# Pressure inside two soap bubbles are 1.01 and 1.02 atmosphere, respectively. The ratio of their volumes is:

1) 4 : 1

2) 2 : 1

3) 0.8 : 1

4) 8 : 1

Solution:

$$P_{in}=P_{0}+\frac{4T}{R_{1}}$$,

$$1.01=1+\frac{4T}{R_{1}}$$,

$$0.01=\frac{4T}{R_{1}}$$,

$$0.02=\frac{4T}{R_{2}}$$,

$$\frac{1}{2}=\frac{R_{2}}{R_{1}}$$,

$$\frac{V_{1}}{V_{2}}=\frac{8}{1}$$