TECHNIQUE: Error-Question Patterns

Open almost any JEE-Main session or NEET paper and there it is — one question on errors, near the front of the physics section, worth as much as the hardest rotation problem. Students who fear it lose four marks in ninety seconds of confused re-derivation; students who have drilled it collect the same marks in thirty seconds, because they know a secret the paper-setter cannot hide: the error question is always the same six questions. That is not folklore; it is a consequence of the syllabus. The measurement unit hands the examiner a small toolbox — absolute and relative errors, the sum rule, the product-and-power rule, least count, the reporting conventions, the pendulum experiment — and each tool admits exactly one MCQ-shaped costume. Six templates, endlessly re-skinned with new letters and new numbers. You built the machinery in the concept modules — the error modules, Significant Figures, Least Count & Instrument Errors. This module is the choreography: the recognition cue for each template, the one-line answer that resolves it, and a drill ladder that runs each one against the clock. By the end, an error question should feel less like a problem and more like a handshake you have practised six times.

Key takeaways

Every error MCQ in the JEE/NEET archive is one of six templates; recognizing which one you are facing is 80% of the solve.
T1, the exponent harvest: for $Z = a^p b^q / c^r$ , the maximum % error is $|p|\,\%a + |q|\,\%b + |r|\,\%c$ — read the exponents, never re-derive.
Sums and differences switch currencies: absolute errors add there, and a small difference of two large numbers carries a huge percentage error.
A single reading is uncertain by its least count; timing $n$ oscillations grows the measured time $n$ -fold while the stopwatch error stays fixed, shrinking the relative error $n$ -fold.
Report errors to one significant figure, ending the value at the same decimal place — the reporting ritual is itself a tested template.
Key idea: name the template first — then harvest exponents (products) or add absolutes (sums), and the one-line answer falls out in 30–60 seconds.

The core insight

Exam error questions come in six recurring templates, each with a fixed one-line resolution — identify the template from its phrasing cue, apply its line ( $|n|$ -weighted relative errors for products and powers, added absolute errors for sums and differences, least count for single readings), and the mark is earned without any fresh derivation.

Where this lives today

The worst-case rules you drill for JEE are the training wheels of uncertainty budgeting, and the budgets now guard extraordinary claims. When a Kibble balance realises the kilogram from Planck's constant, its published result is an itemised T2 table — dozens of contributions, each $|n|$ -weighted, with engineering effort aimed at the largest line; LIGO's calibration errors propagate through the strain formula by the grown-up (quadrature) version of equation (1). The mindset transfers verbatim: find the dominant term, shrink it, repeat. Closer to home, the six patterns reappear in every first-year college lab — and in one distinctly 2026 habitat: machine-learning benchmark papers now quote score ± error across random seeds, and referees reject a 0.2-point "improvement" sitting inside a ±0.5 spread. That is the difference ambush, T5, wearing a conference badge.

The physical picture

Watch a top scorer meet an error question. They do not start computing; they start classifying. "Maximum percentage error in the density" lights one lamp; "which quantity should be measured most carefully" another; "the time for 20 oscillations" a third. Only then do they touch the numbers — and by that point the question is already dead, because each lamp has a one-line answer bolted to it.

Figure

The recognition map — six phrasing cues, six templates, six one-line answers

Why only six? Because the measurement syllabus is short and each item in it can be dressed as an MCQ in essentially one way. A note of honesty before the drills: the templates are tuned for the exams that time-pressure you — JEE-Main, JEE-Advanced, NEET — but a CBSE or ISC student meets the same six patterns in board numericals with "show your working" attached; the recognition pays everywhere, the speed pays in MCQs.

The recognition table

This table is the module. Everything after it is proof, practice and fine print.

template	phrasing cue	the one-line answer
T1 exponent harvest	"maximum % error in $Z = a^p b^q/c^r$ "	$\dfrac{\Delta Z}{Z} = \\|p\\|\dfrac{\Delta a}{a} + \\|q\\|\dfrac{\Delta b}{b} + \\|r\\|\dfrac{\Delta c}{c}$
T2 dominant culprit	"which quantity must be measured most carefully" / "contributes most to the error"	the largest $\\|n\\| \times (\Delta x/x)$ term — watch the small-valued, high-power quantity
T3 least-count percent	"% error in a single reading of…"	$\Delta x = \mathrm{LC}$ , so $\% = \dfrac{\mathrm{LC}}{x} \times 100$
T4 reporting ritual	a table of repeated readings; "report the result"	mean $\bar{x}$ , mean absolute error $\overline{\Delta x}$ ; error to 1 s.f., value cut at the error's decimal place
T5 difference ambush	$a - b$ (or $a + b$ ) of measured values, then "% error of the result"	$\Delta(a \pm b) = \Delta a + \Delta b$ always; for a small difference, divide that by the small gap
T6 oscillation divider	"time for $n$ oscillations is $t \pm \Delta t$ … error in $g$ "	$T = t/n$ , $\Delta T = \Delta t/n$ , then $\dfrac{\Delta g}{g} = \dfrac{\Delta L}{L} + 2\dfrac{\Delta T}{T}$

