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Question

# A student is trying to measure the diameter of a lead shot using a screw gauge. The pitch of the screw gauge is 1 mm and has 50 divisions on its circular scale. 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 the lead shot is placed between the jaws, 3 linear scale divisions are clearly visible while 31st​ division on the circular scale coincides with the reference line. The diameter of the lead shot is:

A

3.62 mm

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B

3.74 mm

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C

3.55 mm

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D

3.50 mm

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Solution

## The correct option is D 3.50 mm We know that, Least Count = (Pitch)(No. of divisions on the head scale) = (1mm)(50) = 0.02 mm The zero error is positive, given by Z.E. = 6 × L.C. = 6 × 0.02 mm = 0.12 mm First part of the measurement, P.S.R = 3 mm Head scale division that coincides with the pitch scale axis, H.S.C = 31 Second part of the measurement, H.S.C × L.C = 31 × 0.02 mm = 0.62 mm Measurement with zero error = P.S.R + H.S.C × L.C = 3 + 0.62 = 3.62 mm Zero error = +0.12 mm [Positive because zero is below the datum line] So, corrected reading = 3.62 – 0.12 = 3.50 mm

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