The key idea
Units and measurement technique appear throughout physics. This guide practises reading scales, converting length, area and volume, and using repeated readings without confusing random uncertainty with systematic error.
Understand it · 01
Choose the instrument before taking the reading
Start with the size of the quantity and the detail you need to distinguish. A metre rule can measure a table; a measuring cylinder can measure the volume of water. The range must include the measurement, while the resolution must be fine enough to make the comparison meaningful. A digital display with many decimal places does not automatically make a measurement accurate: the instrument may still be incorrectly calibrated.
Check what the instrument reads before the measurement. An empty balance should read zero. If it consistently reads above zero, its displayed masses are too large by that offset. Correct the offset or re-zero the balance. Repeating the same biased reading many times will not remove this systematic error, even if the readings agree closely.
Understand it · 02
Read the scale, then explain the method
For length, align the object with the scale and view each endpoint perpendicularly. Looking from an angle can shift the apparent position: this is parallax. You can use an undamaged part of a ruler if its zero end is worn, but the length is then the difference between the endpoint readings. The final scale reading alone is a position, not necessarily the object's length.
For water in a measuring cylinder, place the cylinder on a level surface and read the bottom of the curved meniscus at eye level. Each numbered interval may contain several smaller divisions, so find the value of one division first. Record the unit beside the result and keep sensible precision; a rough scale cannot support an arbitrarily long decimal answer.
Understand it · 03
Reduce the fractional effect of reaction time
Timing a single short oscillation makes your reaction delay a substantial fraction of the result. Instead, use a fixed reference position and time several complete oscillations. Start at zero; one full oscillation returns the object to the same position moving in the same direction. Divide the total time by the number of complete oscillations to obtain the period.
Repeat that whole procedure and calculate the mean. A similar start-and-stop timing uncertainty is a smaller fraction of a longer measured interval. This improves the estimate, but it does not make reaction time vanish, and it cannot repair a slow stopwatch. If a reading is very different, investigate the method or repeat it rather than silently deleting an inconvenient value.
Understand it · 04
Convert the quantity, including its dimensions
Attach the conversion factor to the unit instead of memorising isolated decimal shifts. One centimetre equals 0.01 metre. A square centimetre is a square with both sides 0.01 metre, so its area is 0.0001 square metre. A cubic centimetre has three such dimensions and equals 0.000001 cubic metre. Area and volume conversions therefore need the length factor squared or cubed.
Keep intermediate values until the end of a calculation to avoid accumulating rounding error. Then state a final answer with an appropriate unit and precision. A quick estimate is valuable: converting a length from centimetres to metres should give a smaller numerical value, because each metre contains many centimetres. Unit checking is part of the physics, not a decorative final step.
Another view of the physics
Read the change in the scale reading
Key definitions
- SI base units
- The seven SI base units are metre (m), kilogram (kg), second (s), ampere (A), kelvin (K), mole (mol) and candela (cd). Derived units combine base units; for example, m/s is a unit of speed.
- Scalar quantity
- A quantity that has magnitude (size) only — e.g. speed, distance, mass, energy, time.
- Vector quantity
- A quantity that has both magnitude and direction — e.g. velocity, displacement, force, acceleration, momentum.
Equations, units & conditions
Average from repeats
average = sum of readings ÷ number of readings
Period from timing
T = total time ÷ number of oscillations
Units: s
Worked examples, step by step
Example 1
A student times 20 complete swings of a pendulum at 30.4 s. Calculate the period of the pendulum.
Follow the method
- The period T is the time for one complete swing.
- T = total time ÷ number of swings = 30.4 ÷ 20.
T = 1.52 s
Correct a zero error before averaging
A digital balance reads +0.4 g when empty. With a sample on the balance, three readings are 18.7 g, 18.8 g and 18.9 g. Find the corrected mean mass, assuming the zero offset stays constant.
Follow the method
- Mean displayed mass = (18.7 + 18.8 + 18.9) ÷ 3 = 18.8 g.
- The balance over-reads by 0.4 g, so subtract the offset: 18.8 − 0.4 = 18.4 g.
- Repeating reduces the effect of random variation; it does not remove a persistent zero error.
Corrected mean mass = 18.4 g = 0.0184 kg.
