Two soap bubbles $A$ and $B$ are kept in a closed chamber where the air is maintained at pressure $8 \mathrm{Nm}^{-2}$. The radii of bubbles $A$ and $B$ are $2 \mathrm{~cm}$, respectively. Surface tension of the soap-water used to make bubbles is $0.04$ $\mathrm{Nm}^{-1}$. Find the ratio $\frac{n_B}{n_A}$, where $n_A$ and $n_B$ are the number of moles of air in bubbles $A$ and $B$, respectively. [Neglect the effect of gravity]

Solution

Although not given in the question but we will have to assume that temperatures of $A$ and $B$ are same.
$$ \begin{aligned} & \frac{n_B}{n_A}=\frac{p_B V_B / R T}{p_A V_A / R T}=\frac{p_B V_B}{p_A V_A} \\ &= \frac{p+4 s / r_A \times 4 / 3 \pi\left(r_A\right)^3}{\left(p+4 s / r_B\right) \times 4 / 3 \pi\left(r_B\right)^3} \\ &(s=\text { surface tension) } \end{aligned} $$

Hence, the correct answer is 6.