Find out work done to expend soup bobble to radius R = 5 cm (surface tension of water = 0.1 N/m)

Options

A. `2.8xx10^(-3)J`
B. `6.28xx10^(-3)J`
C. `3.7xx10^(-3)J`
D. `5.8xx10^(-3)J`

Explanation

To find the work done to expand a soap bubble to a radius of \( R = 5 \, \text{cm} \) with a surface tension of water \( \gamma = 0.1 \, \text{N/m} \), we can follow these steps:

### Step 1: Understand the concept of work done on a soap bubble
The work done \( W \) to expand a soap bubble can be calculated using the formula:
\[ W = \gamma \times A \]
where \( \gamma \) is the surface tension and \( A \) is the surface area of the bubble.

### Step 2: Calculate the surface area of the soap bubble
A soap bubble is a sphere, and the surface area \( A \) of a sphere is given by:
\[ A = 4\pi r^2 \]
where \( r \) is the radius of the bubble.

### Step 3: Convert the radius from centimeters to meters
Given that the radius \( R = 5 \, \text{cm} \), we need to convert this into meters:
\[ R = 5 \, \text{cm} = 5 \times 10^{-2} \, \text{m} \]

### Step 4: Substitute the radius into the surface area formula
Now, substituting \( r = 5 \times 10^{-2} \, \text{m} \) into the surface area formula:
\[ A = 4\pi (5 \times 10^{-2})^2 \]
Calculating \( (5 \times 10^{-2})^2 \):
\[ (5 \times 10^{-2})^2 = 25 \times 10^{-4} = 2.5 \times 10^{-3} \]
Thus,
\[ A = 4\pi (2.5 \times 10^{-3}) = 10\pi \times 10^{-3} \]

### Step 5: Calculate the work done
Now substituting \( A \) and \( \gamma \) into the work done formula:
\[ W = \gamma \times A = 0.1 \times (10\pi \times 10^{-3}) \]
Calculating this:
\[ W = 0.1 \times 10 \times \pi \times 10^{-3} = \pi \times 10^{-3} \, \text{J} \]
Using \( \pi \approx 3.14 \):
\[ W \approx 3.14 \times 10^{-3} \, \text{J} \approx 6.28 \times 10^{-3} \, \text{J} \]

### Final Answer
Thus, the work done to expand the soap bubble to a radius of 5 cm is approximately:
\[ W \approx 6.28 \times 10^{-3} \, \text{J} \]