Two fly wheels A and B are mounted side by side with frictionless bearings on a common shaft. Their moments of inertia about the shaft are `5.0kgm^(2)` and `20.0kgm^(2)` respectively. Wheel A is made to rotate at 10 revolution per second. Wheel B, initially stationary, is now coupled to A with the help of a clutch. The rotation speed of the wheels will become

Options

A. `2sqrt(5)` rps
B. 0.5 rps
C. 2 rps
D. None of these

Explanation

To solve the problem, we will use the principle of conservation of angular momentum. Here are the steps:

### Step-by-Step Solution:

1. **Identify Given Data**:
- Moment of inertia of wheel A, \( I_A = 5.0 \, \text{kg m}^2 \)
- Moment of inertia of wheel B, \( I_B = 20.0 \, \text{kg m}^2 \)
- Initial angular velocity of wheel A, \( \omega_A = 10 \, \text{revolutions per second} \)
- Initial angular velocity of wheel B, \( \omega_B = 0 \, \text{revolutions per second} \)

2. **Convert Angular Velocity to Radians per Second**:
- Since \( 1 \, \text{revolution} = 2\pi \, \text{radians} \), we convert \( \omega_A \):
\[ \omega_A = 10 \, \text{rev/s} \times 2\pi \, \text{radians/rev} = 20\pi \, \text{radians/s} \]

3. **Calculate Initial Angular Momentum**:
- The initial angular momentum \( L_{initial} \) is the sum of the angular momentum of both wheels:
\[ L_{initial} = I_A \cdot \omega_A + I_B \cdot \omega_B \]
- Since \( \omega_B = 0 \):
\[ L_{initial} = I_A \cdot \omega_A = 5.0 \, \text{kg m}^2 \cdot 20\pi \, \text{radians/s} = 100\pi \, \text{kg m}^2/\text{s} \]

4. **Final Angular Momentum**:
- When the clutch is engaged, both wheels will rotate together with a common angular velocity \( \omega_f \). The final angular momentum \( L_{final} \) is:
\[ L_{final} = (I_A + I_B) \cdot \omega_f \]
- Substituting the values:
\[ L_{final} = (5.0 + 20.0) \, \text{kg m}^2 \cdot \omega_f = 25.0 \, \text{kg m}^2 \cdot \omega_f \]

5. **Set Initial and Final Angular Momentum Equal**:
- By conservation of angular momentum:
\[ L_{initial} = L_{final} \]
\[ 100\pi = 25.0 \cdot \omega_f \]

6. **Solve for Final Angular Velocity**:
\[ \omega_f = \frac{100\pi}{25.0} = 4\pi \, \text{radians/s} \]
- Convert back to revolutions per second:
\[ \omega_f = \frac{4\pi}{2\pi} = 2 \, \text{rev/s} \]

### Final Answer:
The final rotation speed of the wheels will be \( 2 \, \text{revolutions per second} \).