Question

Pressure inside two soap bubbles are $1.01\text{atm}$ and $1.02\text{atm}$ respectively. The ratio of their volume is-

• A. $4:1$
• B. $0.8:1$
• C. $8:1$
• D. $2:1$

Answer 1

The correct option is C $8:1$

Excess pressure inside the soap bubble is,

$\Delta P=P_{in}-P_{out}=\frac{4T}{R}$
$\Delta P_1=P_{in}-P_{out}=0.01=\frac{4T}{R_1}....\left(1\right)$
$\Delta P_2=P_{in}-P_{out}=0.02=\frac{4T}{R_2}....\left(2\right)$
Dividing, equation $\left(2\right)$ by $\left(1\right)$
$\frac{1}{2}=\frac{R_2}{R_1}\Rightarrow R_1=2R_2$
Volume $V=\frac{4}{3}\pi R^3$
$\therefore \frac{V_1}{V_2}=\frac{R_1^3}{R_2^3}=\frac{8R_2^3}{R_2^3}=\frac{8}{1}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.