In a screw gauge when the circular scale is given five complete rotations it moves linearly by 2.5 mm. If the circular scale has 100 divisions, the least count of screw gauge is _____ mm.

The linear distance moved by the screw in one complete rotation is called the pitch.

Five complete rotations produce a linear advance of $$2.5 \text{ mm}$$.
Therefore, pitch $$= \frac{2.5 \text{ mm}}{5} = 0.5 \text{ mm}$$.

The least count (LC) is defined as
$$LC = \frac{\text{pitch}}{\text{number of divisions on the circular scale}}$$.

Here the circular scale has $$100$$ divisions, so
$$LC = \frac{0.5 \text{ mm}}{100} = 0.005 \text{ mm}$$.

Writing $$0.005$$ in scientific notation:
$$0.005 \text{ mm} = 5 \times 10^{-3} \text{ mm}$$.

Hence, the least count of the screw gauge is $$5 \times 10^{-3} \text{ mm}$$.
Option D which is: $$5 \times 10^{-3}$$.