The seconds hand of a watch has length $6cm$. Speed of the end point and magnitude of difference in velocities at positions $A$ and $B$, as shown in the figure will be

The correct option is A $2\pi \frac{mm}{s},2\sqrt{2}\pi \frac{mm}{s}$
We know that speed is given as
$v=R\omega =R\left(\frac{2\pi }{t}\right)$
$=60mm\times \frac{2\pi }{60}$
$=2\pi mm/s$

Since it performs uniform circular motion, change in velocity can be given as
$|\Delta \overrightarrow{v}|=\sqrt{v^2+v^2-2v\times v\cos \theta }$
$\because$ $\theta ={90}^{\circ }=\frac{\pi }{2}\text{rad}$
$\Rightarrow |\Delta \overrightarrow{v}|=\sqrt{v^2+v^2-2v\times v\cos \left(\frac{\pi }{2}\right)}=\sqrt{2}v$
$\Rightarrow |\Delta \overrightarrow{v}|=2\sqrt{2}\pi mm/s$