Physical Quantities & Measurements

— Physics · AS Level (7408) —

★ Exam Tip: This topic is 10% of your total marks every year (Paper 1 + Paper 2 combined = 10 marks). It's one of the easiest topics — master it and turn a B into an A with barely any effort.

1. Physical Quantities

Definition: A physical quantity is any quantity that can be measured.

Physical quantities split into two branches:

SI base quantitiesMost basic physical quantities — NOT made of any other quantity.
Derived quantitiesMade up of (derived from) SI base quantities.

① The 7 SI base quantities

★ Exam Tip: Easy 1-mark MCQs ask "which of these IS/ISN'T an SI base unit?" Know this table cold.

Base quantity	SI base unit	Symbol
length	metre	m
mass	kilogram	kg
time	second	s
current	ampere	A
thermodynamic temperature	kelvin	K
luminous intensity	candela	cd
amount of substance	mole	mol

Common mistake: the base unit of mass is kilogram, NOT gram — it's the only base unit that already has a prefix (kilo) built in!

Top 5 rows (length, mass, time, current, temperature) — know these perfectly. Bottom 2 (candela, mole) — good to know but lower priority.

② Prefixes

Always work in powers of 10 — never multiply/divide by 1000, 100 etc. by hand. Convert the prefix straight to its power of 10.

Prefix	Symbol	Power of 10
tera	T	×10¹²
giga	G	×10⁹
mega	M	×10⁶
kilo	k	×10³
— (none)	—	×10⁰ = 1
centi	c	×10⁻²
milli	m	×10⁻³
micro	μ	×10⁻⁶
nano	n	×10⁻⁹
pico	p	×10⁻¹²

Note: capital M = mega (10⁶); lowercase m = milli (10⁻³). Easy to mix up!

Example (illustration):

3 km = 3×10³ m | 500 Gg = 500×10⁹ g | 5 nA = 5×10⁻⁹ A

2. Derived Units

Method: to find the SI base units of any quantity — find a formula for it, then express every quantity in that formula in base units, substituting further formulas wherever needed (e.g. acceleration → Δv/Δt → base units), until everything is in kg, m, s, A, K.

★ Exam Tip: Order of units in your answer does not matter (kg m s⁻² is the same as m s⁻² kg to the examiner) — only the powers matter.

① Worked derivations (build these up from scratch)

Force: F = maa = Δv/Δt = (m s⁻¹)/s = m s⁻²F = kg × m s⁻² = kg m s⁻² (N)

Pressure: P = F/AA = l×l = m²P = kg m s⁻² / m² = kg m⁻¹ s⁻²

Pressure (2nd way): P = ρghρ = m/V = kg/m³ ; g = m s⁻² ; h = mP = kg m⁻³ × m s⁻² × m = kg m⁻¹ s⁻² ✓ same!

Charge: Q = ItA s

Note: whichever formula you use for a quantity, the derived SI base units always come out the same — that's how you can check your working.

Work done: W = Fs= kg m s⁻² × m = kg m² s⁻² (J)

Voltage: V = W/Q= kg m² s⁻² / (A s) = kg m² s⁻³ A⁻¹

Resistance: R = V/I= kg m² s⁻³ A⁻¹ / A = kg m² s⁻³ A⁻²

G.P.E = mgh & K.E = ½mv²both = kg × m s⁻² × m = kg m² s⁻² (J)

② Past-paper questions

m16 qp12 q3MCQ · specific heat capacity

Specific heat capacity has unit J kg⁻¹ K⁻¹. What is this in SI base units?

J = kg m² s⁻² (from work done). So J kg⁻¹K⁻¹ = kg m² s⁻² × kg⁻¹ × K⁻¹ = m² s⁻² K⁻¹ (the two kg's cancel).

C ✓ m² s⁻² K⁻¹

Answer: C

s16 qp13 q2MCQ + structured · stress/fracture, thermal energy, pipe flow

(a) Stress to fracture: σ = k√(γE/d), E = Young modulus, d = distance between atoms (m), k has no units. Find base units of γ.

σ = F/A = kg m⁻¹s⁻² (stress = force/area, using l×l for area). E = stress/strain = kg m⁻¹s⁻² (strain has no units). Square both sides:
(kg m⁻¹s⁻²)² = γ · (kg m⁻¹s⁻²)/m ⇒ kg²m⁻²s⁻⁴ = γ × kg m⁻²s⁻² ⇒ γ = kg s⁻²

B ✓ kg s⁻²

Answer: B

(c) E = cA(T₁−T₂)t / L. Find base units of c.

