The work done in increasing the radius of a soap bubble from 4 cm to 5 cm is Joule (given surface tension of soap water to be `25xx10^(-3)N//m)`

Options

A. `0.5657xx10^(-3)`
B. `5.657xx10^(-3)`
C. `56.5xx10^(-3)`
D. `565xx10^(-3)`

Explanation

To find the work done in increasing the radius of a soap bubble from 4 cm to 5 cm, we can follow these steps:

### Step 1: Understand the Problem
We need to calculate the work done in increasing the radius of a soap bubble. The surface tension of the soap solution is given, and we have the initial and final radii of the bubble.

### Step 2: Convert Units
Convert the radii from centimeters to meters:
- Initial radius \( R_i = 4 \, \text{cm} = 4 \times 10^{-2} \, \text{m} \)
- Final radius \( R_f = 5 \, \text{cm} = 5 \times 10^{-2} \, \text{m} \)

### Step 3: Calculate the Change in Surface Area
The surface area \( A \) of a bubble with radius \( R \) is given by:
\[ A = 4\pi R^2 \]
For a soap bubble, there are two surfaces (inside and outside), so the effective surface area \( A \) is:
\[ A = 2 \times 4\pi R^2 = 8\pi R^2 \]

Now, calculate the initial and final surface areas:
- Initial surface area \( A_i = 8\pi (R_i^2) = 8\pi (4 \times 10^{-2})^2 \)
- Final surface area \( A_f = 8\pi (R_f^2) = 8\pi (5 \times 10^{-2})^2 \)

### Step 4: Calculate the Change in Surface Area
The change in surface area \( \Delta A \) is:
\[ \Delta A = A_f - A_i \]
Substituting the values:
\[ \Delta A = 8\pi (5 \times 10^{-2})^2 - 8\pi (4 \times 10^{-2})^2 \]

### Step 5: Simplify the Expression
Factor out \( 8\pi \):
\[ \Delta A = 8\pi \left( (5 \times 10^{-2})^2 - (4 \times 10^{-2})^2 \right) \]
Calculating the squares:
\[ (5 \times 10^{-2})^2 = 25 \times 10^{-4} \]
\[ (4 \times 10^{-2})^2 = 16 \times 10^{-4} \]
Thus,
\[ \Delta A = 8\pi \left( 25 \times 10^{-4} - 16 \times 10^{-4} \right) = 8\pi (9 \times 10^{-4}) \]

### Step 6: Calculate the Work Done
The work done \( W \) is given by:
\[ W = T \cdot \Delta A \]
Substituting the values:
\[ W = (25 \times 10^{-3} \, \text{N/m}) \cdot (8\pi (9 \times 10^{-4})) \]

### Step 7: Final Calculation
Now, calculate \( W \):
\[ W = 25 \times 10^{-3} \cdot 8\pi \cdot 9 \times 10^{-4} \]
Calculating \( 8\pi \approx 25.13 \):
\[ W \approx 25 \times 10^{-3} \cdot 25.13 \cdot 9 \times 10^{-4} \]
\[ W \approx 0.5654 \times 10^{-3} \, \text{J} \]

### Conclusion
The work done in increasing the radius of the soap bubble from 4 cm to 5 cm is approximately \( 0.5654 \times 10^{-3} \, \text{J} \).

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