A simple pendulum is released from $A$ as shown. If $M$ and $l$ represent the mass of the bob and length of the pendulum respectively, the gain in kinetic energy at B is

The correct option is B $\frac{\sqrt{3}Mgl}{2}$
Vertical height $=h=l\cos \theta =l\cos {30}^{\circ }$
Loss of potential energy $=Mgh=Mgl\cos$${30}^{\circ }=\frac{\sqrt{3}}{2}Mgl$

By the conservation of energy
Loss of potential energy = kinetic energy gained
$\therefore$ kinetic energy gained $=\frac{\sqrt{3}}{2}Mgl$