Physics - Micrometer Zero Error and a Disagreement About Sign
Two students use a micrometer screw gauge to measure the diameter of a metal wire. With the jaws closed, the thimble reads 3 divisions below zero on the sleeve; the instrument has a zero error of −0.03 mm. Their readings on the wire are 2.45 mm, 2.47 mm, and 2.46 mm. Student A says the zero error must be added, giving diameters 2.48, 2.50, and 2.49 mm. Student B says it must be subtracted, giving 2.42, 2.44, and 2.43 mm. The wire's nominal diameter is 2.45 mm. The students must report a mean diameter and decide whether their result agrees with the nominal value within the instrument's resolution of 0.01 mm. The thimble has 50 divisions per revolution, and one revolution advances the spindle by 0.50 mm.
1. Determine the correct sign of the zero-error correction and compute the corrected diameters. Explain why the other student's reasoning is wrong.
Read the worked answer
The zero reading is −0.03 mm, meaning that when the true gap is zero the instrument displays 0.03 mm less than zero. Therefore every reading is too small by 0.03 mm, and the correction is +0.03 mm: true = reading − zero error = reading − (−0.03) = reading + 0.03.
Corrected diameters:
- 2.45 + 0.03 = 2.48 mm
- 2.47 + 0.03 = 2.50 mm
- 2.46 + 0.03 = 2.49 mm
Mean = (2.48 + 2.50 + 2.49)/3 = 2.49 mm.
Student B's error is a sign confusion: B sees a negative zero error and subtracts it, which makes the correction negative as well. The rule is not 'negative error means subtract'; the rule is 'subtract the zero reading.' Because the zero reading is already negative, subtracting it adds 0.03 mm.
2. Student B argues that because the zero error is negative, it should be subtracted. Identify the conceptual error and state a general rule that would prevent it.
Read the worked answer
The conceptual error is treating the sign of the zero error as an instruction about the operation. The zero error is the reading shown when the true value is zero. If that reading is −0.03 mm, the instrument under-reads by 0.03 mm, so the true value is larger than the displayed value. The correction is therefore +0.03 mm.
A general rule: true value = instrument reading − zero error. Test it with the closed-jaw case: reading = −0.03 mm, zero error = −0.03 mm, so true = −0.03 − (−0.03) = 0. The rule passes the test. Student B's rule, 'subtract because the error is negative,' fails the same test: −0.03 − (−0.03) is correct only if one applies the formula, not if one reasons from the word 'negative.' The judgement is to check the correction against the closed-jaw case before applying it to measurements.
3. Should the students report 2.49 mm or 2.5 mm? Justify using significant figures and the resolution, and decide whether the result agrees with the nominal 2.45 mm.
Read the worked answer
Report 2.49 mm, not 2.5 mm. The instrument resolution is 0.01 mm, and the individual readings are recorded to 0.01 mm, so the mean can be quoted to 0.01 mm. Writing 2.5 mm would imply a resolution of 0.1 mm and would throw away real information. A reasonable statement is 2.49 ± 0.02 mm, allowing for the resolution and the uncertainty in the zero correction.
The nominal value is 2.45 mm. The difference is 2.49 − 2.45 = 0.04 mm, which is four times the instrument resolution and larger than the stated uncertainty. On these data, the measured diameter does not agree with the nominal value; the wire is slightly thicker than nominal, or the nominal figure is a label rather than a tight specification. The students should not force agreement by choosing Student B's sign or by rounding to 2.5 mm. They should report the corrected mean and note the disagreement.