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Physics · Mechanical Properties of Matter

JEE Main 2026 — 5 April, Evening Shift — Question 20

A copper wire of length 3m3\mathrm{m} is stretched by 3mm3\mathrm{mm} by applying an external force. The volume of the wire is 600×10−6m3600\times 10^{-6}\mathrm{m}^3. The elastic potential energy stored in the wire in stretched condition would be J. (Given Young modulus of copper = 1.1×1011N/m21.1\times 10^{11}\mathrm{N/m}^2)

Answer: 33

Numerical answer — enter this value.

Step-by-step solution

U=12Y(strain)2×volume=12×1.1×1011×(3×10−33)2×600×10−6=33U = \frac12 Y (\text{strain})^2 \times \text{volume} = \frac12 \times 1.1\times10^{11} \times \left(\frac{3\times10^{-3}}{3}\right)^2 \times 600\times10^{-6} = 33 J.

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Exam
JEE Main 2026
Paper
5 April, Evening Shift
Subject
Physics
Chapter
Mechanical Properties of Matter
Topic
Longitudinal Strain and Elastic Potential Energy

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