NTA Abhyas JEE Main2020PhysicsLaws of MotionPractice
A particle of mass m is suspended from a ceiling through a string of length L . The particle moves in a horizontal circle of radius r . The speed ( v ) of the particle and the tension ( T ) in the string is
Options
- Av = 2 r g L 2 - r 2 1/2  and T = 3 m g L L 2 - r 2 1/2
- Bv = r g L 2 - r 2 1/4 and T = m g L L 3 - r 3 1/3
- Cv = 3 r g L 2 - r 2 1/2 and T = m g L L 2 - r 2 1/2
- Dv = r g L 2 - r 2 1/4 and T = m g L L 2 - r 2 1/2
Correct answer
D. v = r g L 2 - r 2 1/4 and T = m g L L 2 - r 2 1/2
Step-by-step solution
The situation is shown in the diagram. The angle θ made by the string with the vertical is given by sin θ = r L ...(i) The forces on the particle are (1) the tension T along the string and (2) the weight mg vertically downward. The particle is moving in a circle with constant speed v . Thus, the radial acceleration towards the centre has magnitude v 2 r . Resolving the forces along the radial direction and applying Newton's second law, T   sin ​ θ   =   m v 2 r   ...(ii