The work done in 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. 2W
C. `sqrt2W`
D. `4^(1//3)W`

Answer

To solve the problem of finding the work done in blowing a soap bubble of volume \(2V\) given that the work done for volume \(V\) is \(W\), we can follow these steps:

### Step-by-Step Solution:

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 (\(\Delta A\)) and the surface tension (\(T\)) of the soap solution. The formula can be expressed as:
\[ W = T \cdot \Delta A \]

2. **Determine the surface area of a sphere**:
For a soap bubble, which can be approximated as a sphere, the surface area \(A\) is given by:
\[ A = 4\pi r^2 \]
where \(r\) is the radius of the bubble.

3. **Relate volume to radius**:
The volume \(V\) of a sphere is given by:
\[ V = \frac{4}{3}\pi r^3 \]
From this equation, we can express the radius in terms of volume:
\[ r = \left(\frac{3V}{4\pi}\right)^{1/3} \]

4. **Find the relationship between work done and volume**:
Since the work done is proportional to the change in surface area, and the surface area is proportional to \(r^2\), we can express the work done as:
\[ W \propto r^2 \]
Substituting \(r\) in terms of \(V\):
\[ W \propto \left(\left(\frac{3V}{4\pi}\right)^{1/3}\right)^2 = \left(\frac{3V}{4\pi}\right)^{2/3} \]
Thus, we can conclude:
\[ W \propto V^{2/3} \]

5. **Calculate work done for volume \(2V\)**:
Now, let’s find the work done \(W_2\) for the volume \(2V\):
\[ W_2 \propto (2V)^{2/3} = 2^{2/3} \cdot V^{2/3} \]
Since we know \(W \propto V^{2/3}\), we can write:
\[ W_2 = 2^{2/3} \cdot W \]

6. **Final expression**:
Therefore, the work done in blowing a soap bubble of volume \(2V\) is:
\[ W_2 = 2^{2/3} \cdot W \]

### Conclusion:
The work done in blowing a soap bubble of volume \(2V\) is \(2^{2/3} \cdot W\).