A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 ${{cm}^{2}}$. What maximum pressure would the smaller piston have to bear?

Given, maximum mass that can be lifted (m) = 3000 kg

Area of cross-section (A) = 425 ${{cm}^{2}}$ = 4.25 x ${{10}^{-2}}$ ${{m}^{2}}$

.ā. Maximum pressure on the bigger piston p = F/A = mg/A

= 3000 x 9.8 / 4.25 x ${{10}^{-2}}$ = 6.92 x ${{10}^{5}}$ Pa

According to Pascals law, the pressure applied on an enclosed liquid is transmitted equally in all directions.

.ā. Maximum pressure on smaller piston = maximum pressure on bigger piston

pā = p =6.92 x ${{10}^{5}}$ Pa