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 1.5 m min^–1, what is the speed of ejection of the liquid through the holes?

 

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^–3 m
Radius of each hole, r = d/2 = 0.5 × 10^–3 m
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 

                             V₂ = A₁V₁ / A₂ 

                                 = 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.