A bob of mass m is suspended at a point O by a light string of length l and left to perform vertical motion (circular) as shown in figure. Initially, by applying horizontal velocity v₀ at the point ' A ', the string becomes slack when, the bob reaches at the point ' D '. The ratio of the kinetic energy of the bob at the points B and C is _________.

Solution

Physics sir · 3 steps

Slack at the top means tension is zero at D, so v_D² = gl. Then energy conservation gives B and C.

1. Condition at D. At the top point D, tension and gravity both point towards O. The string just goes slack there, so T = 0 and gravity alone supplies the centripetal force: mg = mv_D²/l, so v_D² = gl.
2. Speed at A. D is 2l above A. Energy conservation gives v₀² = v_D² + 2g(2l) = 5gl.

   v² = v₀² - 2gh = 5gl - 2gh

   Point	Height above A	v² = 5gl - 2gh
   B (60° from downward vertical)	l - l cos 60° = l/2	5gl - gl = 4gl
   C (60° from upward vertical)	l + l cos 60° = 3l/2	5gl - 3gl = 2gl

3. So. Kinetic energy is proportional to v², so K_B/K_C = 4gl/2gl = 2, option (c).

The idea to keep. For a string, slack at the top means T = 0 there, not v = 0: use vₜₒₚ² = gl. Then measure every height from the lowest point and apply energy conservation.

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