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NTA Abhyas JEE Main2020PhysicsLaws of MotionPractice

The figure below shows a block of mass M connected to an ideal string which passes through a thin fixed smooth pipe. On the other end, a particle of mass m is connected which revolves in a vertical circle of radius r . If the coefficient of friction between M and the surface is μ = 2 3 , then for what minimum value of M , the block of mass m can undergo complete vertical circular motion?

Options

  1. AM min = 6 m
  2. BM min = 9 m
  3. CM min = 3 m
  4. DM min = 15 m

Correct answer

B. M min = 9 m

Step-by-step solution

For the particle to undergo complete vertical circular motion, its speed at the lowest point should be at least v = 5 g r . Also, the tension in the string is maximum when the particle is at the lowest point. The tension at the lowest position of the particle is T = m g + m v 2 r ⇒ T = m g + m 5 g r r = 6 m g The friction force acting on the block should be able to balance the maximum tension. f = T = 6 m g f = 6 m g ≤ μ M g ⇒ M ≥ 6 m 2 3 = 9 m M min = 9 m

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