Two small conducting spheres each of mass `9xx10^(-4)kg` are suspended from the same point by non conducting strings of length `100cm `,They are given equal and similar charges until the strings are equally inclined at `45^(@)` each to the vertical,The charge on each spehre is ..........

Options

A. `1.4xx10^(-6)`
B. `1.6xx10^(-6)`
C. `2xx10^(-6)`
D. `1.96xx10^(-6)`

Explanation

To solve the problem, we need to find the charge on each of the two small conducting spheres that are suspended and inclined at 45 degrees from the vertical.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two small conducting spheres, each with a mass \( m = 9 \times 10^{-4} \, \text{kg} \).
- They are suspended from the same point by non-conducting strings of length \( L = 100 \, \text{cm} = 1 \, \text{m} \).
- The spheres are given equal and similar charges, and they are inclined at \( 45^\circ \) to the vertical.

2. **Free Body Diagram**:
- Each sphere experiences three forces:
- The gravitational force \( mg \) acting downwards.
- The tension \( T \) in the string acting along the string.
- The electrostatic force \( F \) due to the repulsion between the two charged spheres.

3. **Components of Forces**:
- The tension \( T \) can be resolved into two components:
- Vertical component: \( T \cos(45^\circ) \)
- Horizontal component: \( T \sin(45^\circ) \)

4. **Equilibrium in the Vertical Direction**:
- In the vertical direction, the forces must balance:
\[ T \cos(45^\circ) = mg \]
- Substituting \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ T \cdot \frac{1}{\sqrt{2}} = mg \]
- Therefore,
\[ T = mg \sqrt{2} \]

5. **Equilibrium in the Horizontal Direction**:
- In the horizontal direction, the tension's horizontal component equals the electrostatic force:
\[ T \sin(45^\circ) = F \]
- Substituting \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ T \cdot \frac{1}{\sqrt{2}} = F \]

6. **Distance Between the Spheres**:
- The distance \( R \) between the two spheres can be calculated using geometry:
\[ R = 2L \sin(45^\circ) = 2L \cdot \frac{1}{\sqrt{2}} = \sqrt{2}L \]
- Substituting \( L = 1 \, \text{m} \):
\[ R = \sqrt{2} \, \text{m} \]

7. **Electrostatic Force**:
- The electrostatic force \( F \) between the two charges \( q \) is given by Coulomb's law:
\[ F = \frac{k q^2}{R^2} \]
- Substituting \( R = \sqrt{2} \):
\[ F = \frac{k q^2}{(\sqrt{2})^2} = \frac{k q^2}{2} \]

8. **Setting the Forces Equal**:
- From the horizontal equilibrium, we can set the two expressions for \( F \) equal:
\[ T \cdot \frac{1}{\sqrt{2}} = \frac{k q^2}{2} \]
- Substituting \( T = mg \sqrt{2} \):
\[ mg \sqrt{2} \cdot \frac{1}{\sqrt{2}} = \frac{k q^2}{2} \]
- Simplifying gives:
\[ mg = \frac{k q^2}{2} \]

9. **Solving for Charge \( q \)**:
- Rearranging gives:
\[ q^2 = \frac{2mg}{k} \]
- Substituting \( m = 9 \times 10^{-4} \, \text{kg} \), \( g = 10 \, \text{m/s}^2 \), and \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \):
\[ q^2 = \frac{2 \cdot (9 \times 10^{-4}) \cdot 10}{9 \times 10^9} \]
- Simplifying gives:
\[ q^2 = \frac{18 \times 10^{-4}}{9 \times 10^9} = 2 \times 10^{-13} \]
- Taking the square root:
\[ q = \sqrt{2 \times 10^{-13}} = \sqrt{2} \times 10^{-7} \, \text{C} \approx 1.414 \times 10^{-7} \, \text{C} \]

### Final Answer:
The charge on each sphere is approximately \( 1.414 \times 10^{-6} \, \text{C} \) or \( 1.4 \times 10^{-6} \, \text{C} \).