Q 11-08-017NEETJEE MainMedium
A load of $200$ N stretches a wire by $2$ mm, within the elastic limit. The elastic potential energy stored in the wire is
Answer: (A) $0.2$ J
The force grows from zero to $200$ N as the wire stretches, so the average force is half the final value:
$U = \dfrac{1}{2}F\Delta L = \dfrac{1}{2} \times 200 \times 2 \times 10^{-3} = 0.2$ J.
Solution by Sreeraj P, M.Sc Physics