NTA Abhyas JEE Main2020PhysicsWork, Power and EnergyPractice
A spherical ball of mass m is kept at the highest point in the space between two fixed, concentric spheres A and B (see fig.). The smaller sphere A has a radius R and space between the two spheres has a width d . The ball has a diameter very slightly less than d . All surfaces are frictionless. The ball is given a gentle push (towards the right in the figure). The angle made by the radius vector of the ball with the
Options
- Am g   ( 3 cos θ   – 2   )
- Bm g   (   2 cos θ   –   3   )
- C3 m g   (   2 cos θ   –   1   )
- D2 m g   (   3 cos θ   –   1   )
Correct answer
A. m g   ( 3 cos θ   – 2   )
Step-by-step solution
h = R + d 2 1 - cos  θ The velocity of the ball at an angle θ is ν 2 = 2 g h = 2 R + d 2 1 - cos  θ g ...... (i) Let N be the normal reaction (away from the centre) at angle θ . Then, mg  cos  θ - N = m ν 2 R + d 2 Substituting the value of ν 2 from equation (i), we get mg  cos  θ - N = 2 mg 1 - cos  θ N = m g 3 cos  θ - 2