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An aluminum rod of Young’s modulus $Y = 7 \times 10^{10} N / m^{2}$ undergoes elastic strain of 0.04%. The energy per unit volume stored in the rod in S.I units is _________

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answer is B.

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Detailed Solution

$Y = 7 \times 10^{11} N / m^{2} S t r a i n = \frac{0.04}{100} = 4 \times 10^{- 4} \frac{E}{V} = \frac{1}{2} \times Y \times s t r a i n^{2} = \frac{1}{2} \times 7 \times 10^{11} \times 16^{8} \times 10^{- 8} = 56 \times 10^{3} = 56, 000 J / m^{3}$

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