Question

White an expression for the pressure at a depth of d1 meters below the liquid surface

Answer 1

Answer:

$\rho \cdot g$, where $\rho$ is the density of the liquid, and $g$ is the value of gravitational acceleration.

Step-by-step explanation:

Let $\rho$ be the density of a liquid. Let $g$ represent the gravitational acceleration (near the surface of the earth, $g \approx 9.81\; \rm N \cdot kg^{-1}$.)

The pressure $P$ at a depth of $h$ under the surface of this liquid would be

$P = \rho \cdot g \cdot h$.

Here's how to deduce this equation from the definition of pressure.

Pressure is the amount of force on a surface per unit area. For example, if a force of $1\; \rm N$ is applied over a surface with an area of $1\; \rm m^2$, then the pressure on that surface would be $\rm 1\; \rm Pa$ (one Pascal.)

Consider a flat, square object that is horizontally submerged under some liquid at a depth of $h$. Assume that $A$ is the area of that square. The volume of the liquid that sits on top of this square would be $V = A \cdot h$. If the density of that liquid is $\rho$, then the mass of that much liquid would be $m= \rho \cdot V= \rho \cdot A \cdot h$.

The weight of that much liquid would be $W = m \cdot g = \rho \cdot A \cdot h \cdot g$. The liquid on top of that object would exert a force of that size on the object. Since that force is exerted over an area of $A$, the pressure on the object would be

$\displaystyle P = \frac{F}{A} = \frac{\rho \cdot A \cdot h \cdot g}{A} = \rho \cdot h \cdot g$.

In this question, $h = 1\; \rm m$. As a side note, if $\rho$ and $g$ are also in standard units ($\rm kg \cdot m^{-3}$ for $\rho$ and $\rm N \cdot kg^{-1}$ for $g$), then $P$ would be in Pascals ($\rm Pa$, where $1\; \rm Pa = 1\; \rm N \cdot m^{-2}$.)