Question

Pressure inside two soap bubbles are $1.01atm$ and $1.03atm$. Ratio between their volumes is:
A. $27:1$
B. $3:1$
C. $127:101$
D. None of these

Answer 1

Hint:Here we have to use the concept of surface tension.The inclination of liquid surfaces to shrink into the smallest available surface area is surface tension. Surface friction enables insects to float and slip on a water surface, typically denser than water.The results of intermolecular forces at the interface create surface tension. Surface friction depends on the composition of the liquid, temperature and the ambient environment. Liquids were molecules that would have a high surface tension and strong attractive intermolecular forces.

Complete step by step answer:
The surface tension is decreased so that bubbles can form when a soapy dish detergent is applied to water. This eliminates the pull or attraction that the molecules of water exert on each other, decreasing the solution's surface tension.The soap molecules crowd hydrophobic ends to the surface, seeking to escape water, and stay out of the layer of water molecules. Water molecules differ from each other as a result. A reduction in surface tension is caused by the increased distance between the water molecules, causing bubbles to form.

Given, pressure inside first soap bubble, $P_1=1.01atm$
Pressure inside second soap bubble, $P_2=1.03atm$
Inside the first bubble the pressure difference is given by:
$\mathrm{\Delta }{p}_1=1.01-1=0.01atm$
Inside the second bubble the pressure difference is given by:
$\mathrm{\Delta }{p}_2=1.03-1=0.03atm$
If the bubble is of radius $r$ and surface tension $t$ , the excess pressure is given by:
$p={\displaystyle \frac{4t}{r}}$
If $r_1$ and $r_2$ are the radii of first and second bubble then the ratio of pressure is as:
$$\begin{gathered} {\displaystyle \frac{\mathrm{\Delta }{p}_1}{\mathrm{\Delta }{p}_2}}={\displaystyle \frac{{\displaystyle \frac{4t}{r_1}}}{{\displaystyle \frac{4t}{r_2}}}} \\ \Rightarrow {\displaystyle \frac{\mathrm{\Delta }{p}_1}{\mathrm{\Delta }{p}_2}}={\displaystyle \frac{0.01}{0.03}} \end{gathered}$$
Now,
${\displaystyle \frac{r_2}{r_1}}={\displaystyle \frac{1}{3}}$
Now, the ratio of the volume of the bubbles is given by:
$$\begin{gathered} {\displaystyle \frac{v_1}{v_2}}={\displaystyle \frac{{\displaystyle \frac{4}{3}}\pi {r_1}^3}{{\displaystyle \frac{4}{3}}\pi {r_2}^3}} \\ \Rightarrow {\displaystyle \frac{v_1}{v_2}}={\left({\displaystyle \frac{r_1}{r_2}}\right)}^3 \\ \therefore {\displaystyle \frac{v_1}{v_2}}={\displaystyle \frac{27}{1}} \end{gathered}$$
Hence, the ratio of the volume is given as: $v_1:v_2=27:1$

Thus, option A is correct.

Note:Here we have to careful while calculating the ratio as ${\displaystyle \frac{r_1}{r_2}}={\displaystyle \frac{3}{1}}$ and not vice versa otherwise our answer would be wrong. Also we need to subtract the excess pressure inside the soap bubble because pressure inside and outside a soap bubble may be different.The air pressure inside the bubble will still be higher than the outside air pressure. By keeping the bubble as thin as possible, the soap film can minimise its surface area as a result of surface tension.