Question

A wire suspended vertically is stretched by a $20kgf$. Applied to its free end. The increase in length of the wire is $2mm$. The energy stored in the wire is $(g=10m{s}^{-2})$:

• A. $0.1J$
• B. $0.2J$
• C. $0.4J$
• D. $5J$

Answer 1

The correct option is B $0.2J$

So that,

$\begin{array}{c}u/volume=\frac{1}{2}\times stress\times strain\\ \frac{u}{volume}=\frac{1}{2}\times \frac{200}{A}\times \frac{\Delta l}{l}\\ u=\frac{1}{2}\times 200\times \frac{2}{1000}\\ u=0.2j\end{array}$

So,

Option $B$ is correct answer.