Question 142 of 192

A small soap bubble of radius 4cm is trapped inside another bubble of radius 6cm 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$2.4cm$B$12cm$C$4.8cm$D$6cm$

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Answer: (A) $2.4cm$

$\begin{matrix}{P}_{2}-{P}_{1}=\dfrac{4T}{{r}_{1}} ....(1) \\ \\ {P}_{1}-{P}_{0}=\dfrac{4T}{{r}_{2}}.....(2) \\ \\ adding (1) & (2) \\ \\ {P}_{2}-{P}_{0}=\dfrac{4T}{{r}_{1}}+\dfrac{4T}{{r}_{2}} \\ \\ {P}_{2}={P}_{0}+\dfrac{4T}{{r}_{1}}+\dfrac{4T}{{r}_{2}} \\ \\ {P}_{2}={P}_{0}+\dfrac{4T}{6}+\dfrac{4T}{4}\end{matrix}$

for another bubble of radius r

${P}_{2}={P}_{0}+\dfrac{4T}{r}$

$\Rightarrow \dfrac{1}{r}=\dfrac{1}{6}+\dfrac{1}{4}$

$\Rightarrow r=2.4cm$