ExamBro
ExamBro
JEE Mains · Physics · STD 11 - Fluid mechanics
  • JEE Mains 2023
  • Easy
  • MCQ

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 \(g=10\,m / s ^2\)]

Pick an option to check your answer.
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

\(\text { Force }= mg =5000\,g\)

\(\text { Area of cross section }=250\,cm ^2=250 \times 10^{-4}\,m ^2\)

\(\text { max imum pressure }=\frac{\text { Force }}{\text { area of cross section }}\)

\(\quad=\frac{5000\,g }{250 \times 10^{-4}}=\frac{20 \times g }{10^{-4}}=2 \times 10^6\,Pa\)

Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app