Question

A student is trying to measure the diameter of a lead shot using a screw gauge. The pitch of the screw gauge is 1 mm and has 50 divisions on its circular scale. When the two jaws of the screw gauge are in contact with each other, the zero of the circular scale lies 6 divisions below the line of graduation. When the lead shot is placed between the jaws, 3 linear scale divisions are clearly visible while ${31}^{st}$​ division on the circular scale coincides with the reference line. The diameter of the lead shot is:

• A. 3.62 mm
• B. 3.74 mm
• C. 3.55 mm
• D. 3.50 mm

Answer 1

The correct option is D

3.50 mm

We know that, Least Count = $\frac{\left(Pitch\right)}{(No.ofdivisionsontheheadscale)}$ = $\frac{\left(1mm\right)}{\left(50\right)}$ = 0.02 mm

The zero error is positive, given by Z.E. = 6 $\times$ L.C. = 6 $\times$ 0.02 mm = 0.12 mm

First part of the measurement, P.S.R = 3 mm

Head scale division that coincides with the pitch scale axis, H.S.C = 31

Second part of the measurement, H.S.C $\times$ L.C = 31 $\times$ 0.02 mm = 0.62 mm

Measurement with zero error = P.S.R + H.S.C $\times$ L.C = 3 + 0.62 = 3.62 mm

Zero error = +0.12 mm [Positive because zero is below the datum line]

So, corrected reading = 3.62 – 0.12 = 3.50 mm