# A bob of mass m is suspended by inextensible string of length l from a fixed point. The bob is given a speed of $\sqrt{6gl}$ at the bottom point. Find the tension in the string when string deflects through an angle 120 degree from vertical.

$\\ \text{By conservation of energy theorem,}\\ \frac{1}{2} mu^{2}=m g\frac{3l}{2}+\frac{1}{2} m v^{2}\\ Or,\ v=\sqrt{3 g l}\\ At the above point , T+m g \cos 60^{\circ}=\frac{m v^{2}}{l}\\ By putting v=\sqrt{3 g l}, we get T=\frac{5}{2} m g$

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