First slide

Fluid dynamics

Question

Water is flowing through a tube of non-uniform cross-section ratio of the radius at entry and exit end of the pipe is 3 : 2. Then the ratio of velocities at entry and exit of liquid is

Easy

A

4 : 9
B

9 : 4
C

8 : 27
D

1 : 1

Solution

If velocities of water at entry and exit points are v₁ and v₂, then according to equation of continuity,
$A_{1} v_{1} = A_{2} v_{2} \Rightarrow \frac{v_{1}}{v_{2}} = \frac{A_{2}}{A_{1}} = {(\frac{r_{2}}{r_{1}})}^{2} = {(\frac{2}{3})}^{2} = \frac{4}{9}$

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