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 answer is Option 3: 2.14 ± 0.02 mm, π(1.14 ± 0.02) mm².

We will work through this problem systematically — first finding the Least Count, then applying the zero error correction, then computing the mean diameter and its uncertainty, and finally calculating the cross-sectional area with propagated error.

Step 1: Find the Least Count (LC)

The pitch of the main scale is 0.5 mm. One full rotation of the circular scale moves the main scale by 2 divisions.

So the distance moved per full rotation (also called the screw pitch) is:

\text{Screw Pitch}=2\times 0.5\text{mm}=1\text{mm}

The circular scale has 100 divisions, so:

\text{LC}=\frac{\text{Screw Pitch}}{\text{No. of circular scale divisions}}=\frac{1}{100}\text{mm}=0.01\text{mm}

Step 2: Calculate the Zero Error

When the two arms of the gauge touch each other (without wire), the reading should ideally be zero. But the readings are: Main Scale = 0 divisions, Circular Scale = 4 divisions.

\text{Zero Error}=(0\times 0.5)+(4\times 0.01)=0.04\text{mm (positive zero error)}

Step 3: Calculate Raw Readings for Each Attempt

For each measurement, the total raw reading = (Main Scale reading × 0.5 mm) + (Circular Scale reading × LC).

Attempt 1 — Main Scale: 4 divisions, Circular Scale: 20 divisions:

R_1=(4\times 0.5)+(20\times 0.01)=2.00+0.20=2.20\text{mm}

Attempt 2 — Main Scale: 4 divisions, Circular Scale: 16 divisions:

R_2=(4\times 0.5)+(16\times 0.01)=2.00+0.16=2.16\text{mm}

Step 4: Apply Zero Error Correction

Corrected Reading = Raw Reading - Zero Error

d_1=2.20-0.04=2.16\text{mm}

d_2=2.16-0.04=2.12\text{mm}

Step 5: Find Mean Diameter

\bar{d}=\frac{d_1+d_2}{2}=\frac{2.16+2.12}{2}=\frac{4.28}{2}=2.14\text{mm}

Step 6: Find Mean Absolute Error in Diameter

\mathrm{|}{d}_1-\bar{d}\mathrm{|}=\mathrm{|2.16}-\mathrm{2.14|}=0.02\text{mm}

\mathrm{|}{d}_2-\bar{d}\mathrm{|}=\mathrm{|2.12}-\mathrm{2.14|}=0.02\text{mm}

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

Therefore: Diameter = 2.14 ± 0.02 mm

Step 7: Calculate the Cross-Sectional Area

The cross-sectional area of a circular wire is:

A=\pi r^2=\pi {\left(\frac{d}{2}\right)}^2

The radius:

r=\frac{2.14}{2}=1.07\text{mm}

A=\pi \times {1.07}^2=\pi \times 1.1449\approx \pi \times 1.14\text{mm²}

Step 8: Propagate Error into the Area

Since A=\pi r^2, the relative error in A is twice the relative error in r (and since r = d/2, the relative errors of r and d are the same):

\frac{\Delta A}{A}=2\frac{\Delta r}{r}=2\frac{\Delta d}{d}=2\times \frac{0.02}{2.14}\approx 0.01869

\Delta A=\pi \times 1.14\times 0.01869\approx \pi \times 0.0213\approx \pi \times 0.02\text{mm²}

Therefore: Area = π(1.14 ± 0.02) mm²

Final Answer: Diameter = 2.14 ± 0.02 mm and Cross-sectional Area = π(1.14 ± 0.02) mm², which corresponds to Option 3.