Question 70 of 280

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:

A$\dfrac{2}{3}$B$\dfrac{3}{2}$C$\sqrt{\dfrac{3}{2}}$D$\sqrt{\dfrac{2}{3}}$

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Answer: (C) $\sqrt{\dfrac{3}{2}}$

Let a cylinder of mass ${m}_{1}$, length $L$ and radius $R$ then take elementary disc of radius $R$ and trickiness $dx$ at distance of $x$ from axis $OO'$ then moment of inertia about $OO'$ as this element

$dl=\dfrac{{dmR}^{2}}{4}+{dmx}^{2}$
$I=\int dl=\int \dfrac{{dmR}^{2}}{4}+{\int }_{n=L/2}^{n=-L/2}\dfrac{M}{L}dx\times {x}^{2}$
$\Rightarrow I=\dfrac{{MR}^{2}}{4}+\dfrac{{ML}^{2}}{12}$
$\Rightarrow I=\dfrac{M}{4}\times \dfrac{V}{\pi L}+\dfrac{{ML}^{2}}{12}$
$\dfrac{dI}{dL}=-\dfrac{mV}{4{\pi L}^{2}}+\dfrac{M\times 2L}{12}=0$
$\Rightarrow V=\dfrac{2}{3}{\pi L}^{3} \Rightarrow {\pi R}^{2}L=\dfrac{2}{3}{\pi L}^{3} \Rightarrow \dfrac{L}{R}=\sqrt{\dfrac{3}{2}}$$\left(I=\dfrac{MV}{4\pi L}+\dfrac{{ML}^{2}}{12}\right)$