In a screw gauge, there are 100 divisions on the circular scale and the main scale moves by 0 . 5   mm on a complete rotation of the circular scale. The zero of circular scale lies 6 divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs, 4 linear scale divisions are clearly visible while 46 th division the circular scale coincide wit

Step-by-step solution

Least count of screw gauge is = Pitch No .   of   circular   divisions = 0 . 5   mm 100 Least count, L C = 5 × 10 - 3   mm The zero of circular scale lies 6 divisions below the line of graduation. So, zero error is positive. Positive zero error = M S R + C S R ( L C ) = 0   mm + 6 5 × 10 - 3   mm Diameter of wire, d = M S R + C S R   ( L C ) – Positive   zero   error = 4 × 0 . 5   mm + 46 5 × 10 - 3 - 6 5 × 10 - 3   mm = 2 &#