Let the radius of curvature of the common internal film surface of the double bubble formed by two bubbles $\mathrm{A}$ and $\mathrm{B}$ be $\mathrm{r}$.



Excess of pressure as compared to the atmosphere inside $\mathrm{A}$ is

${\mathrm{p}}_1=\frac{4\mathrm{T}}{{\mathrm{r}}_1}=\frac{4\mathrm{T}}{0.03}$

Excess of pressure inside $\mathrm{B}$ is

${\mathrm{p}}_2=\frac{4\mathrm{T}}{{\mathrm{r}}_2}=\frac{4\mathrm{T}}{0.04}$

In the double bubble, the pressure difference between $\mathrm{A}$ and $\mathrm{B}$ on either side of the common surface is

$\frac{4\mathrm{T}}{0.03}-\frac{4\mathrm{T}}{0.04}=\frac{4\mathrm{T}}{\mathrm{r}}$

$⟹ \frac{1}{0.03}-\frac{1}{0.04}=\frac{1}{\mathrm{r}}$

$⟹\mathrm{ }\mathrm{r}=\frac{0.03\times 0.04}{0.01}=0.12 \mathrm{m}$