JEE Main 2025 — Rotational Motion Question with Solution

Moment of Inertia of the Full Disc:

The moment of inertia (M.I.) of the entire disc without any cavity is given by:

$I_1=\frac{1}{2}M{R}^2$

Mass of the Removed Disc:

The mass of the removed disc, which is of radius $\frac{R}{3}$, is calculated as:

$\text{Mass of removed disc}=\left(\frac{M}{\pi R^2}\right)\times {\left(\frac{R}{3}\right)}^2\pi =\frac{M}{9}$

Moment of Inertia of the Removed Disc:

The moment of inertia of the removed disc involves two parts: about its center and due to its position.

About its center:

$\frac{\frac{M}{9}{\left(\frac{R}{3}\right)}^2}{2}=\frac{M{R}^2}{162}$

Due to its position (distance from O to the new center of the small disc):

The distance is $\frac{2R}{3}$, so:

$\frac{M}{9}\times {\left(\frac{2R}{3}\right)}^2=\frac{4M{R}^2}{81}$

Total M.I. of removed disc:

$I_2=\frac{M{R}^2}{162}+\frac{4M{R}^2}{81}=\frac{M{R}^2}{18}$

Moment of Inertia of the Remaining Part:

Subtract the moment of inertia of the removed disc from the full disc:

$I=I_1-I_2=\frac{1}{2}M{R}^2-\frac{M{R}^2}{18}=\frac{9M{R}^2-M{R}^2}{18}=\frac{8M{R}^2}{18}=\frac{4M{R}^2}{9}$

Thus, the value of $x$ is $9$.