Physics - Fluid Question with Solution | TestHub

Excess pressure inside a soap bubble is $\frac{4T}{R}$. Therefore,

$P_1-P_0=\frac{4T}{R_1}=0.01 \mathrm{atm}$

$P_2-P_0=\frac{4T}{R_2}=0.02 \mathrm{atm}$

Dividing,

$\frac{1}{2}=\frac{R_2}{R_1}$

$\Rightarrow R_1=2R_2$

$\Rightarrow \frac{V_1}{V_2}=\frac{R_1^3}{R_2^3}=\frac{8R_2^3}{R_2^3}=\frac{8}{1}$