JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Rotational Mechanics

Let the disc have mass $M$, radius $r$ and uniform surface density. Each removed disc has radius $\frac{r}{4}$, hence mass $m=M{\left(\frac{r/4}{r}\right)}^2=\frac{M}{16}$, and its centre is at distance $\frac{3r}{4}$ from the centre. The axis $A$ passes through the centre perpendicular to the plane, so only each hole's distance from the centre matters. The complete disc has $I=\frac{1}{2}M{r}^2$. For one removed disc, the parallel-axis theorem gives $I_{hole}=\frac{1}{2}m{\left(\frac{r}{4}\right)}^2+m{\left(\frac{3r}{4}\right)}^2=m{r}^2\left(\frac{1}{32}+\frac{9}{16}\right)=\frac{19}{32}m{r}^2$. For two holes, $2\times \frac{19}{32}m{r}^2=\frac{19}{16}\times \frac{M}{16}{r}^2=\frac{19}{256}M{r}^2$. The remainder therefore has $I=\frac{1}{2}M{r}^2-\frac{19}{256}M{r}^2=\frac{128-19}{256}M{r}^2=\frac{109}{256}M{r}^2$, so $X=109$.