`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

Answer

To solve the problem, we need to find the number of divisions on the vernier scale given the least count and the divisions on the main scale.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The main scale (MS) of the vernier calipers is divided into 10 divisions over 1 cm.
- The least count (LC) of the calipers is given as 0.005 cm.

2. **Calculate the Value of One Main Scale Division (MSD)**:
\[ \text{MSD} = \frac{\text{Total length of main scale}}{\text{Number of divisions}} = \frac{1 \text{ cm}}{10} = 0.1 \text{ cm} \]

3. **Use the Formula for Least Count**:
The least count is defined as:
\[ \text{Least Count} = \text{MSD} - \text{VSD} \]
where VSD is the value of one vernier scale division.

4. **Substitute the Values**:
We know that:
\[ 0.005 \text{ cm} = 0.1 \text{ cm} - \text{VSD} \]
Rearranging gives:
\[ \text{VSD} = 0.1 \text{ cm} - 0.005 \text{ cm} = 0.095 \text{ cm} \]

5. **Set Up the Equation for the Number of Vernier Scale Divisions (X)**:
Let \(X\) be the number of divisions on the vernier scale. The value of one vernier scale division can also be expressed as:
\[ \text{VSD} = \frac{X \cdot \text{MSD}}{X} = \frac{X \cdot 0.1 \text{ cm}}{X} \]
Therefore:
\[ 0.095 \text{ cm} = \frac{X \cdot 0.1 \text{ cm}}{X} \]

6. **Solve for X**:
From the equation:
\[ 0.095X = 0.1X - 0.1 \]
Rearranging gives:
\[ 0.1 = 0.1X - 0.095X \]
\[ 0.1 = 0.005X \]
Dividing both sides by 0.005:
\[ X = \frac{0.1}{0.005} = 20 \]

7. **Conclusion**:
The number of divisions on the vernier scale is \(X = 20\).

### Final Answer:
The vernier scale must have **20 divisions**.