The pressure inside two soap bubbles are 1.05 atm and 1.07 atmosphere. The volume of two bubbles are in the ratio of

Options

A. `2:7`
B. `3:7`
C. `1:7`
D. `4:7`

Explanation

To find the ratio of the volumes of two soap bubbles with given internal pressures, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Given Pressures**:
- The pressure inside the first bubble \( P_A = 1.05 \, \text{atm} \)
- The pressure inside the second bubble \( P_B = 1.07 \, \text{atm} \)
- The atmospheric pressure \( P_0 = 1 \, \text{atm} \)

2. **Calculate the Excess Pressures**:
- The excess pressure for the first bubble \( P_{XS_A} \) is given by:
\[ P_{XS_A} = P_A - P_0 = 1.05 \, \text{atm} - 1 \, \text{atm} = 0.05 \, \text{atm} \]
- The excess pressure for the second bubble \( P_{XS_B} \) is given by:
\[ P_{XS_B} = P_B - P_0 = 1.07 \, \text{atm} - 1 \, \text{atm} = 0.07 \, \text{atm} \]

3. **Relate Excess Pressure to Radius**:
- The excess pressure in a soap bubble is related to the radius \( R \) of the bubble by the formula:
\[ P_{XS} = \frac{4T}{R} \]
- Since the surface tension \( T \) is the same for both bubbles, we can write:
\[ P_{XS_A} \propto \frac{1}{R_A} \quad \text{and} \quad P_{XS_B} \propto \frac{1}{R_B} \]

4. **Set Up the Ratio of Excess Pressures**:
- From the proportionality, we can write:
\[ \frac{P_{XS_A}}{P_{XS_B}} = \frac{R_B}{R_A} \]
- Substituting the values of \( P_{XS_A} \) and \( P_{XS_B} \):
\[ \frac{0.05}{0.07} = \frac{R_B}{R_A} \]
- Simplifying gives:
\[ \frac{R_B}{R_A} = \frac{5}{7} \]

5. **Calculate the Ratio of Volumes**:
- The volume \( V \) of a bubble is given by:
\[ V = \frac{4}{3} \pi R^3 \]
- Therefore, the ratio of the volumes of the two bubbles is:
\[ \frac{V_A}{V_B} = \frac{R_A^3}{R_B^3} = \left(\frac{R_A}{R_B}\right)^3 \]
- Since \( \frac{R_B}{R_A} = \frac{5}{7} \), we have:
\[ \frac{R_A}{R_B} = \frac{7}{5} \]
- Thus:
\[ \frac{V_A}{V_B} = \left(\frac{7}{5}\right)^3 = \frac{343}{125} \approx 2.744 \]

6. **Final Result**:
- The ratio of the volumes \( V_A : V_B \) is approximately \( 2.74 : 1 \).