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Q.

The pitch of a screw gauge is 0.5 mm and there are 50 divisions on its circular scale and one main scale division = 0.5 mm. Before starting the measurement, it is found that when jaws of the screw gauge are brought in contact, the zero of the circular scale lies 4 division below the reference line. When a metallic wire is placed between the jaws, five main scale divisions are clearly visible and 18th division on the circular scale coincides with the reference line. The diameter of the wire is

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a

2.62 mm

b

2.72 mm

c

2.64 mm

d

2.68 mm

answer is C.

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Detailed Solution

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Zero error =+4×0.01=+0.04mm

Measured diameter = M. S. R. + L.C. x C. S. R. =  (5×0.5 )mm+ (0.01mm )×18=2.68mm 

Actual diameter = 2.68 – (0.04) = 2.64 mm

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