Two spherical soap bubbles

Question:

Two spherical soap bubbles of radii $r_{1}$ and $r_{2}$ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to:

  1. $\frac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+\mathrm{r}_{2}}$

  2. $\sqrt{\mathrm{r}_{1} \mathrm{r}_{2}}$

  3. $\sqrt{\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}}$

  4. $\frac{\mathrm{r}_{1}+\mathrm{r}_{2}}{2}$


Correct Option: , 3

Solution:

no. of moles is conserved

$\mathrm{n}_{1}+\mathrm{n}_{2}=\mathrm{n}_{3}$

$\mathrm{P}_{1} \mathrm{~V}_{1}+\mathrm{P}_{2} \mathrm{~V}_{2}=\mathrm{P}_{3} \mathrm{~V}$

$\frac{4 \mathrm{~S}}{\mathrm{r}_{1}}\left(\frac{4}{3} \pi \mathrm{r}_{1}^{3}\right)+\frac{4 \mathrm{~S}}{\mathrm{r}_{2}}\left(\frac{4}{3} \pi \mathrm{r}_{2}^{3}\right)=\frac{4 \mathrm{~S}}{\mathrm{r}_{3}}\left(\frac{4}{3} \pi \mathrm{r}_{3}^{3}\right)$

$\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}=\mathrm{r}_{3}^{2}$

$\mathrm{r}_{3}=\sqrt{\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}}$

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