elasticityandthermalexpansion elasticity & thermal expansion...sol. energy stored per unit...

Elasticity and Thermal Expansion

E17-1

ends. Which of the statement is correct(1) Increase in length is inversely proportional to its length L(2) Increase in length is proportional to area of cross-section A(3) Increase in length is inversely proportional to A(4) Increase in length is proportional to Young’s modulus

Ans. (3)

Sol.A

lYA

FLl

1

Q.2 The dimensions of four wires of the same material are given below. In which wire the increase in length will be maximumwhen the same tension is applied(1) Length 100 cm, Diameter 1 mm (2) Length 200 cm, Diameter 2 mm(3) Length 300 cm, Diameter 3 mm (4) Length 50 cm, Diameter 0.5 mm

Ans. (4)

Sol. 2d

L

A

Ll

A

L

A

FY

1.

2d

Ll [As F and Y are constant]

The ratio of 2d

Lis maximum for case (4)

Q.3 The ratio of the lengths of two wires A and B of same material is 1 : 2 and the ratio of their diameter is 2 : 1. They arestretched by the same force, then the ratio of increase in length will be(1) 2 : 1 (2) 1 : 4 (3) 1 : 8 (4) 8 : 1

Ans. (3)

Sol.

d

Ll

AY

FLl

2

1

2

2

1

2

1

d

d

L

L

l

l

8

1

2

1

2

12

Q.4 The area of cross-section of a wire of length 1.1 metre is 1 mm2. It is loaded with 1 kg. If Young’s modulus of copper is

211 /101.1 mN , then the increase in length will be (If )/10 2smg

(1) 0.01 mm (2) 0.075 mm· (3) 0.1 mm (4) 0. 15 mmAns. (3)

Sol. mmmAY

mgLl 1.0

10101.1

1.1101611

Q.5 If the temperature increases, the modulus of elasticity(1) Decreases (2) Increases(3) Remains constant (4) Becomes zero

Ans. (1)

Sol. Because due to increase in temperature intermolecular forces decreases.

Q.6 A force F is needed to break a copper wire having radius R. The force needed to break a copper wire of radius 2R will be(1) F/2 (2) 2F (3) 4F (4) F/4

Ans. (3)

Elasticity & Thermal Expansion

Elastic behaviour, longitudinal stress young modulusQ.1 The length of an iron wire is L and area of cross-section is A. The increase in length is l on applying the force F on its two

EXERCISE-I


E17-2

Sol. Breaking Force Area of cross section of wire (r2)If radius of wire is double then breaking force will become four times.

Q.7 If Young’s modulus for a material is zero, then the state of material should be(1) Solid (2) Solid but powder (3) Gas (4) None of the above

Ans. (2)

Sol. Y is defined for solid only and for powders, 0Y

Q.8 A copper wire and a steel wire of the same diameter and length are connected end to end and a force is applied, whichstretches their combined length by 1 cm. The two wires will have(1) Different stresses and strains(2) The same stress and strain(3) The same strain but different stresses(4) The same stress but different strains

Ans. (4)

Sol.area

ForceStress 1. .

In the present case, force applied and area of cross-section of wires are same, therefore stress has to be the same.

Y

StressStrain

Since the Young’s modulus of steel wire is greater than the copper wire, therefore, strain in case of steel wire is less thanthat in case of copper wire.

Q.9 A graph is shown between stress and strain for a metal. The part in which Hooke’s law holds good is

Stre

ss

A

B

C

D

StrainO

(1) OA (2) AB (3) BC (4) CD

Ans. (1)

Sol. In the region OA, stress strain i.e. Hooke’s law hold good.

