In case of vertical circular motion of a particle by a thread of length r if the tension in the thread is zero at an angle 30^ shown in figure, the velocity at the bottom point (A) of the circular path is ( g= gravitational acceleration)

Step-by-step solution

At the point where tension T = 0 and angle from vertical is 60° (i.e., 30° from horizontal): T + mg 60° = mV^2 Since T = 0 : V^2 = g 2 Using mechanical energy conservation from bottom (A) to the point where T = 0 : 1 2 mu^2 = mg( + 60°) + 1 2 mV^2 u^2 = 2g (1 + 1 2 ) + g 2 = 3g + g 2 = 7g 2 u = 7g 2