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⚛️ Physics  ·  Mechanical Properties of Fluids  ·  NEET & JEE

Two pistons in a hydraulic system have areas 1 cm² and 50 cm². For a load of 500 N on the large piston, the required force on the small piston is 10 N. If the small piston moves down by 50 cm, the large piston moves up by:

Answer: 1 cm.

  • A 1 cm
  • B 50 cm
  • C 100 cm
  • D 0.02 cm

Correct answer: A. 1 cm

Explanation: Volume conservation: A<sub>1</sub> × d<sub>1</sub> = A<sub>2</sub> × d<sub>2</sub>. 1 × 50 = 50 × d<sub>2</sub>. d<sub>2</sub> = 1 cm. Energy is conserved: work in = work out (ideal case).

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 →