Topic: Metrology and Inspection
A screw gauge moves 1 mm when two rotations were given. If the circular scale (near thimble) has 100 divisions marked circumferentially, what will be the least count?
- A. 0.001
- B. 0.020
- C. 0.010
- D. 0.005
Answer: D. 0.005
Explanation: Pitch is the distance advanced by the screw in one complete rotation.
Pitch = $\frac{1~mm}{2~rotations} = 0.5~mm$.
Least Count (LC) = $\frac{Pitch}{\text{Number of circular divisions}} = \frac{0.5~mm}{100} = 0.005~mm$.