Zero error and significant figures in a density determination
A learner is asked to identify a small metal cylinder from its density. The top-pan balance gives a mass of 1.92 g. The diameter is measured with a micrometer screw gauge whose jaws, when fully closed, give a reading of +0.10 mm. Three readings are taken: 6.10 mm, 6.14 mm and 6.12 mm. The length is measured with vernier calipers as 25.0 mm, and these calipers read zero correctly. The learner computes the volume as πr²L, obtains 0.712 cm³, divides to get 2.70 g/cm³, and writes 'the metal is aluminium (2.70 g/cm³)'. A second learner objects that the result looks suspiciously exact and that the uncertainty in the length alone makes the identification unsafe. The teacher asks the class to decide how many significant figures are justified and whether the identification is sound.
1. Calculate the corrected mean diameter, the volume in cm³ and the density. State how many significant figures are justified.
Read the worked answer
Mean observed diameter = (6.10 + 6.14 + 6.12)/3 = 6.12 mm.
A zero reading of +0.10 mm means the instrument adds 0.10 mm to every reading, so the corrected diameter is 6.12 − 0.10 = 6.02 mm. Radius = 3.01 mm = 0.301 cm; length = 2.50 cm.
V = πr²L = π × (0.301)² × 2.50 = 0.712 cm³.
ρ = m/V = 1.92/0.712 = 2.70 g/cm³.
Significant figures: mass and length are quoted to 3 s.f., and the corrected diameter to 3 s.f. The dominant uncertainty is in the diameter: ±0.01 mm on 6.02 mm is 0.17 %, doubled through d² to 0.33 %; the length contributes about 0.2 %. Total uncertainty is roughly 0.5 %, i.e. ±0.015 g/cm³. Three significant figures (2.70) are defensible, but the last digit is not certain, so the honest statement is 2.70 ± 0.02 g/cm³, not an exact 2.70.
2. The uncorrected calculation gives 2.61 g/cm³. Explain the effect of the zero error and assess whether identifying the metal as aluminium is safe.
Read the worked answer
Without the correction, d = 6.12 mm, giving V = 0.735 cm³ and ρ = 2.61 g/cm³.
The zero error inflates the diameter by 0.10/6.02 ≈ 1.7 %. Since volume depends on d², the volume is inflated by about 3.3 %, so the density is too low by about 3.3 %. This is a systematic error: taking more readings and averaging will not remove it, only correcting the zero will.
Correcting moves the answer from 2.61 to 2.70 g/cm³, which matches the accepted value for aluminium. The corrected uncertainty of about ±0.02 g/cm³ comfortably includes 2.70, so the result is consistent with aluminium. It is not proof: an alloy could have the same density, and a density match never identifies a material on its own. The safe wording is 'consistent with aluminium, within the stated uncertainty', not 'therefore aluminium'.
3. Suggest and justify improvements that would make the identification more reliable.
Read the worked answer
The volume is the weak link because the cylinder is small, so percentage errors are large.
- Measure the diameter at several points along the cylinder and at right angles, and average. This checks for taper or out-of-roundness, which a single pair of readings cannot detect.
- Re-check the zero error before and after the measurements. A zero that drifts (for example because the gauge warms in the hand) is a systematic error that averaging cannot fix.
- If identical cylinders are available, stack several end to end and measure the total length. The absolute error in length is then spread over a much larger length, so the percentage error falls.
- Repeat the whole determination and quote a mean with a spread, giving a realistic uncertainty rather than one calculated value.
A displacement method in a measuring cylinder would be worse here: 0.7 cm³ is smaller than the reading uncertainty of a typical 50 cm³ cylinder. Finally, compare with a table of densities (magnesium 1.74, aluminium 2.70, zinc 7.1 g/cm³) and state the conclusion as a range, not a single number.