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AP EAMCET20224 Jul 2022Evening ShiftPhysicsMechanical Properties of FluidsActual

In a hydraulic lift, compressed air exerts a force F on a small piston of radius 3 cm . Due to this pressure the second piston of radius 5 cm lifts a load of 1875 kg . The value of F is (Acceleration due to gravity = 10 m s - 2 )

Options

  1. A1250   N
  2. B125   N
  3. C6750   N
  4. D675   N

Correct answer

C. 6750   N

Step-by-step solution

The principle for hydraulic lifts is based on Pascal's law. Pressure on both the piston will be same, F 1 A = F 2 A ⇒ F 1 π × 3 × 10 - 2 2 = F π × 5 × 10 - 2 2 ⇒ F 9 = 1875 × 10 25 ⇒ F = 6750   N

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