# Moment of inertia of a cylinder of mass M and 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 make for given mass of a material, the ratio $\frac{L}{R}$ for it to have minimum possible I is Option: 1   Option: 2 Option: 3   Option: 4

$I=m\left[\frac{R^{2}}{4}+\frac{L^{2}}{12}\right]$

Given,

$M = \rho (\pi R^2 L)$

So,

$I=M\left[\frac{R^2}{4 }+\frac{M^{2}}{12(\rho \ \pi R^2)^2)}\right] \\ \\ \text{For minimum I} -$

$\frac{dI}{dR} = 0 \\ \\ \Rightarrow M[\frac{2R}{4} + \frac{-4M^2}{12 \rho^2 \pi^2R^5}] = 0 \\ \\ R = (2/3)^{1/6} \times (\frac{M}{\rho \pi })^{1/3} \dots(1)$

$L = \frac{M}{\rho \pi (\frac{2}{3})^{1/3} (\frac{M}{\rho \pi})^{2/3}} \dots(2)$

So, (2) divided by (1) will give

$\frac{L}{R} = \sqrt{\frac{3}{2}}$

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