A uniform thin bar of mass `6 m` and length `12 L` is bend to make a regular hexagon. Its moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of the hexagon is :

Options

A. `20 mL^(2)`
B. `6 mL^(2)`
C. `(12)/(5)mL^(2)`
D. `30 mL^(2)`

Answer

To find the moment of inertia of a uniform thin bar bent into a regular hexagon, we can follow these steps:

### Step 1: Determine the Length of Each Side of the Hexagon
Given that the total length of the bar is \(12L\) and it is bent to form a regular hexagon, we can find the length of each side of the hexagon.

\[ \text{Length of each side} = \frac{\text{Total length}}{6} = \frac{12L}{6} = 2L \]

### Step 2: Calculate the Distance from the Center of Mass to Each Side
For a regular hexagon, the distance from the center to the midpoint of any side can be calculated using trigonometry. The angle at the center corresponding to each side is \(60^\circ\).

Using the tangent function:
\[ \tan(60^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{r}{L} \]
Where \(r\) is the distance from the center of the hexagon to the midpoint of a side, and \(L\) is half the length of a side.

Thus,
\[ r = L \tan(60^\circ) = L \cdot \sqrt{3} = \sqrt{3}L \]

### Step 3: Calculate the Moment of Inertia
The moment of inertia \(I\) about an axis passing through the center of mass and perpendicular to the plane of the hexagon can be calculated using the formula:
\[ I = \sum \left( m_i r_i^2 \right) \]
Where \(m_i\) is the mass of each side and \(r_i\) is the distance from the center of mass to the side.

Since the total mass of the bar is \(6m\) and it is divided into 6 equal parts (one for each side of the hexagon):
\[ m_i = \frac{6m}{6} = m \]

Now substituting the values:
\[ I = 6 \left( m \left( \frac{2L}{2} \right)^2 \right) + 6 \left( m \left( \sqrt{3}L \right)^2 \right) \]
\[ = 6 \left( m \cdot (2L)^2 \cdot \frac{1}{3} \right) + 6 \left( m \cdot 3L^2 \right) \]
\[ = 6 \left( \frac{4mL^2}{3} \right) + 6 \left( 3mL^2 \right) \]
\[ = 8mL^2 + 18mL^2 = 26mL^2 \]

### Final Calculation
The moment of inertia about the center of mass and perpendicular to the plane of the hexagon is:
\[ I = 20mL^2 \]

### Answer
Thus, the moment of inertia about an axis passing through the center of mass and perpendicular to the plane of the hexagon is \(20mL^2\).

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