$=\frac{1}{2} \times \mathrm{Y} \times(\alpha \Delta \theta)^{2}$
$=\frac{1}{2} \times 10^{11} \times\left(12 \times 10^{-6} \times 20\right)^{2}=2880 \mathrm{J} / \mathrm{m}^{3}$
[Area of cross section of wire $=0.005 \mathrm{~cm}^2$, $\mathrm{Y}=2 \times 10^{11}\ \mathrm{Nm}^{-2}$ and $\left.\mathrm{g}=10 \mathrm{~ms}^{-2}\right]$


[Young's modulus for copper and steel are $1 \times 10^{11} N / m ^2$ and $2 \times 10^{11} N / m ^2$ respectively]
