Question:
A small soap bubble of radius 4 cm is trapped inside another bubble of radius 6 cm without any contact. Let \(P_2\) be the pressure inside the inner bubble and \(P_0\), the pressure outside the outer bubble. Radius of another bubble with pressure difference \(P_2-P_0\) between its inside and outside would be :
A small soap bubble of radius 4 cm is trapped inside another bubble of radius 6 cm without any contact. Let \(P_2\) be the pressure inside the inner bubble and \(P_0\), the pressure outside the outer bubble. Radius of another bubble with pressure difference \(P_2-P_0\) between its inside and outside would be :
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A soap bubble has two surfaces, so the excess pressure is \(\frac{4T}{r}\); add the excess pressures of the two bubbles.
Updated On: Nov 11, 2026
- 12 cm
- 2.4 cm
- 6 cm
- 4.8 cm
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The Correct Option is B
Solution and Explanation
Let \(P_1\) be the pressure between the bubbles. Inner bubble: \(P_2-P_1=\frac{4T}{4}\). Outer bubble: \(P_1-P_0=\frac{4T}{6}\).
\(P_2-P_0=4T\left(\frac14+\frac16\right)=4T\cdot\frac{5}{12}=\frac{5T}{3}\).
For a single bubble of radius \(R\): \(\frac{4T}{R}=\frac{5T}{3}\Rightarrow R=\frac{12}{5}=2.4\) cm.
Answer: (2) 2.4 cm
\(P_2-P_0=4T\left(\frac14+\frac16\right)=4T\cdot\frac{5}{12}=\frac{5T}{3}\).
For a single bubble of radius \(R\): \(\frac{4T}{R}=\frac{5T}{3}\Rightarrow R=\frac{12}{5}=2.4\) cm.
Answer: (2) 2.4 cm
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