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JEE Main202431 Jan 2024Morning ShiftPhysicsLaws of MotionActual

In the given arrangement of a doubly inclined plane two blocks of masses M and m are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is 0 . 25 . The value of m , for which M = 10 kg will move down with an acceleration of 2 m s - 2 , is: (take g = 10 m s - 2 and tan 37 ° = 3 4 )

Options

  1. A9 kg
  2. B4 . 5 kg
  3. C6 . 5 kg
  4. D2 . 25 kg

Correct answer

B. 4 . 5 kg

Step-by-step solution

For M block: 10 g sin 53 ° - μ ( 10 g ) cos 53 ° - T = 10 × 2 ⇒ T = 80 - 15 - 20 ⇒ T = 45 N For m block: T – m g sin 37 ° - μ m g cos 37 ° = m × 2 ⇒ 45 = 10 m ⇒ m = 4 . 5 kg

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