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The pitch of a screw gauge having 50 divisions on its circular scale is 1mm. 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 (each one being 1mm) are clearly visible while 31st division on the circular scale coincide with the reference line. The diameter of the wire is (in mm) ______.

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answer is 3.74.

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Detailed Solution

$Δ l = 1 m m$

N = 50 divisions
Zero error = -6 divisions = -0.12 mm
Diameter = measured value + zero correction

$= 3 \times 1 + (6 + 31) \times \frac{1}{50}$

$= 3 + 0.74 = 3.74 m m$

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