Ratio of radius of gyration of a hollow sphere to that of a solid cylinder of equal mass, for moment of Inertia about their diameter axis A B as shown in figure is √(8 / x). The value of x is :

Solution

Physics sir · 4 steps

The sphere is a standard result; the cylinder needs its central-diameter formula plus a parallel-axis shift.

1. Hollow sphere. About a diameter, I₁ = 2/3MR². With I = Mk², this gives k₁² = 2/3R².
2. Cylinder about centre. The cylinder has radius R and length L = 4R. About a diameter through its centre, I_(cm) = M(R²/4 + L²/12) = M(R²/4 + 16R²/12) = 19/12MR².
3. Shift to the end. Axis AB is a diameter of the end face, parallel to the central diameter and L/2 = 2R away from it. Apply the parallel axis theorem.

   I₂ = 19/12MR² + M(2R)² = 67/12MR² ⇒ k₂² = 67/12R²

4. So. k₁/k₂ = √((2/3)/(67/12)) = √(8/67), so x = 67, option (c).

The idea to keep. For a solid cylinder about a diameter, use M(R²/4 + L²/12) through the centre. If the axis sits on an end face, add M(L/2)².

Answer from the NTA final answer key. Solution by PrepWiser.