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

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.