Question

When two soap bubbles of radius ${{r}_{1}}$ and ${{r}_{2}}$ $({{r}_{1}} > {{r}_{2}})$ coalesce, the radius of curvature of common surface is
A . ${{r}_{2}}-{{r}_{1}} \\ $
B . $\dfrac{{{r}_{2}}-{{r}_{1}}}{{{r}_{1}}{{r}_{2}}} \\ $
C . $\dfrac{{{r}_{1}}{{r}_{2}}}{{{r}_{2}}-{{r}_{1}}} \\ $
D . ${{r}_{2}}+{{r}_{1}}$

Answer 1

Hint: Given two soap bubbles with radius ${{r}_{1}}$ and ${{r}_{2}}$ and the two bubbles coalesce means they collide with each other. To solve this question, we use the condition of the excess pressure inside the soap bubble. By taking the two equations with radius ${{r}_{1}}$ and ${{r}_{2}}$ and simplifying it, we get the radius of curvature.

Formula Used:
The excess pressure inside the soap bubble is
$P=\dfrac{4T}{r}$
Where T is the surface tension and R is the radius of the soap bubble.

Complete step by step solution:
Let us consider two soap bubbles of radius ${{r}_{1}}$ and ${{r}_{2}}$ such that $({{r}_{1}}>{{r}_{2}})$

We know the excess pressure inside the soap bubble for radius ${{r}_{1}}$ is
${{P}_{1}}-{{P}_{0}}=\dfrac{4T}{{{r}_{1}}} \\ $
Where T is the surface tension and ${{P}_{1}}$ is the excess pressure inside the first soap bubble and ${{P}_{2}}$ is the excess pressure inside the second soap bubble.
Then ${{P}_{1}}={{P}_{0}}+\dfrac{4T}{{{r}_{1}}} \\ $……………………………………………. (1)
And the excess pressure inside the soap bubble for radius ${{r}_{2}}$is
${{P}_{2}}-{{P}_{0}}=\dfrac{4T}{{{r}_{2}}} \\ $
Then ${{P}_{2}}={{P}_{0}}+\dfrac{4T}{{{r}_{2}}}$…………………………………………. (2)
Let us consider the two bubbles merge
So ${{P}_{1}}-{{P}_{2}}=\dfrac{4T}{r} \\ $

Now we put the value of ${{P}_{1}}$ and ${{P}_{2}}$ in the above equation, we get
${{P}_{0}}+\dfrac{4T}{{{r}_{1}}}-{{P}_{0}}-\dfrac{4T}{{{r}_{2}}}=\dfrac{4T}{r} \\ $
That is $\dfrac{4T}{{{r}_{1}}}-\dfrac{4T}{{{r}_{2}}}=\dfrac{4T}{r} \\ $
Now we take 4T common from LHS, we get
$4T\left( \dfrac{1}{{{r}_{1}}}-\dfrac{1}{{{r}_{2}}} \right)=\dfrac{4T}{r} \\ $
Now we cancel 4T from both sides and get
$\left( \dfrac{1}{{{r}_{1}}}-\dfrac{1}{{{r}_{2}}} \right)=\dfrac{1}{r} \\ $
Then $r=\dfrac{{{r}_{1}}{{r}_{2}}}{{{r}_{2}}-{{r}_{1}}} \\ $
Therefore, the radius of curvature of common surface is $r=\dfrac{{{r}_{1}}{{r}_{2}}}{{{r}_{2}}-{{r}_{1}}}$

Thus, option C is the correct answer.

Note: Surface tension is the film of a liquid which occurs on the surface because of the attraction of the surface particles by the liquid that tries to minimise the surface area of liquid drop. This phenomenon is called surface tension.