In the above question, if the disc executes rotatory motion, its angular acceleration will be:

Options

A. 2.5 `"rad"//"sec"^(2)`
B. 5 `"rad"//"sec"^(2)`
C. 10 `"rad"//"sec"^(2)`
D. 20 `"rad"//"sec"^(2)`

Answer

To find the angular acceleration of the disk executing rotational motion, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Tangential force \( F = 10 \, \text{N} \)
- Radius of the disk \( R = 0.5 \, \text{m} \)
- Moment of inertia \( I = 2 \, \text{kg} \cdot \text{m}^2 \)

2. **Calculate the Torque**:
- Torque \( \tau \) is given by the formula:
\[ \tau = R \times F \]
- Since the force is applied tangentially, we can use:
\[ \tau = R \cdot F = 0.5 \, \text{m} \times 10 \, \text{N} = 5 \, \text{N} \cdot \text{m} \]

3. **Use the Torque-Angular Acceleration Relationship**:
- The relationship between torque, moment of inertia, and angular acceleration is given by:
\[ \tau = I \cdot \alpha \]
- Rearranging this equation to solve for angular acceleration \( \alpha \):
\[ \alpha = \frac{\tau}{I} \]

4. **Substitute the Values**:
- Substitute the calculated torque and the moment of inertia into the equation:
\[ \alpha = \frac{5 \, \text{N} \cdot \text{m}}{2 \, \text{kg} \cdot \text{m}^2} = 2.5 \, \text{rad/s}^2 \]

5. **Final Answer**:
- The angular acceleration of the disk is:
\[ \alpha = 2.5 \, \text{rad/s}^2 \]