Errors in Measurement of Physical Quantities
Senior 2 · Physics
Errors in Measurement of Physical Quantities
Lesson 4 of 4
22 min
Let's begin
Many quantities in physics are not read directly from a scale. They are calculated from other measurements: density from mass and volume, speed from distance and time, area from length and width. Each measurement carries its own uncertainty, and those uncertainties pass into the answer. This lesson shows how to combine them.
Worked example. A classroom is 8.0 m ± 0.1 m long and 6.0 m ± 0.1 m wide. Find its perimeter.
P = 2 × (8.0 + 6.0) = 28.0 m. Each length appears twice, so ΔP = 2 × (0.1 + 0.1) = 0.4 m. So P = 28.0 m ± 0.4 m.
Worked example. A liquid is heated from 20.0 °C ± 0.5 °C to 35.0 °C ± 0.5 °C. Find the temperature rise.
Rise = 35.0 − 20.0 = 15.0 °C. The uncertainty is 0.5 + 0.5 = 1.0 °C. The errors add even though you subtracted the readings. The percentage error jumps to 1.0 ÷ 15.0 × 100% = 6.7%, which is why subtracting two close readings is risky.
Worked example. Find the floor area of the same classroom.
So A = 48 m² ± 1 m².
Worked example. A stone has a mass of 50.0 g ± 0.5 g and a volume of 20.0 cm³ ± 0.5 cm³. Find its density.
So ρ = 2.50 g/cm³ ± 0.09 g/cm³.
Worked example. A square floor tile has sides of 5.0 cm ± 0.1 cm (2%). Area = 5.0² = 25 cm². Squaring doubles the percentage error: 2 × 2% = 4%, so ΔA = 4% of 25 = 1.0 cm². A = 25 cm² ± 1 cm².
For example, 48 m² ± 1.4 m² becomes 48 m² ± 1 m², and 2.50 ± 0.0875 becomes 2.50 ± 0.09.
Tip
+ or −: add the ± values. × or ÷: add the percentages. Power n: multiply the percentage by n
Uncertainties never cancel out. Whatever the operation, the combined uncertainty is always larger than each separate one.
Two jerrycans have masses of 250 g ± 1 g and 120 g ± 1 g when empty. What is their total mass?
Solution: 370 g, with uncertainty 1 + 1 = 2 g, so 370 g ± 2 g.
A moto covers 100.0 m ± 0.5 m in 8.0 s ± 0.2 s. Find its average speed with its uncertainty.
Solution: v = 100.0 ÷ 8.0 = 12.5 m/s. % error = 0.5% + 2.5% = 3.0%. Δv = 3.0% of 12.5 = 0.375, which rounds to 0.4 m/s. So v = 12.5 m/s ± 0.4 m/s.
A cube has sides of 4.0 cm ± 0.1 cm. Find its volume with uncertainty.
Solution: V = 4.0³ = 64 cm³. The side has a 2.5% error, and cubing triples it to 7.5%. ΔV = 7.5% of 64 = 4.8, which rounds to 5 cm³. So V = 64 cm³ ± 5 cm³.
A stretched spring is 75.3 cm ± 0.1 cm long. Unstretched, it was 50.1 cm ± 0.1 cm long. Find the extension.
Solution: 75.3 − 50.1 = 25.2 cm, with uncertainty 0.1 + 0.1 = 0.2 cm, so 25.2 cm ± 0.2 cm.
Summary
Sources and credits
The Rwandan examples, worked examples and practice items are written by Ganzaa.
