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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. 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

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 A B C stressstress s t r a i n

Young’s Modulus A L L+L FF

Shear Modulus xx h A F -F

V Bulk Modulus F F 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 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)YX 2 (b) 2 YX 2 (c) 0.5Y 2 X(d)0.5 YX 2 Illustrative Problem

Potential energy stored = ½ x strain x stress Stress = Strain x Modulus = XY PE = 0.5 YX 2 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). 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 Let the initial and final volumes are v and v + dv. Hence answer is (d).

Class Exercise - 4 The length of a metal wire is when the tension in it is T 1 and is when tension is T 2. The unstrectched length of the wire is

Solution Hence answer is (c). Let the original length

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 × 10 6 atm(b) 1.02 × 10 4 atm (c) 1.02 × 105 atm(d) 1.02 × 10 3 atm

Solution = 100 atm = 1.02 × 10 6 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 10 6 kg weight is applied it. The shear modulus of the cube is (a) 4.9 × 10 11 N/m 2 (b) 9.8 × 10 11 N/m 2 (c) 4.9 × 10 9 N/m 2 (d) 9.8 × 10 9 N/m 2

Solution Hence answer is (a). = 4.9 × 10 11 N/m 2

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). = 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 × 10 7 pa. Bulk modulus of copper is 1.25 × 10 11 N/m 2 (a) 10.24 × 10 2 mm 3 (b) 1.024 × 10 2 mm 3 (c) 10.24 mm 3 (d) 102. 4 × 10 2 mm 3

Solution Hence answer is (c). = 10.24 mm 3

Class Exercise - 9 An aluminum rod, Young’s modulus 7 × 10 9 N/m 2 has a breaking strain of 0.2%. The minimum cross-sectional area of rod in m 2 to support a load of 10 4 N is (a) 1 × 10 –2 (b) 1 × 10 –3 (c) 1.4 × 10 –2 (d) 7.1 × 10 –4

Solution Y = 7 × 10 9 N/m 2 F = 10 4 N 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).

Thank you

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