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JEE Mains · Physics · STD 11 - 9.1 fluid mechanics

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

  1. A \(200 \times 10^{+6}\,Pa\)
  2. B \(20 \times 10^{+6}\,Pa\)
  3. C \(2 \times 10^{+6}\,Pa\)
  4. D \(2 \times 10^{+5}\,Pa\)
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Answer & Solution

Correct Answer

(C) \(2 \times 10^{+6}\,Pa\)

Step-by-step Solution

Detailed explanation

\(\text { Force }= mg =5000\,g\) \(\text { Area of cross section }=250\,cm ^2=250 \times 10^{-4}\,m ^2\) \(\text { max imum pressure }=\frac{\text { Force }}{\text { area of cross section }}\)…
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