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JEE Main 20265 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
5 April
Shift
Evening Shift
Subject
Physics
Chapter
Mechanical Properties of Solids

JEE Main 2026 · 5 April · Evening Shift · Physics · Mechanical Properties of Solids

A copper wire of length 3 m is stretched by 3 mm by applying an...

A copper wire of length 3 m3\text{ m} is stretched by 3 mm3\text{ mm} by applying an external force. The volume of the wire is 600×10−6 m3600 \times 10^{-6}\text{ m}^3. The elastic potential energy stored in the wire in stretched condition would be _______ J.
(Given Young modulus of copper =1.1×1011 N/m2= 1.1 \times 10^{11}\text{ N/m}^2)

Correct answer

0

Step-by-step solution

Given parameters:
L=3 mL = 3\text{ m}
ΔL=3×10−3 m\Delta L = 3 \times 10^{-3}\text{ m}
V=600×10−6 m3V = 600 \times 10^{-6}\text{ m}^3
Y=1.1×1011 N/m2Y = 1.1 \times 10^{11}\text{ N/m}^2

Strain is given by:
Strain=ΔLL=3×10−33=10−3\text{Strain} = \dfrac{\Delta L}{L} = \dfrac{3 \times 10^{-3}}{3} = 10^{-3}

Elastic potential energy stored in the wire is:
U=12×Stress×Strain×VolumeU = \dfrac{1}{2} \times \text{Stress} \times \text{Strain} \times \text{Volume}

Since Stress=Y×Strain\text{Stress} = Y \times \text{Strain}:
U=12×Y×(Strain)2×VU = \dfrac{1}{2} \times Y \times (\text{Strain})^2 \times V

Substituting the values:
U=12×(1.1×1011)×(10−3)2×(600×10−6)U = \dfrac{1}{2} \times (1.1 \times 10^{11}) \times (10^{-3})^2 \times (600 \times 10^{-6})

U=12×1.1×1011×10−6×600×10−6U = \dfrac{1}{2} \times 1.1 \times 10^{11} \times 10^{-6} \times 600 \times 10^{-6}

U=12×1.1×600×10−1U = \dfrac{1}{2} \times 1.1 \times 600 \times 10^{-1}

U=12×66=33 JU = \dfrac{1}{2} \times 66 = 33\text{ J}

Answer: 3333
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