Q 11-09-098JEE MainJEE Main 2023 (8 Apr, Shift 2)Easy
A hydraulic automobile lift is designed to lift vehicles of mass $5000$ kg. The area of cross section of the cylinder carrying load is $250\ \text{cm}^2$. The maximum pressure the smaller piston would have to bear is [Assume $g=10\ \text{m s}^{-2}$]
Answer: (D) $2\times10^6\ \text{Pa}$
By Pascal's law the pressure is the same on both pistons: $P=\dfrac{5000\times10}{250\times10^{-4}}=2\times10^6$ Pa.
Solution by Sreeraj P, M.Sc Physics