Question

Two soap bubbles of radii $r_1$ and $r_2$ equal to 4 cm and 5 cm are touching each other over a common surface $s_1{s}_2$ (shown in figure). Its radius will be

• A. 4 cm
• B. 20 cm
• C. 5 cm
• D. 4.5 cm

Answer 1

Answer: (B)

. 20 cm.

The excess pressure inside any soap bubble is given by the expression,

$P_{Excess}=\frac{4T}{r}$

where $T$ is the surface tension and $r$ is the radius of the bubble.

The excess pressure due to surface tension in first bubble is given by, $P_1=\frac{4T}{r_1}$

The excess pressure due to surface tension in Second bubble is given by, $P_2=\frac{4T}{r_2}$

Now if two bubble are touching each other, then they have a common surface $s_1{s}_2$. The excess pressure due to Surface tension at surface $s_1{s}_2$ is $P=\frac{4T}{R}$

As both surfaces are touching each other, hence the pressure difference at the common surface would be the excess pressure due to tension for the surface $s_1{s}_2$

$⟹P=P_1-P_2$

$\frac{4T}{R}=\frac{4T}{r_1}-\frac{4T}{r_2}$

$\frac{1}{R}=\frac{1}{r_1}-\frac{1}{r_2}$

$⟹R=\frac{r_1{r}_2}{r_2-r_1}$

$R=\frac{5\times 4}{5-4}=20cm$.