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 20 divisions. If the screw gauge has a zero error of +0.05 mm, what is the volume 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 = 20

Second part of the measurement, H.S.C $\times$ L.C = 20 $\times$ 0.01 mm = 0.2 mm

Diameter of the lead shot with zero error = P.S.R + H.S.C $\times$ L.C = 1 + 0.2 = 1.2 mm

Zero error = + 0.05 mm

Corrected reading of diameter = 1.2 – 0.05 = 1.15 mm

So, radius of the lead shot = 0.575 mm

Using the formula, V = $\frac{4\pi r^2}{3}$ = $\frac{4\pi {\left(0.575mm\right)}^2}{3}$ =0.796 ${mm}^3$