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Measurements and units
Syllabus objectives: fundamental and derived quantities · dimensions · measuring instruments and errors
👀 Overview
Measurements and units form the foundation of all physics problems in JAMB. You must know how to identify fundamental and derived quantities, use correct symbols and convert between units, and handle instrument errors accurately.
💡 Fundamental and Derived Quantities
The International System of Units (SI) recognises seven fundamental quantities: length (metre, m), mass (kilogram, kg), time (second, s), electric current (ampere, A), thermodynamic temperature (kelvin, K), amount of substance (mole, mol) and luminous intensity (candela, cd). Every other physical quantity is derived from these, e.g., velocity (m s⁻¹), force (kg·m s⁻² = newton, N), energy (kg·m² s⁻² = joule, J). In the exam you will often be asked to write the dimension of a derived quantity; remember to express it in terms of M, L, T, I, Θ, N, J.
🧠 Dimensions and Dimensional Analysis
A dimension is a symbol that shows how a quantity depends on the fundamental ones. For example, the dimension of pressure is M L⁻¹ T⁻². Dimensional analysis helps you check equations, convert units and derive relationships. The rule: an equation is physically correct only if both sides have identical dimensions. In JAMB, you may be given a formula and asked to find the missing exponent by equating dimensions.
🔍 Measuring Instruments and Their Errors
Instruments are classified as direct (ruler, stop‑watch) or indirect (balance, vernier caliper). Each has a least count – the smallest division readable. Absolute error = measured value – true value; relative error = (absolute error / measured value) × 100 %. Systematic errors arise from mis‑calibration, while random errors come from fluctuations in reading. To minimise error, take several readings and use the mean, and always note the instrument’s least‑count in your answer.
📌 Unit Conversions
Conversion is a matter of multiplying by appropriate factors that equal one. Common conversions in the exam: 1 km = 10³ m, 1 cm = 10⁻² m, 1 mg = 10⁻⁶ kg, 1 h = 3600 s, 1 N = 10⁵ dynes. Remember the prefix table (k, h, da, d, c, m, µ, n). When converting compound units, treat each part separately, e.g., 1 km h⁻¹ = (10³ m)/(3600 s) = 0.2778 m s⁻¹.
✨ Significant Figures and Rounding
The number of significant figures (sig‑figs) reflects the precision of a measurement. Rules: all non‑zero digits are significant; zeros between non‑zero digits are significant; leading zeros are not; trailing zeros are significant only if a decimal point is present. In JAMB, after calculations you must round the final answer to the same number of sig‑figs as the least precise data point.
🔑 Key facts, formulas and definitions
- SI base unit of length
- metre (m) – defined by the distance light travels in vacuum in 1/299 792 458 s
- Least count
- Smallest division on a measuring instrument; error ≤ ±(least count/2)
- Dimension of force
- M L T⁻²
- Relative error formula
- ε% = (Δx / x) × 100
- Conversion factor method
- Multiply by fractions equal to 1 to cancel unwanted units
- Significant figures rule
- Result must have same sig‑figs as the measurement with fewest sig‑figs
- Systematic vs random error
- Systematic – same each time (calibration); Random – varies (reading)
- Standard gravity
- g = 9.81 m s⁻² used for weight calculations in Nigeria
✏️ Worked examples
A student measures the length of a table with a steel ruler (least count 0.5 mm) and records 1.235 m. Find the absolute and relative error.
- Least count = 0.5 mm = 0.0005 m.
- Absolute error ≤ ±(least count/2) = ±0.00025 m.
- Relative error = (0.00025 / 1.235) × 100 ≈ 0.020 %.
Convert 72 km h⁻¹ to m s⁻¹ and state the answer with three significant figures.
- 72 km h⁻¹ = 72 × (10³ m) / (3600 s).
- 72 × 10³ / 3600 = 72 × 0.27778 = 20.0 m s⁻¹.
- Three sig‑figs → 20.0 m s⁻¹.
The period of a simple pendulum is given by T = 2π√(L/g). If L = 0.80 m (±0.001 m) and g = 9.81 m s⁻², find T and its absolute error using differential method.
- T = 2π√(0.80/9.81) = 2π√0.08155 = 2π×0.2856 = 1.795 s.
- dT = (π/√(gL)) dL (since g is constant).
- π/√(gL) = 3.1416 / √(9.81×0.80) = 3.1416 / √7.848 = 3.1416 / 2.801 = 1.122.
- dT = 1.122 × 0.001 = 0.00112 s ≈ ±0.001 s.
- Express with proper sig‑figs → T = 1.795 s ± 0.001 s.
⚠️ Common exam traps
- Confusing mass (kg) with weight (N); always multiply mass by g when asked for weight.
- Ignoring the least‑count error; always add ±(least‑count/2) to the measured value before calculating relative error.
- Using wrong prefix for conversion (e.g., treating 1 mm as 10⁻³ m instead of 10⁻³ m). Double‑check the power of ten.
- Rounding intermediate results; keep extra digits until the final answer, then apply sig‑fig rules.
- Writing dimensions with symbols instead of letters (use M, L, T, I, Θ, N, J, not kg, m, s).
⚡ Quick recap
- Seven SI base quantities → M, L, T, I, Θ, N, J.
- Always express derived units in base units before converting.
- Least count = smallest scale division; error ≤ ±(LC/2).
- Relative error % = (Δx / x) × 100.
- Match sig‑figs of answer to the least precise measurement.
- Check dimensions of every equation – they must be identical on both sides.
🎬 Watch and learn
MEASUREMENTS AND UNITS ( JAMB PHYSICS )
MaxChem5 · YouTube
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10 untimed UTME-style questions with explanations at the end · more drill sizes
ⓘ Study notes written by Sabipage AI from the JAMB syllabus. Not official JAMB material — check anything you are unsure of with your teacher or textbook.
ⓘ Practice questions written by Sabipage AI from the JAMB syllabus. Not official JAMB questions.