Q 11-06-190JEE MainJEE Main 2020 (3 Sep, Shift 1)Medium
Moment of inertia of a cylinder of mass $M$, length $L$ and radius $R$ about an axis passing through its centre and perpendicular to the axis of the cylinder is $I = M\left(\dfrac{R^{2}}{4} + \dfrac{L^{2}}{12}\right)$. If such a cylinder is to be made for a given mass of a material, the ratio $\dfrac{L}{R}$ for it to have minimum possible $I$ is:
Answer: (C) $\sqrt{\dfrac32}$
Fixed mass means fixed volume $V = \pi R^{2}L$, so $R^{2} = \dfrac{V}{\pi L}$ and
$$I = M\left(\frac{V}{4\pi L} + \frac{L^{2}}{12}\right)$$
$\dfrac{dI}{dL} = M\left(-\dfrac{V}{4\pi L^{2}} + \dfrac{L}{6}\right) = 0 \Rightarrow L^{3} = \dfrac{3V}{2\pi} = \dfrac{3R^{2}L}{2}$.
So $L^{2} = \dfrac32R^{2}$ and $\dfrac LR = \sqrt{\dfrac32}$.
Solution by Sreeraj P, M.Sc Physics