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Bernoulli's Principle

An ideal liqu...

Question

An ideal liquid (water) flowing through a tube of non-uniform cross-sectional area, where area at $A$ and $B$ are $40 c m^{2}$ and $20 c m^{2}$ respectively. If pressure difference between $A$ and $B$ is $700 \frac{N}{m^{2}}$ , then volume flow rate is (density of water = $1000 \frac{k g}{m^{3}}$ )

$2720 \frac{c m^{3}}{s}$

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$2420 \frac{c m^{3}}{s}$

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$1810 \frac{c m^{3}}{s}$

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$3020 \frac{c m^{3}}{s}$

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Solution

The correct option is A
$2720 \frac{c m^{3}}{s}$

Step1: Given data.
Cross-sectional area of a tube at point $A$ , $A_{A} = 40 c m^{2}$
Cross-sectional area of a tube at point $B$ , $A_{B} = 20 c m^{2}$
Change in pressure between point $A$ and $B$ , $∆ P = P_{A} - P_{B} = 700 \frac{N}{m^{2}}$
Density of water, $ρ = 1000 k g m^{- 3}$
Step2: Finding the volume flow rate.
We assume that , $V_{A} = v o l u m e f l o w r a t e a t A V_{B} = v o l u m e f l o w r a t e a t B$
From equation of continuity:
$A_{A} V_{A} = A_{B} V_{B}$
$\Rightarrow$ $40 V_{A} = 20 V_{B}$
$\Rightarrow$ $2 V_{A} = V_{B}$
$\Rightarrow$ $V_{A} = \frac{V_{B}}{2}$
Again,
Using Bernoulli's equation:
$P_{A} + \frac{1}{2} ρ V_{A}^{2} = P_{B} + \frac{1}{2} ρ V_{B}^{2}$
$P_{A} - P_{B} = \frac{1}{2} ρ (V_{B}^{2} - V_{A}^{2})$
$∆ P = \frac{1}{2} ρ (V_{B}^{2} - V_{A}^{2})$
$∆ P = \frac{1}{2} ρ (V_{B}^{2} - {(\frac{V_{B}}{2})}^{2})$
$∆ P = \frac{1}{2} ρ (V_{B}^{2} - \frac{V_{B}^{2}}{4})$
$700 = \frac{1}{2} \times 1000 \times \frac{3}{4} V_{B}^{2}$
$V_{B} = \sqrt{\frac{28}{15}} m s^{- 1} = \sqrt{\frac{28}{15}} \times 100 c m s^{- 1}$
Now,
Volume flow rate $= V_{B} \times A_{B}$
$= \sqrt{\frac{28}{15}} \times 100 \times 20$
$= 2732 c m^{3} s^{- 1}$
So, Answer comes nearly $2720 \frac{c m^{3}}{s}$
Hence, option A is correct.

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An ideal liquid (water) flowing through a tube of non-uniform cross-sectional area, where area at A and B are 40cm2 and 20cm2 respectively. If pressure difference between A and B is 700Nm2, then volume flow rate is (density of water = 1000kgm3)

An ideal liquid (water) flowing through a tube of non-uniform cross-sectional area, where area at $A$ and $B$ are $40 c m^{2}$ and $20 c m^{2}$ respectively. If pressure difference between $A$ and $B$ is $700 \frac{N}{m^{2}}$ , then volume flow rate is (density of water = $1000 \frac{k g}{m^{3}}$ )