The pitch of a screw gauge is `1 mm` and there are `50` divisions on its cap. When nothing is put in between the studs, `44th` divisions of the circular scale coincides with the reference line and the line and the zero of the main scale is not visible or zero of circular scale is lying above the reference line. When a glass plate is placed between the studs, the main scale reads three divisions and the circular scale reads `26` divisions. Calculate the thickness of the plate.

Answer

`R_(t) = 3.64 mm`

Explanation

Zero error `= -(50 -44) ((1mm)/(50))=-0.12 mm`
Thickness of plate `= 3+26 xx(1)/(50) + 0.12 mm= 3.64 mm`