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JEE Main 2026 (5 Apr, Shift 2) · Elastic Potential Energy

Elastic Potential Energy question from JEE Main 2026 (5 Apr, Shift 2)

Q1JEE Main 2026 · 5 APR · SHIFT 2Medium

Elastic Potential Energy

A copper wire of length 3 m is stretched by 3 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)

Numerical answer type. Work it out, then open the answer.

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Answer: 33

  1. U=12Yε2×VU = \dfrac{1}{2}Y\varepsilon^2 \times V, with ε=3 mm3 m=10−3\varepsilon = \dfrac{3\text{ mm}}{3\text{ m}} = 10^{-3}.
  2. 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}
  3. U=33 JU = 33\text{ J}
  4. So the answer is 33 J.
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