A whell of mass 4 kg and radius of gyration 0.4 m is making 300 r.p.m. Its moment of inertia is

Options

A. `"6.4 kg m"^(2)`
B. `"0.64 kg m"^(2)`
C. `"0.32 kg m"^(2)`
D. `"64 kg m"^(2)`

Explanation

To find the moment of inertia of the wheel, we can use the formula that relates mass (m) and radius of gyration (k) to the moment of inertia (I):

\[ I = m \cdot k^2 \]

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of the wheel, \( m = 4 \, \text{kg} \)
- Radius of gyration, \( k = 0.4 \, \text{m} \)

2. **Calculate \( k^2 \):**
\[ k^2 = (0.4 \, \text{m})^2 = 0.16 \, \text{m}^2 \]

3. **Substitute the values into the moment of inertia formula:**
\[ I = m \cdot k^2 = 4 \, \text{kg} \cdot 0.16 \, \text{m}^2 \]

4. **Perform the multiplication:**
\[ I = 4 \cdot 0.16 = 0.64 \, \text{kg} \cdot \text{m}^2 \]

5. **Conclusion:**
The moment of inertia of the wheel is:
\[ I = 0.64 \, \text{kg} \cdot \text{m}^2 \]

### Final Answer:
The moment of inertia of the wheel is \( 0.64 \, \text{kg} \cdot \text{m}^2 \).

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