MHTCET 18 Apr 2026 Shift 1 Paper

Concept:For a hollow sphere, the moment of inertia about its centre is $I_{CM} = \frac{2}{3}MR^2$.The parallel axis theorem connects the moment of inertia about a parallel axis at distance $h$ from the centre.Explanation:Let the axis $XY$ be parallel to the central axis and at a perpendicular distance $h$ from the centre.Using the parallel axis theorem:$$I_{XY} = I_{CM} + Mh^2$$Substitute $I_{CM}$:$$I_{XY} = \frac{2}{3}MR^2 + Mh^2$$Given the radius of gyration about $XY$ is $K = R$, the moment of inertia is:$$I_{XY} = MK^2 = MR^2$$Equating both expressions for $I_{XY}$:$$MR^2 = \frac{2}{3}MR^2 + Mh^2$$Cancel $M$ from every term:$$R^2 = \frac{2}{3}R^2 + h^2$$$$h^2 = R^2 - \frac{2}{3}R^2 = \frac{1}{3}R^2$$Taking the square root:$$h = \frac{R}{\sqrt{3}}$$Answer:The distance of the axis from the centre is $h = \frac{R}{\sqrt{3}}$.Therefore, the correct option is A.