The pitch of a screw gauge is 0.5 mm and the head scale is divided in 100 parts. What is the least count of screw gauge?

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Solution

Given : Pitch = 0.5 mm
Number of divisions on circular head = 100
Least count $=$ Pitch/total number of divisions on circular head
$= 0.5 m m / 100$
$= 0.005 m m o r 0.0005 c m$

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The pitch of a scrw gauge is 0.5 mm and the head scale is divided in 100 parts. What is the least count of screw gauge ?

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