The length of a rod is $20\, cm$ and area of cross-section $2\,c{m^2}$. The Young's modulus of the material of wire is $1.4 \times {10^{11}}\,N/{m^2}$. If the rod is compressed by $5\, kg-wt$ along its length, then increase in the energy of the rod in joules will be
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(a) Energy = $\frac{1}{2}Fl = \frac{1}{2} \times F \times \left( {\frac{{FL}}{{AY}}} \right) = \frac{1}{2} \times \frac{{{F^2}L}}{{AY}}$
$ = \frac{1}{2} \times \frac{{{{(50)}^2} \times 20 \times {{10}^{ - 2}}}}{{2 \times {{10}^{ - 4}} \times 1.4 \times {{10}^{11}}}}$$ = 8.57 \times {10^{ - 6}}J$
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