The least count of a screw gauge with 50 divisions on the circular scale and pitch 1 mm is:
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Question

The least count of a screw gauge with 50 divisions on the circular scale and pitch 1 mm is:

Accepted Answer

The correct answer is 0.02 mm. Least count = Pitch / Number of circular scale divisions = $1 	ext{ mm} / 50 = 0.02 	ext{ mm}$ (or 0.002 cm).

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