A screw gauge with a pitch of 0.5 mm and a circular scale with 50 division is used to measure the thickness of a thin sheet of aluminium. Before starting the measurement, it is found that when the two jaws of the screw are brought in constant, 45th division coincides with the main scale line and that the zero of main scale line and that the zero of main scale is barely visible. What is the thickness of the sheet, if the main scale reading is 0.5 mm and 25th division coincides with the main scalel line ?

Options

A. 0.75 mm
B. 0.80mm
C. 0.70 mm
D. 0.50mm

Explanation

To find the thickness of the thin sheet of aluminum using the screw gauge, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Pitch of the screw gauge (P) = 0.5 mm
- Number of divisions on the circular scale (N) = 50
- Main scale reading (MSR) = 0.5 mm
- Coinciding division on the circular scale (C) = 25th division
- Coinciding division when jaws are in contact = 45th division

2. **Calculate the Least Count (LC):**
The least count of the screw gauge is given by the formula:
\[ LC = \frac{P}{N} = \frac{0.5 \text{ mm}}{50} = 0.01 \text{ mm} \]

3. **Calculate the Zero Error:**
The zero error occurs when the jaws of the screw gauge are in contact. The 45th division coincides with the main scale line, which means:
\[ \text{Zero Error} = \text{(Coinciding division - 50)} \times LC = (45 - 50) \times 0.01 \text{ mm} = -0.05 \text{ mm} \]
Since the zero error is negative, it indicates that the reading is less than the actual value.

4. **Calculate the Zero Correction:**
The zero correction is the negative of the zero error:
\[ \text{Zero Correction} = -(\text{Zero Error}) = -(-0.05 \text{ mm}) = 0.05 \text{ mm} \]

5. **Calculate the Reading:**
The reading of the screw gauge is calculated as:
\[ \text{Apparent Reading} = \text{MSR} + (C \times LC) = 0.5 \text{ mm} + (25 \times 0.01 \text{ mm}) = 0.5 \text{ mm} + 0.25 \text{ mm} = 0.75 \text{ mm} \]

6. **Calculate the Actual Reading:**
To find the actual thickness of the aluminum sheet, we add the zero correction to the apparent reading:
\[ \text{Actual Reading} = \text{Apparent Reading} + \text{Zero Correction} = 0.75 \text{ mm} + 0.05 \text{ mm} = 0.80 \text{ mm} \]

### Final Answer:
The thickness of the thin sheet of aluminum is **0.80 mm**.

---