First mechanics final

Rotational Dynamics | 251

18. A disc of moment of inertia 10 kgm2 about its centre rotates steadily about the centre with an angular
velocity of 20 rad s–1. Calculate (i) its rotational energy, (ii) its angular momentum about the centre,

(iii) the number of revolutions per second of the disc.

Solution: Angular velocity (ω) = 20 rad s–1
Moment of inertia of disc (I) = 10 kgm2

(i) Rotational Energy (K.Erot) = ?

We have, KErot = 1 Iω2 = 1 × 10 × 202 = 2000 J.
2 2

(ii) Angular momentum (L) = ?
We have, L = Iω = 10 × 20 = 200 kgm2 s–1

(iii) No. of revolutions per sec i.e. frequently (f) = ?

Since, ω = 2πf

∴ f = ω = 20 = 10 = 3.2 rev/sec
2π 2π π

19. A flywheel has a kinetic energy of 200 J. Calculate the number of revolutions it makes before coming

to rest if a constant opposing couple of 5 Nm is applied to the flywheel. If the moment of inertia of
the flywheel about its centre is 4 kgm2. How long does it take to come to rest?

Solution:

Kinetic energy of flywheel (K.E) = 200 J No. of revolutions (n) = ?
Constant opposing torque (τ) = –5 Nm Moment of inertia (I) = 4 kgm2

Time taken (t) = ? Final angular velocity (ω) = 0

Since, K.E. = 1 Iω02
2

or, 200 = 1 × 4 × ω20
2

∴ ω0 = 10 rad s–1
Also, τ = Iα

or, –5 = 4 × α

∴ α = –1.25 rad s–2
Now, ω2 = ω02 + 2αθ
or, 02 = 102 – 2 × 1.25 × θ

∴ θ = 40 rad.

Again, θ = 2πn

or, n = θ = 2 40 = 6.4 rev
2π × 3.14

The time taken to come to rest is, t = ω – ω0 = 0 – 10 = 8 sec.
α –1.25

20. A disc rolling along a horizontal plane has a moment of inertia 2.5 kgm2 about its centre and a mass

of 5 kg. The velocity along the plane is 2 ms–1. If the radius of the disc is 1 m, find (i) the angular

velocity, (ii) the total energy (rotational and translational) of the disc. [HSEB 2072 Set E]

Solution: Velocity (v) = 2ms–1 radius (r) = 1 m
I = 2.5 kgm2, mass (m) = 5 kg

(i) Angular velocity (ω) = ?

We have v = rω

or, ω = v = 2 = 2 rad s–1
r 1

(ii) Total energy (E) = ?

We have, E = K.Erot + K.Etrans = 1 Iω2 + 1 mv2 = 1 × 2.5 × 22 + 1 × 5 × 22
2 2 2 2

∴ E = 15 Joules

252 | Essential Physics

21. A wheel of moment of inertia 20 kgm2 about its axis is rotated from rest about its centre by a constant

torque τ and the energy gained in 10s is 360 J. Calculate (i) the angular velocity at the end of 10s,

(ii) τ, (iii) the number of revolutions made by the wheel before coming to rest if τ is removed at 10 s

and a constant opposing torque of 4N is then applied to the wheel.

Solution:

Moment of inertia (I) = 20 kgm2 Initial angular velocity (ω0) = 0

Constant torque = τ; Time (t) = 10 secs Energy gained (E) = 360 J

(i) ω = ? after 10 sec

Since, E = 1 Iω2
2

or, ω= 2E 2 × 360
I= 20

∴ ω = 6 rad s–1

(ii) τ = ?

We have, τ = Iα = I (ω – ω0) = 20 × (6 – 0)
t 10

∴ τ = 12 Nm

(iii) No. of revolutions (n) = ?

Constant opposing torque (τ) = –4N

Final angular velocity (ω') = 0
Initial angular velocity (ω0' ) = 6 rad s–1

Now, τ = Iα'

or, –4 = 20 × α'
∴ α' = – 0.2 rads–2
Also, ω'2 = ω'2 + 2 α'θ

or, 0

02 = 62 – 2 × 0.2 × θ

∴ θ = 36 90 rad
0.4=

Now, θ = 2πn

or, n = θ = 2 90 = 14.3 rev
2π × 3.14

22. A uniform rod of length 3m is suspended at one end so that it can move about an axis perpendicular
to its length. The moment of inertia about the end is 6 kgm2 and the mass of the rod is 2 kg. If the rod

is initially horizontal and then released, find the angular velocity of the rod when (i) it is inclined at

30° to the horizontal; (ii) reaches the vertical.

Solution: l C.G. P
O2
Length of rod (l) = 3m x1 = l sinθ
M.I. about end 0 (fixed end), I = 6 kgm2 30° 2

mass of rod (m) = 2 kg C.G.

In the initial horizontal position OP, ω0 = 0
(i) ω = ? when θ = 30°

Let, the C.G. at position OQ i.e. θ = 30° is lowered down through x1 C.G. Q
as shown in figure.
Then, Loss in P.E. = Gain in K.E.

or, mgx1 = 1 Iω2 + 1 mv2
2 2
R
mg2l  1 1 2l ω2
or, sinθ = 2 Iω2 + 2 m

or, mg2l  sinθ = 1 ω2 I + m4l2 . . . (1)
2

Rotational Dynamics | 253

∴ 2 × 10 × 23 sin30° = 12ω2 6 + 2 × 32
4

∴ ω = 1.69 rads–1

(ii) ω = ? when θ = 90°

We have, from equation (1)

mg 2l  sin θ = 1 ω2 I + m4l2
2

or, 2 × 10 × 23 × sin90° = 1 × ω2 × 6 + 2 × 32
2 4

∴ ω = 2.39 rad s–1
23. A recording disc rotates steadily at 45 rev. min–1 on a table. When a small mass 0.02 kg is dropped

gently on the disc at a distance of 0.04 m from its axis and sticks to the disc, the rate of revolution
falls to 36 rev.min–1. Calculate the moment of inertia of the disc about its centre.

Solution:

f1 = 45 rev min–1 = 45 rev s–1 = 3 rev s–1;
60 4

f2 = 36 rev min–1 = 36 rev s–1 = 3 rev s–1
60 5

Moment of inertia of the disc (I1) = ?

Mass dropped (m) = 0.02 kg; Distance from centre (r) = 0.04 m

New moment of inertia (I2) = I1 + mr2

Now, using angular momentum conservation principle, I2ω2 = I1ω1

or, (I1 + mr2) × 2πf2 = I1 × 2πf1

or (I1 + 0.02 × 0.042) × 3 = I1 × 3
5 4

or, I1 + 3.2 × 10–5 = I1 × 5/4

or, I1 45 – 1 = 3.2 × 10–5

or, I1 = 3.2 × 10–5 × 4
∴ I1 = 1.28 × 10–4 kgm2
24. A disc of moment of inertia 0.1 kgm2 about its centre and radius 0.2 m is released from rest on a

plane inclined at 30° to the horizontal. Calculate the angular velocity after it has rolled 2m down the

plane if its mass is 5kg.

Solution: Radius of disc (r) = 0.2 m
M.I. of disc (I) = 0.1 kgm2

ω0 = 0, θ = 30°, ω = ? distance travelled (s) = 2m Sh
mass (m) = 5 kg

Here, Loss in P.E. = Gain in K.E. θ

mgh = 1 Iω2 + 1 mv2
2 2

or, mg ×s sinθ = 1 Iω2 + 1 (rω)2
2 2m

1 Iω2 1 mr2ω2 θ
2 2
or, mg s sinθ = +

or, mg s sinθ = 1 (I + mr2) ω2
2

∴ ω= 2 mg s sinθ
(I + mr2)

= 2 × 5 × 10 × 2 × sin 30° = 18.3 rad s–1
(0.1 + 5 × 0.22)

254 | Essential Physics

25. A flywheel with an axle 1.0 cm in diameter is mounted in frictionless bearings and set in motion by

applying a steady tension of 2N to a thin thread wound tightly round the axle. The moment of inertia

of the system about its axis of rotation is 5.0 × 10–4 kgm2. Calculate (a) the angular acceleration of

the flywheel when 1m of thread has been pulled off the axle, (b) the constant retarding couple which

must then be applied to bring the flywheel to rest in one complete turn, the tension in the thread have

been completely removed.

Solution:

Diameter of axle (d) = 1 cm Radius of the axle (r) = d = 0.5 cm = 5 × 10–3 m
2

Force applied to the axle (F) = T = 2N Moment of inertia of the system (I) = 5 × 10–4 kgm2

Torque on the axle, τ = rF = rT = 5 × 10–3 × 2 = 0.01 Nm

(a) Angular accelerations (α) = ?

τ = Iα

or, 0.01 = 5 × 10–4 × α

∴ α = 20 rad s–2

(b) Constant retarding couple (τ') = ?

Let n be the no. of revolutions made by the axle when 1 m of the thread has been pulled off, then,

2πr × n = 1

1
or, n = 2πr

If θ be the angular displacement, then,

θ = 2πn = 2π × 1 = 1 = 5 1 = 200 rad
2πr r × 10–3

Now, the angular velocity of the flywheel and the axle when 1m of the thread has been pulled off is,

ω2 = ω02 + 2αθ = 02 + 2 × 20 × 200
∴ ω = 89.44 rad s–1

Now to bring the flywheel at rest in one complete turn,

θ' = 1 turn = 2π rad; ω' = 89.44 rad s–1
0

ω' = 0
Now, ω'2 = ω'02 + 2α'θ'
or, 02 = (89.44)2 + 2 × α' × 2π

∴ α' = 8000 rad s–2
– 4π

So, τ' = Iα' = 5 × 10–4 × –840π00 = –32 Nm

The negative sign shows the retarding couple

26. A horizontal disc rotating freely about a vertical axis makes 100 r.p.m. A small piece of wax of mass

10g falls vertically on to the disc and adheres to it at a distance of 9 cm from the axis. If the number

of revolutions per minute is thereby reduced to 90, calculate the moment of inertia of the disc.

Solution:

100 5
Frequency of disc (f1) = 100 rpm = 60 rev/s = 3 rev/s

Mass of wax (m) = 10g = 0.01 kg

Distance of wax from the axis (r) = 9 cm = 0.09m

90 3
New frequency of disc with wax (f2) = 90 rpm = 60 rev/s = 2 rev/s

M.I. of disc without wax (I1) = ? M.I. of disc with wax (I2) = I1 + mr2

From law of conservation of angular momentum,

I2ω2 = I1ω1
or, (I1 + mr2) × 2πf2 = I1 × 2πf1

Rotational Dynamics | 255

or, I1+ mr2 = f1
I1 f2

or, I1 + 0.01 × 0.092 = 5/3 = 10
I1 3/2 9

0.01 × 0.092 10
or, 1 + I1 = 9

0.01 × 0.092 10 1
or, I1 = 9 – 1 = 9

∴ I1 = 9 × 0.01 × 0.092 = 7.3 × 10–4 kgm2

27. The atoms in the oxygen molecule O2 may be considered to be point masses separated by a distance
of 1.2 × 10–10 m. The molecular speed of an oxygen molecule at s.t.p. is 460 ms–1. Given that the

rotational kinetic energy of the molecule is two thirds of its translational kinetic energy. Calculate its

angular velocity at s.t.p. assuming that molecular rotation takes place about an axis through the

centre and perpendicular to the line joining the atoms.

Solution:
Distance between the atoms in O2(r) = 1.2 × 10–10 m
At s.t.p, molecular speed (v) = 460 ms–1

At s.t.p, angular velocity (ω) = ?

The M.I. of O2 about an axis through its centre and perpendicular to the line joining them is

I = m2r2 + m2r 2 = 1 mr2; Where, m = mass of oxygen atom
2

2
Given, (K.E.)rot = 3 (K.E.)trans

or, 1 Iω2 = 2 × 1 (m + m) v2
2 3 2

or, 1 . 1 mr2.ω2 = 2 × 1 × 2mv2
2 2 3 2

or, mr2ω2 = 2 mv2
4 3

or, ω2 = 8 v2
3 r2

∴ ω = v 8 = 1.2 460 × 8 = 6.3 × 1012 rad s–1
r 3 × 10–10 3

28. In the design of a passenger bus, it is proposed to derive the motive power from the energy stored in a
flywheel. The flywheel, which has a moment of inertia of 4.0 × 102 kgm2, is accelerated to its
maximum rate of rotation 3.0 × 103 revolutions per minute by electric motors at stations along the

bus route, (i) calculate the maximum kinetic energy which can be stored in the flywheel. (ii) If, at an

average speed of 36 kilometers per hour, the power required by the bus is 20 KW, what will be the

maximum possible distance between stations on the level?

Solution: fmax = 3 × 103 rev/min = 3 × 103 rev/s = 50 rev/s.
I = 4 × 102 kgm2 60

(i) K.E.max = 1 Iωm2ax = 1 I (2πf max)2= 2π2 I fm2ax = 2 × 3.14 × 4 × 102 × (50)2 = 2 × 107 J
2 2

(ii) v = 36 Kmhr–1 = 36 × 1000 ms–1 = 10 ms–1
60 × 60

P = 20 KW = 20000 watts

Maximum possible distance (s) = ?

Since, K.E. max = F × S

256 | Essential Physics

or, K.E.max = P × S [since, P = F × v]
v

or, S = K.E. max × v = 2 × 107 × 10 = 104 m = 10 km
P 20000

29. A flywheel of moment of inertia 0.32 kgm2 is rotated steadily at 120 rad s–1 by a 50 W electric motor.

(i) Find the kinetic energy and angular momentum of the flywheel. (ii) Calculate the value of the

frictional couple opposing the rotation. (iii) Find the time taken for the wheel to come to rest after the

motor has been switched off.

Solution: ω = 120 rad s–1 P = 50 W
I = 0.32 kgm2

(i) K.E. = ? Angular momentum (L) = ?

We have the relation, K.E. = 1 Iω2 = 1 × 0.32 × 1202 = 2304 J
2 2

and, L = Iω = 0.32 × 120 = 38.4 kgm2s–1

(ii) Frictional couple (τ) = ?

We have the relation, P = τω

or, τ = P = 50 = 5 = 0.42 Nm
ω 120 12

(iii) Time to come to rest (t) = ?

ω' – ω0' 0 –ω – ω
α α α
We have the relation, t = = =

Also, τ = Iα or, α = τ = 0.42 = –1.31 rads–2
I –0.32

Here, τ is negative for frictional couple

∴ w –120
t = –α = –1.31 = 91.4 sec.

Practice Short Questions

1. A ballet dancer stretches her arms to reduce her motion. Explain. [HSEB 2058]

2. A dancer girl is rotating over a turn table with her arms outstretched. If she lower her arms how does this

affect her motion? [HSEB 2057]

3. What happens when a skater, spinning with arms outstretched, pulls her arms in?

4. A ring, a disc and a sphere all of the same radius and mass roll down an incline plane from the same height h.

Which of the three reach bottom first and last?

