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?

The law of continuity states that,

A₁v₁ = A₂v₂

⇒ v₂ = A₁v₁ / A₂ ……….(i)

Where,

‘A₁’ is the area of cross section of the spray pump

‘A₂’ is the area of cross section of the end with the 40 holes

‘v₁’is the speed of flow of fluid inside the tube

‘v₂’ is the speed of flow of the fluid through the holes

Given,

Area of cross section of spray pump, A₁ = 8 cm² = 8 × 10^-4 m²

Diameter of each hole, d = 1 mm = 1 × 10^-3 m

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

Velocity of the liquid flow inside the tube, v₁ = 1.5 m min^-1

⇒ v₁ = 1.5/60 m sec^-1 = 0.025 m/s

Thus,

Area of cross section of each hole, a = πr²

⇒ a = 3.14 × (0.5 × 10^-3 m)²

⇒ a = 0.785 × 10^-6 m²

Total area of cross section of 40 holes, A₂ = 40 × a

⇒ A₂ = 40 × 0.785 × 10^-6 m²

⇒ A₂ = 31.41 × 10^-6 m²

Therefore, from equation (i),

⇒ v_{1 =}

⇒ 0.633 m/s

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