A screw gauge has $50$ divisions on its circular scale. The circular scale is $4$ units ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of $0.5$ mm is noticed on the pitch scale. The nature of zero error involved and the least count of the screw gauge are, respectively:
Answer: (B) Positive, $10\ \mu\text{m}$
Least count $= \dfrac{\text{pitch}}{\text{number of circular divisions}} = \dfrac{0.5\ \text{mm}}{50} = 0.01$ mm $= 10\ \mu\text{m}$.
With the jaws closed, the circular scale zero has already moved $4$ divisions past the reference line, so the gauge reads more than zero when it should read zero. This is a positive zero error (of $4\times10\ \mu\text{m} = 40\ \mu\text{m}$).
Solution by Sreeraj P, M.Sc Physics