A particle of mass `1kg` is placed at a distance of `4m` from the centre and on the axis of a uniform ring mass `5kg` and radius `3m`. The work done to increase the distance of the particle from `4m` to `3sqrt(3)`m is

Explanation

To solve the problem, we need to calculate the work done to move a particle of mass \(1 \, \text{kg}\) from a distance of \(4 \, \text{m}\) to \(3\sqrt{3} \, \text{m}\) from the center of a uniform ring of mass \(5 \, \text{kg}\) and radius \(3 \, \text{m}\).

### Step-by-Step Solution:

1. **Identify the Initial and Final Distances**:
- Initial distance from the center of the ring, \(x_i = 4 \, \text{m}\)
- Final distance from the center of the ring, \(x_f = 3\sqrt{3} \, \text{m}\)

2. **Calculate the Potential Energy**:
The gravitational potential \(V\) at a distance \(x\) from the center of a ring of mass \(M\) is given by the formula:
\[ V = -\frac{GM}{\sqrt{R^2 + x^2}} \]
where \(R\) is the radius of the ring.

3. **Calculate Initial Potential Energy**:
For the initial position \(x_i = 4 \, \text{m}\):
\[ V_i = -\frac{G \cdot 5}{\sqrt{3^2 + 4^2}} = -\frac{5G}{\sqrt{9 + 16}} = -\frac{5G}{5} = -G \]

4. **Calculate Final Potential Energy**:
For the final position \(x_f = 3\sqrt{3} \, \text{m}\):
\[ V_f = -\frac{G \cdot 5}{\sqrt{3^2 + (3\sqrt{3})^2}} = -\frac{5G}{\sqrt{9 + 27}} = -\frac{5G}{\sqrt{36}} = -\frac{5G}{6} \]

5. **Calculate Work Done**:
The work done \(W\) is equal to the change in potential energy:
\[ W = V_f - V_i = \left(-\frac{5G}{6}\right) - \left(-G\right) \]
Simplifying this:
\[ W = -\frac{5G}{6} + G = -\frac{5G}{6} + \frac{6G}{6} = \frac{G}{6} \]

### Final Answer:
The work done to increase the distance of the particle from \(4 \, \text{m}\) to \(3\sqrt{3} \, \text{m}\) is:
\[ W = \frac{G}{6} \, \text{J} \]