A point `P(R sqrt(3),0,0)` lies on the axis of a ring of mass M and radius R. The ring is located in yx-plane with its centre at origin O. A small particle of mass `m` starts from P and reaches O under gravitational attraction only. Its speed at O will be

Options

A. `sqrt((GM)/(R))`
B. `sqrt((Gm)/(R))`
C. `sqrt((GM)/(2R))`
D. `sqrt((2m)/(sqrt(2)R))`

Explanation

To find the speed of the particle of mass `m` when it reaches the origin `O` from point `P(R√3, 0, 0)` under the influence of gravitational attraction from the ring, we can use the principle of conservation of energy.

### Step-by-Step Solution

1. **Identify the Initial and Final Positions**:
- The point `P` is located at coordinates `(R√3, 0, 0)`.
- The origin `O` is at `(0, 0, 0)`.

2. **Calculate the Distance from the Ring to Point P**:
- The distance from point `P` to the center of the ring (origin `O`) is `R√3`.

3. **Determine the Gravitational Potential Energy**:
- The gravitational potential `V` due to a ring of mass `M` at a distance `z` along its axis is given by:
\[ V = -\frac{GM}{\sqrt{R^2 + z^2}} \]
- At point `P`, the distance from the ring is `z = R√3`, so the potential energy at `P` is:
\[ V_P = -\frac{GM}{\sqrt{R^2 + (R\sqrt{3})^2}} = -\frac{GM}{\sqrt{R^2 + 3R^2}} = -\frac{GM}{2R} \]

4. **Calculate the Potential Energy at the Origin**:
- At the origin `O`, the distance from the ring is `z = 0`, so the potential energy at `O` is:
\[ V_O = -\frac{GM}{\sqrt{R^2 + 0^2}} = -\frac{GM}{R} \]

5. **Apply Conservation of Energy**:
- The total mechanical energy is conserved, which means the increase in kinetic energy equals the decrease in potential energy:
\[ \Delta KE = -\Delta PE \]
- The change in potential energy is:
\[ \Delta PE = V_O - V_P = \left(-\frac{GM}{R}\right) - \left(-\frac{GM}{2R}\right) = -\frac{GM}{R} + \frac{GM}{2R} = -\frac{GM}{2R} \]
- Thus, the increase in kinetic energy is:
\[ \Delta KE = \frac{GM}{2R} \]

6. **Relate Kinetic Energy to Speed**:
- The kinetic energy at the origin is given by:
\[ KE = \frac{1}{2} mv^2 \]
- Setting this equal to the increase in kinetic energy:
\[ \frac{1}{2} mv^2 = \frac{GM}{2R} \]
- Solving for `v`:
\[ mv^2 = \frac{GM}{R} \]
\[ v^2 = \frac{GM}{mR} \]
\[ v = \sqrt{\frac{GM}{mR}} \]

### Final Answer:
The speed of the particle at the origin `O` will be:
\[ v = \sqrt{\frac{GM}{mR}} \]