Acceleration of the solid sphere while rolling down without slipping
\(a_{1}=\frac{g \sin \theta}{1+\frac{k^{2}}{r^{2}}}\)
Acceleration developed while slipping down \(a_{2}=g \sin \theta\)
\(\text { Required ratio } \frac{a_{1}}{a_{2}}=\frac{g \sin \theta}{1+\frac{k^{2}}{r^{2}}} / g \sin \theta\)
\(\frac{a_{1}}{a_{2}}=\frac{1}{1+\frac{k^{2}}{r^{2}}}\)
\(\text { For a solid sphere } \frac{k^{2}}{r^{2}}=\frac{2}{5}\)
\(\therefore \text { Ratio of accelerations } \frac{a_{1}}{a_{2}}=\frac{1}{1+\frac{2}{5}}\)
\(=\frac{1}{5+\frac{2}{5}}=\frac{1}{\frac{7}{5}}=\frac{5}{7}\)
\(\therefore a_{1}: a_{2}=5: 7 \)

Moment of inertia of a disc
\(\mathrm{I}_{1}=\frac{M R^{2}}{2}\)
\(\text { Mass of small disc }=\frac{M}{\pi R^{2}} \times \pi \times\left(\frac{R}{2}\right)^{2}\)
\(=\frac{M}{\pi R^{2}} \times \frac{\pi R^{2}}{4}=\frac{M}{4}\)
By the theorem of parallel axis, the moment of inertia of the small disc. About an axis passing through 0 is
\(I_{2} =\frac{1}{2} \times \frac{M}{4}\left(\frac{R}{2}\right)^{2}+\frac{M}{4}\left(\frac{R}{2}\right)^{2} \)
\(=\frac{M}{8} \times \frac{R^{2}}{4}+\frac{M}{4} \times \frac{R^{2}}{4} \)
\(=\frac{M R^{2}}{32}+\frac{M R^{2}}{16}=\frac{M R^{2}+2 M R^{2}}{32} \)
\(I_{2} =\frac{3 M R^{2}}{32} \)
Moment of inertia of the remaining part is I= I₁ - I₂
\(=\frac{M R^{2}}{2}-\frac{3 M R^{2}}{32} \)
\(=\frac{16 M R^{2}-3 M R^{2}}{32}=\frac{13 M R^{2}}{32}\)
\(I =\frac{13 M R^{2}}{32} \)