Workdone to blow a bubble of volume V is W. The workdone is blowing a bubble of volume 2 V will be

Options

A. `(2)^(1//3)` W
B. 2 W
C. `(4)^(1//3)` W
D. 4 W

Explanation

To solve the problem of finding the work done to blow a bubble of volume \(2V\) when the work done to blow a bubble of volume \(V\) is \(W\), we can follow these steps:

### Step 1: Understand the relationship between work done and surface energy
The work done in blowing a bubble is related to the change in surface energy. The change in surface energy (\( \Delta E \)) is given by:
\[ \Delta E = \text{Surface Tension} \times \text{Change in Area} \]

### Step 2: Relate the area of a spherical bubble to its radius
The surface area \(A\) of a spherical bubble is given by:
\[ A = 4\pi r^2 \]
where \(r\) is the radius of the bubble.

### Step 3: Express work done in terms of radius
Since work done is proportional to the change in surface area, we can say:
\[ W \propto A \propto r^2 \]
Thus, work done is proportional to the square of the radius of the bubble.

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

### Step 5: Substitute radius in the work done equation
Since work done is proportional to \(r^2\), we can substitute \(r\):
\[ W \propto \left(\left(\frac{3V}{4\pi}\right)^{1/3}\right)^2 \]
This simplifies to:
\[ W \propto \left(\frac{3V}{4\pi}\right)^{2/3} \]

### Step 6: Relate work done to volume
Thus, we can conclude that:
\[ W \propto V^{2/3} \]
This means that if \(W\) is the work done for volume \(V\), then for volume \(2V\):
\[ W' \propto (2V)^{2/3} \]

### Step 7: Calculate the work done for volume \(2V\)
Now, we compute \(W'\):
\[ W' = k(2V)^{2/3} = k \cdot 2^{2/3} \cdot V^{2/3} \]
Since \(kV^{2/3} = W\), we can express \(W'\) as:
\[ W' = W \cdot 2^{2/3} \]

### Step 8: Final expression
Thus, the work done to blow a bubble of volume \(2V\) is:
\[ W' = W \cdot 2^{2/3} \]

### Conclusion
The work done to blow a bubble of volume \(2V\) is \(W \cdot 2^{2/3}\).