NSEP 2015 Problem 6 | Surface Tension and Pressure

Two air bubble with radius $r_1$ and $r_2$ $(r_2>r_1)$ formed of the same liquid stick to each other to form a common interface. Therefore, the radius of curvature of the common surface is

(a) $\sqrt{r_1 r_2}$ (b) Infinity (c) $\frac{r_2}{r_1}\sqrt{{r_2}^2-{r_2}^2}$ (d) $\frac{r_1 r_2}{r_2 - r_1}$

Concept of Physics H.C. Verma

University Physics by H. D. Young and R.A. Freedman

Fundamental of Physics D. Halliday, J. Walker and R. Resnick

National Standard Examination in Physics(NSEP) 2015-2016

Option-(d) $\frac{r_1 r_2}{r_2 - r_1}$

For a bubble of radius $r$ with double surface and whose inside pressure is $p_{in}$ and outside pressure is $p_{out}$ and also the surface tension is $T$, the relation between pressure and radius is,

$$p_{in} - p_{out} = \Delta p = \frac{4T}{r}$$

When 2 bubble merge on their common interface the pressure difference is Just the difference between the pressure(inside) of both two bubbles. Also, the surface tension remains same.

We know $p_1 - p_0 = \frac{4T}{r_1}$ and $p_2 - p_0 = \frac{4T}{r_2}$, Then,

$$ p_1 - p_2 = \frac{4T}{R} $$

here $R$ is the radius of curvature of the interface.

Hence,

$p_{1}-p_{2}$=$\left(p_{1}-p_{0}\right)-\left(p_{2}-p_{0}\right)$=$4 T\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right)$

This gives us,

$$ \frac{1}{R} = \frac{1}{r_1}-\frac{1}{r_2} \to R= \frac{r_1 r_2}{r_2 - r_1} $$

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