Q. The second’s hand of a watch has length $6\ cm.$ Speed of end point and magnitude of difference of velocities at two perpendicular positions will be
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 $8.88\ mm/s$
Explanation
(d) $v = r\omega = \frac{r \times 2\pi}{T} = \frac{0.06 \times 2\pi}{60} = 6.28mm/s$
Magnitude of change in velocity$\ = \left| \overrightarrow{v_{2}} - \overrightarrow{v_{1}} \right|$
$= \sqrt{v_{1}^{2} + v_{2}^{2}} = 8.88\ mm/s$ $\lbrack As\ v_{1} = v_{2} = 6.28mm/s\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.