Question

Two soap bubbles each with radius $r_1$ and $r_2$ coalesce in vacuum under isothermal conditions to form a bigger bubble of radius $R$. Then $R$ is equal to

• A. $\sqrt{r_1^2+r_2^2}$
• B. $\sqrt{r_1^2-r_2^2}$
• C. $r_1+r_2$
• D. $\frac{\sqrt{r_1^2-r_2^2}}{2}$
• E. $2\sqrt{r_1^2-r_2^2}$

Answer 1

Answer: (A)

By Boyle's law
$pV=$ constant
So, $p_1{V}_1+p_2{V}_2=pV$
where $p_1=\frac{T}{r_1}$
$V_1=\frac{4}{3}\pi r_1^3$
$p_2=\frac{2T}{r_2}$
$V_2=\frac{4}{3}\pi r_2^3$
$p=\frac{2T}{R}$
$V=\frac{4}{3}\pi R^3$
$\frac{2T}{r_1}\times \frac{4}{3}\pi r_1^3+\frac{2T}{r_2}\times \frac{4}{3}\pi r_2^3=\frac{2T}{R}\times \frac{4}{3}\pi R^3$
$2T\times \frac{4}{3}\pi \left(\frac{1}{r_1}\times r^3+\frac{1}{r_2}\times r_2^3\right)=2T\times \frac{4}{3}\pi \left(\frac{1}{R}\times R^3\right)$
$r_1^2+r_2^2=R^2$
$R=\sqrt{r_1^2+r_2^2}$