Question 148 of 280
A circular disc reaches from top to bottom of an inclined plane of length . When it slips down the plane, if takes . When it rolls down the plane then it takes , where is _________
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Answer: 3
For slipping
$\begin{aligned}& \mathrm{a}=\mathrm{g} \sin \theta \\
& \ell=\frac{1}{2} \mathrm{at}^2 \Rightarrow \mathrm{t}=\sqrt{\frac{2 \ell}{\mathrm{g} \sin \theta}}\end{aligned}$
For rolling
$\begin{aligned}& \mathrm{a}^{\prime}=\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{k}^2}{\mathrm{R}^2}}\left[\mathrm{k}=\frac{\mathrm{R}}{\sqrt{2}}\right] \\& \Rightarrow \mathrm{a}^{\prime}=\frac{2 \mathrm{~g} \sin \theta}{3} \\& \ell=\frac{1}{2} \mathrm{a}^{\prime}\left(\mathrm{t}^{\prime}\right)^2 \\& \Rightarrow \mathrm{t}^{\prime}=\sqrt{\frac{6 \ell}{2 \mathrm{~g} \sin \theta}}=\sqrt{\frac{\alpha}{2}} \sqrt{\frac{2 \ell}{\mathrm{g} \sin \theta}} \\
& \Rightarrow \alpha=3\end{aligned}$
$\begin{aligned}& \mathrm{a}=\mathrm{g} \sin \theta \\
& \ell=\frac{1}{2} \mathrm{at}^2 \Rightarrow \mathrm{t}=\sqrt{\frac{2 \ell}{\mathrm{g} \sin \theta}}\end{aligned}$
For rolling
$\begin{aligned}& \mathrm{a}^{\prime}=\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{k}^2}{\mathrm{R}^2}}\left[\mathrm{k}=\frac{\mathrm{R}}{\sqrt{2}}\right] \\& \Rightarrow \mathrm{a}^{\prime}=\frac{2 \mathrm{~g} \sin \theta}{3} \\& \ell=\frac{1}{2} \mathrm{a}^{\prime}\left(\mathrm{t}^{\prime}\right)^2 \\& \Rightarrow \mathrm{t}^{\prime}=\sqrt{\frac{6 \ell}{2 \mathrm{~g} \sin \theta}}=\sqrt{\frac{\alpha}{2}} \sqrt{\frac{2 \ell}{\mathrm{g} \sin \theta}} \\
& \Rightarrow \alpha=3\end{aligned}$