The pitch of a screw gauge is `0.5mm` and there are `50` divisions on circular scale. When there is nothing between the two ends (studs) of screw gauge, `45th` division of circular scale is coincide with screw guage, and in this situation zero of main scale is not visible. When a wire is place between the studs, the linear scale reads `2` divisions and `20th` divisions of circular scale coincides with references line. For this situation mark the correct statement(s).

Options

A. L.C. of the instrument is 0.01 mm
B. Zero correction for the instrument is `-0.45` mm
C. Thickness of wire is 1.65 mm
D. All of the above

Answer

A

Explanation

Leat Count `=(1//2)/(50)=(1)/(100)`
zero correction `=-(1-0.45)=(-0.55mm)`
Reading of wire = `1mm +((20)/(100))mm+0.55mm=2.75mm`