JEE Physics 2026 Question with Solution

Given: A uniform circular disc has mass $M$ and radius $r$. Two circular regions are cut out.

Find: The value of $x$ in

$$I_{\text{remainder}} = \frac{x}{256}Mr^2$$

Use additivity of moment of inertia and treat each hole as removed mass.

Area mass density of the original disc is

$$\sigma = \frac{M}{\pi r^2}$$

The cut-out regions are two circles of radius

$$a = \frac{r}{4}$$

whose centers are at distance

$$d = \frac{3r}{4}$$

from axis $A$.

Mass of one cut-out circle is

$$m = \sigma \pi a^2 = \frac{M}{\pi r^2}\pi\left(\frac{r}{4}\right)^2 = \frac{M}{16}$$

Moment of inertia of one cut-out about its own center is

$$I_{cm} = \frac{1}{2}ma^2 = \frac{1}{2}\left(\frac{M}{16}\right)\left(\frac{r}{4}\right)^2 = \frac{Mr^2}{512}$$

Using the parallel axis theorem, moment of inertia of one cut-out about the main axis is

$$I_{\text{hole}} = I_{cm} + md^2 = \frac{Mr^2}{512} + \frac{M}{16}\left(\frac{3r}{4}\right)^2$$ $$I_{\text{hole}} = \frac{Mr^2}{512} + \frac{9Mr^2}{256} = \frac{Mr^2 + 18Mr^2}{512} = \frac{19Mr^2}{512}$$

Total moment of inertia removed for two holes is

$$2I_{\text{hole}} = 2 \times \frac{19Mr^2}{512} = \frac{19Mr^2}{256}$$

Moment of inertia of the original full disc about axis $A$ is

$$I_{\text{total}} = \frac{1}{2}Mr^2 = \frac{128Mr^2}{256}$$

Therefore, moment of inertia of the remainder is

$$I_{\text{remainder}} = I_{\text{total}} - 2I_{\text{hole}} = \frac{128 - 19}{256}Mr^2 = \frac{109}{256}Mr^2$$

Hence,

$$x = 109$$

So the required numerical value is 109.

The solution lists Correct Answer: 236, but the worked solution clearly gives

$$I_{\text{remainder}} = \frac{109}{256}Mr^2$$

Therefore, the answer derived from the solution working is 109.