5. Flywheels are used in railway engine, why?

6. Two satellites of equal mass are orbiting the earth at different heights. Considering them as particles, do they

have equal moment of inertia?

7. How can a diver diving from a height increases the number of turns of his body in air?

8. A metal ring is melted and a solid sphere is made out of it. What happens to the M.I. about a vertical axis

through the centre?

9. A rope dancer holds an umbrella or pole to balance him, why?

10. Two metal spheres have same mass and radius. If one is hollow and the other is solid, which will have greater

M.I.?

11. If a body is rotating, is it necessarily being acted by an external torque?

12. Force A has a magnitude twice that of force B. Is it possible for B to exert a greater torque on an object than

force A?

13. Why is it harder to open and shut the door if we apply force near the hinge?

14. Why is it more difficult to revolve a stone by tieing it to a longer string than by tieing it to a shorter string?

15. What would happen if wheels of your cycle had no spokes?

16. Why is friction necessary for the disc to roll on the surface in the straight line?

17. A ballet dancer stretches her hands when she wants to come to rest. Why? [HSEB 2070]

Rotational Dynamics | 257

Practice Long Questions

1. Define moment of inertia and angular momentum. Establish a relation between them. [HSEB 2068]

2. What is meant by a couple? Derive an expression for the work done by a couple. [HSEB 2066 old]

3. Define the terms couple and moment of a couple. Derive an expression for the work done by a couple.

[HSEB 2062]

4. Explain the meaning of the term 'moment of inertia'. Show that the quantity 1 Iω2 is the kinetic energy of
2

rotation of a rigid body rotating about an axis with angular velocity ω. [HSEB 2060]

5. Define moment of inertia. How is it related with rotational kinetic energy of a body? [HSEB 2057]

6. What is meant by moment of inertia? How is it related with the rotational kinetic energy of a body?

[HSEB 2051]

7. Explain the principle of moment and moment of inertia of a rigid body. [HSEB 2062, S]

8. State and explain the principle of conservation of angular momentum with example.

9. Obtain an expression for the kinetic energy of a body rolling on a horizontal surface without slipping.

10. Show that the rate of doing work on a rotating body by a torque is equal to the product of the torque and the

angular velocity of the body.

11. g sin θ
What is radius of gyration? Show that the acceleration of a body rolling down an inclined plane is, a = 1 + K2/r2 ,

where θ is, the angle of inclination of the plane, K is the radius of gyration and r is the radius of the body.

12. Show that in rotational motion, power is the product of torque and angular velocity.

13. Define moment of inertia. Obtain an expression for the moment of inertia of a thin and uniform rod about an

axis passing through the centre and perpendicular to its length. [HSEB 2073 C]

14. Define moment of inertia. How is it related with angular momwntum of a body rotating about an axis of rotation?

[NEB 2074]

15. Explain the concept of torque and angular acceleration in the case of a rigid body. Derive a relation between them

[NEB 2074]

Practice Numerical Questions

1. A constant torque of 200 Nm turns a wheel about its centre. The moment of inertia about this axis is 100 kg

m2. Find the kinetic energy gained after 20 revolutions. [Ans: 25132.82J] [HSEB 2063]

2. The angle θ through which a bicycle wheel turns is given by θ(t) = a + bt2 – ct3; where a, b and c are positive

constants such that for t in seconds, θ will be in radians. (a) Calculate the angular acceleration of the wheel as

a function of time. (b) At what time is the angular velocity of the wheel instantaneously not changing?

[Ans: α = 2b – 6ct, t = b/3c]

3. An electric fan is turned off, and its angular velocity decreases uniformly from 500 rev/min to 200 rev/min in
4.00 s. (a) Find the angular acceleration in rev/s2 and the number of revolutions made by the motor in the

4.00s interval. (b) How many more seconds are required for the fan to come to rest if the angular acceleration

remains constant at the value calculated in part (a)? [Ans: α = –1.25 rev/s2 n = 23.3, t = 2.66 secs]

4. A wheel rotates with a constant angular velocity of 6.00 rad/s. (a) Compute the radial acceleration of a point

0.500 m from the axis, using the relation arad = ω2r, (b) Find the tangential speed of the point and compute its
V2
radial acceleration from the relation arad = r . [Ans: 18.0 ms–2, 3.00 ms–1, 18.0 ms–2]

5. The spin cycles of washing machine have two angular speeds, 423 rev/min and 640 rev/min. The internal

diameter of the drum is 0.470 m. (a) What is the ratio of the maximum radial force on the laundry for the

higher angular speed to that for the lower speed? (b) What is the ratio of the maximum tangential speed of the

laundry for the higher angular speed to that for the lower speed? [Ans: 2.29:1, 1.51:1]

6. A roller turns through an angle θ(t) given by θ(t) = γt2 – βt3, where γ = 3.20 rad/s2 and β = 0.500 rad/s3. (a)

Calculate the angular velocity of the roller as a function of time. (b) Calculate the angular acceleration of the

roller as a function of time. (c) What is the maximum positive angular velocity, and at what value of t does it

occur? [Ans: 6.40t – 1.50t2; 6.40 – 3.00t, 6.83 rad s–1, 2.13 sec]

7. A 1.50 kg grinding wheel is in the form of a solid cylinder of radius 0.100m. (a) What constant torque will

bring it from rest to an angular speed of 1200 rev/min in 2.5 s? (b) Through what angle has it turned during

that time? (c) Calculate the work done by the torque. (d) What is the grinding wheel's kinetic energy when it

is rotating at 1200 rev/min? [Ans: 0.38 Nm, 160 rad, 61 J, 60 J]

258 | Essential Physics

8. A solid disk is rolling without slipping on a level surface at a constant speed of 2.00 m/s. How far can it roll

up a 30.0° ramp before it stops? [Ans: 0.612 m]

9. Find the magnitude of the angular momentum of the second hand on a clock about an axis through the centre

of the clock face. The clock hand has a length of 15 cm and a mass of 6g. Take the second hand to be a

slender rod rotating with constant angular velocity about one end. [Ans: 4.71 × 10–6 kg m2 s–1]

10. Energy of 484 J is spent in increasing the speed of a flywheel from 60 r.p.m. to 240 r.p.m. Find the moment

of inertia of the wheel. [Ans: 1.63 kgm2]

MBBS, B.E., B.Sc. Nursing Entrance Preparation Questions

1. A rotating disc has __________ kinetic energy if mass is M and velocity is V MV2
d. 4
a. MV2 b. 1 MV2 c. 3 MV2
2 4

2. When the size of earth is reduced to half, mass remaining the same, the time period of earth rotation will be :

a. 6 hours b. 24 hours c. 12 hours d. 48 hours

3. The body applied with constant torque changes the angular momentum I to final angular momentum 4I in 3
°°

sec, the find torque

a. 3 I b. I c. 4I d. 2I
° ° ° °

4. If the ratio of the earth 's orbit is made one fourth, the duration of year will become

1 1 1 1
a. 2 times b. 4 times c. 8 times c. 16 times

5. If there is a change of angular momentum from 2J to 4J in 4 sec. The the torque is

a. 0.25J b. 0.5J c. 1J c. 2J

6. A shell at rest explodes. The centre of mass of the fragment

a. moves along the parabolic path b. moves along the straight line

c. moves along an elliptical path d. remains at rest
7. A wheel of mass 10kg has a moment of inertia 160kgm2 about its own axis. The radius of gyration is

a. 10m b. 4m c. 5m c. 6m

8. The radius of gyration of a solid disc of mass 1 kg mass and radius 50 cm about an axis through centre of mass

and perpendicular to its face is

a. 25 cm b. 25 2 cm c. 20cm d. 25 6 cm

9. In rotational motion, the physical quantity that imparts angular acceleration is

a. Force b. Torque c. moment of inertia d. angular momentum

10. A sphere of mass 0.5 kg and diameter 1m rolls without sliding with a constant velocity 5 m/s. Calculate the

fraction of total energy with rotational

1 2 5 7
a. 2 b. 7 c. 7 d. 10

11. A small solid sphere rolls without slipping down a 30° inclined plane . It's acceleration is

25 7 16 d. 5
a. 7 b. 25 c. 9

12. the moment of inertia of a uniform circular disc about diameter is I the M.I. about an axis passing through a

point on its rim and perpendicular to plane will be

a. 3I b. 4I c. 5I d. 6I

13. Two solid spheres are made of same material and having the radii in the ratio 1:2. The ratio of moment of

inertia is

a. 1:2 b. 1:4 c. 1:16 d. 1:32

14. Two particles of masses m1 and m2 are at distance x. Then, the center of mass lies at distance from m1:

( )a. m1 m1 m2 m2
m2 m1 + m2 m1 m1 + m2
x ( )b. x ( )c. x ( )d. x

15. The centre of mass of thin triangular plate lies on

a. the centre of its base b. its centroid c. mid – point d. one of its vertices

16. An inclined plane makes an angle 30° with horizontal. A solid sphere rolling down this inclined plane from rest

without slipping has a linear acceleration equal to

g 2g 5 5
a. 3 b. 3 c. 7 g d. 14 g

Answer Key

1. d 2. a 3. b 4. d 5. b 6. d 7. b 8. b 9. b

10. b 11. a 12. b 13. b 14. d 15. b 16. d 17. d 18. a

Ch apter

10 Elasticity

Teaching Manual Physics Grade – XI, Higher Secondary Education Board
Curriculum and Training Division Sanothimi, Bhaktapur

Syllabus:

Elasticity–Hooke's law: Force constant, Verification of Hooke's law; Stress; Strain; Elasticity and
Plasticity; Elastic modulus: Young modulus and its determination, Bulk modulus, Shear modulus,
Poisson's ratio. Elastic potential energy

Objectives:

The objectives of this sub unit is to make the students able to understand the elastic nature of
matter.

Activities (Micro syllabus) :

1. Explain the elastic nature of matter. Define elasticity, plasticity and rigidity of a substance.
Give examples.

2. Explain the meaning of stress and strain. State their units and dimensions.
3. Discuss various types of stress and strain.
4. Draw the stress – strain curve for a body and explain various region on the curve.
5. State and explain Hooke's law. Demonstrate an experimental method to verify Hooke's law.
6. Discuss different types of moduli of elasticity. Explain how the Young's modulus of

elasticity is determined in laboratory.
7. Derive an expression for the energy stored in stretched wire. Define the term energy density

of a stretched wire.
8. Discuss Poisson's ratio.
9. Explain the terms elastic hysteresis and elastic fatigue of a substance.
10. Numerical problems: Focused numerical problems given

(a) As exercise in the University Physics (Ref.1) and
(b) In Advanced level Physics (Ref. 2).

260 | Essential Physics

10.1 Introduction

All elements and compounds generally exists in any one of the three states – solid, liquid and gas. A
matter is a substance which has a certain mass and occupies some volume. A matter is made of molecules
and atoms. The molecules and atoms in a matter are held each other with a force. The force among the
molecules or atoms of a matter is called intermolecular or interatomic force. The interatomic forces are
responsible for the structure of a molecule and the intermolecular forces are responsible for the structure
of a matter.

10.2 Intermolecular Force

The intermolecular force is responsible to F(r)
hold together the atoms or molecules of a A

matter. The intermolecular force is a function r < r0
of intermolecular distance between the two (F is repulsive)

molecules, F(r). Figure 10.1 (A) shows a

graph between F(r) and intermolecular

distance, r. Suppose, a molecule of the matter Equilibrium position (r = r0)
is at point O and another molecule at M, at a
distance r. If the molecule at M is brought B E M r
towards the molecule at O, then it is found O D
that,
r0
a. When two molecules are far from each
r > r0
(F is attractive)

other, i.e. OM, the force between them C
is negligibly small, negative and so, (A)
attractive.
U(r)

b. When these molecules are brought P

closer, the attractive force increases

and becomes maximum at a particular

distance, OE. When the molecules are

brought closer than OE, the attractive

force starts decreasing and becomes r0 Equilibrium position (r = r0) r
OQ S
zero at a distance OB = r0. This is
called equilibrium position and the

distance, r0, is called equilibrium
distance at which the repulsive force

is completely balanced by the

attractive force. R Umin at r = r0
c. When the molecules are brought
(B)
closer than r0, then the intermolecular

force becomes positive and so, [Fig. 10.1. Variation of (A) intermolecular force and (B) Potential energy.]

repulsive. The repulsive force increases with the decrease in the interatomic distance as shown by

part BA of the graph.

The potential energy of a molecule is electrical in origin and is a function of intermolecular distance r, i.e.

U(r). The intermolecular force, F is equal to the negative gradient of the corresponding potential energy

function, i.e,

F = – dU
dr

Elasticity | 261

Figure 10.1 (B) shows a graph between U(r) and the intermolecular distance, r. Along part PQ of the
graph, the P.E. is positive and along part QRS, it is negative. At point R, the potential energy is minimum,
Umin, and the molecule is at equilibrium position (i.e., r = r0), When an external force (deforming force) is
applied to elongate a body, then the molecule are pulled apart and the distance between them becomes
greater than equilibrium distance, r0. Then an attractive force (restoring force) starts acting among the
molecules which brings the molecules back to their normal position when the deforming force is
removed. On the other hand, if the body is compressed, then the molecules come close to each other and
the distance between them becomes less than equilibrium distance, r0. Then a repulsive force (restoring
force) is developed among the molecules which brings the molecules back to their equilibrium position
after the removal of the deforming force.

10.3 Elastic Behaviour of Solids

Each atom or molecule in a solid is surrounded by neighbouring atoms or molecules. They are bounded to

each other by interatomic or intermolecular forces and stay in state of equilibrium position. When the

solid is deformed, the atoms (or molecules) are

displaced from their equilibrium position and their

interatomic (or intermolecular) distance changes.

When the deforming force is removed, the

interactomic or intermolecular forces bring them

back to their original equilibrium positions. Thus,

the body regains its original shape and size. This

restoring mechanism in a solid can be visualized by

taking a model of spring – ball system as shown in

figure 10.2. Here, the spherical balls represent atoms

and the springs represent the interatomic forces. If [Fig. 10.2. Spring-Ball model for the illustration of elastic

we try to displace any ball from its equilibrium behaviour of solids]

position, then the spring system tries to restore the

ball back to its original equilibrium position.

Rigid Body :

A solid body which is not deformed under the action of various forces, is called a rigid body. An ideal

rigid body does not change its shape or size under the action of forces of any magnitude. In a rigid body,

the distance between its molecules remains the same under the action of the applied force. In fact, there is

no body, which is perfectly rigid. The diamond is the nearest approach to a rigid body.

Deforming force :

Sometimes, a force acting an a body produces a change in shape and size of the body, instead of

producing a change in its state of rest or of uniform motion. Such a force is called deforming force, and

the body is said to be deformed.

Elasticity :

The property of a matter by virtue of which it regains its original shape and size, when the deforming

forces have been removed, is called elasticity.

In physics, elasticity stands for opposition to change, contrary to the concept of elasticity in daily life.

