Moment of Inertia & Radius of Gyration question from JEE Main 2018 (16 Apr, Shift 1)

Answer: (a)

1. About z (through centre, perpendicular): $I_z = \frac{1}{2}MR^2$.
2. z' is parallel at distance R: $I_{z'} = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
3. So the answer is (a) $1 : 3$.