Two spherical soap bubbles of radii r₁ and r₂ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to :

When two bubbles unite to form one bubble, the number of moles is conserved.

${\mathrm{n}}_1+{\mathrm{n}}_2={\mathrm{n}}_3$

By using the ideal gas equation,

${\mathrm{P}}_1{\mathrm{V}}_1{\mathrm{RT}}_1+{\mathrm{P}}_2{\mathrm{V}}_2{\mathrm{RT}}_2={\mathrm{P}}_3{\mathrm{V}}_3{\mathrm{RT}}_3$

Due to the process's isothermal condition,

${\mathrm{P}}_1{\mathrm{V}}_1+{\mathrm{P}}_2{\mathrm{V}}_2={\mathrm{P}}_3{\mathrm{V}}_3$

Using the bubble formula's excess pressure,

$4{\mathrm{Sr}}_1\left(\frac{4}{3}{{\mathrm{\pi r}}_1}^3\right)+4{\mathrm{Sr}}_2\left(\frac{4}{3}{{\mathrm{\pi r}}_2}^3\right)+4{\mathrm{Sr}}_1\left(\frac{4}{3}{{\mathrm{\pi r}}_1}^3\right)$

${{\mathrm{r}}_1}^2+{{\mathrm{r}}_2}^2={{\mathrm{r}}_3}^3$

${\mathrm{r}}_3=\sqrt{{{\mathrm{r}}_1}^2+{{\mathrm{r}}_2}^2}$

Hence the correct answer is $\sqrt{{{\mathrm{r}}_1}^2+{{\mathrm{r}}_2}^2}.$