Practice questions with worked solutions
Try each question on paper before opening the solution. These are original academy questions; the suggested marks are a self-checking guide, not an official exam-board mark scheme. Difficulty labels describe this practice set, not predicted grades.
Starter · 3 suggested marks
1. Convert length, area and volume
Convert (a) 36 mm to m, (b) 8.0 cm² to m² and (c) 75 cm³ to m³.
Show solution and self-check: question 1
- Milli means 10⁻³: 36 mm = 36 × 10⁻³ m.
- Square the length conversion: 1 cm² = (10⁻² m)² = 10⁻⁴ m².
- Cube the length conversion: 1 cm³ = (10⁻² m)³ = 10⁻⁶ m³.
(a) 0.036 m; (b) 8.0 × 10⁻⁴ m²; (c) 7.5 × 10⁻⁵ m³.
Self-check · one mark per point
- Correct length conversion and unit.
- Correct area conversion and squared unit.
- Correct volume conversion and cubed unit.
Starter · 3 suggested marks
2. Measure away from a damaged zero
A rod lies along a ruler from the 1.3 cm mark to the 6.8 cm mark. Calculate its length in metres and state how to avoid parallax when reading the ends.
Show solution and self-check: question 2
- Length = final position − initial position = 6.8 − 1.3 = 5.5 cm.
- Convert to metres: 5.5 ÷ 100 = 0.055 m.
- View each endpoint with the eye directly above its scale mark, so the line of sight is perpendicular to the ruler.
Length = 0.055 m; read the ruler directly above each endpoint.
Self-check · one mark per point
- Subtracts the endpoint readings.
- Converts to 0.055 m.
- Explains the perpendicular viewing direction.
Build · 4 suggested marks
3. Time a repeated motion
A student times 15 complete oscillations three times: 23.1 s, 23.4 s and 23.7 s. Find the mean period and explain why timing several oscillations is preferable to timing one.
Show solution and self-check: question 3
- Mean total time = (23.1 + 23.4 + 23.7)/3 = 23.4 s.
- Period = 23.4/15 = 1.56 s.
- A similar reaction-time uncertainty is a smaller fraction of the longer interval.
Mean period = 1.56 s. Timing several oscillations reduces the fractional effect of reaction time.
Self-check · one mark per point
- Calculates mean time 23.4 s.
- Divides by 15 complete oscillations.
- States 1.56 s with its unit.
- Explains the smaller fractional timing uncertainty.
Build · 3 suggested marks
4. Convert an area from its sides
A rectangular tile measures 24 cm by 15 cm. Calculate its area in square metres. A student obtains 3.6 m² by dividing 360 cm² by 100; explain the mistake.
Show solution and self-check: question 4
- Convert the lengths: 24 cm = 0.24 m and 15 cm = 0.15 m.
- Area = 0.24 × 0.15 = 0.036 m².
- For area the factor of 100 must be squared: 360 cm²/10 000 = 0.036 m².
Area = 0.036 m². The student used a length conversion for an area.
Self-check · one mark per point
- Uses consistent metre lengths or the 10 000 area factor.
- Finds 0.036 m².
- Explains that both dimensions must be converted.
Build · 4 suggested marks
5. Find a period from repeated timings
Three measurements for 20 complete oscillations are 28.4 s, 28.8 s and 28.6 s. Calculate the mean period. Explain why timing 20 oscillations helps and why this method cannot correct a clock that runs slow.
Show solution and self-check: question 5
- Mean time = (28.4 + 28.8 + 28.6) ÷ 3 = 28.6 s.
- Period = 28.6 ÷ 20 = 1.43 s.
- A similar start/stop reaction uncertainty is a smaller fraction of the longer total time.
- A slow clock creates a systematic error in every timing; repetition or division does not calibrate the clock.
Mean period = 1.43 s. The fractional timing uncertainty is reduced, but the clock error remains.
Self-check · one mark per point
- Finds mean total time 28.6 s.
- Divides by 20 to obtain 1.43 s.
- Explains the smaller fractional effect of reaction time.
- Identifies the persistent systematic clock error.
Stretch · 4 suggested marks
6. Combine volume measurement with density
Water in a measuring cylinder rises from 42 cm³ to 58 cm³ when a fully submerged metal sample is added. Its mass is 124.8 g. Find the density in g/cm³ and kg/m³. Explain how trapped air bubbles would affect the calculated density.
