In a screw gauge when the circular scale is given five complete rotations it moves linearly by 2.5 mm. If the circular scale has 100 divisions, the least count of screw gauge is mm.
- 1 10-2
- 1 10-3
- 5 10-2
- 5 10-3
Answer
(4) 5 10-3
Solution
Pitch = linear distance moved in one complete rotation
⇒ Pitch = 2.5 mm5=0.5 mm
∵ Least count = PitchNumber of circular scale divisions
∴ LC = 0.5 mm100=0.005 mm=5 10-3 mm
Answer: (4)