Qualitatively, a body is said to be more elastic if it is more rigid. For this reason, steel is more elastic than

rubber.

Restoring force:

When the deforming force is removed, the body (elastic) regains its original configuration i.e. length,

volume and shape. This is so because the molecules of the elastic body try to regain their original

positions. The force developed within the body because of relative molecular displacement is called

internal force or elastic force or restoring force.

262 | Essential Physics

Elastic Bodies :
The bodies which possess the property of elasticity are called elastic bodies. Those bodies which recover
their original configuration immediately and completely after the removal of deforming forces are called
perfectly elastic bodies. Perfectly elastic body is an ideal concept and no material behave as perfectly
elastic. A quartz fibre is the nearest approach to a perfectly elastic body.

Plastic Bodies :
The bodies which do not show any tendency to recover their original configuration after the removal of
deforming forces, are called plastic bodies. Putty, Paraffin wax etc. are familiar examples of plastic
bodies.

Plasticity :
The property of a body (plastic) by virtue of which it remains deformed even after the removal of
deforming forces is called plasticity.

10.4 Stress

When a body is deformed, then the restoring forces are developed uniformly inside the body. The

restoring force per unit area set up in the body, when deformed by the external force, is called stress.

Thus, Stress = Restoring Force

Area

Since, the restoring force set up in the body is equal and opposite to the deforming force, therefore, the

stress may be measured as the deforming force acting per unit area i.e.,

Deforming Force F
Stress = Area = A

Its S.I unit is Nm–2 and C.G.S. unit is Dyne cm–2. The dimensional formula of stress is [ML–1 T–2]. It is a

scalar quantity. The stress and pressure have same units and dimensioned formula.

Types of Stress

Depending up on the ways of application of external forces on a body, the stress are of three types.

L ∆L L
F A
F F

A ∆L
(a) Tensile stress (b) Compression stress

[Fig. 10.3, Normal stress]

i. Normal stress : The deforming force acting per unit area normal to the surface of a body is called

F normal stress. The normal stress is developed when there is a change in length of a body. The

normal stress is of two types: tensile stress and compression stress.
a. Tensile stress : The normal stress is said to be tensile if there is increase in the length or

extension of the body along the direction of applied force as shown in figure 10.3 (a).
b. Compression stress : The normal stress is said to be compressive or compression stress if

there is decrease in length or compression of the body due to the applied force as shown in
figure 10.3 (b).

Elasticity | 263

i. Tangential stress or shearing stress : The tangential force acting x Top Face
per unit area tangential (parallel) to the surface of a body is called F
tangential stress or shearing stress. The tangential stress produces
a change in shape of the body.

Consider a cubical body having its bottom face rigidly fixed as Lθ
shown in figure 10.4. Let F be a tangential force applied on the [Fig. 10.4, Tangential stress]
top face of the body such that the shape of the body changes
without any change in volume. If A be the area of the top face,
then

F F
tangential stress = A

iii. Volume stress or Bulk stress or Hydraulic stress :Consider a V

body immersed in a fluid (liquid or gas). The fluid applies a force

in perpendicular direction at each point of the surface as shown in

figure 10.5. The body is compressed uniformly from all sides, V' F
which leads to decrease in its volume without any change of F

geometrical shape. The body is said to be under volume stress or

bulk stress or hydraulic stress. Thus, the force acting per unit

area perpendicular to the surface of the body is called bulk stress.

So, F

Force F [Fig. 10.5, Bulk Stress]
Bulk stress = Area = A .

10.5 Strain

If a deforming force acts on a body, the body undergoes change in its dimensions and the body is said to
be deformed or strained. The ratio of change in dimension of a body to its original dimension is called
strain, i.e.

change in dimension
Strain = original dimension

Since, the strain is the ratio of two similar quantities it has no unit and dimensions.

Types of Strain

Depending upon the types of deformations i.e., length or volume or shape, the strain are of three types:

i. Longitudinal strain : If the applied force produces a change in length of the body only, then the
strain is called longitudinal strain. It is defined as the ratio of change in length to the original
length of a body produced under the action of the deforming force. Therefore, from figure 10.3 (a),

Longitudinal Strain = change in length = ∆L
original length L

ii. Volumetric strain :It is defined as the ratio of change in volume to the original volume of a body
produced under the action of the deforming force. Therefore, from figure 10.5., we have

Volumetric strain = change in volume = ∆V
original volume V

iii. Shear strain : If the deforming force produces a change in the shape only but not in size, then the
strain produced is called shear strain. The shear strain is produced if the body is under tangential or
shearing stress. It is defined as the angle, θ through which the face of the body originally normal to
the fixed face is turned under the action of shearing stress as shown in figure 10.4, i.e.,

264 | Essential Physics

Tanθ = x
L

Since θ is small,

∴Tanθ ≈ θ.

So, θ= x
L

Here, θ is also called angle of shear.

10.6 Elastic Limit

The upper limit of deforming force upto which the body regains its original form completely after the
removal of the deforming force, is called elastic limit of the body. When the deforming force is increased
beyond the elastic limit, the body loses its property of elasticity and gets permanently deformed.

10.7 Hooke's Law and Force Constant

This is the fundamental law of elasticity which was given by an English experimental physicist Robert
Hooke in 1676. He studied experimentally the relation between extension produced in a wire and the
tension (applied force) in the wire, and summed up his findings in a rule known as Hooke's law.

The Hooke's law states that, "Within elastic limit, the extension produced on an elastic body is
directly proportional to the force producing it". If F be the applied force and x be the extension
produced then,

F∝x

or, F = Kx . . . (1)

Where, K is proportionality constant called force constant or spring constant.

From equation (1),

F . . . (2)
K= x

So, the force constant (or spring constant) of a wire (or a spring) is defined as the applied force per unit
extension. S.I. unit of K is Nm–1.

If F be the restoring force (intermolecular force) developed on the deformed body, then equation (1)
becomes

F = – Kx . . . (3)

The negative sign indicates that the restoring force is developed opposite to the extension produced.

In 1807, Thomas Young modified the law to general form which states that, "Within elastic limit, the
stress is directly proportional to strain". That is, Stress ∝ Strain. (Within elastic limit).

or, Stress = (constant) Strain

Stress . . . (4)
or, Strain = constant (E)

This constant (E) is known as modulus of elasticity or coefficient of elasticity. The value of E depends

on the nature of material and the nature of deformation. Since, strain is unitless and so the unit of E is
same as that of stress i.e. Nm–2 in S.I. system and dyne cm–2 in CGS system. The dimensions of E is,

[ML–1 T–2], same as that of stress.

Elasticity | 265

10.8 Experimental Verification of Hooke's Law

Support

0
10
20
30

Table O
Load applied

(a) (b)

[Fig. 10.6, Experimental verification of Hooke's law]

To verify Hooke's law, a spring is suspended from a rigid support as shown in figure 10.6 (a). A scale pan
and a light pointer P are attached to its lower end. The pointer P can slide over a vertical scale. Note the
reading of the pointer P on the vertical scale when the scale pan is empty. Some weight is added on pan
and then the reading of the pointer P on the scale is noted again. The extension produced in the spring is
given by the difference between these two readings. Now, we go on adding the weights on the scale pan,
step by step, and the corresponding extensions of the spring are noted. When a graph is plotted between
these weights on the pan (load applied) and the extensions produced in the spring, then a straight line
graph is obtained as known in figure 10.6 (b). This verifies that, extension ∝ load applied. This verifies
the Hooke's law.

10.9 Types of Moduli of Elasticity

Corresponding to three different types of strain, there are three different types of modulus of elasticity as

discussed below.

1. Young's Modulus of Elasticity (Y)
The Young's Modulus of elasticity of a material is defined as the ratio of normal stress to the

longitudinal strain, within the elastic limit.

∴ Normal stress
Y = Longitudinal strain

If L, A and L be the length, area of cross– section and extension produced on a wire [Figure

10.3] by a force F then

F/A FL . . . (1)
Y = ∆L = A∆L

–L

Here, the applied force F is normal to the cross–section area A of the wire as shown in figure 10.3.

2. Bulk Modulus of Elasticity (K)

The bulk modulus of elasticity of a body is defined as the ratio of normal stress to volumetric

strain, within the elastic limit.

∴ Normal stress
K = Volumetric strain

266 | Essential Physics

If we apply a force F normally and uniformly over the entire surface area A of a sphere of volume

V such that its volume decreases by ∆V as shown in figure 10.5, then

F ∆V
Normal stress = A and Volumetric strain = – V .

The negative sign shows that the volume decreases with the increase in pressure.

∴ F/A P PV
K = ∆V = – ∆V = – ∆V
–V V

Where, P = F/A is the change in pressure.

Compressibility: The reciprocal of bulk modulus (K) of a material is called its compressibility.
1

Therefore, Compressibility = K
The SI unit of compressibility is N–1m2 or Pa–1.

3. Shear Modulus or Modulus of Rigidity (η)

The shear modulus of a body is defined as the ratio of tangential stress to shear strain, within the

elastic limit.

∴ η = Tangential stress
Shear strain

If F be the tangential force acting parallel to the upper free face of a cubical body having surface
area A and having lower face fixed, as shown in figure 10.4, then

F
Tangential stress = A

If the vertical face of the cubical body shifts through an angle θ, then, Shear strain = θ

Therefore, η = F/A = F
θ Aθ

If the top face or upper face is slided a distance x from its original position as in figure 10.4, then,

Tanθ = x
L

Since, θ is small. ∴ Tanθ ≈ θ = x Support
L

So, η= F FL
Ax
A xL =

The shear modulus, η, has significance only for solid bodies.

10.10 Determination of Young's Modulus of AB
Elasticity

Figure 10.7 shows Vernier apparatus to determine Young's Modulus Y in 0 0
laboratory. Wire A is a reference wire which is suspended from a fixed
support which carries a fixed graduated scale M. At the lower end of A, a M V
fixed load W is kept which keeps the wire straight and free from kinks. Wire
B is the experimental wire which is also suspended from the same support 10 10
close to the reference wire A. A Vernier scale V is attached at the free end of
the experimental wire B which can slide against the main scale M attached to H
the reference wire A. A hanger H is attached at the lower end of the Vernier
[Fig. 10.7, Vernier Apparatus]
1
scale V at which a number of 1 kg or 2 kg weights can be slipped.

Elasticity | 267

At first, the initial length of wire B is noted and its diameter is measured through a micrometer screw
gauge. Let L be its initial length and d be its diameter.At the beginning, the reading of the main scale M

1
and the Verner scale V are noted keeping some initial load on the hanger H. A known weight, say 2 kg

or 1 kg is slipped in to the hanger H so that the elongation takes place on the wire B. Now, the readings
on the scales are noted. The difference between these two readings gives the extension produced due to
the extra weight kept on H. The weight on H is gradually increased and the extension produced is noted

every time.

From these data, extension versus load curve is plotted which should be a straight line passing through the
origin as shown in figure 10.8.

Now, Slope = Tanθ = ∆F
∆L

Since, Stress = ∆F
A

∆L
and, strain = L

Therefore, Young's modulus, F
Stress ∆F/A

Y = Strain = ∆L

L

Y = ∆F L ∆F
∆L ×A

L θ
Y = Slope × A . . . (1) ∆L

πd2 O ∆L
Where, A = 4 , is the area of cross–section of the wire
Extension

∆F = Load = Mg. [Fig. 10.8, Graph of force verses extension]

Here, all the quantities on R.H.S of equation (1) are known and hence Y can be calculated.

10.11 Energy Stored in a Stretched Wire Support

When a wire is stretched, the internal restoring forces appear and L
the work has to be done against these forces. The work done on
stretching the wire against the internal restoring forces is stored in Al
the wire in the form of elastic potential energy. F
Consider a wire having initial length L and cross–sectional area A,
is fixed on a rigid support at one end and is stretched by an [Fig. 10.9, Energy Stored in a wire due to strain]
external force F applied at the other free end as shown in figure
10.9. If the extension is x, then

x
Longitudinal strain = L ,

F
and, Normal stress = A
So, Young's modulus Y is

268 | Essential Physics

Normal stress F/A FL
Y = Longitudinal strain = x = Ax

L

YAx . . . (1)
or, F = L

The small work done dW by the external force F in a small extension dx is

dW = Fdx

YAx . . . (2)
Using (1), dW = L dx

The total work done W by the external force F in an extension from x = 0 to x = l is,

l YAx YA l
∫ ∫ ∫l YA x2 l
W = dW = L dx = L 0 x dx = L 2 0
00

∴ 1 YAl2 . . . (3)
W= 2 L

This work done is stored in the wire in the form of its elastic potential energy (U)

∴ 1 YAl2 . . . (4)
U= 2 L

1 YLAl l
Equation (4) can be written as, U = 2

1 . . . (5)
U=2 Fl

1
Therefore, elastic potential Energy = 2 × stretching force × extension

Energy Density ( ρe): Support
The elastic potential energy per unit volume of the wire is called its energy D
density.

Elastic P.E. U
So, Energy Density = Volume = V

1
2 Fl 1
or, ρe = A.L = 2 AF ×  Ll  ∆D

∴ ρe = 1 stress × strain . . . (6) L
2

10.12 Poisson's Ration (σ)

When a wire is stretched by a deforming force,, then it undergoes change in length in ∆L
the direction of the force and its diameter also changes in the direction perpendicular F
to the force. The change in dimension per unit original dimension of the wire in the

direction of force is called longitudinal strain (α) and the change in dimension per

unit original dimension in the direction perpendicular to the force is called lateral

strain (β). Experimentally, it has been found that the lateral strain (β) is directly
proportioned to longitudinal strain (α), within the elastic limit. That is,

Lateral strain ∝ longitudinal strain.

or, β ∝ α

[Fig. 10.10, Poision's ratio]

Elasticity | 269

or, β = σα

∴ σ = β . . . (1)
α

Where σ is the proportionality constant called Poisson's ratio. So, the Poisson's ratio (σ) is defined as the
ratio of lateral strain to the longitudinal strain within the elastic limit.

From figure 10.10, we have

Longitudinal strain, α = change in length ∆L
original length =L

and Lateral strain, β = change in diameter = ∆D
original diameter D

Therefore,

Poisson's ratio σ = β ∆D/D
α= ∆L/L

or, σ= L × ∆D . . . (2)
D ∆L

σ is a unitless and dimensionless quantity.

10.13 Relation Between Stress and Strain (Stress – Strain
Curve)

Plastic deformation d S a = Proportional limit
bc b = elastic limit (upper yield point)
c = lower yield point
a d = breaking stress
s = fracture point
ob = elastic deformation
bc = elastic and plastic deformation
bd = plastic deformation
oo' = permanent set

Permanent set

O O' Strain

[Fig. 10.11, Stress-strain curve]

Figure 10.11 shows a typical stress – strain curve for a metal wire such as copper or soft iron. From point
o to a, the curve is a straight line, the stress is directly proportional to strain (Hooke's law). The slope of
oa is equal to Young's modulus, Y. The stress at point a is called proportional limit. From point a to b,
the stress is no longer proportional to strain and Hooke's law is not obeyed. When the stress is increased
beyond a up to b, a large strain is produced but the wire returns to its original length when the load is
removed. In region ob, the material shows elastic behaviour and the end point b is called the upper
yield point. The stress at the upper yield point b is called the elastic limit.

