Two springs of spring constants 1500 N/m and 3000 N/m respectively are stretched with the same force. The ratio of potential energies will be

The correct option is C. 2 : 1

The potential energy of spring $U=\frac{1}{2}k{x}^2$

where k = spring constant

x = elongation/compression of the spring

Also spring force $F=kx\Rightarrow x=\frac{F}{k}$

Therefore, $U=\frac{1}{2}k(\frac{F}{k}{)}^2=\frac{1}{2}\frac{F^2}{k}$

For the same force, $U\propto \frac{1}{k}$

$\Rightarrow \frac{U_1}{U_2}=\frac{k_2}{k_1}=\frac{3000}{1500}=2:1$.