In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is:

We begin by recalling the basic relation for a screw gauge: the pitch is the linear distance travelled by the screw in one complete rotation.

We are told that $$5$$ complete rotations shift the screw by $$0.25\ \text{cm}.$$ Hence, the pitch is obtained by simple division:

$$\text{Pitch} \;=\; \frac{0.25\ \text{cm}}{5} \;=\; 0.05\ \text{cm}.$$

Next, the least count (L.C.) of a screw gauge is given by the standard formula

$$\text{Least Count} \;=\; \frac{\text{Pitch}}{\text{Number of circular scale divisions}}.$$

With $$100$$ circular (head) scale divisions, we substitute:

$$\text{L.C.} \;=\; \frac{0.05\ \text{cm}}{100} \;=\; 0.0005\ \text{cm}.$$

The reading for the thickness of the wire has two contributions.

1. Main‐scale reading (MSR): We have a reading of $$4$$ main‐scale divisions. On the main scale, one division equals the pitch $$\bigl(0.05\ \text{cm}\bigr).$$ Hence

$$\text{MSR} \;=\; 4 \times 0.05\ \text{cm} \;=\; 0.20\ \text{cm}.$$

2. Circular‐scale reading (CSR): The screw shows $$30$$ circular divisions. Each circular division corresponds exactly to the least count. Therefore

$$\text{CSR} \;=\; 30 \times 0.0005\ \text{cm} \;=\; 0.015\ \text{cm}.$$

Adding the two contributions gives the total measured thickness of the wire:

$$\text{Total thickness} \;=\; \text{MSR} + \text{CSR}$$ $$= 0.20\ \text{cm} + 0.015\ \text{cm}$$ $$= 0.215\ \text{cm}.$$

Zero error is stated to be negligible, so no further correction is necessary.

Hence, the correct answer is Option D.