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JEE Main 2023 (8 Apr, Shift 1) · Elastic Potential Energy

Elastic Potential Energy question from JEE Main 2023 (8 Apr, Shift 1)

Q4JEE Main 2023 · 8 APR · SHIFT 1Easy

Elastic Potential Energy

An aluminium rod with Young's modulus $Y = 7.0 \times 10^{10}\text{ N/m}^2$ undergoes elastic strain of $0.04\%$ . The energy per unit volume stored in the rod in SI unit is:

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Answer: (a)

$u = \dfrac{1}{2}Y\varepsilon^2$ with $\varepsilon = 0.04\% = 4 \times 10^{-4}$ .
$u = \dfrac{1}{2} \times 7 \times 10^{10} \times 16 \times 10^{-8} = 5600\text{ J/m}^3$
So the answer is (a) 5600.

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