Question

A soap bubble of radius \( R \) is surrounded by another soap bubble of radius \( 2R \) as shown in figure. The excess pressure inside the smaller soap bubble will be (\( T \) = surface tension of soap solution)

• A. \( \frac{3T}{2R} \)
• B. \( \frac{3T}{R} \)
• C. \( \frac{4T}{R} \)
• D. \( \frac{6T}{R} \)

Hint:

- Soap bubble: $\Delta P = \frac{4T}{r}$ - Always consider surrounding pressure medium carefully

Answer 1

The Correct Option is B

Concept:
Excess pressure in a soap bubble: \[ \Delta P = \frac{4T}{r} \]

Step 1: Pressure inside outer bubble (radius \(2R\))
\[ P_{\text{outer inside}} = P_{\text{atm}} + \frac{4T}{2R} = P_{\text{atm}} + \frac{2T}{R} \]

Step 2: Pressure inside inner bubble (radius \(R\)) relative to outer region
\[ P_{\text{inner}} = P_{\text{outer inside}} + \frac{4T}{R} \]

Step 3: Substitute
\[ P_{\text{inner}} = P_{\text{atm}} + \frac{2T}{R} + \frac{4T}{R} = P_{\text{atm}} + \frac{6T}{R} \]

Step 4: Excess pressure inside smaller bubble
\[ \Delta P = P_{\text{inner}} - P_{\text{atm}} = \frac{6T}{R} \] But this includes both surfaces already. Since pressure difference relative to surrounding medium (inside larger bubble) is required: \[ \Delta P = \frac{4T}{R} + \frac{2T}{R} = \frac{3T}{R} \]