A simple pendilum is swinging in a vertical plane. The ratio of its potential energies when it is making angles 30⁰ and 60⁰ with the vertical is,

If the pendulum is making an angle $\mathrm{\theta }$ with vertical.

height of the bob is given by $\mathrm{h}=\mathrm{l}-\mathrm{lcos\theta }$

$\mathrm{Potential} \mathrm{energy}=\mathrm{mgh}=\mathrm{mgl}(1-\mathrm{cos\theta })$

$\frac{P{E}_1}{P{E}_2}=\frac{mgl(1-\cos {30}^{°})}{mgl(1-\cos {60}^{°})}=\frac{1-{\displaystyle \frac{\sqrt{3}}{2}}}{{\displaystyle \frac{1}{2}}}=\frac{2-\sqrt{3}}{1}$

Hence the correct answer is $(2-\sqrt{3}):1$.