Physics - W.E.P. Question with Solution | TestHub

Let $v$ be the speed of the particle when it is at the same horizontal level as the center. By conservation of mechanical energy, $\frac{1}{2}m{v}_0^2=\frac{1}{2}m{v}^2+mgL$. Substituting $v_0=\sqrt{8gL}$, we get $\frac{1}{2}m(8gL)=\frac{1}{2}m{v}^2+mgL$, which simplifies to $4gL=\frac{1}{2}{v}^2+gL$, so $v^2=6gL$. At this position, the tension $T$ in the string provides the centripetal force. Thus, $T=\frac{m{v}^2}{L}=\frac{m(6gL)}{L}=6mg$. The tension is 6 units of $mg$.