First slide

Fluid dynamics

Question

An ideal fluid flows in the pipe as shown in the figure. The pressure in the fluid at the bottom $P_{2}$ is the same as it is at the top $P_{1}$ . If the velocity of the top $v_{1}$ = 2 m/s. Then the ratio of areas $A_{1}$ . $A_{2}$ is

Moderate

A

2 : 1
B

4 : 1
C

8 : 1
D

4 : 3

Solution

Using equation of continuity, we have
$v_{2} = \frac{A_{1}}{A_{2}} v_{1}$
From Bernoulli's theorem,
$p_{1} + ρ g h_{1} + \frac{1}{2} ρ v_{1}^{2}$
$p_{2} + ρ g h_{2} + \frac{1}{2} ρ v_{2}^{2}$ $\Rightarrow$ $g (h_{1} - h_{2}) = \frac{1}{2} (v_{2}^{2} - v_{1}^{2})$
$\Rightarrow 60 = (\frac{A_{1}^{2}}{A_{2}^{2}} - 1) v_{1}^{2}$ $\Rightarrow \frac{A_{1}}{A_{2}} = \frac{4}{1}$

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