A) 4 times that in the other pipe
B) \[{\scriptstyle{}^{1}/{}_{4}}\] times that in the other pipes
C) 2 times that in the other pipe
D) \[{\scriptstyle{}^{1}/{}_{2}}\] times that in the other pipes
Correct Answer: A
Solution :
As, AV = constant Note to the wind that velocity in pipe is inversely proportional to the area of X-section of the pipe: \[\Rightarrow \] \[\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{{{A}_{2}}}{{{A}_{1}}}=\frac{d_{2}^{2}}{d_{1}^{2}}\] \[\Rightarrow \] \[\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{{{4}^{2}}}{{{2}^{2}}}=4\] \[\Rightarrow \] \[{{V}_{1}}=4{{V}_{2}}\]You need to login to perform this action.
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