Marks on main scale of vernier caliper are made at 1mm separation. There are 25 equal divisions on vernier scale and the same matches with 20 divisions on main scale. What is the least cound of given vernier caliper?

Options

A. `0.1 mm`
B. `0.25 mm`
C. `0.15 mm`
D. `0.2mm`

Explanation

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

### Step 1: Identify the value of one main scale division (MSD)
The problem states that the marks on the main scale of the vernier caliper are made at 1 mm separation. Therefore, the value of one main scale division (MSD) is:
\[ \text{1 MSD} = 1 \text{ mm} \]

### Step 2: Determine the value of one vernier scale division (VSD)
It is given that 25 divisions on the vernier scale match with 20 divisions on the main scale. We can express this relationship mathematically:
\[ \text{20 MSD} = \text{25 VSD} \]
To find the value of one vernier scale division (VSD), we can rearrange the equation:
\[ \text{VSD} = \frac{20 \text{ MSD}}{25} \]
Substituting the value of MSD:
\[ \text{VSD} = \frac{20 \times 1 \text{ mm}}{25} = \frac{20}{25} \text{ mm} = 0.8 \text{ mm} \]

### Step 3: Calculate the least count (LC) of the vernier caliper
The least count of the vernier caliper is defined as the difference between one main scale division and one vernier scale division:
\[ \text{Least Count (LC)} = \text{1 MSD} - \text{1 VSD} \]
Substituting the values we found:
\[ \text{LC} = 1 \text{ mm} - 0.8 \text{ mm} = 0.2 \text{ mm} \]

### Conclusion
The least count of the given vernier caliper is:
\[ \text{Least Count} = 0.2 \text{ mm} \]