A bob of a stationary pendulum is given a sharp hit to impart it a horizontal speed of $\sqrt{3gl}.$ Find the angle rotated by the string before it becomes slack.

The correct option is C $\theta =co{s}^{-1}(-\frac{1}{3})$
$V=\sqrt{3gl}$

$\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2=-mgh$

$v^2=u^2-2g\left(1+\cos \theta \right)$

$\Rightarrow v^2=3gl-2gl\left(1+\cos \theta \right)$--------------$\left(1\right)$

Again$,$

$\frac{m{v}^2}{l}=mg\cos \theta$

$v^2=\lg \cos \theta$

from equation $\left(1\right)$ and $\left(2\right),$ we get

$3gl-2gl-2gl\cos \theta =gl\cos \theta$

$3\cos \theta =1\Rightarrow \cos \theta =\frac{1}{3}$

$\theta ={\cos }^{-1}\left(\frac{1}{3}\right)$

So$,$ angle rotated before the string becomes slack

$={180}^0-{\cos }^{-1}\left(\frac{1}{3}\right)={\cos }^{-1}\left(\frac{1}{3}\right)$

Hence,

option $\left(C\right)$ is correct answer.