Two seperate soap bubbles of radii `3 xx10^(-3)` m and `2 xx10^( -3)` m respectively, formed of same liquid (surface tension `6.5 xx10^(-2)` N/m) come together to form a double bubble. The radius of interface of doubl bubble is

Options

A. `6xx 10^(-3)` m
B. `4 xx 10^(-3)` m
C. `1.5 xx 10^(-3)` m
D. `0.66 xx 10^(-3)` m

Explanation

To solve the problem of finding the radius of the interface of a double bubble formed by two soap bubbles of different radii, we can follow these steps:

### Step 1: Understand the Problem
We have two soap bubbles with radii \( R_1 = 3 \times 10^{-3} \) m and \( R_2 = 2 \times 10^{-3} \) m, formed from the same liquid with a surface tension \( T = 6.5 \times 10^{-2} \) N/m. We need to find the radius \( R \) of the interface of the double bubble formed when these two bubbles come together.

### Step 2: Write the Excess Pressure Formula
The excess pressure inside a soap bubble is given by the formula:
\[ P = \frac{4T}{R} \]
where \( P \) is the excess pressure, \( T \) is the surface tension, and \( R \) is the radius of the bubble.

### Step 3: Calculate the Excess Pressures
For the two bubbles, we can write the excess pressures as:
- For the first bubble (radius \( R_1 \)):
\[ P_1 = \frac{4T}{R_1} \]
- For the second bubble (radius \( R_2 \)):
\[ P_2 = \frac{4T}{R_2} \]

### Step 4: Set Up the Pressure Equation
When the two bubbles come together, the pressure difference between the two bubbles can be expressed as:
\[ P_2 - P_1 = \frac{4T}{R} \]
where \( R \) is the radius of the new interface bubble.

### Step 5: Substitute the Pressures
Substituting the expressions for \( P_1 \) and \( P_2 \):
\[ \frac{4T}{R_2} - \frac{4T}{R_1} = \frac{4T}{R} \]

### Step 6: Simplify the Equation
We can factor out \( 4T \) from both sides:
\[ \frac{1}{R_2} - \frac{1}{R_1} = \frac{1}{R} \]

### Step 7: Substitute the Values
Substituting the values of \( R_1 \) and \( R_2 \):
\[ \frac{1}{2 \times 10^{-3}} - \frac{1}{3 \times 10^{-3}} = \frac{1}{R} \]

### Step 8: Find a Common Denominator
Finding a common denominator (which is \( 6 \times 10^{-3} \)):
\[ \frac{3}{6 \times 10^{-3}} - \frac{2}{6 \times 10^{-3}} = \frac{1}{R} \]
This simplifies to:
\[ \frac{1}{6 \times 10^{-3}} = \frac{1}{R} \]

### Step 9: Solve for \( R \)
Taking the reciprocal gives us:
\[ R = 6 \times 10^{-3} \text{ m} \]

### Conclusion
The radius of the interface of the double bubble is \( R = 6 \times 10^{-3} \) m.

---