6.5 Pressure in Liquids
We have learnt in lower classes that liquids exert pressure in all directions. Moreover, liquid pressure increases with depth.
Let us determine the pressure at a certain depth of a liquid. Figure 6.8 shows a container of liquid. Consider an area A in the liquid at depth h. The force acting on this area is equal to the weight of the liquid column over surface A. The volume of this liquid is V = Ah. If ρ is the density of the liquid, then the mass m of the liquid column will be:
m = ρV = ρAh
Therefore, the force acting on area A will be:
F = mg = ρAhg
The pressure P at area A will be:
P = F / A = ρAhg / A
P = ρgh .................... (6.4)
Equation (6.4) shows that pressure in a liquid increases with depth. The value of pressure depends on the depth and density of the liquid.
Pressure produces force at right angle to the surface. A force or its component that is parallel to the surface does not contribute to pressure. The pressure, by definition, is only contributed by the normal component of the force. If there is a hole in the surface of the liquid container, the liquid spurts at right angles to the surface before curving downward due to gravity.
Example 6.2
Calculate the pressure of a column of mercury 76 cm high. Density of mercury is 13.6 × 10³ kg m⁻³.
Solution:
Density ρ = 13.6 × 10³ kg m⁻³ Height h = 76 cm = 76 × 10⁻² m g = 10 m s⁻²
As Pressure = ρgh:
P = 13.6 × 10³ kg m⁻³ × 10 m s⁻² × 76 × 10⁻² m
P = 1.034 × 10⁵ N m⁻²
P = 1.034 × 10⁵ Pa
Example 6.3
A cylindrical water tank 2 m deep has been built on the top of a building 20 m high. What will be the pressure of water at the ground floor when the tank is full? Density of water is 1000 kg m⁻³. Take g = 10 m s⁻².
Solution:
Height h = 2 + 20 = 22 m Density ρ = 1000 kg m⁻³ g = 10 m s⁻²
P = ρgh = 22 m × 1000 kg m⁻³ × 10 m s⁻²
P = 220000 Pa = 2.2 × 10⁵ Pa