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

Options

A. `6.28 and 0 mm//s`
B. `8.88 and 4.44 mm//s`
C. `8.88 and 6.28 mm//s`
D. `6.28 and 8.88mm//s`

Answer

D

Explanation

`v=r omega =(rxx2pi)/(T)=(0.06xx2pi)/(60)=6.28 mm//s` magnitude of change in velocity `=|vecv_(n)-vecv_(1)|`
`sqrt(v_(1)^(2)+v_(2)^(2))=8.88 mm//s (As v_(1)=v_(2)=6.28 mm//s)`