Q 11-06-109JEE MainJEE Main 2023 (24 Jan, Shift 2)Medium
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 $R'=\dfrac R2$ and length $L'=\dfrac L2$ is carved out of the original cylinder. If $I_2$ is the moment of inertia of the carved out portion of the cylinder, then $\dfrac{I_1}{I_2}=$ ______. (Both $I_1$ and $I_2$ are about the axis of the cylinder)
Numerical value type. Enter your answer.
Answer: 32
The carved part has $\dfrac14$ of the cross-section and $\dfrac12$ the length, so its mass is $\dfrac M8$.
$$I_1=\frac12MR^2,\qquad I_2=\frac12\cdot\frac M8\cdot\frac{R^2}{4}=\frac{MR^2}{64}$$
$\dfrac{I_1}{I_2}=32$.
Solution by Sreeraj P, M.Sc Physics