First slide

Fluid dynamics

Question

The cylindrical tube of a spray pump has a cross-section of 6 ${cm}^{2}$ one of which has 50 holes each of diameter I mm. If the liquid flow inside the tube is 1.2 m per minute, then the speed of ejection of the liquid through the holes is

Moderate

A

9 : 4
B

3 : 2
C

$\sqrt{3} : \sqrt{2}$
D

$\sqrt{2} : \sqrt{3}$

Solution

Here, area of cross section of tube
$a_{1} = 6 {cm}^{2} = 6 \times 10^{- 4} m^{2}$
Number of holes = 50
Diameter of each hole, D = 1 mm = $10^{- 3}$ m
Radius of hole, r = $\frac{D}{2} = \frac{1}{2} \times 10^{- 3} = 5 \times 10^{- 4} m$
Area of cross-section of each hole
= ${πr}^{2} = π {(5 \times 10^{- 4})}^{2} m^{2}$
Total area of cross-section of 50 holes,
$a_{2} = 50 \times π {(5 \times 10^{- 4})}^{2} m^{2}$
Speed of liquid inside the tube,
$v_{1} = 1.2 m \min^{- 1}$
$= \frac{1.2}{60} {ms}^{- 1} = 0.02 {ms}^{- 1}$
Let the velocity of ejection of the liquid through the hole $v_{2}$
As $a_{1} v_{1} = a_{2} v_{2}$
or $v_{2} = \frac{a_{1} v_{1}}{a_{2}} = \frac{6 \times 10^{- 4} \times 0.02}{50 \times π {(5 \times 10^{- 4})}^{2}} = 0.31 {ms}^{- 1}$

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