Question
Derive an expression for Pressure due to the Liquid column.

Answer

A tall beaker filled with water to form a liquid column
Area of the cross-section at bottom $= A$
Height of liquid column $=h$
The density of the liquid $=\rho$
Thrust at bottom of liquid column $(F)=$ Weight of liquid.
$F = mg \ldots(1)(\because m -$ mass of liquid $)$
Mass, $m =\rho \times V$
Volume of liquid columñ, $V=$ Area of cross-section $(A) \times$ height $(h)$
$V = Ah$
Substitute (3) in (2) $n _{ s }=\rho Ah$
Substitute (4) in (1) $F =\rho$ Ahg
Pressure $( P )=\frac{\text { Thrust }( F )}{\operatorname{Area}( A )}=\frac{\rho A h g }{ A }$
$\therefore P =$ hpg - This is the expression for pressure due to the liquid column.

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