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### 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 $$h$$. 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
$F=\mathit{mg}\to 1$

We can get the mass of liquid by multiplying volume of the liquid by its density.
$m=\mathrm{\rho }V\to 2$

Volume of the liquid column,
$\mathit{Volume}=\mathit{Area}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{liquid}\phantom{\rule{0.147em}{0ex}}\mathit{column}×\mathit{Height}$
$V=A\phantom{\rule{0.147em}{0ex}}h\to 3$

Substitute equation $$3$$ in $$2$$,
$m=\mathrm{\rho }\mathit{Ah}\to 4$

Substitute equation $$4$$ in $$1$$,
$\mathit{Force}=\mathit{mg}=\mathrm{\rho }\mathit{Ahg}$

Now, apply it in the pressure formula
$\mathit{Pressure}=\frac{\mathit{Thrust}\left(F\right)}{\mathit{Area}\left(A\right)}=\frac{\mathrm{\rho }\mathit{Ahg}}{A}=\mathrm{\rho }\mathit{hg}$

Therefore, Pressure due to liquid column $$=$$ $\mathrm{\rho }\mathit{hg}$

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 $$1000$$ $\mathit{kg}{m}^{-3}$
To find: Pressure
Pressure $$=$$ $\mathrm{\rho }\mathit{hg}$
Apply the known values,
$\begin{array}{l}\mathit{Pressure}=\mathrm{\rho }\mathit{hg}\\ =1000×0.9×9.8\\ =8820\phantom{\rule{0.147em}{0ex}}\mathit{Pa}\end{array}$