Discuss the variation of pressure with depth or pressure produced due to fluid depth $\mathrm{h}$ and density of fluid $\rho $.

Suppose a fluid of density $\rho$ is in a static equilibrium in a container as shown in figure.

Consider a cylindrical element of fluid having area of base $A$ and height $h$. The pressure at points $1$ and$ 2$ are $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$ respectively.

The force acting on liquid column is as below :

$(1)$ The forces on point-1 $\mathrm{F}_{1}=\mathrm{P}_{1} \mathrm{~A}$ (downward)

$(2)$ The force on point- $2 \mathrm{~F}_{2}=\mathrm{P}_{2} \mathrm{~A}$ (upward)

$(3)$ Weight of fluid column $\mathrm{W}=m g$

$\mathrm{W}=\mathrm{A} h \rho g \quad \text { (downward) }$

Fluid column is in equilibrium hence downward forces = upward forces

$\mathrm{F}_{1}+\mathrm{W}=\mathrm{F}_{2}$

$\therefore \mathrm{F}_{2}-\mathrm{F}_{1}=\mathrm{W}$

$\therefore \mathrm{P}_{2} \mathrm{~A}-\mathrm{P}_{1} \mathrm{~A}=\mathrm{A} h \rho g$

$\therefore \mathrm{P}_{2}-\mathrm{P}_{1}=h \rho g$

$\therefore \mathrm{P}_{2}=\mathrm{P}_{1}+h \rho g$

Equation $(1)$ shows that pressure difference depends on the vertical distance $(h)$ between the points ( $1$ and $2$), mass density of the fluid ( $\rho$ ) and acceleration due to gravity ( $g$ ) but it does not depends on the area of cross section.

If the effect of gravitation is neglected

$\mathrm{P}_{2}-\mathrm{P}_{1}=h \rho g=0$

$\therefore \quad \mathrm{P}_{2}=\mathrm{P}_{1}$

It indicates that if the effect of gravitation is neglected, the pressure at every point in fluid is same.