Moment of Inertia & Radius of Gyration question from JEE Main 2024 (5 Apr, Shift 1)

Answer: (c)

1. Hollow sphere about a diameter: $K_1^2 = \frac{2}{3}R^2$.
2. Solid cylinder (radius $R$, length $L = 4R$), axis through centre perpendicular to its length: $M\left(\frac{R^2}{4} + \frac{L^2}{12}\right) = \frac{19}{12}MR^2$.
3. AB passes through the end face, $\frac{L}{2} = 2R$ away: $I = \frac{19}{12}MR^2 + 4MR^2 = \frac{67}{12}MR^2$.
4. $\frac{K_1^2}{K_2^2} = \frac{2/3}{67/12} = \frac{8}{67}$
5. So the answer is (c) $x = 67$.