Question

The following figure shows $p_1$ and $p_2$ being the pressures at the elevations $y_1$ and $y_2$ in the liquid.



If $\rho$ and $g$ are the density of the liquid and the acceleration due to gravity respectively, then:

Answer 1

Let the total height of the liquid in the container be $h$.


The depth of the point $1$ $=d_1=h-y_1$
The depth of the point $2$ $=d_2=h-y_2$

Pressure at the point $1:$ $p_1=\rho g(h-y_1)$
Pressure at the point $2:$ $p_2=\rho g(h-y_2)$

$⟹p_1-p_2=$ $\rho g(h-y_1)-\rho g(h-y_2)$
$=\rho gh-\rho g{y}_1-\rho gh+\rho g{y}_2$
$=-\rho g{y}_1+\rho g{y}_2=\rho g(y_2-y_1)$

$\therefore p_1-p_2=\rho g(y_2-y_1)$