The pitch of a screw gauge is `0.5mm` and there are `50` divisions on circular scale. When there is nothing between the two ends (studs) of screw gauge, `45th` division of circular scale is coincide with screw guage, and in this situation zero of main scale is not visible. When a wire is place between the studs, the linear scale reads `2` divisions and `20th` divisions of circular scale coincides with references line. For this situation mark the correct statement(s).

Options

A. Least count of the instrument is `0.01mm`
B. Zero correction for the instruement is `+0.45mm`
C. Thickness of wire is `1.65mm`
D. All of the above

Explanation

To solve the problem step by step, we need to analyze the information provided and calculate the necessary values.

### Step 1: Determine the Least Count of the Screw Gauge
The least count (LC) of a screw gauge can be calculated using the formula:
\[ \text{Least Count} = \frac{\text{Pitch}}{\text{Number of divisions on Circular Scale}} \]
Given:
- Pitch = 0.5 mm
- Number of divisions on Circular Scale = 50

Calculating the least count:
\[ \text{Least Count} = \frac{0.5 \text{ mm}}{50} = 0.01 \text{ mm} \]

### Step 2: Identify the Zero Error
When nothing is placed between the studs, the 45th division of the circular scale coincides with the reference line, and the zero of the main scale is not visible. This indicates a positive zero error.

To calculate the zero error:
\[ \text{Zero Error} = \text{Division coinciding} \times \text{Least Count} \]
\[ \text{Zero Error} = 45 \times 0.01 \text{ mm} = 0.45 \text{ mm} \]

### Step 3: Calculate the True Reading with the Wire
When the wire is placed between the studs:
- Main Scale Reading (MSR) = 2 divisions = 2 mm
- Circular Scale Reading (CSR) = 20 divisions

Calculating the circular scale contribution:
\[ \text{Circular Scale Contribution} = \text{CSR} \times \text{Least Count} = 20 \times 0.01 \text{ mm} = 0.20 \text{ mm} \]

### Step 4: Calculate the True Reading
The true reading (TR) can be calculated using the formula:
\[ \text{True Reading} = \text{Main Scale Reading} + \text{Circular Scale Reading} + \text{Zero Correction} \]
Where the zero correction is positive due to the positive zero error:
\[ \text{True Reading} = 2 \text{ mm} + 0.20 \text{ mm} + 0.45 \text{ mm} \]
Calculating:
\[ \text{True Reading} = 2 + 0.20 + 0.45 = 2.65 \text{ mm} \]

### Conclusion
The correct statements based on the calculations are:
1. The least count of the screw gauge is 0.01 mm.
2. The zero correction is +0.45 mm.
3. The thickness of the wire is 2.65 mm.

### Final Answer
All statements regarding the screw gauge measurements are correct.

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