The distance between two consecutive threads on the screw of a screw gauge is 0.5 mm. The number of divisions on the circular scale is 100. a wire is placed between of the studs of the screw gauge. Find the diameter of the wire if the pitch scale shows 14th division and 40th circular division coincides with the base line.
The given apparatus was detected to have negative zero error. The 90th division on the circular scale of coincides with reference line, when the studs are in contact.

Explanation

1 MSD =1 mm
`" Pitc " = (5 MSD)/(5) =1 MSD =1mm`
Least count `= (P)/(N)`
`N=(1 mm)/(1 mu m) =100`
N=500 , P =(N) `xx` (L.C.) =500 `xx 1 mu m = 0.5` mm
Pitch =0.5 mm