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MHT CET202618 April 2026Morning ShiftPhysicsMechanical Properties of FluidsActual

A piston of cross-sectional area 2.5 10⁻² m ^2 is used in a hydraulic lift to exert a force of 250 N on water. The cross-sectional area of the other piston which supports a car of mass 3000 kg is ( g = 9.8 m/s ^2 )

Options

  1. A1.96 m ^2
  2. B2.94 m ^2
  3. C3.92 m ^2
  4. D5.88 m ^2

Correct answer

B. 2.94 m ^2

Step-by-step solution

Using Pascal's law, the pressure applied on the first piston is equal to the pressure on the second piston. F₁ A₁ = F₂ A₂ Given F₁ = 250 N , A₁ = 2.5 10⁻² m ^2 , m = 3000 kg , and g = 9.8 m/s ^2 . The force on the second piston is F₂ = mg = 3000 9.8 = 29400 N . Substituting the values: 250 2.5 10⁻² = 29400 A₂ 10000 = 29400 A₂ A₂ = 29400 10000 = 2.94 m ^2 Answer: 2.94 m ^2

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