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A metal rod of Young’s modulus $2 \times 10^{10} {Nm}^{- 2}$ undergoes an elastic strain of $0.02 %$ . The energy per unit volume stored in the rod in $j o u l e / m^{3}$ is

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Detailed Solution

$u = \frac{1}{2} \times Y \times ε^{2}$

$\Rightarrow u = \frac{1}{2} \times 2 \times 10^{10} \times {(0.0002)}^{2} = 400 \frac{J}{m^{3}}$

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