C) Intensity of the characteristic X-rays depends on the electrical power given to the X-ray tube (D) Cut-off wavelength of the continuous X-rays depends on the energy of the electrons in the Xrays tube [JEE 2008] Paragraph for Question No. 28 to 30 In mixture of H–He+ gas (He+ is singly ionized He atom), H atoms and He+ ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to He– ions (by collision). Assume that the Bohr model of atom is exactly valid. [JEE 2008] 28.The quantum number n of the state finally poulated in He+ ions is (A) 2 (B) 3 (C) 4 (D) 5 29.The wavelength of light emitted in the visible region by He+ ions after collisions with H atoms is (A) 6.5 × 10–7 m (B) 5.6 × 10–7m –7 (C) 4.8 × 10 m (D) 4.0 × 10–7m 30.The ratio of the kinetic energy of the n = 2 electron for the H atom to that of He+ ion is 1 4 (C) 1

(A)

1 2 (D) 2

(B)

Paragraph for Questions 31 to 33 When a particle is restricted to move along xaxis between x = 0 and x = a, where a is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x = 0 and x = a. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear

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MODERN PHYSICS p2 momentum as E = . Thus, the energy of the 2m particle can be denoted by a quantum number ‘n’ taking values 1, 2, 3 … (n = 1, called the ground state) corresponding to the number of loops in the standing wave. Use the model described above to answer the following three questions for a particle moving in the line x = 0 to x = a.Take h = 6.6 × 10–34 Jsand e= 1.6 × 10–19 C. [JEE 2009]

31.The allowed energy for the particle for a particular value of n is proportional to (A) a–2 (B) a–3/2 –1 (C) a (D) a2 32.If the mass of the particle is m = 1.0 × 10–30 kg and a = 6.6 nm, the energy of the particle in its ground state is closest to (A) 0.8 meV (B) 8 meV (C) 80 meV (D) 800 meV 33.The speed of the particle, that can take discrete values, is proportional to (A) n –3/2 (B) n –1 1/2 (C) n (D) n 34.Photoelectric effect experiments are performed using three different metal plates p, q and r having work functions p = 2.0 eV, q = 2.5 eV and r = 3.0 eV, respectively. A light beam containing wavelengths of 550 nm, 450 nm and 350 nm with equal intensities illuminates each of the plates. The correct I–V graph for the experiment is [JEE 2009] I

I

36.A diatomic molecule has moment of inertia I. By Bohr's quantization condition its rotational energy in the nth level (n = 0 is not allowed ) is

(A)

1 n2

 h2   2   8 I   

1  h2  (B) n  2   8 I 

 h2  (C) n 2   8 I 

2  2 h (D) n  2   8 I 

37.It is found that the excitation frequency from ground to the first excited state of rotation for 4 11 the CO molecule is close to  10 Hz. Then the  moment of inertia of CO molecule about its center of mass is close to (Take h = 2 × 10–34 Js) (A) 2.76 × 10–46 kg m2 (B) 1.87 × 10–46 kg m2 (C) 4.67 × 10–47 kg m2 (D) 1.17 × 10–47 kg m2 38.In a CO molecule, the distance between C (mass = 12 a.m.u.) and O (mass = 16 a.m.u.) 5 – 27 where 1 a.m.u. =  10 kg, is close to 3 –10 (A) 2.4 × 10 m (B) 1.9 × 10–10 m –10 (C) 1.3 × 10 m (D) 4.4 × 10–11 m 39.The wavelength of the first spectral line in o

the Balmer series of hydrogen atom is 6561 A . The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is [JEE-2011]

p q

(A)

(B)

r

V

V

o

o

(A) 1215 A

(B) 1640 A

o

o

(C) 2430 A

(D) 4687 A

I

I

r q

(C)

p

(D)

r

p q

V

35.An  - particle and a proton are accelerated from rest by a potential difference of 100 V. After this, their de Broglie wavelengths are a and p p respectively. The ratio , to the nearst integer,,  is [JEE 2010] Paragraph for questions 36 to 38 The key feature of Bohr's theory of spectrum of hydrogen atoms is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition. [JEE 2010]

40.A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in free-space. It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is A × 10z ( where 1 < A < 10). The value of 'Z' is [JEE-2011] 41.A proton is fired from very far away towards a nucleus with charge Q = 120 e, where e is the electronic charge. It makes a closest approach of 10 fm to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start -27 is: (take the proton mass, mp = (5/3) x 10 kg; -15

h/e = 4.2 x 10

J.s/C;

1 9 = 9 X 10 m/F; 1 fm 4 o

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MODERN PHYSICS

EXERCISE - I

ANSWER KEY

1.

