Physics · Mechanical Properties of Matter

JEE Main 2025 — 2 April, Morning Shift — Question 69

A steel wire of length 2 m and Young's modulus $2.0 \times 10^{11} \mathrm{Nm}^{-2}$ is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and $10^{-3}$ respectively, then the elastic potential energy density of the wire is $\qquad$ $\times 10^{5}$ (in SI units).

Answer: 25

Numerical answer — enter this value.

Step-by-step solution

$U=\frac{1}{2} \times y \times(\text { longitudinal strain })^{2}$

$\sigma=\left|\frac{\text { transverse strain }}{\text { Iongitudinal strain }}\right|$

Longitudinal strain $=\frac{10^{-3}}{0.2}=5 \times 10^{-3}$

$U=\frac{1}{2} \times 2 \times 10^{11} \times 25 \times 10^{-6}$

$=25 \times 10^{5} \mathrm{~J} / \mathrm{m}^{3}$

Answer key and solution verified before publishing.

Practise Mechanical Properties of Matter

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 2 April, Morning Shift
Subject: Physics
Chapter: Mechanical Properties of Matter
Topic: Longitudinal Strain and Elastic Potential Energy