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

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11200

5600

8400

2800

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{Energy}{Vol} = \frac{1}{2} \times stress \times strain \\ = \frac{1}{2} Y {strain}^{2} \\ = \frac{1}{2} \times 7 \times 10^{10} \times {(\frac{0.04}{100})}^{2} \\ = 5600 J / m^{3} \end{array}$

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