Q.10 The strain-stress curves of three wires of different materials are shown in the figure. P, Q and R are the elastic limits of the

wires. The figure shows that

Q

R

P

O

Stra

in

Stress

(1) Elasticity of wire P is maximum (2) Elasticity of wire Q is maximum(3) Tensile strength of R is maximum (4) None of the above is true

Ans. (4)

Sol. As stress is shown on x-axis and strain on y-axis

So we can say thatslope

1

tan

1cot

Y

So elasticity of wire P is minimum and of wire R is maximum


E17-3

Q.11 The graph shows the behaviour of a length of wire in the region for which the substance obeys Hook’s law. P and Q

represent

Q

P

(1) P = applied force, Q = extension(2) P = extension, Q = applied force(3) P = extension, Q = stored elastic energy(4) P = stored elastic energy, Q = extension

Ans. (3)

Sol. Graph between applied force and extension will be straight line because in elastic range,

Applied force extensionbut the graph between extension and stored elastic energy will be parabolic in nature

As 22/1 kxU or 2xU .

Q.12 The points of maximum and minimum attraction in the curve between potential energy (U) and distance (r)of a diatomic

molecules are respectively

P

Qr

R

T

U

S

(1) Sand R (2) T and S (3) R and S (4) S and T

Ans. (4)

Sol. Attraction will be minimum when the distance between the molecule is maximum.

Attraction will be maximum at that point where the positive slope is maximum becausedx

dUF

(1) Too less (2) Greater than all matters(3) Less than all matters (4) Zero

Ans. (2)

Q.14 The ratio of lengths of two rods A and B of same material is 1 : 2 and the ratio of their radii is 2 : 1, then the ratio of modulusof rigidity of A and B will be(1) 4 : 1 (2) 16 : 1 (3) 8 : 1 (4) 1 : 1

Ans. (4)Sol. Modulus of rigidity is the property of material.

Q.15 Two wires A and B of same length and of the same material have the respective radii 1r and 2r . Their one end is fixed with

a rigid support, and at the other end equal twisting couple is applied. Then the ratio of the angle of twist at the end of Aand the angle of twist at the end of B will be

(1) 22

21

r

r(2) 2

1

22

r

r(3) 4

1

42

r

r(4) 4

2

41

r

r

Ans. (3)

Tangential stress and strain, shear modulus

Q.13 Modulus of rigidity of diamond is


E17-4

Sol. Twisting couplel

rC

2

4

If material and length of the wires A and B are equal and equal twisting couple are applied then

4

1

r

4

1

2

2

1

r

r

Q.16 Modulus of rigidity of a liquid(1) Non zero constant(2) Infinite(3) Zero(4) Can not be predicted

Ans. (3)

Q.17 For a given material, the Young’s modulus is 2.4 times that of rigidity modulus. Its Poisson’s ratio is(1) 2.4 (2)1.2 (3)0.4 (4)0.2

Ans. (4)

Sol. )1(2 Y 1.

2.012.1)1(24.2

Q.18 The upper end of a wire of radius 4 mm and length 100 cm is clamped and its other end is twisted through an angle of 30°.

Then angle of shear is(1) 12° (2) 0.12° (3)1.2° (4) 0.012°

Ans. (2)

Sol. Angle of shearL

r oo 12.030

100

104 1

Q.19 Shearing stress causes change in(1) Length (2) Breadth (3) Shape (4) Volume

Ans. (3)

Q.20 When compared with solids and liquids, the gases have(1) Minimum volume elasticity(2) Maximum volume elasticity(3) Maximum Young’s modulus(4) Maximum modulus of rigidity

Ans. (1)

Sol. A small change in pressure produces a large change in volume.

