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( R^2 \over 4 + L^2 \over 12 ) . If such a cylinder is to be made for a given mass of a material, the ratio L \over R for it to have minimum possible I is

Rotational Motion PYQ — JEE Main 2020 · 3 Sep · Shift 1 · Q63 with Solution

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Correct answer: B

For a solid cylinder of mass $M$ , length $L$ , and radius $R$ , about an axis through its centre and perpendicular to its own axis:

I = M\left(\frac{R^2}{4} + \frac{L^2}{12}\right)

We need to minimize $I$ for a given mass of material.

For a cylinder of uniform density $\rho$ ,

M = \rho \times \text{Volume} = \rho \pi R^2 L

Since $M$ and material are fixed, $\rho$ is fixed, so:

R^2 L = \text{constant}

Let

R^2L = k

Then

L = \frac{k}{R^2}

Substitute $L = \dfrac{k}{R^2}$ into the expression for $I$ :

I = M\left(\frac{R^2}{4} + \frac{1}{12}\left(\frac{k}{R^2}\right)^2\right)

I = M\left(\frac{R^2}{4} + \frac{k^2}{12R^4}\right)

To minimize $I$ , minimize

f(R) = \frac{R^2}{4} + \frac{k^2}{12R^4}

\frac{df}{dR} = \frac{R}{2} - \frac{4k^2}{12R^5}

\frac{df}{dR} = \frac{R}{2} - \frac{k^2}{3R^5}

For minimum,

\frac{R}{2} - \frac{k^2}{3R^5} = 0

\frac{R}{2} = \frac{k^2}{3R^5}

3R^6 = 2k^2

But $k = R^2L$ , so

k^2 = R^4L^2

Substitute:

3R^6 = 2R^4L^2

Divide by $R^4$ :

3R^2 = 2L^2

\frac{L^2}{R^2} = \frac{3}{2}

Hence,

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

The correct option is:

\boxed{\sqrt{\frac{3}{2}}}

So, Option B is correct.

Stored correct answer: B

Our derived answer: B

They agree.