Q 11-06-240JEE MainJEE Main 2017 (2 Apr)Hard
The moment of inertia of a uniform cylinder of length $l$ and radius $R$ about its perpendicular bisector is $I$. What is the ratio $l/R$ such that the moment of inertia is minimum?
Answer: (B) $\sqrt{\dfrac32}$
For a fixed mass $m$ (fixed volume $V = \pi R^2 l$ of the same material),
$$I = m\left(\frac{R^2}{4} + \frac{l^2}{12}\right),\qquad R^2 = \frac{V}{\pi l}$$
$$I = m\left(\frac{V}{4\pi l} + \frac{l^2}{12}\right)$$
For a minimum,
$$\frac{dI}{dl} = m\left(-\frac{V}{4\pi l^2} + \frac{l}{6}\right) = 0 \;\Rightarrow\; \frac{V}{\pi} = \frac{2l^3}{3}$$
Putting back $V/\pi = R^2 l$: $R^2 l = \dfrac{2l^3}{3}$, so
$$\frac{l^2}{R^2} = \frac32 \;\Rightarrow\; \frac{l}{R} = \sqrt{\frac32}$$
Solution by Sreeraj P, M.Sc Physics