A bob of heavy mass $m$ is suspended by a light string of length $l$. The bob is given a horizontal

Question

A bob of heavy mass $m$ is suspended by a light string of length $l$. The bob is given a horizontal velocity $v_0$ as shown in figure. If the string gets slack at some point $P$ making an angle $	heta$ from the horizontal, the ratio of the speed $v$ of the bob at point $P$ to its initial speed $v_0$ is

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(b) : Let speed at $P$ be $v$. \ $ \begin{aligned} & P M=l \sin 	heta \ & h=l+l \sin 	heta \end{aligned} $ \ Using conservation of energy, \ $ \begin{aligned} & \frac{1}{2} m v_0^2+0=m g h+\frac{1}{2} m v^2 \ & \frac{1}{2} v_0^2=g l(1+\sin 	heta)+\frac{1}{2} v^2 \end{aligned} $ \ At $P, T=0$ So, $T-m g \cos \left(90^{\circ}+	hetaight)=\frac{m v^2}{l}$ $ m g \sin 	heta=\frac{m v^2}{l} \ $ or $\frac{
u^2}{\sin 	heta}=g l$ \ Putting in (i), $ \begin{aligned} & \frac{1}{2} v_0^2=\frac{2 \cdot v^2}{2 \cdot \sin 	heta}(1+\sin 	heta)+\frac{1}{2} v^2 \ & \frac{1}{2} v_0^2=\frac{v^2}{2}\left[\frac{2}{\sin 	heta}(1+\sin 	heta)+1ight] \ & v_0^2=v^2\left[\frac{2+2 \sin 	heta+\sin 	heta}{\sin 	heta}ight] \ & v_0^2=v^2 	imes\left[\frac{2+3 \sin 	heta}{\sin 	heta}ight] \ & \frac{v^2}{v_0^2}=\frac{\sin 	heta}{2+3 \sin 	heta} \ & \frac{v}{v_0}=\sqrt{\frac{\sin 	heta}{213 \sin 	heta}} \end{aligned} $