Moment of Inertia & Radius of Gyration question from JEE Main 2026 (23 Jan, Shift 2)

Answer: 109

1. Each hole: radius $\frac{r}{4}$, mass $\frac{M}{16}$, centre $\frac{3r}{4}$ from axis A.
2. One hole: $\frac{1}{2}\frac{M}{16}\frac{r^2}{16} + \frac{M}{16}\frac{9r^2}{16} = \frac{19Mr^2}{512}$
3. Two holes: $\frac{19Mr^2}{256}$
4. $I = \frac{128Mr^2}{256} - \frac{19Mr^2}{256} = \frac{109}{256}Mr^2$, so $x = 109$.