There are 100 divisions on the circular scale of a screw gauge of pitch $1\ \text{mm}$. With no measuring quantity in between the jaws, the zero of the circular scale lies 5 divisions below the reference line. The diameter of a wire is then measured using this screw gauge. It is found that 4 linear scale divisions are clearly visible while 60 divisions on circular scale coincide with the reference line. The diameter of the wire is:
Answer: (C) $4.55\ \text{mm}$
Least count $= \dfrac{1\ \text{mm}}{100} = 0.01\ \text{mm}$.
When the zero of the circular scale lies below the reference line with the jaws closed, the zero error is positive: $+5\times0.01 = +0.05\ \text{mm}$.
Observed reading $= 4 + 60\times0.01 = 4.60\ \text{mm}$.
Corrected diameter $= 4.60 - 0.05 = 4.55\ \text{mm}$.
Solution by Sreeraj P, M.Sc Physics