The main scale of a vernier callipers has n divisions/cm. n divisions of the vernier scale coincide with (n-1) divisions of main scale. The least count of the vernier callipers is,

Options

A. `(1)/(n(n+1))cm`
B. `(1)/((n+1)(n-1))cm`
C. `(1)/(n)cm`
D. `(1)/(n^2)cm`

Explanation

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

### Step 1: Understand the definitions
- **Main Scale Division (MSD)**: The smallest division on the main scale of the vernier calipers.
- **Vernier Scale Division (VSD)**: The smallest division on the vernier scale.
- **Least Count (LC)**: The smallest measurement that can be taken using the instrument, calculated as:
\[ LC = 1 \text{ MSD} - 1 \text{ VSD} \]

### Step 2: Determine the value of 1 MSD
Given that there are \( n \) divisions per centimeter on the main scale, we can express 1 MSD as:
\[ 1 \text{ MSD} = \frac{1 \text{ cm}}{n} = \frac{10 \text{ mm}}{n} \quad (\text{since } 1 \text{ cm} = 10 \text{ mm}) \]

### Step 3: Determine the value of 1 VSD
From the problem, we know that \( n \) divisions of the vernier scale coincide with \( (n-1) \) divisions of the main scale. Therefore, the length of \( n \) VSD can be expressed as:
\[ n \text{ VSD} = (n-1) \text{ MSD} \]
Thus, we can express 1 VSD as:
\[ 1 \text{ VSD} = \frac{(n-1) \text{ MSD}}{n} = \frac{(n-1) \cdot \frac{10 \text{ mm}}{n}}{n} = \frac{10(n-1)}{n^2} \text{ mm} \]

### Step 4: Calculate the Least Count
Now we can substitute the values of 1 MSD and 1 VSD into the Least Count formula:
\[ LC = 1 \text{ MSD} - 1 \text{ VSD} \]
Substituting the values we found:
\[ LC = \frac{10 \text{ mm}}{n} - \frac{10(n-1)}{n^2} \]
To simplify this expression, we can find a common denominator:
\[ LC = \frac{10n - 10(n-1)}{n^2} = \frac{10n - 10n + 10}{n^2} = \frac{10}{n^2} \text{ mm} \]

### Final Answer
Thus, the least count of the vernier calipers is:
\[ LC = \frac{10}{n^2} \text{ mm} \]

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