A particle is suspended by a light vertical inelastic string of length 1 from a fixed support. At its equilbrium position it is projected horizontally with a speed `sqrt(6 g l)`. Find the ratio of the tension in the string in its horizontal position to that in the string when the particle is vertically above the point of support.

Answer

To solve the problem, we need to find the ratio of the tension in the string when the particle is in a horizontal position (let's call this \( T_A \)) to the tension in the string when the particle is vertically above the point of suspension (let's call this \( T_B \)).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- A particle is suspended by a light vertical inelastic string of length \( L \).
- At equilibrium, it is projected horizontally with a speed \( v_0 = \sqrt{6gL} \).

2. **Finding the Velocity at Points A and B**:
- When the particle reaches point A (horizontal position), it has fallen a distance \( L \).
- When the particle reaches point B (topmost position), it has fallen a distance \( 2L \).

3. **Applying Work-Energy Theorem**:
- The work done by gravity when moving from the initial position to point A:
\[ -mgL = \frac{1}{2}mV_A^2 - \frac{1}{2}m(\sqrt{6gL})^2 \]
\[ -mgL = \frac{1}{2}mV_A^2 - 3mgL \]
\[ \Rightarrow \frac{1}{2}mV_A^2 = 2mgL \Rightarrow V_A^2 = 4gL \]
\[ V_A = \sqrt{4gL} = 2\sqrt{gL} \]

- The work done by gravity when moving from the initial position to point B:
\[ -mg(2L) = \frac{1}{2}mV_B^2 - \frac{1}{2}m(\sqrt{6gL})^2 \]
\[ -2mgL = \frac{1}{2}mV_B^2 - 3mgL \]
\[ \Rightarrow \frac{1}{2}mV_B^2 = mgL \Rightarrow V_B^2 = 2gL \]
\[ V_B = \sqrt{2gL} \]

4. **Finding Tension at Point A**:
- At point A, the only force providing centripetal acceleration is the tension \( T_A \):
\[ T_A = \frac{mV_A^2}{L} = \frac{m(4gL)}{L} = 4mg \]

5. **Finding Tension at Point B**:
- At point B, both the tension \( T_B \) and the weight \( mg \) act towards the center:
\[ T_B + mg = \frac{mV_B^2}{L} \]
\[ T_B + mg = \frac{m(2gL)}{L} = 2mg \]
\[ T_B = 2mg - mg = mg \]

6. **Finding the Ratio of Tensions**:
- Now we can find the ratio of the tensions:
\[ \frac{T_A}{T_B} = \frac{4mg}{mg} = 4 \]

### Final Answer:
The ratio of the tension in the string in its horizontal position to that in the string when the particle is vertically above the point of support is \( \boxed{4} \).