The pitch of a screw gauge having 50 divisions on its circular scale is 1 mm. When the two jaws of the screw gauge are in contact with each other, the zero of the circular scale lies 6 divisions below

Question

The pitch of a screw gauge having 50 divisions on its circular scale is 1 mm. When the two jaws of the screw gauge are in contact with each other, the zero of the circular scale lies 6 divisions below the line of graduation. When a wire is placed between the jaws, 3 linear scale divisions are clearly visible while 31st division on the circular scale coincides with the reference line. Find diameter of the wire.

TestHub · Accepted Answer

LC = 50 1 mm ​ = 0.02 mm The zero lies below the reference line, so the zero error is positive. e = 6 × 0.02 = 0.12 mm Measured diameter = 3 + 31 × 0.02 = 3.62 mm Actual diameter = 3.62 − 0.12 = 3.50 mm ≈ 3.5 mm

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