The length of a wire is $1.0\, m$ and the area of cross-section is $1.0 \times {10^{ - 2}}\,c{m^2}$. If the work done for increase in length by $0.2\, cm$ is $0.4\, joule$, then Young's modulus of the material of the wire is
Medium
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(c) $W = \frac{1}{2}\frac{{YA{l^2}}}{L}$ $⇒$ $0.4 = \frac{1}{2} \times \frac{{Y \times {1^{ - 6}} \times {{(0.2 \times {{10}^{ - 2}})}^2}}}{1}$
$ Y$$ = 2 \times {10^{11}}N/{m^2}$
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