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

Explanation

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 `P_(1)andP_(2)` respectively.
The force acting on liquid column is as below :
(1) The forces on point `-1F_(1)=P_(1)A` (downward)
(2) The force on point `-2F_(2)=P_(2)A` (upward)
(3) Weight of fluid column `W=mgW=Ahrhog` (downward)
Fluid column is in equilibrium hence downward forces = upward forces
`F_(1)+W=F_(2)`
`thereforeF_(2)-F_(1)=W`
`thereforeP_(2)A-P_(1)A=Ahrhog`....(1)
`thereforeP_(2)-P_(1)=hrhog`
`thereforeP_(2)=P_(1)+hrhog` ...(2)
Equation (1) shows that pressure difference depends on the vertical distance (h) between the points (1and 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
`P_(2)-P_(1)=hrhog=0`
`thereforeP_(2)=P_(1)`
It indicates that if the effect of gravitation is neglected , the pressure at every point in fluid is same.
Let there be point 1 on the surface of container. The pressure on the surface is atmospheric pressure `P_(a).P_(1)=P_(a)andP_(2)=PthereforeP-P_(a)=hrhogorP=P_(a)+hrhog` ... (3)
P is called absolute pressure.
Equation (3) shows that the pressure P at depth below the surface of liquid open to the atmosphere is greater than atomspheric pressure by an amount `hrhog`.
The excess of pressure , `P-P_(a)`, at depth h is called a gauge pressure at that point.