Propagating Errors: Products, Quotients & Powers

There is a question JEE has asked, in some costume, nearly every year for two decades: the mass, radius and length of a cylinder are measured with errors of 1%, 2% and 1% — what is the maximum percentage error in its density? Students who learned a vague "errors add" mantra answer 4%. The mark scheme says 6%. The missing 2% hides in one squared symbol: density is $\rho = m/(\pi r^2 l)$ , and that little ² silently doubles the radius error before it reaches your answer. That is the whole drama of this module. When quantities are added or subtracted, absolute uncertainties add — the previous module's story. But the moment quantities are multiplied, divided or raised to powers — how nearly every formula in physics is built — the bookkeeping switches currency: percentage errors add, and every power $n$ is an amplifier with gain $|n|$ . One rule, one amplifier, and you can put an honest error bar on any formula a paper throws at you: $g$ from a pendulum, resistivity from a wire, viscosity from a falling sphere. The stakes are real in both exam systems this module serves: JEE-Main and NEET ask the "maximum percentage error" question as a fifteen-second formula read; CIE A-Level practicals make you run the arithmetic on your own data. The rule takes five minutes; the traps — the squared radius, the negative power, the error counted twice — are where the marks live.

Key takeaways

For products and quotients, relative (percentage) errors add: $z = ab$ or $a/b$ gives $\frac{\Delta z}{z} = \frac{\Delta a}{a} + \frac{\Delta b}{b}$ — in the worst case they never cancel.
A power is an amplifier: $z = a^n$ gives $\frac{\Delta z}{z} = |n|\,\frac{\Delta a}{a}$ — squaring doubles a percentage error, a square root halves it, the sign of $n$ never matters.
Master formula: $z = \dfrac{a^p\, b^q}{c^r}$ has maximum relative error $p\frac{\Delta a}{a} + q\frac{\Delta b}{b} + r\frac{\Delta c}{c}$ — all contributions positive, denominator's included.
Pure numbers ( $\pi$ , 2, $4\pi^2$ ) carry zero error; only measured quantities contribute.
Exams mean the maximum (worst-case) error, where everything adds; real labs add independent errors in quadrature.
Key idea: multiplication passes errors through as percentages, so the relative error of a product is the sum of the relative errors of its factors, each multiplied by the magnitude of its power.

The core insight

When quantities combine by multiplication, division or powers, each factor's relative error — weighted by the magnitude of its exponent — adds to give the maximum relative error of the result: $\frac{\Delta z}{z} = p\frac{\Delta a}{a} + q\frac{\Delta b}{b} + r\frac{\Delta c}{c}$ for $z = a^p b^q / c^r$ .

Where this lives today

Every accredited laboratory on the planet runs the professional version of this module daily: the GUM's quadrature rules govern how a calibration lab states a standard resistor's uncertainty and how CERN quotes the Higgs boson's mass — each a propagated budget in which one amplified term usually dominates. The most famous open budget is Newton's constant: CODATA lists $G$ with relative uncertainty around $2.2 \times 10^{-5}$ , thousands of times cruder than other fundamental constants, because $G$ enters torsion-balance formulas through geometry factors that amplify every dimensional error — the squared-radius problem at research grade. The same arithmetic runs the semiconductor industry's 2026 roadmap: a fab overlays circuit layers to within about a nanometre across a 300 mm wafer, every stage tracked as a propagated percentage budget so engineers know which dictator term to attack next. From pendulum lab to chip fab, "find the biggest term in the error sum and fix that" is the working strategy of precision.

The physical picture

Think of a formula as a pipeline. Raw measurements enter on one side, each carrying a tag stating its percentage error; the formula multiplies, divides and squares them; one number flows out. What does the output tag read? The answer has a beautiful shape: percentage errors travel through multiplication unchanged — a factor known to 1% contributes 1% of doubt whether it multiplies or divides — but wherever the pipeline raises a quantity to a power, the tag passes through an amplifier: $r^2$ doubles the tag on $r$ , $\sqrt{c}$ halves the tag on $c$ . At the output, all the amplified tags add up.

Figure

The error pipeline — how percentage errors flow through the density formula

Why do percentages travel so cleanly, when absolute errors do not? Because multiplication is scaling. If your radius reading is 2% too big, then $r^2$ is about 4% too big and the density comes out about 4% too small — whether the radius was millimetres or metres. A percentage slip in any factor produces the same percentage slip in the product. Absolute errors have no such privilege: 0.5 mm matters enormously to a 5 mm wire and not at all to a 5 m rod. And why does a power multiply the error rather than just pass it along? Geometry answers best. Take a square of side $a$ and let the side be uncertain by $\Delta a$ . The area can grow along two edges at once:

Figure

Why squaring doubles the percentage error — the growing square

Two strips, each of area $a\,\Delta a$ — that pair is the factor 2 in the power rule; a cube would grow on three faces. See powers as repeated multiplication of the same uncertain number — an error striking identically each time, never cancelling itself — and the entire module reduces to one sum.

The rules, one at a time

The vocabulary comes from the previous modules: for a measured quantity $a$ with absolute error $\Delta a$ , the relative error is the ratio $\Delta a / a$ , times 100 for the percentage. Suppose two independent measurements combine as a product or quotient. In the worst case both conspire — numerator high, denominator low — pushing the result as far from truth as the errors allow. Working that worst case through (Derivation A) gives the central rule:

\frac{\Delta z}{z} = \frac{\Delta a}{a} + \frac{\Delta b}{b} \qquad \text{for } z = ab \ \text{ or } \ z = \frac{a}{b} \qquad (1)

where: $\Delta z / z$ — the maximum relative error of the result; $\Delta a / a$ , $\Delta b / b$ — the relative errors of the factors. Read it physically: it does not matter whether $b$ multiplies or divides — an uncertain denominator corrupts the result exactly as much as an uncertain numerator, so both enter with a plus sign. A speed $v = d/t$ built from a distance good to 1% and a time good to 2% is trustworthy to 3%, never 1%; the "errors cancel in a ratio" instinct is precisely what the plus signs forbid. Powers come next; the growing-square picture already told us the answer. Raising a measurement to the power $n$ multiplies its relative error by $|n|$ :

\frac{\Delta z}{z} = |n|\,\frac{\Delta a}{a} \qquad \text{for } z = a^n \qquad (2)

where: $n$ — any exponent: positive, negative, or fractional. The absolute-value bars are load-bearing: for $z = 1/b = b^{-1}$ , a denominator that reads low pushes the result high — the error is as large as ever, merely pointing the other way, and "maximum error" cares only about size. Fractional powers run in reverse gear: $\sqrt{a} = a^{1/2}$ halves a percentage error, which is why square-root formulas forgive sloppy inputs and squaring formulas punish them. Every exam formula is some assembly of these two rules; write the general case once and recognise it forever. For a quantity built as a product of powers,

z = \frac{a^p\, b^q}{c^r} \quad \Longrightarrow \quad \frac{\Delta z}{z} = p\,\frac{\Delta a}{a} + q\,\frac{\Delta b}{b} + r\,\frac{\Delta c}{c} \qquad (3)

where: $p$ , $q$ , $r$ — the magnitudes of the exponents, all entering positively even though $c$ sits in the denominator. This one line answers most error questions ever set: read each exponent off the formula, multiply it by that quantity's percentage error, add everything. Constants like $\pi$ or $4\pi^2$ are pure numbers with zero error — they scale $z$ but contribute nothing to its uncertainty. The router:

the formula does	what adds	rule
$a + b$ , $a - b$	absolute errors: $\Delta a + \Delta b$	previous module (sums)
$ab$ , $a/b$	relative errors: $\frac{\Delta a}{a} + \frac{\Delta b}{b}$	equation (1)
$a^n$	relative error, amplified: $\|n\| \frac{\Delta a}{a}$	equation (2)
$a^p b^q / c^r$	weighted relative errors, all positive	equation (3)
$\pi$ , numeric constants	nothing	zero error

Derivation AWhy relative errors add for products and quotients

We want the maximum possible error in $z = ab$ when each factor is known only to within its absolute error, and we want to see why the answer comes out in relative terms. The strategy is direct: build the worst case by pushing both factors to the same extreme, expand the product, and discard what is genuinely negligible. Watch the one approximation made — dropping the doubly-small term — because it is also the honest limit of the whole method.

Step 1. Push both factors high: $z_{\max} = (a + \Delta a)(b + \Delta b) = ab + a\,\Delta b + b\,\Delta a + \Delta a\,\Delta b$ .

Step 2. Drop the last term. $\Delta a \, \Delta b$ is a small quantity times a small quantity: if both errors are 1%, their product is 1% of 1% — one part in ten thousand, invisible next to the two first-order terms. Why we may drop it: error analysis is bookkeeping of small quantities to first order; a second-order term is smaller than the doubt we already carry in the error estimates themselves.

Step 3. The maximum absolute error is then $\Delta z = z_{\max} - ab = a\,\Delta b + b\,\Delta a$ . Divide through by $z = ab$ : the mixed products untangle, giving $\frac{\Delta z}{z} = \frac{\Delta a}{a} + \frac{\Delta b}{b}$ — the relative errors add.

Step 4. For the quotient $z = a/b$ , the worst case pairs a high numerator with a low denominator: $z_{\max} = \frac{a + \Delta a}{b - \Delta b}$ . Using $\frac{1}{1 - x} \approx 1 + x$ for small $x = \Delta b/b$ , this becomes $z_{\max} \approx \frac{a}{b}\left(1 + \frac{\Delta a}{a}\right)\left(1 + \frac{\Delta b}{b}\right)$ , and multiplying out to first order lands on exactly the same sum as Step 3. Why the denominator adds rather than subtracts: dividing by too small a number makes the result too big — the numerator's excess and the denominator's deficit push $z$ in the SAME direction, so their effects reinforce. *Verified: take $a = 100 \pm 1$ (1%) and $b = 50 \pm 1$ (2%). Exactly, $z_{\max} = 101 \times 51 = 5151$ against $z = 5000$ : excess $151/5000 = 3.02\%$ , and the rule predicts $1\% + 2\% = 3\%$ — the 0.02% gap is precisely the dropped second-order term.*

Go deeper · Derivation B · 1 minThe power rule, including roots and negative powers

Derivation BThe power rule, including roots and negative powers

We want $\frac{\Delta z}{z} = |n| \frac{\Delta a}{a}$ for $z = a^n$ , and we want it for every exponent an exam can produce: positive integers, fractions like the square root, and negative powers. The route is to lean entirely on Derivation A — a power is a repeated product — and then to pull roots and reciprocals out of the same machine by reading it backwards. The subtle point to watch: the factors in $a \cdot a$ are not independent measurements, and that is exactly why the errors reinforce rather than partially cancel.

Step 1. Squares first. $z = a^2$ is the product $a \cdot a$ , so rule (1) applies with both factors the same: $\frac{\Delta z}{z} = \frac{\Delta a}{a} + \frac{\Delta a}{a} = 2\,\frac{\Delta a}{a}$ . Why cancellation is impossible here: with two independent factors, one might read high while the other reads low. But $a \cdot a$ uses the SAME reading twice — if it is 2% high, it is 2% high in both slots, every time. Identical errors cannot cancel each other.

Step 2. Any positive integer follows by repetition: $a^3 = a^2 \cdot a$ adds one more copy, $\frac{\Delta z}{z} = 3\frac{\Delta a}{a}$ , and in general $n$ copies of the same relative error give $n\,\frac{\Delta a}{a}$ — the three growing faces of the cube in the second figure.

Step 3. Fractional powers, read backwards. Let $z = \sqrt{a}$ , so that $z^2 = a$ . Apply Step 1 to the left side: $2\,\frac{\Delta z}{z} = \frac{\Delta a}{a}$ , hence $\frac{\Delta z}{z} = \frac{1}{2}\frac{\Delta a}{a}$ . The square root is a de-amplifier with gain one half, and the same inversion gives gain $1/3$ for cube roots — consistent with $|n|$ for $n = \tfrac12, \tfrac13$ .

Step 4. Negative powers. For $z = b^{-1} = 1/b$ , the worst case is dividing by the low extreme: $z_{\max} = \frac{1}{b - \Delta b} \approx \frac{1}{b}\left(1 + \frac{\Delta b}{b}\right)$ , so $\frac{\Delta z}{z} = \frac{\Delta b}{b}$ — full size, opposite direction. Combining with Step 2, $z = b^{-n}$ carries $n\,\frac{\Delta b}{b}$ . The magnitude of the exponent is all that survives: $\frac{\Delta z}{z} = |n|\frac{\Delta a}{a}$ . Verified: $a = 100 \pm 2$ (2%). Squaring exactly: $102^2 = 10404$ , excess $404/10000 = 4.04\%$ against the predicted $4\%$ . Rooting exactly: $\sqrt{102} = 10.0995$ , excess $0.0995/10 = 1.0\%$ against the predicted $1\%$ . Both check to first order.

One honest aside before the drill work. These add-everything rules compute the maximum error — the worst case in which every measurement conspires against you, the convention JEE, NEET and A-Level mark schemes all assume. Professional labs, following the international Guide to the Expression of Uncertainty in Measurement, instead add independent errors in quadrature — square each term, sum, take the root — because independent random errors rarely all strike the same way. Quadrature gives the smaller, realistic estimate; the linear sum a guaranteed ceiling. For every exam this module serves: add linearly, and keep quadrature filed away as the way real labs live.

Limiting cases

case	result	meaning
one factor exact ( $\Delta b = 0$ )	$\frac{\Delta z}{z} = p\,\frac{\Delta a}{a}$	constants and exact values drop out of the sum — $\pi$ contributes nothing
$n = 1$	$\frac{\Delta z}{z} = \frac{\Delta a}{a}$	a lone factor passes its error straight through, gain 1
$n = \tfrac12$ (square root)	error halved	roots forgive: a 4% error under a square root emerges as 2%
$n = -1$ (reciprocal)	error unchanged in size	dividing flips the direction of the error, never its magnitude
one error dominates (e.g. $2\%$ vs $0.1\%$ )	$\frac{\Delta z}{z} \approx$ dominant term	the error budget is a dictatorship — improve the worst measurement first
errors not small ( $\gtrsim 10\%$ )	first-order rule underestimates	the dropped $\Delta a\,\Delta b$ terms wake up; the method assumes small errors

The dominant-term row is the practical soul of the subject. If $g = 4\pi^2 L/T^2$ is built from $L$ known to 0.1% and $T$ to 0.5%, the period contributes $1\%$ — ten times the length's share — so a better metre rule is pointless and better timing is everything. Reading a budget to find its dictator is how experimenters decide what to improve, and exactly the reasoning JEE-Advanced likes to probe.

Common trapFor a product you add the absolute errors — no, that mixes currencies

The addition-subtraction rule ( $\Delta z = \Delta a + \Delta b$ ) is drilled first, so under pressure it leaks into product questions: students compute $\Delta \rho$ by adding $\Delta m$ , $\Delta r$ , $\Delta l$ in raw units — dimensional nonsense, grams plus centimetres, hiding inside symbols until the final line exposes it. Currencies must match the operation: sums and differences trade in absolute errors; products, quotients and powers trade in relative errors. Convert every input to a percentage the moment you see multiplication, keep all bookkeeping in percent, and convert back to an absolute error only at the very end if asked. The reverse leak is just as damaging: relative errors on a subtraction, where the correct absolute-error sum can make the relative error of a small difference explode.

Worked examples

Board level — density of a cylinder, the full method

The mass of a solid cylinder is measured as $m = 4.00 \pm 0.04$ g, its radius as $r = 0.250 \pm 0.005$ cm, and its length as $l = 6.00 \pm 0.06$ cm. Find the density and the maximum percentage error in it, and report the result with its absolute uncertainty.

Step 1. Convert every input to a percentage — the currency for a formula made of products: $\frac{\Delta m}{m} = \frac{0.04}{4.00} = 1\%$ ; $\frac{\Delta r}{r} = \frac{0.005}{0.250} = 2\%$ ; $\frac{\Delta l}{l} = \frac{0.06}{6.00} = 1\%$ .

Step 2. Write the formula and read off the exponents: $\rho = \dfrac{m}{\pi r^2 l}$ has $p = 1$ on $m$ , $r = 2$ on $r$ , and $1$ on $l$ ; $\pi$ is exact.

Step 3. Apply equation (3): $\frac{\Delta \rho}{\rho} = 1\% + 2(2\%) + 1\% = \boxed{6\%}$ . The radius contributes four of the six points — the square is the story.

Step 4. Compute the central value: volume $V = \pi r^2 l = \pi (0.250)^2 (6.00) = 1.178\ \mathrm{cm^3}$ , so $\rho = \frac{4.00}{1.178} = 3.395\ \mathrm{g/cm^3}$ .

Step 5. Convert back to an absolute error: $\Delta \rho = 0.06 \times 3.395 = 0.20\ \mathrm{g/cm^3}$ . Report $\boxed{\rho = 3.4 \pm 0.2\ \mathrm{g/cm^3}}$ — uncertainty to one significant figure, value cut at the same decimal place.

Competitive level — g from a pendulum (JEE-Main / NEET pattern)

In a simple-pendulum experiment, the length is measured as $L = 1.000$ m with a metre scale of least count 1 mm, and the time for 20 oscillations is $t = 40.0$ s measured with a stopwatch of least count 0.2 s. Find the maximum percentage error in the value of $g$ .

Step 1. Rearrange $T = 2\pi\sqrt{L/g}$ to $g = 4\pi^2 L / T^2$ . Exponents: $1$ on $L$ , $2$ on $T$ (denominator — magnitude counts); $4\pi^2$ is exact.

Step 2. Percentage errors of the measured quantities: $\frac{\Delta L}{L} = \frac{0.001}{1.000} = 0.1\%$ ; $\frac{\Delta t}{t} = \frac{0.2}{40.0} = 0.5\%$ .

Step 3. The period is $T = t/20$ ; dividing by an exact 20 scales the value but not the relative error, so $\frac{\Delta T}{T} = \frac{\Delta t}{t} = 0.5\%$ — which is exactly why experimenters time 20 oscillations: the same 0.2 s doubt, diluted over 40 s.

Step 4. Assemble: $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\,\frac{\Delta T}{T} = 0.1\% + 2(0.5\%) = \boxed{1.1\%}$ . (Here $g = 4\pi^2/(2.00)^2 = \pi^2 = 9.87\ \mathrm{m/s^2}$ , so $\Delta g \approx 0.11\ \mathrm{m/s^2}$ .) Timing note: 60–90 seconds. The only decisions are the exponent on $T$ and realising $\Delta T/T = \Delta t/t$ ; everything else is arithmetic. If you catch yourself computing $\Delta L + \Delta t$ , stop — wrong currency.

Competitive level — the pure pattern question

A physical quantity is given by $P = \dfrac{a^2\, b^3}{\sqrt{c}\; d}$ . The percentage errors in the measurement of $a$ , $b$ , $c$ and $d$ are 1%, 2%, 4% and 2% respectively. Find the maximum percentage error in $P$ .

Step 1. Read exponent magnitudes straight off the formula: $2$ on $a$ , $3$ on $b$ , $\tfrac12$ on $c$ , $1$ on $d$ . Denominator placement is irrelevant to the error — every term enters with a plus.

Step 2. Weighted sum, equation (3): $\frac{\Delta P}{P} = 2(1\%) + 3(2\%) + \tfrac12(4\%) + 1(2\%) = 2\% + 6\% + 2\% + 2\% = \boxed{12\%}$ .

Step 3. Sanity-read the budget: $b$ alone contributes half the total — in a real experiment, remeasuring $b$ (not the worse-looking $c$ , already tamed to 2% by its square root) is what would rescue the result. Timing note: 30–45 seconds, entirely mechanical. This template appears in JEE-Main archives nearly verbatim year after year; the only trap is rushing the $\tfrac12$ on the square root.

Exam traps

The un-doubled square. The single most-paid-for mistake in this topic: reading $r^2$ but adding $\frac{\Delta r}{r}$ once. Underline every exponent before summing; the exponent is the multiplicity.

Sign-of-the-power confusion. $z = a^{-2}$ contributes $2\frac{\Delta a}{a}$ , not $-2\frac{\Delta a}{a}$ and not zero. If a term in your maximum-error sum is negative, you have made an error about errors.

Currency mixing. Adding absolute errors where the formula multiplies, or percentages where it subtracts. The router table is the antidote: identify the operation, pick the currency, convert only at the borders.

The diameter-radius shuffle. Measuring diameter $d$ and using $r = d/2$ leaves the relative error unchanged (exact 2 changes nothing) — do not halve it; and $r^2$ built from that one reading still carries $2\frac{\Delta d}{d}$ , per Step 1 of Derivation B.

Quoting the error wrong at the finish. Mark schemes want the uncertainty to 1 significant figure and the value cut at the matching decimal place: $3.4 \pm 0.2$ , never $3.395 \pm 0.204$ — a free mark kept or thrown away in the last five seconds (see the Significant Figures module).

Board conventions. JEE, NEET and CIE A-Level all mean linear worst-case addition by "maximum error"; CIE additionally asks which quantity dominates the budget when the question turns to improving the experiment. Where a value of $g$ is needed, use exactly what the paper supplies ( $9.81$ for CIE unless stated; JEE/NEET state $9.8$ or $10$ per paper) — a supplied constant is exact and never enters your error sum.

Origins · 1 minPierre-Simon Laplace and how this was discovered

Pierre-Simon Laplace

1749–1827 · Paris

Laplace made error a subject of mathematics rather than embarrassment. As the leading celestial mechanician of his age, he faced the defining data problem of the 1700s: observations of the same planet, by different astronomers with different instruments, never quite agreed. His probability memoirs asked the modern question — given scattered, error-laden readings, what is the most believable true value, and how wrong could it be? His 1810 central limit theorem explained why heaps of small independent errors pile into the same bell curve whatever their origins, giving error analysis its universal law. The propagation rules here are the worst-case, exam-friendly cousins of the machinery Laplace built.

History

1774

— Laplace's first memoir on the probability of causes: reasoning backwards from scattered data to the true value begins.

1805

— Legendre publishes the method of least squares for reconciling conflicting measurements.

1809

— Gauss's Theoria Motus derives the normal law of errors and ties least squares to it.

1810

— Laplace proves the central limit theorem: sums of many small independent errors are bell-curved, whatever their source.

1993

— The BIPM's Guide to the Expression of Uncertainty in Measurement (GUM) standardises error propagation for every accredited lab on Earth.

Going further

Texts. NCERT Physics Part 1, Ch. 1 — the JEE/NEET canonical statement of maximum-error rules · H.C. Verma, Concepts of Physics Vol. 1, Ch. 2 · CIE 9702 syllabus, Practical Assessment appendix · J.R. Taylor, An Introduction to Error Analysis, Ch. 3 — the quadrature story, one level up.

Provenance. Standard first-order maximum-error conventions per NCERT Ch. 1 and CIE 9702 practical guidance. A verifier should recompute: $101 \times 51 = 5151$ vs $5000 \to 3.02\%$ vs $3\%$ ; $102^2 = 10404 \to 4.04\%$ ; $\sqrt{102} = 10.0995 \to 1.0\%$ ; density $1 + 2(2) + 1 = 6\%$ , $\rho = 4.00/1.178 = 3.395 \to 3.4 \pm 0.2\ \mathrm{g/cm^3}$ ; pendulum $0.1 + 2(0.5) = 1.1\%$ , $g = \pi^2 = 9.87$ ; pattern $2(1) + 3(2) + \tfrac12(4) + 2 = 12\%$ .

Prerequisites. Absolute, Relative & Percentage Errors · Propagating Errors: Sums & Differences (the sibling module — absolute-error currency). Leads to. Least Count & Instrument Errors · Error-Question Patterns (the technique drill built on this module) · Significant Figures, Rounding & Reporting.

Why do errors ADD when I divide? Shouldn't dividing two too-big numbers cancel the errors?

Sometimes it does — if numerator and denominator both happen to read high, a ratio partially self-corrects. But maximum error asks about the worst case: numerator high while denominator reads low, both slips pushing the quotient the same way. Since you cannot know which way each error actually went (that is what "error" means), the honest ceiling adds the two relative errors. The lucky-cancellation scenario is real but unbankable — and averaging over both situations is exactly what the professional quadrature rule captures.

Is the rule exact, or an approximation? My exact worst-case calculation gives a slightly bigger number.

It is a first-order approximation, and your observation is correct: expanding $(1 + \frac{\Delta a}{a})(1 + \frac{\Delta b}{b})$ exactly leaves the cross-term, so the true worst case for 1% and 2% inputs is 3.02%, not 3.00%. For few-percent errors the discrepancy is far smaller than the errors themselves, so everyone drops it. The rule only degrades when individual errors reach tens of percent — at which point you should be redesigning the measurement, not propagating it.

When do I add in quadrature instead of adding directly?

Ask which convention owns the paper in front of you. JEE, NEET, boards and CIE A-Level all mean the linear worst-case sum when they say "maximum error" — use equations (1)–(3) as written. Quadrature ( $\sqrt{\text{sum of squares}}$ ) is the international-lab convention for independent random errors, which rarely all conspire; it always gives a smaller number and becomes your default in degree-level lab work. Writing it in an exam that asked for maximum error loses the mark, because it answers a different question.

Why does the same measurement squared count twice, but two different measurements each count once? Both are two factors.

Independence is the difference. Two different measurements might err in opposite directions — the worst case adds them, but cancellation was at least possible. The same reading used twice has no such freedom: if your radius is 2% high, $r \times r$ is 2% high in both slots, guaranteed. The worst case and the only case coincide, giving exactly $2\frac{\Delta r}{r}$ — and relabeling one measurement as $d/2 \times d/2$ never manufactures independence.

Why do we time 20 oscillations instead of one? The stopwatch error is the same 0.2 s either way.

Exactly — and that is the trick. The absolute error stays 0.2 s, but spread over a 40 s measurement instead of 2 s, the relative error shrinks from 10% to 0.5%. Since $T = t/20$ divides by an exact number, that diluted percentage passes through unchanged, and $g$ (which doubles it) inherits 1% instead of 20%. The craft lesson: when a least count is fixed, measure the largest multiple you can and divide by an exact count — the same idea as measuring the thickness of 100 pages rather than one.

The question gives errors in some quantities and least counts in others — what do I do with the least counts?

Convert them to the same currency before summing. A least count is the instrument's smallest step, and for a single reading it is the absolute error to assign: a metre scale with 1 mm divisions contributes $\Delta L = 1$ mm (some boards use half a division — follow yours). Divide by the measured value and it joins the sum as a relative error like any other term. The next module, Least Count & Instrument Errors, covers where instrument uncertainties come from.