In the above question, the value of its angular velocity after 2 seconds will be:

Options

A. 2.5 rad/sec
B. 5 rad/sec
C. 10 rad/sec
D. 20 rad/sec

Answer

To find the angular velocity after 2 seconds when a tangential force is applied to a ring, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Force \( F = 10 \, \text{N} \)
- Radius \( R = 0.5 \, \text{m} \)
- Moment of Inertia \( I = 2 \, \text{kg} \cdot \text{m}^2 \)
- Time \( t = 2 \, \text{s} \)

2. **Calculate the Torque**:
The torque \( \tau \) due to the tangential force is given by the formula:
\[ \tau = F \times R \]
Substituting the values:
\[ \tau = 10 \, \text{N} \times 0.5 \, \text{m} = 5 \, \text{N} \cdot \text{m} \]

3. **Relate Torque to Angular Momentum**:
The net torque is also related to the change in angular momentum:
\[ \tau = \frac{dL}{dt} = \frac{L_f - L_i}{t} \]
Where \( L_f \) is the final angular momentum and \( L_i \) is the initial angular momentum. Since the system starts from rest, \( L_i = 0 \).

4. **Substitute Angular Momentum**:
Angular momentum is defined as:
\[ L = I \cdot \omega \]
Therefore, we can rewrite the equation as:
\[ \tau = \frac{I \cdot \omega_f - I \cdot \omega_i}{t} \]
Since \( \omega_i = 0 \):
\[ \tau = \frac{I \cdot \omega_f}{t} \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ \omega_f = \frac{\tau \cdot t}{I} \]

6. **Substituting the Values**:
Substitute \( \tau = 5 \, \text{N} \cdot \text{m} \), \( t = 2 \, \text{s} \), and \( I = 2 \, \text{kg} \cdot \text{m}^2 \):
\[ \omega_f = \frac{5 \, \text{N} \cdot \text{m} \times 2 \, \text{s}}{2 \, \text{kg} \cdot \text{m}^2} = \frac{10}{2} = 5 \, \text{rad/s} \]

7. **Final Result**:
The angular velocity after 2 seconds is:
\[ \omega_f = 5 \, \text{rad/s} \]

### Conclusion:
The value of the angular velocity after 2 seconds is \( 5 \, \text{rad/s} \).

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