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
- A1 2   kg
- B1 3   kg
- C1 12   kg
- 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