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The main scale of a vernier calliper has nn divisions/cm. nn divisions of the vernier scale coincide with (n−1)(n-1) divisions of the main scale. The least count of the vernier calliper is:

A

1(n+1)(n−1) cm\frac{1}{(n+1)(n-1)}\text{ cm}

B

1n cm\frac{1}{n}\text{ cm}

C

1n2 cm\frac{1}{n^2}\text{ cm}

D

1n(n+1) cm\frac{1}{n(n+1)}\text{ cm}

Step-by-Step Solution

  1. Find the value of 1 Main Scale Division (MSD): Since there are nn divisions per cm on the main scale, the length of 1 MSD=1n cm1 \text{ MSD} = \frac{1}{n}\text{ cm}.
  2. Find the relation between MSD and VSD: It is given that nn divisions of the vernier scale (VSD) coincide with (n−1)(n-1) divisions of the main scale (MSD). Therefore, n VSD=(n−1) MSDn \text{ VSD} = (n-1) \text{ MSD}, which implies 1 VSD=n−1n MSD1 \text{ VSD} = \frac{n-1}{n} \text{ MSD}.
  3. Calculate the Least Count (LC): The least count of a vernier calliper is defined as the difference between one main scale division and one vernier scale division. LC=1 MSD−1 VSD\text{LC} = 1 \text{ MSD} - 1 \text{ VSD} LC=1 MSD−n−1n MSD=(1−n−1n) MSD=1n MSD\text{LC} = 1 \text{ MSD} - \frac{n-1}{n} \text{ MSD} = \left(1 - \frac{n-1}{n}\right) \text{ MSD} = \frac{1}{n} \text{ MSD}.
  4. Substitute the value of 1 MSD: LC=1n×(1n cm)=1n2 cm\text{LC} = \frac{1}{n} \times \left(\frac{1}{n}\text{ cm}\right) = \frac{1}{n^2}\text{ cm}.
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