A point `P` lies on the axis of a fixed ring of mass `M` and radius `a`, at a distance `a` from its centre `C`. A small particle starts from `P` and reaches `C` under gravitational attraction only. Its speed at `C` will be.

Options

A. zero
B. `sqrt((2GM)/a)`
C. `sqrt((2GM)/a(sqrt(2)-1))`
D. `sqrt((2GM)/a(1-1/(sqrt(2))))`

Explanation

To solve the problem, we will use the principle of conservation of mechanical energy. The total mechanical energy (potential energy + kinetic energy) of the particle remains constant as it moves under the influence of gravitational attraction.

### Step-by-Step Solution:

1. **Identify the Initial and Final Points**:
- Initial Point (P): The particle starts from point P, which is at a distance `a` from the center `C` of the ring along the axis.
- Final Point (C): The particle reaches the center `C` of the ring.

2. **Write the Conservation of Energy Equation**:
- The total mechanical energy at point P (initial) is equal to the total mechanical energy at point C (final).
\[ \text{Initial Potential Energy} + \text{Initial Kinetic Energy} = \text{Final Potential Energy} + \text{Final Kinetic Energy} \]

3. **Calculate Initial Potential Energy**:
- The gravitational potential energy \( U \) of the particle of mass \( m \) at point P due to the ring of mass \( M \) is given by:
\[ U_P = -\frac{GMm}{\sqrt{a^2 + a^2}} = -\frac{GMm}{\sqrt{2a^2}} = -\frac{GMm}{a\sqrt{2}} \]
- The initial kinetic energy \( K_P \) is zero since the particle starts from rest:
\[ K_P = 0 \]

4. **Calculate Final Potential Energy**:
- At point C (the center of the ring), the potential energy \( U_C \) is:
\[ U_C = -\frac{GMm}{a} \]
- The final kinetic energy \( K_C \) is given by:
\[ K_C = \frac{1}{2}mv^2 \]

5. **Set Up the Equation**:
- Using the conservation of energy:
\[ U_P + K_P = U_C + K_C \]
Substituting the values:
\[ -\frac{GMm}{a\sqrt{2}} + 0 = -\frac{GMm}{a} + \frac{1}{2}mv^2 \]

6. **Simplify the Equation**:
- Cancel \( m \) from all terms (assuming \( m \neq 0 \)):
\[ -\frac{GM}{a\sqrt{2}} = -\frac{GM}{a} + \frac{1}{2}v^2 \]
- Rearranging gives:
\[ \frac{1}{2}v^2 = \frac{GM}{a} - \frac{GM}{a\sqrt{2}} \]

7. **Factor Out Common Terms**:
- Factor out \( \frac{GM}{a} \):
\[ \frac{1}{2}v^2 = \frac{GM}{a}\left(1 - \frac{1}{\sqrt{2}}\right) \]

8. **Solve for \( v \)**:
- Multiply both sides by 2:
\[ v^2 = \frac{2GM}{a}\left(1 - \frac{1}{\sqrt{2}}\right) \]
- Taking the square root:
\[ v = \sqrt{\frac{2GM}{a}\left(1 - \frac{1}{\sqrt{2}}\right)} \]

### Final Answer:
The speed of the particle at point C will be:
\[ v = \sqrt{\frac{2GM}{a}\left(1 - \frac{1}{\sqrt{2}}\right)} \]