C

2.

B

3.

D

4.

C

5. A

6.

A

7.

C

8.

D

9.

C

10. D

11. B

12. C

13. C

14. A

15. C

16. B

17. B

18. B

19. A

20. B

21. B

22. C

23. C

24. A

25. D

26. B

27. A

28. A

29. B

30. C

31. A

32. C

33. D

34. A

35. D

36. C

37. D

38. B

39. C

40. A

41. A

42. A

43. C

44. C

45. C

46. C

47. D

48. C

49. B

50. C

51. A

52. A

53. B

54. B

55. A

56. C

57. C

58. C

59. D

60. C

61. C

62. B

63. D

64. B

65. A

66. A

67. D

68. D

EXERCISE - II

ANSWER KEY 1. 6. 11. 16.

A,C A,C,D A A,D

2. 7. 12. 17.

B, C A, D A,B A,B

3. 8. 13. 18.

D A,C,D A,C,D A,B,C

4. B 9. A 14. A,C

5. B 10. B 15. B

EXERCISE - III

ANSWER KEY 1.

885

2.

3.

(a) 0.6 volt,

(b) 2.0 mA

4.

when the potential is steady, photo electric emission just stop when hv = (3 + 1) eV = 4.0 eV

5.

5.76 × 10–11 A 6.

9.

–h/e Et2

13. 7 : 36

(a) 2.25 eV,

15/8 V

(b) 4.2 eV,

7.

5 10 20 , 16 80 

(c) 2.0 eV, 0.5 eV

8.

2 eV, 6.53 × 10–34 J-s

10. 8.6 MeV

11. 487.06 nm

12. 4.26 m/s, 13.2 eV

14. 22.8 nm

 1 2 15.    1 2

16. 18 / (5R)

17. 1.257 × 10–23 Am2

18. 2.48 × 10–12 m

20. (i) 5, 16.5 eV, 36.4 A, 340 eV, –680 eV,

19. 5

h 1.06 × 10–11 m 2

21. z = 3, n = 7

22. 54.4 eV

23. n = 3, 3 : 1

26. fb = fa + f

27. 62.5 × 10–12, 192.5 × 10–12

24. 0004

25. 0.88 eV

28. 6210 eV

29. during combination = 3.365 eV; after combination = 3.88 eV (5  3) & 2.63 eV (4  3) 30. 69.5 keV

31. (i) KE = 3.4 eV, (ii)  = 6.66 Å

32. 0.61 Å

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MODERN PHYSICS

EXERCISE - IV

ANSWER KEY 1.

(a) 105 s–1 ;

3. 5.

(i) 2; (ii) 1.99 eV, 0.760 V

8.

(i) 1.33 × 1016 photons/m2-s ; 0.096 Å (ii) 2.956 × 1015 photons/m2s ; 0.0213 A

(b) 286.18 ;

15

10. 6.8 eV, 5 × 10

2.

23.04×10–19J ; (iii) 4  1, 4  3 4 6. 1.1 × 1012 7.

9.

(iii) 1.06 volt

(d) 111 s

(i) 5/16 photon/sec,

3.1 × 106 m/s 489.6 eV, 25.28 Å (i)4125 Å, (ii) 13.2 A

(ii) 5/1600 electrons/sec

Hz

11. (i) Allowed values of energy of neutron = 6.36 eV and 0.312 eV; Allowed values of energy of He atom = 17.84 eV and 16.328 eV, (ii) 18.23 × 1014 Hz, 9.846 × 1015 Hz, 11.6 × 1015 Hz 12. n = 6, Z = 3

13. KE  151 eV, dleast = 0.5 Å

14. 7.4 eV, 4.7 Volts

EXERCISE - V

ANSWER KEY 1.

(a) C, (b) A

2.

(a) n = 2, z = 4 ; G.S.E. –217.6 eV; min. energy = 10.58 eV ;

(b) 6.25 × 1019 per sec, 0.5 eV 6. A 7. B 10. D 11. A

3. D 4. D 5. A 8. 3, 4052.3 nm 9. 5 × 107, 2000 N/C, 23 eV 12. z = 42

I A 4A

13.

–5

2

I = 2 × 10 W/m I = 10 W/m –5

2

14. A

15. B

19. B

20.

