The second's hand of watch has length 6 cm. Speed of end point and magnitude of difference of velocitie at two perpendicular positions will be

Options

A. `2 pi` & `0 mm//s`
B. `2sqrt(2)pi` & `4.44 mm//s`
C. `2sqrt(2) pi` & `2pi mm//s`
D. `2pi` & `2sqrt(2) mm//s`

Answer

D

Explanation

`l=6 cm, v=?, omega=(2pi)/60=pi/30 rad//s`
So `v=omegal=pi/30xx6=pi/5 cm//s=2pi mm//s`
Difference `=sqrt(2) pi/5 cm//s=2sqrt(2) pi mm//s`