Moment of Inertia & Radius of Gyration question from JEE Main 2017

Answer: (b)

1. $I = m\left(\frac{R^2}{4} + \frac{l^2}{12}\right)$ with fixed volume $V = \pi R^2 l$, so $R^2 = \frac{V}{\pi l}$.
2. $I = m\left(\frac{V}{4\pi l} + \frac{l^2}{12}\right)$; $\frac{dI}{dl} = 0 \Rightarrow \frac{V}{4\pi l^2} = \frac{l}{6}$
3. $l^3 = \frac{3V}{2\pi} = \frac{3R^2 l}{2} \Rightarrow l^2 = \frac{3}{2}R^2$
4. $\frac{l}{R} = \sqrt{\frac{3}{2}}$, so the answer is (b).