Correct answer option is (B)

Loss in kinetic energy $=$ Gain in potential energy

$\begin{array}{rl}& \Rightarrow \frac{1}{2}{\mathrm{mv}}^2=\mathrm{mg}\ell (1-\cos \theta )\\ & \Rightarrow \frac{{\mathrm{v}}^2}{\ell }=2\mathrm{g}(1-\cos \theta )\end{array}$

Acceleration at lowest point $=\frac{{\mathrm{v}}^2}{\ell }$

Acceleration at extreme point $=g\sin \theta$

$\begin{array}{rl}& \text{ Hence, }\frac{{\mathrm{v}}^2}{\ell }=\mathrm{g}\sin \theta \\ & \therefore \sin \theta =2(1-\cos \theta )\\ & \Rightarrow \tan \frac{\theta }{2}=\frac{1}{2}\Rightarrow \theta =2{\tan }^{-1}\left(\frac{1}{2}\right)\end{array}$