Question

A screw gauge with pitch $1$ mm and $100$ divisions on circular scale is used to measure the thickness of a glass-slab. When there is no object between the studs and the faces of the screw gauge touch each other, the ${40}^{th}$ division of circular scale coincides with the reference line. Now, the glass place is held between the studs and reading of linear scale is found to be$5$ and that of circular scale is $26$. What is the thickness of the glass plate? It is given that zero of linear scale is not visible when the faces touch each other.

Answer 1

Least count = $\frac{\text{Pitch}}{\text{CSD}}=\frac{1\text{mm}}{100}=0.01\text{mm}$

Zero error$=–40\times 0.01=–0.4mm$

Reading of screw gauge $=5mm+26\times 0.01=5.26mm$

Actual reading $=5.26mm–(–0.4mm)$

$=5.66mm$