Question

A spherical soap bubble of radius $1\text{cm}$ is formed inside another of radius $3\text{cm}$. The radius of a single soap bubble which maintains the same pressure difference as inside the smaller and outside the larger soap bubble is

• A. $1.33\text{cm}$
• B. $0.75\text{cm}$
• C. $7.5\text{cm}$
• D. $13.3\text{cm}$

Answer 1

The correct option is B $0.75\text{cm}$

Pressure outside the bigger drop = $P_1$
Pressure inside the bigger drop = $P_2$
Radius of bigger drop, $r_1=3\text{cm}$
So, excess pressure, $\begin{array}{}{P}_2-P_1=\frac{4T}{r_1}=\frac{4T}{3}& \text{(1)}\end{array}$
Pressure inside small drop = $P_3$
$\therefore$ Excess pressure, $\begin{array}{}{P}_3-P_2=\frac{4T}{r_2}=\frac{4T}{1}& \text{(2)}\end{array}$
Pressure difference between inner side of small drop and outer side of bigger drop,
Adding $\left(1\right)$ and $\left(2\right)$, we get
$\Rightarrow P_3-P_1=\frac{4T}{3}+\frac{4T}{1}=\frac{16T}{3}$
Now, this pressure difference should exist in a single drop of radius $r$
$\therefore \frac{4T}{r}=\frac{16T}{3}$ or $r=\frac{3}{4}\text{cm}=0.75\text{cm}$