The pitch and the number of divisions, on the circular scale, for a given screw gauge are $0.5\ \text{mm}$ and $100$ respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies $3$ divisions below the mean line.
The readings of the main scale and the circular scale, for a thin sheet, are $5.5\ \text{mm}$ and $48$ respectively, the thickness of this sheet is:
Answer: (C) $5.725\ \text{mm}$
Least count $= \dfrac{0.5}{100} = 0.005\ \text{mm}$.
The zero of the circular scale is below the mean line, so the zero error is positive: $+3\times0.005 = +0.015\ \text{mm}$.
Observed reading $= 5.5 + 48\times0.005 = 5.740\ \text{mm}$.
$$\text{Thickness} = 5.740 - 0.015 = 5.725\ \text{mm}$$
Solution by Sreeraj P, M.Sc Physics