A screw gauge has pitch 1 mm and 50 circular divisions. With no object, the 44th circular division coincides with the reference line and the main-scale zero is not visible. With a glass plate, the main scale reads 3 mm and the 26th circular division coincides. Find the plate thickness in mm.

Question

G1 Physics · Accepted Answer

Least count = 1/50 = 0.02 mm. Since the main-scale zero is not visible at closure, the zero error is negative: −(50−44)×0.02 = −0.12 mm. Observed plate reading = 3 + 26×0.02 = 3.52 mm. Corrected thickness = 3.52−(−0.12)=3.64 mm.