Surface tension of a soap bubble is `2.0times10^(-2)Nm^(-1)`.Work done to increase the radius of soap bubble from `3.5cm` to `7cm` will be : Take `[pi=(22)/(7)]`

Options

A. `9.24 times 10^(-4)J`
B. `18.48 times 10^(-4)J`
C. `0.72 times 10^(-4)J`
D. `5.76 times 10^(-4)J`

Answer

To find the work done to increase the radius of a soap bubble from 3.5 cm to 7 cm, we can follow these steps:

### Step 1: Understand the relationship between surface tension and energy
The energy associated with a soap bubble is given by the formula:
\[ U = S \times A \]
where \( U \) is the energy, \( S \) is the surface tension, and \( A \) is the surface area of the bubble. Since a soap bubble has two surfaces (inside and outside), the effective area is doubled:
\[ U = 2S \times A = 2S \times 4\pi r^2 = 8\pi S r^2 \]

### Step 2: Calculate the initial and final energy
We need to find the change in energy (\( \Delta U \)) as the radius changes from \( r_1 = 3.5 \, \text{cm} \) to \( r_2 = 7 \, \text{cm} \).

Convert the radii from centimeters to meters:
\[ r_1 = 3.5 \, \text{cm} = 0.035 \, \text{m} \]
\[ r_2 = 7 \, \text{cm} = 0.07 \, \text{m} \]

Now, calculate the initial and final energy:
\[ U_1 = 8\pi S r_1^2 = 8 \times \frac{22}{7} \times (2.0 \times 10^{-2}) \times (0.035)^2 \]
\[ U_2 = 8\pi S r_2^2 = 8 \times \frac{22}{7} \times (2.0 \times 10^{-2}) \times (0.07)^2 \]

### Step 3: Calculate \( U_1 \) and \( U_2 \)
Calculating \( U_1 \):
\[ U_1 = 8 \times \frac{22}{7} \times (2.0 \times 10^{-2}) \times (0.035)^2 \]
\[ = 8 \times \frac{22}{7} \times (2.0 \times 10^{-2}) \times 0.001225 \]
\[ = 8 \times \frac{22}{7} \times 2.0 \times 10^{-2} \times 0.001225 \]
\[ = 8 \times \frac{22 \times 2.0 \times 0.001225}{7} \times 10^{-2} \]
\[ = 8 \times \frac{0.0541}{7} \times 10^{-2} \]
\[ = 0.0613 \times 10^{-2} \text{ J} \]

Calculating \( U_2 \):
\[ U_2 = 8 \times \frac{22}{7} \times (2.0 \times 10^{-2}) \times (0.07)^2 \]
\[ = 8 \times \frac{22}{7} \times (2.0 \times 10^{-2}) \times 0.0049 \]
\[ = 8 \times \frac{22 \times 2.0 \times 0.0049}{7} \times 10^{-2} \]
\[ = 8 \times \frac{0.2156}{7} \times 10^{-2} \]
\[ = 0.2465 \times 10^{-2} \text{ J} \]

### Step 4: Calculate the work done
Now, we find the work done (\( W \)):
\[ W = \Delta U = U_2 - U_1 \]
\[ = (0.2465 \times 10^{-2}) - (0.0613 \times 10^{-2}) \]
\[ = 0.1852 \times 10^{-2} \text{ J} \]

### Final Result
The work done to increase the radius of the soap bubble from 3.5 cm to 7 cm is approximately:
\[ W \approx 1.852 \times 10^{-3} \text{ J} \]