Question

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. $2\pi$ & $0$ mm/s
• B. $2\sqrt{2}\pi$ & $44$mm/s
• C. $2\sqrt{2}\pi$ & $2\pi$ mm/
• D. $2\pi$ & $2\sqrt{2}\pi$ mm/s

Answer 1

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

Angular displacement$\left(\theta \right)$ made by the second's hand in 60 seconds$=2\pi$

Angular velocity$=\omega =\theta /t=2\pi /60$ rad/sec.

Velocity$=v=r\times \omega =2\pi$ $mm/s$

Let $v_a$ be the velocity of the hand in one position and $v_b$ be the velocity of hand in second position which is perpendicular to the first position

$\Rightarrow |v_a-v_b|=\sqrt{v_a^2+v_b^2-2v_a{v}_b\cos \theta }$

Putting $v_a=2\pi ,v_b=2\pi ,\theta ={90}^o$, we get

$\Rightarrow |v_a-v_b|=2\sqrt{2}\pi$ mm/s

Hence, correct answer is option $D$