Two full turns of the circular scale of a screw gauge having 50 divisions cover a distance of 1   mm . Further, it is found that the screw gauge has a zero error of - 0 . 02   mm . While measuring the diameter of a thin wire, a student notes the main scale reading of 4   mm and the number of circular scale divisions in line with the main scale as 37 . The diameter of the wire is,

Step-by-step solution

Distance cover in 1 full turn will be 1   mm 2 = 0 . 5   mm . Least count of screw gauge, L C = Distance   cover   in   1   full   turn number   of   division   in   circular   scale = 0 . 5   mm 50 = 0 . 01   mm Reading of the screw gauge, R =   main   scale   reading + Circular   scale   reading × LC R = 4   mm + 37 × 0 . 01   mm = 4 . 37   mm Reading without zero error, R ' = R - zero   er