Two springs of spring constants 1500 N /m and 3000 N /m respectively are stretched with the same force. They will have potential energy in the ratio

The correct option is 2:1.

Step-by-step derivation:

The potential energy ($U$) stored in a stretched spring with spring constant $k$ and extension $x$ is given by the formula:
$U=\frac{1}{2}k{x}^2$

According to Hooke's Law, the restoring force $F$ applied to stretch a spring is related to its extension by:
$F=kx$
This can be rewritten to express extension in terms of force and spring constant:
$x=\frac{F}{k}$

Substituting the value of $x$ into the potential energy formula gives:
$U=\frac{1}{2}k{\left(\frac{F}{k}\right)}^2=\frac{F^2}{2k}$

Since both springs are stretched with the same force ($F$ is constant), the potential energy is inversely proportional to the spring constant:
$U\propto \frac{1}{k}$

Therefore, the ratio of their potential energies is:
$\frac{U_1}{U_2}=\frac{k_2}{k_1}$

Given the spring constants:
$k_1=1500\text{N/m}$
$k_2=3000\text{N/m}$

Substitute these values into the ratio:
$\frac{U_1}{U_2}=\frac{3000}{1500}=\frac{2}{1}$

Thus, the ratio of their potential energies is 2:1.