JEE Main 2026 — Jan 22, Shift 2 — Physics solution

JEE Main2026Jan 22, Shift 2Physics

Two masses $m$ and $2 m$ are connected by a light string going over a pulley (disc) of mass $30 m$ with radius $r = 0.1$ m. The pulley is mounted in a vertical plane and it is free to rotate about its axis. The 2m mass is released from rest and its speed when it has descended through a height of 3.6 m is _____ m/s. (Assume string does not slip and $g = 10$ m/s $^{2}$ )

Your answer

Solution

Let $v$ be the speed of either mass after $2 m$ has descended 3.6 m. The pulley (disc) has $I = \frac{1}{2} (30 m) R^{2}$ and rotates with $ω = v / R$ .

Energy conservation: net loss in PE = total KE.

Net PE loss $= (2 m) g (3.6) - m g (3.6) = m g (3.6)$ (since $m$ rises by the same amount $2 m$ descends).

Total KE $= \frac{1}{2} m v^{2} + \frac{1}{2} (2 m) v^{2} + \frac{1}{2} \cdot \frac{30 m R ^{2}}{2} \cdot \frac{v ^{2}}{R ^{2}}$

$= \frac{1}{2} m v^{2} + m v^{2} + \frac{15}{2} m v^{2} = 9 m v^{2}$ .

$9 m v^{2} = m g (3.6) \Rightarrow v^{2} = \frac{g \cdot 3.6}{9} = \frac{10 \times 3.6}{9} = 4$ .

$v = 2$ m/s.

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