### 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.