A ring of radius `R =8m` is made of a highly dense-material Mass of the ring is `m_(R) = 2.7 xx 10^(9)kg` distributed uniformly over its circumference. A particle of mass (dense) `m_(p) = 3 xx 10^(8)kg` is palced on the axis of the ring speed (in cm/sec) of the particle at the instant when it passes through centre of ring .

Answer

To solve the problem, we will use the principle of conservation of energy. The gravitational potential energy of the particle when it is at a distance from the ring will be converted into kinetic energy as it passes through the center of the ring.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Radius of the ring, \( R = 8 \, \text{m} \)
- Mass of the ring, \( m_R = 2.7 \times 10^9 \, \text{kg} \)
- Mass of the particle, \( m_p = 3 \times 10^8 \, \text{kg} \)
- Initial distance of the particle from the center of the ring, \( x = 6 \, \text{m} \)

2. **Calculate the Gravitational Potential Energy (U) at Distance x:**
The gravitational potential energy of the particle at a distance \( d \) from the center of the ring is given by:
\[ U = -\frac{G m_R m_p}{d} \]
where \( d = \sqrt{x^2 + R^2} \).

Substituting the values:
\[ d = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, \text{m} \]
Thus, the potential energy at this distance is:
\[ U = -\frac{(6.67 \times 10^{-11}) (2.7 \times 10^9) (3 \times 10^8)}{10} \]

3. **Calculate the Gravitational Potential Energy at the Center of the Ring (U_c):**
At the center of the ring, the distance \( d \) becomes \( R \):
\[ U_c = -\frac{G m_R m_p}{R} \]
Thus,
\[ U_c = -\frac{(6.67 \times 10^{-11}) (2.7 \times 10^9) (3 \times 10^8)}{8} \]

4. **Use Conservation of Energy:**
The change in potential energy equals the kinetic energy gained by the particle:
\[ K.E. = U_c - U \]
Since \( K.E. = \frac{1}{2} m_p v^2 \), we can set up the equation:
\[ \frac{1}{2} m_p v^2 = U_c - U \]

5. **Solve for v (speed of the particle):**
Rearranging the equation gives:
\[ v^2 = \frac{2 (U_c - U)}{m_p} \]
Taking the square root gives:
\[ v = \sqrt{\frac{2 (U_c - U)}{m_p}} \]

6. **Substituting the Values:**
Calculate \( U \) and \( U_c \) using the values substituted in steps 2 and 3, and then find \( v \).

7. **Convert the Speed to cm/sec:**
Since the final answer is required in cm/sec, multiply the result by 100.

### Final Calculation:
1. Calculate \( U \) and \( U_c \) using the gravitational constant \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \).
2. Substitute into the equation for \( v \) and compute the final answer.