Two soap bubbles A and B have radii `r_(1) and r_(2)` respectively. If `r_(1) lt r_(2)` than the excess pressure inside

Options

A. bubbles A and B will be equal
B. bubbles A will be less than that in bubbles B
C. bubbles A will be greater than that in bubble B
D. bubbles A and B will be zero.

Answer

To solve the problem of comparing the excess pressure inside two soap bubbles A and B with radii \( r_1 \) and \( r_2 \) respectively, we can use the formula for the excess pressure inside a soap bubble. The excess pressure \( \Delta P \) inside a soap bubble is given by the formula:

\[ \Delta P = \frac{4T}{r} \]

where \( T \) is the surface tension of the soap solution and \( r \) is the radius of the bubble.

### Step-by-Step Solution:

1. **Identify the Radii**:
We have two soap bubbles, A and B, with radii \( r_1 \) and \( r_2 \) respectively. Given that \( r_1 < r_2 \).

2. **Apply the Excess Pressure Formula**:
For bubble A, the excess pressure \( \Delta P_1 \) is:
\[ \Delta P_1 = \frac{4T}{r_1} \]
For bubble B, the excess pressure \( \Delta P_2 \) is:
\[ \Delta P_2 = \frac{4T}{r_2} \]

3. **Compare the Excess Pressures**:
Since \( r_1 < r_2 \), we can analyze the relationship between \( \Delta P_1 \) and \( \Delta P_2 \):
- Since \( r_1 \) is smaller than \( r_2 \), \( \frac{1}{r_1} > \frac{1}{r_2} \).
- Therefore, \( \Delta P_1 > \Delta P_2 \).

4. **Conclusion**:
The excess pressure inside bubble A (with radius \( r_1 \)) is greater than the excess pressure inside bubble B (with radius \( r_2 \)). Thus, we conclude:
\[ \Delta P_1 > \Delta P_2 \]

### Final Answer:
The correct option is that the excess pressure in bubble A will be greater than that in bubble B.