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JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion

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 \( g=10 m/s^{2} \))

  1. A 1
  2. B 2
  3. C 3
  4. D 4
Verified Solution

Answer & Solution

Correct Answer

(B) 2

Step-by-step Solution

Detailed explanation

Using energy conservation \( \frac{1}{2}mv^{2}+\frac{1}{2}2mv^{2}+\frac{1}{2}\frac{30mR^{2}}{2}\times\frac{v^{2}}{R^{2}}=mgh \) \( 9\,mv^{2}=mgh \) \( v=\sqrt{\frac{gh}{9}}=\sqrt{\frac{10\times3.6}{9}} \) \( v=\sqrt{4}=2\,m/s \)
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