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Water is flowing through a tube of non–uniform cross–section. If the radius of the tube at the entrance and exist is 3 : 2, then the ratio of velocity of liquid entering and leaving the tube is

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8 : 27

1 : 1

4 : 9

9 : 4

answer is B.

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Detailed Solution

$A v = constant$

$\Rightarrow$ $π R^{2} v = constant$

$\Rightarrow$ $R^{2} v = constant$

$R_{1}^{2} v_{1} = R_{2}^{2} v_{2}$

$\Rightarrow$ $\frac{v_{1}}{v_{2}} = {(\frac{R_{2}}{R_{1}})}^{2} = {(\frac{2}{3})}^{2} = \frac{4}{9}$

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