Explanation

(d) The potential energy of a stretched spring is
$U = \frac{1}{2}kx^{2}$
Here, k=spring constant, x=elongation in spring.
But given that, the elongation is 2 cm.
So $U = \frac{1}{2}K(2)^{2}$
Or $U = \frac{1}{2}k \times 4$ ...(i)
If elongation is 10 cm then potential energy
$U^{'} = \frac{1}{2}k(10)^{2}$
Or $U^{'} = \frac{1}{2}k \times 100$ …(ii)
On dividing Eq. (ii) by Eq. (i), We have
$\frac{U'}{U} = \frac{\frac{1}{2}k \times 100}{\frac{1}{2}k \times 4}$
Or $\frac{U^{'}}{U} = 25\ \ \ \Rightarrow U^{'} = \ 25U$