# Question #8ae4b

Mar 25, 2017 Given that an ideal fluid flows in the tube as shown figure.

The pressure of the fluid at the bottom $\left({P}_{2}\right)$ is equal to the pressure of the fluid at the top $\left({P}_{1}\right)$ i.e. $\left({P}_{1} = {P}_{2}\right)$ )

The difference in height of the top and bottom ${h}_{1} - {h}_{2} = 3 m$

The velocity of the fluid at the top ${v}_{1} = 2 m \text{/} s$

The velocity of the fluid at the bottom ${v}_{2} = \text{unknown}$

The area of cross-section of the pipe at the top ${A}_{1}$

The area of cross-section of the pipe at the bottom ${A}_{2}$

We are to find out the ratio of ${A}_{1} : {A}_{2}$

Now applying Bernoulli principle

$\text{Energy per unit volume at the top"= "Energy per unit volume at the bottom}$

we have the following equation

${P}_{1} + \frac{1}{2} \rho {v}_{1}^{2} + \rho g {h}_{1} = {P}_{2} + \frac{1}{2} \rho {v}_{2}^{2} + \rho g {h}_{2}$

where

$\rho = \text{density of the fluid}$

$\mathmr{and} g = \text{acceleration due to pop gravity"=10m"/} {s}^{2}$

Inserting given values in the above equation we get

${\cancel{P}}_{2} + \frac{1}{2} \cancel{\rho} {2}^{2} + \cancel{\rho} 10 {h}_{1} = {\cancel{P}}_{2} + \frac{1}{2} \cancel{\rho} {v}_{2}^{2} + \cancel{\rho} 10 {h}_{2}$

$\implies 2 + 10 {h}_{1} = \frac{1}{2} {v}_{2}^{2} + 10 {h}_{2}$

$\implies 2 + 10 {h}_{1} - 10 {h}_{2} = \frac{1}{2} {v}_{2}^{2}$

$\implies 2 + 10 \left({h}_{1} - {h}_{2}\right) = \frac{1}{2} {v}_{2}^{2}$

$\implies 2 + 10 \times 3 = \frac{1}{2} {v}_{2}^{2}$

$\implies {v}_{2}^{2} = 64$

$\implies {v}_{2} = \sqrt{64} = 8 m \text{/} s$

Now applying principle of continuity of flow of ideal fluid we can write

${A}_{1} / {A}_{2} = {v}_{2} / {v}_{1} = \left(8 m \text{/"s)/(2m"/} s\right) = \frac{4}{1}$ which is option (2)