A uniform solid cylinder with radius R and length L has moment of inertia \(\mathrm{I}_{1}\), about the axis of cylinder. A concentric solid cylinder of radius \(\mathrm{R}^{\prime}=\frac{\mathrm{R}}{2}\) and length \(\mathrm{L}^{\prime}=\frac{\mathrm{L}}{2}\) is carved out of the original cylinder. If \(\mathrm{I}_{2}\) is the moment of inertia of the carved out portion of the cylinder then \(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\) (Both \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\) are about the axis of the cylinder)

JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Rotational Mechanics

A uniform solid cylinder with radius R and length L has moment of inertia

I_{1}

, about the axis of cylinder. A concentric solid cylinder of radius

R^{'} = \frac{R}{2}

and length

L^{'} = \frac{L}{2}

is carved out of the original cylinder. If

I_{2}

is the moment of inertia of the carved out portion of the cylinder then

\frac{I _{1}}{I _{2}} =

(Both

I_{1}

and

I_{2}

are about the axis of the cylinder)

Answer: 32

The moment of inertia of a solid cylinder about its axis is

I = \frac{1}{2} m R^{2}

, with mass

m = ρ π R^{2} ℓ

. For the full cylinder,

I_{1} = \frac{1}{2} (ρ π R^{2} L) R^{2} = \frac{1}{2} ρ π R^{4} L

. The carved concentric cylinder has radius

\frac{R}{2}

and length

\frac{L}{2}

, so its mass is

ρ π (\frac{R}{2})^{2} \frac{L}{2} = \frac{ρ π R ^{2} L}{8}

and

I_{2} = \frac{1}{2} \cdot \frac{ρ π R ^{2} L}{8} \cdot (\frac{R}{2})^{2} = \frac{ρ π R ^{4} L}{64}

. Dividing,

\frac{I _{1}}{I _{2}} = \frac{1/2}{1/64} = 32

JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Integer answer