There are 100 divisions on the circular scale of a screw gauge of pitch 1 mm. With no measuring quantity in between the jaws, the zero of the circular scale lies 5 divisions below the reference line. The diameter of a wire is then measured using this screw gauge. It is found that 4 linear scale divisions are clearly visible while 60 divisions on circular scale coincide with the reference line. The diameter of the wire is :

We need to find the diameter of a wire measured using a screw gauge that has a positive zero error. Since the least count of a screw gauge is given by $$\text{LC} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm}$$, we have determined the smallest measurable length increment.

The problem states that with no measuring quantity between the jaws, the zero of the circular scale lies 5 divisions below the reference line. From this we infer a positive zero error, which is calculated as $$\text{Zero error} = +5 \times \text{LC} = +5 \times 0.01 = +0.05 \text{ mm}$$. Since a positive zero error indicates that the instrument reads more than the actual value, we must subtract it from the measured reading.

Next, the observed reading consists of a main scale reading and a circular scale reading. For the main scale reading, 4 linear scale divisions are visible, and since the pitch is 1 mm, each linear division equals 1 mm. Hence $$\text{MSR} = 4 \times 1 = 4 \text{ mm}$$. The circular scale reading corresponds to the 60th division coinciding with the reference line, giving $$\text{CSR} = 60 \times \text{LC} = 60 \times 0.01 = 0.60 \text{ mm}$$. Altogether, the observed reading is $$\text{Observed reading} = \text{MSR} + \text{CSR} = 4 + 0.60 = 4.60 \text{ mm}$$.

Finally, applying the zero error correction yields $$\text{Corrected reading} = \text{Observed reading} - \text{Zero error} = 4.60 - 0.05 = 4.55 \text{ mm}$$. The correct answer is Option (3): 4.55 mm.