Question

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

• A. \( 1 : 4 \)
• B. \( 1 : 1 \)
• C. \( 1 : 2 \)
• D. \( 1 : 8 \)

Hint:

For soap bubbles, excess pressure varies inversely with radius, while volume varies as cube of radius.

Answer 1

The Correct Option is D

Step 1: Write excess pressure formula for soap bubble.
For a soap bubble, excess pressure is \[ \Delta P = \frac{4T}{R} \] where \( T \) is surface tension and \( R \) is radius.
Step 2: Use given condition.
\[ \Delta P_1 = 2 \Delta P_2 \] \[ \frac{4T}{R_1} = 2 \left(\frac{4T}{R_2}\right) \]
Step 3: Solve for radii ratio.
\[ \frac{1}{R_1} = \frac{2}{R_2} \Rightarrow R_2 = 2R_1 \]
Step 4: Use volume relation.
Volume of a bubble \[ V \propto R^3 \]
Step 5: Find ratio of volumes.
\[ \frac{V_1}{V_2} = \left(\frac{R_1}{R_2}\right)^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \]
Step 6: Conclusion.
The ratio of volumes is \( 1 : 8 \).