The radii of two soap bubbles are $r_1$ and $r_2$. In isothermal condition they combine with each other to…
The radii of two soap bubbles are $r_1$ and $r_2$. In isothermal condition they combine with each other to form a single bubble. The radius of resultant bubble is
$\mathrm{R}=\frac{\mathrm{r}_1+\mathrm{r}_2}{2}$
$R=r_1\left(r_1 r_2+r_2\right)$
$\mathrm{R}=\sqrt{\mathrm{r}_1^2+\mathrm{r}_2^2}$
$\mathrm{R}=\mathrm{r}_1+\mathrm{r}_2$
Solution
Under isothermal condition, $\mathrm{T}$ is constant.
This means the surface energy of the bubbles before combining will be equal to the surface energy after combining.
$\begin{aligned}
& \quad \text { i.e. } 4 \pi r_1^2 T+4 \pi r_2^2 T=4 \pi R^2 T \\
& \Rightarrow r_1^2+r_2^2=R^2 \\
& \therefore \quad R=\sqrt{r_1^2+r_2^2}
\end{aligned}$