A) \[3.8\times {{10}^{-3}}cm\text{/}s\]
B) \[4.2\times {{10}^{-7}}cm\text{/}s\]
C) \[4.4\times {{10}^{-3}}cm\text{/}s\]
D) \[4.4\times {{10}^{-5}}cm\text{/}s\]
Correct Answer: C
Solution :
As between the time 6 : 00 am and 6 : 30 am the angular change is \[180{}^\circ \] The linear distance between the initial and final position of the tip is equal to the diameter of the clock. In this case, displacement \[=2\times 4=8\,cm\] Also, number of seconds between 6:00 am and 6:30 am = 60 x 30 = 1800 s Thus, average velocity in this interval of time displacement \[{{v}_{av}}=\frac{\text{displacement}}{\text{time}}=\frac{8}{1800}=4.4\times {{10}^{-3}}cm\text{/}s\]
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