JEE Physics 2025 Question with Solution

Given: A uniform disc has mass $M$ and radius $R$. A smaller disc of radius $R/2$ is removed. The axis is perpendicular to the plane and passes through the center of the original disc.

Find: The moment of inertia of the remaining portion about the same axis.

Use subtraction of moments of inertia together with the parallel-axis theorem.

For the original full disc,

$$I_{\text{full}} = \frac{1}{2}MR^2$$

The mass of the removed small disc is proportional to its area:

$$m' = M \cdot \frac{\pi (R/2)^2}{\pi R^2} = \frac{M}{4}$$

Its moment of inertia about its own center is

$$I'_{\text{cm}} = \frac{1}{2}m'\left(\frac{R}{2}\right)^2 = \frac{1}{2}\cdot \frac{M}{4}\cdot \frac{R^2}{4} = \frac{MR^2}{32}$$

The distance between the center of the original disc and the center of the removed disc is

$$d = \frac{R}{2}$$

So, by the parallel-axis theorem, the moment of inertia of the removed part about the original center is

$$I'_{\text{about origin}} = I'_{\text{cm}} + m'd^2$$ $$I'_{\text{about origin}} = \frac{MR^2}{32} + \frac{M}{4}\cdot \frac{R^2}{4} = \frac{MR^2}{32} + \frac{MR^2}{16} = \frac{3MR^2}{32}$$

Hence the moment of inertia of the remaining part is

$$I_{\text{remaining}} = I_{\text{full}} - I'_{\text{about origin}} = \frac{1}{2}MR^2 - \frac{3}{32}MR^2$$ $$I_{\text{remaining}} = \frac{16}{32}MR^2 - \frac{3}{32}MR^2 = \frac{13}{32}MR^2$$

Therefore, the correct option is D.

Given: Original disc moment of inertia about its center is

$$I_{\text{original}} = \frac{MR^2}{2}$$

The removed portion has mass $M/4$ and radius $R/2$.

Find: Moment of inertia of the remaining part.

Write the required expression as

$$I = \frac{MR^2}{2} - \left[ \frac{\frac{M}{4}\left(\frac{R}{2}\right)^2}{2} + \frac{M}{4}\left(\frac{R}{2}\right)^2 \right]$$

Now simplify the terms inside the bracket.

$$\frac{\frac{M}{4}\left(\frac{R}{2}\right)^2}{2} = \frac{M}{4}\cdot \frac{R^2}{4}\cdot \frac{1}{2} = \frac{MR^2}{32}$$ $$\frac{M}{4}\left(\frac{R}{2}\right)^2 = \frac{M}{4}\cdot \frac{R^2}{4} = \frac{MR^2}{16}$$

Combine them:

$$\frac{MR^2}{32} + \frac{MR^2}{16} = \frac{MR^2}{32} + \frac{2MR^2}{32} = \frac{3MR^2}{32}$$

Substitute back:

$$I = \frac{MR^2}{2} - \frac{3MR^2}{32}$$

Using a common denominator,

$$I = \frac{16MR^2}{32} - \frac{3MR^2}{32} = \frac{13MR^2}{32}$$

Therefore, the moment of inertia of the remaining part is $\frac{13}{32}MR^2$.