# Variation of Pressure with Depth

Consider a fluid at rest in a container. In Fig. 1 point 1 is at height h above a point 2. The pressures at points 1 and 2 are $P_1$  and $P_2$ respectively. Consider a cylindrical element of fluid having area of base A and height h. As the fluid is at rest the resultant horizontal forces should be zero and the resultant vertical forces should balance the weight of the element.

Fig.1 Fluid under gravity. The effect of gravity is illustrated through pressure on a vertical cylindrical column.

The forces acting in the vertical direction are due to the fluid pressure at the top ($P_1A$) acting downward, at the bottom ($P_2A$) acting upward. If mg is weight of the fluid in the cylinder we have

($P_2 −P_1$) A = mg   (1)

Now, if ρ is the mass density of the fluid, we have the mass of fluid to be m = ρV= ρhA so that

$P_2 −P_1$ = ρgh      (2)

Pressure difference depends on the vertical distance h between the points (1 and 2), mass density of the fluid ρ and acceleration due to gravity g. If the point 1 under discussion is shifted to the top of the fluid (say, water), which is open to the atmosphere, $P_1$  may be replaced by atmospheric pressure (Pa ) and we replace $P_2$ by P. Then Eq. (2 gives

P = $P_a$  + ρgh       (3)

Thus, the pressure P, at depth below the surface of a liquid open to the atmosphere is greater than atmospheric pressure by an amount ρgh. The excess of pressure, P − $P_a$, at depth h is called a gauge pressure at that point.

Fig 2 Illustration of hydrostatic paradox. The three vessels A, B and C contain different amounts of liquids, all upto the same height.

The area of the cylinder is not appearing in the expression of absolute pressure in Eq. (3). Thus, the height of the fluid column is important and not cross-sectional or base area or the shape of the container. The liquid pressure is the same at all points at the same horizontal level (same depth).

The result is appreciated through the example of hydrostatic paradox. Consider three vessels A, B and C [Fig.2] of different shapes. They are connected at the bottom by a horizontal pipe. On filling with water, the level in the three vessels is the same, though they hold different amounts of water. This is so because water at the bottom has the same pressure below each section of the vessel.

Example 1

What is the pressure on a swimmer 10 m below the surface of a lake?

Here h = 10 m and ρ = 1000 kg.m$^{-3}$.

Take g = 10 m.s$^{–2}$

From Eq. (3)

P = P$_a$  + ρgh = 1.01 × 10$^5$  Pa + 1000 kg.m$^{–3}$ × 10 m.s$^{–2}$ × 10 m

= 2.01 × 10$^5$  Pa

≈ 2 atm

This is a 100% increase in pressure from surface level. At a depth of 1 km, the increase in pressure is 100 atm! Submarines are designed to withstand such enormous pressures.