A brass rod of cross-sectional area $1\,c{m^2}$ and length $0.2\, m$ is compressed lengthwise by a weight of $5\, kg$. If Young's modulus of elasticity of brass is $1 \times {10^{11}}\,N/{m^2}$ and $g = 10\,m/{\sec ^2}$, then increase in the energy of the rod will be
Medium
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(b) $U = \frac{1}{2} \times \frac{{{{{\rm{(stress)}}}^{\rm{2}}}}}{Y} \times {\rm{volume}}$$=$ $\frac{1}{2} \times \frac{{{F^2} \times A \times L}}{{{A^2} \times Y}}$
$=$$\frac{1}{2} \times \frac{{{F^2}L}}{{AY}} = \frac{1}{2} \times \frac{{{{(50)}^2} \times 0.2}}{{1 \times {{10}^{ - 4}} \times 1 \times {{10}^{11}}}}$$=$ $2.5 \times {10^{ - 5}}J$
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