The surface tension of a soap solution is `2xx10^(-2) N//m`. To blow a bubble of radius 1cm, the work done is

Options

A. `4pixx10^(-6) J`
B. `8pixx10^(-6) J`
C. `12 pixx10^(-6) J`
D. `16 pi xx10^(-6) J`

Answer

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

### Step 1: Understand the Concept of Surface Tension
Surface tension (\(T\)) is defined as the force per unit length acting along the surface of a liquid. It can also be understood in terms of work done to increase the surface area of the liquid.

### Step 2: Formula for Work Done
The work done (\(W\)) to increase the surface area of a liquid against surface tension is given by:
\[ W = T \times \Delta A \]
where \(\Delta A\) is the change in surface area.

### Step 3: Calculate the Surface Area of the Bubble
For a bubble, which has two surfaces (inside and outside), the total surface area (\(A\)) of a bubble with radius \(r\) is:
\[ A = 4\pi r^2 \]
Since there are two surfaces, the effective area for the work done is:
\[ \Delta A = 2 \times 4\pi r^2 = 8\pi r^2 \]

### Step 4: Substitute the Radius
Given that the radius \(r = 1 \, \text{cm} = 0.01 \, \text{m}\), we can substitute this value into the area formula:
\[ \Delta A = 8\pi (0.01)^2 = 8\pi (0.0001) = 8\pi \times 10^{-4} \, \text{m}^2 \]

### Step 5: Substitute Surface Tension Value
Now substitute the surface tension value \(T = 2 \times 10^{-2} \, \text{N/m}\) into the work done formula:
\[ W = T \times \Delta A = (2 \times 10^{-2}) \times (8\pi \times 10^{-4}) \]

### Step 6: Calculate Work Done
Now calculate the work done:
\[ W = 16\pi \times 10^{-6} \, \text{J} \]

### Final Result
Thus, the work done to blow a bubble of radius 1 cm is:
\[ W = 16\pi \times 10^{-6} \, \text{J} \approx 5.0265 \times 10^{-5} \, \text{J} \]