\((i)\) For solid sphere, the moment of inertia about the diameter is \(I _{ s }=\frac{2}{5} MR ^2\)

Now \(I = MK ^2\) for any body, where \(K\) is radius of gyration of that body.

so \(MK ^2=\frac{2}{5} MR ^2 \Rightarrow K = R \sqrt{2 / 5}\)

\((ii)\) The moment of inertia of disc about an axis passing through its centre and perpendicular to plane is

\(I _{ d }=\frac{ MR ^2}{2}= MK ^2 \Rightarrow K = R \sqrt{1 / 2}\)

Now acceleration of any body which is rolling on an inclined plane is

\(a=\frac{g \sin \theta}{1+K^2 / R^2}\)

For same R, the acceleration of the body depends only on radius of gyration \(K\), [see eq\((iii)\)] so solid sphere will reach earlier to bottom of an inclined plane than disc.