21. v = 1.546 × 1018 Hz 22. A,C

23. n = 24

24. B

25. A

26. B

27. B

28. C

29. C

30. A

31. A

32. B

33. D

34. A

35. 3

36. D

37. B

38. C

39. A

40. 7

–2V

17. C

16. 0.55 eV

vP

18. A

2

41. 7

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MODERN PHYSICS - II THEORY AND EXERCISE BOOKLET

CONTENTS S.NO.

TOPIC

PAGE NO.

1. Nucleus ............................................................................................................. 3 2. Mass Defect ...................................................................................................... 4 3. Binding Energy .................................................................................................. 4 4. Radioactivity ................................................................................................... 4 – 5 5. Radioactive Decay .......................................................................................... 5 – 8 6. Nuclear stability ................................................................................................ 8 7. Nuclear Force .................................................................................................... 8 8. Radioactivity Decay ....................................................................................... 9 – 11 9. Nuclear Fission ................................................................................................ 12 10. Nuclear Fusion ............................................................................................... 12 11. Exercise -I .................................................................................................. 13 – 22 12. Exercise - II ................................................................................................ 23 – 25 13. Exercise - III ................................................................................................ 26 - 31 14. Exercise - IV .................................................................................................. 32 15. Exercise - V ................................................................................................ 33 - 38 16. Answer key ................................................................................................ 39 – 40

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MODERN PHYSICS - 2

Syllabus : Atomic nucleus; Alpha, beta and gamma radiations; Law of radioactive decay; Decay constant; Half-life and mean life; Binding energy and its calculation; Fission and fusion processes; Energy calculation in these processes.

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MODERN PHYSICS - 2

NUCLEAR PHYSICS

1.

It is the branch of physics which deals with the study of nucleus. NUCLEUS : (a) Discoverer : Rutherford (b) Constituents : neutrons (n) and protons (p) [collectively known as nucleons] 1. Neutron : It is a neutral particle. It was discovered by J. Chadwick. Mass of neutron, mn = 1.6749286 × 10–27 kg 2. Proton : It has a charge equal to +e. It was discovered by Goldstein. Mass of proton, mp = 1.6726231 × 10–27 kg (mp (c)

~ 

mn)

Representation : XA

Z

where X Z A

or    =

A Z

X

symbol of the atom Atomic number = number of protons Atomic mass number = total number of nucleons. no. of protons + no. of neutrons.

Atomic mass number : It is the nearest integer value of mass represented in a.m.u (atomic mass unit) 1 a.m.u =

1 [mass of one atom of 6C12 atom at rest and in ground state] 12

1.6603 × 10–27 kg ; 931.478 MeV/c2 mass of proton (mo) = mass of neutron (mn) = 1 a.m.u. Some definitions : (1) Isotopes : The nuclei having the same number of protons but different number of neutrons are called isotopes. (2) Isotones : Nuclei with the same neutron number N but different atomic number Z are called isotones. (3) Isobars : The nuclei with the same mass number but different atomic number are called isobars (d) Size of nucleus : Order of 10–15 m (fermi) Radius of nucleus ; R = R0A1/3 where R0 = 1.1 × 10–15 m (which is an empirical constant) A = Atomic mass number of atom. (e) Density : density 

Am p Amp 3mp mass    3  1.67  10 –27 3   3  1017 kg / m 3 4 volume 4 4  R 3 1/ 3 3 0 R (R 0 A ) 4  3.14  (1.1 10 –15 )3 3 3

Nuclei of almost all atoms have almost same density as nuclear density is independent of the mass number (A) and atomic number (Z).

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MODERN PHYSICS - 2

MASS DEFECT It has observed that there is a differene between expected mass and actual mass of a nucleus. Mexpected = z mp + (A – Z) mn Mobserved = Matom – Zme It is found that Mobserved < Merxpected Hence, mass defect is defined as Mass defect = Mexpected – Mobserved m = [Zmp + (A – Z)mn – [Matom – Zme]

3.

BINDING ENERGY It is the minimum energy required to break the nucleus into its constituent particles. or Amount of energy released during the formation of nucleus by its constituent particles and bringing them from infinite separation. Binding Energy (B.E.) = mc2 BE = m (in amu) × 931.5 MeV/amu = m × 931.5 MeV = m × 931 MeV Note : If binding energy per nucleon is more for a nucleus then it is more stable. For Example

If

 B.E1   B.E 2     A  A  1   2

   

then nucleus 1 would be more stable. 3.1

• •

4.

