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. $\frac{\sqrt{r_1^2 - r_2^2}}{2}$

Answer 1

Answer: (A)

$\sqrt{r_1^2 + r_2^2}$

Answer 2

By Boyle's law
$p V =$ constant
So, $p_{1} V_{1}+p_{2} V_{2} =p V$
where $p_{1}=\frac{T}{r_{1}}$
$V_{1}=\frac{4}{3} \pi r_{1}^{3}$
$p_{2} =\frac{2 T}{r_{2}}$
$V_{2} =\frac{4}{3} \pi r_{2}^{3}$
$p =\frac{2 T}{R}$
$V =\frac{4}{3} \pi R^{3}$
$\frac{2 T}{r_{1}} \times \frac{4}{3} \pi r_{1}^{3}+\frac{2 T}{r_{2}} \times \frac{4}{3} \pi r_{2}^{3}=\frac{2 T}{R} \times \frac{4}{3} \pi R^{3}$
$2 T \times \frac{4}{3} \pi\left(\frac{1}{r_{1}} \times r^{3}+\frac{1}{r_{2}} \times r_{2}^{3}\right)=2 T \times \frac{4}{3} \pi\left(\frac{1}{R} \times R^{3}\right)$
$r_{1}^{2}+r_{2}^{2}=R^{2}$
$R=\sqrt{r_{1}^{2}+r_{2}^{2}}$