Question

A wire suspended vertically from one of its ends stretched by attaching a weight of $200N$ to the lower end. The weight stretches the wire by $1mm$. Then the elastic energy stored in the wire is:

• A. $0.2J$
• B. $10J$
• C. $20J$
• D. $0.1J$

Answer 1

The correct option is D $0.1J$

Given : $F=200N$ $\Delta L=1mm=0.001$ m

Elastic energy $E=\frac{1}{2}\times Stress\times Strain\times Volume$

$\therefore$ $E=\frac{1}{2}\times \frac{F}{A}\times \frac{\Delta L}{L}\times AL=\frac{F\Delta L}{2}$

OR $E=\frac{200\times 0.001}{2}=0.1$ J