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 division 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 coincide with the reference line. The diameter of the wire is :
Text Solution
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Δ λ = 1 mm
N = 50 division
zero error = –6 Divisions
= – 0.12 mm
Diameter = Measured value + zero correction
= 3 × 1 + (6 + 31) × 
= 3 + 0.74 = 3.74 mm
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