Ans.

Here, the mass of the cylinder is M, radius is a and length is l. If \( \rho \) be the density of material of the cylinder, then the mass of the cylinder is \( M=\pi{a^2}l\rho \). or, \( a^2l=constant=k \)(say)

The moment of inertia of this cylinder about an axis passing through the centre and perpendicular to the length of the cylinder is

\( \displaystyle{I=M\left(\frac{a^2}{4}+\frac{l^2}{12}\right)} \)
[To know the derivation of the moment of inertia of the cylinder about an axis passing through the centre and perpendicular to its length (CLICK HERE) ]

or, \( \displaystyle{I=M\left(\frac{k}{4l}+\frac{l^2}{12}\right)=f(t)} \)

For, \( I \) to be a minimum,

\( \frac{dI}{dl}=0 \).

or, \( M\left[-\frac{k}{4l^2}+\frac{l}{6}\right]=0 \)

or, \( \displaystyle{\frac{k}{4l^2}=\frac{l}{6}} \)

or, \( \displaystyle{\frac{l}{a}=\frac{\sqrt{3}}{\sqrt{2}}} \)