$\mathrm{A)\; 2}\pi cm/s$

$\mathrm{B)\; 4}\pi cm/s$

$\mathrm{C)\; 16}\pi cm/s$

$\mathrm{D)\; 8}\pi cm/s$

Solution: the correct answer is $\mathrm{D)\; 8}\pi cm/s$

$ l=1m=100cm~and~s=16cm $

$ l\times (2\theta )=s $

$ 2\theta =\frac{s}{l}=0.16 $

$ \theta =0.08 $

Applying the Law of Conservation of Mechanical Energy and using the lowest point on the journey as a reference for gravitational potential energy: –

Energy at the extreme=Energy at the mean, which is the lowest.

$ mg(l-lcos\theta )=\frac{1}{2}m{{v}^{2}} $

$ v=\sqrt{2gl\left( 1-\cos \theta  \right)} $

$ v=\sqrt{4gl{{\sin }^{2}}\frac{\theta }{2}} $

$ v=2\sqrt{gl}\sin \frac{\theta }{2} $

$ \theta \,is\,very\,small,\,\sin \frac{\theta }{2}\approx \frac{\theta }{2} $

$ v=2\sqrt{gl}\frac{\theta }{2} $

$ v=\theta \sqrt{gl}=0.08\sqrt{{{\pi }^{2}}\times 1} $

$ v=8\pi \,cm/s $