`1 cm` of main scale of a vernier callipers is divided into `10` divisions. The least count of the callipers is `0.005 cm`, then the vernier scale must have.

Options

A. 10 divisions
B. 20 divisions
C. 25 divisions
D. 50 divisions

Explanation

To solve the problem, we need to find out how many divisions are there in the vernier scale of the vernier calipers given the information provided.

### Step-by-Step Solution:

1. **Identify the Main Scale Division (MSD)**:
- We know that 1 cm of the main scale is divided into 10 divisions.
- Therefore, 1 main scale division (MSD) = \( \frac{1 \text{ cm}}{10} = 0.1 \text{ cm} \).

2. **Given Least Count (LC)**:
- The least count of the vernier calipers is given as \( 0.005 \text{ cm} \).

3. **Calculate the Vernier Scale Division (VSD)**:
- The formula for the least count is:
\[ \text{Least Count} = \text{1 MSD} - \text{1 VSD} \]
- Rearranging this gives:
\[ \text{1 VSD} = \text{1 MSD} - \text{Least Count} \]
- Substituting the values we have:
\[ \text{1 VSD} = 0.1 \text{ cm} - 0.005 \text{ cm} = 0.095 \text{ cm} \]

4. **Set Up the Relationship Between VSD and MSD**:
- Let \( x \) be the number of vernier scale divisions.
- The relationship between the vernier scale divisions and the main scale divisions is:
\[ x \text{ VSD} = (x - 1) \text{ MSD} \]
- Substituting the values:
\[ x \cdot 0.095 \text{ cm} = (x - 1) \cdot 0.1 \text{ cm} \]

5. **Solve for \( x \)**:
- Expanding the equation:
\[ 0.095x = 0.1x - 0.1 \]
- Rearranging gives:
\[ 0.1x - 0.095x = 0.1 \]
\[ 0.005x = 0.1 \]
- Dividing both sides by \( 0.005 \):
\[ x = \frac{0.1}{0.005} = 20 \]

6. **Conclusion**:
- The number of divisions in the vernier scale is \( x = 20 \).