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NTA Abhyas JEE Main2020PhysicsLaws of MotionPractice

Block A of mass m and block B of mass 2 m are placed on a fixed triangular wedge by means of a massless, inextensible string and a frictionless pulley as shown in the figure. The wedge is inclined at 45 ° to the horizontal on both sides. The coefficient of friction between block A and the wedge is 2 / 3 and that between block B and the wedge is 1 / 3 . If the blocks A and B are released from rest, find the accelerati

Options

  1. A0
  2. B1
  3. C2
  4. D3

Correct answer

A. 0

Step-by-step solution

Acceleration of block A: Maximum friction force that can be obtained at A is (f max ) A =μ A (mg cos 45 0 ) = 2 3 (mg /√2) = 2 m g 3 Similarly, (f max ) B = μ B (2mg cos 45°) = 1 3 ( 2 m g 2 ) = 2 m g 3 Therefore, maximum value of friction that can be obtained on the system is 2 2 m g 3 Net pulling force on the system is F = F 1 -F 2 = 2 m g 2 = m g 2 = m g 2 .... (ii) From Eqs. (i) and (ii), we can see that Net pulling force < f max Therefore, the system will not move or the acceleration of block A will be zero.

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