Question

A soap bubble of radius 3 cm is formed inside the another soap bubble of radius 6 cm. The radius of an equivalent soap bubble which has the same excess pressure as inside the smaller bubble with respect to the atmospheric pressure is _________ cm.

Hint:

For soap bubbles "nested" inside each other, the effective curvature is the sum of the individual curvatures: \(\frac{1}{R_{eq}} = \frac{1}{r_1} + \frac{1}{r_2}\).

Answer 1

Correct Answer: 2

Step 1: Understanding the Concept:
The excess pressure inside a soap bubble of radius \(r\) is given by \(\Delta P = \frac{4T}{r}\), where \(T\) is the surface tension. When one bubble is inside another, the pressures add up sequentially.
Step 2: Key Formula or Approach:
1. \(P_{in} - P_{out} = \frac{4T}{r}\)
2. Total excess pressure relative to atmosphere = \(\sum \frac{4T}{r_i}\)
Step 3: Detailed Explanation:
Let \(P_0\) be atmospheric pressure.
Let \(r_1 = 6 \text{ cm}\) (outer bubble) and \(r_2 = 3 \text{ cm}\) (inner bubble).

1. Excess pressure inside the larger bubble (gap between bubbles) relative to atmosphere:
\[ P_{gap} - P_0 = \frac{4T}{r_1} \]
2. Excess pressure inside the smaller bubble relative to the gap:
\[ P_{inner} - P_{gap} = \frac{4T}{r_2} \]
3. Total excess pressure inside the smaller bubble relative to atmosphere:
\[ \Delta P_{total} = (P_{inner} - P_{gap}) + (P_{gap} - P_0) = \frac{4T}{r_2} + \frac{4T}{r_1} \]
\[ \Delta P_{total} = 4T \left( \frac{1}{3} + \frac{1}{6} \right) = 4T \left( \frac{2+1}{6} \right) = 4T \left( \frac{1}{2} \right) = \frac{4T}{2} \]

For an equivalent bubble of radius \(R_{eq}\) to have the same excess pressure:
\[ \frac{4T}{R_{eq}} = \frac{4T}{2} \implies R_{eq} = 2 \text{ cm} \]
Step 4: Final Answer:
The radius of the equivalent bubble is 2 cm.