Q.21 The isothermal elasticity of a gas is equal to(1) Density (2) Volume (3) Pressure (4) Specific heat

Ans. (3)

Sol. Isothermal elasticity PKi

Q.22 The adiabatic elasticity of a gas is equal to

(1) density (2) volume (3) pressure (4) specific heat

Ans. (3)

Sol. Adiabatic elasticity PKa

Pressure and volumetric strain, bulk modulus of elasticity


E17-5

Q.23 The only elastic modulus that applies to fluids is(1) Young’s modulus (2) Shear modulus(3) Modulus of rigidity (4) Bulk modulus

Ans. (4)

Q.24 If the volume of the given mass of a gas is increased four times, the temperature is raised from 27°C to 127°C. Theelasticity will become(1) 4 times (2) 1/4 times (3) 3 times (4) 1/3 times

Ans. (4)

Sol. From the ideal gas equation2

22

1

11

T

VP

T

VP

3

1

300

400

4

1

1

2

2

1

1

2

1

2

T

T

V

V

P

P

E

E

31

2

EE

i.e. elasticity will become3

1times.

Q.25 A ball falling in a lake of depth 200 m shows 0.1% decrease in its volume at the bottom. What is the bulk modulus of thematerial of the ball

(1) 28 /106.19 mN (2) 210 /106.19 mN

(3) 210 /106.19 mN (4) 28 /106.19 mN

Ans. (1)

Q.26 For a constant hydraulic stress on an object, the fractional change in the object’s volume

V

Vand its bulk modulus (2)

are related as

(1) BV

V

(2)

BV

V 1

(3)

2BV

V

(4)

2

BV

V

Ans. (2)

Sol.VV

pB

/

1.

V

V

B

1p[ constant]

force

Exte

nsio

n

I. It will be easier to compress this rubber than expand itII. Rubber does not return to its original length after it is stretchedIII. The rubber band will get heated if it is stretched and releasedWhich of these can be deduced from the graph(1) III only (2) II and III (3) I and III (4) I only

Ans. (1)

Sol. Area of hysterisis loop gives the energy loss in the process of stretching and unstretching of rubber band and this loss

will appear in the form of heating.

Elastic potential energy

Q.27 The diagram shows a force-extension graph for a rubber band. Consider the following statements


E17-6

Q.28 If x longitudinal strain is produced in a wire of Young’s modulus y, then energy stored in the material of the wire per unitvolume is

(1) 2yx (2) 22 yx (3) xy 2

2

1(4)

2

2

1yx

Ans. (4)

Sol. Energy stored per unit volume StrainStress2

1

22

2

1(Strain)modulussYoung'

2

1xY

Q.29 If the potential energy of a spring is V on stretching it by 2 cm, then its potential energy when it is stretched by 10 cm willbe(1) V/25 (2) 5V (3) V/5 (4)25V

Ans. (4)

Sol.2

2

1l

L

YAU

1. 2lU

252

1022

1

2

1

2

l

l

U

U 12 25UU

i.e. potential energy of the spring will be 25 V

Q.30 Two wires of same diameter of the same material having the length l and 2l. If the force F is applied on each, the ratio of

the work done in the two wires will be

(1) 1 : 2 (2) 1 : 4 (3) 2 : 1 (4) 1 : 1Ans. (1)

FlW2

1 1. lW (F is constant)

2

1

22

1

2

1 l

l

l

l

W

W

Q.31 A 5 metre long wire is fixed to the ceiling. A weight of 10 kg is hung at the lower end and is 1 metre above the floor. Thewire was elongated by 1 mm. The energy stored in the wire due to stretching is(1) Zero (2) 0.05 joule (3) 100 joule (4) 500 joule

Ans. (2)

Sol. mgllFW2

1

2

1 1.

J05.010110102

1 1

Q.32 If the tension on a wire is removed at once, then

(1) It will break(2) Its temperature will reduce(3) There will be no change in its temperature(4) Its temperature increases

Ans. (4)

Sol. Due to tension, intermolecular distance between atoms is increased and therefore potential energy of the wire is increasedand with the removal of force interatomic distance is reduced and so is the potential energy. This change in potentialenergy appears as heat in the wire and thereby increases the temperature.


