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 ) $
