Animated Solution for Physics - Properties of Solids and Liquids: A soap bubble of radius 3 cm is formed inside the another soap bubble of radius 6 cm. The radius of an equivalent soap bubble which has the same excess pressure as inside the smaller bubble with respect to the atmospheric pressure is ........ cm.

Have you ever blown a soap bubble and watched it float gracefully in the air? It’s a beautiful display of physics in action. The delicate film of soapy water is held together by surface tension, a force that acts like a stretched elastic sheet, constantly trying to minimize the surface area of the liquid.

Because the surface is pulling inward, the air inside the bubble gets squeezed. This means the pressure inside a soap bubble is always slightly higher than the atmospheric pressure outside. This difference is what we call the excess pressure.

To understand exactly how much extra pressure is inside, we use the Young-Laplace equation. For a single soap bubble in the air, the excess pressure $\Delta p$ is given by:

Here, $S$ is the surface tension of the soap solution, and $r$ is the radius of the bubble. You might wonder, why is there a $4$ in the numerator instead of a $2$? It’s because a soap bubble is essentially a hollow shell of liquid. It has two free surfaces: an inner surface in contact with the trapped air, and an outer surface in contact with the atmosphere. Each surface contributes $\frac{2S}{r}$ to the pressure, giving a total of $\frac{4S}{r}$.

A Pressure Hierarchy

Now, imagine a fascinating scenario: a smaller soap bubble formed inside a larger one. This is exactly what our problem presents. We have an outer bubble with a radius $r_A=6\text{ cm}$ and an inner bubble with a radius $r_B=3\text{ cm}$.

How does the pressure work here? It builds up like a hierarchy. The outer bubble maintains a pressure that is $\frac{4S}{r_A}$ higher than the outside atmosphere. The inner bubble, in turn, maintains a pressure that is $\frac{4S}{r_B}$ higher than its immediate surroundings (which is the air trapped between the two bubbles).

Therefore, the total excess pressure inside the innermost bubble, relative to the outside atmospheric pressure, is the sum of both pressure jumps:

The question asks us to find the radius of an equivalent soap bubble that would have the exact same total excess pressure as our concentric system. Let's call the radius of this hypothetical bubble $R_{\text{eq}}$.

By definition, the excess pressure of this single equivalent bubble would be:

Since we want this to equal the total excess pressure of our two-bubble system, we can set up our master equation:

Look closely at the master equation. The term $4S$ is common to every part of the equation. Assuming the bubbles are made of the same soap solution (so $S$ is constant), we can divide the entire equation by $4S$. This leaves us with a beautifully elegant geometric relationship:

Does this formula look familiar? It is mathematically identical to the formula for calculating the equivalent resistance of two resistors connected in parallel! Just as adding resistors in parallel decreases the overall resistance, adding concentric bubbles increases the total pressure, which corresponds to a smaller equivalent radius.

Now, all that's left is to substitute our given values into this elegant relation. We know $r_A=6\text{ cm}$ and $r_B=3\text{ cm}$.

To add these fractions, we find a common denominator, which is $6$:

Simplifying the fraction gives us:

Finally, taking the reciprocal of both sides reveals our answer:

The equivalent bubble has a radius of $2\text{ cm}$. Notice how this radius is smaller than both $3\text{ cm}$ and $6\text{ cm}$, perfectly aligning with our parallel resistor analogy. This is a classic, high-yield concept in physics that beautifully connects fluid mechanics with geometric harmonic sums!