case \((i)\) Transitional motion

acceleration \(=g \sin \theta\)

case \((ii)\) acceleration \(=\frac{ g \sin \theta}{1+\frac{1}{ MR ^2}}\)

\(I = MR ^2\)

\(a =\frac{ g \sin \theta}{2}\)

Now using \(\rho=u t+\frac{1}{2} at ^2\)

\(a _1 t _1^2= a _2 t _2^2 \quad \theta\) and \(\rho\) same for both are and \(v =0\)

\(\frac{ t _1{ }^2}{ t _2^2}=\frac{ a _2}{ a _1}=\frac{ g \sin \theta}{2 g \sin \theta}\)

\(\frac{t_1}{t_2}=\frac{1}{\sqrt{2}}\)

\(\therefore \frac{t_1}{t_2}=\frac{1}{\sqrt{2}}\)