Question

Two separate bubbles (radii 0.002 m and 0.004 m) formed of the same soap solution (surface tension = 0.07 N/m) comes together to form a double bubble. The internal film common to both the bubbles has radius $R$. The value of $R$ (in mm) is (Answer upto two digit after the decimal point)

Answer 1

Excess pressure is always present on concave side. So interface is concave towards small bubble.
$p_1=p_o+\frac{4T}{r_1}{p}_2=p_o+\frac{4T}{r_2}$
Excess pressure$=p_1-p_2=4T\left(\frac{1}{r_1}-\frac{1}{r_2}\right]$
This excess pressure for the interface is also equal to $\frac{4T}{R}$.
$\because 4T\left(\frac{1}{r_1}-\frac{1}{r_2}\right]=\frac{4T}{R}R=\frac{r_1{r}_2}{r_1-r_2}=\frac{(0.002)(0.004)}{0.004-0.002}=0.004m=4mm$