A particle is rotated in verticla circle by connecting it to string fixed. The minimum speed of the particle when the string is horizontal for which the particle will complete the circle is

Options

A. `sqrt(gl)`
B. `sqrt(2gl)`
C. `sqrt(3gl)`
D. `sqrt(5gl)`

Explanation

To find the minimum speed of a particle at the horizontal position in a vertical circle that allows it to complete the circle, we can use the principles of energy conservation and dynamics.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a particle attached to a string that rotates in a vertical circle. We need to find the minimum speed of the particle when the string is horizontal (let's call this point B) so that it can complete the circular motion.

2. **Identify Key Points**:
- Let point A be the lowest point of the circle and point B be the horizontal position of the string.
- At point A, the particle has a certain speed (let's denote it as \( v_A \)).
- At point B, we will denote the speed as \( v_B \).

3. **Using Conservation of Energy**:
- The total mechanical energy at point A should equal the total mechanical energy at point B.
- The potential energy at point A (lowest point) is taken as zero, and the height at point B is \( L \) (the radius of the circle).
- The kinetic energy at point A is \( \frac{1}{2} m v_A^2 \).
- The potential energy at point B is \( mgh = mgL \).
- The kinetic energy at point B is \( \frac{1}{2} m v_B^2 \).

4. **Setting Up the Equation**:
\[ \text{Total Energy at A} = \text{Total Energy at B} \]
\[ 0 + \frac{1}{2} m v_A^2 = mgL + \frac{1}{2} m v_B^2 \]

5. **Simplifying the Equation**:
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v_A^2 = gL + \frac{1}{2} v_B^2 \]

6. **Rearranging the Equation**:
\[ v_A^2 = 2gL + v_B^2 \]

7. **Finding Minimum Speed at the Top (Point A)**:
- For the particle to just complete the circle, at the topmost point (point C), the centripetal force must be provided by the weight of the particle. Thus:
\[ mg = \frac{mv_C^2}{L} \]
\[ v_C^2 = gL \]

8. **Substituting \( v_C \) Back**:
- We know that \( v_A^2 = 2gL + v_B^2 \) and at the top \( v_C = \sqrt{gL} \):
\[ v_C^2 = gL \]
- Substitute \( v_B = v_C \):
\[ v_A^2 = 2gL + gL = 3gL \]
\[ v_A = \sqrt{3gL} \]

9. **Conclusion**:
- The minimum speed of the particle when the string is horizontal (point B) for it to complete the circle is \( v_B = \sqrt{3gL} \).

### Final Answer:
The minimum speed of the particle when the string is horizontal for it to complete the circle is \( \sqrt{3gL} \).