11.6 Screw gauge solved examples

  Basic Mathematics and Measurements

Question: When a screw gauge with a $L.C.$ of $0.01\ mm$ is used to measure the thickness of a lamina, the reading on the sleeve is found to be $0.5\ mm$ and the reading on the thimble is found to be $27$ division. Find the thickness of lamina in $cm$.

Solution: The following details are given,

$M.S.R.=0.5\ mm$
$n=27$
$L.C.=0.01\ mm$

The reading of screw gauge $(R)$ is written as,
$$R = M.S.R. + n \times L.C.$$$$R = \left( {0.5 + 27 \times 0.01} \right)\,mm$$$$R = \left( {0.5 + 0.27} \right)\,mm$$$$R = 0.77\,mm$$or$$R = 0.077\,cm$$

Question: A screw gauge with a pitch of $0.5\ mm$ and a circular scale with $50$ divisions is used to measure the thickness of a thin sheet of aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the $45^{th}$ division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is $0.5\ mm$ and the $25^{th}$ division coincides with the main scale line ?

Solution: Screw gauge pitch is $0.5\ mm$ and has $50$ circular scale divisions.
$$L.C. = \frac{{{\text{Pitch}}}}{{{\text{Number of circular scale divisions}}}}$$$$L.C. = \frac{{0.5\,mm}}{{50}} = 0.01\,mm$$
For zero error,