Moment of Inertia & Radius of Gyration question from MHT CET 2026 (13 May, Shift 2)

Answer: (c)

1. Disc, tangent in its plane: $\frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2 \Rightarrow K_1^2 = \frac{5}{4}R^2$.
2. Ring, tangent in its plane: $\frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2 \Rightarrow K_2^2 = \frac{3}{2}R^2$.
3. $\frac{K_1}{K_2} = \sqrt{\frac{5/4}{3/2}} = \sqrt{\frac{5}{6}}$
4. So the answer is (c) $\sqrt{5} : \sqrt{6}$.