Physics - Fluid Question with Solution | TestHub

Pressure difference $=\frac{4T}{r}$ For smaller soap. ${\mathrm{p}}_{\mathrm{atm}}-{\mathrm{p}}_{i1}=\frac{4\mathrm{T}}{\mathrm{r}}$ For bigger soap, ${\mathrm{p}}_{\mathrm{atm}}-{\mathrm{p}}_{i2}=\frac{4\mathrm{T}}{2\mathrm{r}}$ As the pressure inside smaller bubble is greater than pressure inside bigger bubble, so ar fown from smaller to bigger and thu's radius of the smaller bubble will decrease and that of the bigger bubble will increase.