### Video Transcript

The diagram shows a U-shaped tube
containing both water and oil, which are immiscible. Atmospheric pressure acts equally
on the tops of the columns of water and oil. How does the ratio of the densities
of the oil to the water, π sub oil to π sub water, compare to the heights β one
and β two? (A) π sub oil to π sub water
equals β two divided by β one. (B) π sub oil to π sub water
equals β one divided by β one plus β two. (C) π sub oil to π sub water
equals β one plus β two divided by β one times β two. (D) π sub oil to π sub water
equals β one divided by β two. (E) π sub oil to π sub water
equals β two divided by β one plus β two.

In our diagram, we indeed see this
U-shaped tube filled with two fluids, oil and water. The column of oil has a height
equal to β two. And if we follow a horizontal line
from the bottom of the column of oil to the water, then the height of the column of
water above this line on the opposite side of the tube is β one. Our question asks us about the
ratio of the densities of these fluids. A fluidβ²s density π and its height
β are connected through this equation for the pressure generated by such a
fluid. This equation tells us, for
example, that the pressure generated by our column of oil, that is, the downward
pressure exerted here due to the oil, is equal to the oilβ²s density times the
acceleration due to gravity multiplied by the height of the column of oil.

Because oil and water are
immiscible, that is, they donβ²t mix, thereβ²s a clear boundary here between the two
fluids. If we look on the other side of our
U-shaped tube at that same elevation, we can notice something interesting. The amount of water below the pink
line on the right side of our tube, that is, this amount of water here, is exactly
equal to the amount of water below that pink horizontal line on the other side of
our tube. Therefore, the pressure along this
dashed pink horizontal line is constant. Specifically, the pressure here on
the left-hand side is identical to the pressure here on the right-hand side.

Letβ²s clear some space on screen to
work. And we can notice that above this
point on the left side of our tube is a column of oil of height β two. That column creates a pressure,
weβ²ll call it π sub oil, of the density of oil times the acceleration due to
gravity times β two. If we then follow this horizontal
line to that same elevation point on the right side of our tube, we see that this
point has a column of water above it of height β one. This also produces a pressure,
weβ²ll call it π sub water. And it equals the density of water
times π times β one.

As we saw earlier, the pressure at
any point inside the U-shaped tube along the pink line is the same. That means that the pressure due to
the column of oil of height β two equals the pressure due to the column of water of
height β one. Therefore, we can say that π sub
oil times π times β two equals π sub water times π times β one. Notice that the acceleration due to
gravity is common to both sides of this equation. Therefore, if we divide both sides
by π, that factor will cancel out.

Weβ²re now getting close to our
answer. Recall that we want to solve for
the ratio of π sub oil to π sub water. If we divide both sides of this
remaining equation by π sub water times β two, then on the left-hand side, the
height β two cancels, and on the right π sub water cancels out. That leaves us with this result
that the density of oil divided by the density of water equals β one divided by β
two. Reviewing our answer options, we
see that this agrees with option (D). This then is our answer. And note that the height β one is
smaller than the height β two. This tells us that the density of
our oil is less than the density of water.