Potential energy per unit volume \(=\frac{1}{2} \times\) stress \(\times\) strain

\(=\frac{1}{2} \times \frac{F}{A} \times \frac{\Delta L}{L}\)

So, Potential energy \(=\) potential energy per unit volume \(\times\) volume

\(=\frac{1}{2} \times \frac{F \cdot \Delta L}{A \cdot L} \times A \cdot L\)   \(\{\) Volume \(=\) Length \(\times\) cross-sectional area \(\}\)

\(\Delta U=\frac{1}{2} \cdot F \cdot \Delta L\)  \(\left\{\begin{array}{l}F=10 \times 10 \,N \\ \Delta L=10 \,mm =10 \times 10^{-3} \,m \end{array}\right.\)

Substituting values

\(\Delta U=\frac{1}{2} \times 100 \times \frac{10}{1000}\)

\(\Delta U=0.5 \,J\)