Question

The pressure inside the two soap bubbles is $1.01$ and $1.02$ atmosphere. When they are in air , the ratio of their volumes is (both have same surface tension):

• A. $8$
• B. $4$
• C. $16$
• D. $64$

Answer 1

The correct option is A $8$

Pressure difference between outside and inside is
$P_{diff}=\frac{4T}{R}$
Given,
$P_{1diff}=1.01-1=0.01=\frac{4T}{R_1}$
$P_{2diff}=1.02-1=0.02=\frac{4T}{R_2}$
$\therefore \frac{R_1}{R_2}=\frac{0.02}{0.01}$
$\therefore \frac{V_1}{V_2}=(\frac{R_1}{R_2}{)}^3=(\frac{0.02}{0.01}{)}^3=8$