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 P₂ is the same as it is at the top P₁. If the velocity of the top v₁ = 2 m/s. Then the ratio of areas A₁:A₂ is

Moderate

A

4:1
B

2:1
C

8:1
D

4:3

Solution

$U \sin g e q u a t i o n o f c o n t i n u i t y w e h a v e v_{2} = \frac{A_{1}}{A_{2}} v_{1}$
$F r o m B e r n o u l l i' s t h e o r e m, p_{1} + ρ g h_{1} + \frac{1}{2} ρ v_{1}^{2} = p_{2} + ρ g h_{2} + \frac{1}{2} ρ v_{2}^{2}$
$\begin{array}{l} \Rightarrow g (h_{1} - h_{2}) = \frac{1}{2} (v_{2}^{2} - v_{1}^{2}) \\ \Rightarrow g (h_{1} - h_{2}) = \frac{1}{2} (\frac{v_{2}^{2}}{v_{1}^{2}} - 1) v_{1}^{2} \\ \Rightarrow 60 = (\frac{A_{1}^{2}}{A_{2}^{2}} - 1) v_{1}^{2} (∵ v o l u m e r a t e = A_{1} V_{1} = A_{2} V_{2}) \\ \Rightarrow 60 = (\frac{A_{1}^{2}}{A_{2}^{2}} - 1) 2^{2} \\ \Rightarrow \frac{A_{1}}{A_{2}} = \frac{4}{1} \end{array}$

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