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 a material, the ratio $$\frac{L}{R}$$ for it to have minimum possible $$I$$ is:

We are told that a solid cylinder of mass $$m$$, length $$L$$ and radius $$R$$ has its moment of inertia about an axis through its centre and perpendicular to the cylinder axis given by

$$I \;=\; m\left(\frac{R^{2}}{4} + \frac{L^{2}}{12}\right).$$

The problem says that a fixed mass of material is to be used. For a uniform material the mass is the product of density and volume. If $$\rho$$ is the density, then

$$m \;=\; \rho \,(\text{Volume}) \;=\; \rho\;(\pi R^{2}L).$$

Since $$m$$ and $$\rho$$ are both fixed, the product $$R^{2}L$$ must be a constant. Let us write this constant as

$$k \;=\; \frac{m}{\rho\pi}, \qquad\text{so that}\qquad R^{2}L \;=\; k.$$

Solving for the length we obtain

$$L \;=\; \frac{k}{R^{2}}.$$

We now substitute this expression for $$L$$ into the formula for the moment of inertia.

$$\begin{aligned} I &= m\left(\frac{R^{2}}{4} + \frac{L^{2}}{12}\right) \\ &= m\left(\frac{R^{2}}{4} + \frac{1}{12}\left(\frac{k}{R^{2}}\right)^{2}\right) \\ &= m\left(\frac{R^{2}}{4} + \frac{k^{2}}{12\,R^{4}}\right). \end{aligned}$$

Because the factor $$m$$ is a positive constant, minimising $$I$$ is equivalent to minimising the function

$$f(R) \;=\; \frac{R^{2}}{4} \;+\; \frac{k^{2}}{12\,R^{4}}.$$

To find the minimum we differentiate with respect to $$R$$ and set the derivative to zero.

Using the power-rule derivative $$\frac{d}{dR}(R^{n}) = nR^{n-1},$$ we have

$$\frac{df}{dR} \;=\; \frac{2R}{4} \;-\; \frac{4k^{2}}{12\,R^{5}} \;=\; \frac{R}{2} \;-\; \frac{k^{2}}{3\,R^{5}}.$$

Setting $$\dfrac{df}{dR}=0$$ gives

$$\frac{R}{2} \;=\; \frac{k^{2}}{3\,R^{5}}.$$ Multiplying both sides by $$6R^{5}$$ we get

$$3\,R^{6} \;=\; 2k^{2},$$

hence

$$R^{6} \;=\; \frac{2k^{2}}{3} \quad\Longrightarrow\quad R^{3} \;=\; k\sqrt{\frac{2}{3}}.$$ (The positive root is taken because radius is positive.)

We now compute the desired ratio $$\dfrac{L}{R}.$$ Since $$L = \dfrac{k}{R^{2}},$$ we have

$$\frac{L}{R} \;=\; \frac{k}{R^{3}}.$$ Substituting $$R^{3} = k\sqrt{\dfrac{2}{3}},$$ we get

$$\frac{L}{R} \;=\; \frac{k}{k\sqrt{\dfrac{2}{3}}} \;=\; \frac{1}{\sqrt{\dfrac{2}{3}}} \;=\; \sqrt{\frac{3}{2}}.$$

Hence, the correct answer is Option C.