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⚛️ Physics  ·  Mechanical Properties of Fluids  ·  NEET & JEE

Terminal velocity of a sphere of radius r falling in a fluid of viscosity η (Stokes law) is:

Answer: v t = 2r²(ρ-σ)g/9η.

  • A v<sub>t</sub> = 2r²(ρ-σ)g/9η
  • B v<sub>t</sub> = 6πηrv
  • C v<sub>t</sub> = r(ρ-σ)/η
  • D v<sub>t</sub> = 4πr³ρg/3

Correct answer: A. v<sub>t</sub> = 2r²(ρ-σ)g/9η

Explanation: At terminal velocity, weight - buoyancy = Stokes drag: (4/3)πr³(ρ-σ)g = 6πηr v<sub>t</sub>. Solving: v<sub>t</sub> = 2r²(ρ-σ)g/9η.

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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