Video: Recalling the Formula Relating Fluid Pressure at a Point, Fluid Density, and Point Depth

Which of the following formulas correctly shows the relationship between the pressure exerted by a fluid, the height β„Ž of the fluid above the point at which pressure is measured, the density 𝜌 of the fluid, and the gravitational acceleration rate 𝑔? [A] 𝑃 = π‘”β„Ž/𝜌 [B] 𝑃 = πœŒπ‘”/β„Ž [C] 𝑃 = 𝑔/πœŒβ„Ž [D] 𝑃 = πœŒπ‘”β„Ž [E] 𝑃 = β„Ž/πœŒπ‘”

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Video Transcript

Which of the following formulas correctly shows the relationship between the pressure exerted by a fluid, the height β„Ž of the fluid above the point at which pressure is measured, the density 𝜌 of the fluid, and the gravitational acceleration rate 𝑔? A) 𝑃 is equal to 𝑔 times β„Ž divided by 𝜌. B) 𝑃 is equal to 𝜌 times 𝑔 divided by β„Ž. C) 𝑃 is equal to 𝑔 divided by 𝜌 times β„Ž. D) 𝑃 is equal to 𝜌 times 𝑔 times β„Ž. E) 𝑃 is equal to β„Ž divided by 𝜌 times 𝑔.

So in this exercise, we’re trying to find the correct mathematical relationship between four quantities: pressure, density, height, and gravity. We’re told that one of our five answer options shows this correct relationship. To figure out which one it is, there are two ways we can go about it. The first way β€” and certainly the quickest way β€” is to recall off the top of our head the equation for the pressure produced by a fluid. But let’s say that we’re not able to recall that equation. Well, there is still a way we can solve for the correct answer. The way we can do it is by looking at the units on either side of each one of these equations.

Here’s what we mean by that. We can see that each one of these five equations has four terms in it. Each one has pressure, it has height, β„Ž, it has density, 𝜌, and it has acceleration due to gravity, 𝑔. Now we can take each one of these terms and we can look at the units that that term involves. For example, the units of pressure are pascals, abbreviated Pa. And a pascal we can recall is equal to a newton of force spread over a square meter of area. And we can further recall that because a Newton is equal to a kilogram meter per second squared, then that means the units of pressure are kilograms meters per second squared per meter squared, which simplifies in the most basic of base units to a kilogram per meter second squared.

The base unit of height on the other hand is meters. The base unit of density is kilograms per cubic meter. And we can recall that the units of the acceleration due to gravity are meters per second squared. Now here’s the reason we’ve gone through all this. If we look at the left-hand side of each one of our five candidate equations, we see they all involve the pressure 𝑃 by itself. We’ve seen from our analysis that the units of pressure are kilograms per meter second squared. This means that these are the units on the left-hand side of all of our candidate equations.

And for an equation to be true, that is for it to be accurate, that means that the units on the right-hand side must match these. They must also be kilograms per meter second squared. If they’re not, then that means the two sides of the equation can’t be equal. And that means that that particular equation doesn’t correctly represent the relationship between these four values. So here is what we’ll do. We’ll go through these equations one by one, starting with equation A. And we’ll analyse the units on the right-hand side. If those units match up to the units of pressure, then we’ll mark that equation as a possibly correct solution.

Starting out with equation A, the right-hand side here has 𝑔 times β„Ž divided by 𝜌. Looking over at the units of each one of these terms, we see that the units of 𝑔 are meters per second squared, the units of β„Ž are meters, and the units of density are kilograms per cubic meter. In the numerator of this fraction, these two factors of meters combine. And then if we multiply this fraction by meters cubed divided by meters cubed, then that term meters cubed cancels out in our denominator. And we have meters to the fifth divided by second squared divided by kilograms or simply meters to the fifth divided by second squared kilograms. When we compare these units to the units we have for pressure, we see that there’s not a match, which means that option A is not a reasonable candidate for the correct mathematical relationship.

Moving on to the right-hand side of option B. Here, we have units of kilograms per cubic meter for 𝜌 multiplied by meters per second squared for 𝑔 divided by meters for β„Ž. This numerator simplifies to kilograms per meter squared second squared. And then if we divide both numerator and denominator by meters, we have a final result of kilograms per cubic meter second squared. This is close to the units for pressure, but it’s not quite right. Therefore, option B isn’t our answer either.

Moving to the right-hand side of option C. Here we have meters per second squared, the units of 𝑔, divided by kilograms per cubic meter, the units of 𝜌, multiplied by meters, the units of height. A meter divided by meters cubed is equal to one over meters squared. And then, if we multiply top and bottom of this fraction by the inverse of the denominator, in other words multiply it by meter squared per kilogram, then the meter squared per kilogram, the denominator, cancels out with kilograms per meter squared. We’re left with a final result of meters cubed per kilogram second squared, also not equal to the units of pressure. So we cross option C off our list.

Looking at the right side of equation D. Here we have units of kilograms per cubic meter multiplied by meters per second squared multiplied by meters. Combining all these factors of meters, we get a result of kilograms per meter second squared. And we see that this is a match for the units of pressure. So let’s put a star by option D. And we’ll keep that in mind as a possible answer choice.

Looking finally at the right-hand side of option E. Here we have units of meters divided by kilograms per cubic meter multiplied by meters per second squared. This denominator simplifies to kilograms per meter squared second squared. And if we multiply both top and bottom by meters squared second squared, this term cancels out in our denominator, leaving us with meters cubed second squared per kilogram. We see this is not a match for the units of pressure. And therefore, option E is off the table. And this means that option D is our answer. The pressure exerted by a fluid is equal to the density of that fluid multiplied by 𝑔 multiplied by β„Ž.

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