E (energy) = kg m² s⁻². Rearranging: c = EL / (A(T₁−T₂)t) = (kg m²s⁻² × m)/(m² × K × s) = kg m s⁻³ K⁻¹

Answer: kg m s⁻³ K⁻¹

Pipe-flow questionStructured · viscosity constant

V/t = πPr⁴/(8Cl). P = pressure difference, r = pipe radius, l = pipe length. Find base units of C.

V/t = m³/s. Rearrange: C = Pr⁴/(8l × (V/t)) = (kg m⁻¹s⁻² × m⁴)/(m × m³s⁻¹) = kg m⁻¹⁺⁴⁻¹⁻³ s⁻²⁺¹ = kg m⁻¹ s⁻¹

Answer: kg m⁻¹ s⁻¹ (same answer type keeps appearing on this topic!)

3. Homogeneity of an Equation

RHS derived units = LHS derived units

Definition: every physics equation is homogeneous — the units on both sides must match. If they don't, it isn't a valid physics equation.

Trick: to find the units of an unknown constant, write the base units of every OTHER quantity in the equation, then rearrange to make the constant the subject. Numbers/π/constants with no physical meaning get no units.

① Worked examples

Resistive forceF = crv (sphere in air)

F is resistive force on a sphere of radius r moving at speed v; c is a constant. Find SI base units of c.

kg m s⁻² = (c) × m × m s⁻¹ ⇒ c = kg m s⁻² / (m × m s⁻¹) = kg m⁻¹ s⁻¹

Sphere in fluidF = 6πDRv

F is frictional force, R = radius, v = speed, D depends on the fluid. Show base units of D are kg m⁻¹s⁻¹.

6π is a pure number → no units. kg m s⁻² = (D) × m × m s⁻¹ ⇒ D = kg m⁻¹ s⁻¹

Pipe flowV/t = πPr⁴/(8Cl)

Determine base units of C (nozzle-size constant).

m³/s = (kg m⁻¹s⁻² × m⁴)/(C × m) ⇒ C = kg m⁻¹⁺⁴⁻¹s⁻² × s = kg m⁻¹ s⁻¹

s16 qp12 q2Star luminosity & wind turbine

MCQ: L = 4πr²σT⁴, σ has units W m⁻²K⁻⁴ (W = power = kg m²s⁻³). Find SI base units of L.

L = m² × (kg m²s⁻³ m⁻²K⁻⁴) × K⁴ = kg m² s⁻³

Answer: C (kg m² s⁻³)

(b)(i) Wind turbine power P = CL²ρv³ (L = blade length, ρ = air density, v = wind speed). Show C has no units.

kg m²s⁻³ = (C) × m² × kg m⁻³ × (m s⁻¹)³ = C × m² kg m⁻³ m³s⁻³ = C × kg m²s⁻³ ⇒ C has no units ✓

Watch out: two different questions on this page both use the letter C for a different constant each — always re-derive from that question's own equation, never assume C means the same thing twice!

Gas nozzlem/t = kC√(ρP)

k has no units; C depends on nozzle size; ρ = gas density, P = pressure. Determine base units of C.

kg/s = C × √(kg m⁻³ × kg m⁻¹s⁻²) = C × √(kg² m⁻⁴ s⁻²) = C × kg m⁻²s⁻¹
⇒ C = (kg s⁻¹)/(kg m⁻²s⁻¹) = m²

② Checking if a given equation IS homogeneous

Substitute base units on both sides and confirm they match — you don't need to solve for anything.

Is s = ut + ½at² homogeneous?

LHS: m. RHS: (m s⁻¹)(s) + (m s⁻²)(s²) = m + m = m. LHS = RHS ✓ homogeneous

4. Linear Law

y = mx + c m = gradient = riserun = y₂ − y₁x₂ − x₁ c = y-intercept

gradient m, intercept c

h = ½at² ⇒ plotting h vs t² gives gradient ½a

① How to linearise ANY equation

Force the equation into the shape y = mx + c: whichever quantity/term you choose for the y-axis and x-axis, take the whole power with it — never split a squared term.

