One main scale division of a vernier callipers is $$a$$ cm and $$n^{th}$$ division of the vernier scale coincide with $$(n-1)^{th}$$ division of the main scale. The least count of the callipers in mm is:

We are given that one main scale division (MSD) of the vernier callipers is $$a$$ cm, and that $$n$$ divisions of the vernier scale coincide with $$(n-1)$$ divisions of the main scale.

From the given condition: $$n \times \text{VSD} = (n-1) \times \text{MSD}$$, which gives $$\text{VSD} = \frac{(n-1)}{n} \times a$$ cm.

The least count is $$\text{LC} = 1 \text{ MSD} - 1 \text{ VSD} = a - \frac{(n-1)}{n} \, a = a\left(1 - \frac{n-1}{n}\right) = a \times \frac{1}{n} = \frac{a}{n}$$ cm.

Converting to millimeters: $$\text{LC} = \frac{a}{n} \times 10 = \frac{10a}{n}$$ mm.