A screw gauge with a pitch of `0.5mm` and a circular scale with `50` divisions is used to measure the thicknes of a thin sheet of Aluminium. Before starting the measurement, it is found that wen the jaws of the screw gauge are brought in cintact, the `45^(th)` division coincide with the main scale line and the zero of the main scale is barely visible. what is the thickness of the sheet if the main scale readind is `0.5 mm` and the `25th` division coincide with the main scale line?

Options

A. (a) `0.70mm`
B. (b) `0.50mm`
C. ( c ) `0.75mm`
D. (d) `0.80mm`

Explanation

To solve the problem step by step, we will follow the necessary calculations and logical deductions based on the information provided.

### 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

Substituting the values:
\[ \text{Least Count} = \frac{0.5 \text{ mm}}{50} = 0.01 \text{ mm} \]

### Step 2: Identify the Zero Error
From the problem, we know that when the jaws of the screw gauge are brought into contact, the 45th division of the circular scale coincides with the main scale line. This indicates a zero error.

The zero error can be calculated as:
\[ \text{Zero Error} = \text{Division coinciding} \times \text{Least Count} = 45 \times 0.01 \text{ mm} = 0.45 \text{ mm} \]
Since the zero of the main scale is barely visible, this indicates a negative zero error:
\[ \text{Zero Error} = -0.45 \text{ mm} \]

### Step 3: Calculate the Main Scale Reading (MSR)
The main scale reading (MSR) is given as:
\[ \text{MSR} = 0.5 \text{ mm} \]

### Step 4: Calculate the Circular Scale Reading (CSR)
The problem states that the 25th division of the circular scale coincides with the main scale line. Therefore, the circular scale reading can be calculated as:
\[ \text{CSR} = \text{Division coinciding} \times \text{Least Count} = 25 \times 0.01 \text{ mm} = 0.25 \text{ mm} \]

### Step 5: Calculate the Final Reading
The final reading can be calculated using the formula:
\[ \text{Final Reading} = \text{MSR} + \text{CSR} - \text{Zero Error} \]
Substituting the values:
\[ \text{Final Reading} = 0.5 \text{ mm} + 0.25 \text{ mm} - (-0.45 \text{ mm}) \]
\[ \text{Final Reading} = 0.5 \text{ mm} + 0.25 \text{ mm} + 0.45 \text{ mm} = 1.20 \text{ mm} \]

### Conclusion
The thickness of the sheet of Aluminium is:
\[ \text{Thickness} = 1.20 \text{ mm} \]