In a vernier calliper, when both jaws touch each other, zero of the vernier scale shifts towards left and its $$4^{th}$$ division coincides exactly with a certain division on main scale. If 50 vernier scale divisions equal to 49 main scale divisions and zero error in the instrument is $$0.04 \text{ mm}$$ then how many main scale divisions are there in $$1 \text{ cm}$$?

We need to find how many main scale divisions are in 1 cm for a vernier caliper.

First, the least count LC is defined as one main scale division (MSD) minus one vernier scale division (VSD), and since 50 VSD equal 49 MSD, we have 1 VSD = 49/50 MSD. Moreover, the zero error equals the number of coinciding VSD multiplied by the least count.

Expressing the least count in terms of MSD gives $$\text{LC} = 1\,\text{MSD} - 1\,\text{VSD} = 1\,\text{MSD} - \frac{49}{50}\,\text{MSD} = \frac{1}{50}\,\text{MSD}$$

Since the zero of the vernier has shifted left (indicating a positive zero error) and the fourth division on the vernier coincides with a main scale division, the zero error is four times the least count: $$\text{Zero error} = 4 \times \text{LC} = 4 \times \frac{1}{50}\,\text{MSD} = \frac{4}{50}\,\text{MSD}$$

We are told this zero error measures 0.04 mm, so $$\frac{4}{50}\,\text{MSD} = 0.04\,\text{mm} \implies \text{MSD} = 0.04 \times \frac{50}{4} = 0.04 \times 12.5 = 0.5\,\text{mm}$$

Hence, the number of main scale divisions in 1 cm (10 mm) is $$\frac{10\,\text{mm}}{0.5\,\text{mm}} = 20$$

The correct answer is Option (3): 20.