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A metal rod of Young's modulus 2 × 10¹⁰ Nm^–2 undergoes an elastic strain of 0.02% the energy per unit volume stored in the rod in joule/m³ is

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Detailed Solution

Y = 2 x 10¹⁰ N / m²
$\frac{∆ l}{l} = \frac{0.02}{100} = 2 x 10^{- 4}; \frac{W}{vol} = \frac{1}{2} x {(strain)}^{2} Y = \frac{1}{2} x 2 x 10^{10} x 4 x 10^{- 8} = 400 J / m^{3}$

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