Moment of Inertia & Radius of Gyration question from JEE Main 2022 (28 Jul, Shift 1)

Answer: 3

1. Each disc has radius $r = \frac{a}{2}$. About a diameter, $I = \frac{1}{4}Mr^2 = \frac{Ma^2}{16}$.
2. Top and bottom discs: $OO'$ is their diameter, so $2 \times \frac{Ma^2}{16}$.
3. Side discs: $OO'$ is a tangent in their plane, $I = \frac{1}{4}Mr^2 + Mr^2 = \frac{5}{4}Mr^2 = \frac{5Ma^2}{16}$ each.
4. $I = \frac{2Ma^2}{16} + \frac{10Ma^2}{16} = \frac{3}{4}Ma^2$
5. So $x = 3$.