A hydraulic automobile lift is designed to lift cars with a maximum mass of `3000Kg`. The area of cross section of the piston carrying the load is `425cm^(2)`. What maximum pressures would the smaller piston have to bear?

Explanation

m=3000 kg
`A=425 cm^(2)=425 xx 10^(-4) m^(2)`
Maximum pressure which the larger piston can bear
`P=F/=(mg)/(A)=(3000 xx 9.8)/(425 xx 10^(-4)) Pa`
`=6.92 xx 10^(5)Pa`
As the pressure is transmitted equally in all directions. Therefore, the maximum presssure which the smaller piston can bear is `6.92 xx 10^(5)Pa.`