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. `5m//s`
B. `10m//s`
C. `20m//s`
D. `30m//s`

Explanation

To solve the problem, we need to determine the minimum speed of a charge projected from infinity to reach the center of two concentric rings with given charges. Here’s a step-by-step solution:

### Step 1: Understanding the System
We have two concentric rings:
- Smaller ring (radius \( R \)) with charge \( +q \)
- Larger ring (radius \( 4R \)) with charge \( -8q \)

A particle with mass \( 10 \, \text{g} \) (or \( 0.01 \, \text{kg} \)) and charge \( -q \) is projected from infinity towards the center of the rings.

### Step 2: Electric Field Calculation
The electric field \( E \) at a distance \( x \) along the axis of a ring is given by the formula:
\[ E = \frac{k \cdot Q \cdot x}{(R^2 + x^2)^{3/2}} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge on the ring, and \( R \) is the radius of the ring.

For the smaller ring:
\[ E_1 = \frac{k \cdot q \cdot x}{(R^2 + x^2)^{3/2}} \]

For the larger ring:
\[ E_2 = \frac{k \cdot (-8q) \cdot x}{(16R^2 + x^2)^{3/2}} \]

### Step 3: Finding the Point Where Electric Field is Zero
To find the point where the electric field is zero, we set \( E_1 + E_2 = 0 \):
\[ \frac{k \cdot q \cdot x}{(R^2 + x^2)^{3/2}} + \frac{-8k \cdot q \cdot x}{(16R^2 + x^2)^{3/2}} = 0 \]

This simplifies to:
\[ \frac{1}{(R^2 + x^2)^{3/2}} = \frac{8}{(16R^2 + x^2)^{3/2}} \]

Cross-multiplying and simplifying leads to:
\[ (16R^2 + x^2)^{3/2} = 8(R^2 + x^2)^{3/2} \]

### Step 4: Solving for \( x \)
After simplification, we find:
\[ x^2 = 4R^2 \implies x = 2R \]

### Step 5: Energy Conservation
At infinity, the particle has kinetic energy and no potential energy. As it approaches the center, it will have potential energy due to the electric fields of the rings and kinetic energy.

The total energy at infinity (when projected with speed \( v \)) is:
\[ \frac{1}{2} mv^2 \]

At \( x = 2R \), the potential energy \( U \) is given by the contributions from both rings:
\[ U = U_1 + U_2 \]
Where:
- \( U_1 = -\frac{kq(-q)}{\sqrt{R^2 + (2R)^2}} = -\frac{kq^2}{\sqrt{5}R} \)
- \( U_2 = -\frac{k(-8q)(-q)}{\sqrt{(4R)^2 + (2R)^2}} = -\frac{8kq^2}{\sqrt{20}R} = -\frac{4kq^2}{\sqrt{5}R} \)

Thus,
\[ U = -\frac{kq^2}{\sqrt{5}R} - \frac{4kq^2}{\sqrt{5}R} = -\frac{5kq^2}{\sqrt{5}R} = -\frac{kq^2\sqrt{5}}{R} \]

### Step 6: Setting Up the Energy Equation
Setting the total energy at infinity equal to the total energy at \( x = 2R \):
\[ \frac{1}{2} mv^2 = -\frac{kq^2\sqrt{5}}{R} \]

### Step 7: Substitute Known Values
Given \( \frac{kq^2}{R} = \frac{2\sqrt{5}}{3} \), we can substitute:
\[ \frac{1}{2} mv^2 = -\frac{2\sqrt{5}}{3} \cdot \sqrt{5} = -\frac{10}{3} \]

### Step 8: Solve for \( v \)
\[ \frac{1}{2} mv^2 = \frac{10}{3} \]
\[ mv^2 = \frac{20}{3} \]
\[ v^2 = \frac{20}{3 \cdot 0.01} = \frac{2000}{3} \]
\[ v = \sqrt{\frac{2000}{3}} \approx 20 \, \text{m/s} \]

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