Moment of Inertia & Radius of Gyration question from MHT CET 2024 (11 May, Shift 2)

Answer: (a)

1. $I = \frac{MR^2}{2}$, so $MR^2 = 2I$.
2. Axis tangential to rim (distance $R$): $I_1 = I + MR^2 = I + 2I = 3I$
3. Axis midway (distance $\frac{R}{2}$): $I_2 = I + \frac{MR^2}{4} = I + \frac{I}{2} = \frac{3I}{2}$
4. $I_1 : I_2 = 3I : \frac{3I}{2} = 2 : 1$
5. So the answer is (a) $2 : 1$.