Question

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

• A. 102 : 101
• B. $(102)^{3}:(101)^{3}$
• C. 8 : 1
• D. 2 : 1

Answer 1

Answer: (C)

8 : 1

Answer 2

Excess pressure $\Delta P = P_{in} - P_{\text{out}}$ $= 1.01 \mathrm {~atm} - 1 \mathrm {~atm}$

$= 0.01atm$ and similarly $\Delta P_{2} = 0.02atm$

and volume of air bubble $V = \frac{4}{3}\pi r^{3}$ $\therefore V \propto r^{3} \propto \frac{1}{(\Delta P)^{3}}$ [as $\Delta P \propto \frac{1}{r}$ or $r \propto \frac{1}{\Delta P}$]

∴ $\frac{V_{1}}{V_{2}} = \left( \frac{\Delta P_{2}}{\Delta P_{1}} \right)^{3} = \left( \frac{0.02}{0.01} \right)^{3} = \left( \frac{2}{1} \right)^{3} = \frac{8}{1}$