Question

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

• A. $2\pi$ & $0mm/s$
• B. $2\sqrt{2}\pi$ & $4.44mm/s$
• C. $2\sqrt{2}\pi$ & $2\pi mm/s$
• D. $2\pi$ & $2\sqrt{2\pi }mm/s$

Answer 1

The correct option is D $2\pi$ & $2\sqrt{2\pi }mm/s$

The correct option is D

We have

$l=6cm$

So,

$v=r\omega$

$=r\times \frac{2\pi }{T}$

$=\frac{0.06\times 2\pi }{60}$

$=6.288mm{s}^{-1}$

And the magnitude of thye velocity is change :

$|\overrightarrow{V_2}-\overrightarrow{V_1}|=\sqrt{V_1^2=V_2^2}$

$=\sqrt{{6.27}^2+{6.28}^2}$

$=8.888m{s}^{-1}$

$=6.288mm{s}^{-1}\&8.888m{s}^{-1}$ is similar to

$2\pi$ & $2\sqrt{\pi }mm/s$