Explanation:
(i) For solid sphere, the moment of inertia about the diameter is \({I_s} = \frac{2}{5}M{R^2}\)
Now \(I = M{K^2}\) for any body, where \(K\) is radius of gyration of that body.
so \(M{K^2} = \frac{2}{5}M{R^2} \Rightarrow K = R\sqrt {2/5} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\)
(ii) The moment of inertia of disc about an axis passing through its centre & perpendicular to plane is
\({I_d} = \frac{{M{R^2}}}{2} = M{K^2} \Rightarrow K = R\sqrt {1/2} {\rm{ }}\,\,\,\,\,\,\,\,\left( 2 \right)\)
Now acceleration of any body which is rolling on an inclined plane is
\(a = \frac{{g\sin \theta }}{{1 + {K^2}/{R^2}}}{\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 3 \right)\)
For same \(R\), the acceleration of the body depends only on radius of gyration \(K\) [see eq(3)] so solid sphere will reach earlier to bottom of an inclined plane than disc.
