A screw gauge has divisions on its circular scale. The circular scale is ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of is noticed on the pitch scale. The nature of zero error involved and the least count of the screw gauge, are respectively:
Positive,
Step 1. Given data :
Displacement =
Screw gauge divisions =
Step 2. Find the least count of the screw gauge:
And the nature of the zero error is positive because it is units ahead of pitch scale marking.
Hence, option (C) is correct.