A vernier callipers has 20 divisions on the vernier scale, which coincides with $$19^{th}$$ division on the main scale. The least count of the instrument is $$0.1$$ mm. One main scale division is equal to ______ mm.

We are given a vernier callipers where 20 divisions on the vernier scale coincide with 19 divisions on the main scale, and the least count is 0.1 mm. We need to find the value of one main scale division (MSD).

The given condition is $$20 \text{ VSD} = 19 \text{ MSD}$$ which implies $$1 \text{ VSD} = \frac{19}{20} \text{ MSD}$$.

The least count (LC) is defined as the difference between one main scale division and one vernier scale division: $$\text{LC} = 1 \text{ MSD} - 1 \text{ VSD}$$.

Substituting the expression for VSD from the previous relation gives $$\text{LC} = 1 \text{ MSD} - \frac{19}{20} \text{ MSD} = \text{MSD} \left(1 - \frac{19}{20}\right) = \frac{1}{20} \text{ MSD}$$.

Since the least count is given as $$\text{LC} = 0.1$$ mm, one can write $$0.1 = \frac{1}{20} \times \text{MSD}$$ which yields $$\text{MSD} = 0.1 \times 20 = 2 \text{ mm}$$.

Therefore, one main scale division is equal to 2 mm. The correct answer is Option (2): 2.