JEE Mains · Physics · STD 11 - 9.2 surface tension
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....... \(cm\)
- A \(6\)
- B \(12\)
- C \(4.8\)
- D \(2.4\)
Answer & Solution
Correct Answer
(D) \(2.4\)
Step-by-step Solution
Detailed explanation
Clearly from figure, \({P_2} = {P_0} + \frac{{4T}}{6} + \frac{{4t}}{4}\) \(or,\,\,\,{P_2} - {P_0} = \frac{{4T}}{6} + \frac{{4T}}{4}\) \(...\left( i \right)\) Let \(r\) be the radius of bubble with pressure difference \({P_2} - {P_0}\,so,\) \({P_2} - {P_0} = \frac{{4T}}{r}\)…
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