Two thin wires, Wire-1 of diameter 0.650 mm and Wire-2 of unknown diameter d are given.

To obtain the value of d, the diameters of the two wires are measured with a screw gauge. The screw gauge has a pitch of 0.5 mm and there are 100 divisions on the circular scale (CS). The smallest division on the linear scale (LS) is 0.5 mm. The table shows the readings of LS and CS for the measurements. The value of d (in µm) is:

	Readings
	LS (mm)	CS
Wire-1	0.5	42
Wire-2	1.5	95

The correct answer is 1915.

Step 1: Understand the least count of the screw gauge.

The pitch of the screw gauge is given as 0.5 mm, and the circular scale has 100 divisions.

The Least Count (LC) of a screw gauge is calculated as:

$\text{Least Count (LC)}=\frac{\text{Pitch}}{\text{Total Circular Scale Divisions}}$

$\text{LC}=\frac{0.5\text{mm}}{100}=0.005\text{mm}$

Step 2: Determine the zero error of the screw gauge.

The measured diameter using a screw gauge is given by the formula:

$\text{Measured Diameter}=\text{Linear Scale Reading (LS)}+(\text{Circular Scale Reading (CS)}\times \text{LC})-\text{Zero Error (ZE)}$

For Wire-1:

Given actual diameter of Wire-1, $d_1=0.650\text{mm}$.

From the table, for Wire-1: $\text{LS}=0.5\text{mm}$ and $\text{CS}=42$.

Observed Reading for Wire-1:

${\text{Reading}}_1=0.5\text{mm}+(42\times 0.005\text{mm})=0.5\text{mm}+0.210\text{mm}=0.710\text{mm}$

Now, accounting for Zero Error (ZE):

$d_1={\text{Reading}}_1-\text{ZE}$

$0.650\text{mm}=0.710\text{mm}-\text{ZE}$

$\text{ZE}=0.710\text{mm}-0.650\text{mm}=+0.060\text{mm}$

Step 3: Calculate the diameter of Wire-2 ($d$).

From the table, for Wire-2: $\text{LS}=1.5\text{mm}$ and $\text{CS}=95$.

Observed Reading for Wire-2:

${\text{Reading}}_2=1.5\text{mm}+(95\times 0.005\text{mm})=1.5\text{mm}+0.475\text{mm}=1.975\text{mm}$

Applying the Zero Error correction to find actual diameter $d$:

$d={\text{Reading}}_2-\text{ZE}$

$d=1.975\text{mm}-0.060\text{mm}=1.915\text{mm}$

Step 4: Convert the diameter to micrometers (μm).

Since $1\text{mm}=1000\text{μm}$:

$d=1.915\times 1000\text{μm}=1915\text{μm}$

Thus, the value of $d$ in μm is 1915.