NTA Abhyas JEE Main2020PhysicsWork, Power and EnergyPractice
A ring of mass m free to slide on a fixed smooth horizontal rod is attached to a particle of mass M kg by a inextensible string of length l . Initially, both M and m are at rest and the string is vertical. A horizontal velocity v 0 is imparted to the particle. The maximum height up to which block will rise w.r.t its initial position is (M = 2m)
Options
- Av 0 2 2 g
- Bv 0 2 4 g
- Cv 0 2 6 g
- Dv 0 2 8 g
Correct answer
C. v 0 2 6 g
Step-by-step solution
The reduced mass of the system is m reduced = m × 2 m m + 2 m = 2 m 3 ∆ K = 1 2 2 m 3 v 0 2 = 2 m g h ∴ h = 1 2 2 m 3 v 0 2 2 m g = v 0 2 6 g