Q115 / 200 · NEET 2023 (Manipur) · Physics

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

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

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

Hence, the correct answer is option: 'B'.

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