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

(N/A) Suppose a fluid of density $\rho$ is in static equilibrium in a container.
Consider a cylindrical element of fluid having area of base $A$ and height $h$. The pressure at points $1$ and $2$ are $P_{1}$ and $P_{2}$ respectively.
The forces acting on the liquid column are as follows:
$(1)$ The force at point $1$ is $F_{1} = P_{1} A$ (downward).
$(2)$ The force at point $2$ is $F_{2} = P_{2} A$ (upward).
$(3)$ The weight of the fluid column is $W = mg = A h \rho g$ (downward).
Since the fluid column is in equilibrium,the downward forces must equal the upward forces:
$F_{1} + W = F_{2}$
$P_{1} A + A h \rho g = P_{2} A$
Dividing by $A$ on both sides:
$P_{2} = P_{1} + h \rho g$
This equation shows that the pressure difference depends on the vertical distance $h$ between the points,the mass density of the fluid $\rho$,and the acceleration due to gravity $g$,but it does not depend on the area of the cross-section $A$.
If the effect of gravitation is neglected $(g = 0)$,then $P_{2} = P_{1}$,indicating that the pressure at every point in the fluid is the same.