QJEEM17Ph59f9: JEE Main 2017 Physics - Class 11 JEEM17Ph59f9 Solution: The moment of inertia of a uniform cylinder of length and…

Question

Verified Solution • 2025–26 Session moderate

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 \over R}

such that the moment of inertia is minimum?

Accepted Answer

The volume of the cylinder V =

\pi {R^2}l

\therefore

{R^2} = {V \over {\pi l}}

We know, moment of inertia of a uniform cylinder of length

l

and radius R about its perpendicular bisector is,

I = {{M{l^2}} \over {12}} + {{M{R^2}} \over 4}

[ Putting

{R^2} = {V \over {\pi l}}

in this equation]

\Rightarrow

I = {{M{l^2}} \over {12}} + {{MV} \over {4\pi l}}

Here

I

is a function of

l

as M and V are constant.

I

will be maximum or minimum when

{{{dI} \over {dl}}}

= 0.

\Rightarrow {{Ml} \over 6} - {{MV} \over {4\pi {l^2}}} = 0

\Rightarrow {{Ml} \over 6} = {{MV} \over {4\pi {l^2}}}

\Rightarrow {l \over 6} = {{\pi {R^2}l} \over {4\pi {l^2}}}

[ as

{V = \pi {R^2}l}

]

\Rightarrow {{{R^2}} \over {{l^2}}} = {4 \over 6}

\Rightarrow {l \over R} = \sqrt {{3 \over 2}}

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