E17-7

Q.33 A wire is suspended by one end. At the other end a weight equivalent to 20 N force is applied. If the increase in lengthis 1.0 mm, the increase in energy of the wire will be(1) 0.01 J (2) 0.02 J (3) 0.04 J (4) 1.00 J

Ans. (1)

Sol. Increase in energy J01.0101202

1 3

Q.34 The Young’s modulus of a wire is Y. If the energy per unit volume is E, then the strain will be

(1)Y

E2(2) EY2 (3) EY (4)

Y

E

Ans. (1)

Sol. Energy per unit volume2strain)(Y

2

1

Y

E2strain

Q.35 A wire of length L and cross-sectional area A is made of a material of Young’s modulus Y. It is stretched by an amount x.

The work done is

(1)L

YxA

2(2)

L

AYx 2

(3)L

AYx

2

2

(4)L

AYx 22

Ans. (3)

of the wire is rigidly fixed. If coefficient of linear expansion of the wire C /108 6 and Young’s modulus

211 /102.2 mNY and its temperature is increased by 5°C, then the increase in the tension of the wire will be

(1) 4.2 N (2) 4.4 N (3) 2.4 N (4) 8.8 NAns. (4)

Sol. Increase in tension of wire YA

N8.851010102.2108 42116

Q.37 A wire of cross-sectional area 23 mm is first stretched between two fixed points at a temperature of 20°C. Determine the

tension when the temperature falls to 10°C. Coefficient of linear expansion 1510 C and 211 /102 mNY

(1) 20 N (2) 30 N· (3) 60 N (4) 120 N

Ans. (3)

Sol. tYAF N60)1020(10103102 5611

Q.38 The coefficient of linear expansion of brass and steel are 1 and 2 . If we take a brass rod of length 1l and steel rod of

length 2l at 0°C, their difference in length )( 12 ll will remain the same at a temperature if

(1) 1221 ll (2) 212

221 ll (3) 2

221

21 ll (4) 2211 ll

Ans. (4)

Sol. )1( 222 lL 1. and )1( 111 lL

)()()( 11221212 llllLL

Thermal expension

Q.36 A force of 200 N is applied at one end of a wire of length 2 m and having area of cross-section 10 2 cm 2 . The other end


E17-8

Now )()( 1212 llLL so, 01122 ll

Q.39 A rod is fixed between two points at 20°C. The coefficient of linear expansion of material of rod is C /101.1 5 and

Young’s modulus is mN /102.1 11 . Find the stress developed in the rod if temperature of rod becomes 10°C

(1) 27 /1032.1 mN (2) 215 /1010.1 mN

(3) 28 /1032.1 mN (4) 26 /1010.1 mN

Ans. (1)

Sol. Thermal stress = Y

)1020(101.1102.1 511 27 /1032.1 mN


E17-9

Y = 9 × 1010 N/m2

A

F= Y

F =AY

=(2x10–3)2 x 9 × 109 x100

1=x 4 x 10–6 x 9 × 107 = 360N

Q.2 The load versus elongation graph for four wires of the same materials is shown in the figure. The thinnest wire isrepresented by the line :

Sol. /

A/F

= Y

2rYF

1

Y

F= r2

Y & are same for all then

For same load r

1

Sol. F =h

xA = 0.4 × 1011 × 1 × .005 ×

1

1002. 2

= 4×104 N

EXERCISE-II

Q.3 A square brass plate of side 1.0 m and thickness 0.005 m is subjected to a force F on its smaller opposite edges, causing adisplacement of 0.02 cm. If the shear modulus of brass is 0.4 × 1011 N/m2, the value of the force F is(1) 4 × 103 N (2)400 N (3*) 4 × 104 N (4)1000 N

Ans. (3)

(1) OC (2) OD (3*) OA (4) OBAns. (3)

Q.1 The diameter of a brass rod is 4 mm and Young’s modulus of brass is 9 × 1010 N/m2. The force required to stretch it by 0.1%of its length is :

(1*) 360 N (2)36 N (3) 144

× 103 N (4) 36

× 105N

Ans. (1)Sol. d = 4mm


E17-10

Ans. (C)

