Deduce an expression for the pressure at a depth inside a liquid.

Consider a vessel containing a liquid with density ρ. Let the liquid be stationary. In order to calculate pressure at a depth h, consider a horizontal circular surface PQ with area A at depth h below the free surface XY of the liquid as shown below.

The pressure on surface PQ will be due to the weight of the liquid column above the surface PQ, (i.e., the liquid contained in cylinder PQRS of height h with PQ as it's base and top face RS lying on the free surface XY of the liquid).

Thrust exerted on the surface PQ

= Weight of the liquid column PQRS

= Volume of liquid column PQRS x density x g

= (Area of base PQ x height) x density x g

= (A x h) ρ x g = A h ρ g

This thrust is exerted on the surface PQ of area A. Therefore, pressure

P = $\dfrac{\text{Thrust on surface}}{\text{Area of surface}}$ = $\dfrac{\text{A h ρ g}}{\text{A}}$ = h ρ g

Hence, Pressure = depth x density of liquid x acceleration due to gravity = h ρ g