Question

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

• A. $0.2\text{J}$
• B. $10\text{J}$
• C. $20\text{J}$
• D. $0.1\text{J}$

Answer 1

The correct option is D $0.1\text{J}$

The elastic energy stored
$U=\frac{1}{2}\times F\times \Delta l$
here, $F=200\text{N}$ and $\Delta l=1\times {10}^{-3}\text{m}$
$=\frac{1}{2}\times 200\times {10}^{-3}J=0.1\text{J}$