Question

Air is pushed in a soap bubble to increase its radius from 'R' to '2R'. In this case, the pressure inside the bubble

• A. does not change
• B. decreases
• C. becomes zero
• D. increases

Hint:

Remember: smaller bubbles are tightly bound and highly pressurized, while larger bubbles relax and drop their internal pressure tension down as they expand ($P \propto 1/R$). Thus, larger size means lower internal pressure!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to determine the behavior of the internal pressure inside a spherical soap bubble when its radius is blown up from $R$ to $2R$.

Step 2: Detailed Explanation:
The gauge pressure or excess pressure ($P_{\text{excess}}$) inside a spherical soap bubble suspended in air is created by surface tension forces across its two surface interfaces, governed by: $$ P_{\text{excess}} = P_{\text{inside}} - P_{\text{outside}} = \frac{4T}{r} $$ where $T$ is the surface tension constant of the soap solution and $r$ is the bubble radius. This reveals that excess internal pressure is inversely proportional to the bubble's radius ($P_{\text{excess}} \propto \frac{1}{r}$). When more air is blown inside, expanding the radius from $R \rightarrow 2R$, the excess internal pressure drop updates via: $$ P_{\text{final}} = P_{\text{outside}} + \frac{4T}{2R} $$ Because the denominator term increases, the total value of the internal pressure decreases.

Step 3: Final Answer:
The pressure inside the bubble decreases, which corresponds to option (B).