Question

The pressure inside two soap bubbles in air are $1.02\text{atm}$ and $1.03\text{atm}$ respectively. The ratio of their volumes is -

• A. $102:103$
• B. $9:4$
• C. $(103{)}^3:(102{)}^3$
• D. $27:8$

Answer 1

The correct option is D $27:8$

Outside pressure $=1\text{atm}$
Pressure inside first bubble $=1.02\text{atm}$
Pressure inside second bubble $=1.03\text{atm}$
Excess pressure -
$\Delta P_1=1.02-1=0.02\text{atm}$
$\Delta P_2=1.03-1=0.03\text{atm}$

We know,
$\Delta P=\frac{2T}{r}\Rightarrow \Delta P\propto \frac{1}{r}$
$\Rightarrow \frac{r_1}{r_2}=\frac{\Delta P_2}{\Delta P_1}=\frac{0.03}{0.02}=\frac{3}{2}$
Since $V=\frac{4}{3}\pi r^3$,
$\therefore \frac{V_1}{V_2}={\left(\frac{r_1}{r_2}\right)}^3={\left(\frac{3}{2}\right)}^3$
or $\frac{V_1}{V_2}=\frac{27}{8}$