In a garage, a 1200 kg car stands on a large piston of area 0.6 m².

In a liquid, $P = \rho gh$ at a depth $h$. Water: $\rho = 1000\ \text{kg/m}^3$. Total pressure $=$ air pressure ($1.0 \times 10^{5}\ \text{Pa}$) $+ \rho gh$. A liquid passes a pressure on equally (Pascal's law).

Pressure $=$ thrust $\div$ area, $P = \dfrac{F}{A}$, in pascals ($1\ \text{Pa} = 1\ \text{N/m}^2$). $1\ \text{cm}^2 = 1 \times 10^{-4}\ \text{m}^2$.