A steel wire of length 2 m and $1.2\times {10}^{-7}{\mathrm{m}}^2$ in cross-sectional area is stretched by a force of 36 N. Calculate the work done $\left(\text{ in }\times {10}^{-2}\mathrm{J}\right)$ stretching the wire $\left(\mathrm{Y}=1.8\times {10}^{11}\mathrm{N}/{\mathrm{m}}^2\right)$.

The stress in steel wire $\mathrm{\sigma }=\frac{\mathrm{F}}{\mathrm{A}}=\frac{36}{1.2\times {10}^{-7}}=3\times {10}^8\mathrm{N}/{\mathrm{m}}^2$

As Young's modulus of elasticity, $Y=\frac{\text{ Stress }}{\text{ Strain }}$

Strain $=\frac{\text{ stress }}{\mathrm{Y}}=\frac{3\times {10}^8}{1.8\times {10}^{11}}=\frac{5}{3}\times {10}^{-3}$

Work done $=\frac{1}{2}\left(\text{ stress }\right)\left(\text{ strain }\right)\text{ volume }$

                  $=\frac{1}{2}\times 3\times {10}^8\times \frac{5}{3}\times {10}^{-3}\left(2\times 1.2\times {10}^{-7}\right)=6\times {10}^{-2}\text{ Joule. }$