1. Moment of inertia about the perpendicular bisector

For a uniform solid cylinder of mass $M$, length $l$, and radius $R$, the moment of inertia about an axis through its center and perpendicular to its own axis is

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

2. What is being minimized?

The question says the moment of inertia about this axis is $I$, and asks for the ratio $\dfrac{l}{R}$ such that the moment of inertia is minimum.

If $M$ were also allowed to vary independently, then $I$ could be reduced trivially by reducing $M$. So the meaningful interpretation is that the cylinder is made of a given material and has fixed volume.

Thus,

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

Since density is constant,

$M=\rho \pi R^2 l.$

Substitute into $I$:

$I=\frac{1}{12}(\rho \pi R^2 l)(3R^2+l^2).$

So we minimize

$f(R,l)=R^2l(3R^2+l^2)$

subject to

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

Let

$R^2l = k.$

Then

$I \propto k(3R^2+l^2).$

Since $k$ is constant, we only need to minimize

$3R^2+l^2$

with the constraint

$R^2l=k.$

3. Use the constraint

From

$R^2l=k \implies l=\frac{k}{R^2}.$

Substitute into the expression to minimize:

$g(R)=3R^2+\left(\frac{k}{R^2}\right)^2=3R^2+\frac{k^2}{R^4}.$

4. Differentiate and set to zero

$\frac{dg}{dR}=6R-4\frac{k^2}{R^5}.$

For minimum,

$6R-\frac{4k^2}{R^5}=0$

$6R^6=4k^2$

$3R^6=2k^2.$

But since $k=R^2l$, we have

$k^2=R^4l^2.$

So

$3R^6=2R^4l^2$

$3R^2=2l^2$

$\frac{l^2}{R^2}=\frac{3}{2}$

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

5. Check with options

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

This matches Option B.

6. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.