JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Rotational Mechanics

Two masses m and 2m are connected by a light string going over a pulley (disc) of mass 30m 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}

)

Answer: 2

Solution

As the

2 m

mass falls a height

h

the

m

mass rises the same

h

, so the net loss of potential energy is

(2 m - m) g h = m g h

. This appears as the translational kinetic energy of the two blocks plus the rotational kinetic energy of the disc pulley, with

I = \frac{1}{2} (30 m) r^{2}

and

ω = \frac{v}{r}

m g h = \frac{1}{2} (2 m) v^{2} + \frac{1}{2} m v^{2} + \frac{1}{2} (\frac{1}{2} 30 m r^{2}) \frac{v ^{2}}{r ^{2}} = \frac{3}{2} m v^{2} + \frac{15}{2} m v^{2} = 9 m v^{2} .

Hence

v^{2} = \frac{g h}{9} = \frac{10 \times 3.6}{9} = 4

, giving

v = 2 m/s

. The speed after descending

3.6 m

2 m/s

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Integer answer

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JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Rotational Mechanics

Solution

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