A body crosses the topmost point of a vertical circle with a critical speed. Its centripetal acceleration, when the string is horizontal will be

Options

A. `6g`
B. `3g`
C. `2g`
D. `g`

Explanation

To solve the problem, we need to find the centripetal acceleration of a body when it is at the horizontal position after crossing the topmost point of a vertical circle with a critical speed.

### Step-by-Step Solution:

1. **Understanding Critical Speed at the Topmost Point**:
- At the topmost point of the vertical circle, the critical speed \( V_c \) is given by the formula:
\[ V_c = \sqrt{gL} \]
where \( g \) is the acceleration due to gravity and \( L \) is the length of the string (or radius of the circle).

2. **Finding the Speed at the Horizontal Position**:
- When the body moves from the topmost point to the horizontal position, it converts potential energy into kinetic energy.
- The change in height from the topmost point to the horizontal position is \( L \) (the radius of the circle).
- We can use the energy conservation principle:
\[ \text{Initial Kinetic Energy} + \text{Initial Potential Energy} = \text{Final Kinetic Energy} + \text{Final Potential Energy} \]
- At the topmost point:
- Initial Kinetic Energy = \( \frac{1}{2} m V_c^2 = \frac{1}{2} m gL \)
- Initial Potential Energy = \( mg(2L) \) (since it's at height \( 2L \))
- At the horizontal position:
- Final Kinetic Energy = \( \frac{1}{2} m V^2 \)
- Final Potential Energy = \( mgL \) (since it's at height \( L \))
- Setting up the equation:
\[ \frac{1}{2} m gL + mg(2L) = \frac{1}{2} m V^2 + mgL \]
- Simplifying:
\[ \frac{1}{2} gL + 2gL = \frac{1}{2} V^2 + gL \]
\[ \frac{5}{2} gL = \frac{1}{2} V^2 + gL \]
\[ \frac{5}{2} gL - gL = \frac{1}{2} V^2 \]
\[ \frac{3}{2} gL = \frac{1}{2} V^2 \]
\[ V^2 = 3gL \]

3. **Calculating Centripetal Acceleration**:
- The centripetal acceleration \( a_c \) is given by the formula:
\[ a_c = \frac{V^2}{L} \]
- Substituting \( V^2 = 3gL \):
\[ a_c = \frac{3gL}{L} = 3g \]

### Final Answer:
The centripetal acceleration when the string is horizontal is \( 3g \).