If in a Vernier callipers 10 VSD coincides with 8 MSD, then the least count of Vernier calliper is [given 1 MSD = 1mm]

Options

A. `1xx10^(-4)m`
B. `2xx10^(-4)m`
C. `1xx10^(-3)m`
D. `8xx10^(-4)m`

Answer

To find the least count of a Vernier caliper given that 10 Vernier Scale Divisions (VSD) coincide with 8 Main Scale Divisions (MSD), we can follow these steps:

### Step 1: Understand the relationship between VSD and MSD
We know from the problem that:
- 10 VSD = 8 MSD

### Step 2: Calculate the value of 1 VSD
To find the value of one Vernier Scale Division (VSD), we can express it in terms of Main Scale Division (MSD):
\[ 1 \text{ VSD} = \frac{8 \text{ MSD}}{10} = 0.8 \text{ MSD} \]
Since we are given that 1 MSD = 1 mm, we can substitute this value:
\[ 1 \text{ VSD} = 0.8 \text{ mm} \]

### Step 3: Calculate the least count of the Vernier caliper
The least count (LC) of the Vernier caliper is given by the formula:
\[ \text{Least Count} = \text{Value of 1 MSD} - \text{Value of 1 VSD} \]
Substituting the values we have:
\[ \text{Least Count} = 1 \text{ mm} - 0.8 \text{ mm} = 0.2 \text{ mm} \]

### Step 4: Convert the least count to meters (if required)
To express the least count in meters:
\[ 0.2 \text{ mm} = 0.2 \times 10^{-3} \text{ m} = 2 \times 10^{-4} \text{ m} \]

### Final Answer
Thus, the least count of the Vernier caliper is:
\[ \text{Least Count} = 0.2 \text{ mm} \text{ or } 2 \times 10^{-4} \text{ m} \]

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