The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5 $$\mu$$m diameter of a wire is:

We have a screw gauge whose main scale (linear scale) is exactly like the scale on a ruler engraved along the barrel. The distance through which the screw advances when it completes one full rotation is called the pitch. Here the statement “least count of the main scale is 1 mm” means that one full turn shifts the screw by $$1\;\text{mm}$$, so the pitch is

$$\text{Pitch}=1\;\text{mm}.$$

The overall precision of a screw gauge is governed by its least count (L.C.). The standard formula connecting the least count, the pitch and the total number of equal divisions on the circular scale is first stated:

$$\text{Least count}=\frac{\text{Pitch}}{\text{Number of divisions on circular scale}}.$$

Let the required number of circular-scale divisions be denoted by $$n$$. Substituting the pitch value we obtain

$$\text{L.C.}=\frac{1\;\text{mm}}{n}.$$

According to the problem, we wish to resolve a diameter as small as $$5\;\mu\text{m}$$. This means the least count must be at most $$5\;\mu\text{m}$$. First convert this target precision into millimetres because the pitch is already in millimetres.

We know $$1\;\mu\text{m}=10^{-6}\;\text{m}=10^{-3}\;\text{mm}.$$ Hence

$$5\;\mu\text{m}=5\times10^{-3}\;\text{mm}=0.005\;\text{mm}.$$

Now we equate the least count expression to this value:

$$\frac{1\;\text{mm}}{n}=0.005\;\text{mm}.$$

To solve for $$n$$, we multiply both sides by $$n$$ and then divide both sides by $$0.005\;\text{mm}$$, yielding

$$1\;\text{mm}=0.005\;\text{mm}\times n$$

$$\Rightarrow n=\frac{1}{0.005}.$$

Carrying out the division, we first rewrite the denominator in fractional form:

$$0.005=5\times10^{-3}=\frac{5}{1000}=\frac{1}{200}.$$

Hence

$$n=\frac{1}{0.005}=200.$$

This integer represents the minimum number of equidistant divisions that must be etched on the circular scale so that every turn (or fraction thereof) can measure changes as small as $$5\;\mu\text{m}$$.

Hence, the correct answer is Option D.