When the stress is increased beyond b up to c, the wire doesn't regain its original length on the removal of
stress and it follows the dotted line in the figure 10.11. The length at zero stress is now greater than the
original length; the wire has undergone an irreversible deformation called a permanent set (oo'). Along
bc, the wire shows both elastic and plastic behaviour. The point c is called lower yield point. Further
increase of load beyond c produces a large strain for a small increase in stress up to point d. The stress at
point d is called breaking stress. The behaviour of the wire from b to d is called plastic deformation.

270 | Essential Physics

Beyond d, even a stress less than that at c, may continue to stretch the wire until it breaks. At point s, the
fracture takes place and point s is called fracture point.
The materials such described above, which elongate considerably and undergo plastic deformation until
they break are called ductile (lead, copper, silver, gold, soft iron etc). But, if the fracture occurs soon after
the elastic limit is crossed are called brittle substances (steel, piano string, glass etc).

10.14 Elastic After Effect

Elastic bodies regain their original configuration after the removal of external deforming forces. Some
elastic bodies recover their original configuration immediately and other few take some time to recover
their original configuration. This temporary delay in regaining the original configuration by an
elastic body after the removal of external deforming force is called elastic after effect. The elastic
after effect is very small for quartz and phosphor bronze but has very large value for glass fibres. Due to
this reason, the suspensions made from quartz or phosphor bronze are used in making coil galvanometers.

10.15 Elastic Fatigue

Elastic fatigue is the property of an elastic body by virtue of which its behaviour becomes less elastic
under the repeated application of the deforming forces. An elastic body gets tired under the application of
alternate cycles of stress and strain continuously for a long time. Because of the tiredness, the elastic
behaviour of the body changes and it will have much more elongation even for a small stress. Therefore,
the bridges are declared unsafe after a long use.

10.16 Elastic Hysteresis

Figure 10.12 shows a stress–strain curve for increasing and Increasing stress
decreasing stress for rubber. The stress is not proportional to strain
but the behaviour is elastic because the material returns to its initial Decreasing stress
length on the removal of the load. The material follows different
curves for increasing and decreasing stress. This phenomenon O Strain
is called elastic hysteresis. It is a consequence of elastic after [Fig. 10.12, Elastic hysteresis]
effect. The strain persists even after the stress is removed. The
strain lags behind the stress. The work done by the material while it
returns back to its initial shape is less than that required to perform
it. This difference is stored inside the material during the process of
expansion and contraction, which is equal to the area in the
hysteresis curves Thus, the material such as rubber, which has large
elastic hysteresis, is very useful as vibration absorbers.

Boost for Objectives

• A body is rigid if the external forces applied on it do not produce any deformation. Diamond is most rigid.
• The property of restoring (recovering) the original shape and size of deformed body after removal of

deforming force is called elasticity.

• A body is said to be perfectly elastic when it regains its original shape and size completely after removal of
deforming force.

• A body is said to be perfectly plastic when it does not regain its original shape and size at all.
• Quartz, fibre, phosphor, bronze are best elastic while mud, wax, plastic are best plastic.
• Y for perfectly plastic body is zero.
• Y is defined for solids only.
• B is defined for all solid, liquid and gaseous states.
• Bulk modulus measures compressibility while Young's modulus measures extensibility.
• Bsolids > Bliquids > Bgases
• Shear modulus (η) is defined for solids only.

Elasticity | 271

• η measures rigidity.

• Modulus of elasticity (Y, B and η) are independent of dimensions of body, strain and stress but depend on the

nature of material.

• For Solids, Y = 3η

• Elasticity is due to intermolecular cohesive force.

• For a wire under tension by two equal and opposite forces, each of magnitude F, the stress in F/A (NOT

2F/A)

• Hammering and rolling increases but annealing decreases elasticity.

• Elastic coefficients are Tensors (NOT Scalars)

• The properties of matter like elasticity, viscosity and surface tension arise due to intermolecular force

(cohesive forces).

• Stress and pressure have same units and dimensions, but the pressure is always normal to the surface of a

body whereas stress may be parallel or normal to the surface of the body.

• When a beam is bent, strain produced is longitudinal (both compression and extension strain).

• The area under the stress strain graph represents the work done per unit volume.

• Steel is more elastic than copper, hence used to make springs.[Hint: Y ∝ 1
extension]

• Glass is more elastic than rubber.

• If the temperature of a gas, enclosed in a vessel of rigid material is increased by ∆θ, then change in pressure is

∆P = Kγ∆θ.

• Interatomic and intermolecular force vary with distance as F ∝ 1 .
r6

• Molecular forces are electrical in origin.

Short Questions with Answers

1. Explain in terms of breaking stress, why elephant has thicker legs as compared to human beings?
Ans:
[HSEB Model, 2069 B, 2066]
2.
Ans: Breaking force (load)
We have, Breaking stress = Cross–section area
3.
Ans: The maximum stress that can be applied to a material without breaking (or fracture) is called breaking

stress, which is inversely proportional to the cross – section area. If the area increases, then the stress

decreases. So, the elephant has thicker legs as compared to human beings to keep the stress on its leg

below the breaking stress.

Compare the mechanical properties of a steel cable, made by twisting many thin wires together,

with those of solid steel rod of the same diameter. [HSEB 2069 B]

A steel cable, made by twisting many thin wires together, has more shear strength than a solid steel rod

of the same diameter. So, the mechanical strength increases in twisting thin wires. Moreover, a steel

cable is more flexible, more stronger and has higher resistance to metal fatigue than the solid steel rod

of same diameter.

Two wires A and B have equal lengths and are made of same material. If the diameter of wire A is

twice that of wire B. Which wire has the greater extension for a given load? [HSEB 2064]

F/A Fl Fl
We have, Y = e/l = eA = e(πd2/4)

4Fl . . . (1)
or, e = πd2Y

Here, l and Y are same for both wires A and B. The extension in wires A and B are,
dA2
4Fl 4Fl ∴ eB = dB2
eA = πdA2Y and eB = πdB2Y eA

eB = (2dB)2 [ since, dA = 2dB]
eA dB2

eB = 4 ⇒ eB = 4eA.
eA

So, the extension produced on wire B is four times that of the wire A for a given load.

272 | Essential Physics

4. Why are rubbers used as vibration absorber? [HSEB 2063, 2059]
Ans:
Elastic hysteresis in rubber is large so that work done by a deforming force in deformation is more than
5.
Ans: the work done by the rubber in regaining its original configuration. The difference between these two

6. energies is absorbed by the rubber and appears as heat. When a rubber is placed between a vibrating

Ans: body and floor, the rubber is compressed and mechanical energy is released in each cycle of vibration.

7. The K.E. of vibration is converted in the form of heat and, the rubber and the floor are heated.
Ans:
8. Differentiate between elasticity and plasticity. [HSEB 2062]
Ans:
9. The property of a body by virtue of which it regains its original configuration on the removal of the
Ans:
10. deforming force is called elasticity. The materials like steel, copper, iron possess elasticity.
Ans:
11. The plasticity is the property of a body by virtue of which a deformed body doesn't regain its original
Ans:
configuration on the removal of the deforming force. Paraffin wax is an example of a plastic body.
12.
Ans: Nobody is perfectly elastic or plastic.

Explain why steel is said to be more elastic than rubber. [HSEB 2061]

OR

Explain which one is more elastic rubber or steel? [HSEB 2058]

stress F/A FL
We have, Y = strain = e/L = Ae .

If steel and rubber wires, having equal lengths (L) and same cross–sectional area (A) are stretched by

applying same stretching force (F), then extension produced in rubber (eR) is more than in steel (eS).

Since, Y ∝ 1 ⇒ Ys ∝ 1 and YR ∝ 1
e eS eR

∴ YS = eR > 1 ⇒ YS > YR
YR eS

That is, Y for steel is more than that for rubber. Hence, steel is more elastic than rubber.

Why are bridges declared unsafe after long use? [HSEB 2057, S 2071]

When bridges are used for a long time, due to alternate cycles of stress and strain, they get fatigued

(tired). In the fatigued condition, the strain produced for a given stress in the bridges will be large which

may result in the destruction of the bridges. Hence the bridges are declared unsafe after long use.

Why is concrete with steel reinforcing rods embedded in it stronger than plain concrete?

The steel is more elastic than the materials of a plain concrete. When steel reinforcing rods are

embedded in a plain concrete, it becomes more elastic and its rigidity also increases. Therefore, the

concrete with steel reinforcing rods are stronger than the plain concrete.

Water is more elastic than air, why?

Volume elasticity is the reciprocal of compressibility. We know, air is more compressible than water.

For same stress, volume strain is more in air than in water. So, water is more elastic than air.

Which of the three states: solid, liquid or gas; has the least volume elasticity?

1
We know, volume elasticity = Compressibility . For same stress, compressibility is maximum for a gas

than solid or liquid. So, the gas has the least value of volume elasticity.

How will you justify that stone is more rigid than iron? [HSEB 2065]

A substance is said to be more rigid if it has higher value of the modulus of elasticity (shearing

modulus). Since, Shear modulus (η) = Shear stress . The shear strain in stone is less than that in iron for
Shear stain

same stress i.e. stone offers more resistance to the deforming force than iron. Therefore, the shear

modulus (η) for stone is higher than that for iron and hence stone is more rigid than iron.

Does the temperature affect elasticity? [HSEB 2063 S]

Yes. When temperature increases then the elasticity of a material decreases. For example, Young's

modulus (Y) of a material is given by

Stress F/A F L F L F
Y = Strain = l/L = A . l = A . Lα∆θ = Aα∆θ

For constant AF and α, Y∝ 1 . This shows that Y decreases for increase in temperature (∆θ).
∆θ

Elasticity | 273

13. How is Young's modulus related with linear expansivity of a metallic wire?

Ans: We have, Y = AF 1 . For constant stress AF and ∆θ, Y ∝ 1
.α∆θ α

Hence, the Young's modulus (Y) is inversely proportional to the linear expansively (α) of a material
wire.

14. Why do spring balances show wrong readings after they have been used for a long time?

Ans : This is due to elastic fatigue. When a spring balance has been used for a long time , then the spring
develops an elastic fatigue. The spring of such a balance takes comparatively longer time to regain its
original configuration. As a result, the spring balance shows wrong readings.

15. Why are springs made of steel and not of copper?

Ans: The Young's modulus of steel is more than that of copper. So, for a given deforming force, steel spring
is stretched or compressed lesser than copper spring and regains the original configuration quickly on
the removal of the deforming force. Due to this reason. Springs are made of steel and not of copper.

16. Why does a wire get heated when it is bent back and forth?

Ans: When a wire is bent back and forth, the deformation on it becomes beyond the elastic limit. The work
done against the interatomic forces during deformations is no longer stored totally in the form of
potential energy. The deformation affects the crystalline structure of the wire and work done is
converted in the form of heat energy.

17. What is the value of the modulus of perfectly rigid body?

Volume stress
Ans : Infinity. Since, the modulus of elasticity is, Modulus of elasticity = Volume strain . For a perfectly rigid

body there is no change in shape or size by applying an external force. So, its volume strain is zero and
hence its modulus of elasticity is infinity.

18. Stress and pressure are both forces per unit area. Then in what respect does stress differ from
pressure?

Ans: Pressure is a special case of stress. The stress is the internal force per unit area which may or may not be
normal to the surface. Stress can change both shape and size of a body. But, pressure is the external
force per unit area which always acts normal to the surface of a body. It changes size but doesn't change
shape of a body.

19. Is it possible to double the length of a metallic wire by applying a force over it?
Ans: No. It is not possible. We know, the strain is only of the order of 10–3 within the elastic limit. A metallic

wire actually breaks much before it is stretched to double of its length.

20. What is elastic limit and breaking stresss? [HSEB 2072 D]

Ans: The upper limit of deforming force upto which the body regains its original form completely after the
removal of the deforming force, is called elastic limit of the body. When the deforming force is
increased beyond the elastic limit, the body loses its properly of elasticity and gets permanently
deformed.

Again, Breaking stress is the maximum stress that can be applied to a material without breaking (or
fracture). If the deformning force is increasesd beyond the yield point of a material, then the material
suddenly breaks. The stress that appears in the material at the yield point (breaking point) is known as
breaking stress.

21. Explain the physical meaning of Poisson's ratio. [HSEB 2072 E]

Ans: We have, Poisson's ratio, σ = lateral strain(β) .
longitudinal strain (α)

Poisson's ratio (σ) gives the decrease in lateral size per unit increase in length of a wire within elastic limit.

274 | Essential Physics

22. What happens in the Young's modulus of elasticity of a material when the load hanging on it is
doubled?

OR

Will the Young's modulus of elasticity change if the load hanging on it is doubled? Why?

[NEB 2074]

Ans: The Young's modulus of a material is a constant quantity and it is an independent with the stress applied
and dimensions of the material of the wire. We have, Young's modulus Y is,
stress
Y = strain = constant, for a given material of a wire. Thus, the Young's modulus of elasticity of a
material remains same although the load hanging on it is doubled.

Numerical Examples

1. A wire of length 2.5 m and area of cross -section 1 × 10-6 m2 has a mass of 15 kg hanging on it.

Whaty is the extension produced? How much is the energy stored in the extended wire if Young's

modulus of wire is 2 × 1011 Nm–2. [NEB 2074]

Solution:
Length of wire (L) = 2.5m
Area of cross - section (A) = 1 × 10–6 m2

Hanging mass (m) = 15 kg

Extension produced (x) = ?
Energy stored in the extended wire (E) = ?
Young's modulus (Y) = 2 × 1011 Nm-2

g = 10ms-2
We know,

stress F/A FL mgL
Y = strain = x/L = xA = xA

mgL
or, x = YA

or, 15 × 10 × 2.5 = 1.875 × 10-3
x = 2 × 1011 × 1 ×10-6

∴ x = 1.875 × 10-3m.

Again, energy stored (E) is,

11 = 1 × 15 × 10 ×1.875 ×10-3
E = 2 Fx = 2 mgx 2

∴ E = 0.14 Joule.

So, extension produced and energy stored in the wire are 1.875 × 10-3 m and 0.14 Joules respecti

2. How much force is required to punch a hole 1cm in diameter in a steel sheet 5mm thick whose

shearing strength is 2.76 × 108 Nm–2. [HSEB 2069, B]

Solution:
Given :Diameter of hole (d) = 1cm = 0.01m, Thickness of sheet (t) = 5mm = 5 × 10–3m, Force required

(F) = ?

