Moment of Inertia & Radius of Gyration question from JEE Main 2025 (7 Apr, Shift 1)

Answer: 199

1. Disc A: PQ is its diameter, $I_A = I = \frac{1}{4}MR^2$, so $MR^2 = 4I$.
2. Solid sphere B touches PQ: $\frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$.
3. Shell C touches PQ: $\frac{2}{3}MR^2 + MR^2 = \frac{5}{3}MR^2$.
4. $I_{total} = I + 4I\left(\frac{21+25}{15}\right) = I + \frac{184}{15}I = \frac{199}{15}I$, so $x = 199$.