###### Advertisements

###### Advertisements

Derive an equation for the total pressure at a depth ‘h’ below the liquid surface.

###### Advertisements

#### Solution

In order to understand the increase in pressure with depth below the water surface, consider a water sample of cross-sectional area in the form of a cylinder. Let h_{1} and h_{2} be the depths from the air-water interface to level 1 and level 2 of the cylinder, respectively. Let F_{1} be the force acting downwards on level 1 and F_{2} be the force acting upwards on level 2, such that, F_{1} = P_{1} A and F_{2} = P_{2}A. Let us assume the mass of the sample to be m and under equilibrium condition, the total upward force (F_{2}) is balanced by the total downward force (F_{1} + mg), in other words, the gravitational force will act downward which is being exactly balanced by the difference between the force F_{2} – F_{1}.

F_{2} – F_{1} = mg …… (1)

where m is the mass of the water available in the sample element. Let ρ be the density of the water then, the mass of water available in the sample element is

m = ρV = ρA (h_{2} – h_{1})

V = A (h_{2} – h_{1})

Hence, gravitational force, F_{G} = ρA (h_{2} – h_{1}) g

On substituting the W value in equation (1)

F_{2} = F_{1} + mg ⇒ P_{2}A = P_{1}A + ρA (h_{2} – h_{1}) g

Cancelling out A on both sides,

**Sample of water with base area A in a static fluid with its forces in equilibrium**

P_{2} = P_{1} + ρ(h_{2} – h_{1}) g .........…(2)

**The pressure (P) at a depth (h) below the water surface**

If we choose level 1 at the surface of the liquid (i.e., air-water interface) and level 2 at a depth ‘h’ below the surface, then the value of h_{1} becomes zero (h_{1} = 0) and in turn, P_{1} assumes the value of atmospheric pressure (say P_{a}). In addition, the pressure (P_{2}) at a depth becomes P.

Substituting these values in the equation, we get

P = P_{a} + ρgh …… (3)

which means, the pressure at a depth h is greater than the pressure on the surface of the liquid, where Pa is the atmospheric pressure which is equal to 1.013 × 10^{5} Pa.

If the atmospheric pressure is neglected or ignored then

P = ρgh ….. (4)

for a given liquid, ρ is fixed and g is also constant, then the pressure due to the fluid column is directly proportional to vertical distance or height of the fluid column. This implies the height of the fluid column is more important to decide the pressure and not the cross-sectional or base area or even the shape of the container.

** Illustration of hydrostatic parodox**

Let us consider three vessels of different shapes A, B and C as shown in the figure. These vessels are connected at the bottom by a horizontal pipe. When they are filled with a liquid (say water), it occupies the same level even though the vessels hold different amounts of water. It is true because the liquid at the bottom of each section of the vessel experiences the same pressure.

#### APPEARS IN

#### RELATED QUESTIONS

State Pascal’s law in fluids.

State Archimedes principle.

What do you mean by upthrust or buoyancy?

State the law of floatation.

State and prove Pascal’s law in fluids.

State and prove Archimedes principle.

We can cut vegetables easily with a sharp knife as compared to a blunt knife. Why?