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NEET2021PhysicsRotational MotionActual

A uniform rod of length 200   cm and mass 500   g is balanced on a wedge placed at 40   cm mark. A mass of 2   kg is suspended from the rod at 20   cm and another unknown mass m is suspended from the rod at 160   cm mark as shown in the figure. Find the value of m such that the rod is in equilibrium. ( g = 10   m   s - 2

Options

  1. A1 2   kg
  2. B1 3   kg
  3. C1 12   kg
  4. D1 6   kg

Correct answer

C. 1 12   kg

Step-by-step solution

Balancing torque about the wedge, we get 2 g × 20 = 0 . 5 g × 60 + m g × 120 ⇒ 2 = 1 . 5 + 6 m ⇒ m = 0 . 5 6 = 1 12   kg

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