Let `r` be the radius vector of a particle in motion about some reference point and `r` its modulus. Similarly, `v` be the velocity vector and `v` its modulus. Then

Options

A. `v!=(dr)/(dt)`
B. `v=(dr)/(dt)`
C. `v=|(dr)/(dt)|`
D. `|dr|!=dr`

Answer

To solve the problem, we need to analyze the motion of a particle in circular motion and the relationships between the radius vector \( \mathbf{r} \), its modulus \( r \), the velocity vector \( \mathbf{v} \), and its modulus \( v \).

### Step-by-Step Solution:

1. **Understanding the Radius Vector**:
The radius vector \( \mathbf{r} \) is defined as the vector drawn from a reference point (the center of the circular path) to the position of the particle. In circular motion, the magnitude of this radius vector \( r \) remains constant, but its direction changes as the particle moves along the circular path.

**Hint**: Remember that in circular motion, the radius is constant, but the direction of the radius vector changes continuously.

2. **Differentiating the Radius Vector**:
Since the radius \( r \) is constant, the differential change in the radius vector can be expressed as:
\[ d\mathbf{r} = d\mathbf{r}(t) \]
However, since the magnitude \( r \) is constant, the differential change in magnitude \( dr \) is zero:
\[ dr = 0 \]

**Hint**: Consider how the radius vector behaves as the particle moves. What happens to its length?

3. **Velocity Vector**:
The velocity vector \( \mathbf{v} \) is defined as the rate of change of the radius vector with respect to time:
\[ \mathbf{v} = \frac{d\mathbf{r}}{dt} \]
However, since \( \mathbf{r} \) is changing direction but not magnitude, we need to consider that the velocity vector is not simply the derivative of the radius vector in terms of its magnitude.

**Hint**: Think about how velocity relates to the change in position over time, especially in circular motion.

4. **Magnitude of Velocity**:
The modulus of the velocity vector \( v \) is given by:
\[ v = |\mathbf{v}| = \left| \frac{d\mathbf{r}}{dt} \right| \]
This means that the speed \( v \) is the magnitude of the change in the radius vector over time.

**Hint**: Remember that speed is a scalar quantity, while velocity is a vector quantity.

5. **Conclusion**:
From the above analysis, we can conclude that:
- The modulus of the radius vector \( r \) is constant, hence \( dr \neq 0 \).
- The velocity vector \( \mathbf{v} \) is not equal to \( \frac{d\mathbf{r}}{dt} \) in terms of its magnitude because the direction of \( \mathbf{r} \) is changing, even though its magnitude remains constant.
- Therefore, the correct conclusion is that \( v \neq \frac{dr}{dt} \) and \( v = |\mathbf{v}| = \left| \frac{d\mathbf{r}}{dt} \right| \).

### Final Answer:
The incorrect option is \( v = \frac{dr}{dt} \).