C

To determine the speed of the particle at the moment it passes through the center of the ring, we can use energy conservation principles.

Step 1: Calculate the gravitational potential energy (U) of the particle when it is at a distance of x_0 = 6 m from the center of the ring. The gravitational potential energy is given by:
$$ U = - \frac{G m_R m_p}{r} $$
where \( G \) is the gravitational constant (approximately \( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)), \( m_R \) is the mass of the ring, \( m_p \) is the mass of the particle, and \( r \) is the distance from the center of the ring to the particle (which is 6 m).

Step 2: Calculate U at x_0 = 6 m.
\( m_R = 2.7 \times 10^9 \, \text{kg} \)
\( m_p = 3 \times 10^8 \, \text{kg} \)

We can substitute these values into our equation:
$$ U = - \frac{(6.674 \times 10^{-11}) (2.7 \times 10^9)(3 \times 10^8)}{6} $$
Step 3: Calculate U to find the value.
$$ U = - \frac{(6.674 \times 10^{-11}) (2.7 \times 10^9)(3 \times 10^8)}{6} = - \frac{5.3959 \times 10^7}{6} \approx -8.993 \times 10^6 \, \text{J} $$

Step 4: As the particle moves towards the center of the ring, it will convert this potential energy into kinetic energy (K). At the center, all potential energy will become kinetic energy, which is given by:
$$ K = \frac{1}{2} m_p v^2 $$
Where \( v \) is the speed of the particle. Setting the kinetic energy equal to the magnitude of potential energy gives:
$$ \frac{1}{2} m_p v^2 = |U| \approx 8.993 \times 10^6 $$

Step 5: Rearranging for v, we have:
$$ v^2 = \frac{2 |U|}{m_p} $$
$$ v = \sqrt{\frac{2 \cdot 8.993 \times 10^6}{3 \times 10^8}} \approx \sqrt{\frac{17986 \times 10^6}{3 \times 10^8}} \approx \sqrt{59.95} \approx 7.74 \, \text{m/s} $$

Step 6: Convert the speed to cm/sec:
$$ v \approx 7.74 \times 100 = 774 \, \text{cm/s} $$
Therefore, the speed of the particle at the moment it passes through the center of the ring is approximately 774 cm/sec.