Elasticity | 275

We have, Shearing strength = Force = Force thickness
Area Circumference ×

∴ Force = Shearing strength × circumference × thickness

or, F = Shearing strength ×πd × t
or, F = 2.76 × 108 × 3.14 × 0.01 × 5 × 10–3

∴ F = 4.33 × 104N
3. Find the work done in stretching a wire of cross–sectional area 10–2cm2 and 2m long through 0.1mm,

if Y for the material of wire is 2 × 1011Nm–2. [HSEB 2068, Old Q.]

Solution: Length of wire (L) = 2m
Given: Cross-sectional area (A) = 10–2 cm2 = 10–6 m2 Y = 2 × 1011Nm–2

Extension (e) = 0.1mm = 0.1 × 10–3m,

Work done (W) = ?

stress F/A FL
Since, Y = strain = e/L = eA

∴ YAe
F= L

1 1 YAe YAe2 2 × 1011 × 10–6 × (0.1 × 10–3)2
So, Work done (W) = 2 × F × e = 2 × L × e = 2L =
2×2

∴ W = 5 × 10–4 J.
4. A uniform steel wire of density 7800 kgm–3 weighs 16 gram and is 250 cm. It lengthens by 1.2mm

when a load of 8kg is applied. Calculate the value of Young's modulus for the steel and the energy

stored in the wire. [HSEB 2067 S, 2070]

Solution:

Given: Density of the wire (ρ) = 7800kgm–3 Mass of the wire (M) = 16gm= 16 × 10–3kg

Original length of the wire (L) = 250 cm = 2.50m

Applied load on the wire (m) = 8kg Extension (e) = 1.2mm = 1.2 × 10–3m

Young's modulus (Y) = ? Energy stored (E) = ?

Here, Mass of wire = Density × volume of wire

M=ρ×V=ρ×A×L

∴ M
A = ρL

stress F/A FL (mg)L (mg)L2ρ (8 × 10) × (2.50)2 × 7800
Now, Y = strain = e/L = eA = M = e.M = 1.2 × 10–3 × 16 × 10–3

e. ρL

∴ Y = 2.03 × 1011 Nm–2.

Also, the energy stored in the wire,

11 1 × (8 × 10) × 1.2 × 10–3
E=2×F×e=2 × (mg) × e = 2

∴ E= 0.048 Joules.
5. A steel wire of density 8000kg m–3 weighs 24gm and is 250cm long. It lengthens by 1.2mm when

stretched by a force of 80N. Calculate the Young's modulus for the steel and the energy stored in the

wire. [HSEB 2064]
Solution:

Given: density of the wire (ρ) = 8000 kgm–3 Mass of the wire (M) = 24gm = 24 × 10–3kg

Original length of the wire (L) = 250cm = 2.50m. Force applied (F) = 80N

Extension (e) = 1.2mm = 1.2 × 10–3m Young's modulus (Y) = ?

Energy stored (E) = ?

We have, M = ρ × V = ρ × A × L

276 | Essential Physics

∴ A= M
ρL

stress F/A FL FL FL2ρ 80 × (2.50)2 × 8000
Now, Y = strain = e/L = eA = M = e.M = 1.2 × 10–3 × 24 × 10–3

e. ρL

∴ Y = 1.4 × 1011 Nm–2

Also, 1 1 × 80 × 1.2 × 10–3
Energy stored, E = 2 ×F×e=2

∴ Ε = 0.048 Joules
6. A copper wire of diameter 0.5 mm is stretched between two points at 250C. Calculate the increase in

tension in the wire if the temperature falls 00C. Given Young's modulus of copper

wire = 1.2 × 1011Nm–2, linear expansivity of copper = 18 × 10–6K–1. [HSEB 2063]

Solution: Young's modulus (Y) = 1.2 × 1011Nm–2
Diameter (d) = 0.5mm = 0.5 × 10–3m

Change in temperature (∆θ) = 25°C – 0°C = 25°C Linear expansivity (α) = 18 × 10–6K–1

Increase in tension (T) = ?

We have,

Y = stress = T/A = TL
strain ∆L A∆L

L

∴ T = YA.∆L
L

Also, ∆L = α L∆θ
∴
Τ = YA.αL∆θ = YA α ∆θ = Y π4d2 α ∆θ
or, L

T = 1.2 × 1011 × 3.14 × (0.5 × 10–3)2 × 18 × 10–6 × 25
4

∴ T = 10.6N.

7. A force of 25N is applied to the end of a wire 3m long and produces an extension of 0.25mm. If the

diameter of the wire is 2mm, calculate the stress on the wire, its strain and the value of Young's

modulus. [T.U. 2039]

Solution :

Given: F = 25N L = 3m

Extension (e) = 0.25mm = 0.25 × 10–3m

Diameter (d) = 2mm = 2 × 10–3m

Stress on the wire = ?

Strain on the wire = ?

Y for the wire = ?

Since, F F 4F = 3.14 4 × 25 = 7.96 × 106 Nm–2
Stress = A = πd2 = πd2 × (2 × 10–3)2

4

Again, Strain = e = 0.25 × 10–3 = 8.33 × 10–5.
L 3

Also, Y = Stress = 7.96 × 106 = 9.6 × 1010 Nm–2
Strain 8.33 × 10–5

Hence, the stress, strain and Young's modulus of the wire are 7.96 × 106 Nm–2, 8.33 × 10–5 and 9.6 ×
1010 Nm–2 respectively.

8. A specimen of oil having an initial volume 600cm3 is subjected to a pressure increase of 3.6 × 106 Pa,
and volume is found to decrease by 0.45 cm3. What is the bulk modulus of the material ? The

compressibility?

Elasticity | 277

Solution:
Given: Initial volume of oil (v) = 600cm3= 600 × 10–6m3

Increase in pressure (∆P) = 3.6 × 106 Pa
Decrease in volume of oil (∆V) = 0.45cm3 = 0.45 × 10–6m3

Bulk modulus (B) = ? Compressibility (K) = ?

Since,

∆P ∆P.V 3.6 × 106 × 600 × 10–6
B = ∆V = ∆V = 0.45 × 10–6

V

∴ B = 4.8 × 109 Pa

1
Now, Compressibility = Bulk modulus

∴ 11 = 2.1 × 10–10 Pa–1
K = B = 4.8 × 109

9. The rubber cord of a catapult has a cross-sectional area 1.00mm2 and a total unstretched length

10.0cm. It is stretched to 12cm and then released to project a missile of mass = 5.0g. Calculate the
velocity of projection, taking the Young's modulus for the rubber as 5.0 × 108 Nm–2.

Solution:
Given: Cross-sectional area (A) = 1mm2= 10–6m2.

Ustretched length (L) = 10cm = 0.10m Extension (e) = 12 – 10 = 2cm = 0.02m
Mass of the missile (m) = 5g = 5 × 10–3kg Y = 5 × 108 Nm–2

Velocity of projection (v) = ?

From energy conservation,

K.E. of missile = Energy stored in the rubber cord.

1 mv2 = 1 Fe
2 2

Fe
or v = m = ?

Stress F/A FL
Now, Y = Strain = e/L = eA

YAe 5 × 108 × 10–6 × 0.02 = 100N
F= L =
0.10

Fe 100 × 0.02
So, v = m =
5 × 10–3

∴ v = 20 ms–1
10. A braided nylon rope 2.5cm diameter has a breaking strength of 1.24 × 105N. Find the breaking

strength of similar rope 1.25cm in diameter.

Solution:

First case:

Diameter of rope (d1) = 2.5cm = 2.5 × 10–2m
Breaking strength (F1) = 1.24 × 105N

Since, Breaking stress = Breaking strength (F1)
Cross–sectional area (A1)

or, Breaking stress = F1 = 4F1 . . . (1)
πd12 πd21

4

or, 4 × 1.24 × 105 = 2.52 × 1010 Nm–2
Breaking stress = 3.14 × (2.5 × 10–2)2

Second Case :
Breaking strength (F2) = ?

278 | Essential Physics

Diameter of the rope (d2) = 1.25cm = 1.25 × 10–2m.
Since, From equation (1), we have

Breaking stress = 4F2
πd22

or, 2.52 × 1010 = 3.14 × 4F2 × 10–2)2
(1.25

∴ F2 = 3.1 × 106N
11. A uniform steel wire of density 8000kgm–3 weighs 20gm and is 2.5m long. It lengthens by 1mm when

stretched by a force of 80N. Calculate the value of the Young's modulus of steel and the energy

stored in the wire. [HSEB 2073 C]

Solution: Mass of wire (m) = 20gm = 0.020kg
Given, Density of wire (ρ) = 8000kgm–3 Extension (e) = 1mm = 10–3m

Length of wire (L) = 2.5m

Streching fore (F) = 80N Young's modulus of wire (Y) = ?

Energy strored (E) = ?

ρ = m m
volume = A.L

or, A =ρmL

FL FLρL FL2ρ 80 × 2.52 × 8000
Now, Y = eA = em = m = 10.3 × 0.020

∴ Y = 2 × 1011Nm–2

And, energy stored,

11 × 80 × 10–3 = 0.04 Joules
E = 2. F.e = 2

12. Calculate the work done in stretching a steel wire 100cm in length and cross-sectional area 0.030cm2

when a load of 100N is slowly applied before the elastic limit is reached. [HSEB 2072 C, 2070 D]

Solution:

Given, Work done (W) = ? Length (L) = 100cm = 1m
Cross-sectional area (A) = 0.030cm2 = 0.030 × 10–4m2 Load (F) = 100N
For steel, Y = 2 × 1011 Nm–2

FL
We have, Y = Ae

∴ FL
e = YA

Now, work done, W = 1 F.e = 1 F. YFLA F2L = 2 × 2 × (100)2 × 1 × 10–4 = 8.33 × 10–3J
2 2 = 2YA 1011 × 0.030

Important Numerical Problems

1. A circular steel wire 2.00m long must stretch no more than 0.25cm when a tensile force of 400N is

applied to each end of the wire. What minimum diameter is required for the wire.

Solution: Elongation (e) = 0.25m = 0.25 × 10–2m
Length of wire (L) = 2m

Tensile force (F) = 400N Minimum diameter (d) = ?
Young's modulus for steel (Y) = 20 × 1010 Nm–2

Stress F/A FL
Since, Y = Strain = e/L = eA

FL
or, A = Ye

Elasticity | 279

πd2 400 × 2
or, 4 = 20 × 1010 × 0.25 × 10–2

or, d2 = 400 × 2 × 4
20 × 1010 × 0.25 × 10–2 × 3.14

∴ d = 1.43 × 10–3 m.
2. A metal rod that is 4.00m long and 0.50cm2 in cross-sectional area, is found to stretch 0.20cm under

a tension of 5000N. What is Young's modulus for this metal?

Solution:

Length of metal rod (L) = 4m Elongation (e) = 0.20cm = 0.20 × 10–2m
Cross-sectional area (A) = 0.50cm2= 0.50 × 10–4m2

Tension (F) = 5000N

Young's modulus of the metal (Y) = ?

We have, Stress F/A = FL = 0.50 × 5000 ×4 × 10–4 = 2.0 × 1011 Nm–2
Y = Strain = e/L eA 10–4 × 0.20

3. In a material testing laboratory, a metal wire made from a new alloy is found to break when a tensile

force of 90.8N is applied perpendicular to each end. If the diameter of the wire is 1.84mm, what is the

breaking stress of the alloy?

Solution:

Tensile force (F) = 90.8N
Diameter of the wire (d) = 1.84mm= 1.84 × 10–3m

Breaking stress = ?

F F 4F = 3.14 4 × 90.8 10–3)2 = 3.41 × 107 Pa.
Therefore, breaking stress = A = πd2 = πd2 × (1.84 ×

4

4. A steel cable with cross-sectional area 3.00 cm2 has an elastic limit of 2.40 × 108 Pa. Find the

maximum upward acceleration that can be given a 1200kg elevator supported by the cable if the

stress in not to exceed one - third of the elastic limit. [HSEB 2070]

Soution: Elastic limit = 2.40 × 108 Pa.
Cross-section area (A) = 3cm2 = 3 × 10–4 m2

Maximum upward acceleration (a) = ? Mass of the elevator (M) = 1200kg

1
Stress = 3 (elastic limit)

or F = 1 (2.40 × 108)
A 3

or, F = 1 (2.40 × 108) × 3 × 10–4
3

∴ F = 24000N.

Now, for upward motion of the elevator,

F – Mg = Ma

or, 24000 – 1200 × 9.8 = 1200 × a
∴ a = 10.2 ms–2

5. A 12.0kg mass, fastened to the end of an aluminum wire with an unstretched length of 0.50m, is

whirled in a vertical circle with a constant angular speed of 120 rev/min. The cross-sectional area of
the wire is 0.014 cm2. Calculate the elongation of the wire when the mass is (a) at the lowest point of
the path and (b) at the highest point of its path.(Young's modulus of aluminum wire = 7.0 × 1010 Pa)

Solution:

Mass (M) = 12kg Original length (L) = 0.50m

120
f = 120 rev/min = 60 = 2 rev/sec.

∴ ω = 2πf = 4π rad/sec
Cross-sectional area (A) = 0.014 cm2 = 0.014 × 10–4 m2

280 | Essential Physics

(a) Elongation at bottom (e1) = ?
(b) Elongation at top (e2) = ?

Y for aluminum = 7 × 1011 Pa

(a) At the bottom of the vertical circle, the total stretching force (tension) is,

F1 = Mrω2 + Mg = MLω2 + Mg = 12 × 0.50 × (4π)2 + 12 × 9.8= 1.06 × 103N

Now, Y = F1/A = F1L
e1/L e1A

or, e1 = F1L = 7 1.06 × 103 × 0.50
YA × 1010 × 0.014 × 10–4

∴ e1 = 5.4 × 10–3m

(b) At the top of the vertical circle, the total stretching force (tension) is,

F2 = Mrω2 – Mg = MLω2 – Mg = 12 × 0.50 × (4π)2 – 12 × 9.8= 0.830 × 103N
103 ×
Now, e2 = F2L = 9 0.830 × 0.014 0.50 = 4.2 × 10–3m.
YA × 109 × × 10–4

6. A brass rod with a length of 1.40m and a cross-sectional area of 2.00 cm2 is fastened end - to - end to
a nickel rod with length L and cross-sectional area 1.00 cm2. The compound rod is subjected to equal
and opposite pulls of magnitude 4.00 × 104 N at its ends. (a) Find the length L of the nickel rod if the

elongations of the two rods are equal. (b) What is the stress in each rod? (c) What is the strain in
each rod? (Young's modulus for brass = 9.0 × 1010 Pa and nicked = 21 × 1010 Pa).

Solution:

Given: For Brass rod For Nickel rod

LB = 1.40m LN = L = ?
AB = 2 cm2 = 2 × 10–4 m2 AN = 1 cm2 = 10–4m2
YB = 9 × 1010 Pa YN = 21 × 1010 Pa
Force at each end (F) = 4 × 104 N.

eB = eN (Given).

(a) LN = L = ?

