The Porous Stone: Reporting a Density You Cannot Trust
A Senior Secondary 2 candidate is asked to find the density of an irregular stone during a physics practical. She uses a balance reading to 0.1 g and a 100 cm³ measuring cylinder whose scale is marked in 2 cm³ intervals. The stone balances at 78.4 g. She pours water into the cylinder and reads the bottom of the meniscus at 40 cm³, then lowers the stone on a thread and reads 50 cm³. Her partner reads the second value as 52 cm³ because his eye is above the level of the meniscus. The teacher says the accepted density of the rock type is between 2.6 and 2.9 g/cm³. In her report the candidate writes 'density = 7.840 g/cm³' and claims an accuracy of ±0.1 g/cm³, matching the balance. The stone is porous and two small air bubbles cling to it as it enters the water. The candidate must decide what her measurement is actually worth before she submits the report.
1. Calculate the density from the readings and determine the percentage uncertainty contributed by the volume measurement. State how many significant figures the final answer is entitled to, and criticise the reported value of 7.840 g/cm³.
Read the worked answer
Step 1 - Volume. The cylinder reads to the nearest 2 cm³, so each reading carries about ±1 cm³. Volume = 50 − 40 = 10 cm³, with uncertainty ≈ √(1² + 1²) ≈ ±1.4 cm³.
Step 2 - Density. ρ = m/V = 78.4 ÷ 10.0 = 7.84 g/cm³.
Step 3 - Which error dominates. Mass: 0.1 ÷ 78.4 ≈ 0.13 %. Volume: 1.4 ÷ 10 ≈ 14 %. The volume dominates by roughly a hundredfold, so the balance's precision is irrelevant to the quality of the answer.
Step 4 - Significant figures. 14 % of 7.84 is about 1.1 g/cm³. Writing 7.840 g/cm³ claims agreement to 0.001 g/cm³, a thousand times better than the data allow. The honest report is 7.8 g/cm³, ±1 g/cm³ - and even the second figure is generous.
Step 5 - The deeper problem. The value lies far outside 2.6–2.9 g/cm³, and the candidate's claimed ±0.1 g/cm³ cannot explain that gap. A discrepancy that large is systematic error, not random error, so the accuracy claim is wrong twice over.
2. Explain how the trapped air bubbles and the porosity of the stone push the result in opposite directions, and recommend two changes to the method that would make the volume uncertainty small enough to be useful.
Read the worked answer
Opposite pushes.
- Bubbles: each bubble occupies space inside the cylinder, so the level rises more than the stone alone would cause. Measured V is too large, so ρ comes out too low.
- Absorption: a porous stone drinks water from the cylinder, so the level rises less. Measured V is too small, so ρ comes out too high.
Since 7.84 g/cm³ sits above the accepted range, absorption (plus air trapped inside the pores) must be outweighing the bubbles. Note also that the partner's parallax reading of 52 cm³ would give V = 12 cm³ and ρ = 6.5 g/cm³ - a 17 % swing from one careless eye position.
Improvements.
- Use an overflow (displacement) can emptying into a narrow 10 cm³ cylinder with 0.1 cm³ graduations, or a burette. Uncertainty falls from ±14 % to around ±1 %.
- Seal the surface: soak the stone in water until it stops absorbing, or give it a thin wax coat, before the measurement. Repeat with two or three stones so the displaced volume is large compared with the reading error.