l त्रिज्या के ऊध्र्वाधर वृत्त में उच्चतम ब

$T+m g=\frac{m v_{1}^{2}}{l}$v

There is certain velocity so called as critical velocity/minimum velocity (v) of object at highest point below which string become slack le. tension T vanishes (T=0).

$m g=\frac{m v_{1}^{2}}{l}$

$v_{1}=\sqrt{g l}$

The decrease in potential energy between top $-$position and bottom position is

$m g l-(-m g l)=2 m g l$

This must be equal to the increase in kinetic energy, when particle move from highest point

i.e.

$\frac{1}{2} m v_{2}^{2}-\frac{1}{2} m v_{1}^{2}$

Using law of conservation of energy.

$2 m g l=\frac{1}{2} m v_{2}^{2}-\frac{1}{2} m v_{1}^{2}$

$2 m g l=\frac{1}{2} m v_{2}^{2}-\frac{1}{2} m g l$

$4 m g l=m v_{2}^{2}-m g l$

$v_{2}^{2}=5 g l$

$v_{2}=\sqrt{5 g l}$