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99 Percentile Qs Bank for JEE MainPhysicsRotational Motion

A solid ball rolls down a parabolic path ABC from a height h as shown in the figure. The portion AB of the path is rough while BC is smooth. How high will the ball climb in BC ?

Options

  1. AH ⁡ = 5 7 h ⁡
  2. BH ⁡ = 5 2 h ⁡
  3. CH ⁡ = 7 5 h ⁡
  4. DH ⁡ = 3 7 h ⁡

Correct answer

A. H ⁡ = 5 7 h ⁡

Step-by-step solution

Using the conservation of mechanical energy between A and B At B, total kinetic energy = m g h Here, m = mass of the ball The ratio of rotational to translational kinetic energy would be, K ⁡ R ⁡ K ⁡ T ⁡ = 2 5 ∴ K ⁡ R ⁡ = 2 7 mg h ⁡ and K ⁡ T ⁡ = 5 7 mgh In portion BC , friction is absent. Therefore, rotational kinetic energy will remain constant and translational kinetic energy will convert into potential energy. Hence, if H be the height to which ball c

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