Question

Two soap bubbles have diameter in the ration 2 : 3. Compare the excess of pressure inside these bubbles.

Answer 1

Excess pressure inside the bubble

$P_{ex}=\frac{4T}{R}$

$\therefore \frac{P_{e{x}_1}}{P_{e{x}_2}}=\frac{\frac{4T}{R_1}}{\frac{4T}{R_2}}=\frac{R_2}{R_1}=\frac{D_2}{D_1}$

Given, $\frac{D_1}{D_2}=\frac{2}{3},\therefore \frac{D_2}{D_1}=\frac{3}{2}$

Hence $\frac{P_{e{x}_1}}{P_{e{x}_2}}=\frac{3}{2}$

or $P_{e{x}_1}:P_{e{x}_2}=3:2$