We have a thin spherical of radius `R`. Half of the sphere carrying surface change density `sigma_(1)` and rest half carrying `sigma_(2)`. Magnitude of force of electromagnetic interaction between hemispherical parts carrying surface charge density `sigma_(1)` and rest of the hemispherical portion is `f_(0)=alphaxx10^(4)` Newton. Here `alpha` is an integer. Value of `alpha` will be.
Given `sigma_(1)sqrt(epsilon_(0))xx10^(2)cm//m^(2),sigma_(2)=4sigma_(1),R=sqrt(3/(pi)m)`

Options

A. `3`
B. `4`
C. `5`
D. `6`

Explanation

To solve the problem, we need to calculate the electromagnetic interaction force between two hemispherical parts of a sphere with different surface charge densities. Let's break down the solution step by step.

### Step 1: Understand the given parameters
We have:
- Radius of the sphere, \( R = \sqrt{\frac{3}{\pi}} \) m
- Surface charge density of the first hemisphere, \( \sigma_1 = \sqrt{\epsilon_0} \times 10^2 \) cm/m²
- Surface charge density of the second hemisphere, \( \sigma_2 = 4\sigma_1 \)

### Step 2: Convert units
Convert \( \sigma_1 \) from cm/m² to m²:
\[ \sigma_1 = \sqrt{\epsilon_0} \times 10^2 \text{ cm/m}^2 = \sqrt{\epsilon_0} \times 10^2 \times 10^{-4} \text{ m/m}^2 = \sqrt{\epsilon_0} \times 10^{-2} \text{ m/m}^2 \]

### Step 3: Calculate the area of one hemisphere
The area \( A \) of a hemisphere is given by:
\[ A = 2\pi R^2 \]
Substituting \( R = \sqrt{\frac{3}{\pi}} \):
\[ A = 2\pi \left(\sqrt{\frac{3}{\pi}}\right)^2 = 2\pi \cdot \frac{3}{\pi} = 6 \text{ m}^2 \]

### Step 4: Calculate the force of interaction
The force \( F_0 \) between the two hemispheres can be calculated using the formula:
\[ F_0 = \frac{1}{2} \cdot \frac{A}{\epsilon_0} \cdot \sigma_1 \cdot \sigma_2 \]
Substituting \( \sigma_2 = 4\sigma_1 \):
\[ F_0 = \frac{1}{2} \cdot \frac{6}{\epsilon_0} \cdot \sigma_1 \cdot (4\sigma_1) = \frac{1}{2} \cdot \frac{6}{\epsilon_0} \cdot 4\sigma_1^2 \]
\[ F_0 = \frac{12\sigma_1^2}{2\epsilon_0} = \frac{6\sigma_1^2}{\epsilon_0} \]

### Step 5: Substitute \( \sigma_1 \) into the force equation
Substituting \( \sigma_1 = \sqrt{\epsilon_0} \times 10^{-2} \):
\[ F_0 = \frac{6(\sqrt{\epsilon_0} \times 10^{-2})^2}{\epsilon_0} = \frac{6 \cdot \epsilon_0 \cdot 10^{-4}}{\epsilon_0} = 6 \cdot 10^{-4} \text{ N} \]

### Step 6: Express the force in terms of \( \alpha \)
Given that \( F_0 = \alpha \times 10^4 \text{ N} \):
\[ 6 \cdot 10^{-4} = \alpha \times 10^4 \]
To find \( \alpha \):
\[ \alpha = 6 \]

### Conclusion
Thus, the value of \( \alpha \) is \( 6 \).