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 Question:
Mechanical force per unit area in a soap bubble refers to the excess pressure due to surface tension. We first need to find the relationship between the radii.
Step 2: Key Formula or Approach:
1. Volume conservation: \( V = 64v \).
2. Mechanical force per unit area (Excess pressure) \( P = \frac{4T}{R} \).
Step 3: Detailed Explanation:
Volume of big bubble \( V = \frac{4}{3} \pi R^3 \).
Volume of 64 small bubbles \( 64 \times \frac{4}{3} \pi r^3 \).
By conservation of volume: \( R^3 = 64 r^3 \Rightarrow R = 4r \).
Mechanical force per unit area \( P \propto \frac{1}{\text{radius}} \).
Ratio:
\[ \frac{P_{big}}{P_{small}} = \frac{4T/R}{4T/r} = \frac{r}{R} \]
Substituting \( R = 4r \):
\[ \frac{P_{big}}{P_{small}} = \frac{r}{4r} = \frac{1}{4} \]
Step 4: Final Answer:
The ratio is \( 1 : 4 \).