Question

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

• A. $2:1$
• B. $8:1$
• C. $108:101$
• D. $(102)^{2}:(103)^{2}$

Answer 1

Answer: (B)

$8:1$

Answer 2

From the question we have Excess pressure inside Ist bubble is $=(1.01-1)$ atmosphere $=0.01$ atmosphere Excess pressure inside IInd bubble is $=(1.02-1)=0.02$ atmosphere While we know excess pressure are given by $\frac{4 T}{r_{1}}$ and $\frac{4 T}{r_{2}}$ respectively So, $\frac{\frac{4 T}{r_{1}}}{\frac{4 T}{r_{2}}}=\frac{0.01}{0.02}$ or $\frac{r_{2}}{r_{1}}=\frac{1}{2}$ Let volumes of first and second bubbles are $V_{1}$ and $V_{2}$ respectively So, $\frac{V_{1}}{V_{2}}=\frac{\frac{4}{3} \pi r_{1}^{3}}{\frac{4}{3} \pi r_{2}^{3}}$ $=\left(\frac{r_{1}}{r_{2}}\right)^{2}=\left(\frac{2}{1}\right)^{3}=\frac{8}{1}$ Hence, $V_{1}: V_{2}=8: 1$