Assertion $$A$$ : If in five complete rotations of the circular scale, the distance travelled on the main scale of the screw gauge is 5 mm and there are 50 total divisions on a circular scale, then the least count is 0.001 cm.
Reason $$R$$ : Least Count = $$\frac{\text{Pitch}}{\text{Total divisions on circular scale}}$$
In the light of the above statements, choose the most appropriate answer from the options given below.

We recall the basic definitions for a screw gauge.

First, the pitch is the distance moved by the spindle on the main scale in one complete rotation of the circular scale. Algebraically,

$$\text{Pitch}=\frac{\text{Distance travelled on main scale}}{\text{Number of complete rotations}}.$$

In the given situation, the spindle advances $$5\ \text{mm}$$ on the main scale when the circular scale is rotated completely five times. Substituting these data, we have

$$\text{Pitch}=\frac{5\ \text{mm}}{5}=1\ \text{mm}.$$

Next, the least count (L.C.) of the screw gauge is obtained from the relation

$$\text{Least Count}=\frac{\text{Pitch}}{\text{Total divisions on the circular scale}}.$$

This formula is exactly what the Reason $$R$$ states.

We are told that the circular scale possesses $$50$$ total divisions. Substituting the numerical values,

$$\text{L.C.}=\frac{1\ \text{mm}}{50}=0.02\ \text{mm}.$$

To express this least count in centimetres we use the conversion $$1\ \text{mm}=0.1\ \text{cm}.$$ Hence,

$$0.02\ \text{mm}=0.02\times0.1\ \text{cm}=0.002\ \text{cm}.$$

So the calculated least count equals $$0.002\ \text{cm}$$ and not $$0.001\ \text{cm}$$ as claimed in Assertion $$A$$.

Therefore, Assertion $$A$$ is incorrect, whereas Reason $$R$$ is a correct and relevant statement.

Hence, the correct answer is Option A.