In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division= 0.05 mm, then the least count of the vernier callipers is__________ mm.

We need to find the least count of the vernier callipers.

50 vernier scale divisions (VSD) = 48 main scale divisions (MSD)

1 MSD = 0.05 mm

Find the value of 1 VSD:

$$1 \text{ VSD} = \frac{48}{50} \text{ MSD} = \frac{48}{50} \times 0.05 = 0.048 \text{ mm}$$

Calculate the least count:

$$\text{Least Count} = 1 \text{ MSD} - 1 \text{ VSD} = 0.05 - 0.048 = 0.002 \text{ mm}$$

Therefore, the least count is Option 3: 0.002 mm.