Moment of Inertia & Radius of Gyration question from MHT CET 2026 (20 Apr, Shift 1)

Answer: (b)

1. Ring A about diameter: $I_A = \frac{1}{2}MR^2$. Ring B about tangent in its plane: $I_B = \frac{3}{2}MR^2$.
2. Equal KE: $\frac{1}{2}I_A\omega_A^2 = \frac{1}{2}I_B\omega_B^2 \Rightarrow \frac{\omega_B}{\omega_A} = \sqrt{\frac{I_A}{I_B}} = \sqrt{\frac{1}{3}}$
3. So the answer is (b) $1 : \sqrt{3}$.