A) 0.1
B) 0.2
C) 0.3
D) 0.4
E) 0.5
Correct Answer: D
Solution :
From the equation of continuity, we have Av = rate of flow (R) Given, \[R=200\,c{{m}^{3}}{{s}^{-1}},\]\[A=0.5\,{{m}^{2}}\] \[=0.5\times {{({{10}^{-3}})}^{2}}mm\] and \[R=200\times {{(10)}^{3}}mm{{s}^{-1}}\] \[\therefore \] \[v=\frac{200\times {{(10)}^{2}}}{0.5\times {{10}^{-6}}}\] \[=400\times {{10}^{-6}}\times {{10}^{3}}\] \[=0.4\,mm\,{{s}^{-1}}\]
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