Physics - Fluid Question with Solution | TestHub

According to given figure,Let the radius of curvature of the common internal film surface of the double bubble formed by two bubbles be $r$. Excess of pressure as compared to atmosphere inside $A$ is $(4T/a)$. In double bubble, the pressure difference between $A$ and $B$ on either side of the common surface $=\frac{4T}{b}-\frac{4T}{a}$ This will be equal to ( $4\mathrm{T}/\mathrm{r}$ ) $\therefore \frac{4T}{r}=\frac{4T}{b}-\frac{4T}{a}$ or $\frac{1}{r}=\left(\frac{1}{b}-\frac{1}{a}\right)=\frac{a-b}{ba}$ or $r=\frac{ab}{a-b}$