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....... \(cm\)
Pick an option to check your answer.
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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