Kinematics Question 745

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 [RPET 1997]

Options:

A) 6.28 and 0 mm/s

B) 8.88 and 4.44 mm/s

C) 8.88 and 6.28 mm/s

D) 6.28 and 8.88 mm/s

Show Answer

Answer:

Correct Answer: D

Solution:

$ v=r\omega =\frac{r\times 2\pi }{T}=\frac{0.06\times 2\pi }{60}=6.28mm/s $ Magnitude of change in velocity =$ |\overrightarrow{v _{2}}-\overrightarrow{v _{1}}| $

$ =\sqrt{v _{1}^{2}+v _{2}^{2}}=8.88mm/s $

$ ( \text{As }v _{1}=v _{2}=6.28mm/s ) $



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