Question
Two soap bubbles have radii 2 cm and 3 cm respectively. What would be the radius of curvature of the surface formed at the contact surface when the two bubbles are brought near each other?
✓
Answer
The value of pressure excess at common surface will be equal to the differene in pressure excess of the two bubbles
$\begin{aligned}i.e. ~\frac{4 T}{r} & =\frac{4 T}{r_1}-\frac{4 T}{r_2} \\\frac{1}{r} & =\frac{1}{r_1}-\frac{1}{r_2} \\\frac{1}{r} & =\frac{r_2-r_1}{r_1 r_2} \\r & =\frac{r_1-r_2}{r_2-r_1}\end{aligned}$
where $r$ is the radius of curvature of the common surface.
$\text { Hence, } \quad r=\frac{2 \times 3}{3-2}=6 cm$
$\begin{aligned}i.e. ~\frac{4 T}{r} & =\frac{4 T}{r_1}-\frac{4 T}{r_2} \\\frac{1}{r} & =\frac{1}{r_1}-\frac{1}{r_2} \\\frac{1}{r} & =\frac{r_2-r_1}{r_1 r_2} \\r & =\frac{r_1-r_2}{r_2-r_1}\end{aligned}$
where $r$ is the radius of curvature of the common surface.
$\text { Hence, } \quad r=\frac{2 \times 3}{3-2}=6 cm$
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