A) 4 times that in the other pipe
B) \[\frac{1}{4}\] times that in the other pipe
C) 2 times that in the other pipe
D) \[\frac{1}{2}\] times that in the other pipe
Correct Answer: A
Solution :
\[{{d}_{A}}=2\,cm\]and\[{{d}_{B}}=4\ cm\] \[\therefore \ \ {{r}_{A}}=1\ cm\] and \[{{r}_{B}}=2\ cm\] From equation of continuity, av = constant \ \[\frac{{{v}_{A}}}{{{v}_{B}}}=\frac{{{a}_{B}}}{{{a}_{A}}}=\frac{\pi {{({{r}_{B}})}^{2}}}{\pi {{({{r}_{A}})}^{2}}}={{\left( \frac{2}{1} \right)}^{2}}\Rightarrow {{v}_{A}}=4{{v}_{B}}\]
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