A hydraulic automobile lift is designed to lift cars with a maximum mass of 300 kg. the area of cross-section of the piston carrying the load is `425cm^(3)`. What maximum pressure would smaller piston have to bear?

Explanation

According to Pascal's law
`P_(1)=P_(2)=(F_(2))/(A_(2))=(3000xx10)/(425xx10^(-4))`
`Pa=3000/425xx10^(5) Pa=7.06xx10^(5)Pa`