If I, `alpha` and `tau` are the moment of inertia, angular acceleration and torque respectively of a body rotating about an axis with angular velocity `omega` then,

Options

A. `tau=Iomega`
B. `tau=Ialpha`
C. `tau=Iomega`
D. `tau=alpha/I`

Answer

To solve the problem, we need to establish the relationship between torque (τ), moment of inertia (I), angular acceleration (α), and angular momentum (L) for a body rotating about an axis with angular velocity (ω).

### Step-by-Step Solution:

1. **Understanding Angular Momentum**:
The angular momentum (L) of a rotating body is given by the formula:
\[ L = I \cdot \omega \]
where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.

2. **Differentiating Angular Momentum**:
To find the relationship involving torque, we differentiate the angular momentum with respect to time (t):
\[ \frac{dL}{dt} = \frac{d}{dt}(I \cdot \omega) \]

3. **Applying the Product Rule**:
Since the moment of inertia \(I\) is constant for a rigid body, we can apply the product rule of differentiation:
\[ \frac{dL}{dt} = I \cdot \frac{d\omega}{dt} \]
Here, \(\frac{d\omega}{dt}\) represents the angular acceleration (\(\alpha\)).

4. **Relating Torque to Angular Acceleration**:
By definition, the rate of change of angular momentum is equal to the torque (\(τ\)):
\[ \tau = \frac{dL}{dt} \]
Substituting the expression we derived:
\[ \tau = I \cdot \alpha \]

5. **Conclusion**:
Thus, we have established the relationship:
\[ \tau = I \cdot \alpha \]
This indicates that torque is equal to the product of moment of inertia and angular acceleration.

### Final Answer:
The correct relationship is:
\[ \tau = I \cdot \alpha \]