A spherical soap bubble of radius 1 cm is formed inside another of radius 3 cm the radius of single soap bubble which maintains the same pressure difference as inside the smaller and outside the larger soap bubble is ____cm

Options

A. `0.75 cm`
B. `0.75 m`
C. `7.5 cm`
D. `7.5 m`

Answer

To solve the problem, we need to find the radius of a single soap bubble that maintains the same pressure difference as the pressure difference between the inside of a smaller bubble (radius 1 cm) and the outside of a larger bubble (radius 3 cm).

### Step-by-Step Solution:

1. **Understand the Pressure Difference Formula**:
The excess pressure inside a soap bubble is given by the formula:
\[ \Delta P = \frac{4T}{R} \]
where \( T \) is the surface tension and \( R \) is the radius of the bubble.

2. **Calculate the Excess Pressure for the Smaller Bubble**:
For the smaller bubble (radius \( r = 1 \) cm):
\[ \Delta P_{inside} = \frac{4T}{1} = 4T \]

3. **Calculate the Excess Pressure for the Larger Bubble**:
For the larger bubble (radius \( R = 3 \) cm):
\[ \Delta P_{outside} = \frac{4T}{3} \]

4. **Determine the Total Pressure Difference**:
The total pressure difference between the inside of the smaller bubble and the outside of the larger bubble is:
\[ \Delta P = \Delta P_{inside} - \Delta P_{outside} = 4T - \frac{4T}{3} \]
To simplify this, we can find a common denominator:
\[ \Delta P = 4T - \frac{4T}{3} = \frac{12T}{3} - \frac{4T}{3} = \frac{8T}{3} \]

5. **Set Up the Equation for the New Bubble**:
Let \( r_n \) be the radius of the new single soap bubble. The excess pressure for this new bubble is:
\[ \Delta P_{new} = \frac{4T}{r_n} \]

6. **Equate the Pressure Differences**:
Since we want the pressure difference to be the same, we set:
\[ \frac{4T}{r_n} = \frac{8T}{3} \]

7. **Solve for \( r_n \)**:
Cancel \( 4T \) from both sides:
\[ \frac{1}{r_n} = \frac{2}{3} \]
Taking the reciprocal gives:
\[ r_n = \frac{3}{2} = 1.5 \text{ cm} \]

### Final Answer:
The radius of the single soap bubble which maintains the same pressure difference is **1.5 cm**.