Answer:
Option D
Explanation:
MI of a solid cylinder about its perpendicular bisector of length is
$I= m(\frac{l^{2}}{12}+\frac{R^{2}}{4})$
$\Rightarrow I=\frac{mR^{2}}{4}+\frac{ml^{2}}{12}=\frac{m^{2}}{4\pi\rho l}+\frac{ml^{2}}{12}$
[ $\therefore \rho\pi R^{2}l=m$ ]
For l to be maximum,
$\frac{\text{d}I}{\text{dl}}=-\frac{m^{2}}{4\pi\rho}(\frac{1}{l^{2}})+\frac{ml}{6}=0$
$\Rightarrow \frac{m^{2}}{4\pi\rho}=\frac{ml^{3}}{6}=0$
Now, putting $m=\rho \pi R^{2}l$
$\therefore l^{3}=\frac{3}{2\pi\rho}.\rho\pi R^{2}l$
$\frac{l^{2}}{R^{2}}=\frac{3}{2}$
$\frac{l}{R}=\sqrt{\frac{3}{2}}$
