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JEE Mains · Physics · STD 11 - Surface tension
  • JEE Mains 2018
  • Hard
  • MCQ

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....... \(cm\)

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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

\({P_2} - {P_0}\,so,\)

\({P_2} - {P_0} = \frac{{4T}}{r}\)                                           \(...(ii)\)

\(From\,e{q^n}\,\left( i \right)\,and\,\left( {ii} \right),\)

\(\frac{{4T}}{r} = \frac{{4T}}{6} + \frac{{4T}}{4}\)

\(\frac{1}{r} = \frac{1}{6} + \frac{1}{4} \Rightarrow r = 2.4\,cm\)

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