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?

We have a solid uniform cylinder of length $$l$$, radius $$R$$ and density $$\rho$$. Its mass is

$$M=\rho \,(\text{volume})=\rho\,\pi R^{2}l.$$

The question asks for the moment of inertia about an axis that passes through the centre of mass and is perpendicular to the axis of the cylinder (that is, through the mid-point of the length and lying in the plane of the end faces). For a solid cylinder, the standard formula for the moment of inertia about such an axis is

$$I=\frac{1}{12}M\left(3R^{2}+l^{2}\right).$$

We are told that the cylinder is uniform, so $$\rho$$ is constant, and we wish to find the shape (that is, the ratio $$l/R$$) for which this moment of inertia is minimum while the volume (and therefore the mass) is fixed. Therefore we keep

$$V=\pi R^{2}l=\text{constant}.$$

Because $$M=\rho V$$ and $$V$$ is fixed, the factor $$M/12$$ in $$I$$ is itself constant. Hence minimising $$I$$ is the same as minimising the purely geometric quantity

$$F(R,l)=3R^{2}+l^{2}$$

subject to the volume constraint $$\pi R^{2}l=V.$$ We now use the constraint to eliminate one variable. Solving for $$l$$ gives

$$l=\frac{V}{\pi R^{2}}.$$

Substituting this expression for $$l$$ in $$F$$ we obtain a function of the single variable $$R$$:

$$F(R)=3R^{2}+\left(\frac{V}{\pi R^{2}}\right)^{2}=3R^{2}+\frac{V^{2}}{\pi^{2}R^{4}}.$$

To locate the minimum we differentiate with respect to $$R$$ and set the derivative equal to zero. Writing $$K=\dfrac{V^{2}}{\pi^{2}}$$ for brevity, we have

$$F(R)=3R^{2}+\frac{K}{R^{4}},$$

so

$$\frac{dF}{dR}=6R-\frac{4K}{R^{5}}.$$

Setting $$\dfrac{dF}{dR}=0$$ for an extremum, we get

$$6R-\frac{4K}{R^{5}}=0 \;\;\Longrightarrow\;\; 6R=\frac{4K}{R^{5}} \;\;\Longrightarrow\;\; 6R^{6}=4K.$$

Dividing by 2:

$$3R^{6}=2K \;\;\Longrightarrow\;\; R^{6}=\frac{2}{3}K.$$

Now restore $$K=\dfrac{V^{2}}{\pi^{2}}$$:

$$R^{6}=\frac{2}{3}\,\frac{V^{2}}{\pi^{2}}.$$

Next we find the corresponding length $$l$$ using the constraint again:

$$l=\frac{V}{\pi R^{2}}.$$

For convenience, first compute $$V$$ in terms of $$R$$ from the relation for $$R^{6}$$. Take square roots on both sides:

$$R^{3}=\left(\frac{2}{3}\right)^{1/2}\frac{V}{\pi} \;\;\Longrightarrow\;\; V=\pi R^{3}\sqrt{\frac{3}{2}}.$$

Substituting this $$V$$ back into $$l=\dfrac{V}{\pi R^{2}}$$ gives

$$l=\frac{\pi R^{3}\sqrt{\dfrac{3}{2}}}{\pi R^{2}} =R\sqrt{\frac{3}{2}}.$$

Therefore the required ratio is

$$\frac{l}{R}=\sqrt{\frac{3}{2}}.$$

Because $$\dfrac{d^{2}F}{dR^{2}}=6+\dfrac{20K}{R^{6}}$$ is positive at this value of $$R$$, the extremum is indeed a minimum.

Hence, the correct answer is Option B.