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

as mass is constant \(\Rightarrow m =\rho V =\) constant

\(V = constant\)

\(\pi^{2} R l=\) constant \(\Rightarrow R ^{2} L =\) constant

\(2 RL + R ^{2} \frac{ dL }{ dR }=0\)\(...(2)\)

From equation \((1)\)

\(\frac{ dI }{ clR }= M \left(\frac{2 R }{4}+\frac{2 L }{12} \times \frac{ dI }{ dr }\right)=0\)

\(\frac{ R }{2}+\frac{ L }{6} \frac{ dI }{ dR }=0\)

Substituting value of \(\frac{ dI }{ d R }\) from eqution \((2)\)

\(\frac{ R }{2}+\frac{ L }{6}\left(\frac{-2 L }{ R }\right)=0\)

\(\frac{ R }{2}=\frac{ L^{2}}{3 R } \Rightarrow \frac{ L }{ R }=\sqrt{\frac{3}{2}}\)