To solve this problem, we can use the principles of energy conservation and centripetal force.
### Given:
- A particle is rotating in a vertical circle.
- The length of the string is \( l \).
- We need to find the minimum speed of the particle when the string is horizontal for it to complete the circle.
### Solution:
1. **Position Considerations:**
- Let the particle be at the lowest point of the circle initially. At this point, the potential energy is minimum, and kinetic energy is maximum.
- When the string is horizontal, the particle is at a height \( l \) above the lowest point.
- At the highest point, the particle is at a height \( 2l \) above the lowest point.
2. **Energy Conservation Principle:**
- To just complete the circle, the speed at the highest point must be such that the centripetal force requirement is met by the tension and gravitational force. The minimum condition occurs when the tension is zero at the topmost point (all centripetal force provided by gravity).
At the highest point:
- Centripetal force \( F_c = \frac{mv^2}{l} = mg \).
- Therefore, the minimum speed at the highest point \( v_{\text{min}} = \sqrt{gl} \).
3. **Using Energy Conservation:**
- Total mechanical energy at the horizontal position should be equal to the total mechanical energy at the top position for the particle to complete the circle.
- Let the minimum speed at the horizontal position be \( v_h \). At the horizontal position:
\[
\text{Kinetic Energy} = \frac{1}{2}mv_h^2
\]
\[
\text{Potential Energy} = mgl
\]
- At the highest point (top of the circle):
\[
\text{Kinetic Energy} = \frac{1}{2}mv_{\text{min}}^2 = \frac{1}{2}m(gl) = \frac{mgl}{2}
\]
\[
\text{Potential Energy} = mg(2l) = 2mgl
\]
- According to energy conservation:
\[
\frac{1}{2}mv_h^2 + mgl = \frac{mgl}{2} + 2mgl
\]
\[
\frac{1}{2}mv_h^2 + mgl = \frac{5mgl}{2}
\]
\[
\frac{1}{2}mv_h^2 = \frac{5mgl}{2} - mgl
\]
\[
\frac{1}{2}mv_h^2 = \frac{3mgl}{2}
\]
\[
mv_h^2 = 3mgl
\]
\[
v_h^2 = 3gl
\]
\[
v_h = \sqrt{3gl}
\]
### Conclusion:
The minimum speed of the particle when the string is horizontal for which the particle will complete the circle is \( \sqrt{3gl} \).
Therefore, the correct answer is **(C) \( \sqrt{3gl} \)**.
