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 \(\theta\) from the horizontal, the ratio of the speed v of the bob at point \(P\) to its initial speed \(v_0\) is:

\(\text {C.O.M.E. } \frac{1}{2} mv_0^2=mg \ell(1+\sin \theta)+\frac{1}{2} mv_{p}^2 \ldots \)
\( \text {At } pt P Tp+mg^2 \sin \theta=\frac{mv_{p}^2}{\ell} \left(\text { as } T_{p}=0\right) \)
\( mg \sin \theta=\frac{mv_{p}^2}{\ell} \Rightarrow mv_{p}^2=mg \ell \sin \theta \ldots \ldots \text { (ii) }\)
From (i) & (ii)
\(\begin{array}{l}\frac{1}{2} m v_0^2=m g \ell(1+\sin \theta)+\frac{1}{2} m g \ell \sin \theta \\ v_0^2=2 g \ell(1+\sin \theta)+g \ell \sin \theta \\ v_0=\sqrt{2 g \ell+3 g \ell \sin \theta} \ldots .(i i i) \\ v _{ p }=\sqrt{ g \ell \sin \theta} \\ v _0=\sqrt{2 g \ell+3 g \ell \sin \theta} \\ \frac{ v _{ p }}{ v _0}=\sqrt{\frac{\sin \theta}{2+3 \sin \theta}}\end{array}\)