Two identical thin rings each of radius R are coaxially placed at a distance R. If the rings have a uniform mass distribution and each has masses `2m` and `4m` respectively, then the work done in moving a mass `m` from centre of one ring to that of the other is

Options

A. zero
B. `(sqrt(2)Gm^(2))/(R)(1-sqrt(2))`
C. `(Gm^(2))/(sqrt(2)R)(sqrt(2)-1)`
D. `(Gm^(2))/(sqrt(2)R)`

Explanation

To find the work done in moving a mass \( m \) from the center of one ring to the center of the other, we will calculate the gravitational potential energy at both positions and find the difference.

### Step-by-Step Solution:

1. **Identify the Configuration**:
- We have two identical thin rings, each of radius \( R \).
- The first ring has a mass \( m_1 = 2m \) and the second ring has a mass \( m_2 = 4m \).
- The distance between the centers of the two rings is \( R \).

2. **Gravitational Potential due to a Ring**:
- The gravitational potential \( U \) at a distance \( z \) from the center of a ring of mass \( M \) and radius \( R \) is given by:
\[ U = -\frac{GM}{\sqrt{R^2 + z^2}} \]
- Here, \( G \) is the gravitational constant.

3. **Calculate the Potential at the Center of the First Ring**:
- At the center of the first ring (where the mass \( m \) starts), the distance \( z = 0 \):
\[ U_1 = -\frac{G \cdot 2m}{R} \]

4. **Calculate the Potential at the Center of the Second Ring**:
- To find the potential at the center of the second ring, we need to consider the distance from the center of the first ring to the center of the second ring, which is \( R \).
- The distance from the center of the second ring to the mass \( m \) is \( R \):
\[ U_2 = -\frac{G \cdot 4m}{R} \]

5. **Calculate the Work Done**:
- The work done \( W \) in moving the mass \( m \) from the center of the first ring to the center of the second ring is given by the change in potential energy:
\[ W = U_2 - U_1 \]
- Substituting the values:
\[ W = \left(-\frac{G \cdot 4m}{R}\right) - \left(-\frac{G \cdot 2m}{R}\right) \]
- Simplifying:
\[ W = -\frac{G \cdot 4m}{R} + \frac{G \cdot 2m}{R} = -\frac{G \cdot 2m}{R} \]

6. **Final Result**:
- The work done in moving the mass \( m \) from the center of one ring to the center of the other is:
\[ W = -\frac{2Gm}{R} \]