$=\frac{1}{2} \times(\text { Force }) \times(\Delta l)$

$\Delta l_{1}=\frac{\mathrm{FL}_{1}}{\mathrm{Y}_{\mathrm{cu}} \mathrm{A}}=0.8 \mathrm{mm} \quad \& \mathrm{\Delta l}_{2}=\frac{\mathrm{FL}_{2}}{\mathrm{Y}_{\mathrm{steel}} \mathrm{A}}=0.2 \mathrm{mm}$

Net $\Delta l=\Delta l_{1}+\Delta l_{2}=1 \mathrm{mm}=10^{-3} \mathrm{m}$

Elastic potential energy $=\frac{1}{2} \times 500 \times 10^{-3}=?$

$=\frac{1}{4} J$