The Young's modulus of the material of wire is 2 × 10 ^ 10 Nm ^ - 2 . If the elongation strain is 1% , then the energy stored in the wire per unit volume in Jm ^ - 3 is

Correct answer: A. 10 ^ 6 Jm ^ - 3 Per unit volume energy stored = (1)/(2) × Y × (strain)2 = (1)/(2) × Y × ((l)/(L))2 given l = L × 1% or l = (L)/(100) Thus per unit volume energy stored = (1)/(2) × 2 × 10 ^ 10 × ((L)/(100L)) ^ 2 = 10 ^ 6 Jm ^ - 3

JEE Main

Exam: JEE Main
Subject: Physics
Chapter: Mechanical Properties of Solids

JEE Main · Physics · Mechanical Properties of Solids

The Young's modulus of the material of wire is 2 × 10 ^ 10 Nm ^ - 2...

The Young's modulus of the material of wire is

2 \times {10}^{10}{Nm}^{- 2} .

If the elongation strain is

\text{1%}

, then the energy stored in the wire per unit volume in

{Jm}^{- 3}

Options

A ${10}^{6}{Jm}^{- 3}$
B ${10}^{8}{Jm}^{- 3}$
C ${2 \times 10}^{6}{Jm}^{- 3}$
D ${2 \times 10}^{8}{Jm}^{- 3}$

Correct answer

A. ${10}^{6}{Jm}^{- 3}$

Step-by-step solution

Per unit volume energy stored

= \dfrac{1}{2} \times Y \times \left(strain\right)2 = \dfrac{1}{2} \times Y \times \left(\dfrac{l}{L}\right)2

given

l = L \times 1%

l = \dfrac{L}{100}

Thus per unit volume energy stored

= \dfrac{1}{2} \times 2 \times {10}^{10} \times {\left(\dfrac{L}{100L}\right)}^{2}

= {10}^{6}{Jm}^{- 3}

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