Variation of binding energy per nucleon with mass number : The binding energy per nucleon first increases on B.E. an averge and reaches a maximum of about 8.7 A MeV for A. 50  80. For still heavier nuclei, the 56 26Fe binding energy per nucleon 8.8 MeV slowly decreases as A increases. Binding energy per nucleon is maximum for 26Fe56, which is equal to 8.8 MeV. Binding energy per nucleon is more for medium nuclei than for heavy 56 nuclei. Hence, medium nuclei are highly stable. A The heavier nuclei being unstable have tendency to split into medium nuclei. This process is called Fission. The Lighter nuclei being unstable have tendency to fuse into a medium nucleus. This process is called Fusion. RADIOACTIVITY : It was discovered by Henry Becquerel. Spontaneous emission of radiations (, , ) from unstable nucleus is called radioactivity. Substances which shows radioactivity are known as radioactive substance. Radioactivity was studied in detail by Rutherford. In radioactive decay, an unstable nucleus emits  particle or  particle. After emission of  or  the remaining nucleus may emit - particle, and converts into more stable nucleus. - particle : It is a doubly charged helium nucleus. It contains two protons and two neutrons. Mass of  - particle = Mass of 2He4 atom – 2me = 4 mp Charge of  - particle = + 2e 394,50-Rajeev Gandhi Nagar Kota, Ph. No. : 93141-87482, 0744-2209671 IVRS No : 0744-2439051, 52,53, www. motioniitjee.com, [email protected]

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MODERN PHYSICS - 2

-particle : (a) – (electron) : Mass = me ; Charge = – e + (b)  (positron) Mass = me ; Charge = + e positron is an antiparticle of electron. Antiparticle : A particle is called antiparticle of other if on collision both can annihilate (destroy completely) and converts into energy. For example : (i) electron (–e, me) and positron (+e, me) are anti particles, (ii) neutrino () and antineutrino (  ) are anti particles. -particle : They are energetic photons of energy of the order of Mev and having rest mass zero. 5.

RADIOACTIVE DECAY (DISPLACEMENT LAW) : 5.1  - decay : Z

XA 

Z–2

Y A – 4  2 He 4  Q

Q value : It is definied as energy released during the decay process. Q value = rest mass energy of reactans – rest mass energy of products. This energy is available in the form of increase in K.E. of the products

Y

A zX

+ 2He4 + Q

Z-electrons Z-electrons  particle (it has charge + 2e)

Mx = mass of atom zXA My = mass of atom z – 2YA – 4 MHe = mass of atom 2He4 Q value = [(Mx – Zme) – {(My – (Z – 2)me) + (MHe – 2me)}] c2 = [Mx – My – MHe] c2 Considering actual number of electrons in  - decay Q value = [Mx – (My + 2me) – (MHe – 2me)] c2 = [Mx – My – MHe]c2 Let,

Calculation of Kinetic energy of final products : As atom X was initially at rest and no external forces are acting, so final momentum also has to be zero. Hence both Y and  - particle will have same momentum in magnitude but in opposite direction. p

p

Y



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MODERN PHYSICS - 2

p 2  p Y

2

2mT = 2 mY TY

Q = Ty + T

(Here we are representing T for kinetic energy)

mT = mYTy

T 

mY Q ; m  m Y

TY 

m Q m  m Y

T 

A–4 Q A

TY 

4 Q A

;

From the above calculation, one can see that all the particles emitted should have same kinetic energy. Hence, if they are passed through a region of uniform magnetic field having direction perpendicular to velocity, they should move in a circle of same radius. r

 r

v

B

mv mv 2Km   qB 2eB 2eB

Experimental Observation : Experimentally it has been observed that all the -particles do not move in the circle of same radius, but they move in 'circles having different radii. This shows that they have different kinetic energies. But it is also observed that they follow circular paths of some fixed values of radius i.e. yet the energy of emitted -particles is not same but it is quantized. The reason behind this is that all the daughter nuclei produced are not in their ground state but some of the daughter nuclei may be produced in their excited states and they emits photon to aquire their ground state. Y* (excited state)    Q

X

Y + photon ( particle)

The only difference between Y and Y* is that Y* is in excited state and Y is in ground state. Let, the energy of emited - particles be E  Q = T + TY + E where Q = [Mx – MY – MHe] c2 T + TY = Q – E T 

5.2

mY (Q – E ) ; m  m Y

TY 

m (Q – E ) m  mY

– – decay : z

X A  Z 1 Y A  –1 e 0  Q

–1

e 0 can also be written as

–1

0

Here also one can see that by momentum and energy conversion, we will get Te 

mY Q me  m Y ,

TY 

me Q me  m Y

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MODERN PHYSICS - 2

as me