Question

In air, a charged soap bubble of radius 'R' breaks into 27 small soap bubbles of equal radius 'r'. Then the ratio of mechanical force acting per unit area of big soap bubble to that of a small soap bubble is

• A. $\frac{9}{1}$
• B. $\frac{1}{81}$
• C. $\frac{1}{3}$
• D. $\frac{3}{1}$

Answer 1

Answer: (A)

$\frac{9}{1}$

Answer 2

The problem involves understanding how the electrostatic pressure changes when a large charged soap bubble breaks into smaller ones. Here's a breakdown:

1. Volume Conservation:

   The volume of the big bubble is $V_{\text{big}} = \frac{4}{3}\pi R^3$. The total volume of 27 small bubbles is $27 \times \frac{4}{3}\pi r^3$. Equating both volumes:

   $$\frac{4}{3}\pi R^3 = 27\cdot \frac{4}{3}\pi r^3 \implies r^3 = \frac{R^3}{27} \implies r = \frac{R}{3}$$

2. Electrostatic Pressure (Force/Area):

   For a charged conductor, the pressure due to surface charge density is given by $P = \frac{\sigma^2}{2\epsilon_0}$ with $\sigma = \frac{Q}{4\pi R^2}$. Thus, pressure on the big bubble is proportional to $P_{\text{big}} \propto \frac{Q^2}{R^4}$.

   When the bubble breaks, assuming charge is uniformly distributed, each small bubble gets charge $q = \frac{Q}{27}$, and its pressure is proportional to:

   $$P_{\text{small}} \propto \frac{q^2}{r^4} = \frac{\left(\frac{Q}{27}\right)^2}{\left(\frac{R}{3}\right)^4} = \frac{Q^2}{729}\times\frac{81}{R^4} = \frac{Q^2}{9\,R^4}$$

3. Ratio of Pressures:

   $$\frac{P_{\text{big}}}{P_{\text{small}}} = \frac{\frac{Q^2}{R^4}}{\frac{Q^2}{9R^4}} = 9$$

Thus, the ratio of mechanical force acting per unit area of the big bubble to that of a small bubble is $\frac{9}{1}$.