There are n coplanar vectors each of magnitude m and each vector is inclined to the preceeding vector at an angle `2pi//n`. Then the magnitude of their resultant is

Options

A. n m
B. `m^(2)`
C. m/n
D. zero

Explanation

To solve the problem of finding the magnitude of the resultant of n coplanar vectors, each of magnitude m and inclined to the preceding vector at an angle of \( \frac{2\pi}{n} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Vectors**:
- We have n coplanar vectors, each with a magnitude \( m \).
- Each vector is inclined to the preceding vector at an angle \( \theta = \frac{2\pi}{n} \).

2. **Visualizing the Arrangement**:
- If we arrange these vectors in a circular formation, the first vector can be placed along the positive x-axis.
- The second vector will then be at an angle \( \theta \) from the first, the third vector at \( 2\theta \), and so on, until the nth vector, which will be at an angle \( (n-1)\theta \).

3. **Total Angle Calculation**:
- The total angle for n vectors is:
\[ n \cdot \theta = n \cdot \frac{2\pi}{n} = 2\pi \]
- This indicates that the vectors complete a full circle.

4. **Applying the Polygon Law of Vector Addition**:
- According to the polygon law of vector addition, if the vectors form a closed polygon, the resultant vector is zero.
- Since the vectors are arranged in a circular pattern and return to the starting point, the head and tail of the initial and final vectors coincide.

5. **Conclusion**:
- Therefore, the magnitude of the resultant vector is:
\[ R = 0 \]

### Final Answer:
The magnitude of the resultant of the n coplanar vectors is \( 0 \).

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