instrument use (screw gauge)

Question

Pitch of a screw gauge is 0.5 mm and circular scale has 50 divisions. Least count is:

RankGuru · Accepted Answer

The Least Count ($LC$) of a Screw Gauge is defined as:
$LC = \frac{	ext{Pitch}}{	ext{Total number of divisions on circular scale}}$
Given:
- Pitch = $0.5 	ext{ mm}$
- Divisions = $50$
$LC = \frac{0.5 	ext{ mm}}{50} = 0.01 	ext{ mm}$.

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instrument use (screw gauge)

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