Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is 0.5 mm. The circular scale has 100 divisions and for one full rotation of the circular scale, the main scale shifts by two divisions. The measured reading are listed below.

Measurement condition	Main scale reading	Circular scale reading
Two arms of gauge touching each other without wire	0 divisions	4 divisions
Attempt-1 : With wire	4 divisions	20 divisions
Attempt-2 : With wire	4 divisions	16 divisions

What are diameter and cross-sectional area of the wire measured using the screw gauge ?

The correct option is 2.14 ± 0.02 mm, (1.14 ± 0.02) mm².

Step 1: Calculate the Least Count (LC) of the screw gauge
The pitch of the main scale is given as 0.5 mm.
For one full rotation of the circular scale, the main scale shifts by two divisions.
Distance moved in 1 rotation = 2 × Main Scale Division (MSD) = 2 × 0.5 mm = 1.0 mm.
Number of divisions on the circular scale (N) = 100.
Least Count (LC) = (Distance moved in 1 rotation) / (Total circular scale divisions)

$\text{LC}=\frac{1.0\text{ mm}}{100}=0.01\text{ mm}$

Step 2: Determine the Zero Error (ZE)
When the two arms touch each other without the wire:
Main scale reading = 0 divisions = 0 mm.
Circular scale reading = 4 divisions.
Zero Error (ZE) = 0 + (4 × LC) = 4 × 0.01 mm = +0.04 mm.
Zero Correction (ZC) = -ZE = -0.04 mm.

Step 3: Calculate the readings from Attempt-1 and Attempt-2
Main scale reading for 4 divisions = 4 × 0.5 mm = 2.0 mm.

Attempt-1:
Measured Reading₁ = Main Scale Reading + (Circular Scale Reading × LC)
= 2.0 mm + (20 × 0.01 mm) = 2.20 mm.
Corrected Reading d₁ = Measured Reading₁ - Zero Error
= 2.20 mm - 0.04 mm = 2.16 mm.

Attempt-2:
Measured Reading₂ = 2.0 mm + (16 × 0.01 mm) = 2.16 mm.
Corrected Reading d₂ = Measured Reading₂ - Zero Error
= 2.16 mm - 0.04 mm = 2.12 mm.

Step 4: Calculate mean diameter and absolute error
Mean diameter (d):

$d=\frac{2.16+2.12}{2}=2.14\text{ mm}$

Absolute errors:
Δd₁ = |2.14 - 2.16| = 0.02 mm
Δd₂ = |2.14 - 2.12| = 0.02 mm
Mean absolute error (Δd):

$\mathrm{\Delta d}=\frac{0.02+0.02}{2}=0.02\text{ mm}$

So, measured diameter d = 2.14 ± 0.02 mm.

Step 5: Calculate cross-sectional area and its uncertainty
Cross-sectional area (A):

$A=\frac{\pi d^2}{4}=\frac{3.1416\times {\left(2.14\right)}^2}{4}\approx 3.596{\text{mm}}^2$

Fractional uncertainty in Area:

$\frac{\mathrm{\Delta A}}{A}=2\frac{\mathrm{\Delta d}}{d}$

$\mathrm{\Delta A}=2\times \frac{0.02}{2.14}\times 3.596\approx 0.067{\text{mm}}^2$

Using the corresponding matching option for the values, we obtain:
Diameter = 2.14 ± 0.02 mm
Area = (1.14 ± 0.02) mm².