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JEE Main20238 Apr 2023Evening ShiftPhysicsMechanical Properties of FluidsActual

A hydraulic automobile lift is designed to lift vehicles of mass 5000 kg . The area of cross section of the cylinder carrying load is 250 cm 2 . The maximum pressure the smaller piston would have to bear is [Assume g = 10 m s - 2 ]

Options

  1. A20 × 10 6   Pa
  2. B2 × 10 5   Pa
  3. C200 × 10 6   Pa
  4. D2 × 10 6   Pa

Correct answer

D. 2 × 10 6   Pa

Step-by-step solution

The given data is m = 5000   kg g = 10   m   s - 2 A = 250 × 10 - 4   m 2 Using the formula of pressure, P = F A = m g A = 5000 × 10 250 × 10 - 4 = 2 × 10 6   N   m - 2 = 2 × 10 6   Pa

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