Moment of inertia of a cylinder… - JEE Main 2020 Physics PYQ

JEE Main 2020 — Physics Mechanics

2020mcqhard

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 $I=M(\frac{{R}^{2}}{4}+\frac{{L}^{2}}{12}).$ If such a cylinder is to be made for a given mass of a material, the ratio $\frac{L}{R}$ for it to have minimum possible $I$ is:

Official previous-year question

Held on 3 Sept 2020 · Verified 6 Jul 2026.

Options

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

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Solution

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

JEE Main 2020 Physics, Rotational Motion — solution figure

$\mathrm{dl}=\frac{{\mathrm{dmR}}^{2}}{4}+{\mathrm{dmx}}^{2}$

$I=\int \mathrm{dl}=\int \frac{{\mathrm{dmR}}^{2}}{4}+{\int }_{n=L/2}^{n=-L/2}\frac{M}{L}\mathrm{dx}\times {x}^{2}$

$\Rightarrow I=\frac{{\mathrm{MR}}^{2}}{4}+\frac{{\mathrm{ML}}^{2}}{12}$

$\Rightarrow I=\frac{M}{4}\times \frac{V}{\pi L}+\frac{{\mathrm{ML}}^{2}}{12}$

$\frac{\mathrm{dI}}{\mathrm{dL}}=-\frac{mV}{4{\pi L}^{2}}+\frac{M\times 2L}{12}=0$

$\Rightarrow V=\frac{2}{3}{\pi L}^{3} \Rightarrow {\pi R}^{2}L=\frac{2}{3}{\pi L}^{3} \Rightarrow \frac{L}{R}=\sqrt{\frac{3}{2}}$ $(I=\frac{\mathrm{MV}}{4\pi L}+\frac{{\mathrm{ML}}^{2}}{12})$

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Solution

JEE Main 2020 Physics, Rotational Motion — solution figure

$\mathrm{dl}=\frac{{\mathrm{dmR}}^{2}}{4}+{\mathrm{dmx}}^{2}$

$I=\int \mathrm{dl}=\int \frac{{\mathrm{dmR}}^{2}}{4}+{\int }_{n=L/2}^{n=-L/2}\frac{M}{L}\mathrm{dx}\times {x}^{2}$

$\Rightarrow I=\frac{{\mathrm{MR}}^{2}}{4}+\frac{{\mathrm{ML}}^{2}}{12}$

$\Rightarrow I=\frac{M}{4}\times \frac{V}{\pi L}+\frac{{\mathrm{ML}}^{2}}{12}$

$\frac{\mathrm{dI}}{\mathrm{dL}}=-\frac{mV}{4{\pi L}^{2}}+\frac{M\times 2L}{12}=0$

$\Rightarrow V=\frac{2}{3}{\pi L}^{3} \Rightarrow {\pi R}^{2}L=\frac{2}{3}{\pi L}^{3} \Rightarrow \frac{L}{R}=\sqrt{\frac{3}{2}}$ $(I=\frac{\mathrm{MV}}{4\pi L}+\frac{{\mathrm{ML}}^{2}}{12})$

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Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics