`1 cm` on the main scale of a vernier callipers is divided into `10 equal` parts. If `10` divisions of vernier coincide with `8` small divisions of main scale, then the least count of the calliper is.

Options

A. 0 .01 cm
B. 0.02 cm
C. 0.05 cm
D. 0.005 cm

Explanation

To find the least count of the vernier calipers, we can follow these steps:

### Step 1: Determine the value of one main scale division (MSD)
Since `1 cm` on the main scale is divided into `10 equal parts`, we can calculate:
\[ \text{1 MSD} = \frac{1 \text{ cm}}{10} = \frac{10 \text{ mm}}{10} = 1 \text{ mm} \]

### Step 2: Relate the vernier scale divisions to the main scale divisions
According to the problem, `10 divisions of the vernier scale coincide with 8 divisions of the main scale`. Therefore, we can express this relationship as:
\[ 10 \text{ Vernier Scale Divisions (VSD)} = 8 \text{ Main Scale Divisions (MSD)} \]

### Step 3: Calculate the value of one vernier scale division (VSD)
From the relationship established in Step 2, we can find the value of one vernier scale division:
\[ 1 \text{ VSD} = \frac{8 \text{ MSD}}{10} = \frac{8}{10} \text{ MSD} = \frac{8}{10} \times 1 \text{ mm} = 0.8 \text{ mm} \]

### Step 4: Calculate the least count (LC)
The least count of the vernier calipers is given by the formula:
\[ \text{Least Count (LC)} = \text{1 MSD} - \text{1 VSD} \]
Substituting the values we found:
\[ \text{LC} = 1 \text{ mm} - 0.8 \text{ mm} = 0.2 \text{ mm} \]

### Final Answer
Thus, the least count of the vernier calipers is:
\[ \text{Least Count} = 0.2 \text{ mm} \]

---