Question

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

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

Answer 1

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

$W=200\text{N},$
Elongation, $x={10}^{-3}\text{m}$
We know that,
Elastic energy, $U=\frac{1}{2}Fx$
$\Rightarrow U=\frac{1}{2}\times 200\times {10}^{-3}=0.1\text{J}$