The amount of work done in forming a soap bubble of radius `r = 1 cm` (surface tension `T = 3 xx 10^(-2) N//m`) is

Options

A. `37.38 mu J`
B. `40.20 muJ`
C. `75.36 mu J`
D. `20.10 mu J`

Answer

To find the amount of work done in forming a soap bubble of radius \( r = 1 \, \text{cm} \) with surface tension \( T = 3 \times 10^{-2} \, \text{N/m} \), we can follow these steps:

### Step 1: Understand the Concept
A soap bubble has two surfaces (inner and outer). The work done in forming the bubble is related to the change in surface area and the surface tension of the liquid.

### Step 2: Formula for Work Done
The work done \( W \) in forming a soap bubble can be calculated using the formula:
\[ W = T \times \Delta A \]
where \( T \) is the surface tension and \( \Delta A \) is the change in surface area.

### Step 3: Calculate the Change in Surface Area
For a soap bubble of radius \( r \), the surface area \( A \) is given by:
\[ A = 4\pi r^2 \]
Since there are two surfaces, the total surface area for the bubble is:
\[ \Delta A = 2 \times A = 2 \times 4\pi r^2 = 8\pi r^2 \]

### Step 4: Substitute the Values
Now, substitute \( r = 1 \, \text{cm} = 0.01 \, \text{m} \):
\[ \Delta A = 8\pi (0.01)^2 = 8\pi (0.0001) = 0.0008\pi \, \text{m}^2 \]

### Step 5: Calculate the Work Done
Now substitute \( T \) and \( \Delta A \) into the work done formula:
\[ W = T \times \Delta A = (3 \times 10^{-2}) \times (0.0008\pi) \]
Calculating this gives:
\[ W = 3 \times 10^{-2} \times 0.0008 \times \pi \approx 3 \times 10^{-2} \times 0.0008 \times 3.14 \approx 7.54 \times 10^{-5} \, \text{J} \]
Converting this to microjoules:
\[ W \approx 75.36 \, \mu J \]

### Final Answer
The amount of work done in forming the soap bubble is approximately \( 75.36 \, \mu J \).

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