Common mistake: for v² = u² + 2as, plotting v against s is wrong — you must plot the whole v² against s (you can't take a square root partway through a linear-law question).

K.E. = ½mv² ⇒ plot K.E. (y) vs mv² (x)gradient = ½ ; passes through origin (c = 0)

v² = u² + 2as ⇒ plot v² (y) vs s (x)gradient = 2a ; y-intercept = u²

★ Exam Tip: examiners will always tell you which axis to use / give you the graph — you are never expected to "guess" the layout from scratch. Just identify what's on y, what's on x, then read off gradient/intercept.

Tip (spotting a linear-law question): look for two unusual quantities you've never graphed together before that form a straight line — that's the giveaway.

② Past-paper questions

s21 qp13 q7MCQ · falling ball

Ball dropped from height h, hits ground after time t. Graph of h vs t² is a straight line through origin, gradient G. Which expression gives acceleration?

h = ½at² ⇒ y=h, x=t² ⇒ G = ½a ⇒ a = 2G

Answer: C (2G)

m21 qp12 q14MCQ · cylindrical rods

Metal rods, same radius r, length l, mass m. Graph of m vs l is a straight line, gradient G. Which expression gives density?

m = ρV = ρπr²l ⇒ y=m, x=l ⇒ gradient G = ρπr² ⇒ ρ = G/πr²

Answer: C (G/πr²)

w20 qp12 q22MCQ · waves in deep water

v² = gλ/2π. Student measures λ and f. Which graph gives a straight line through the origin?

Sub v = fλ: (fλ)² = gλ/2π ⇒ f²λ² = gλ/2π ⇒ f² = (g/2π)×(1/λ). So y = f², x = 1/λ.

D ✓ f² against 1/λA,B,C ✗ don't isolate the λ term as a clean x with the whole f² kept on y

Answer: D

s21 qp13 q30MCQ · diffraction grating

dsinθ = nλ. Graph plotted to find wavelength λ. Which y-axis/x-axis/how-to-find-λ combination is correct?

n = (1/λ)·dsinθ ⇒ y = n, x = dsinθ, gradient m = 1/λ ⇒ λ = 1/gradient

B ✓ y: n, x: dsinθ, λ = 1/gradient

Answer: B

5. Deriving Equations (proportionality method)

This is how physicists actually find formulas — by varying one quantity at a time in an experiment and seeing what changes.

simple pendulum: length l, bob mass m, period T

Experiment finds: T increases when l increases; T unaffected by mass m; T decreases when g decreases.

So: T ∝ l, and T ∝ 1/g (or g^y). Since we don't know the power, write T ∝ lˣ, T ∝ gʸ ⇒ combine:

T = K lˣ gʸ (K = constant, no units — a pure number has no physical dimension)

Homogeneity: s = mˣ · (m s⁻²)ʸ ⇒ s¹ = m^(x+y) s^(−2y) Compare m: 0 = x + y Compare s: 1 = −2y ⇒ y = −½ ⇒ x = +½ T = K l^1/2 g^−1/2 ⇒ T = K√(l/g)

Trick: always leave the "subject" quantity (the one you're finding a formula for, here T) with no power assigned — push every unknown power onto the other side. It doesn't matter which side you choose, but keep it consistent for less messy algebra.

① Worked example: Electrical power

Experiment shows: P increases when current I increases; P increases when resistance R increases.

P ∝ Iˣ Rʸ ⇒ P = kIˣRʸ. Power units: P = W/t = kg m²s⁻²/s = kg m²s⁻³ kg m²s⁻³ = Aˣ⁻²ʸ · kgʸ · m²ʸ · s⁻³ʸ (using R = kg m²s⁻³A⁻²) Compare kg: y=1. Compare m: 2=2y ✓. Compare s: −3=−3y ✓. Compare A: 0=x−2y ⇒ x=2 P = kI²R (constant k = 1, confirmed by the well-known formula)

② Worked example: Kinetic energy

Experiment shows: K.E. increases when v increases; K.E. increases when m increases.

K.E. ∝ mʸvˣ ⇒ K.E. = Kmʸvˣ. kg m²s⁻² = kgʸ · (m s⁻¹)ˣ Compare kg: y=1. Compare m: 2=x. Compare s: −2=−x ⇒ x=2 ✓ consistent K.E. = Kmv², and K.E. vs mv² graph gives gradient K = ½ (found from linear law / experiment)

Note: plotting the y-quantity against the full x-term (mv²) and finding the gradient is exactly how the constant K in front (here ½) is discovered experimentally — deriving equations and linear law are two halves of the same technique.

