Young modulus of elasticity of brass is $10^{11} N / m^{2}$ . The increase in its energy on pressing a rod of length $0.2 m$ and cross–sectional area $4 c m^{2}$ made of brass with a force of $40 N$ along its length, will be $(i n μ J)$

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Detailed Solution

Energy $U = \frac{1}{2 Y} {(s t r e s s)}^{2} v o l u m e$

Stress $= \frac{F}{A} = \frac{40}{4 \times 10^{- 4}} = 10^{5} N / m^{2}$

Volume $= A l$

$U = \frac{1}{2 x 10^{11}} \times {(10^{5})}^{2} \times 4 \times 10^{- 4} \times 0.2 = 4 μ J$

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