JEE Physics 2025 Question with Solution

Given: Mass of the original disc is $M$, radius is $R$, and a small disc of radius $\frac{R}{3}$ is removed. The required moment of inertia of the remaining part about the axis through $O$ and perpendicular to the plane is $\frac{4}{x}MR^2$.

Find: The value of $x$.

Since the question says as shown in figure and the extracted figure is not available, the solution indicates the intended case is non-concentric (tangential) removal. Therefore the removed small disc has its centre at a distance $\frac{2R}{3}$ from $O$.

For the original disc,

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

Because the disc is uniform, the mass of the removed small disc is proportional to area:

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

Moment of inertia of the removed disc about its own centre is

$$I_{\text{cm}} = \frac{1}{2}m\left(\frac{R}{3}\right)^2 = \frac{1}{2}\cdot \frac{M}{9} \cdot \frac{R^2}{9} = \frac{MR^2}{162}$$

Using the parallel axis theorem for distance $\frac{2R}{3}$,

$$I_{\text{removed}} = I_{\text{cm}} + m\left(\frac{2R}{3}\right)^2$$ $$I_{\text{removed}} = \frac{MR^2}{162} + \frac{M}{9}\cdot \frac{4R^2}{9}$$ $$I_{\text{removed}} = \frac{MR^2}{162} + \frac{4MR^2}{81} = \frac{MR^2}{18}$$

Hence the moment of inertia of the remaining part is

$$I_{\text{remaining}} = I_{\text{original}} - I_{\text{removed}} = \frac{1}{2}MR^2 - \frac{MR^2}{18}$$ $$I_{\text{remaining}} = \frac{4}{9}MR^2$$

Now compare with the given form:

$$\frac{4}{x}MR^2 = \frac{4}{9}MR^2$$

So,

$$\frac{4}{x} = \frac{4}{9}$$ $$x = 9$$

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

Given: the solution discusses both concentric and non-concentric removal.

If the removed disc were concentric, then

$$I_{\text{removed}} = \frac{MR^2}{162}$$

and

$$I_{\text{remaining}} = \frac{1}{2}MR^2 - \frac{MR^2}{162} = \frac{40}{81}MR^2$$

This would give

$$\frac{4}{x} = \frac{40}{81}$$ $$x = \frac{81}{10}$$

which does not match the intended numerical answer.

The second approach in the solution explicitly notes that because the question refers to a figure, the intended geometry is the tangentially removed disc, for which the parallel axis theorem must be used. That produces

$$I_{\text{remaining}} = \frac{4}{9}MR^2$$

and hence $x = 9$.

So the correct extracted answer is $9$, and the first approach is incomplete because it ignores the offset shown in the missing figure.