③ Past-paper questions

Rock densityρ = Mrⁿ/(kL)

n is an integer, k = 2.094 (no units). Use SI base units to show n = −2.

kg m⁻³ = kg·(m)ⁿ/m ⇒ kg m⁻³ = kg m⁽ⁿ⁻¹⁾ ⇒ compare m: −3 = n−1 ⇒ n = −2

w20 qp12 q2MCQ · waves on a wire

v = T^pμ^q, T = tension (force), μ = mass per unit length. Find p and q.

m s⁻¹ = (kg m s⁻²)ᵖ(kg m⁻¹)^q = kg^(p+q) m^(p−q) s⁻²ᵖ
kg: p+q=0 m: p−q=1 s: −2p=−1 ⇒ p=½ ⇒ q=−½

Answer: C (p=½, q=−½)

s18 qp12 q2 & w17 qp13 q2MCQ · drag coefficient / turbine power (no-units constants)

C_d = 2F/(vⁿρA), C_d has no units. Find n. (Second question: P = kρAvⁿ, k has no units, find n.)

Drag: no units = kg m s⁻² / (mⁿs⁻ⁿ · kg m⁻³ · m²) ⇒ m⁰s⁰ = m⁽²⁻ⁿ⁾ s⁽ⁿ⁻²⁾ ⇒ n = 2
Turbine: kg m²s⁻³ = kg m⁻³·m²·mⁿs⁻ⁿ ⇒ 2 = −1+n ⇒ n = 3

Answers: s18 qp12 q2 → B (n=2); w17 qp13 q2 → C (n=3)

Watch out: these two questions look almost identical but give different values of n — always redo the algebra for the specific equation given, don't pattern-match from memory!

6. Precision & Accuracy

Precisionmeasure of the spread/range of your measured values
Accuracymeasure of how close the measured values are to the true value

Trick: Range = highest value − lowest value. Smaller range ⇒ more precise. To judge accuracy, compare the average of your readings to the true value — closer average ⇒ more accurate.

Example (illustration):

Expt 1: 2,4,8,1,5,6,3 → range = 8−1 = 7 → imprecise.
Expt 2: 7.8,8.0,8.2,8.4,7.7,7.9 → range = 8.4−7.7 = 0.7 → precise.
True value = 9.81: an experiment averaging 6.9 is inaccurate; one averaging 9.7 is accurate.

① The four target/distribution diagrams

precise, accurate

precise, inaccurate

accurate, imprecise (symmetric spread)

imprecise, inaccurate (skewed — avg shifts away from true value)

Note: a lopsided (asymmetric) distribution shifts the average away from the true value even if it peaks near it — that data set is called inaccurate because of the skew.

② Precision of a measuring INSTRUMENT

Here precision ≠ spread of your data — it means: an instrument is more precise if it has smaller graduations / can measure smaller values. e.g. micrometer (0.01 mm divisions) is more precise than vernier calipers (0.1 mm divisions).

Common mistake: you can NEVER say one instrument is "more accurate" than another — each gives an accurate reading for its own precision. A vernier reading 2.1 mm and a micrometer reading 2.06 mm (rounds to 2.1) are both accurate; the micrometer is just more precise.

③ Past-paper questions

DefinitionsStructured

Describe (i) high precision (ii) low accuracy for a set of measurements.

(i) The data set will have less range/spread. (ii) The average of the measured values is very far from the true value.

w17 qp13 q4MCQ · vernier vs micrometer

Aluminium sheet marked 3.2 mm. Vernier gives 0.1 mm readings; micrometer gives 0.01 mm. Which statement MUST be correct?

We can't say either is "more accurate" — but the micrometer's smaller divisions definitely make it more precise.

B ✓ the micrometer gives a more precise measurementA,C ✗ can't compare accuracy between different instruments

Answer: B

s16 qp23 q2 & s16 qp11 q4Structured + MCQ

(a) Describe effects of systematic/random errors on micrometer readings. (b) Distinguish precision vs accuracy.