Since, YB = FLB . . . (1);
eB AB

and YN = FLN = FL . . . (2)
eN AN eNAN

Dividing equation (1) by equation (2), we get

YB = FLB/eB AB = LB × AN [ eB = eN]
YN FL/eN AN L × AB
21 × 1010 × 1.40 × 10–4
∴ L = YN × LB × AN = 9 × 1010 × 2 × 10–4 = 1.63m.
YB AB

(b) Stress in each rod = ? 4 × 104
2 × 10 –4
Stress for brass = FB = F = = 2 × 108 Pa.
AB AB

Stress for nickel = FN = F 4 × 104 = 4 × 108 Pa
AN AN = 10–4
stress for brass 2 × 108
Strain in brass = YB = 9 × 1010 = 2.2 × 10–3

Stress for nickel 4 × 108 = 10–3
Strain in nickel = YN = 9 × 1010 1.9 ×

7. What stress would cause a wire to increase in length by one - tength of one percent if the Young's
modulus for the wire is 12 × 1010 Nm–2? What force would produce this stress if the diameter of the

wire is 0.56 mm?

Solution :
Given, Young's modulus (Y) = 12 × 1010 Nm–2

Diameter of the wire (d) = 0.56mm = 0.56 × 10–3m

Elasticity | 281

Stress = ?

Force (F) = ?

1 11 L
Extension (e) = 10 (1 % of L) = 10 × 100 × L = 1000

Stress
Since, Y = Strain

or, Stress = Y × Strain = Y × Le = 12 × 1010 × L/1L000 = 12 × 1010 × 1 = 12 × 109 Nm–2
1000

F F 4F
Again, stress = A = πd2 = πd2

4

or, 12 × 109 = 3.14 × 4×F 10–3)2
(0.56 ×

∴ F= 29.5N
8. A wire 2m long and cross-sectional area 10–6 m2 is stretched 1mm by a force of 50N in the elastic

region. Calculate (i) the strain, (ii) the Young's modulus, (iii) the energy stored in the wire.

Solution :

Given, Length of wire (L) = 2m Cross-section area (A) = 10–6 m2
Force (F) = 50N
Extension (e) = 1mm = 10–3m
10–3
(i) e =2 = 5 × 10–4
Strain = L

(ii) Young's modulus (Y) = ?

We have the relation, F/A F 50 = 1011 Nm–2
Y = Strain = A × strain = 10–6 × 5 × 10–4

(iii) 1 1 × 50 × 10–3 = 0.025 Joules.
Energy stored = 2 Fe = 2

9. A force of 20N is applied to the ends of a wire 4m long, and produces an extension of 0.24mm. If the

diameter of the wire is 2mm, calculate the stress on the wire, its strain, and the value of the Young's

modulus.

Solution:

Given, F = 20N L = 4m
extension (e) = 0.24mm = 0.24 × 10–3m diameter (d) = 2mm = 2 × 10–3m

stress = ?, strain = ?, Y = ?

we know that, stress = F = F = 4F = 3.14 4 × 20 = 6.4 × 106 Nm–2
A πd2 πd2 × (2 × 10–3)2

4

Strain = e = 0.24 × 10–3 = 6 × 10–5
L 4

and, Y = Stress = 6.4 × 106 = 1.1 × 1011 Nm–2
Strain 6 × 10–5

10. What force must be applied to a steel wire 6m long and diameter 1.6mm to produce an extension of
1mm? (Young's modulus for steel = 2.0 × 1011 Nm–2).

Solution: d = 1.6mm = 1.6 × 10–3m
Y = 2.0 × 1011 Nm–2
Given, L = 6m
Extension (e) = 1mm = 10–3m

F=?

We have,

Stress F/A FL
Y = Strain = e/L = eA

∴ YAe Y πd2 2 × 1011 × 3.14 × (1.6 × 10–3)2 × 10–3
F = L = L 4 .e = = 67N
6×4

282 | Essential Physics

11. Find the extension produced in a copper wire of length 2m and diameter 3mm when a force of 30N is
applied. (Young modulus for copper = 1.1 × 1011 Nm–2).

Solution : d = 3mm = 3 × 10–3m;
L = 2m Y = 1.1 × 1011 Nm–2
F = 30N

Extension (e) = ?
F/A FL

We know that the relation,Y = e/L = eA

∴ FL FL 4FL = 1.1 × 4 × 30 × 2 10–3)2 = 7.7 × 10–5m.
e = YA = πd2 = Y × πd2 1011 × 3.14 × (3×
Y× 4

12. A spring is extended by 30mm when a force of 1.5N is applied to it. Calculate the energy stored in the

spring when hanging vertically supporting a mass of 0.20kg if the spring was unstretched before

applying the mass. Calculate the loss in potential energy of the mass. Explain why these values differ.

Solution: F1 = 1.5N m = 0.20 kg
e1 = 30mm = 30 × 10–3m

e2 = ? when F2 = mg = 0.20 × 10 = 2N

Since, F ∝ e YAe
[since, F = L ]

∴ F2 = e2
F1 e1

or, e2 = F2 × e1 = 2 × 30 × 10–3 = 40 × 10–3m.
F1 1.5

11 × 2 × 40 × 10–3 = 0.04 J.
Energy stored = 2 F2e2 = 2

Loss in P.E. of the mass = mge2 = 0.20 × 10 × 40 × 10–3 = 0.08 J.

This shows that the loss in P.E. of the mass is twice the energy gained by the wire. Half of the energy is

stored in the from the elastic potential energy and remaining energy is dissipated in the form of heat in

the stretched wire.

13. If the Young modulus for steel is 2.00 × 1011 Nm–2. Calculate the work done in stretching a steel wire
100 cm in length and of cross - sectional area 0.030 cm2 when a load of 100N is slowly applied

without the elastic limit being reached.

Solution:
Y = 2 × 1011 Nm–2 ; Work done (W) = ?

Original length (L) = 100 cm = 1m
Cross - section area (A) = 0.030 cm2 = 0.030 × 10–4 m2

Load (F) = 100N
Stress F/A FL

We have, Y = Strain = e/L = eA

∴ e = FL = 2 × 100 × 1 × 10–4 = 1.67 × 10–4m
YA 1011 × 0.030

Now, 1 Fe = 1 × 100 × 1.67 × 10–4 = 8.3 × 10–3 Joules.
Work done (W) = Energy stored = 2 2

14. A cylindrical copper wire and a cylindrical steel wire, each of length 1.5m and diameter 2mm, are

joined at one end to form a composite wire 3m long. The wire is loaded until its length becomes

3.003m. Calculate the strains in the copper and steel wires, and the force applied to the wire. (Young
modulus for copper = 1.2 × 1011 Nm–2, Y for steel = 2.0 × 1011 Nm–2).

Solution: dcu = dst = 2mm = 2 × 10–3m
Given, Lcu = Lst = 1.5m
Total extended length = 3.003m;

Total extension (e) = 3.003 – (1.5 + 1.5)

or, e = 0.003m

Elasticity | 283

or, ecu + est = 0.003m . . . (1)
F/A FL 4FL

Now, Y = e/L = eA = e × πd2

4FL
or, e = Y × πd2

4FLcu

ecu Ycu × π 2 Yst
est dcu Ycu
∴ = =
4FLst

Yst × π 2
dst

∴ ecu = 2 × 1011 = 5 . . . (2)
est 1.5 × 1011 3

From equations (1) and (2), we get

5
3 est + est = 0.003

or, est = 0.003 × 3 = 1.125 × 10–3m

8

From equation (2), we get

5 × 1.125 × 10–3 = 1.875 × 10–3m
ecu = 3

Now, Strain in copper = ecu = 1.875 × 10–3 = 1.25 × 10–3
And, Lcu 1.5

Strain in steel = est = 1.125 × 10–3 = 0.75 × 10–3
Lst 2

Again, Force applied (F) = Fcu = Fst = ?
YAe Ye × πd2

We know that the relation, F = L = 4L

∴ Fcu = Ycu ecu × πdcu2 = 1.2 × 1011 × (1.875 × 10–3) × 3.14 × (2 × 10–3)2 = 471N
4Lcu 4 × 1.5

So, F = Fcu = Fst = 471N.
15. A wire of length 3.0m and cross-sectional area 1.0 × 10–6 m2 has a mass of 15 kg hung on it. What is

the stress produced in the wire? (g = 9.8 ms–2) . If the Young's modulus for the material is 2.0 × 1011

Nm–2, what is the extension, x, produced? When extended, how much energy is stored in the wire?

Solution: A = 1.0 × 10–6m2 m = 15kg, g = 9.8 ms–2
Given, L = 3.0m; Y = 2.0 × 1011 Nm–2
Stress = ?,

F mg = 15 × 9.8 = 1.47 × 108 Nm–2
Therefore, Stress = A =A 10–6

Extension produced (x) = ?

Stress Stress Stress × L
Therefore, Y = Strain = x/L = x

or, 2 × 1011 = 1.47 × 108 × 3
x

∴ x = 1.47 × 108 × 3 = 2.2 × 10–3m
2 × 1011

Again, Energy stored (E) = ?

1 1 1 × (15 × 9.8) × 2.2 × 10–3 = 0.16 Joules.
Therefore, E = 2 Fx = 2 (mg) × x =2

16. Calculate the minimum tension with which platinum wire of diameter 0.1mm must be mounted

between two points in stout invar frame if the wire is to remain taut when the temperature rises 100K.

284 | Essential Physics

Platinum has linear expansivity 9 × 10–6 K–1 and Young modulus 17 × 1010 Nm–2. The thermal

expansion of invar may be neglected.

Solution: d = 0.1mm = 0.1 × 10–3m
Given, Minimum tension (T) =?

Temperature rise (∆θ) = 100k Young's modulus (Y) = 17 × 1010 Nm–2
Linear expansivity (α) = 9 × 10–6k–1

We have,

Stress
Y = Strain

F/A FL YAe = YA .(αL∆θ)
= e/L = eA =L L

∴ F = YA α ∆θ

= Y.π4d2 .α ∆θ

= 17 × 1010 × 3.14 × (0.1 × 10–3)2 ×9 × 10–6 × 100 = 1.2N.
4

17. A railway track, uses long welded steel rails which are prevented from expanding by friction in the
clamps. If the cross-sectional area of each rail is 75cm2, what is the elastic energy stored per
kilometer of track when its temperature is raised by 100C? (Linear expansivity of steel = 1.2 × 10–5 K–
1, Young modulus for steel = 2 × 1011 Nm–2).

Solution : ∆θ = 100C
Given, A = 75 cm2 = 75 × 10–4 m2 Y = 2 × 1011 Nm–2
α = 1.2 × 10–5 K–1

Elastic energy stored per km of track = ?

∴ length (L) = 1km = 1000m

We have,

F = YA α ∆θ and, e = L2 – L1 = L1 α ∆θ = L α ∆θ
∴ Energy stored,

1
E = 2 Fe

= 1 (YA α ∆θ) (L α ∆θ) = 1 YAL (α∆θ)2
2 2

1 × 2 × 1011 × 75 × 10–4 × 1000 × (1.2 × 10–5 × 10)2
=2

= 1.1 × 104 Joules.

18. A vertical brass rod of circular section is loaded by placing a 5kg weight on top of it. If its length
is 50 cm, its radius of cross-section 1 cm, and the Young modulus of the material 3.5 × 1010 Nm–2,

find (a) the contraction of the rod, (b) the energy stored in it. [HSEB 2070 C]

Solution :

Given, m = 5kg L = 50 cm = 0.5m 5 kg
r = 1 cm = 0.01m Y = 3.5 × 1010 N m–2

(a) Contraction (e) = ?

Stress
Since, Y = Strain

F/A FL L
= e/L = eA

∴ FL mgL
e = YA = Y(πr2)

5 × 10 × 0.5
= 3.5 × 1010 × 3.14 × (0.01)2
= 2.3 × 10–6m.

Elasticity | 285

(b) Energy stored (E) = ?

We know that, E = 1 F e = 1 (mg) e= 1 × 5 × 10 × 2.3 × 10–6= 5.7 × 10–5 Joules.
2 2 2

19. A uniform wire of unstretched length 2.49m is attached to two points A and B which are 2.0m apart

and in the same horizontal line. When a 5kg mass is attached to the midpoint C of the wire, the

equilibrium position of C is 0.75m below the line AB. Neglecting the weight of the wire and taking
the Young modulus of its material to be 2 × 1011 Nm–2, find (i) the strain in the wire, (ii the stress in

the wire, (iii) the energy stored in the wire. A 1m C 1m B
Solution:

Given, unstretched length (L) = 2.49m Tcosθ
Tcosθ
AB = 2m AC = BC = 1m TT
θθ
m = 5kg CO = 0.75m O

Y = 2 × 1011 Nm–2 mg

i. Strain = ?

In ∆BOC, OB = BC2 + CO2 = 12 + 0.752 = 1.25m

So, Stretched length = AO + OB = 2OB = 2 × 1.25 = 2.50m

∴ Extension (e) = 2.50m – 2.49m = 0.01m

So, Strain = e = 0.01 = 4.02 × 10–3
L 2.49

ii. Stress = Y × Strain
= 2 × 1011 × 4.02 × 10–3 = 8.04 × 108 Nm–2.

1
iii. Energy stored (E) = 2 Fe

1
= 2 T e. Where, T is tension on the wire.

In equilibrium position,

Tcosθ + Tcosθ = mg
or, 2Tcosθ = mg

mg
or, T = 2cosθ

mg
= CO

2 × OB

mg × OB
= 2 ×CO

5 × 10 × 1.25
= 2 × 0.75 = 41.67N

1
Now, E = 2 Te

1
= 2 × 41.67 × 0.01= 0.208 Joules.

20. The rubber cord of a catapult is pulled back until its original length has been doubled. Assuming that
the cross-sectional of the cord is 2mm square, and that Young modulus for rubber is 107 Nm–2,

calculate the tension in the cord. If the two arms of the catapult are 6cm apart, and the unstretched

length of the cord is 8cm, what is the stretching force?
Solution:

Given, Unstretched length (L) = 8cm

Stretched length = 2L

So, Extension (e) = 2L – L = L = 8cm

Cross - sectional area (A) = 2mm square = (2mm)2 = 4mm2 = 4 × 10–6 m2

Y = 107 Nm–2 AB = 6cm

286 | Essential Physics

Tension in the wire (T) = ? A 3 cm C 3 cm B
Here, C is the midpoint of AB Tcosθ
∴ AC = CB = 3cm

Stress T/A TL TL T Tcosθ T
Since, Y = Strain = e/L = eA = LA = A T
or, T = YA = 107 × 4 × 10–6
θθ
∴ T = 40N

Now, Stretching force (F) = ? O
Here, mg
F = mg = Tcosθ + Tcosθ

= 2Tcosθ = 2Τ × OC = 2T × OB2 – BC2
OB OB

Here, OB + OA = 2L

OB + OB = 2L

∴ ΟΒ = 2L = L= 8cm
2

82 – 32
So, F = 2 × 40 × 8 = 74N.

