A simple pendulum bob is given a horizontal velocity $\sqrt{\frac{7}{2}\mathrm{g}\mathcal{l}}$  at the bottom. The string slackens after swinging through an angle $\mathrm{\theta }$ with the vertical. The $\mathrm{\theta }$ is

From the figure,

Tension is zero at B,

$\frac{m{v}^2}{l}=mg\sin \alpha$

By conservation of energy,

$\frac{1}{2}m{v}^2+mg(l+l\sin \alpha )=\frac{1}{2}m{u}^{{ }_2}$

$\frac{m{v}^2}{l}+2mg(1+\sin \alpha )=\frac{7}{2}mg$

$\sin \alpha =\frac{1}{2}$

$$\begin{gathered} \alpha ={30}^{°} \\ \Rightarrow \theta ={90}^{°}+{30}^{°} \\ \Rightarrow \theta ={120}^{°} \end{gathered}$$

Hence the correct answer is ${120}^{°}.$