In an experiment to find out the diameter of wire using screw gauge, the following observation were noted:
(a) Screw moves 0.5 mm on main scale in one complete rotation
(b) Total divisions on circular scale = 50
(c) Main scale reading is 2.5 mm
(d) 45th division of circular scale is in the pitch line
(e) Instrument has 0.03 mm negative error
Then the diameter of wire is:

We need to determine the correct diameter of the wire using the observations recorded from the screw gauge experiment.

The total measurement from a screw gauge is calculated using the formula:

$$\text{Total Reading} = \text{Main Scale Reading (MSR)} + \text{Circular Scale Reading (CSR)} - \text{Zero Error}$$

Step 1: Calculate the Least Count (LC) of the Screw Gauge

The least count is the smallest distance the screw can measure accurately, which is defined as the pitch divided by the total number of divisions on the circular scale.

• Pitch: The distance moved on the main scale in one complete rotation = $$0.5\text{ mm}$$.
• Circular scale divisions: $$50$$

$$LC = \frac{\text{Pitch}}{\text{Total Circular Scale Divisions}} = \frac{0.5\text{ mm}}{50} = 0.01\text{ mm}$$

Step 2: Calculate the Circular Scale Reading (CSR)

The circular scale reading is obtained by multiplying the coinciding division line on the pitch axis by the least count.

• Coinciding division: $$45^{\text{th}}\text{ division}$$

$$\text{CSR} = 45 \times LC = 45 \times 0.01\text{ mm} = 0.45\text{ mm}$$

Step 3: Account for the Zero Error

The instrument is given to have a negative error:

• Zero Error: $$-0.03\text{ mm}$$

When computing the true reading, a negative zero error must be subtracted, which mathematically amounts to adding its absolute magnitude to the observed reading.

Step 4: Calculate the Total Diameter of the Wire

Given the Main Scale Reading ($\text{MSR}$) is $$2.5\text{ mm}$$, we substitute all the values into our main formula:

$$\text{Diameter} = \text{MSR} + \text{CSR} - \text{Zero Error}$$

$$\text{Diameter} = 2.5\text{ mm} + 0.45\text{ mm} - (-0.03\text{ mm})$$

$$\text{Diameter} = 2.5\text{ mm} + 0.45\text{ mm} + 0.03\text{ mm} = 2.98\text{ mm}$$

Therefore, the correct diameter of the wire is 2.98 mm, which corresponds to Option C.