A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle $60°$ with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is $\left(2-\sqrt{3}\right)/\sqrt{10} \mathrm{s}$, then the height of the top of the inclined plane (in meters) is __________. Take $g=10\mathrm{ }\mathrm{m}\mathrm{ }{\mathrm{s}}^{-2}$.

Solution

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

Hence, the correct answer is 0.75.