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JEE Mains · Physics · STD 11 - Surface tension
  • JEE Mains 2014
  • Hard
  • MCQ

Two soap bubbles coalesce to form a single bubble. If \(V\) is the subsequent change in volume of contained air and \(S\) change in total surface area, \(T\) is the surface tension and \(P\) atmospheric pressure, then which of the following relation is correct?

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Answer & Solution

Correct Answer

(B) \(3PV+4ST = 0\)

Step-by-step Solution

Detailed explanation

Let \(P_i\) and \(R_i\) be the inside pressure and radius of the ith soap bubble respectively.

\(\therefore {P_1} = P + \frac{{4T}}{{{R_1}}}.\,\,\,{P_2} = P + \frac{{4T}}{{{R_2}}}\,\,and\,\,{P_3} = P + \frac{{4T}}{{{R_3}}}\)

\(Also\,{P_1}{V_1} + {P_2}{V_2} = {P_3}{V_3}\)

\(\therefore \left( {P + \frac{{4T}}{{{R_1}}}} \right)\frac{{4\pi }}{3}R_1^3 + \left( {P + \frac{{4T}}{{{R_2}}}} \right)\frac{{4\pi }}{3}R_2^3\)

\( = \left( {p + \frac{{4T}}{{{R_3}}}} \right)\frac{{4\pi }}{3}R_3^3\)

\(P\left( {\frac{{4\pi }}{3}R_1^3 + \frac{{4\pi }}{3}R_2^3 - \frac{{4\pi }}{3}R_3^2} \right)\)

\( + \frac{{4T}}{3}\left( {4\pi R_1^2 + 4\pi R_2^2 - 4\pi R_3^2} \right) = 0\)

\(P\left( {{V_1} + {V_2} - {V_3}} \right) + \frac{{4T}}{3}\left( {{S_1} - {S_2} - {S_3}} \right) = 0\)

\(PV + \frac{{4T}}{3}S = 0\,\,\,\,\,\,\, \Rightarrow \,\,\,\,3PV + 4ST = 0\)

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