Two full turns of the circular scale of a screw gauge cover a distance of 1mm on its main scale. The total number of divisions on the circular scale is 50. Further, it is found that the screw gauge has a zero error of - 0.03 mm. While measuring the diameter of a thin wire, a student notes the main scale reading of 3 mm and the number of circular scale divisions in line with the main scale is 35. The diameter of the wire is

The correct option is 3.38 mm.

To find the diameter of the thin wire, we first need to determine the pitch and the least count of the screw gauge.

Step 1: Find the pitch of the screw gauge
The pitch is defined as the distance moved by the screw per complete rotation of the circular scale. We are given that 2 full turns cover a distance of 1 mm on the main scale. Therefore, the pitch (p) is:
$p=\frac{1\mathrm{mm}}{2}=0.5\mathrm{mm}$

Step 2: Find the least count of the screw gauge
The least count (LC) is the pitch divided by the total number of divisions on the circular scale (N). Here, N = 50:
$\mathrm{LC}=\frac{p}{N}=\frac{0.5\mathrm{mm}}{50}=0.01\mathrm{mm}$

Step 3: Calculate the observed reading
The main scale reading (MSR) is 3 mm.
The circular scale reading (CSR) corresponds to the 35th division. The circular scale value is:
$\mathrm{CSR}=35\times \mathrm{LC}=35\times 0.01\mathrm{mm}=0.35\mathrm{mm}$
Thus, the observed (measured) reading is:
$\mathrm{Observed\; Reading}=\mathrm{MSR}+\mathrm{CSR}=3\mathrm{mm}+0.35\mathrm{mm}=3.35\mathrm{mm}$

Step 4: Apply the zero correction to find the true diameter
The screw gauge has a zero error of -0.03 mm. The formula for the correct reading is:
$\mathrm{True\; Reading}=\mathrm{Observed\; Reading}-\mathrm{Zero\; Error}$
Substituting the values:
$\mathrm{True\; Diameter}=3.35\mathrm{mm}-(-0.03\mathrm{mm})=3.35\mathrm{mm}+0.03\mathrm{mm}=3.38\mathrm{mm}$

Therefore, the diameter of the wire is 3.38 mm.