Work done in increasing the size of a soap bubble from a radius of 3cm to 5cm is nearly (Surface tension of soap solution `=0.03Nm^-1`)

Options

A. ` 4 pi m J `
B. ` 0.2 pi m J `
C. `2 pi m J `
D. `0.4 pi mJ `

Explanation

To find the work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Work Done**:
The work done (W) in increasing the size of a soap bubble is given by the formula:
\[ W = 2T \times \Delta A \]
where \( T \) is the surface tension and \( \Delta A \) is the change in surface area.

2. **Calculate the Change in Surface Area**:
The surface area \( A \) of a sphere is given by:
\[ A = 4\pi R^2 \]
For the initial radius \( r = 3 \) cm and the final radius \( R = 5 \) cm, the change in surface area \( \Delta A \) is:
\[ \Delta A = A_{final} - A_{initial} = 4\pi (5^2) - 4\pi (3^2) \]
Simplifying this gives:
\[ \Delta A = 4\pi (25 - 9) = 4\pi (16) = 64\pi \, \text{cm}^2 \]

3. **Convert the Area to Square Meters**:
Since we need to work in SI units, convert \( \Delta A \) from cm² to m²:
\[ \Delta A = 64\pi \, \text{cm}^2 \times \left(\frac{1 \, \text{m}^2}{10^4 \, \text{cm}^2}\right) = \frac{64\pi}{10^4} \, \text{m}^2 \]

4. **Substitute Values into the Work Done Formula**:
Now substitute \( T = 0.03 \, \text{N/m} \) and \( \Delta A = \frac{64\pi}{10^4} \, \text{m}^2 \) into the work done formula:
\[ W = 2T \times \Delta A = 2 \times 0.03 \times \frac{64\pi}{10^4} \]
Simplifying this gives:
\[ W = 0.06 \times \frac{64\pi}{10^4} = \frac{3.84\pi}{10^3} \, \text{J} \]

5. **Convert to Millijoules**:
To express the work done in millijoules:
\[ W = 3.84\pi \, \text{mJ} \]
Approximating \( \pi \approx 3.14 \):
\[ W \approx 3.84 \times 3.14 \approx 12.06 \, \text{mJ} \]

6. **Final Approximation**:
Therefore, the work done in increasing the size of the soap bubble is approximately:
\[ W \approx 0.4\pi \, \text{mJ} \quad \text{(approximately 12.06 mJ)} \]

### Final Answer:
The work done in increasing the size of the soap bubble from a radius of 3 cm to 5 cm is nearly **0.4π millijoules**.