Five particles of mass 2 kg are attached to the rim of a circular disc of radius 0.1 m & negligible mass. Moment of inertia of the system about an axis passing through the centre of the disc & perpendicular to its plane is

Options

A. `1 "kg-m"^(2)`
B. `0.1 "kg-m"^(2)`
C. `2 "kg-m"^(2)`
D. `0.2 "kg-m"^(2)`

Explanation

To find the moment of inertia of the system about an axis passing through the center of the disc and perpendicular to its plane, we can follow these steps:

### Step 1: Identify the parameters
- Mass of each particle, \( m = 2 \, \text{kg} \)
- Number of particles, \( n = 5 \)
- Radius of the disc, \( r = 0.1 \, \text{m} \)

### Step 2: Understand the formula for moment of inertia
The moment of inertia \( I \) for point masses about an axis is given by the formula:
\[ I = \sum_{i=1}^{n} m_i r_i^2 \]
where \( m_i \) is the mass of each particle and \( r_i \) is the distance from the axis of rotation.

### Step 3: Apply the formula
Since all particles have the same mass and are at the same distance from the axis:
\[ I = n \cdot m \cdot r^2 \]
Substituting the values:
\[ I = 5 \cdot 2 \, \text{kg} \cdot (0.1 \, \text{m})^2 \]

### Step 4: Calculate \( r^2 \)
Calculate \( (0.1 \, \text{m})^2 \):
\[ (0.1)^2 = 0.01 \, \text{m}^2 \]

### Step 5: Substitute and calculate \( I \)
Now substituting back into the equation:
\[ I = 5 \cdot 2 \cdot 0.01 \]
\[ I = 10 \cdot 0.01 = 0.1 \, \text{kg m}^2 \]

### Step 6: Conclusion
Thus, the moment of inertia of the system about the given axis is:
\[ I = 0.1 \, \text{kg m}^2 \]

### Final Answer
The correct option is option 2: \( 0.1 \, \text{kg m}^2 \).

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