21. A copper wire, 200cm long and 1.22 mm diameter, is fixed horizontally to two rigid supports 200 cm

long. Find the mass in grams of the load which, when suspended at the midpoint of the wire,
produces a sag of 2cm at that point. Young modulus for copper = 12.3 × 1010 Nm–2.

Solution:

Given, unstretched length,

L = AB = 200cm = 2m AC = CB = 100cm = 1m

Sag produced, CO = 2cm = 0.02m;
Y = 12.3 × 1010 Nm–2

Diameter (d) = 1.22 mm = 1.22 × 10–3m

Mass suspended (m) = ?

Here, in equilibrium, mg = Tcosθ + Tcosθ = 2Τcosθ

2Tcosθ …(1)
or, m = g = ?

Now, extension (e) = (AO + OB) – AB = (OB + OB) – AB = 2OB – L= 2 OC2 + CB2 – L

or, e = 2 0.022 + 12 – 2 A 1m C 1m B
∴ e = 4 × 10–4m

Stress T/A TL Tcosθ
Now, Y = Strain = e/L = eA
Tcosθ
YeA T T
or, T = L

Ye πd2 θθ
=L × 4
O
12.3 × 1010 × 4 × 10–4 3.14 × (1.22 × 10–3)2
∴ T = 2 × 4 = 28.75N

Again, cosθ = CO mg
OB

= CO = 0.02 12 = 0.02
OC2 + BC2 0.022 +

2Tcosθ
From (1), m = g

2 × 28.75 × 0.02
= 10 = 0.115kg.

∴ m = 0.115 kg

Elasticity | 287

Practice Short Questions

1. What is the difference between elastic and plastic body?

2. The material in human bones and elephant bones is essentially the same but an elephant has much thicker

legs. Explain why, in terms of breaking stress.

3. Electric power lines are sometimes made by using wires with a steel core, and a copper and steel twisted

together, why?

4. What is elastic after effect?

5. What is elastic fatigue?

6. Why is the surface of a liquid in the earth's gravitational field always horizontal?

7. Two identical springs of steel and copper are equally stretched? On which more work will have to be done?

8. What is the value of the modulus for an incompressible liquid?

9. What do you understand by yield point?

10. What is elastic limit? What happens if it is exceeded?

11. Metals wires after being heavily loaded do not regain their lengths completely. Explain, why?

12. Do liquids possess rigidity?

13. A cable is cut to half of its original length. How does it affect the maximum load that the wire can support?

14. Steel bridges are declared unsafe after a few decades of use. Why? [HSEB 2069 S]

Practice Long Questions

1. Define the Young modulus. Describe an experimental method to determine this quantity for a

material in the form of wire.

2. Derive an expression for the energy density of a body in a stretched wire. Define the term energy

density of a body under strain. [HSEB 2067, 2069 A]

3. Hooke's law for a tensile stress can be written as Fx = Kx, where x is the object's change length and
k is the force constant.

(a) What is the force constant of a rod of length l0, cross-sectional area A, and Young's
modulus Y ?

(b) In terms of l0, A and Y, how much work is required to stretch the object a distance x?
4. What are the proportionality limit and the elastic limit? Derive an expression for the energy stored

in a stretched wire? [HSEB 2069 B]

5. Define stress and strain. Derive an expression for energy stored in a stretched wire. [HSEB 2068]

6. State Hooke's law. How would you verify it experimentally? [HSEB 2058]

7. What does the term strain refer to? What are the different types of strain? Find an expression for

the energy stored in a wire. [HSEB 2062]

8. Why is energy stored in a wire? Prove that the elastic energy stored per unit volume of a stretched

1 [HSEB 2061]
wire is equal to 2 × stress × strain.

9. What is Poisson's Ratio? Derive an expression for the energy stored in a stretched wire?

HSEB 2071 C]

10. Define Young's modulus, bulk modulus, modulus of rigidity and Poisson's ratio. [HSEB 2071 D]

11. What are the proportional limit and the elastic limit? Derive an expression for the energy stored in

a stretched wire? [HSEB 2069 B]

Practice Numerical Questions

1. A uniform steel wire of density 7800 kgm–3 weighs 169g and is 250cm long. It lengthens by1.2mm

when stretched by a force of 80N. Calculate the energy stored in the wire. [Y for steel = 2 × 1011

Nm–2] [Ans: 0.4992 J] [HSEB 2053]

288 | Essential Physics

2. A relaxed biceps muscle requires a force of 25.0 N for an elongation of 3.0 cm; the same muscle

under maximum tension requires a force of 5000 N for the same elongation. Find Young's

modulus for the muscle tissue under each of these conditions if the muscle is assumed to be a
uniform cylinder with length 0.200 m and cross sectional area 50.0 cm2.

[Ans: 3.33 × 104 Nm–2, 6.67 × 106 Nm-1]

3. A circular steel wire 2.00 m long must stretch no more than 0.25 cm when a tensile force of 400 N

is applied to each end of the wire. What minimum diameter is required for the wire?
[Ans: 1.42 × 10–4 m]

4. Two rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m long and 1.50

cm in diameter. The combination is subjected to a tensile force with magnitude 400 N. For each

rod, what is (a) the strain? (b) the elongation?
[Ans: (i) 1.13 × 10–5, 1.88 × 10–5 (b) 8.47 × 10–6 m, 1.41 × 10–5 m.]

5. A nylon rope used by mountainaineers elongates 1.10 m under the weight of an 65.0 kg climber. If

the rope is 45.0 m in length and 7.0 mm in diameter, what is Young's modulus for this material?
[Ans: 6.77 × 108 Nm–2]

6. In construction a large mobile, an artist hangs an aluminum sphere of mass 6.0 kg from a vertical
steel wire 0.50 m long and 2.5 × 10–3 cm2 in cross-sectional area. On the bottom of the sphere he

attaches a similar steel wire, from which he hangs a brass cube of mass 10.0 kg. For each wire,

compute (a) the tensile strain and (b) the elongation.

[Ans: upper: 3.1 × 103; lower: 2.0 × 10–3; (b) upper : 1.6 mm, lower : 0.98 mm]

7. A specimen of oil having an initial volume of 600 cm3 is subjected to a pressure increase of 3.6 ×

106 Pa, and the volume is found to decrease by0.45 cm3. What is the bulk modulus of the material?

To compressibility? [Ans: 4.8 × 109 Pa, 2.1 × 10–10 Pa–1]

8. A steel cable with cross-sectional area 3.00 cm2 has an electic limit of 2.40 × 108 Pa. Find the

maximum upward acceleration that can be given a 1200 kg elevator supported by the cable if the

stress is not to exceed one-third of the elastic limit. [Ans: 10.2 ms–2]

9. A wire 2 m long and cross-sectional are a 10–6 m2 is stretched 1 mm by a force of 50 N in the

elastic region. Calculate (i) the strain (ii) the Young modulus, (iii) the energy stored in the wire.

[Ans: 1 , 1011 Nm–2, 0.025 J]
200

10. A force of 20 N is applied to the ends of a wire 4 m long, and produces an extension of 0.24 mm. If

the diameter of the wire is 2 mm, calculate the stress on the wire, its strain, and the value of the

Young modulus. [Ans: 6.4 × 106 Nm–2, 6 × 10–5, 1.1 × 1011 N m–2]

11. What force must be applied to a steel wire 6 m long and diameter 1.6 mm to produce an extension

of 1 mm? (Young modulus for steel = 2.0 × 1011 N m–2) [Ans: 67 N]

12. A spring is extended by 30 mm when a force of 1.5 N is applied to it. Calculate the energy stored

in the spring when hanging vertially supporting a mass of 0.20 kg if the spring was unstretched

before applying the mass. Calculate the loss in potential energy of the mass. [Ans: 0.04 J, 0.08 J]

13. What stress would cause a wire to increase in length by one-tenth of one percent if the Young

modulus for the wire is 12 × 1010 N m–2? What force would produce this stress if the diameter of

the wire is 0.56 mm? [Ans: 1.2 × 108 Nm–2, 29.6 N]

14. A wire of length 3.0 m and cross-sectional area 1.0 × 10–6 m2 has a mass of 15 kg hug on it. What

is the stress produced in the wire? (g = 9.8 m–2). If the Young's modulus for the material is 2.0 ×

1011 Nm–2. What is the extension x produced? When extended how much energy is stored in the

wire? [Ans: 2.2 mm, 0.162 J]

15. A vertical brass rod of circular section is loaded by placing a 5 kg weight on top of it. If its length
is 50 cm, its radius of cross-section 1 cm, and the Young modulus of the material 3.5 × 1010 N m–2,

find (a) the contraction of the rod, (b) the energy stored in it.
[Ans: (a) 2.27 × 10–6 m (b) 5.57 × 10–5 J]

Elasticity | 289

MBBS, B.E., B.Sc., Nursing Entrance Preparation Questions

1. In order to elongate a given wire to double its length , one requires a stress which is

a. of infinite magnitude b. equal to young's modulus

c. double the young's modulus d. half of young 's modulus

2. The young modulus of a material is 2 × 101 0 Nm–2. If its length is doubled then find the force applied on it
if it has area 100m2.

a. 4×1012N b. 2×1010N c. 2×1012N d. 1018N

3. A mental rod of young's modulus of elasticity 2×1011Nm–2 undergoes an elastic strain of 0.05. The energy
stored per unit volume of the rod in Jm–3 is

a. 2.5×108 b. 0.250 c. 0.500 d. 0.750

4. A steel wire of length 2.5m and cross- sectional area 0.80×10–6 m2 has a young's modulus of 2.0×1011 pa.
The upper end of the wire is fixed and a load of 8N is tied at the other end. The extension in mm of the
wire is

a. 0.125 b. 0.0250 c. 0.500 d. 0.750

5. Steel is more elastic than rubber. This means for same applied stress, the strain will be

a. more for steel b. less for steel

c. move or less for steel d. equal for steel and rubber

6. P.E. in a sting when stretched by 2cm is U. It's P.E. when stretched by 10cm is

U U c. 25U d. 5U
a. 25 b. 5

7. Which one of the following substances possesses the highest elasticity?

a. rubber b. glass c. copper d. steel

8. The breaking force for a wire of radius r of given material is F .The breaking force for the wire of same
material of radius 2 r is

a. F b. 2F c. 4F F
c. 4

9. A cable is replaced by another wire of same length and material but twice the diameter. The elongation for
a given load will be

a. 4 time b. twice c. one half c. one forth

10. If a metallic wire is stretched a little beyond the elastic limit and relapsed, It will:

a. lose its elastic property completely b. not contract at all

c. contract only up to its length at elastic limit

d. contract but its final length will be greater than its initial length.

11. When a spring is extended by loading, the strain produced is

a. longitudinal b. volumetric c. shearing d. none

12. Bulk modulus for a perfectly elastic body is

a. zero b. infinity c. 1 d. finite

13. Young's modulus for a perfectly elastic body is

a. zero b. infinity c. 1 d. finite

14. Young's modulus for a perfectly plastic body is

a. zero b. infinity c. 1 d. finite

15. The extensibility of a wire of young 's modulus Y varies

a. Y 1 c. Y2 d. Y°
b. Y

290 | Essential Physics

16. With a rise of temperature, The young's modulus

a. increases b. decreases c. remains unchanged d. change erratically

17. According to the Hooke's law of elasticity; if stress is increased, the ratio of stress to strain

a. increases b. decreases

c. become zero d. remains constant

18. The breaking force of a wire depends upon

a. length of wire b. radius of the wire

c. material of wire d. both (b) and (c)

19. The breaking stress of wire depends upon

a. length of the wire b. radius of the wire

c. material of the wire d. shape of the cross- section

20. Two identical wires of A and B are taken and equal stretching forces are applied on them along the length.
It is observed that A elongates more than B then

a. B is more elastic than A b. A is more elastic than B

c. A and B are equally elastic d. A is elastic and B is plastic

Answer Key

1. b 2. c 3. a 4. a 5. a 6. c 7. d 8. c 9. d 10. d
11. c 12. b 13 b 14. a 15. b 16. b 17. d 18. d 19. c 20. a

Ch apter

11 Periodic Motion

Teaching Manual Physics Grade – XI, Higher Secondary Education Board
Curriculum and Training Division Sanothimi, Bhaktapur

Syllabus:

Periodic motion – Oscillatory motion; Circle of reference; Equation of Simple Harmonic Motion
(SHM); Energy in SHM; Application of SHM; Motion of a body suspended from coiled spring,
angular SHM; simple pendulum; Damped oscillation; Forced oscillation and resonance.

Objectives:

The main objective of this subunit is to make the students able to identify and describe simple
harmonic motion.

Activities (Micro syllabus) :

1. Discuss periodic and oscillatory motion.
2. Define simple harmonic motion with examples.
3. Explain how motion around the circle is related to SHM. Obtain the expression for

acceleration and time period.
4. Give equation of SHM and explain characteristics of SHM.
5. Show that the motion of mass suspended in vertical and horizontal spring is SHM.
6. Define simple pendulum. Show that motion of a bob of a simple pendulum is SHM.
7. Prove total energy in SHM is conserved.
8. Discuss free, damped and forced oscillation.
9. Find the condition for resonance.
10. Numerical problems : Focused numerical problems given:

(a) As exercise in the University Physics (Ref. 1) and
(b) In Advanced Level Physics (Ref. 2) and also in Physics for XI (Ref. 3).

292 | Essential Physics

11.1 Introduction

When a body moves along a straight line, then its motion is translatory motion, which we have discussed
in previous chapters. In this chapter, we discuss the vibratory motion of a physical system.

11.2 Periodic motion

The motion which repeats itself after equal intervals of time is called a periodic motion. The interval of
time is called the time period of the periodic motion. For examples, the motion of the pendulum of a wall
clock, oscillations of a mass suspended from a spring, the motion of a planet around the sun, the motion
of the hands of a clock etc.

11.3 Oscillatory Motion

If a body moves back and forth (to and fro) repeatedly about a mean position, then it is said to possess
oscillatory or vibrating motion. For examples, the motion of the pendulum of a wall clock, oscillations
of a mass suspended from a spring, the oscillations of the hands of a walking person etc.

An oscillatory motion is always periodic but a periodic motion may not be oscillatory. For example,
the motion of earth round the sun is periodic but not oscillatory. An oscillatory motion can be expressed
in terms of sine or cosine function or their combination. So, the oscillatory motion is also called the
harmonic motion.

11.4 Simple Harmonic Motion (SHM)

A simple harmonic motion is defined as an oscillatory motion about a fixed point (mean position) in
which the restoring force is always directly proportional to the displacement from that point and is always
directed towards that point.

If x be the small displacement from the mean position of a particle and F be the restoring force acting on
it, then

F∝x

or, F = – Kx . . . (1)

Where K is a proportionality constant known as force constant. The negative sign indicates that the
restoring force F is developed opposite to the displacement from the mean position.

From Newton's second law of motion.

F = ma . . . (2)

Where, m = mass of the particle and a = acceleration

From equations (1) and (2), we get

ma = – Kx

K . . . (3)
or, a = –m x

Here, K and m are constants.

