The following Topics and Sub-Topics are covered in this chapter and are available on MSVgo:
Now let us find how the liquid column’s pressure is based on the depth of the liquid from the free surface.
Here let us consider the liquid with density in a large container. There is a cylinder with a surface XY(Area A) at a depth of h from the free surface AB, as shown in the figure.
Source: From the book Concise Physics Part 1(Class 9)
The next step is to find the thrust of the liquid in the cylinder VWXY of height h exerted on its bottom surface XY.
Thrust exerted on surface XY= Weight of the liquid in the liquid column VWXY
= Volume of the liquid column VWXY density acceleration due to gravity
= (Area of bottom surface XY height) g
Thrust exerted on surface XY =(A h) g
Now pressure applied on the liquid column =
Pressure applied on the liquid column=h g
Pressure applied on the liquid column= depth of the liquid from the free surface density of liquid acceleration due to gravity
At a particular point on Earth, g is constant. So the pressure on the liquid will only depend on its depth h from the free surface and density of the liquid. The pressure of the liquid is independent of the containers’ shape and size. It is also independent of the area occupied by this container.
But it will depend on the atmospheric pressure applied to the liquid at a given point. Atmospheric pressure is the pressure applied within the atmosphere present on Earth.
Total pressure applied on any liquid at depth h = Atmospheric Pressure + Pressure exerted by the liquid column
Pascal’s law will explain the transmission of pressure in liquids. We have seen in the above section that the pressure exerted by a liquid of density at a depth of h from the free surface is given as,
The pressure exerted by the liquid = h
It implies that the difference in pressure between stationary points in the liquid should only depend on the vertical height difference between the free surface and these points. It would imply that if we decrease the pressure at some point in the liquid, the pressure must drop by at the same amount at all other points in the liquid.
Pascal’s law states that when the liquid is confined in a container, the pressure exerted by this liquid will equally transmit without reducing its magnitude across all directions.
Hydraulic machines are a perfect example of the transmission of pressure in liquids, as explained in Pascal’s law.
Buoyancy is a force exerted on an object partially or entirely immersed in a stationary fluid. It is a result of the pressure acting on the opposite sides of the object.
Archimedes Principle states that the buoyant force applied on the partially or wholly immersed object is equivalent to the fluid’s weight displaced by the object. This force will always act in the upward direction from the centre of mass of the complete fluid. Archimedes of Greece derived this principle, hence the name.
Mathematically it can be written as Buoyant force=Weight of the fluid displaced by the object.
Suppose when an object is immersed entirely in the fluid, and it displaces volume V of fluid.
Mass of the fluid displaced= Density X Volume
Mass of the fluid displaced =
Weight of the fluid displaced = Mass of the fluid X Acceleration due to gravity
Weight of the fluid displaced =
We can write Archimedes principle as,
Where is the buoyant force
The flotation principle states that when an object floats on the liquid, the buoyant force acting on the object is equal to its weight.
It is important to remember that the pressure exerted by any liquid will depend only on the density of the liquid and acceleration due to gravity at that particular point. Also, the buoyant force exerted on a fluid is equal to the weight of the fluid displaced.