Question

While performing an experiment with a screw gauge to measure the diameter of a wire, a student finds that when the two jaws of the screw gauge are in contact with each other, the zero of the circular scale lies 5 divisions above the line of graduation. The pitch of the screw gauge is 1 mm and has 100 divisions on its circular scale. While measuring the diameter of the wire, 4 linear scale divisions are clearly visible while ${63}^{rd}$ division on the circular scale coincides with the reference line. What is the diameter of the wire?

Answer 1

We know that, Least Count = $\frac{\left(Pitch\right)}{(No.ofdivisionsontheheadscale)}$ = $\frac{\left(1mm\right)}{\left(100\right)}$ = 0.01mm

The zero error is negative, given by Z.E. = – 5 $\times$ L.C = – 5 $\times$ 0.01 mm = – 0.05 mm

First part of the measurement, P.S.R = 4 mm

Head scale division that coincides with the pitch scale axis, H.S.C = 63

Second part of the measurement, H.S.C $\times$ L.C = 63 $\times$ 0.01 mm = 0.63 mm

Measurement with zero error = P.S.R + H.S.C $\times$ L.C = 4 + 0.63 = 4.63 mm

So, corrected reading = 4.63 + 0.05 = 4.68 mm