Question

In air, a charged soap bubble of radius \( R \) breaks into 64 small soap bubbles of equal radius \( r \). The ratio of mechanical force per unit area of big soap bubble to that of a small bubble is

• A. 1 : 4
• B. 4 : 1
• C. 2 : 1
• D. 1 : 2

Hint:

The pressure on a soap bubble is inversely proportional to its radius. If the radius of a bubble is reduced, the pressure increases.

Answer 1

The Correct Option is B

Step 1: Understanding the surface tension and pressure relation.
The mechanical force per unit area (pressure) on a soap bubble is given by:
\[ P = \frac{4T}{r}, \]
where \( T \) is the surface tension of the soap film and \( r \) is the radius of the bubble.

Step 2: Pressure for large and small bubbles.
For the large bubble of radius \( R \), the pressure is:
\[ P_{\text{big}} = \frac{4T}{R}. \]
For a small bubble of radius \( r \), the pressure is:
\[ P_{\text{small}} = \frac{4T}{r}. \]

Step 3: Relating the radius of the small bubbles.
Since the total volume of the soap remains constant, the volume of the large bubble is equal to the total volume of the 64 small bubbles:
\[ \frac{4}{3} \pi R^3 = 64 \times \frac{4}{3} \pi r^3. \]
Simplifying:
\[ R^3 = 64r^3 \quad \Rightarrow \quad R = 4r. \]

Step 4: Finding the ratio of pressures.
Now, the ratio of the pressure of the big bubble to the small bubble is:
\[ \frac{P_{\text{big}}}{P_{\text{small}}} = \frac{\frac{4T}{R}}{\frac{4T}{r}} = \frac{r}{R} = \frac{1}{4}. \]
Final Answer:
Thus, the ratio of the pressure is:
\[ \boxed{4:1}. \]