A particle is rotated in a vertical circle by connecting it to a string of length `l` and keeping the other end of the 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 solve the problem of finding the minimum speed of a particle at the point where the string is horizontal (point B) for it to complete a vertical circle, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the System**:
- A particle is attached to a string of length \( l \) and rotates in a vertical circle. The string is fixed at one end, and the particle moves in a circular path.

2. **Identify Key Points**:
- Let point A be the top of the circle and point B be the point where the string is horizontal.

3. **Minimum Speed at the Top of the Circle**:
- For the particle to complete the vertical circle, at the topmost point (point A), it must have a minimum speed \( V_A \). The minimum speed required at the top of the circle is given by:
\[ V_A = \sqrt{5gl} \]
- This is derived from the balance of forces at the top of the circle, where the gravitational force provides the necessary centripetal force.

4. **Conservation of Energy**:
- We can use the conservation of mechanical energy to relate the speeds at points A and B.
- The total mechanical energy at point A (potential energy + kinetic energy) must equal the total mechanical energy at point B.

5. **Energy at Point A**:
- At point A (top of the circle):
- Potential Energy (PE) = 0 (taking this as the reference point)
- Kinetic Energy (KE) = \( \frac{1}{2} m V_A^2 = \frac{1}{2} m (5gl) \)
- Therefore, the total energy at point A is:
\[ E_A = 0 + \frac{1}{2} m (5gl) = \frac{5}{2} m gl \]

6. **Energy at Point B**:
- At point B (horizontal position):
- Potential Energy (PE) = \( mg \cdot l \) (since it is at height \( l \))
- Kinetic Energy (KE) = \( \frac{1}{2} m V^2 \) (where \( V \) is the speed at point B)
- Therefore, the total energy at point B is:
\[ E_B = mg \cdot l + \frac{1}{2} m V^2 \]

7. **Setting Energies Equal**:
- By conservation of energy, we have:
\[ E_A = E_B \]
- Substituting the expressions for \( E_A \) and \( E_B \):
\[ \frac{5}{2} m gl = mg \cdot l + \frac{1}{2} m V^2 \]

8. **Simplifying the Equation**:
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{5}{2} gl = gl + \frac{1}{2} V^2 \]
- Rearranging gives:
\[ \frac{5}{2} gl - gl = \frac{1}{2} V^2 \]
\[ \frac{3}{2} gl = \frac{1}{2} V^2 \]

9. **Solving for V**:
- Multiply both sides by 2:
\[ 3gl = V^2 \]
- Taking the square root gives:
\[ V = \sqrt{3gl} \]

### Final Answer:
The minimum speed of the particle at point B for it to complete the circle is:
\[ V = \sqrt{3gl} \]