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Two separate air bubbles of radii \mathrm{r_1} and \mathrm{ r_2\left(r_2>r_1\right)} formed of the same liquid come together to form a double bubble. The radius of the internal film surface common to both bubbles is

Option: 1

\mathrm{\frac{r_1 r_2}{r_2-r_1}}


Option: 2

\mathrm{\frac{r_1 r_2}{r_2+r_1}}


Option: 3

\mathrm{\frac{1}{2}\left(r_1+r_2\right)}


Option: 4

\mathrm{\left(r_2-r_1\right)}


Answers (1)

best_answer

Let r be the radius of the common surface and \mathrm{\sigma} the surface tension of the liquid. When the bubbles come together, the difference of the excess pressures on either side of the common surface must be equal, i.e.

                                     \mathrm{ \frac{4 \sigma}{r_1}-\frac{4 \sigma}{r_2}=\frac{4 \sigma}{r} }

which gives \mathrm{\quad r=\frac{r_1 r_2}{r_2-r_1}} , which is choice (a).

Posted by

Ritika Kankaria

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