Moment of Inertia & Radius of Gyration question from NEET 2006 (AIPMT)

Answer: (d)

1. About the central axis normal to the disc: $I_{cm} = \frac{1}{2}MR^2$.
2. The new axis is parallel and touches the rim, so $d = R$.
3. $I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$
4. So the answer is (d) $\frac{3}{2}MR^2$.