Physics

Properties of Matter

Session

Session Objectives

Session Objective

Elasticity – Concept

Stress and Strain

Hooke's law and Young's modulus

Elastic potential energy in a strained body

Poisson's ratio

Shear modulus

Bulk modulus

A

F F

Force per unit area is stress.

AF

Stress

Fractional change in shape brought by application of stress is called strain.

Stress and Strain

Hooke’s Law

Hooke’s law is valid for metals, within limit of proportionality and states that for a body, applied stress is proportional to strain

Stress Strain

StressE cons tant

Strain

Questions

According to Hooke’s law of elasticity,if stress is increased,the ratio of stress to strain

(a) Increase (b) decrease

(c) Becomes zero (d) Remains Constant

Illustrative Problem

Solution :

Hooke’s laws states that

“stress upon strain is constant for a particular material”

Elasticity

Proportionality Limit

Elastic Limit

Fracture point

AB

C

str

e

s

ss t r a i n

Young’s Modulus

tensile stress F AY = =

tensile strain ΔL L

A

L L+L

FF

Shear Modulus

hxAF

strainshearstressshear

x

h

A F

-F

V

Bulk Modulus

VVP

VVAF

strainvolumestressvolume

B

FF V-V

Poisson Ratio

The ratio of the lateral strain to longitudinal strain is constant for a given material. This is called Poisson’s ratio.

• It is represented by .

• It has no units and dimension.

Strain Energy

ExtensionForce21

U

strainstress21

u

Strained Energy

Strain energy per unit volume

Questions

If in a wire of Young’s modulus Y, longitudinal strain X is produced,then the value of potential energy stored in per unit volume will be

(a)YX2 (b) 2 YX2

(c) 0.5Y2X (d)0.5 YX2

Illustrative Problem

Potential energy stored

= ½ x strain x stress

Stress = Strain x Modulus

= XY

PE = 0.5 YX2

Solution

Class Test

Class Exercise - 1

The bulk modulus of a perfectly rigid body is equal to

(a) zero

(b) infinite

(c) unity

(d) a non-zero constant

Solution

Hence answer is (b).

FA.

for perfectly rigid body is zero. Hence, is infinite.

Class Exercise - 2

If a body is strained, its internal energy

(a) decreases

(b) Increases

(c) remains unchanged

(d) changes randomly

Solution

Upon straining, body gets elastic potential energy. Thus, total internal energy increases.

Hence answer is (b).

Class Exercise - 3

A material has Poisson ratio 0.50 of a uniform rod if it suffers a longitudinal strain of 2 × 10-3, Then percentage change in volume is

(a) 0.6 (b) 0.4

(c) 0.2 (d) zero

Solution

rr

3

rror 0.5

2 10

3r10

r

Let the initial and final volumes are v and v + dv.

2v dv r r 2v r 2 r r

3 3v r2 2 10 2 10 0

v r

Hence answer is (d).

Class Exercise - 4

12

The length of a metal wire is when the tension in it is T1 and is when tension is T2. The unstrectched length of the wire is

1 21 2

1 2 2 1 1 2 2 1

2 1 2 1

(a) (b)2

T T T T(c) (d)

T T T T

Solution

Hence answer is (c).

Let the original length

1 1 1Under tension T ,

2 2 2Under tension T ,

1 1

1 2

T Ty

A A

1 2

1 2

T T

1 2

1 2

T T

1 2 2 1

2 1

T TT T

Class Exercise - 5

A spherical ball contracts in volume by 0.0098% when it is subjected to a pressure of 100 atm. Calculate its bulk modulus.

(a) 1.02 × 106 atm (b) 1.02 × 104 atm

(c) 1.02 × 105 atm (d) 1.02 × 103 atm

Solution

v 0.0098v 100

p 100v 0.0098v 100

= 100 atmp

= 1.02 × 106 atm

Hence answer is (a).

Class Exercise - 6

A metal cube of side 0.1 m has its upper face displaced by 0.2 mm when a shearing stress of 106 kg weight is applied it. The shear modulus of the cube is

(a) 4.9 × 1011 N/m2 (b) 9.8 × 1011 N/m2

(c) 4.9 × 109 N/m2 (d) 9.8 × 109 N/m2

Solution

Hence answer is (a).

6

821

F 10 9.8Stress 9.8 10 N

A 10

330.2 10

strain 2 10L 0.1

8

3

stress 9.8 10shear modulus

strain 2 10

= 4.9 × 1011 N/m2

Class Exercise - 7

A rope of 1 cm in diameter breaks if the tension in it exceeds 50 N. The maximum tension that may be given to a similar rope of diameter 2 cm is

(a) 500 N (b) 250 N

(c) 1,000 N (d) 2,000 N

Solution

Hence answer is (d).

FY

A.

YAF

2F A r 2F dia

21 1

2 2

F dF d

2 22

2 11

d 2F F 500

d 1

= 500 × 4 = 2000 N

Class Exercise - 8

What is the volume change of a solid copper cube, 40 mm on each edge, when subjected to a pressure of 2 × 107pa. Bulk modulus of copper is 1.25 × 1011 N/m2

(a) 10.24 × 102 mm3 (b) 1.024 × 102 mm3

(c) 10.24 mm3 (d) 102. 4 × 102 mm3

Solution

Hence answer is (c).

7p 2 10 40 40 40v vv

73

11

2 10 40 40 40v mm

1.25 10

= 10.24 mm3

Class Exercise - 9

An aluminum rod, Young’s modulus 7 × 109 N/m2 has a breaking strain of 0.2%. The minimum cross-sectional area of rod in m2 to support a load of 104 N is

(a) 1 × 10–2 (b) 1 × 10–3

(c) 1.4 × 10–2 (d) 7.1 × 10–4

Solution

Y = 7 × 109 N/m2

0.2Strain

100

F = 104 N

410stress AY

0.2strain100

44 2

9

10 100A 7.1 10 m

0.2 7 10

Hence answer is (d).

Class Exercise - 10

Two wires of same material and length but diameters in the ratio 1 : 2 are stretched by the same force. The potential energy per unit volume of two wires will be in the ratio

(a) 16 : 1 (b) 4 : 1

(c) 1 : 4 (d) 1 : 1

Solution

Hence answer is (a).

1Energy Stress Strain

2

1 Stress

Stress2 Y

21 (Stress)

2 Y

21 1 F

2 Y A

2 42

1 1E

rr

4 41 2

2 1

E r 216 : 1

E r 1

Thank you