# At what depth below the surface of oil, relative density 0.8, will the oil produce a pressure of 120 kN/m2? What depth of water is this equivalent to?

##### 1 Answer

That pressure will be felt at a depth of **15.3 meters**.

#### Explanation:

You need to know two things in order to solve this problem, what *relative density* is and what the formula that establishes a relationship between depth and pressure looks like.

**Relative density** (you'll sometimes see this being referred to as *specific gravity*) is simply the ratio between the density of a substance, in your case oil, and the density of a reference substance, which in your case I assume it's water, at specified conditions.

#color(blue)(d = rho_"oil"/rho_"water")#

Most of the time, relative density is compared with water's density at

This means that the density of oil will be

#rho_"oil" = d * rho_"water"#

#rho_"oil" = 0.8 * "1000 kg/m"""^3 = "800 kg/m"""^3#

The relationship between depth and pressure is given by the formula

#color(blue)(P = rho * g * h)" "# , where

*gravitational acceleration*;

Rearrange to solve for

#h = P/(rho * g) = (120 * 10""^3color(red)(cancel(color(black)("kg")))/(color(red)(cancel(color(black)("m"))) * color(red)(cancel(color(black)("s"^2)))))/(9.8color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"^2))) * 800 color(red)(cancel(color(black)("kg")))/"m"^color(red)(cancel(color(black)(3)))) = color(green)("15.3 m")#

To find the depth at which this pressure would be produced in water, simply replace the density of the oil with that of water

#h = P/(rho * g) = (120 * 10""^3color(red)(cancel(color(black)("kg")))/(color(red)(cancel(color(black)("m"))) * color(red)(cancel(color(black)("s"^2)))))/(9.8color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"^2))) * 1000 color(red)(cancel(color(black)("kg")))/"m"^color(red)(cancel(color(black)(3)))) = color(green)("12.2 m")#