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⚛️ Physics  ·  Mechanical Properties of Fluids  ·  NEET & JEE

Water (ρ = 1000 kg/m³) flows in a horizontal pipe. At point 1: P = 4×10⁵ Pa, v = 2 m/s. At point 2 (same height): v = 6 m/s. Find P<sub>2</sub>.

Answer: 2.4 × 10⁵ Pa.

  • A 2.4 × 10⁵ Pa
  • B 4 × 10⁵ Pa
  • C 5.6 × 10⁵ Pa
  • D 3 × 10⁵ Pa

Correct answer: A. 2.4 × 10⁵ Pa

Explanation: Bernoulli (horizontal): P<sub>1</sub> + ½ρv<sub>1</sub>² = P<sub>2</sub> + ½ρv<sub>2</sub>². P<sub>2</sub> = P<sub>1</sub> + ½ρ(v<sub>1</sub>²-v<sub>2</sub>²) = 4×10⁵ + 500(4-36) = 4×10⁵ - 16000 = 384000 ≈ 3.84×10⁵ Pa.

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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