JEE Main 2022 — Rotational Motion Question with Solution

The portion of the strings between the ceiling and the cylinder are at rest. Hence, the points of the cylinder where the strings leave it are at rest. The cylinder is thus rolling without slipping on the strings. 

Now, from energy conservation

$\mathrm{mgh}=\frac{1}{2}m{v}^2+\frac{1}{2}\mathrm{I}{\omega }^2$

where, $I$ is moment of inertia of cylinder and $\omega =\frac{v}{R}$ is angular velocity.

Then, we have

 $$\begin{gathered} mgh=\frac{1}{2}m{v}^2+\frac{1}{2}\frac{m{R}^2}{2}{\omega }^2 \\ \Rightarrow gh=\frac{1}{2}{v}^2+\frac{1}{2}\frac{R^2}{2}{\left(\frac{v}{R}\right)}^2 \end{gathered}$$

$$\begin{gathered} 10h=\frac{16}{2}+\frac{16}{4} \\ \Rightarrow h=1.2 \mathrm{m}=120 \mathrm{cm} \end{gathered}$$