NTA Abhyas JEE Main2020PhysicsWork, Power and EnergyPractice
The initial configuration of the system is as shown in the figure. The string and the pulley are light. The bob is released and the string starts wrapping around the pulley. (The pulley is held in place by a force applied at the centre) Find the rate r at which the length of the string wrapped around the pulley increases (as a function of θ ).
Options
- Ar = R l - Rθ 2 g l - Rθ sinθ
- Br = R l - Rθ 2 g l - Rθ sinθ + Rcosθ
- Cr = R l - Rθ 2 g l - Rθ sinθ + R 1 - sin θ
- Dr = R l - Rθ 2 g l - Rθ sinθ + R 1 - cosθ
Correct answer
D. r = R l - Rθ 2 g l - Rθ sinθ + R 1 - cosθ
Step-by-step solution
At any instant of time, the length of the string wrapped on the pulley is Rθ and the remaining length L = l - Rθ The rate r = R dθ dt = Rω = R v l - Rθ , where v is the speed of the particle The height through which the particle has come down h = R 1 - cosθ + l - Rθ sinθ ⇒ v = 2 g R 1 - cosθ + l - Rθ sinθ So, r = R l - Rθ 2 g l - Rθ sinθ + R 1 - cosθ