Pressure inside two soap bubbles are 1.01 atm and 1.02 atm, respectively. The ratio of their volume

Q: Why do we subtract 1 atm from the given pressures? Can't we just divide 1.01 by 1.02?

The given pressures ($1.01 \text{ atm}$ and $1.02 \text{ atm}$) are absolute internal pressures. The formula $\frac{4T}{R}$ equals the *excess* pressure, which is the difference between the internal and external pressure. If you don't subtract the $1 \text{ atm}$ of the outside air, your equation is physically meaningless.

Q: Why is the formula $\frac{4T}{R}$ and not $\frac{2T}{R}$?

A soap bubble is essentially a hollow shell of liquid. It has two liquid-air interfaces: one on the inside and one on the outside. Surface tension acts on both of these surfaces, effectively doubling the inward force compared to a solid liquid drop (which only has one outer surface and uses $\frac{2T}{R}$).

Q: Does the actual value of the surface tension $T$ matter for this problem?

No, it doesn't! Because we are taking a ratio of the two bubbles (which are presumably made of the same soap solution), the surface tension $T$ is a constant that perfectly cancels out during the division.

Q: Why is the volume ratio the cube of the radius ratio?

The volume of a sphere is geometrically proportional to the cube of its radius ($V = \frac{4}{3}\pi R^3$). When you divide the volume of one sphere by another, the constants ($\frac{4}{3}\pi$) cancel out, leaving exactly $\left(\frac{R_1}{R_2}\right)^3$.

Q: What if the external pressure wasn't exactly 1 atm?

If the external pressure was different (say, inside a pressurized chamber) but still the same for both bubbles, you would subtract that specific external value to find the excess pressure. The mathematical method of taking the ratio would remain completely identical.

Question

Pressure inside two soap bubbles are $1.01 	ext{ atm}$ and $1.02 	ext{ atm}$, respectively. The ratio of their volume is

Options:

(A) $4 : 1$

(B) $0.8 : 1$

(C) $8 : 1$

(D) $2 : 1$

Sigma Percentile · Accepted Answer

### The Magic of Soap Bubbles

Have you ever wondered why soap bubbles are perfectly spherical, or why they eventually pop? The answer lies in a fascinating physical phenomenon called **surface tension**. A soap bubble is essentially a thin layer of water trapped between two layers of soap molecules. This thin film acts like a stretched elastic balloon, constantly trying to shrink to the smallest possible surface area.

Because the bubble is trying to collapse inward, it compresses the air trapped inside. This means the pressure inside a soap bubble is always slightly higher than the atmospheric pressure outside. This difference is known as the **excess pressure**.

### The Excess Pressure Formula

For a soap bubble, which has two surfaces (an inner surface and an outer surface), the surface tension acts on both. The formula for the excess pressure $\Delta p$ is given by:

$$ \Delta p = p_i - p_o = \frac{4T}{R} $$

where $p_i$ is the internal pressure, $p_o$ is the external atmospheric pressure, $T$ is the surface tension of the soap solution, and $R$ is the radius of the bubble.

### Analyzing the Given Problem

In our problem, we are given two soap bubbles with internal pressures of $1.01 	ext{ atm}$ and $1.02 	ext{ atm}$.

To find the excess pressure, we must subtract the standard external atmospheric pressure, which is $1 	ext{ atm}$.

For the first bubble:
$$ \Delta p_1 = 1.01 	ext{ atm} - 1 	ext{ atm} = 0.01 	ext{ atm} $$

For the second bubble:
$$ \Delta p_2 = 1.02 	ext{ atm} - 1 	ext{ atm} = 0.02 	ext{ atm} $$

### The Mathematical Execution

Now, we can set up a ratio using our excess pressure formula. Since both bubbles are made of the same soap solution, their surface tension $T$ is identical.

$$ \frac{\Delta p_1}{\Delta p_2} = \frac{\frac{4T}{R_1}}{\frac{4T}{R_2}} $$

Substituting our calculated excess pressures:

$$ \frac{0.01}{0.02} = \frac{R_2}{R_1} $$

Simplifying this fraction gives us:

$$ \frac{1}{2} = \frac{R_2}{R_1} \implies \frac{R_1}{R_2} = \frac{2}{1} $$

This tells us that the first bubble has a radius twice as large as the second bubble. Notice the inverse relationship: the bubble with the **smaller excess pressure** actually has the **larger radius**!

### The Final Step: From Radii to Volumes

The question asks for the ratio of their volumes. Assuming the bubbles are perfect spheres, the volume $V$ is given by $V = \frac{4}{3}\pi R^3$.

When we take the ratio of the two volumes, the constant $\frac{4}{3}\pi$ cancels out perfectly:

$$ \frac{V_1}{V_2} = \frac{\frac{4}{3}\pi R_1^3}{\frac{4}{3}\pi R_2^3} = \left(\frac{R_1}{R_2}ight)^3 $$

Now, we simply substitute the ratio of the radii we found earlier:

$$ \frac{V_1}{V_2} = \left(\frac{2}{1}ight)^3 = \frac{8}{1} $$

### Conclusion

The ratio of their volumes is **$8:1$**. This problem beautifully illustrates how linear relationships in one dimension (radius) translate into cubic relationships in three dimensions (volume). Always remember to subtract the atmospheric pressure to find the true excess pressure, and you'll never fall into the trap of directly dividing the absolute internal pressures!

Animated Solution for Physics - Properties of Solids and Liquids: Pressure inside two soap bubbles are $1.01 atm$ and $1.02 atm$ , respectively. The ratio of their volume is

Select Answer:

Visualized Solution

Visualizing the Setup

Formula for $Δ p$

Excess Pressure for Bubble 1

Excess Pressure for Bubble 2

Calculating $\frac{R _{1}}{R _{2}}$

Formula for $\frac{V _{1}}{V _{2}}$

Final Answer

Drops vs Bubbles

The Sigma Insight: Surface Tension and Capillary Action

The Magic of Soap Bubbles

The Excess Pressure Formula

Analyzing the Given Problem

The Mathematical Execution

The Final Step

Conclusion

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Animated Solution for Physics - Properties of Solids and Liquids: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Pressure inside two soap bubbles are 1.01 atm and 1.02 atm, respectively. The ratio of their volume is