Question

The thimble of a screw gauge has $50$ division. The spindle advances $1mm$ when screw is turned through $2$ revaluation. Find:
(1) the pitch of the screw gauge
(2) the least count gauge.

Answer 1

Given,

Number of circular

division = 50 divisions

Linear distance moved

in two complete rotation = 1 mm

Therefore,

(i) Pitch of the screw gauge

pitch = $\frac{1}{2}\times 1$mm

= 0.5 mm

(ii) Least count of the screw gauge

Least

count = $\frac{pitch}{noofcirculardiv}$

= $\frac{0.5}{50}$

= $\frac{5}{10}\times 5$

= 0.01 mm