A body of mass 100 g is moving in circular path of radius 2 m on vertical plane as shown in figure. The velocity of the body at point A is 10 m / s. The ratio of its kinetic energies at point B and C is :
(Take acceleration due to gravity as 10 m / s²)

Solution

Physics sir · 3 steps

Find how high B and C sit above A, then let energy conservation hand you both kinetic energies.

1. Energy at A. With m = 0.1 kg and v_A = 10 m s⁻¹, K_A = 1/2 × 0.1 × 10² = 5 J. Gravity is the only force doing work, so K = K_A - mgh at any point, where h is the height above A.
2. Heights from angles. A point at angle θ from the downward vertical OA is at height h = R(1 - cos θ). B is at 30°. C is a further 90° round, so it is at 120° from OA.

   Point	Angle from OA	Height h = R(1-cos θ)	K = 5 - mgh (with mg = 1 N)
   A	0°	0	5 J
   B	30°	2(1 - √3/2) = 2 - √3 m	5 - (2 - √3) = 3 + √3 J
   C	120°	2(1 + 1/2) = 3 m	5 - 3 = 2 J

   K_B/K_C = (3 + √3)/2

3. So. The ratio of kinetic energies at B and C is (3+√3)/2, which is option (d).

The idea to keep. In vertical circles, measure every angle from the lowest point and use h = R(1 - cos θ). Beyond 90° the cosine is negative, so the height goes above the centre.

Answer from the NTA final answer key. Solution by PrepWiser.