Convert it to a percentage(0.002 ÷ 0.482) × 100 = 0.4%absolute uncertainty ÷ value × 100
Adding or subtractingΔQ = ΔA + ΔB
Add the absolute uncertainties.
mm + mmMultiplying or dividing%ΔQ = %ΔA + %ΔB
Add the percentage uncertainties.
% + %Raising to a power n%ΔQ = |n| × %ΔA
Multiply the percentage uncertainty by the power.
n × %
Exam diagrams for this topic1 figure to inspect and practiseQuestions, hints and marking points in one compact subsection.Open subsectionHide subsection
See it. Read it. Work it.
These figures come from GioPhysics practice papers. Open one, decode the drawing, work the guided questions, then follow its link to the full paper question.
01
InspectRead the figure comment.
02
TraceFollow labels, arrows and axes.
03
AnswerWork one part at a time.
04
CheckReveal hints and marking points.
IGCSE
01Fig. 10.1Physical quantities and measurement techniques · DensityIGCSEOpen diagramClose diagram
Figure comment
Fig. 10.1Two measuring cylinders drawn side by side, each with graduations up the wall and a curved meniscus. The first holds water alone, its surface at the 50.0 cm³ mark. The second holds the same water with the stone lowered in on a thread that runs up and out of the cylinder; the stone lies fully submerged near the base and the surface now stands at the 68.0 cm³ mark.
Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.
Reading cue. The 50.0 cm³ is water alone and the 68.0 cm³ is water with the stone in, so the quantity everything turns on is the difference between the two levels, and it is printed on neither scale.
aDetermine Determine the volume of the stone from the two readings in Fig. 10.1.
recall2 marks
Check answer · 2 marks
volume = 68.0 - 50.0 (1)
= 18.0 cm³ (1)
bDetermine A second stone, identical to the first, is lowered into the same cylinder on a thread beside it. Determine the new reading of the water surface, and state one condition that must hold for your answer to be correct.
routine3 marks
Check answer · 3 marks
the second stone displaces a further 18.0 cm³ (1)
new reading = 68.0 + 18.0 = 86.0 cm³ (1)
condition: both stones must be completely below the surface, and the water must not reach the top of the cylinder (1)
cExplain The cylinder in Fig. 10.1 is graduated in divisions of 1 cm³, so each of the two levels can be judged only to within half a division. Explain why the volume of the stone is known far less precisely than either reading, supporting your answer with figures.
demanding3 marks
Check answer · 3 marks
each reading may be out by up to 0.5 cm³, so their difference may be out by up to 1 cm³ (1)
1 cm³ in 18.0 cm³ is about 6% (1)
0.5 cm³ in 68.0 cm³ is less than 1%, so subtracting two large readings to obtain a small difference magnifies the error (1)
dSuggest A larger stone is to be measured, but lowering it into the cylinder drawn in Fig. 10.1 would take the water above the top graduation. Suggest a change to the method that still uses the same cylinder, and explain why the volume obtained is still correct.
top of the paper3 marks
Check answer · 3 marks
pour out some water first, so that the starting level is much lower, for example 20 cm³ rather than 50.0 cm³ (1)
enough water must remain to cover the stone completely once it is lowered in (1)
only the difference between the two readings is used, and that difference does not depend on the starting level (1)
Transfer challenge
A metal statue is far too large for any measuring cylinder. It is lowered on a thread into an overflow can that has been filled until water just stops running from the spout, and the water pushed out is collected in a measuring cylinder, which then reads 240 cm³. Determine the volume of the statue, and explain why the can must be left to stop dripping before the statue is lowered in.
Check answer · 4 marks
volume of statue = 240 cm³ (1)
the water pushed out has the same volume as the part of the statue below the surface, so the statue must be fully submerged (1)
if the can is still over-full, water standing above the level of the spout runs out on its own (1)
that extra water would be collected as well and the volume obtained would be too large (1)
