Solution:
Solution
Excess pressure for soap bubble, $$\Delta P=\dfrac{4T}{R}$$ where $$T=$$ surface tension.
Here, $$\displaystyle P_A=P_0+\frac{4T}{R_A}=8+\frac{4(0.04)}{0.02}=16 N/m^2$$
and $$\displaystyle P_B=P_0+\frac{4T}{R_B}=8+\frac{4(0.04)}{0.04}=12 N/m^2$$
As $$PV=nRt , n\propto PV$$
Thus, $$\displaystyle \frac{n_A}{n_B}=\frac{P_AV_A}{P_BV_B}=\frac{16\times (4/3)\pi(0.02)^3}{12\times (4/3)\pi(0.04)^3}=1/6$$
$$\displaystyle \therefore \frac{n_B}{n_A}=6$$