Question

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

Answer 1

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

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 = (F)/(A) = (\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)`.)