Correct answer is : 3
If disc slips on inclined plane then it's acceleration
\(
\begin{aligned}
\mathrm{a}_1=\mathrm{g} \sin \theta
\end{aligned}
\)
\(
\begin{aligned}
\therefore \mathrm{l}=\frac{1}{2} \mathrm{a}_1 \mathrm{t}_1{ }^2
\end{aligned}
\)
\(
\begin{aligned}
\mathrm{t}_1=\sqrt{\frac{2 \mathrm{l}}{\mathrm{a}_1}}
\end{aligned}
\) ........(1)
If disc rolls on inclined plane its acceleration
\(
\begin{aligned}
\mathrm{a}_2=\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{I}}{\mathrm{mR}^2}}=\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{mR}^2}{2 \mathrm{mR}^2}}
\end{aligned}
\)
\(
\begin{aligned}
\mathrm{a}_2=\frac{2}{3} \mathrm{~g} \sin \theta
\end{aligned}
\)
Now \(\mathrm{l}=\frac{1}{2} \mathrm{a}_2 \mathrm{t}_2{ }^2\)
\(
\mathrm{t}_2=\sqrt{\frac{2 \mathrm{l}}{\mathrm{a}_2}}
\) .......(2)
Now, given that \(\mathrm{t}_1=\mathrm{t} \ \&\ \mathrm{t}_2=\left(\frac{\alpha}{2}\right)^{1 / 2} \mathrm{t}\)
From equation (1) & (2)
\(
\frac{\mathrm{t}_2}{\mathrm{t}_1}=\sqrt{\frac{\mathrm{a}_1}{\mathrm{a}_2}}=\sqrt{\frac{3}{2}} \Rightarrow \mathrm{t}_2=\sqrt{\frac{3}{2}} \mathrm{t}_1
\)
On Comparing \(\alpha=3\)
