The work done is blowing a soap bubble of volume `V` is `W`. The work done in blowing a soap bubble of volume `2V` is

Options

A. `W`
B. `2^((2)/(3))W`
C. `3^((2)/(3))W`
D. `2W`

Explanation

To find the work done in blowing a soap bubble of volume \(2V\), we can follow these steps:

### Step 1: Understand the relationship between work done and surface area
The work done \(W\) in blowing a soap bubble is related to the change in surface area and the surface tension \(T\) of the soap solution. The formula for the work done is given by:
\[ W = T \times \Delta A \]
where \(\Delta A\) is the change in surface area.

### Step 2: Calculate the surface area for a bubble of volume \(V\)
For a soap bubble of volume \(V\), the volume is related to the radius \(r\) of the bubble by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
From this, we can express the radius \(r\) in terms of volume \(V\):
\[ r = \left(\frac{3V}{4\pi}\right)^{1/3} \]

### Step 3: Calculate the surface area for the bubble of volume \(V\)
The surface area \(A\) of a soap bubble is given by:
\[ A = 4\pi r^2 \]
Substituting the expression for \(r\):
\[ A = 4\pi \left(\left(\frac{3V}{4\pi}\right)^{1/3}\right)^2 = 4\pi \left(\frac{3V}{4\pi}\right)^{2/3} \]

### Step 4: Calculate the work done for volume \(V\)
The work done \(W\) for volume \(V\) can be expressed as:
\[ W = T \times 4\pi \left(\frac{3V}{4\pi}\right)^{2/3} \]

### Step 5: Calculate the work done for volume \(2V\)
Now, for volume \(2V\), we can follow the same steps:
1. The new radius \(r'\) for volume \(2V\) is:
\[ r' = \left(\frac{3(2V)}{4\pi}\right)^{1/3} = 2^{1/3} \left(\frac{3V}{4\pi}\right)^{1/3} \]
2. The new surface area \(A'\) is:
\[ A' = 4\pi (r')^2 = 4\pi \left(2^{1/3} \left(\frac{3V}{4\pi}\right)^{1/3}\right)^2 = 4\pi \cdot 2^{2/3} \left(\frac{3V}{4\pi}\right)^{2/3} \]
3. The work done \(W'\) for volume \(2V\) is:
\[ W' = T \times A' = T \times 4\pi \cdot 2^{2/3} \left(\frac{3V}{4\pi}\right)^{2/3} \]

### Step 6: Relate \(W'\) to \(W\)
From the previous steps, we can see that the work done is proportional to the surface area, which is proportional to \(V^{2/3}\). Therefore, we can write:
\[ \frac{W'}{W} = \left(\frac{2V}{V}\right)^{2/3} = 2^{2/3} \]
Thus, the work done for volume \(2V\) is:
\[ W' = W \cdot 2^{2/3} \]

### Final Answer
The work done in blowing a soap bubble of volume \(2V\) is:
\[ W' = 2^{2/3} W \]