Question

At what depth below the surface of oil, relative density \[0.8\] , will the oil produce a pressure of \[120kN{m^{ - 2}}\] ? What depth of water is this equivalent to?

Answer 1

Hint:To answer the problem, you must first understand relative density and the formula that provides a relationship between depth and pressure. Relative density is defined as the ratio of a substance's density to the standard, which is either air for gases and water for liquids or solids.

Complete step by step answer:
At specified conditions, relative density (also known as specific gravity) is simply the ratio between the density of a material, in your example oil, and the density of a reference substance, which I suppose is water in your case.
$d = \dfrac{{{\rho _{oil}}}}{{{\rho _{water}}}}$
Most of the time, relative density is compared to the density of water at \[{4^ \circ }C\] , which is \[1000{\text{ }}kg{m^{ - 3}}\]
As a result, the density of oil will be
$
  {\rho _{oil}} = d \cdot {\rho _{water}} \\
  {\rho _{oil}} = 0.8 \times 1000kg{m^{ - 3}} \\
  \therefore {\rho _{oil}} = 800kg{m^{ - 3}} \\
 $
The formula describes the relationship between depth and pressure.
$P = \rho \cdot g \cdot h$ where,
$P - $ the pressure produced at the depth \[h\]
The gravitational acceleration is denoted by the letter \[g\] .
 $\rho - $ is the density of the liquid.
Rearrange to find a solution for $h$ always retain in mind $\dfrac{N}{{{m^2}}}$ is equivalent to $\dfrac{{Kg}}{{m \cdot {s^2}}}$ and don't forget that you're dealing with kilonewtons, not Newtons.
$
  h = \dfrac{P}{{\rho \cdot g}} \\
  h = \dfrac{{120 \cdot {{10}^3}\dfrac{{kg}}{{m \cdot {s^2}}}}}{{9.8\dfrac{m}{{{s^2}}} \cdot 800\dfrac{{kg}}{{{m^3}}}}} \\
  \therefore h = 15.3m \\
 $
Simply substitute the density of the oil with that of water to get the depth at which this pressure would be produced in water.
$
  h = \dfrac{P}{{\rho .g}} \\
  h = \dfrac{{120 \cdot {{10}^3}\dfrac{{kg}}{{m \cdot {s^2}}}}}{{9.8\dfrac{m}{{{s^2}}} \cdot 1000\dfrac{{kg}}{{{m^3}}}}} \\
  \therefore h = 12.2m \\
 $

Note:It should be mentioned that relative density, also known as specific density, is the ratio of a substance's density to the density of water at . Because the ratio's numerator and denominator have the same units, they cancel each other out. As a result, there are no units for relative density.