JEE Main 2026 — Rotational Motion Question with Solution

Let the speed of the masses be $v$ and the angular velocity of the pulley be $\omega =v/r$ since the string does not slip.
The pulley is a disc of mass $M=30m$ and radius $r$. Its moment of inertia is $I=\frac{1}{2}M{r}^2=\frac{1}{2}(30m)r^2=15m{r}^2$.
Using the principle of conservation of mechanical energy, the loss in potential energy of the system equals the gain in kinetic energy.
When the mass $2m$ descends by $h=3.6$ m, the mass $m$ ascends by $h=3.6$ m.
Loss in potential energy $\Delta U=(2m)gh-(m)gh=mgh$.
Gain in kinetic energy $\Delta K=\frac{1}{2}(2m)v^2+\frac{1}{2}(m)v^2+\frac{1}{2}I{\omega }^2$.
Substituting $I$ and $\omega$: $\Delta K=\frac{3}{2}m{v}^2+\frac{1}{2}(15m{r}^2)(\frac{v}{r}{)}^2=\frac{3}{2}m{v}^2+\frac{15}{2}m{v}^2=9m{v}^2$.
Equating $\Delta U=\Delta K$: $mgh=9m{v}^2$.
$v^2=\frac{gh}{9}$.
Given $g=10$ m/s${}^2$ and $h=3.6$ m:
$v^2=\frac{10\times 3.6}{9}=\frac{36}{9}=4$.
$v=\sqrt{4}=2$ m/s.