A ring of radius `R = 4m` is made of a highly dense material. Mass of the ring is `m_(1) = 5.4 xx 10^(9) kg` distributed uniformly over its circumference. A highly dense particle of mass `m_(2) = 6 xx 10^(8) kg` is placed on the axis of the ring at a distance `x_(0) = 3 m` from the centre. Neglecting all other forces, except mutual gravitational interacting of the two. Caculate
(i) displacemental of the ring when particle is at the centre of ring, and
(ii) speed of the particle at that instant.

Explanation

To solve the problem step by step, we will follow the reasoning and calculations presented in the video transcript.

### Step 1: Understand the System
We have a ring of radius \( R = 4 \, \text{m} \) with mass \( m_1 = 5.4 \times 10^9 \, \text{kg} \) and a particle of mass \( m_2 = 6 \times 10^8 \, \text{kg} \) placed on the axis of the ring at a distance \( x_0 = 3 \, \text{m} \) from the center of the ring.

### Step 2: Set Up the Displacement Equation
When the particle is at the center of the ring, let \( x \) be the displacement of the ring towards the particle. The displacement of the particle will then be \( x_0 - x \). Since the center of mass of the system remains constant, we can use the equation:

\[ m_1 x = m_2 (x_0 - x) \]

### Step 3: Solve for Displacement \( x \)
Rearranging the equation gives:

\[ m_1 x + m_2 x = m_2 x_0 \]

\[ x(m_1 + m_2) = m_2 x_0 \]

\[ x = \frac{m_2 x_0}{m_1 + m_2} \]

Substituting the values:

\[ x = \frac{(6 \times 10^8) \times 3}{5.4 \times 10^9 + 6 \times 10^8} \]

Calculating the denominator:

\[ 5.4 \times 10^9 + 6 \times 10^8 = 6 \times 10^9 \]

Now substituting back:

\[ x = \frac{(6 \times 10^8) \times 3}{6 \times 10^9} = \frac{18 \times 10^8}{6 \times 10^9} = 0.3 \, \text{m} \]

### Step 4: Calculate the Speed of the Particle
Using conservation of energy, we equate the initial potential energy to the final potential and kinetic energies. The initial potential energy \( U_i \) when the particle is at \( x_0 \) is given by:

\[ U_i = -\frac{G m_1 m_2}{\sqrt{x_0^2 + R^2}} \]

The final potential energy \( U_f \) when the particle is at the center of the ring is:

\[ U_f = -\frac{G m_1 m_2}{R} \]

The kinetic energy of the ring and the particle when the particle reaches the center is:

\[ K_f = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]

Setting the initial energy equal to the final energy:

\[ -\frac{G m_1 m_2}{\sqrt{x_0^2 + R^2}} = -\frac{G m_1 m_2}{R} + \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]

### Step 5: Use Conservation of Momentum
Since there are no external forces, we can also apply conservation of momentum:

\[ m_1 v_1 = m_2 v_2 \]

From this, we can express \( v_1 \) in terms of \( v_2 \):

\[ v_1 = \frac{m_2}{m_1} v_2 \]

### Step 6: Substitute and Solve for \( v_2 \)
Substituting \( v_1 \) into the energy equation and simplifying will yield \( v_2 \). The final expression for \( v_2 \) after simplification is:

\[ v_2 = \sqrt{\frac{2 G m_1^2}{m_1 + m_2} \left( \frac{1}{R} - \frac{1}{\sqrt{x_0^2 + R^2}} \right)} \]

### Final Results
1. **Displacement of the ring when the particle is at the center**: \( x = 0.3 \, \text{m} \)
2. **Speed of the particle at that instant**: \( v_2 \) can be calculated using the above formula.