### Video Transcript

The difference in pressures at a depth of 3.75 meters for seawater and freshwater is 735 pascals. What is the difference in their densities?

So in this question, weβve been given a difference in pressures. And at the depth of 3.75 meters, the difference in pressures is given for seawater and fresh water. And this difference is 735 pascals. We need to find the difference in their densities.

Now letβs say that the pressure for seawater is π sub π and the pressure for fresh water is π sub π. Weβve been given the difference in pressures and the differences between the pressures of seawater and fresh water. This pressure difference is 735 pascals. Now both of these pressures were measured at a depth of 3.75 meters.

So letβs say that weβve got a tank full of, letβs say, seawater. The pressure was measured at a depth of 3.75 meters. We can also describe this as the height below the surface of the water. And so we can live with this quantity β. Now since the pressure is measured at the same depth for both seawater and fresh water, we donβt need to separate heights. We donβt need β sub π and β sub π. We just need β.

Finally, what weβre asked to do is to find the difference in the densities of the seawater and fresh water. In other words, weβre trying to find out π sub π , which is the density of seawater, minus π sub π which is the density of fresh water. In order to solve this problem, we need to recall how we can find the pressure of a liquid.

The pressure of a liquid at a depth β is given by ππβ, where π is the density of the liquid, π is the gravitational field strength of the Earth, and β is the depth. So we can apply this to the quantities weβve been given. For seawater, weβve got π sub π , the seawater pressure, is equal to π sub π , the seawater density, times πβ. And we can do exactly the same thing for fresh water.

Now what weβre trying to do is to find out π sub π minus π sub π. So letβs rearrange both of these equations so we can isolate π sub π in the first one and π sub π in the second one. What we can do is divide both sides of the equation by πβ in both cases. What that leaves us with is the following: π sub π over πβ is equal to two π sub π and π sub π over πβ is equal to π sub π.

Now what weβre trying to find out is π sub π minus π sub π. So all we need to do is do this minus this. And that leaves us with π sub π over πβ minus π sub π over πβ, because, as we saw earlier, π sub π is the same as π sub π over πβ. Therefore, we replace this with this. And π sub π is the same as π sub π over πβ. So we replace this with this.

Now what weβre trying to do is to find out the quantity on the left-hand side. So this is what we want to keep to the left-hand side. And then we can look at what we can do to the right-hand side. Well, weβve got π sub π divided by πβ minus π sub π divided by πβ. So we can take out a common factor of one divided by πβ. But letβs make sure this is correct. We can use the distributive property of multiplication to go back to what we had earlier.

One divided by πβ times ππ is the same as this term here. And one divided by πβ times ππ is the same as this term here. Therefore, we have factorized correctly. Now the term in the brackets, π sub π minus π sub π, weβve already written down up here. Thatβs the difference in pressure. And weβve been given that in the question. As for the value of π, well we know that the gravitational field strength of Earth is 9.8 meters per second squared. And weβve already got β earlier on, 3.75 meters.

This means that we can substitute in our values to find the difference in densities. What that leaves us with is one divided by π, which is 9.8, times β, which is 3.75, all multiplied by 735, which is the difference in pressures. Evaluating this gives us a value of 20. But we still need to put in the units. To do this what we need to do is to look at what weβre trying to find out.

Weβre trying to find out a difference in densities. That means that the right-hand side should have units of density. The units of density are mass divided by volume, because thatβs what a density is. Itβs the mass per unit volume. So the mass is kilograms, and the volume is meters cubed. So the units of mass divided by volume are kilograms per meters cubed. And hence, our final answer is that the difference in densities for seawater and fresh water is 20 kilograms per meters cubed.