question_answer

The length of second's hand in a watch is 1 cm. The change in velocity of its tip in 15 seconds is      [MP PMT 1987, 2003]

A)             Zero

B)                      \[\frac{\pi }{30\sqrt{2}}cm/\sec \]

C)             \[\frac{\pi }{30}cm/\sec \]

D)                      \[\frac{\pi \sqrt{2}}{30}cm/\sec \]

Correct Answer: D

Solution :

                In 15 second's hand rotate through\[90{}^\circ \].             Change in velocity \[\left| \overrightarrow{\Delta v} \right|=2v\sin (\theta /2)\] \[=2(r\omega )\sin (90{}^\circ /2)\]\[=2\times 1\times \frac{2\pi }{T}\times \frac{1}{\sqrt{2}}\] \[=\frac{4\pi }{60\sqrt{2}}=\frac{\pi \sqrt{2}}{30}\frac{cm}{\sec }\]       [As T = 60 sec]