Question

The excess pressure inside the first soap bubble of radius \(R_1\) is three times that inside the second soap bubble of radius \(R_2\). The ratio of volumes of the first bubble to second bubble is

• A. \(1:3\)
• B. \(1:6\)
• C. \(1:27\)
• D. \(1:9\)

Hint:

Excess pressure goes as 1/R and volume as R^3.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The excess pressure inside a soap bubble is \(\Delta P=\dfrac{4T}{R}\), so \(\Delta P\propto\dfrac1R\).

Step 2: Radius ratio:
\(\Delta P_1=3\Delta P_2\) means \(\dfrac1{R_1}=\dfrac3{R_2}\), so \(R_1=\dfrac{R_2}3\).

Step 3: Volume ratio:
Volume goes as \(R^3\):
\[ \frac{V_1}{V_2}=\left(\frac{R_1}{R_2}\right)^3=\frac1{27} \]

Step 4: Choose:
Option (C), ratio \(1:27\).

Final Answer:
The volume ratio is 1 : 27. \[ \boxed{1:27} \]