Question

When two soap bubbles of radii \( a \) and \( b \) (\( b>a \)) coalesce, the radius of curvature of the common surface is:

• A. \( \frac{ab}{b - a} \)
• B. \( \frac{ab}{a + b} \)
• C. \( \frac{b - a}{ab} \)
• D. \( \frac{a + b}{ab} \)

Hint:

When two soap bubbles merge, the radius of curvature of the common interface is determined by the difference in pressures inside the bubbles, which is related to their respective radii and surface tension.

Answer 1

The Correct Option is A

The internal pressure of a soap bubble is quantified by the formula:\[P = P_0 + \frac{4T}{R},\]where \( T \) denotes surface tension and \( R \) represents the bubble's radius.Step 1: For two bubbles with radii \( a \) and \( b \), their respective pressures are:\[P_1 = P_0 + \frac{4T}{a}, \quad P_2 = P_0 + \frac{4T}{b}.\]Step 2: The pressure differential across the shared interface is:\[P_1 - P_2 = \frac{4T}{R}.\]Step 3: Substituting the expressions for \( P_1 \) and \( P_2 \) yields:\[\frac{4T}{a} - \frac{4T}{b} = \frac{4T}{R}.\]Step 4: Simplification results in:\[\frac{1}{a} - \frac{1}{b} = \frac{1}{R}.\]Step 5: Solving for \( R \) gives:\[\frac{1}{R} = \frac{b - a}{ab}, \quad R = \frac{ab}{b - a}.\] Final Answer:\[\boxed{\frac{ab}{b - a}}.\]