A long spring is stretched by \(2 \mathrm{~cm}\) and its potential energy is \(U\). If the spring is stretched by \(10 \mathrm{~cm}\); its potential energy will be

The potential energy of a stretched spring is
\(U=\frac{1}{2} k x^{2}\)
Here, \(k=\) spring constant, \(x=\) elongation in spring. But given that, the elongation is \(2 \mathrm{~cm}\).
So, \(U=\frac{1}{2} k(2)^{2}\)
\(\Rightarrow \quad U=\frac{1}{2} k \times 4 ...(i)\)
If elongation is \(10 \mathrm{~cm}\) then potential energy
\(\begin{aligned}U^{\prime} &=\frac{1}{2} k(10)^{2} \\U^{\prime} &=\frac{1}{2} k \times 100 ...(ii)\end{aligned}\)
On dividing Eq. (ii) by Eq. (i), we have
\(\frac{U^{\prime}}{U}=\frac{\frac{1}{2} k \times 100}{\frac{1}{2} k \times 4}\)
or \(\frac{U^{\prime}}{U}=25 \Rightarrow U^{\prime}=25 U\)