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⚛️ Physics  ·  Mechanical Properties of Fluids  ·  NEET & JEE

In a horizontal Venturi meter, the throat (narrow section) has area A<sub>t</sub> = A/2 where A is pipe area. Fluid density ρ, velocity at pipe entrance v. The height difference Δh in the manometer (density ρ_m) is:

Answer: Δh = 3ρv²/(2(ρ_m - ρ)g).

  • A Δh = 3ρv²/(2(ρ_m - ρ)g)
  • B Δh = ρv²/2ρ_m g overall
  • C Δh = v²/2g in most cases
  • D Δh = 4ρv²/ρ_m g under typical conditions

Correct answer: A. Δh = 3ρv²/(2(ρ_m - ρ)g)

Explanation: By continuity: v<sub>t</sub> = 2v. ΔP = ½ρ(4v² - v²) = 3ρv²/2. Manometer: ΔP = (ρ_m - ρ)gΔh. So Δh = 3ρv²/(2(ρ_m-ρ)g).

Continuity Equation: Narrow Pipe → Faster FlowA₁, v₁ (wide, slow)A₂, v₂ (narrow, fast)A₁v₁ = A₂v₂ (continuity)Smaller area → higher speed → (by Bernoulli) LOWER pressure

Since the same volume of fluid must pass every cross-section per second (continuity), the fluid speeds up where the pipe narrows; Bernoulli's equation then says this faster-moving fluid has lower pressure - the principle behind a venturi meter and an aircraft wing's lift.

Concept context

Pressure, buoyancy, Bernoulli equation, viscosity, and surface tension.

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