Systematic: readings above/below true value by definite amounts. Random: scattering in readings about the true value.
Precision: the smallest graduation on the instrument. Accuracy: how close the reading is to the true value.

s16 qp11 q4: which case reduces systematic error? → A (zeroing a voltmeter needle before use). Repeating & averaging (B–D) only fixes random error.

w16 qp11 q4MCQ · calibration graph

Voltmeter readings are larger than true values, with a systematic error that varies with voltage. Which calibration graph (true value vs meter reading)?

"Larger than true, varying error" ⇒ the meter-reading line must sit consistently to the right of and diverge from the true y=x line as voltage grows — a straight line offset from the origin/diagonal that isn't just a constant shift.

Answer: A

w18 qp12 q5Structured · X,Y,Z instruments

Three instruments X, Y, Z measure a liquid's volume V (true value shown). Distributions given. How many are precise, how many accurate?

X: wide spread but centred on V ⇒ imprecise, accurate. Y: narrow spread, NOT centred on V ⇒ precise, inaccurate. Z: spread never ends (skewed, tailing off) ⇒ imprecise, inaccurate (average pulled away from V by the long tail of large values).

Answer: A (1 precise, 1 accurate)

s19 qp12 q6MCQ · random errors on a graph

p and q directly proportional; graph of q vs p plotted. Which graph shows random errors?

Random error ⇒ points scattered on both sides of a straight best-fit line through the origin (not a curve, not systematically off to one side).

Answer: C

s20 qp11 q5MCQ · ruler expansion

Ruler used correctly but at a higher temperature than calibration ⇒ ruler expands. What error type, and how improved?

Every reading is wrong by the same proportion always ⇒ systematic. Repeating and averaging improves precision, not accuracy (it can't fix a systematic error).

Answer: D

s20 qp12 q4MCQ · I–V graph

I vs V readings plotted; all points lie neatly ON a straight best-fit line, but the line does not pass through the origin (it cuts the V-axis to the right of 0). Best description of the errors?

Points exactly on the line ⇒ no random error. Line has an intercept (doesn't pass through 0,0) ⇒ systematic error present.

Answer: D (systematic only)

w21 qp11 q5MCQ · classifying 4 error sources

1: meter read from different angles. 2: meter always 5% too high. 3: needle sometimes sticks slightly. 4: meter has a zero error. Classify as random/systematic.

1 & 3 vary unpredictably ⇒ random. 2 & 4 are fixed/definite every time ⇒ systematic.

Answer: B (random: 1,3 · systematic: 2,4)

7. Errors

Systematic errorcauses measured value to deviate from the true value by a fixed amount, always (same direction every time)
Random errorcauses measurements to be above or below the true value by random amounts

① Types & fixes

Error	Examples (cause)	Fix	Graph clue
Random	parallax error (reading scale from different angles); timing oscillations (human reflex varies swing to swing)	repeat & average	scatter plot — points don't all lie on the line
Systematic	zero error; wrongly calibrated scale; reaction time	improve experimental method	line has a non-zero x- or y-intercept

Trick — reaction time: time many oscillations at once (e.g. 20 swings on one stopwatch run) instead of one, then divide by 20. The one fixed reaction-time delay gets divided down too, becoming negligible.

Common mistake: "precision of an instrument" (its smallest graduation) is NOT an error — it's a limitation, and shows up as an uncertainty, not a random or systematic error.

② Reading errors/random vs systematic off a graph

no sys, no random — line through origin, points on it

systematic only — y-intercept, points on the line

random only — through origin, points scattered off line

both — intercept AND scattered points

✓ How to read ANY scatter-graph error question

Points scattered off the best-fit line ⇒ random error present
Best-fit line has a non-zero intercept (when it should pass through 0) ⇒ systematic error present

8. Uncertainties

e.g. 2.0 cm ± 0.5 cm principal value (to same d.p. as uncertainty) ± uncertainty (to 1 significant figure)

Definition: the range within which your measured value lies.

★ Exam Tip: ALWAYS quote uncertainty to 1 significant figure, and the principal (main) value to the same number of decimal places as the uncertainty. Never write "±0.5" and "2.317" together — round to "2.3 ± 0.5".

① Combining uncertainties

Operation	Rule
C = A + B or C = A − B	ΔC = ΔA + ΔB (uncertainties always ADD — even for subtraction!)
C = A × B or C = A / B	ΔC/C = ΔA/A + ΔB/B (fractional/% uncertainties add)
C = AˣBʸ/Dᶻ	ΔC/C = x(ΔA/A) + y(ΔB/B) + z(ΔD/D) (multiply each fractional unc. by its power)

Trick: % uncertainty = fractional uncertainty × 100. You can multiply the 100 through term-by-term, or work entirely in % from the start — same answer either way.

