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.

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

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

\(4 m g l=m v_{2}^{2}-m g l\)

\(v_{2}^{2}=5 g l\)

\(v_{2}=\sqrt{5 g l}\)