Measuring Instruments And Least Count

Question

A screw gauge of pitch 0.5 mm has 50 circular divisions and a positive zero error of three divisions; while measuring a wire it reads two main-scale divisions and twenty-eight circular divisions, so what is the corrected diameter?

RankGuru · Accepted Answer

A screw gauge measures small thicknesses by converting rotation into linear advance, and its least count equals the pitch divided by the number of circular divisions. Here least count = 0.5 mm / 50 = 0.01 mm, while each main-scale division equals the pitch of 0.5 mm. The observed reading is the main-scale contribution plus the circular contribution: two main divisions give 2 × 0.5 = 1.00 mm, and twenty-eight circular divisions give 28 × 0.01 = 0.28 mm, so the observed diameter is 1.28 mm. A positive zero error means the instrument reads too high by three divisions, that is +3 × 0.01 = 0.03 mm, which must be subtracted: corrected diameter = 1.28 - 0.03 = 1.25 mm. The choice 1.31 mm wrongly adds the zero-error correction. The choice 1.28 mm ignores the zero error entirely. The choice 1.22 mm subtracts too much by doubling the correction. This procedure follows the NCERT treatment of zero-error correction. As a check, a positive zero error must always reduce the final reading, consistent with our subtraction.

Measuring Instruments And Least Count

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