`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. (a)10 divisions
B. (b)20 divisions
C. (c)25 divisions
D. (d)50 divisions

Explanation

To solve the problem step by step, we will determine how many divisions the vernier scale must have based on the information given.

### Step 1: Understand the relationship between main scale divisions and centimeters
We know that 1 cm of the main scale is divided into 10 divisions. Therefore, we can calculate the value of one main scale division (MSD):
\[ \text{1 MSD} = \frac{1 \text{ cm}}{10} = 0.1 \text{ cm} \]

### Step 2: Use the formula for least count
The least count (LC) of the vernier calipers is given by the formula:
\[ \text{LC} = \text{1 MSD} - \text{1 Vernier Scale Division (VSD)} \]
We are given that the least count is 0.005 cm. Therefore, we can set up the equation:
\[ 0.005 \text{ cm} = 0.1 \text{ cm} - \text{1 VSD} \]

### Step 3: Calculate the value of one Vernier Scale Division
Rearranging the equation from Step 2 gives:
\[ \text{1 VSD} = 0.1 \text{ cm} - 0.005 \text{ cm} = 0.095 \text{ cm} \]

### Step 4: Set up the relationship for X Vernier Scale Divisions
Let \( X \) be the number of divisions on the vernier scale. The total length of \( X \) Vernier Scale Divisions can be expressed as:
\[ X \text{ VSD} = X \times \text{1 VSD} = X \times 0.095 \text{ cm} \]
We also know that \( X \) Vernier Scale Divisions is equal to \( X - 1 \) Main Scale Division (MSD):
\[ X \text{ VSD} = (X - 1) \text{ MSD} \]
Substituting the values, we have:
\[ X \times 0.095 \text{ cm} = (X - 1) \times 0.1 \text{ cm} \]

### Step 5: Solve for X
Now, we can solve for \( X \):
\[ 0.095X = 0.1(X - 1) \]
Expanding the right side:
\[ 0.095X = 0.1X - 0.1 \]
Rearranging gives:
\[ 0.1X - 0.095X = 0.1 \]
\[ 0.005X = 0.1 \]
Dividing both sides by 0.005:
\[ X = \frac{0.1}{0.005} = 20 \]

### Conclusion
Thus, the vernier scale must have **20 divisions**.