# Question Video: Determining the Densities of Liquids in a Liquid Column Manometer Physics

From the given figure, calculate the ratio between the density of liquid B and the density of liquid A.

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### Video Transcript

From the given figure, calculate the ratio between the density of liquid B and the density of liquid A.

In this question, we are shown a figure of a liquid column manometer, which is used to measure the pressure difference between the two arms of the U-shaped tube. Notice that liquid B is on the left-hand side of the manometer while liquid A begins beneath liquid B and continues to the other end. Liquid B is at a height of two ℎ above the dotted line. And liquid A reaches one ℎ above the same line. We are asked to study this image and calculate the ratio between the density of liquid B and the density of liquid A.

To answer this question, let’s first recall that for a liquid column manometer with just a single liquid in it, if the height of the liquid is different in the left and right tubes, then this implies there is a different pressure being exerted on either side. Specifically, if we call the pressure exerted on the left 𝑃 L and the pressure exerted on the right 𝑃 R, then we can recall that the difference between these quantities 𝑃 L minus 𝑃 R is given by 𝜌𝑔Δℎ. Here, 𝜌 is the density of the liquid. 𝑔 is the gravitational field strength of the Earth. And Δℎ is the difference in liquid height between the two sides.

Coming back to the tube given to us in the question, we can see that the height of liquid A is different on both sides. We can see that the height on the right is a distance ℎ higher than that on the left. So we can say that the difference in pressure between the surfaces of liquid A, 𝑃 L minus 𝑃 R, is given by 𝜌 A 𝑔ℎ. 𝜌 A is the density of liquid A. 𝑃 L is the pressure exerted on the left surface of liquid A, which is here. And 𝑃 R is the pressure exerted on the right surface of liquid A, which is here. Let’s keep this equation in mind for later. And let’s make some more space on the screen.

Now, in this particular scenario, the difference in height for liquid A is not caused by a difference in gas pressure on either side. Instead, it’s caused by the pressure exerted by liquid B on the left, which is not present on the right. We can break this down in the following way. Both sides of the tube are open, so we can assume that this gas is air and exerts atmospheric pressure on either side. So on the right side of the tube, the pressure on the surface of liquid A is just atmospheric pressure. We’ll label this as 𝑃 subscript atm. On the left side of the tube, the surface of liquid A experiences atmospheric pressure as this side of the tube is also open to the atmosphere. But it also experiences some pressure exerted by liquid B. We’re calling this pressure 𝑃 subscript B.

Hence, we can say that the pressure on the left surface of liquid A is equal to the atmospheric pressure plus the pressure due to liquid B. But what is the pressure exerted by liquid B on the surface of liquid A? Well, we can recall that within a column of liquid, which is what we have here with liquid B, we can calculate the pressure at any depth using a similar formula to one we’ve seen earlier. The pressure exerted at a point within a liquid, if that point is a distance 𝑥 beneath the surface, is given by 𝜌𝑔𝑥, where 𝜌 is the density of the liquid. And the pressure is exerted at this point due to the weight of all the liquid above it. This is why the deeper down we go into a liquid, the more pressure is exerted.

Now if we use this equation to calculate the pressure exerted at a point at the base of the column of liquid B, then this will also tell us the pressure exerted by liquid B onto the surface of liquid A. This is because the base of liquid B is in contact with the top surface of liquid A. So we can say that the pressure exerted by liquid B onto the surface of liquid A is given by 𝑃 B is equal to 𝜌 B 𝑔 times two ℎ. 𝜌 B of course is the density of liquid B. And we use two ℎ since this is the height of liquid B. Next, we can substitute this back into our equation from earlier. So it now reads 𝑃 L is equal to 𝑃 atm plus two 𝜌 B 𝑔ℎ. The pressure exerted on the left surface of liquid A is equal to the atmospheric pressure plus the pressure exerted by the weight of liquid B.

Finally, let’s go back to the equation we derived earlier: 𝑃 L is equal to 𝑃 R plus 𝜌 A 𝑔ℎ. We can rearrange it by subtracting 𝑃 R from both sides. This gives us an expression for 𝑃 L. We’ve now got two expressions for 𝑃 L. And equating them, we find 𝜌 A 𝑔ℎ plus 𝑃 R is equal to 𝑃 atm plus two 𝜌 B 𝑔ℎ. Recalling that the pressure on the right surface, 𝑃 R, is just the same as atmospheric pressure, 𝑃 atm, we can cancel these terms and be left with 𝜌 A 𝑔ℎ is equal to two 𝜌 B 𝑔ℎ. This is useful because we’re trying to find the ratio between densities of liquid B and liquid A. In other words, we’re trying to find 𝜌 B divided by 𝜌 A. And we have both quantities in our equation here. So let’s rearrange it.

First, we can divide both sides of the equation by 𝑔ℎ, meaning they cancel on both sides. Then, let’s divide both sides by two 𝜌 A. On the left, the factor of 𝜌 A cancels. And on the right, the factor of two cancels. We’re therefore left with our final answer: one-half is equal to 𝜌 B over 𝜌 A. The ratio between the density of liquid B and the density of liquid A is one-half.