Explanation

(d) Moment of inertia of the cylinder about an axis perpendicular to the axis of the cylinder and passing through the centre is
$I = M\left( \frac{R^{2}}{4} + \frac{L^{2}}{12} \right)$ …(i)
If $\rho$ is volume density of the cylinder, then
$M = \left( \pi R^{2}L \right)\rho =$constant …(ii)
$\therefore\ \ L = \frac{M}{\pi R^{2}\rho}$
Put in Eq. (i),
$I = M\left( \frac{R^{2}}{4} + \frac{M^{2}}{12\pi^{2}R^{4}\rho^{4}} \right)$
For $I$ to be minimum, $\frac{\partial I}{\partial R} = 0$
$\frac{\partial I}{\partial R} = M\left( \frac{R}{2} - \frac{M^{2}}{3\pi^{2}R^{5}} \right) = 0$
$\frac{R}{2} = \frac{M^{2}}{3\pi^{2}\rho^{2}R^{5}}\ \ or\ \ \ R^{6} = \frac{2M^{2}}{3\pi^{2}\rho^{2}}$
Using Eq. (ii), $R^{6} = \frac{2\pi^{2}R^{4}L^{2}\rho^{2}}{3\pi^{2}\rho^{2}}$
or $R^{2} = \frac{2}{3}L^{2}\ \ or\ \ \frac{L^{2}}{R^{2}} = \frac{3}{2}$
or $\frac{L}{R} = \sqrt{3/2}$