Zero Error, Significant Figures and the Density of a Wire
In a Class XII practical examination a candidate must determine the density of a uniform metallic wire. The mass is measured on a digital balance of least count 0.01 g and is recorded as 4.62 g. The length is measured with a metre scale of least count 1 mm and recorded as 25.0 cm. The diameter is measured with a screw gauge of pitch 0.5 mm having 50 divisions on its circular scale. With the jaws closed and the main scale reading zero, the 47th circular division coincides with the reference line. With the wire held between the jaws, the main scale reads 2.5 mm and the 22nd circular division lies on the reference line. The candidate writes the final answer as 3.1114 g cm^-3. The examiner asks the candidate to justify the sign of the zero-error correction, to state how many figures the result can honestly carry, and to name the single change in procedure that would most improve the reliability of the measurement.
1. Find the corrected diameter of the wire and explain how the sign of the zero-error correction is decided.
Read the worked answer
Least count = pitch / number of circular divisions = 0.5 mm / 50 = 0.01 mm.
Zero error. With the jaws closed the instrument ought to read zero, but the 47th circular division sits on the reference line. The circular-scale zero has therefore stopped 3 divisions short of the reference line, i.e. the instrument reads less than the true value; this is a negative zero error:
zero error = -(50 - 47) x 0.01 mm = -0.03 mm.
Observed diameter = MSR + (circular division x LC) = 2.5 mm + 22 x 0.01 mm = 2.72 mm.
Correction. Corrected reading = observed reading - zero error = 2.72 - (-0.03) = 2.75 mm. The plus sign is forced by the physics, not chosen: because the gauge under-reads, the true diameter must be larger than the observed value. Equivalently, r = 1.375 mm = 0.1375 cm.
2. Calculate the density and its percentage uncertainty, and state the result with an honest number of significant figures.
Read the worked answer
Volume V = pi r^2 L = pi (0.1375 cm)^2 (25.0 cm) = pi (0.018906)(25.0) = 1.485 cm^3.
Density = m/V = 4.62 / 1.485 = 3.111 g cm^-3 (unrounded arithmetic value 3.1114).
Uncertainty budget (fractional errors add, with the diameter counted twice because r is squared):
- mass: 0.01/4.62 = 0.22 %
- diameter: 2 x (0.01/2.75) = 0.73 %
- length: 0.1/25.0 = 0.40 %
Total = 0.22 + 0.73 + 0.40 = 1.35 %, so the absolute uncertainty is 0.0135 x 3.11 = 0.04 g cm^-3.
Result: density = (3.11 +/- 0.04) g cm^-3. Since the very first digit of the uncertainty is in the hundredths place, the answer can be quoted only to two decimal places, i.e. three significant figures. The candidate's 3.1114 g cm^-3 claims six figures for a measurement that is uncertain in the second decimal place.
3. Which single change in the procedure would most improve the reliability of the result? Justify your choice numerically, and comment on whether the zero error should also be treated as a random error.
Read the worked answer
Compare the contributions to the total fractional error. The diameter term (0.73 %) is the largest and is doubly weighted, so the highest-leverage change is to measure the diameter better, e.g. with a screw gauge of least count 0.005 mm or by using a travelling microscope, or to average many diameters taken at different points along the wire and at right angles to each other. Halving the diameter error would drop the total from 1.35 % to about 0.99 %.
Replacing the metre scale by a vernier calliper helps much less: it removes 0.40 % and leaves about 0.95 %.
However, if measurement time and effort are counted, lengthening the specimen is the cheapest single change: measuring 100 cm instead of 25.0 cm on the same metre scale cuts the length term from 0.40 % to 0.10 %, and a longer specimen also averages out diameter fluctuations along the wire.
The zero error is a systematic error, not a random one. Averaging repeated readings will not remove it; only calibrating and correcting for it will. Its own uncertainty (roughly +/-0.01 mm in locating the coinciding division) is random and does add to the diameter budget.