Sol.V

V=

B

P= 11

5

1025.1

101

= 8×10–7

Sol. K =

AY, K' =

2/

AY4

= 8K

2

2

K2

1

K82

1

2

U

U = 16 J

Q.7 A steel scale is to be preparjed such that the millimeter intervals are to be accurate within 6 × 10–5 mm. The maximum

= 12 × 10-6 k-1

thenL = LT6 × 10–5 mm = (1mm) (12 × 10–6)TT = 5°C

L= 25 cm,A= 0.8 × 10-4 cm2

T = 10°C,= 10-5 °C-1, Y = 2×1010N2

then

L

L

= T =

F

AYF = AYt= (10–5)(0·8 × 10–4) × (2 × 1010) × 10= 160 N

Ans. (1)Sol. Given

Q.8 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 sameamount 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.)(1) 160 N (2)360 N (3)106 N (4)260 N

Ans. (3)Sol. Given L = 1 mm,L = 6 × 10-5 mm

temperature variation from the temperature of calibration during the reading of the millimeter marks is(= 12 × 10–6 k–1)(1) 4.0 ºC (2) 4.5 ºC (3)5.0 ºC (4) 5.5 ºC

Q.6 If work done in stretching a wire by 1mm is 2J, the work necessary for stretching another wire of same material, but withdouble the radius and half the length by 1mm in joule is -(1) 1/4 (2)4 (3)8 (4*) 16

Ans. (4)

Q.5 Expansion during heating –(1) occurs only in a solid (2) increases the density of the material(3*) decreases the density of the material (4) occurs at the same rate for all liquids and solids.

Ans. (3)Sol. On heating volume of substance increases while mass of the substance remains the same. Hence the density will decrease

Q.4 Ametal block is experiencing an atmospheric pressure of 1 × 105 N/m2, when the same block is placed in a vacuum chamber,the fractional change in its volume is (the bulk modulus of metal is 1.25 × 1011 N/m2)(1) 4 × 10–7 (2) 2 × 10–7 (3*) 8 × 10–7 (4) 1 × 10–7


E17-11

Q.9 Two rods of different materials having coefficients of thermal expansion 1,

2and Young’s moduli Y

1, Y

2respectively 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. If1:

2= 2 : 3, the thermal stresses developed in the two rods are equal provided Y

1: Y

2is equal

to

L2= L + L

2t

1 1 1

2 2 2

Stress Y L t L.

Stress L Y L t

1 =1

2

Y2

3 Y 2

3

Y

Y

2

1

Q.10 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

dI = 2CMRdR = 2CMR [RT] = 2IT

Q.11 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 :

Y

Sol. F =AYL

L

=AYYT

f = K

F AY TK

A

f

Y

Sol.L

L

=t = – 20

means read more so actual is less

Q.12 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 :

(1) 25 cm (2) <25 cm (3) > 25 cm (4) can not say

Ans. (2)

YAns. (2)

(1)Y

(2)

(3) (4)Y

Ans. (3)

Sol. I = CMR2

1(1) I T (2) IT (3) 2 IT (4) 3 IT

2

Ans. (3)

Sol. L1= L + L

1t

(1) 2 : 3 (2) 1 : 1 (3) 3 : 2 (4) 4 : 9


E17-12

L = LT

0·075 = 201(100)

0·045 = 202(100)

Let for third rod L1and L

2= 20 – L

1

So L3

=L1+L

2

0·06 = L1

1100 + (20 – L

1)

2100

L1

= 10 cm.

f = coafficient of cubical expansion

Spere=

l’

3

266.5 1.527

1 35f4 7

3 2

f= 8.3 × 10–4/.c

Q.15 A liquid with coefficient of volume expansion is filled in a container of a material having the coefficient of linear expansion .f the liquid overflows on heating, then –

> 3.

Q.16 Two rods having length 1and

2, made of materials with the linear coefficient of expansion

1and

2, were welded together.

