NTA Abhyas JEE Main2020PhysicsLaws of MotionPractice
A bead of mass m can slide without friction along a vertical ring of radius R . One end of a spring of force constant k = 3 m g R is connected to the bead and the other end is fixed at the centre of the ring. Initially, the bead is at the point A and due to a small push it starts sliding down the ring. If the bead momentarily loses contact with the ring at the instant when the spring makes an angle of 60 ° with the v
Options
- A5 R 9
- B3 R 4
- C5 R 6
- D4 R 7
Correct answer
C. 5 R 6
Step-by-step solution
Let's say the bead loses contact with the ring at point B , then N 2 = 0 and m g cos 60 ° + k x = m v 2 R m g 2 + k x = m v 2 R Applying the conservation of mechanical energy principle between A & B m v 2 2 = m g R 1 - cos 60 ° = m g R 2 ⇒ m v 2 R = m g m g 2 + k x = m g ⇒ x = m g 2 k = R 6 The natural length of the spring l = R - x = R - R 6 = 5 R 6