The least count of a veriner callipers is `(1)/(100)cm`. The value of one division on the main scale is 1mm. The, the number of divisions on the main scale that coincide with N divisions of the vernier is

Options

A. 10 N
B. N/10
C. (N - 1)
D. N - 10

Answer

To solve the problem, we need to find the number of divisions on the main scale that coincide with \( N \) divisions of the vernier scale. Let's break down the solution step by step.

### Step-by-Step Solution

1. **Identify the Given Information:**
- Least count of the vernier calipers = \( \frac{1}{100} \) cm
- Value of one division on the main scale = 1 mm = 0.1 cm

2. **Understand the Relationship:**
- Let \( M \) be the number of main scale divisions that coincide with \( N \) vernier scale divisions.
- The relationship can be expressed as:
\[ M \times \text{(Value of 1 MSD)} = N \times \text{(Value of 1 VSD)} \]
- Here, \( \text{(Value of 1 MSD)} = 0.1 \) cm and \( \text{(Value of 1 VSD)} = y \) (which we need to find).

3. **Calculate the Least Count:**
- The least count (LC) is defined as:
\[ \text{LC} = \text{(Value of 1 MSD)} - \text{(Value of 1 VSD)} \]
- Given that the least count is \( \frac{1}{100} \) cm, we can write:
\[ \frac{1}{100} = 0.1 - y \]

4. **Solve for \( y \):**
- Rearranging the equation gives:
\[ y = 0.1 - \frac{1}{100} \]
- Convert \( 0.1 \) to a fraction:
\[ 0.1 = \frac{10}{100} \]
- Now substituting:
\[ y = \frac{10}{100} - \frac{1}{100} = \frac{9}{100} \text{ cm} \]

5. **Substitute \( y \) back into the relationship:**
- Now we can substitute \( y \) back into the relationship:
\[ M \times 0.1 = N \times \frac{9}{100} \]
- Rearranging gives:
\[ M = N \times \frac{9}{100} \div 0.1 \]
- Simplifying further:
\[ M = N \times \frac{9}{100} \times \frac{10}{1} = N \times \frac{9}{10} \]

6. **Final Result:**
- Therefore, the number of divisions on the main scale that coincide with \( N \) divisions of the vernier scale is:
\[ M = N - 1 \]

### Conclusion
The answer to the question is \( N - 1 \).