Show solution and self-check: question 6
- Sample volume = 58 − 42 = 16 cm³.
- Density = mass ÷ volume = 124.8 ÷ 16 = 7.8 g/cm³.
- 1 g/cm³ = 1000 kg/m³, so the density is 7800 kg/m³.
- Attached bubbles displace extra water. The measured volume is too large, so mass divided by that volume gives a density that is too small.
Density = 7.8 g/cm³ = 7800 kg/m³. Trapped bubbles cause an underestimate.
Self-check · one mark per point
- Uses the rise of 16 cm³ as the volume.
- Finds 7.8 g/cm³.
- Converts to 7800 kg/m³.
- Links extra displaced volume to a density underestimate.
Common mistakes & how to avoid them
- Measuring the period of a pendulum with one swing. Time 20 oscillations and divide by 20 — one swing makes your reaction time a huge fraction of the reading.
- Forgetting to convert units before substituting: 250 g must become 0.25 kg, 40 cm must become 0.40 m. Examiners set numbers deliberately to punish this.
- Reading a measuring cylinder from the top of the meniscus, or with your eye above the level. Read the bottom of the meniscus, at eye level.
- Writing an answer with no unit. A correct number with a missing or wrong unit routinely loses the final mark.
Understand the why
Questions, explained
Start with the short answer, then follow the reasoning. Each explanation has a permanent link you can share with a classmate or return to when revising.
Why do you time 20 oscillations instead of one?
Timing 20 complete oscillations makes a similar start-and-stop reaction uncertainty a smaller fraction of the measured interval. Divide the total time by 20 to estimate one period, then repeat the procedure to check consistency. This does not correct a stopwatch that runs slow.
The useful comparison is uncertainty divided by the measured time. A reaction delay of a few tenths of a second matters greatly in a one-second interval but much less in a twenty-second interval. Dividing the longer interval also divides its absolute timing uncertainty when expressing the answer as one period.
Count complete cycles from a fixed reference position, returning in the same direction. Starting at zero avoids counting the initial position as a completed cycle. Twenty is a convenient choice, not a universal requirement: select enough cycles to reduce timing error without letting the motion change significantly during the measurement.
Compare the fractional timing uncertainty
A pendulum takes 36.0 s for 20 oscillations. Assume the uncertainty in the whole timed interval is ±0.2 s. Find the period and compare its fractional timing uncertainty with timing one 1.8 s oscillation.
- Period = 36.0/20 = 1.80 s.
- Fractional uncertainty for the long timing = 0.2/36.0 ≈ 0.00556, or 0.56%.
- For one oscillation: 0.2/1.8 ≈ 0.111, or 11%. The same assumed interval uncertainty has a much larger fractional effect.
Period = 1.80 s; the timing contribution is about 0.56%, compared with about 11% for one oscillation.
Do I need to memorize unit conversions for IGCSE physics?
You need to use the prefixes and unit conversions required by your specification, including converting time, mass and length consistently. Understanding the conversion factor is more reliable than remembering a decimal shift, especially for squared and cubed units such as cm² and cm³.
Begin with what the prefix means: milli is one thousandth, centi one hundredth and kilo one thousand. A quantity does not change when its unit changes; only the numerical value does. A length therefore has a larger number when expressed in millimetres than when expressed in metres.
Apply that factor to every dimension. A square centimetre has two sides measured in centimetres, so its metre conversion is squared. Do not assume every calculation must use SI: consistent grams and cubic centimetres can give density in g/cm³. Use SI when the equation's other quantities require it, such as newtons and square metres for pascals.
Convert a small contact area
Convert 12 cm² to m² before using p = F/A.
- 1 cm = 0.01 m, so 1 cm² = (0.01)² m² = 0.0001 m².
- 12 cm² = 12 × 0.0001 = 0.0012 m².
12 cm² = 1.2 × 10⁻³ m².
What is the difference between a scalar and a vector?
A scalar has magnitude only; a vector has magnitude and direction. Mass, time and speed are scalars, while force, displacement and velocity are vectors. Direction changes how vectors combine, so opposite vectors can cancel even when their individual magnitudes are non-zero.
Consider walking away from a starting point and returning. Your travelled distance continues to grow throughout the journey, but your final displacement is zero. Both descriptions are correct because they answer different questions: route length versus change in position. Their different behaviour is not simply a change of unit.
