A wire $2 \,m$ in length suspended vertically stretches by $10 \,mm$ when mass of $10 \,kg$ is attached to the lower end. The elastic potential energy gain by the wire is $...... J ($take $g=10 \,m / s ^2 )$
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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
$\Delta U=\frac{1}{2} \times 100 \times \frac{10}{1000}$
$\Delta U=0.5 \,J$
$=\frac{1}{2} \times$ 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
$\Delta U=\frac{1}{2} \times 100 \times \frac{10}{1000}$
$\Delta U=0.5 \,J$
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