Pressure inside two soap bubbles are 1.01 and 1.02 atmospheres. Ratio between their volumes is

The correct option is D. 8 : 1.

Given,

Outside pressure = 1 atm

Pressure inside first bubble = 1.01 atm

Pressure inside second bubble = 1.02 atm

Excess pressure $△P_1$ = 1.01 - 1 = 0.01 atm

Excess pressure $△P_2$ = 1.02 - 1 = 0.02 atm

We know that, $△P=\frac{4T}{r}$, where $T$ is the surface tension and $R$ is the radius of the bubble.

$\therefore △P\propto \frac{1}{r}\Rightarrow r\propto \frac{1}{△P}\Rightarrow \frac{r_1}{r_2}=\frac{△P_2}{△P_1}=\frac{0.02}{0.01}=\frac{2}{1}$

Since $V=\frac{4}{3}\pi r^3$

$\therefore \frac{V_1}{V_2}=⟮\frac{r_1}{r_2}{⟯}^3=⟮\frac{2}{1}{⟯}^3=\frac{8}{1}$