A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle…

Solution

$a_c=\frac{g\sin \text{θ}}{1+\frac{I_c}{M{R}^2}}$
$a_{\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}}=\frac{g\sin \text{θ}}{2}$
$a_{\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}}=\frac{2g\sin \text{θ}}{3}$
$\frac{h}{\sin \text{θ}}=\frac{1}{2}\left(\frac{g\sin \text{θ}}{2}\right)t_1^2\Rightarrow t_1=\sqrt{\frac{4h}{g {\sin }^2\text{θ}}}=\sqrt{\frac{16h}{3g}}$
$\Rightarrow \frac{h}{\sin \text{θ}}=\frac{1}{2}\left(\frac{2g\sin \text{θ}}{3}\right)t_2^2\Rightarrow t_2=\sqrt{\frac{3h}{g {\sin }^2\text{θ}}}=\sqrt{\frac{4h}{g}}$
$\Rightarrow \sqrt{\frac{16h}{3g}}-\sqrt{\frac{4h}{g}}=\frac{2-\sqrt{3}}{\sqrt{10}}$
$\sqrt{h}\left[\frac{4}{\sqrt{3}}-2\right]=2-\sqrt{3}$
$\sqrt{h}=\frac{\left(2-\sqrt{3}\right)\sqrt{3}}{\left(4-2\sqrt{3}\right)}=\frac{\sqrt{3}}{2}\Rightarrow h=\frac{3}{4}=0.75 \mathrm{m}$

Asked in: JEE Advanced 2018 (Paper 1)

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