When a force is applied on a wire of uniform cross-sectional area $3 \times {10^{ - 6}}\,{m^2}$ and length $4m$, the increase in length is $1\, mm.$ Energy stored in it will be $(Y = 2 \times {10^{11}}\,N/{m^2})$
Medium
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(c) $U = \frac{1}{2} \times \frac{{YA{l^2}}}{L}$$ = \frac{1}{2} \times \frac{{2 \times {{10}^{11}} \times 3 \times {{10}^{ - 6}} \times {{(1 \times {{10}^{ - 3}})}^2}}}{4}$
$ = 0.075\;J$
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