Circular scale of a screw gauge moves through 4 divisions of main scale in one rotation. If the number of divisions on the circular scale is 200 and each division of the main scale is 1 mm., the least count of the screw gauge is

Options

A. 0.01 mm
B. 0.02 mm
C. 0.03 mm
D. 0.04 mm

Answer

To find the least count of the screw gauge, we can follow these steps:

### Step 1: Understand the terms
- **Pitch**: The distance moved by the screw gauge in one complete rotation.
- **Least Count**: The smallest measurement that can be accurately read on the screw gauge.

### Step 2: Calculate the Pitch
The pitch can be calculated using the formula:
\[ \text{Pitch} = \frac{\text{Total movement in main scale divisions}}{\text{Number of rotations}} \]

From the problem, we know:
- The circular scale moves through 4 divisions of the main scale in one rotation.
- Each division of the main scale is 1 mm.

Thus, the pitch is:
\[ \text{Pitch} = 4 \text{ mm (in one rotation)} \]

### Step 3: Calculate the Least Count
The least count can be calculated using the formula:
\[ \text{Least Count} = \frac{\text{Pitch}}{\text{Number of divisions on the circular scale}} \]

From the problem:
- The number of divisions on the circular scale is 200.

Now substituting the values:
\[ \text{Least Count} = \frac{4 \text{ mm}}{200} = 0.02 \text{ mm} \]

### Conclusion
The least count of the screw gauge is:
\[ \text{Least Count} = 0.02 \text{ mm} \]

### Final Answer
The least count of the screw gauge is **0.02 mm**.

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