The pitch of a screw gauge is 0.5 mm and there are 100 divisions on its circular scale. The instrument reads + 2 divisions, when no object is placed between the studs which are firmly in contact. In measuring the diameter of wire, there are 4 divisions on the pitch scale and `83^(rd)` division of the of the circular scale coincides with the base line / reference line. Then the diameter of the wire is

Options

A. 4.05 mm
B. 4.405 mm
C. 4.415 mm
D. 4.425 mm

Answer

To solve the problem, 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 = 0.5 mm
- Number of divisions on circular scale = 100

Calculating the least count:
\[ \text{Least Count} = \frac{0.5 \text{ mm}}{100} = 0.005 \text{ mm} \]

### Step 2: Calculate the Error
The error in the reading is given as +2 divisions. To find the actual error in millimeters, we multiply the number of divisions by the least count:
\[ \text{Error} = \text{Number of divisions in error} \times \text{Least Count} \]
Calculating the error:
\[ \text{Error} = 2 \times 0.005 \text{ mm} = 0.01 \text{ mm} \]

### Step 3: Read the Measurements
When measuring the diameter of the wire, we have:
- Main scale reading (MSR) = 4 divisions (which corresponds to 4 mm since each division is 1 mm)
- Circular scale reading (CSR) = 83rd division

### Step 4: Calculate the Total Reading
The total reading can be calculated using the formula:
\[ \text{Total Reading} = \text{Main Scale Reading} + (\text{Circular Scale Reading} \times \text{Least Count}) - \text{Error} \]
Substituting the values:
\[ \text{Total Reading} = 4 \text{ mm} + (83 \times 0.005 \text{ mm}) - 0.01 \text{ mm} \]
Calculating the circular scale contribution:
\[ 83 \times 0.005 \text{ mm} = 0.415 \text{ mm} \]
Now substituting this back into the total reading:
\[ \text{Total Reading} = 4 \text{ mm} + 0.415 \text{ mm} - 0.01 \text{ mm} \]
\[ \text{Total Reading} = 4.415 \text{ mm} - 0.01 \text{ mm} = 4.405 \text{ mm} \]

### Final Answer
The diameter of the wire is:
\[ \text{Diameter of the wire} = 4.405 \text{ mm} \]

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