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 $L / R$ for it to have minimum possible I is

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Rotational Mechanics

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 (\frac{R ^{2}}{4} + \frac{L ^{2}}{12})

. If such a cylinder is to be made for a given mass of a material, the ratio

L / R

for it to have minimum possible I is

Answer: (B)

For a fixed mass of material the mass

M = π R^{2} L ρ

is constant, so

R^{2} = \frac{M}{π ρ L}

. Writing the moment of inertia as

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

and setting

\frac{d I}{d L} = 0

gives

- \frac{M}{4 π ρ L ^{2}} + \frac{L}{6} = 0

, i.e.

\frac{M}{π ρ L} = \frac{2 L ^{2}}{3}

, so

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

and

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

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Single correct option