A solid sphere rolls down from the top of an inclined plane. On reaching the bottom of the plane, its velocity is ' V₁ '. When the same sphere slides down from the top of the same plane of same height, its velocity on reaching the bottom is ' V₂ '. The ratio V₁ : V₂ is (neglect friction)

Step-by-step solution

When a solid sphere rolls down an inclined plane of height h , by conservation of mechanical energy: mgh = 1 2 mV₁^2 + 1 2 I ^2 For a solid sphere, the moment of inertia is I = 2 5 mr^2 and for pure rolling, = V₁ r . mgh = 1 2 mV₁^2 + 1 2 ( 2 5 mr^2 ) ( V₁ r )^2 mgh = 1 2 mV₁^2 + 1 5 mV₁^2 = 7 10 mV₁^2 V₁ = 10gh 7 When the sphere slides down the same plane without friction, it only acquires translational kinetic energy. By conservation of mechanical energy: mgh = 1 2 mV₂^2 V₂ = 2gh The ratio of the velocities is: V