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Bernoulli's Principle

An ideal flui...

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 is $v_{1} = 2 m/s$ . Then the ratio of areas $\frac{A_{1}}{A_{2}}$ is

2 : 1

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4 : 1

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8 : 1

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4 : 3

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Solution

The correct option is B $4 : 1$
Given,
$P_{1} = P_{2}$
$v_{1} = 2 m / s$
By continuity equation,
mass flow rate is same
$A_{1} v_{1} = A_{2} v_{2}$ ( $∵ ρ_{1} = ρ_{2}$ )
$⟹ \frac{A_{1}}{A_{2}} = \frac{v_{2}}{v_{1}}$ ----- (1)
Now by using bernoulli's equation,
$P V + \frac{1}{2} m v^{2} + m g h = c o n t .$
For per unit mass, equation becomes
$\frac{P V}{m} + \frac{1}{2} \frac{m v^{2}}{m} + \frac{m g h}{m} = c o n t .$
Now applying equation at inlet and outlet,
$\frac{P_{1}}{ρ} + \frac{v_{1}^{2}}{2} + g h_{1} = \frac{P_{2}}{ρ} + \frac{v_{2}^{2}}{2} + g h_{2}$
as, $P_{1} = P_{2}$ and $h_{1} - h_{2} = 3$

$⟹ \frac{2^{2}}{2} + 10 \times 3 = \frac{v_{2}^{2}}{2}$
$⟹ 4 + 60 = v_{2}^{2}$
$v_{2} = 8 m / s$
put $v_{1}$ and $v_{2}$ in equation (1) we get,
$\frac{A_{1}}{A_{2}} = \frac{8}{2} = \frac{4}{1}$

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