This question involves conservation of energy including translational and rotational kinetic energy, a key concept in JEE Main, JEE Advanced and IIT JEE.
Loss of gravitational potential energy of the system is
$$ \Delta U = (2m - m) g h = mgh $$
$$ \Delta U = m \times 10 \times 3.6 = 36m $$
This energy is converted into kinetic energy of both masses and rotational kinetic energy of the pulley.
Translational kinetic energy of the two masses is
$$ K_{\text{trans}} = \frac{1}{2} m v^2 + \frac{1}{2} (2m) v^2 = \frac{3}{2} m v^2 $$
Moment of inertia of the pulley (disc) is
$$ I = \frac{1}{2} (30m) r^2 = 15 m r^2 $$
Rotational kinetic energy of the pulley is
$$ K_{\text{rot}} = \frac{1}{2} I \omega^2 = \frac{1}{2} (15 m r^2) \left(\frac{v}{r}\right)^2 = \frac{15}{2} m v^2 $$
Total kinetic energy is
$$ K_{\text{total}} = \frac{3}{2} m v^2 + \frac{15}{2} m v^2 = 9 m v^2 $$
Using energy conservation,
$$ 36m = 9 m v^2 $$
$$ v^2 = 4 $$
$$ v = 2\ \text{m/s} $$
Hence, the required speed is $2\ \text{m/s}$.
Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.