Calculate the radius of a bubble which is formed when two goap bubbles having radii 4 cm and 5cm coalesce under isothermal conditions.

Answer

To solve the problem of finding the radius of a bubble formed when two soap bubbles with radii 4 cm and 5 cm coalesce under isothermal conditions, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two soap bubbles with radii \( R_1 = 4 \) cm and \( R_2 = 5 \) cm.
- We need to find the radius \( R \) of the new bubble formed when these two bubbles coalesce.

2. **Using the Isothermal Condition**:
- Under isothermal conditions, the pressure-volume relationship for the soap bubbles can be expressed as:
\[ P_1 V_1 + P_2 V_2 = P V \]
- For soap bubbles, the pressure \( P \) inside a bubble is given by:
\[ P = \frac{4T}{R} \]
- Therefore, we can express the pressures for the two bubbles as:
\[ P_1 = \frac{4T}{R_1} \quad \text{and} \quad P_2 = \frac{4T}{R_2} \]

3. **Calculating the Volumes**:
- The volume \( V \) of a sphere (soap bubble) is given by:
\[ V = \frac{4}{3} \pi R^3 \]
- Thus, the volumes of the two bubbles are:
\[ V_1 = \frac{4}{3} \pi R_1^3 \quad \text{and} \quad V_2 = \frac{4}{3} \pi R_2^3 \]

4. **Substituting into the Equation**:
- Substitute \( P_1 \), \( P_2 \), \( V_1 \), and \( V_2 \) into the isothermal condition equation:
\[ \frac{4T}{R_1} \cdot \frac{4}{3} \pi R_1^3 + \frac{4T}{R_2} \cdot \frac{4}{3} \pi R_2^3 = \frac{4T}{R} \cdot \frac{4}{3} \pi R^3 \]
- Simplifying this, we can cancel \( \frac{4}{3} \pi \) and \( 4T \) from both sides:
\[ \frac{R_1^3}{R_1} + \frac{R_2^3}{R_2} = \frac{R^3}{R} \]
- This simplifies to:
\[ R_1^2 + R_2^2 = R^2 \]

5. **Calculating the New Radius**:
- Now substitute \( R_1 = 4 \) cm and \( R_2 = 5 \) cm into the equation:
\[ 4^2 + 5^2 = R^2 \]
- Calculate \( 4^2 = 16 \) and \( 5^2 = 25 \):
\[ 16 + 25 = R^2 \]
\[ 41 = R^2 \]
- Taking the square root gives:
\[ R = \sqrt{41} \text{ cm} \]

6. **Final Answer**:
- The radius of the bubble formed is \( R \approx 6.4 \) cm (since \( \sqrt{41} \approx 6.4 \)).