How much work will be done in increasing the diameter of a soap bubble from `2cm` to `5cm`? Surface tension solution is `3.0xx10^(-2)N//m.`

Options

A. `3.96 xx 10^4 J`
B. `2.58 xx 10^4 J`
C. `1.46 xx 10^4 J`
D. `1.88 xx 10^4 J`

Explanation

To solve the problem of how much work is done in increasing the diameter of a soap bubble from 2 cm to 5 cm with a surface tension of \(3.0 \times 10^{-2} \, \text{N/m}\), we will follow these steps:

### Step 1: Understand the Problem
A soap bubble has two surfaces (inner and outer), so when we change the diameter, we need to consider the change in surface area for both surfaces.

### Step 2: Convert Diameters to Radii
The initial diameter is 2 cm, so the initial radius \( r_1 \) is:
\[ r_1 = \frac{2 \, \text{cm}}{2} = 1 \, \text{cm} = 0.01 \, \text{m} \]
The final diameter is 5 cm, so the final radius \( r_2 \) is:
\[ r_2 = \frac{5 \, \text{cm}}{2} = 2.5 \, \text{cm} = 0.025 \, \text{m} \]

### Step 3: Calculate the Change in Surface Area
The surface area \( A \) of a sphere is given by the formula:
\[ A = 4 \pi r^2 \]
The change in surface area \( \Delta A \) when changing from \( r_1 \) to \( r_2 \) is:
\[ \Delta A = 2 \times (A_2 - A_1) = 2 \times (4 \pi r_2^2 - 4 \pi r_1^2) \]
\[ = 8 \pi (r_2^2 - r_1^2) \]

### Step 4: Substitute the Values
Now, substituting the values of \( r_1 \) and \( r_2 \):
\[ \Delta A = 8 \pi \left((0.025)^2 - (0.01)^2\right) \]
Calculating \( (0.025)^2 \) and \( (0.01)^2 \):
\[ (0.025)^2 = 0.000625 \quad \text{and} \quad (0.01)^2 = 0.0001 \]
Thus,
\[ \Delta A = 8 \pi (0.000625 - 0.0001) = 8 \pi (0.000525) \]
Calculating \( 8 \pi (0.000525) \):
\[ \Delta A \approx 8 \times 3.14 \times 0.000525 \approx 0.0088 \, \text{m}^2 \]

### Step 5: Calculate the Work Done
The work done \( W \) is given by:
\[ W = T \times \Delta A \]
Where \( T \) is the surface tension. Substituting the values:
\[ W = 3.0 \times 10^{-2} \, \text{N/m} \times 0.0088 \, \text{m}^2 \]
Calculating this gives:
\[ W \approx 3.0 \times 10^{-2} \times 0.0088 \approx 2.64 \times 10^{-4} \, \text{J} \]

### Final Answer
The work done in increasing the diameter of the soap bubble from 2 cm to 5 cm is approximately:
\[ W \approx 2.64 \times 10^{-4} \, \text{J} \]