Question

A screw gauge has a pitch of 1 mm and there are 100 divisions on the circular scale. The instrument reads +2 divisions when nothing is put in between its jaws. In measuring the thickness of a metal sheet, there are 8 divisions on the main scale and ${83}^{rd}$​ division coincides with the reference line. What is the thickness of the sheet?

Answer 1

Least Count of the screw gauge, L.C = $\frac{Pitch}{No.ofdivisionsonthecircularscale}$ = $\frac{1mm}{100}$ = 0.01 mm

First part of the measurement, P.S.R = 8 mm

Head scale division that coincides with the pitch scale axis, H.S.C = 83

Second part of the measurement, H.S.C $\times$ L.C = 83 $\times$ 0.01 mm = 0.83 mm

Measurement with zero error = P.S.R + H.S.C $\times$ L.C = 8 + 0.83 = 8.83 mm

Zero error = + (2 $\times$ 0.01) mm = 0.02 mm

So, corrected reading = 8.83 – 0.02 = 8.81 mm