The pitch of a screw gauge is 1mm and there are 50 divisions on its cap. When the two studs are firmly in contact, the zero of the circular scale lies 6 divisions below the line of graduation. When a wire is held firmly in between the studs, 3 pitch scale divisions are clearly visible, while `31^(st)` division on the circular scale coincide with the reference line. The diameter of the wire is.

Options

A. 3.62 mm
B. 3.50 mm
C. 3.74 mm
D. 3.50 mm

Answer

To solve the problem of finding the diameter of the wire using the screw gauge, we will follow these steps:

### Step 1: Determine the Least Count of the Screw Gauge
The least count (LC) of the screw gauge can be calculated using the formula:
\[ \text{Least Count} = \frac{\text{Pitch}}{\text{Number of divisions on the circular scale}} \]
Given:
- Pitch = 1 mm
- Number of divisions on the circular scale = 50

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

### Step 2: Identify the Zero Error
The zero error is determined by how much the zero of the circular scale is below the reference line when the two studs are in contact. Given that the zero lies 6 divisions below the line:
\[ \text{Zero Error} = -(\text{Number of divisions below the line}) \times \text{Least Count} \]
Substituting the values:
\[ \text{Zero Error} = -6 \times 0.02 \text{ mm} = -0.12 \text{ mm} \]

### Step 3: Read the Measurement with the Wire
When the wire is held between the studs:
- The number of pitch scale divisions visible = 3
- The circular scale division that coincides with the reference line = 31

The total reading can be calculated as:
\[ \text{Total Reading} = \text{Main Scale Reading} + \text{Circular Scale Reading} \times \text{Least Count} \]
Substituting the values:
\[ \text{Total Reading} = 3 + (31 \times 0.02) \text{ mm} \]
Calculating the circular scale contribution:
\[ 31 \times 0.02 = 0.62 \text{ mm} \]
Thus:
\[ \text{Total Reading} = 3 + 0.62 = 3.62 \text{ mm} \]

### Step 4: Calculate the Corrected Diameter
To find the corrected diameter, we need to account for the zero error:
\[ \text{Corrected Diameter} = \text{Total Reading} - \text{Zero Error} \]
Substituting the values:
\[ \text{Corrected Diameter} = 3.62 \text{ mm} - (-0.12 \text{ mm}) = 3.62 + 0.12 = 3.74 \text{ mm} \]

### Final Answer
The diameter of the wire is:
\[ \text{Diameter} = 3.74 \text{ mm} \]