The pitch of a screw gauge having `50` divisions on its circular scale is `1 mm` When the two jaws of the screw gauge are in contact with each other, the zero of the circular scale lies `6` divisions below the line of gradution. when a wire is placed between the jaws, `3` linear scale divisions are clearly visible while `31` division on the circular scale coincides with the reference line. Find diameter of the wire.

Explanation

To find the diameter of the wire using the screw gauge, we will follow these steps:

### Step 1: Calculate the Least Count of the Screw Gauge
The least count (LC) of a screw gauge is calculated using the formula:
\[ \text{Least Count} = \frac{\text{Pitch}}{\text{Number of Circular Scale Divisions}} \]
Given:
- Pitch = 1 mm
- Number of Circular Scale Divisions = 50

Substituting the values:
\[ \text{Least Count} = \frac{1 \text{ mm}}{50} = 0.02 \text{ mm} \]

### Step 2: Determine the Error of the Instrument
The error is determined based on the position of the zero of the circular scale when the jaws are in contact. Since the zero lies 6 divisions below the line of graduation, the error is positive.

Calculating the error:
\[ \text{Error} = +6 \times \text{Least Count} = +6 \times 0.02 \text{ mm} = +0.12 \text{ mm} \]

### Step 3: Calculate the Observed Reading
When the wire is placed between the jaws, the observed reading is calculated from the linear scale and the circular scale.

Given:
- Linear Scale Divisions visible = 3
- Circular Scale Division coinciding = 31

Calculating the observed reading:
\[ \text{Observed Reading} = \text{Linear Scale Reading} + \text{Circular Scale Reading} \]
Where:
\[ \text{Circular Scale Reading} = 31 \times \text{Least Count} = 31 \times 0.02 \text{ mm} = 0.62 \text{ mm} \]
Thus,
\[ \text{Observed Reading} = 3 \text{ mm} + 0.62 \text{ mm} = 3.62 \text{ mm} \]

### Step 4: Calculate the True Reading
The true reading is calculated by subtracting the error from the observed reading:
\[ \text{True Reading} = \text{Observed Reading} - \text{Error} \]
Substituting the values:
\[ \text{True Reading} = 3.62 \text{ mm} - 0.12 \text{ mm} = 3.50 \text{ mm} \]

### Step 5: Conclusion
The diameter of the wire is:
\[ \text{Diameter of the wire} = 3.50 \text{ mm} \]