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 \(g=10\,m / s ^2\)]
- A \(200 \times 10^{+6}\,Pa\)
- B \(20 \times 10^{+6}\,Pa\)
- C \(2 \times 10^{+6}\,Pa\)
- D \(2 \times 10^{+5}\,Pa\)
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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