Moment of Inertia & Radius of Gyration question from MHT CET 2026 (19 May, Shift 2)

Answer: (c)

1. Ring A: $YY'$ is its diameter, $I_A = \frac{1}{2}MR^2$.
2. Rings B, C: $YY'$ is a tangent in their plane, $I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$ each.
3. $I = \frac{1}{2}MR^2 + 2\times\frac{3}{2}MR^2 = \frac{7}{2}MR^2$
4. So the answer is (c) $\frac{7MR^2}{2}$.