A disc is rotating with angular velocity `omega`. A force F acts at a point whose position vector with respect to the axis of rotation is r. The power associated with torque due to the force is given by

Options

A. `(r xxF). omega`
B. `(r xx F) xx omega`
C. `r xx (F. omega)`
D. `r. (F xx omega)`

Explanation

To solve the problem, we need to find the power associated with the torque due to the force acting on a rotating disc. Let's break down the steps to arrive at the solution.

### Step-by-Step Solution:

1. **Understand the System**:
- We have a disc rotating with an angular velocity \( \omega \).
- A force \( F \) is applied at a point whose position vector with respect to the axis of rotation is \( R \).

2. **Define Torque**:
- The torque \( \tau \) due to the force \( F \) acting at point \( R \) is given by the cross product:
\[ \tau = R \times F \]

3. **Define Power**:
- The power \( P \) associated with the torque is defined as the dot product of the torque vector and the angular velocity vector:
\[ P = \tau \cdot \omega \]

4. **Substitute Torque into Power Equation**:
- Substitute the expression for torque into the power equation:
\[ P = (R \times F) \cdot \omega \]

5. **Final Expression**:
- Thus, the power associated with the torque due to the force is:
\[ P = R \times F \cdot \omega \]

6. **Identify the Correct Option**:
- The correct expression for power \( P \) is \( R \times F \cdot \omega \), which matches with option 1.

### Conclusion:
The power associated with the torque due to the force is given by:
\[ P = R \times F \cdot \omega \]
Thus, the correct answer is option 1.