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NEET 2020 (Phase 1) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from NEET 2020 (Phase 1)

Q48NEET 2020 · PHASE 1Easy

Least Count, Vernier Callipers & Screw Gauge

A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale. The pitch of the screw gauge is:

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Answer: (b)

LC $= \dfrac{\text{pitch}}{\text{no. of circular divisions}}$
Pitch $= 0.01\text{ mm}\times 50 = 0.5$ mm
So the answer is (b) $0.5$ mm.

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