A uniform thin bar of mass 6 kg and length 2.4 meter is bent to make an equilateral hexagon. The moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of hexagon is _____ `xx 10^(-1) kg m^2`

Answer

To find the moment of inertia of a uniform thin bar bent into an equilateral 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 2.4 meters and it is bent into an equilateral hexagon with 6 sides, we can calculate the length of each side.

\[ \text{Length of each side} = \frac{\text{Total length}}{\text{Number of sides}} = \frac{2.4 \, \text{m}}{6} = 0.4 \, \text{m} \]

### Step 2: Calculate the mass of each side
The total mass of the bar is 6 kg. Since the mass is uniformly distributed, the mass of each side of the hexagon can be calculated as follows:

\[ \text{Mass of each side} = \frac{\text{Total mass}}{\text{Number of sides}} = \frac{6 \, \text{kg}}{6} = 1 \, \text{kg} \]

### Step 3: Use the formula for the moment of inertia of a hexagon
The moment of inertia \(I\) of a regular hexagon about an axis passing through its center and perpendicular to its plane can be calculated using the formula:

\[ I = \frac{3}{2} m a^2 \]

where \(m\) is the total mass of the hexagon and \(a\) is the distance from the center to the midpoint of a side. For a regular hexagon, this distance \(a\) can be calculated as:

\[ a = \frac{s}{\sqrt{3}} \]

where \(s\) is the length of each side.

### Step 4: Calculate the distance \(a\)
Using the length of each side \(s = 0.4 \, \text{m}\):

\[ a = \frac{0.4}{\sqrt{3}} \approx 0.2309 \, \text{m} \]

### Step 5: Calculate the moment of inertia
Now substituting the values into the moment of inertia formula:

\[ I = \frac{3}{2} \times 6 \, \text{kg} \times (0.2309 \, \text{m})^2 \]

Calculating \( (0.2309)^2 \):

\[ (0.2309)^2 \approx 0.0537 \, \text{m}^2 \]

Now substituting back:

\[ I = \frac{3}{2} \times 6 \times 0.0537 \approx 0.48 \, \text{kg m}^2 \]

### Step 6: Convert to the required format
To express this in the format requested in the question:

\[ I \approx 4.8 \times 10^{-1} \, \text{kg m}^2 \]

### Final Answer
Thus, the moment of inertia about an axis passing through the center of mass and perpendicular to the plane of the hexagon is:

\[ \boxed{4.8 \times 10^{-1} \, \text{kg m}^2} \]