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 readings are listed below.

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

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

The distance between successive main-scale divisions is given as $$0.5\text{ mm}$$.
For one complete rotation of the circular scale the sleeve advances by two such divisions, so the pitch (advance per rotation) is

$$\text{Pitch}=2\times 0.5\text{ mm}=1.0\text{ mm}$$

The circular scale has $$100$$ equal divisions; therefore the least count (smallest readable length) is

$$\text{Least count (LC)}=\frac{\text{Pitch}}{\text{No. of circular divisions}}=\frac{1.0\text{ mm}}{100}=0.01\text{ mm}$$

Zero reading (with jaws in contact)
Main-scale reading = $$0$$ division   ( = $$0\text{ mm}$$ )
Circular-scale reading = $$4$$ divisions

Observed length with no wire = $$4 \times \text{LC}=4\times 0.01=0.04\text{ mm}$$

Since the screw gauge reads $$0.04\text{ mm}$$ when the true separation is $$0$$, it possesses a positive zero error of $$+0.04\text{ mm}$$.
Hence the zero correction (to be applied to every reading) is

$$\text{Zero correction}=-0.04\text{ mm}$$

Attempt 1 (with wire)
Main-scale reading = $$4$$ divisions $$\Rightarrow 4\times 0.5 =2.00\text{ mm}$$
Circular-scale reading = $$20$$ divisions $$\Rightarrow 20\times 0.01 =0.20\text{ mm}$$

Observed diameter $$d_1 = 2.00+0.20 = 2.20\text{ mm}$$
Corrected diameter $$d_1^{\prime}=2.20-0.04 = 2.16\text{ mm}$$

Attempt 2 (with wire)
Main-scale reading = $$4$$ divisions $$\Rightarrow 2.00\text{ mm}$$
Circular-scale reading = $$16$$ divisions $$\Rightarrow 0.16\text{ mm}$$

Observed diameter $$d_2 = 2.00+0.16 = 2.16\text{ mm}$$
Corrected diameter $$d_2^{\prime}=2.16-0.04 = 2.12\text{ mm}$$

Mean diameter

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

Uncertainty in diameter
The two corrected readings differ by $$0.04\text{ mm}$$, so the spread about the mean is $$\pm 0.02\text{ mm}$$. Therefore

$$d = 2.14 \pm 0.02 \text{ mm}$$

Cross-sectional area
For a circular wire $$A=\dfrac{\pi d^{2}}{4}$$.

$$A=\frac{\pi (2.14\text{ mm})^{2}}{4} =\pi \times 1.1449\text{ mm}^2 \;\approx\; \pi(1.14\text{ mm}^2)$$

Uncertainty in area
Fractional error: $$\dfrac{\Delta d}{d}= \dfrac{0.02}{2.14}=0.00935$$.
Since $$A\propto d^{2}$$, fractional error in area is twice this value:

$$\frac{\Delta A}{A}=2\times 0.00935=0.0187$$

Absolute error: $$\Delta A = 0.0187 \times 1.14\text{ mm}^2 \approx 0.02\text{ mm}^2$$

Thus

$$A = \pi\bigl(1.14 \pm 0.02\bigr)\text{ mm}^2$$

Therefore, the measured values are

Diameter = $$2.14 \pm 0.02\text{ mm}$$,   Area = $$\pi(1.14 \pm 0.02)\text{ mm}^2$$

Hence the correct choice is
Option C which is: 2.14 ± 0.02 mm, $$\pi$$(1.14 ± 0.02) mm$$^2$$.