In the above question, the change in angular momentum of disc in first 2 seconds in Nm second will be:

Options

A. 2.5
B. 5
C. 10
D. 20

Answer

To solve the question regarding the change in angular momentum of the disc in the first 2 seconds, we will follow these steps:

### Step-by-Step Solution:

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

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

3. **Calculate the Angular Acceleration:**
- The relationship between torque and angular acceleration (\( \alpha \)) is given by:
\[ \tau = I \alpha \]
- Rearranging for \( \alpha \):
\[ \alpha = \frac{\tau}{I} \]
- Substituting the values:
\[ \alpha = \frac{5 \, \text{N m}}{2 \, \text{kg m}^2} = 2.5 \, \text{rad/s}^2 \]

4. **Calculate the Final Angular Velocity after 2 seconds:**
- The final angular velocity (\( \omega_f \)) can be calculated using the equation:
\[ \omega_f = \omega_0 + \alpha T \]
- Since the initial angular velocity (\( \omega_0 \)) is 0 (the disc starts from rest):
\[ \omega_f = 0 + (2.5 \, \text{rad/s}^2) \times (2 \, \text{s}) = 5 \, \text{rad/s} \]

5. **Calculate the Change in Angular Momentum:**
- The change in angular momentum (\( \Delta L \)) is given by:
\[ \Delta L = L_f - L_i \]
- Since the initial angular momentum (\( L_i \)) is 0 (the disc starts from rest):
\[ \Delta L = L_f \]
- The final angular momentum (\( L_f \)) is given by:
\[ L_f = I \omega_f \]
- Substituting the values:
\[ L_f = 2 \, \text{kg m}^2 \times 5 \, \text{rad/s} = 10 \, \text{kg m}^2/\text{s} \]

6. **Final Result:**
- The change in angular momentum of the disc in the first 2 seconds is:
\[ \Delta L = 10 \, \text{N m s} \]