Theory:

A tall beaker is filled with liquid so that it forms a liquid column. The area of the cross-section at the bottom is \(A\). The density of the liquid is denoted by \(ρ\), and the height of the liquid column is \(h\). In other words, the depth of the water from the top-level surface is \(h\).
 
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Pressure due to a liquid column
 
We know that thrust at the bottom of the column is equal to the weight of the liquid. It is mathematically represented as
F=mg1
 
We can get the mass of liquid by multiplying volume of the liquid by its density.
m=ρV2
 
Volume of the liquid column,
Volume=Areaoftheliquidcolumn×Height
V=Ah3
 
Substitute equation \(3\) in \(2\),
m=ρAh4
 
Substitute equation \(4\) in \(1\),
Force=mg=ρAhg
 
Now, apply it in the pressure formula
Pressure=Thrust(F)Area(A)=ρAhgA=ρhg
 
Therefore, Pressure due to liquid column \(=\) ρhg
 
The above expression shows that pressure in a liquid column is determined by depththe density of the liquid and theacceleration due to gravity. However, the final expression for pressure does not have the term area \(A\) in it. Thus, pressure in a liquid depends on depth only.
 
Example:
 
1. Calculate the pressure exerted by a column of water of height \(0.9\) \(m\). (Density of water \(=\) \(1000\) \(kg\) \({m}^-3\)).
  
Given:
 
The height or depth of the column is \(0.9\).
Density of the water is \(1000\) kgm3
To find: Pressure
Pressure \(=\) ρhg
Apply the known values,
Pressure=ρhg=1000×0.9×9.8=8820Pa