The amount of elastic potential energy per unit volume (in SI unit) of a steel wire of length 100 ~cm to stretch it by 1 ~mm is (if Young's modulus of the wire =2.0 × 10^ 11 ~N ~m^ -2 )

Correct answer: B. 10^5 We know that, elastic potential energy per unit volume =(1)/(2) × Stress × Strain Also, Y= Stress Strain Elastic potential energy per unit volume =(1)/(2) × Young's modulus ×( Strain )^2 aligned & =(1)/(2) × 2 × 10^ 11 ×[(1)/(1000)]^2 & =(1)/(2) × 2 × 10^ 11 × 10^ -6 =10^5( J / m^3) aligned

NEET 2023

Exam: NEET
Year: 2023
Subject: Physics
Chapter: Mechanical Properties of Solids

NEET 2023 · Physics · Mechanical Properties of Solids

The amount of elastic potential energy per unit volume (in SI unit)...

The amount of elastic potential energy per unit volume (in SI unit) of a steel wire of length

100 \mathrm{~cm}

to stretch it by

1 \mathrm{~mm}

is (if Young's modulus of the wire

=2.0 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}

)

Options

A $10^7$
B $10^5$
C $10^{11}$
D $10^{17}$

Correct answer

B. $10^5$

Step-by-step solution

We know that, elastic potential energy per unit volume

=\frac{1}{2} \times

Stress

\times

Strain Also,

Y=\frac{\text{ Stress }}{\text{ Strain }} \Rightarrow \quad

Elastic potential energy per unit volume

=\frac{1}{2} \times

Young's modulus

\times(\text{ Strain })^2

\begin{aligned} & =\frac{1}{2} \times 2 \times 10^{11} \times\left[\frac{1}{1000}\right]^2 \\ & =\frac{1}{2} \times 2 \times 10^{11} \times 10^{-6}=10^5\left(\mathrm{ J} / \mathrm{m}^3\right)\end{aligned}

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