Common mistake: reading a length off a scale (e.g. a ruler with ±0.5 cm uncertainty) needs the uncertainty added twice — once for each end-point you read — even though it's a subtraction (6±0.5) − (2±0.5) = 4 ± 1 cm, not ± 0.5 cm.

ΔC = absolute uncertainty | ΔC/C = fractional uncertainty | (ΔC/C)×100 = percentage uncertainty

② Worked example

A = 8.5 cm ± 0.1 cm, B = 9.5 cm ± 0.1 cm.

C = A+B = 18.0ΔC = 0.1+0.1 = 0.2C = 18.0 ± 0.2 cm

C = B−A = 1.0ΔC = 0.1+0.1 = 0.2 (still ADD)C = 1.0 ± 0.2 cm

C = A×B = 80.75ΔC/C = 0.1/8.5 + 0.1/9.5 ⇒ ΔC = 1.8C = 81 ± 2 cm

C = A/B = 0.895ΔC/C = 0.1/8.5 + 0.1/9.5 ⇒ ΔC = 0.02C = 0.90 ± 0.02

③ Past-paper questions

Pendulum → gg = 4π²L/T²

T = (1.93 ± 0.03) s, L = (92 ± 1) cm. (a) % uncertainty in T and L. (b) % uncertainty in g. (c) g calculated as 9.751 m s⁻²: why is this wrong, and what's the correct value with uncertainty?

(a) %unc(T) = 0.03/1.93×100 = 1.6%. %unc(L) = 1/92×100 = 1.1%
(b) Δg/g = ΔL/L + 2(ΔT/T) = 1.1 + 2(1.6) = 4.3%
(c)(i) 9.751 has more sig figs than the least-precise measurement used to calculate it — can't claim that much precision.
(c)(ii) 4.3% = Δg/9.751 ×100 ⇒ Δg = 0.42 ≈ 0.4 ⇒ g = 9.8 ± 0.4 m s⁻²

Pyramid densityρ = 3m/(hx²)

m=19.5g (±2%), x=4.0cm (±5%), h=4.8cm (±4%). (i) absolute uncertainty in x. (ii) % uncertainty in ρ.

(i) 5% of 4.0 = 0.2 cm
(ii) Δρ/ρ ×100 = %unc(m) + %unc(h) + 2×%unc(x) = 2 + 4 + 2(5) = 16%

s19 qp21 q1v² = kP/ρ

v=3.3×10²ms⁻¹(±3%), P=9.9×10⁴Pa(±2%), ρ=1.29kgm⁻³(±4%). (i) calculate k. (ii) k with absolute uncertainty.

(i) k = v²ρ/P = (3.3×10²)²(1.29)/(9.9×10⁴) = 1.4
(ii) Δk/k ×100 = 2×%unc(v) + %unc(ρ) + %unc(P) = 2(3)+4+2 = 12%. Δk = 0.12×1.4 = 0.168 ≈ 0.2 ⇒ k = 1.4 ± 0.2

w16 qp21 q1m = πd³ρ/6

ρ=8100 kg m⁻³(±5%), m=7.5 kg(±4%). (i) find diameter d. (ii) d with absolute uncertainty.

(i) d = ∛(6m/πρ) = ∛(6×7.5/(π×8100)) = 0.12 m
(ii) d is under a cube root ⇒ each fractional unc. gets ×(1/3): Δd/d = ⅓(Δm/m)+⅓(Δρ/ρ) = ⅓(4)+⅓(5) = 3%. Δd = 0.03×0.121 = 0.0036 ⇒ d = 0.121 ± 0.004 m

w17 qp22 q1σ = 4F/(πd²) — wire stress

d = 0.40±0.02 mm, F = 25.0±0.5 N. (i) show σ=1.99×10⁸ N m⁻². (ii) % uncertainty in σ. (iii) σ with absolute uncertainty.

(i) σ = 4(25)/(π(0.40×10⁻³)²) = 1.99×10⁸ N m⁻²
(ii) Δσ/σ×100 = %unc(F) + 2×%unc(d) = [0.5/25 + 2(0.02/0.4)]×100 = 12%
(iii) Δσ = 12/100 × 1.99×10⁸ = 0.2×10⁸ ⇒ σ = (2.0 ± 0.2)×10⁸ N m⁻²

Note: the board left part (a) — naming a suitable instrument for the wire diameter and how to reduce random error — unanswered; the expected answers are a micrometer screw gauge, and reducing random error by repeating the measurement at different points/orientations along the wire and averaging.

s16 qp13 q4MCQ · y = kML³/(wt³)

L=(80.0±0.2)cm, M=(60±1)g, t=(6.0±0.1)mm, w=(23.0±0.5)mm. Which measurement contributes MOST to the uncertainty in y?

%unc: L→0.2/80×3=0.75%; M→1/60=1.67%; t→0.1/6×3=5.0%; w→0.5/23=2.17%. The power matters — t is cubed, so even a modest % uncertainty gets tripled.

Answer: C (t) — biggest contributor once its power-3 is applied

9. Measuring Instruments

★ Exam Tip: mostly rote memorisation — learn the range and uncertainty of each instrument. Length instruments are asked about most often.

Quantity	Instrument	Range	Uncertainty
Length	metre rule	1 m	0.5 mm
	vernier calipers	10 cm / 100 mm	0.1 mm
	micrometer screw gauge	5 cm / 50 mm	0.01 mm
	measuring tape	> 1 m	0.5 cm
Mass	electronic balance ✓ (top-pan / lever balance also exist)	—	—
Angle	protractor	—	0.5°
Time	analogue stopwatch	—	0.1 s
	digital stopwatch	—	0.01 s
	ticker tape	—	0.02 s
Temperature	alcohol thermometer · mercury thermometer · digital thermometer · thermocouple
Voltage	voltmeter · CRO (also analyses waveforms)
Current	ammeter

Trick — where does "uncertainty = ½ smallest division" come from? On an analogue scale you can confidently read to the middle of the smallest division (e.g. 3.5 mm between the 3 mm and 4 mm lines) but no finer — so the uncertainty is half the smallest division.

Common mistake: comparing a value that "seems more accurate" (e.g. micrometer's extra decimal place) isn't automatically better — the vernier calipers giving 2.1 mm is just as valid within ITS OWN precision; you can't force decimal places an instrument's graduations don't allow.

Note: vernier calipers can uniquely measure BOTH the inside and outside diameter of a pipe (inside + outside jaws) — the micrometer cannot reach inside a pipe.

Trick: the micrometer is the most precise but has the smallest range. If you need 0.01 mm precision but the object is 7 cm long, you can't use it — you must trade precision for range and use vernier calipers instead.

① Past-paper questions

m17 qp12 q5MCQ · choosing an instrument

Need to measure ~10 cm to a precision of 0.01 cm (0.1 mm). Which instrument?

Metre rule only reaches 0.5 mm; micrometer's range (5 cm) is too short; vernier calipers reach 10 cm range at 0.1 mm precision — fits perfectly.

Answer: D (vernier calipers)

s17 qp12 q5MCQ · micrometer reading → area

Micrometer: main scale reads 3.5 mm (0.5 mm pitch shown as "3" + half), thimble reads 0.31 mm. Find cross-sectional area.

Diameter = 3.5 + 0.31 = 3.81 mm ⇒ Area = π(3.81/2)² = 11.4 mm²

Answer: B

w18 qp11 q4MCQ · micrometer with zero error

Diagram 1 (wire in jaws): reading = 2.59 mm. Diagram 2 (jaws fully closed, wire removed): reading = −0.14 mm (zero error). Find true diameter.

True diameter = reading − zero error = 2.59 − (−0.14) = 2.45 mm — remember to subtract a negative zero error, i.e. add it back.

Answer: B

w19 qp11 q4MCQ · choosing instruments for wire & pipe

Measure a wire (~1 mm diameter) and internal diameter of a pipe (~2 cm) accurately. Which instruments?

1 mm wire needs micrometer's fine 0.01 mm precision. A 2 cm internal diameter needs the vernier's inside jaws (a micrometer can't reach inside a pipe).

Answer: C (wire → micrometer, pipe → calipers)

10. Scalars & Vectors

Scalarcompletely specified by magnitude + unit only. e.g. distance, speed, mass, time, temperature
Vectorneeds a direction as well as magnitude + unit to be fully specified

Example (illustration):

2 kg bucket + 5 kg water = 7 kg total — masses (scalars) just add directly, no direction needed.

Resultant = a single vector that has the same effect as two (or more) vectors combined

① Vectors along the SAME line of action

Just add algebraically using a sign convention (+ one way, − the other).

same direction: R = 3+5 = +8 N

opposite: R = 3−8 = −5 N

② Vectors along DIFFERENT lines of action

head-to-tail rule: R = √(3²+5²), tanθ=5/3

parallelogram law: diagonal from the shared tail = resultant

Note: the parallelogram law is really just the head-to-tail rule in disguise — copy one vector to the tip of the other and you get the same diagonal resultant either way.

③ Resolution of a vector

opposite of θ ⇒ sinθ; adjacent to θ ⇒ cosθ

Trick (very important, used everywhere): Don't memorise "horizontal = cos, vertical = sin" — that only works when θ is measured from the horizontal! Instead always check: is this side opposite the angle, or adjacent to it?
opposite ⇒ (resultant)×sinθ | adjacent ⇒ (resultant)×cosθ

Tip: after resolving, redraw both components acting from the same starting point as the original vector (not tip-to-tail) — this avoids mistakes later when taking moments.

★ Exam Tip: scale diagrams are valid but slow and rarely required — exams almost always accept (and expect) the maths/resolution method, since most of the paper is calculation-based anyway.

④ Worked example — adding two inclined vectors

Add a 10 N force at 70° to the horizontal, to a 500 N force at 20° to the vertical.

original set-up

Horizontal (→ +): Rx = 10cos70 + 500sin20 ≈ 174 NVertical (↓ +): Ry = 500cos20 − 10sin70 ≈ 460 NR = √(174² + 460²), tanθ = 460/174

⑤ Past-paper questions

Aircraft in windStructured

Aircraft velocity in still air = 95 m s⁻¹ west. Wind velocity = 28 m s⁻¹, from 65° south of east (i.e. 65° to the aircraft's line, opposing it partly). Find resultant velocity magnitude.

Rx = 95 + 28cos65 ≈ 106.8. Ry = 28sin65 ≈ 25.4.
R = √(Rx² + Ry²) = √(106.8² + 25.4²) ≈ 110 m s⁻¹ (2 s.f.)

s16 qp11 q1MCQ · resultant of 3 wire forces

Three wires pull horizontally on a vertical pole (one set of arrows shown). Which vector diagram shows the resultant R using head-to-tail?

Chain all three force arrows head-to-tail in order; the resultant R is the single arrow from the very first tail to the very last head.

s16 qp12 q3MCQ · vector subtraction X = Y − Z

X and Y are drawn to scale. If X = Y − Z, which diagram best represents Z?

Rearrange: Z = Y − X = Y + (−X). Draw Y, then add the reverse of X head-to-tail, and Z is the vector closing the triangle.

Quick Recap — Full Chapter Cheat-Sheet

✓ Everything on one page

SI base quantities (7): length(m), mass(kg), time(s), current(A), temperature(K), luminous intensity(cd), amount of substance(mol)
Derived units: find a formula → substitute base units → simplify. Same quantity always gives the same base units, whichever formula you use.
Homogeneity: LHS units = RHS units always, in every real physics equation.
Linear law: force into y = mx + c; take whole powers onto one axis; gradient & intercept read off the graph.
Deriving equations: y ∝ xⁿ (unknown power) → combine → use homogeneity to solve for each power.
Precision = small spread/range (repeatability). Accuracy = close to true value.
Systematic error: fixed deviation, always same direction → fix by improving method/calibration. Shows as an intercept on a graph.
Random error: scatter, unpredictable → fix by repeating & averaging. Shows as scattered points off the line.
Uncertainties: + or − ⇒ ADD absolute uncertainties. × or ÷ ⇒ ADD fractional/% uncertainties (× the power if any). Quote unc. to 1 s.f., value to matching d.p.
Instruments: metre rule (1m, 0.5mm) · vernier calipers (10cm, 0.1mm, inside+outside) · micrometer (5cm, 0.01mm, most precise) · measuring tape (>1m, 0.5cm)
Vectors: same line → algebraic sum. Different lines → head-to-tail / parallelogram, or resolve into components (opposite→sin, adjacent→cos) and recombine with Pythagoras.