A travelling microscope has 20 divisions per cm on the main scale while its Vernier scale has total 50 divisions and 25 Vernier scale divisions are equal to 24 main scale divisions, what is the least count of the travelling microscope?

We have a travelling microscope whose main scale has 20 divisions per cm. This means each main scale division (MSD) is $$\dfrac{1}{20}$$ cm $$= 0.05$$ cm.

We are told that 25 Vernier scale divisions (VSD) coincide with 24 main scale divisions. Therefore, $$25 \text{ VSD} = 24 \text{ MSD}$$, which gives $$1 \text{ VSD} = \dfrac{24}{25} \text{ MSD}$$.

The least count (LC) of a Vernier instrument is the difference between one main scale division and one Vernier scale division: $$\text{LC} = 1\text{ MSD} - 1\text{ VSD} = 1\text{ MSD} - \dfrac{24}{25}\text{ MSD} = \dfrac{1}{25}\text{ MSD}$$.

Now substituting the value of one MSD: $$\text{LC} = \dfrac{1}{25} \times 0.05 \text{ cm} = \dfrac{0.05}{25} = 0.002 \text{ cm}$$.

Hence, the correct answer is Option C.