Question

Pressure at a certain depth in a pond is ${{6atm}}$. Find the depth of the pond in meters approximated to the nearest integer.

Answer 1

When a fluid exerts a force in a unit area, it is known as pressure. It is generally measured in ${{N}}.{{{m}}^{ - 2}}$ or Pascal. Some other units are ${{bar}},{{atm}},$ etc. We know that the pressure is the product of density, acceleration due to gravity and the height or depth.

Complete step by step answer:
When a fluid exerts a force in a unit area, it is known as pressure. It is generally measured in ${{N}}.{{{m}}^{ - 2}}$ or Pascal. Some other units are ${{bar}},{{atm}},$ etc. We know that the pressure is the product of density, acceleration due to gravity and the height or depth.
Complete step by step solution:
When an object is kept in water, there will be an external force on the object which is known as fluid pressure. Pressure is directly proportional to the density and depth of the fluid. The equation connecting the pressure, density and depth is given below:
${{P}} = \rho {{gh}}$, where ${{P}}$ is the pressure, ${{g}}$ is the acceleration due to gravity $\left( {9.8{{m}}.{{{s}}^{ - 2}}} \right)$, $\rho$ is the density of fluid and ${{h}}$ is the depth of the fluid. There will be an atmospheric pressure. So the total pressure will be the sum of atmospheric pressure and fluid pressure.
In the given problem, it is given that the total pressure is ${{6atm}}$. First we have to calculate the fluid pressure. We know that the atmospheric pressure is ${{1atm}}$. Thus the fluid pressure can be calculated by subtracting atmospheric pressure from total pressure.
So fluid pressure ${{P}} = 6{{atm}} - 1{{atm = 5atm}}$.
From this fluid pressure, we can calculate the depth of the pond.
i.e. Fluid pressure, ${{P = }}\rho {{gh}}$, where $\rho = 1000{{kg}}.{{{m}}^{ - 3}}$ which is the density of water.
${{g = 9}}{{.8m}}{{.}}{{{s}}^{ - 2}}$
Substituting all these values, we get
${{5atm = }}1000{{kg}}.{{{m}}^{ - 3}} \times {{9}}{{.8m}}{{.}}{{{s}}^{ - 2}} \times {{h}}$
We know that $1{{Pa}} = 1{{kg}}.{{{m}}^{ - 1}}.{{{s}}^{ - 2}}$. So we have to convert the pressure in ${{atm}}$ to ${{Pa}}$ for the convenience.
$1{{atm}} = 101325{{Pa}} \Leftrightarrow {{5atm}} = 5 \times 101325{{Pa}}$
Substituting this value, we get
${{5}} \times {{101325Pa = }}1000{{kg}}.{{{m}}^{ - 3}} \times {{9}}{{.8m}}{{.}}{{{s}}^{ - 2}} \times {{h}}$
By solving the above equation, we get the value of ${{h}}$.
i.e. ${{h = }}\dfrac{{{{5}} \times {{101325Pa}}}}{{1000{{kg}}.{{{m}}^{ - 3}} \times {{9}}{{.8m}}{{.}}{{{s}}^{ - 2}}}} = \dfrac{{506625{{Pa}}}}{{980{{kg}}.{{{m}}^{ - 2}}.{{{s}}^{ - 2}}}} = 51.69{{m}}$
Since $1{{Pa}} = 1{{kg}}.{{{m}}^{ - 1}}.{{{s}}^{ - 2}}$, it gets cancel out in the numerator and denominator. Thus only ${{m}}$ is left as a unit.

Thus the depth of pond is $51.69{{m}}$.

Note: In a fluid, pressure can act in all directions. We know that the atmospheric pressure is $1.01 \times {10^5}{{Pa}}$ or ${{1atm}}$. At any depth, the fluid pressure will not change. It will be the same in all directions. Moreover, it does not depend on the shape of the container.