There are two concentric and coplanar non-conducting rings of radii `R` and `4R`. The charge is distributed uniormly on both rings. The charge on smaller ring is `q` and charge on large ring is `-8q`. A particle of mass `10g` and charge `-q` is projected along the axis from infinity. What is the minimum speed of charge at infinity to reach the common centre of rings (take `(Kq^(2))/R=(2sqrt(5))/3J`)

Options

A. a. `5m//s`
B. b. `10m//s`
C. c. `20m//s`
D. d. `30m//s`

Explanation

To solve the problem, we need to find the minimum speed of a charge projected from infinity to reach the common center of two concentric non-conducting rings. We will use the principles of electrostatics and energy conservation.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two rings: a smaller ring with radius \( R \) and charge \( q \), and a larger ring with radius \( 4R \) and charge \( -8q \).
- A particle with mass \( 10 \, \text{g} \) (or \( 0.01 \, \text{kg} \)) and charge \( -q \) is projected from infinity.

2. **Finding the Electric Field**:
- The electric field \( E \) due to a ring of charge at a point along its axis is given by:
\[ E = \frac{kQx}{(R^2 + x^2)^{3/2}} \]
- For the smaller ring (charge \( q \)):
\[ E_1 = \frac{kqx}{(R^2 + x^2)^{3/2}} \]
- For the larger ring (charge \( -8q \)):
\[ E_2 = -\frac{k(8q)x}{(16R^2 + x^2)^{3/2}} \]

3. **Setting the Electric Fields Equal**:
- We need to find the point \( x \) where the net electric field is zero:
\[ E_1 + E_2 = 0 \]
\[ \frac{kqx}{(R^2 + x^2)^{3/2}} = \frac{k(8q)x}{(16R^2 + x^2)^{3/2}} \]
- Canceling \( kq \) and simplifying gives:
\[ \frac{x}{(R^2 + x^2)^{3/2}} = \frac{8x}{(16R^2 + x^2)^{3/2}} \]

4. **Solving for \( x \)**:
- Cross-multiplying and simplifying leads to:
\[ 8(16R^2 + x^2)^{3/2} = (R^2 + x^2)^{3/2} \]
- After some algebra, we find that \( x = 2R \).

5. **Using Conservation of Energy**:
- The total mechanical energy at infinity (where potential energy is zero) must equal the total mechanical energy at the point \( x = 2R \):
\[ \frac{1}{2} mv^2 = U \]
- The potential energy \( U \) at \( x = 2R \) is given by:
\[ U = k \left( \frac{q(-q)}{\sqrt{R^2 + (2R)^2}} + \frac{-8q(-q)}{\sqrt{(4R)^2 + (2R)^2}} \right) \]
- Plugging in values:
\[ U = -\frac{kq^2}{\sqrt{5R^2}} + \frac{8kq^2}{\sqrt{20R^2}} = -\frac{kq^2}{R\sqrt{5}} + \frac{4kq^2}{R\sqrt{5}} = \frac{3kq^2}{R\sqrt{5}} \]

6. **Substituting Known Values**:
- We know \( \frac{kq^2}{R} = \frac{2\sqrt{5}}{3} \):
\[ U = \frac{3 \cdot \frac{2\sqrt{5}}{3}}{\sqrt{5}} = 2 \]
- Therefore, equating energies:
\[ \frac{1}{2} mv^2 = 2 \implies v^2 = \frac{4}{m} \]

7. **Calculating Minimum Speed**:
- Substituting \( m = 0.01 \, \text{kg} \):
\[ v^2 = \frac{4}{0.01} = 400 \implies v = 20 \, \text{m/s} \]

### Final Answer:
The minimum speed of the charge at infinity to reach the common center of the rings is \( 20 \, \text{m/s} \).