The equivalent coefficients of linear expansion for the obtained rod :-

21

1221

21

2211

21

2211

21

2112

f=

1+

2+ (

1

1+

2

2)T

f= (

1+

2)

T1

21

2211

αα.

Ans. (3)Sol.

1(1 +

1T) +

2(1 +

2T) =

f

(1)

(2)

(3*)

(4)

(1*) > 3 (2) < 3 (3) = 3 (4) none of these

Ans. (1)

Sol. V>V

e

Q.14 Asphere of diameter 7 cm and mass 266.5 gm floats in a bath of a liquid.As the temperature is raised, the sphere just beginsto 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 :(1) 8.486 × 10–4 per °C (2) 8.486 × 10–5 per °C (3) 8.486 × 10–6 per °C (4) 8.486 × 10–3 per °C

Ans. (1)Sol. Given

Ans. (2)

Sol. Given L = 20 cm,L1=0.075 cm,L

2= 0.045 cm

Q.13 Arod 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.

(1) 20 cm (2) 10 cm (3)15 cm (4) 18 cm


E17-13

Volume increases but mass remains same.

Q.18 Ametal ball immersed in Alcohol weights W1at 0°C and W

2at 50°C. The coefficient of cubical expansion of the metal ()

m

Vm

<Val

So completely Immersed

m<

AlSo W

2> W

1[ Displaced mass of alchol is less]

Q.19 The volume thermal expansion coefficient of an ideal gas at constant pressure is

2T

1

Sol. PV = nRT V =P

nRT

V =T

VT

So , =T

1

Sol.L

L

× 100 = 1 = 100t = 100(T

2– T

1)

A

A

× 100 = 200t = 2%

Q.21 If two rods of length L and 2L having coefficients of linear expansion and 2 respectively are connected so that totallength becomes 3L, the average coefficient of linear expansion of the composition rod equals :

5 5

(3L)nett = Lt = (2L) (2)t

net

4 5

3 3

(4) none of these3

(3)2

(2)2

3(1)

Ans. (3)

Sol. L =L1+L

2

Ans. (2)

Q.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

?(1) 1% (2) 2% (3) 3% (4) 4%

(Here T = absolute temperature of gas)Ans. (3)

1(4)(1) T (2) T2 (3*)

T

Ans. (3)Sol.

m<

Al

m>>

ac

is less than that of alcohol ()Al

. Assuming that density of metal is large compared to that of alcohol, it can be shown that(1) W

1> W

2(2) W

1= W

2(3) W

1< W

2(4) any of (1), (2) or (3)

Ans. (4)Sol.

oil=

vessel D.

Q.17 An open vessel is filled completely with oil which has same coefficient of volume expansion as that of the vessel. Onheating both oil and vessel,(1) the vessel can contain more volume and more mass of oil(2) the vessel can contain same volume and same mass of oil(3) the vessel can contain same volume but more mass of oil(4) the vessel can contain more volume but same mass of oil


E17-14

Q.22 Two large holes are cut in a metal sheet. If this is heated, distances AB and BC, (as shown)

V0x

= 20A; V0y

= 30ANow at time T y read 120°CSo. V’

0y=A(120) = 30A(1 +

mT)

and V’0x

= Ah = 20A(1 + m

T)

Dividing120 30

h 20

h = 80.

Ans. (4)Sol. at 0°C

Q.23 Two thermometers x and y have fundamental intervals of 80º and 120º. When immersed in ice, they show the reading of 20ºand 30º. If y measures the temperature of a body as 120º, the reading of x is :(1) 59º (2)65º (3)75º (4)80º

Ans. (1)

Sol. On heating the expansion will take place hence both the distances will increase.

(1*) both will increase (2) both will decrease

(3) AB increases, BC decreases (4) AB decreases, BC increases

elasticityandthermalexpansion elasticity & thermal expansion...sol. energy stored per unit...

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