Question

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?

Answer 1

The maximum force, which the bigger piston can bear, `F = 300 kgf = 3000 xx 9.8 N`
`:.` Maximum pressure on the bigger piston, `P=F/A = (300 xx 9.8)/(425xx10^(-4)) = 6.92 xx 10^(5)Pa`
Since the liquid transmits pressure equally, therefore the maximum pressure the smaller piston can bear is `6.92 xx 10^(5)Pa`.