For vectors along one line, choose one direction as positive and give the opposite direction a negative sign. Equal positive and negative forces then sum to zero. For vectors at right angles, ordinary addition of magnitudes is not appropriate; use a scale drawing or the mathematical method required by your course.
Does repeating measurements remove a zero error?
No. Repetition can reduce the effect of random variation, but a consistent zero error shifts every reading in the same direction. Check the instrument's unloaded reading, then re-zero it or apply the correct offset if that offset remains stable throughout the measurements.
An average can sit very close to the wrong value. If an empty balance reads a positive mass, every sample reading includes that extra contribution. Taking ten readings instead of three improves your knowledge of the displayed mean but does not identify how far the whole set has been displaced.
The correction has the opposite sign to the zero error: subtract a positive offset and add the magnitude of a negative offset. Check again after the experiment if drift is possible. A fault that changes unpredictably cannot be repaired by applying one constant correction, and should prompt investigation or use of another instrument.
Correct a positive offset
An empty balance reads +0.3 g. Three displayed sample masses are 18.6 g, 18.7 g and 18.8 g. Calculate the corrected mean mass.
- Displayed mean = (18.6 + 18.7 + 18.8)/3 = 18.7 g.
- Corrected mean = 18.7 − 0.3 = 18.4 g.
Corrected mean mass = 18.4 g, assuming the offset stayed at +0.3 g.
How do you avoid parallax error when reading a ruler?
Place your eye directly above the scale mark being read, with the line of sight perpendicular to the ruler. Keep the object close to the scale and read both endpoints carefully. Viewing from the side can shift an elevated endpoint's apparent position against the marks.
Parallax arises when the object or pointer and the scale are at different distances from your eye. Looking from another angle changes their apparent alignment. Moving the ruler closer to the object reduces that separation, while looking straight down addresses the viewing-angle problem. Merely repeating a consistently angled reading need not remove the error.
A ruler with a damaged zero can still be useful. Align one endpoint with an undamaged mark and subtract its reading from the other endpoint. Read each end separately from above rather than staying above the middle of a long object. Record the difference as a length, not the final coordinate on the scale.
How can I measure the thickness of one sheet of paper accurately?
Measure the thickness of a counted stack of similar sheets and divide by the number of sheets. A stack is easier to resolve with a ruler than one sheet. Keep the sheets aligned and gently flattened so large air gaps do not dominate the measurement.
This method magnifies the quantity you want to measure before dividing back down. If a ruler reading has roughly the same absolute uncertainty for a single sheet and for a stack, that uncertainty is a much smaller fraction of the stack height. The calculation gives the average thickness of the chosen sheets.
Counting errors, mixed paper types, uneven edges and trapped air are separate limitations. Measure at several positions and investigate inconsistent readings. Do not compress the stack so strongly that you change the paper itself, and do not confuse a book's total thickness with the thickness of its pages if its covers and binding are included.
Use a stack to resolve a thin sheet
A stack of 200 sheets, excluding covers, measures 24 mm thick. Estimate the mean thickness of one sheet.
- Mean thickness = total thickness/number of sheets = 24/200 = 0.12 mm.
- 0.12 mm = 0.00012 m.
Mean sheet thickness = 0.12 mm = 1.2 × 10⁻⁴ m.
Your learning checklist
Use these goals to check your understanding. Requirements vary by specification, tier and exam year; the official references below define your full course.
- Use rulers, measuring cylinders, balances and clocks to measure length, volume, mass and time
- Take repeat readings and average — including timing multiple oscillations or intervals
- Select a measuring instrument with an appropriate range and resolution; check for zero error
- Know the SI units of every quantity on the syllabus and convert between prefixes (km, cm, mm, ms)
Sources & how to cite these notes
Original explanations, illustrations and practice by Dr Desouky Physics Academy. Official specifications guide the course scope; questions and suggested marks on this page are our own revision material.
- Cambridge IGCSE Physics 0625 — 2026–2028 syllabus
- Pearson Edexcel International GCSE Physics 4PH1 — specification
Reference this page
Dr Desouky Physics Academy. IGCSE Physics: Measurements and units. Updated 2026-09-25. https://www.drdesouky.com/revision/igcse-physics/measurements-and-units
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