Calculate the work done in blowing a soap bubble of radius 0.1m surface tension being 0.03 `Nm^(-1)`. What additional work will be performed in further blowing it so that its radius is doubled ?

Explanation

To solve the problem of calculating the work done in blowing a soap bubble of radius 0.1 m with a surface tension of 0.03 Nm^(-1), we will follow these steps:

### Step 1: Understand the Formula for Work Done
The work done (W) in blowing a soap bubble is given by the formula:
\[ W = 2 \times \text{Surface Tension} \times \text{Area} \]
For a soap bubble, the area (A) is the surface area of the bubble, which can be calculated using the formula for the surface area of a sphere:
\[ A = 4 \pi r^2 \]

### Step 2: Calculate the Surface Area of the Bubble
Given the radius \( r = 0.1 \) m, we can calculate the surface area:
\[ A = 4 \pi (0.1)^2 \]
Calculating \( (0.1)^2 \):
\[ (0.1)^2 = 0.01 \]
Now substituting back:
\[ A = 4 \pi (0.01) = 0.04 \pi \]

### Step 3: Substitute Values into the Work Done Formula
Now substituting the area into the work done formula:
\[ W = 2 \times \sigma \times A = 2 \times 0.03 \times (0.04 \pi) \]
Calculating \( 2 \times 0.03 = 0.06 \):
\[ W = 0.06 \times (0.04 \pi) \]

### Step 4: Calculate the Value of Work Done
Using \( \pi \approx 3.14 \):
\[ W = 0.06 \times (0.04 \times 3.14) = 0.06 \times 0.1256 = 0.007536 \]
Thus, the work done is:
\[ W \approx 7.54 \times 10^{-3} \text{ Joules} \]

### Step 5: Calculate Additional Work for Doubling the Radius
Now, we need to find the additional work done when the radius is doubled. The new radius \( r_2 = 2 \times r_1 = 0.2 \) m.

The additional work done is given by:
\[ W_{\text{additional}} = 8 \pi \sigma (r_2^2 - r_1^2) \]
Calculating \( r_2^2 \) and \( r_1^2 \):
\[ r_2^2 = (0.2)^2 = 0.04, \quad r_1^2 = (0.1)^2 = 0.01 \]
Thus,
\[ r_2^2 - r_1^2 = 0.04 - 0.01 = 0.03 \]

### Step 6: Substitute Values into the Additional Work Formula
Now substituting the values into the additional work formula:
\[ W_{\text{additional}} = 8 \pi \sigma (0.03) \]
Calculating:
\[ W_{\text{additional}} = 8 \times 3.14 \times 0.03 \times 0.03 \]
Calculating \( 8 \times 3.14 = 25.12 \):
\[ W_{\text{additional}} = 25.12 \times 0.03 = 0.7536 \times 0.03 = 0.02256 \]
Thus, the additional work done is:
\[ W_{\text{additional}} \approx 2.26 \times 10^{-2} \text{ Joules} \]

### Final Answers
1. The work done in blowing the soap bubble is approximately \( 7.54 \times 10^{-3} \) Joules.
2. The additional work done in further blowing it to double the radius is approximately \( 2.26 \times 10^{-2} \) Joules.