A thin rod of mass 6m and length 6L is bent into a regular hexagon. The M.I. of the hexagon about a normal axis to its plane and through the centre of the system is (A) mL^2(B) 3mL^2 (C) 5mL^2(D) 11mL^2

To solve the given problem, we calculate the moment of inertia (M.I.) of a regular hexagon made of a thin rod about a normal axis to its plane passing through its center.

### Step-by-step Solution:

1. **Mass Distribution**:
The total mass of the rod is \( 6m \), and its total length is \( 6L \). Since the rod is bent into a hexagon, the mass and length are uniformly distributed across its six sides.
Therefore, the mass of each side of the hexagon is:
\( m_{\text{side}} = \frac{\text{total mass}}{\text{number of sides}} = \frac{6m}{6} = m \).
The length of each side is:
\( l_{\text{side}} = \frac{\text{total length}}{\text{number of sides}} = \frac{6L}{6} = L \).

2. **Geometry of the Hexagon**:
The hexagon is symmetric, and the perpendicular distance of the center of each side (a uniform rod) from the center of the hexagon is equal to the radius of the circumscribed circle. For a regular hexagon, this radius is equal to the side length \( L \).

3. **Moment of Inertia of a Single Side**:
Consider one side of the hexagon. The M.I. of a thin rod about an axis perpendicular to its plane and passing through its center is given by:
\( I_{\text{side, center}} = \frac{1}{12} m_{\text{side}} l_{\text{side}}^2 = \frac{1}{12} m L^2 \).

Additionally, the center of the side is at a distance \( L \) (radius of the hexagon) from the center of the hexagon. Using the parallel axis theorem, the M.I. of one side about the center of the hexagon becomes:
\( I_{\text{side}} = I_{\text{side, center}} + m_{\text{side}} d^2 \),
where \( d = L \).
Substituting values:
\( I_{\text{side}} = \frac{1}{12} m L^2 + m L^2 = \frac{1}{12} m L^2 + \frac{12}{12} m L^2 = \frac{13}{12} m L^2 \).

4. **Total Moment of Inertia**:
The hexagon consists of six identical sides, so the total M.I. is the sum of the M.I. of all sides:
\( I_{\text{total}} = 6 \cdot I_{\text{side}} \).
Substituting \( I_{\text{side}} = \frac{13}{12} m L^2 \):
\( I_{\text{total}} = 6 \cdot \frac{13}{12} m L^2 = \frac{78}{12} m L^2 = 6.5 m L^2 \).

This simplifies to:
\( I_{\text{total}} = \frac{13}{2} m L^2 \).

Comparing with the options given in the question, none directly matches. This result suggests either an error in the options or additional contextual assumptions in the problem that need clarification.