Question

The cylindrical tube of a spray pump has a cross-section of 8.0 cm2 one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5m min -1. What is the speed of ejection of the liquid through the holes?

Answer 1

Area of cross-section of the spray pump, A1 = 8 cm2 = 8 × 10 - 4 m2
Number of holes, n = 40
Diameter of each hole, d = 1 mm = 1 × 10 - 3m
Radius of each hole, r = $\frac{d}{2}$ = 0.5 × 10 - 3m
Area of cross-section of each hole, a = πr2 = π(0.5 × 10 - 3)2 m2
Total area of 40 holes, A2 = n × a

= 40 × π (0.5 × 10–3) 2 m2

= 31.41 × 10–6 m2

Speed of flow of liquid inside the tube, V1 = 1.5 m / min = 0.025 m / s
Speed of ejection of liquid through the holes = V2
According to the law of continuity, we have

A1V1 = A2V2

$V_2 =\frac{ A_1V_1 }{ A_2}$

$= \frac{8 × 10 ^{- 4} × 0.025 }{ 31.61 × 10 ^{- 6} }$

= 0.633 m / s
Therefore, the speed of ejection of the liquid through the holes is 0.633 m / s.