∴ a∝x

Equation (3) shows that the acceleration in SHM is always directly proportional to the displacement from
the mean position and negative sign indicates that it is always directed towards the mean position.

Periodic Motion | 293

11.5 Circle of Reference and Equation of SHM

Suppose a particle is moving with uniform speed along the circumferences of a circle of radius r. The
circle is called reference circle and the particle is called reference particle or generating particle.

YY

M P +r B
θ

yr T 3T

X' O θ M' X A 24 A t
x
TC T

4

–r
D

Y' [Fig. 11.1 (b) Graphical representation of
motion of particle p]
[Fig. 11.1(a) Circle of reference]

Let P be the position of the reference particle at any instant t. Let M be the foot of perpendicular drawn
from the position P on the diameter YOY' as shown in figure 11.1 (a).

When the particle P moves from X to Y, then the foot of perpendicular M moves from O to Y.

When the particle P moves from Y to X', then M moves from Y to O.

Similarly, when the reference particle P moves from X' to Y', the foot M moves from O to Y'.

When P moves from Y' to X, then M moves from Y' to O.

Thus, when the reference particle P completes one revolution along the circle, the foot of perpendicular M
completes one vibration about the mean position O along the diameter YOY'.

The motion of foot of perpendicular M along the diameter YOY' is simple harmonic motion (SHM) and
the motion of particle P along the circle is uniform circular motion. The points A, B, C, D and A in
fig.11.1(b) correspond to respective points X, Y, X', Y' and X in figure11.1 (a).

The centre of circle O is called mean position and ends of diameter Y or Y' are called extreme positions
of SHM.

A simple harmonic motion (SHM) possess the following characteristics:
1. Displacement (y) : The distance of foot of perpendicular M from the mean position O, is called

displacement (y) in SHM. Let θ be the angular displacement of the reference particle P at time t as
shown in reference circle.

Then, in ∆ OPM,

sinθ = OM = y
OP r

or, y = rsinθ . . . (1)
∴ y = rsinωt

Where, θ = ωt and ω is constant angular velocity. Equation (1) is the displacement equation for

SHM.

If the foot of perpendicular M' of the particle is taken on the diameter XOX', then

x = rcosθ

∴ x = rcosωt . . . (2)

2. Amplitude (ymax or r):
We have, the displacement equation in SHM is,

y = r sinωt = rsinθ

294 | Essential Physics

Here, y is maximum if sinθ = 1.
∴ ymax = r
So, maximum displacement of the particle from the mean position is called amplitude or
displacement amplitude in SHM which is equal to the radius r of the reference circle. Obviously,
the displacement is maximum at extreme positions Y or Y'.
3. Velocity (v) : The velocity in SHM at an instant is defined as the rate of change of displacement at
that instant. So,

dy
v = dt

or, v= d (rsinωt) [ y = r sinωt]
dt

or, v = rωcosωt . . . (3)

or, v = rω 1 – sin2ωt

or, v = ω r2 – (rsinωt)2

or, v = ω r2 – y2 . . . (4) [ y = rsinωt.]

Equation (3) and (4) are the equations for velocity in SHM. These equations show the velocity (v)

is not uniform in SHM.

At mean position O, y = 0, ∴ v = rω (maximum value)

At extreme positions Y or Y', y = r, ∴ v = 0 (minimum value)

So, a particle in SHM has maximum velocity at mean position and minimum velocity (i.e. zero)

at the extreme positions.

4. Accelerations (a) : The rate of change of velocity is called acceleration (a). So,

dv = d (rω cosωt) [ v = r ω cosωt]
a = dt dt

or, a = rω d (cosωt) = rω (–sinωt)ω
dt

or, a = –ω2 (rsinωt)

∴ a = –ω2y . . . (5) [ y = rsinωt, from equation (1)]

This is the equation for the acceleration in SHM. This equation (5) shows that acceleration in

SHM is directly proportional to the displacement from the mean position and negative sign shows

that it is always directed towards the mean position.

At the mean position 0, y = 0, ∴ a = 0 (minimum value)

At extreme positions Y or Y', y = r , ∴ a = – ωr2 (maximum value)

Thus, a particle in SHM has zero acceleration at the mean position and maximum acceleration

at the extreme positions.

5. Time period (T) : The time taken by the particle to complete one oscillation is called time period

in SHM (T).

The magnitude of acceleration in SHM is given by,

a = ω2y

or, ω2 = a
y

or, ω= a
y

or, 2π = a
T y

or, T = 2π y
a

Periodic Motion | 295

∴ T = 2π displacement . . . (6)
acceleration

In this time T, the foot of perpendicular M makes one complete oscillation along the diameter

YOY' i.e. from O to Y, back to O, then to Y' and finally again back to O.

6. Frequency (f) : The number of oscillations made in one second in called frequency (f) in SHM.

In T seconds, the particle makes 1 oscillation.

1
In 1 second, the particle makes T oscillations

∴ frequency = 1
Time peiod

or, f = 1 = 1 acceleration . . . (7)
T 2π displacement

7. Wavelength (λ): The linear distance travelled by the particle in one oscillation is called

wavelength (λ) of the particle executing SHM. If T be the time period of oscillation and v be the

velocity, then

λ = vT . . . (8)

8. Phase :The phase of a particle executing SHM at any instant gives the position and direction of

motion of the particle with respect to its mean position. It is measured in terms of fraction of time

period T or the fraction of angle 2π.

In the equation, y = r sin(ωt + φ0),
(ωt + φ0) → phase of the particle in SHM,
φ0 → phase constant or phase angle.
φ tells us the position from where the time was considered.

Let, ωt + φ0 be denoted by φ.
Then, φ = ωt + φ0
or, φ – φ0 = ωt

So, phase change in time t is, φ – φ0 = ωt = 2Tπt

The phase change in time T is, φ – φ0 = 2π T = 2π.
T

This shows that the phase change in T seconds will be 2π radian which actually means 'no change

in phase'. Therefore, the time period (T) can also be defined as the time interval during which the

phase of the vibrating particle changes by 2π.

9. Graphical Representation of Displacement, Velocity and Acceleration in SHM:

The displacement (y), velocity (v) and acceleration (a) equations in SHM are,

y = rsinωt = rsin2Tπ t, v = rωcosωt = rωcos 2π t
T

and, a = –ω2y = – ω2rsinωt = –ω2rsin2Tπ t

At t = 0, y = 0, v = rω and a = 0

At T v = 0 and a = –ω2r
t = 4 , y = r,

At T v = –rω and a = 0
t = 2 , y = 0,

At 3T v = 0 and a = ω2r
t = 4 , y = –r,

296 | Essential Physics

At t = T, y = 0, v = rω and a = 0

y

+r

Ot

–r (A)
v

+rω

Ot

–rω (B)
a

ω2r

Ot

–ω2r

(C)

T T 3T T
4 2 4

[Fig. 11.2, Graphical reprsentation of (A) Displacement, (B) Velocity and (C) Acceleration in SHM]

11.6 Applications of Simple Harmonic Motion

1. Simple Pendulum Support S

A simple pendulum is a heavy paint mass suspended by a θ T
weightless, inextensible and flexible string from a rigid l
support which can oscillate freely in a vertical plane. In
actual practice, it consists of a metallic bob suspended from a A
light cotton thread whose one end is fixed to a rigid support. yθ
The point on the rigid support by which the string is O
suspended is called point of suspension. The distance
between the point of suspension and the centre of gravity of mg
the pendulum bob is called the effective length of simple
pendulum. [Fig. 11.3, Simple Pendulum]

Consider a simple pendulum having mass of the bob m and
effective length l as shown in figure 11.3. Point S is the point
of suspension and point O is the equilibrium or mean
position of the pendulum. Here, A represents the displaced

position of the pendulum at any time such that, ∠ASO = θ is

Periodic Motion | 297

the angular displacement and arc OA = y is the linear displacement of the bob.

In the displaced position A, the forces acting on the pendulum bob are:

(i) weight mg of the bob acting vertically downward, and

(ii) tension T in the string along its length and directed towards the point of suspension S.

The weight mg of the bob can be resolved in to two rectangular components:

(a) mgcosθ opposite to the tension T along the string, and

(b) mgsinθ perpendicular to the string i.e., along the mean position O.

The mgcosθ component is balanced by the tension T in the string and mgsinθ component acts

towards the mean position O which provides restoring force to the pendulum and tries to bring the

pendulum bob back to the mean position O. So, restoring force F is,

F = –mgsinθ . . . (1)

The negative sign indicates the opposing nature of restoring force. If 'a' is the acceleration of the

bob, then

F = ma . . . (2)

From equations (1) and (2), we get

ma = mgsinθ

or, a = –gsinθ . . .. (3)

For small θ, sinθ ≈ θ

∴ a = – gθ . . . (4)

Since, θ = arc OA y
radius OS = l

So, from equation (4), we get

y
a = -g l

or, a = – gl y . . . (5)

For a pendulum at a given place gl  is a constant.

∴ a∝y

Equation (5) shows that acceleration of a simple pendulum is directly proportional to the

displacement from the mean position and negative sign shows that it is always directed towards

the mean position. So, the motion of a simple pendulum is simple harmonic motion (SHM).

We know, for a simple harmonic motion,

a = –ω2y . . . (6)

Comparing equations (5) and (6), we get

ω2 = g
l

or, ω= g
l

or, 2π = l
T g , where T is time period.

∴ T = 2π l . . . (7)
g

Equation (7) is the expression for time period (T) of a simple pendulum. This equation shows that

the time period of a simple pendulum is independent with the amplitude of oscillation and mass of

the bob, but depends on its effective length and the value of g at a place.

298 | Essential Physics

Second's pendulum:
A simple pendulum whose time period is two seconds is called a second's pendulum.

Since, T = 2π l
g

For a second's pendulum, T = 2 seconds.

∴ 2 = 2π l
g

Squaring, 4 = 4π2gl

g . . . (8)
or, l = π2
At sea level g = 9.8m/s2.

∴ l = 9.8 = 0.993m = 99.3cm ≈ 100cm
3.142

So, the length of a second's pendulum is about 100cm.

Drawbacks of simple pendulum:

i. The point of suspension may not be perfectly rigid.

ii. The heavy point mass can't be realized in practice.

iii. The weightless and perfectly inextensible string can't be available in practice.

iv. For large angular displacement, the formula for time period doesn't hold true.

v. The motion may not be perfectly linear.

2. Oscillation of a Mass in Horizontal Spring

Suppose, one end of a massless spring S is attached to a rigid support and other free end to a mass

m. Let the mass spring system lies horizontally on a frictionless table as shown in figure 11.4. Let

O be the mean position of the mass corresponding to natural s
length of the spring. At this position O, the force on the mass F m
= 0.
o l=0
Let the mass is pulled right to the position A through a distance F=0
l and F be the restoring force developed in the spring. Then,
from Hooke's law, sF

F∝l o m
sF A
or, F = –Kl . . . (1) l

Where, K is force constant or spring constant of the spring. m
The negative sign indicates that the restoring force acts
opposite to the displacement. When the mass m is released at Bl o
point A, it moves towards the mean position O where it gains
kinetic energy. The mass then moves left from mean position O [Fig. 11.4, Horizontal oscillation

and reached to the position B. The spring S is compressed and of a mass-spring system]

restoring force F on the mass acts opposite to the direction of motion i.e., towards the mean

position O. The mass m keeps on oscillating horizontally about the mean position O.

If 'a' be the acceleration produced on the mass, then from Newton's second law,

F = ma . . . (2)

From equations (1) and (2), we get

ma = –Kl

Periodic Motion | 299

or, a = – mK l . . . (3)

K
For a mass spring system, m is a constant.

∴ a∝l

Equation (3) shows that, for a horizontal mass spring system, acceleration is directly proportional

to the displacement and the negative sign indicates that it is always directed towards the mean
position. So, motion of a horizontal mass spring system is simple harmonic motion (SHM).

Since, for a simple harmonic motion,

a = –ω2l . . . (4).

Where l = extension i.e. y.

Comparing equations (3) and (4), we get

ω2 = K
m

or, ω= K
m

or, 2π = K
T m

∴ T = 2π m . . . (5)
K

Equation (5) is the required expression for time period of oscillation, which shows that it depends
on mass m and spring constant K.

3. Oscillation of a Mass in Vertical Spring

Let a mass m is suspended on a rigid support by the help of a massless spring as shown in figure
11.5. When the mass m is attached to the lower end of the spring, then it elongates through a
distance l. The restoring force produced on the spring is,

F1 = –Kl . . . (1)

Here, F1 = mg, and K is force constant of the spring. Now, the spring is at the position A which is

the equilibrium or mean position of the loaded spring.

Let the mass is pulled down to the position B through a small
distance y, then the restoring force F2 developed on the spring
is,

F2 = –K(l + y) . . . (2)

Now, if the mass is released from position B, then the O

resultant restring force F which causes the oscillation is, F1 l A
F2 y
F = F2 – F1
mg B
or, F = –K(l + y) + Kl

∴ F = –Ky . . . (3) mg

If 'a' be the acceleration produced on the mass, then from [Fig. 11.5, Vertical oscillation
of a mass-spring system.]
Newton's second law,

F = ma . . . (4)

From equations (3) and (4), we get

or, ma = – Ky

300 | Essential Physics

or, a = – mK y . . . (5)

K
Where, m is a constant. Hence, the motion of a vertical mass spring system (loaded spring) is

simple harmonic motion (SHM).

Since, for a SHM,

a = –ω2y . . . (6)

Comparing equations (5) and (6), we get

ω2 = K
m

or, ω= K
m

2π K
or, T = m

∴ T = 2π m . . . (7)
K

This is the required expression of time period (T) of oscillation of a vertical mass spring system.
This shows that the time period of oscillation depends on mass m and spring constan K, but it is
independent with value of acceleration due to gravity (g).

11.7 Angular Simple Harmonic Motion

Simple harmonic motion is also found in rotational motion. For example: the motion of balanced wheel of
a mechanical watch, the oscillation of a heavy rod suspended from a rigid support at its centre of gravity
with the help of a spring in a horizontal plane etc. In these examples, a restoring torque τ is developed in
the spring, which is directly proportional to the angular displacement θ.That is,

τ∝θ S

or, τ = -Kθ . . . (1)

Where, K is torsion constant of the spring.

Since, τ = Iα . . . (2)

Where, I = moment of inertia about the axis of rotation θ
O
and α = angular acceleration

From equations (1) and (2), we get

Iα = –Kθ [Fig. 11.6, Oscillating rod in a

∴ α = – KI  θ . . . (3) horizontal plane about a vertical
axis]
Here, KI  is a constant for a rotating body about the given axis of

rotation. Hence, the angular motion of such a body (e.g. balance wheel of a mechanical watch) is said

to be the angular SHM.

The equation (3) is in the form of equation of angular SHM,

α = –ω2θ . . . (4)

Comparing equations (3) and (4), we get

ω2 = K
I