2023 A/L Physics Past Paper · Question 34 (English Medium)

Answer: (5) $R_A = R_B$, $V_A < V_B$

What the question asks

What is in the figure:

Two soap bubbles, $A$ small and $B$ big, at the two ends of a narrow tube.

A tap in the middle of the tube. At first it is closed, then it is opened.

We want: the relation between the final radii and between the final volumes of $A$ and $B$.

What each letter means:

$R_A$, $R_B$ ($\text{m}$): the radii of the bubbles.

$V_A$, $V_B$ ($\text{m}^3$): the volumes of the bubbles.

$T$ ($\text{N m}^{-1}$): the surface tension of the soap solution.

$p_0$ ($\text{Pa}$): the pressure of the air outside.

$p_A$, $p_B$ ($\text{Pa}$): the pressures of the air inside the bubbles.

Relevant theory

If 'r' is the radius of the bubble, 'T' the surface tension of the soap solution 'p_1' the pressure inside the bubble and 'p_2' the air pressure outside it.

$p_1 - p_2 = \dfrac{4T}{r}$

The first quote tells how the excess pressure inside a soap bubble comes from its surface tension. The second quote gives the result, $p_1 - p_2 = \dfrac{4T}{r}$. A soap bubble has two surfaces, so it is $4T$ and not $2T$.

So the smaller the radius, the bigger the pressure inside.

For example, blow up a balloon. The first puff is the hardest, and it gets easier when the balloon is bigger. So a small balloon pushes back harder than a big one.

Join a small balloon and a big balloon by a tube. The small one pushes its air into the big one.

Applying it to this question

How the two bubbles are joined:

When the tap is open, the air in $A$ and the air in $B$ are joined by the tube. Air goes from the higher pressure to the lower pressure.

The soap film of each bubble is stretched by the air inside it. The film pulls inwards, and the air pushes outwards.

$p_A = p_0 + \dfrac{4T}{R_A}$

$p_B = p_0 + \dfrac{4T}{R_B}$

$p_A > p_B$

Reasoning and calculation

What happens, stage by stage:

First, $p_A$ is bigger than $p_B$, so air goes from $A$ to $B$.

Then $A$ gets smaller. A smaller $R_A$ makes $p_A$ even bigger. And $B$ gets bigger, which makes $p_B$ smaller. The difference grows, so the air keeps going from $A$ to $B$.

This stops only when the pressures are equal, $p_A = p_B$. By the two formulas that needs $R_A = R_B$.

The bubble $A$ cannot be a whole sphere again, because it is the one that shrank. It becomes a part of a sphere of the same radius as $B$, at the end of the tube. So its volume is smaller.

Reading the options

Now match it with the five options.

We found equal radii and a smaller volume for $A$. That is option (5).

(1) and (2) have different radii. (3) has equal volumes. (4) has $R_A$ bigger. None of these match.

Why the other options fail

$R_A$ would still be smaller, so $p_A$ would still be bigger and the air would keep flowing.

Two spheres of different radii cannot have equal volumes.

$A$ cannot become a whole sphere equal to $B$, because it is the bubble that empties.

$A$ starts with the higher pressure, so air leaves it and $R_A$ does not become bigger than $R_B$.

Checks

Does it make sense?

At the end the two pressures are equal, so no air flows any more. That is what equilibrium means.

The bubble $A$ gave away its air, so its volume is smaller than the volume of $B$.

Conclusion

$R_A = R_B,\ V_A < V_B$

The radii are equal, $R_A = R_B$, and $V_A < V_B$. That is option (5).

The answer is (5).

The quotes give the excess pressure $\dfrac{4T}{r}$ in a soap bubble. That air goes from the higher pressure to the lower one, and that $A$ ends as a part of a sphere with the radius of $B$ because it cannot become bigger, is our reasoning from the formula and from the question saying that no bubble breaks.

Teacher's note

A smaller bubble has a higher pressure inside, $p = p_0 + \dfrac{4T}{R}$. Air goes from the small bubble to the big one.

It stops only when the radii are equal, and then the small bubble is a part of a sphere.

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