Question

The excess pressure inside the first soap bubble is three times that inside the second bubble. Then, the ratio of the volume of the first bubble to that of the second bubble will be

• A. $1:27$
• B. $3:1$
• C. $1:3$
• D. $1:9$

Answer 1

The correct option is A $1:27$

Let us suppose, radius of first and second soap bubble is $r_1$ and $r_2$ respectively. Also suppose, $T$ is surface tension.
Now, excess pressure inside first bubble,
$p_1=\frac{4T}{r_1}$ .......(1)
and excess pressure inside second bubble,
$p_2=\frac{4T}{r_2}$ .......(2)
Given, $p_1=3p_2$
$\frac{4T}{r_1}=\frac{3\times 4T}{r_2}$ $[$ from (1) and (2) $]$
$\Rightarrow \frac{r_1}{r_2}=\frac{1}{3}$ .......(3)
Ratio of their volumes,
$\frac{V_1}{V_2}=\frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3}$
$\Rightarrow \frac{V_1}{V_2}={\left(\frac{r_1}{r_2}\right)}^3=\frac{1}{27}$ $[$ from (3) $]$