The pitch of a screw gauge is `1 mm` and there are `100` divisions on its circular scale. When nothing is put in between its jaws, the zero of the circular scale lies `6` divisions below the reference line. When a wire a placed between the jaws, `2` linear scale divisions are clearly visible while ` 62` divisions on circular scale coincide with the reference line. Determine the diameter of the wire.

Explanation

To determine 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 the screw gauge can be calculated using the formula:
\[ \text{Least Count} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} \]
Given:
- Pitch = 1 mm
- Number of divisions on circular scale = 100

Substituting the values:
\[ \text{Least Count} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm} \]

### Step 2: Determine the Positive Error in the Instrument
The zero of the circular scale lies 6 divisions below the reference line, indicating a positive error. The error (e) can be calculated as:
\[ e = \text{Number of divisions below reference line} \times \text{Least Count} \]
Substituting the values:
\[ e = 6 \times 0.01 \text{ mm} = 0.06 \text{ mm} \]

### Step 3: Calculate the Observed Reading
The observed reading (OR) is given by the formula:
\[ \text{Observed Reading} = \text{Main Scale Reading} + (\text{Circular Scale Reading} \times \text{Least Count}) \]
From the problem:
- Main Scale Reading (MSR) = 2 divisions = 2 mm
- Circular Scale Reading (CSR) = 62 divisions

Substituting the values:
\[ \text{Observed Reading} = 2 \text{ mm} + (62 \times 0.01 \text{ mm}) = 2 \text{ mm} + 0.62 \text{ mm} = 2.62 \text{ mm} \]

### Step 4: Calculate the True Value
The true value (TV) is calculated by subtracting the error from the observed reading:
\[ \text{True Value} = \text{Observed Reading} - \text{Error} \]
Substituting the values:
\[ \text{True Value} = 2.62 \text{ mm} - 0.06 \text{ mm} = 2.56 \text{ mm} \]

### Conclusion
The diameter of the wire is:
\[ \text{Diameter of the wire} = 2.56 \text{ mm} \]

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