The second's hand of a watch has 6, cm length | vedclass

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(D) Given: length of the second's hand $l = 6\, cm = 60\, mm$.Time period $T = 60\, s$.Angular velocity $\omega = \frac{2\pi}{T} = \frac{2\pi}{60} = \

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$2\pi$ and $2\sqrt{2}\pi\, mm/s$

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(D) Given: length of the second's hand $l = 6\, cm = 60\, mm$.Time period $T = 60\, s$.Angular velocity $\omega = \frac{2\pi}{T} = \frac{2\pi}{60} = \frac{\pi}{30}\, rad/s$.Speed of the tip $v = \omega l = \frac{\pi}{30} 	imes 60\, mm/s = 2\pi\, mm/s$.At two perpendicular positions,the velocity vectors are $\vec{v}_1$ and $\vec{v}_2$,where $|\vec{v}_1| = |\vec{v}_2| = v$.The magnitude of the difference in velocities is $|\Delta \vec{v}| = |\vec{v}_1 - \vec{v}_2| = \sqrt{v^2 + v^2} = v\sqrt{2}$.Substituting $v = 2\pi\, mm/s$,we get $|\Delta \vec{v}| = 2\pi 	imes \sqrt{2} = 2\sqrt{2}\pi\, mm/s$.

The second's hand of a watch has $6\, cm$ length. The speed of its tip and the magnitude of the difference in velocities of its tip at any two perpendicular positions will be respectively:

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