Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is $0.5\,\text{mm}$. 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: without wire (two arms touching) main scale $0$ division, circular scale $4$ divisions; Attempt-1 with wire main scale $4$ divisions, circular scale $20$ divisions; Attempt-2 with wire main scale $4$ divisions, circular scale $16$ divisions. What are the diameter and cross-sectional area of the wire measured using the screw gauge?

Step 1: Calculate the pitch and least count of the screw gauge.

The pitch of the main scale is $0.5\,\text{mm}$. Since one full rotation of the circular scale shifts the main scale by two divisions, the screw pitch is:

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

The circular scale has $N = 100$ divisions. The least count ($\text{LC}$) is:

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

Step 2: Determine the zero error.

With no wire, the main scale reading is $0$ and the circular scale reading is $4$ divisions.

$$e = 0 + 4 \times 0.01\,\text{mm} = +0.04\,\text{mm}$$

Step 3: Calculate the corrected diameter for Attempt-1.

The observed reading for Attempt-1 is:

$$\text{Observed}_1 = 4 \times 0.5\,\text{mm} + 20 \times 0.01\,\text{mm} = 2.20\,\text{mm}$$

Subtracting the zero error gives the corrected diameter $d_1$:

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

Step 4: Calculate the corrected diameter for Attempt-2.

The observed reading for Attempt-2 is:

$$\text{Observed}_2 = 4 \times 0.5\,\text{mm} + 16 \times 0.01\,\text{mm} = 2.16\,\text{mm}$$

Subtracting the zero error gives the corrected diameter $d_2$:

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

Step 5: Find the mean diameter and its absolute error.

The mean diameter $d$ is:

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

The absolute error in diameter $\Delta d$ is:

$$\Delta d = \frac{|d_1 - d_2|}{2} = \frac{0.04\,\text{mm}}{2} = 0.02\,\text{mm}$$

Thus, the measured diameter is:

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

Step 6: Calculate the cross-sectional area and its absolute error.

The cross-sectional area $A$ is:

$$A = \pi \left(\frac{d}{2}\right)^2 = \pi (1.07\,\text{mm})^2 = \pi (1.14)\,\text{mm}^2$$

Using the fractional error relation for $A \propto d^2$:

$$\frac{\Delta A}{A} = 2\frac{\Delta d}{d}$$ $$\Delta A = 2 \times \frac{0.02}{2.14} \times \pi (1.14)\,\text{mm}^2 = \pi (0.02)\,\text{mm}^2$$

Thus, the measured cross-sectional area is:

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

Answer: Diameter $d = 2.14 \pm 0.02\,\text{mm}$ and cross-sectional area $A = \pi(1.14 \pm 0.02)\,\text{mm}^2$.