Expression for the pressure: At a point at depth h below the surface of a liquid of uniform density the pressure due to the liquid is due to the weight per unit area of a liquid column of height A above that pointIn Fig. 22 to find the pressure due to the liquid at point P. consider the cylindrical liquid column, of cross section A and height, above that pointThe weight of this liquid column
=volume density × acceleration duetogravity
=area height × density × acceleration due togravity
=Ahpg
Pressure due to the liquid column at depth h
= weight of the liquid column / cross-section area
= Ahpg / A
= hpg
If the free surface of the liquid is open to the atmosphere, the pressure on the surface is the atmospheric pressure pe Then, the absolute pressure within the liquid at a depth his p=po+ hpg
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