A wheel of moment of inertia 2.5 `kgm^(2)` has an initial angular velocity of 40 rad `s^(-1)`. A constant torque of 10 Nm acts on the wheel. The time during which the wheel is accelerated to 60 rad `s^(-1)` is:

Options

A. 4s
B. 6s
C. 5s
D. 2.5s

Explanation

To solve the problem step by step, we will follow the principles of rotational motion, specifically using the relationship between torque, moment of inertia, and angular acceleration.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Moment of Inertia (I) = 2.5 kg·m²
- Initial Angular Velocity (ω_initial) = 40 rad/s
- Final Angular Velocity (ω_final) = 60 rad/s
- Torque (τ) = 10 Nm

2. **Use the Torque Formula:**
The relationship between torque (τ), moment of inertia (I), and angular acceleration (α) is given by:
\[ \tau = I \cdot \alpha \]
Rearranging this gives us:
\[ \alpha = \frac{\tau}{I} \]

3. **Calculate Angular Acceleration (α):**
Substitute the values of τ and I into the equation:
\[ \alpha = \frac{10 \, \text{Nm}}{2.5 \, \text{kg·m}^2} = 4 \, \text{rad/s}^2 \]

4. **Use the Angular Acceleration to Find Time (t):**
We know that angular acceleration can also be expressed in terms of the change in angular velocity over time:
\[ \alpha = \frac{\omega_{final} - \omega_{initial}}{t} \]
Rearranging this gives us:
\[ t = \frac{\omega_{final} - \omega_{initial}}{\alpha} \]

5. **Substitute Values to Find Time (t):**
Substitute ω_final, ω_initial, and α into the equation:
\[ t = \frac{60 \, \text{rad/s} - 40 \, \text{rad/s}}{4 \, \text{rad/s}^2} = \frac{20 \, \text{rad/s}}{4 \, \text{rad/s}^2} = 5 \, \text{s} \]

6. **Conclusion:**
The time during which the wheel is accelerated to 60 rad/s is **5 seconds**.

### Final Answer:
**t = 5 seconds**

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