Question

If $P_1$ and $P_2$ are pressures at two points in the liquid at elevations $x_1$ and $x_2$, respectively, and $\rho$ is the density of the liquid, then:

• A. $P_1-P_2=\rho g(x_2-x_1)$
• B. $P_1-P_2=\rho g(x_1-x_2)$
• C. $P_1+P_2=-\rho g(x_2-x_1)$
• D. $P_1+P_2=-\rho g(x_1-x_2)$

Answer 1

The correct option is A $P_1-P_2=\rho g(x_2-x_1)$

Let the total height of the liquid in the container be $h$.
The depth of the point $1$ $=d_1=h-x_1$
The depth of the point $2$ $=d_2=h-x_2$

Pressure at the point $1:$ $P_1=\rho g(h-x_1)$
Pressure at the point $2:$ $P_2=\rho g(h-x_2)$

$⟹P_1-P_2=$ $\rho g(h-x_1)-\rho g(h-x_2)$
$=\rho gh-\rho g{x}_1-\rho gh+\rho g{x}_2$
$=-\rho g{x}_1+\rho g{x}_2=\rho g(x_2-x_1)$

$\therefore P_1-P_2=\rho g(x_2-x_1)$