Question

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 (take $\left. g=10\, m / s ^{2}\right)$

• A. 0.5 J
• B. 5 J
• C. 50 J
• D. 500 J

Answer 1

Answer: (A)

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 }
$\Delta U=\frac{1}{2} \cdot F \cdot \Delta L$
$F=10 \times 10 \,N$
$\Delta L=10\, mm =10 \times 10^{-3} m$
Substituting values
$\Delta U=\frac{1}{2} \times 100 \times \frac{10}{1000} $
$\Delta U=0.5\, J$