Kinematics Question 726
Question: The length of second’s hand in a watch is 1 cm. The change in velocity of it’s tip in 15 seconds is[MP PMT 1987, 2003]
Options:
A) Zero
B)$ \frac{\pi }{30\sqrt{2}}cm/\sec $
C) $ \frac{\pi }{30}cm/\sec $
D)$ \frac{\pi \sqrt{2}}{30}cm/\sec $
Show Answer
Answer:
Correct Answer: D
Solution:
In 15 second’s hand rotate through$ 90{}^\circ $ .
Change in velocity $ | \overrightarrow{\Delta v} |=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]
