A hollow sphere of radius R is suspended from a thin rod. If the sphere is twisted by a small angle about the wire axis and released, simple harmonic oscillation (SHM) will ensure with a period of τ 1 . Now, if the hollow sphere is replaced by a solid sphere of radius and mass equal to that of a hollow sphere, the SHM will have a period τ 2 . The ratio of time periods τ 1 / τ 2 is

Step-by-step solution

Let the torsional coefficient of thin rod be k . Time period of oscillation of mass due to twisting of thin rod is given by τ = 2 π I k , where I is the moment of inertia of mass hanging from the thin rod. For hollow sphere, τ 1 = 2 π 2 3 M R 2 k       . . . 1 For solid sphere, τ 2 = 2 π 2 5 M R 2 k       . . . 2 Using 1 and 2 , we get τ 1 τ 2 = 2 3 2 5 = 5 3