Question

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

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

Answer 1

The correct option is A 0.1 J

Work done in stretching wire

$=\frac{1}{2}\frac{YA{L}^2}{L}=\frac{1}{2}F.l$

Where

$L=$ length of wire

$l=$ increase in length

We know that

$E=\frac{1}{2}\times F\times AL$

$=\frac{1}{2}\times 200\times {10}^{-3}$

$=0.1J$

Hence $\left(A\right)$ option is correct.