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⚛️ Physics  ·  Mechanical Properties of Fluids  ·  NEET & JEE

The critical velocity above which flow becomes turbulent in a tube of diameter d is approximately (Reynolds number ≈ 2000):

Answer: v c = 2000η/(ρd).

  • A v<sub>c</sub> = 2000η/(ρd)
  • B v<sub>c</sub> = 2000ρd
  • C v<sub>c</sub> = η/(ρd)
  • D v<sub>c</sub> = ρd/η

Correct answer: A. v<sub>c</sub> = 2000η/(ρd)

Explanation: Re = ρvd/η. For Re = 2000 (laminar-turbulent transition): v<sub>c</sub> = 2000η/(ρd). For water in a 1 cm pipe: v<sub>c</sub> ≈ 0.2 m/s.

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.

Read the full Mechanical Properties of Fluids notes →