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]



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक