Question

A screw gauge with a least count of 0.01 mm is used to measure the diameter of a lead shot. The reading on the pitch scale is found to be 1 mm and the reading on the head scale is 40 divisions. If the screw gauge has a zero error of – 0.05 mm, what is the surface area of the lead shot?

Answer 1

First part of the measurement, P.S.R = 1 mm

Head scale division that coincides with the pitch scale axis, H.S.C = 40

Second part of the measurement, H.S.C $\times$ L.C = 40 $\times$ 0.01 mm = 0.4 mm

Diameter of the lead shot with zero error = P.S.R + H.S.C $\times$ L.C = 1 + 0.4 = 1.4 mm

Zero error = – 0.05 mm

Corrected reading of diameter = 1.4 + 0.05 = 1.45 mm

So, radius of the lead shot = 0.725 mm

Using the formula, A = 4 $\pi$ $r^2$ = 4 $\times$ 3.14 $\times$ 0.725 mm $\times$ 0.725 mm = 6.602 ${mm}^2$