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?

\(\begin{array}{l}
I = \frac{{m{R^2}}}{4} + \frac{{m{l^2}}}{{12}}\\
I = \frac{m}{4}\left[ {{R^2} + \frac{{{l^2}}}{3}} \right]\\
Let\,V\, = Volume\,of\,cylinder\, = \,\pi {R^2}l\\
\,\, = \frac{m}{4}\left[ {\frac{V}{{\pi l}} + \frac{{{l^2}}}{3}} \right] \Rightarrow \frac{{dI}}{{dl}} = \frac{m}{4}\left[ {\frac{{ - V}}{{\pi {l^2}}} + \frac{{2l}}{3}} \right] = 0
\end{array}\)

\(\begin{array}{l}
\frac{V}{{\pi {l^2}}} = \frac{{2l}}{3} \Rightarrow V = \frac{{2\pi {l^3}}}{3}\\
\pi {R^2}l = \frac{{2\pi {l^3}}}{3} \Rightarrow \frac{{{l^2}}}{{{R^2}}} = \frac{3}{2}\,or,\,\frac{l}{R} = \sqrt {\frac{3}{2}} 
\end{array}\)