Question 15 of 18intermediate🔧 ApplyNumerical3 marks

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?

Correct Answer

The speed of ejection of the liquid through the holes is 0.64 m/s (or approximately 2/π m/s).

Exercise: EXERCISES | Q: 9.16 | (Page 22)
For More Understanding

Explanation

This problem applies the equation of continuity for incompressible fluid flow. The volume flow rate must remain constant throughout the tube and holes system. The key insight is that the total cross-sectional area of all 40 holes is much smaller than the tube area, causing the liquid to speed up significantly as it exits through the holes.

Solution Steps

  1. Step 1: Convert given values to SI units. Cross-sectional area of tube A₁ = 8.0 cm² = 8.0×108.0 \times 10⁻⁴ m². Speed of liquid in tube v₁ = 1.5 m min⁻¹ = 1.5/60 m/s = 0.025 m/s.

  2. Step 2: Calculate area of one hole. Diameter d = 1.0 mm = 1.0×101.0 \times 10⁻³ m. Radius r = d/2 = 0.5×100.5 \times 10⁻³ m. Area of one hole = πr² = π × (0.5×100.5 \times 10⁻³)² = π × 0.25×100.25 \times 10⁻⁶ m².

  3. Step 3: Calculate total cross-sectional area of 40 holes. Total area A₂ = 40 × π × 0.25×100.25 \times 10⁻⁶ = π × 10⁻⁵ m².

  4. Step 4: Apply equation of continuity (A₁v₁ = A₂v₂) for steady flow. The volume flow rate remains constant.

  5. Step 5: Calculate ejection speed. v₂ = A₁v₁/A₂ = (8.0×108.0 \times 10⁻⁴ × 0.025)/(π × 10⁻⁵) = (2.0×102.0 \times 10⁻⁵)/(π × 10⁻⁵) = 2/π ≈\approx 0.64 m/s.

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