The table rests on one algebraic fact (the $|n|$ -weighted addition behind T1/T2) and one experimental fact (the pendulum logic behind T6). Both deserve a proper derivation — memorized lines you cannot rebuild are lines you will misremember under pressure.

Derivation AThe master rule — relative errors add with exponent weights

We want the engine behind T1 and T2: for $Z = a^p b^q / c^r$ the worst-case relative error is the exponent-weighted sum of the input relative errors, with all weights taken positive. The strategy is pure algebra — no calculus — using the fact that a factor measured fractionally wrong scales the product by the same fraction, applied repeatedly for powers. Watch two things: where the worst case enters (we choose the signs of the individual errors against ourselves), and why the denominator's exponent gets a modulus rather than a minus sign.

Step 1. One factor first. If $a$ is really $a(1 + \varepsilon_a)$ with $\varepsilon_a = \Delta a/a$ small, any product containing $a$ once is scaled by exactly $(1+\varepsilon_a)$ : fractional errors pass through multiplication unchanged.

Step 2. Integer powers are repeated products. $a^2$ picks up $(1+\varepsilon_a)^2 = 1 + 2\varepsilon_a + \varepsilon_a^2$ ; for a 1% error, $\varepsilon_a^2$ is one part in $10^4$ , negligible beside $2\varepsilon_a$ , so $a^2$ carries relative error $2\varepsilon_a$ — and by the same expansion $a^p$ carries $p\,\varepsilon_a$ . Why we may drop $\varepsilon^2$ : exam errors are a few percent; their squares are hundredths of a percent, far below the option spacing. This is also why the rule fails for errors of tens of percent — see the limiting cases.

Step 3. Fractional powers come free. If $w = a^{1/2}$ , then $w^2 = a$ , so by Step 2 the relative error of $a$ is twice that of $w$ : $\varepsilon_a = 2\varepsilon_w$ , giving $\varepsilon_w = \tfrac12 \varepsilon_a$ . The exponent rule $|n|\varepsilon$ holds for $n = \tfrac12$ , and the same inversion argument works for any rational power.

Step 4. Division is a negative power. $Z = 1/c$ means that if $c$ reads 1% low, $Z$ comes out 1% high: the error does not cancel, it flips direction. Since the question asks the maximum error, we let every factor err in whichever direction inflates $Z$ — which is why every exponent enters with a modulus:

\frac{\Delta Z}{Z}\bigg|_{\max} = |p|\frac{\Delta a}{a} + |q|\frac{\Delta b}{b} + |r|\frac{\Delta c}{c} \qquad (1)

where: $\Delta x/x$ — each quantity's relative error (pure numbers, unit-free); $p, q, r$ — the exponents as they appear in the formula, sign discarded. Read it physically: an exponent is an error amplifier — squaring a quantity doubles its percentage contribution, square-rooting halves it, and living in the denominator changes nothing but the direction of the damage. Verified: $102^2 = 10404$ , i.e. a 2% high input gives 4.04% ≈ 4% high square; $1/0.98 = 1.0204$ , a 2% low denominator gives 2.04% ≈ 2% high quotient — both match the rule to the promised first order.

Go deeper · Derivation B · 1 minThe oscillation divider — why 20 swings beat a better stopwatch

Derivation BThe oscillation divider — why 20 swings beat a better stopwatch

We derive the T6 line and, with it, the most examined experimental idea in the unit: timing many oscillations is a free upgrade in precision. The setup is the standard pendulum determination of $g$ from $T = 2\pi\sqrt{L/g}$ . The strategy: solve for $g$ , apply the master rule of Derivation A, then see what timing $n$ oscillations does to each term. Watch which errors are fixed by the apparatus and which shrink when you change procedure.

Step 1. Solve for the target: $g = 4\pi^2 L / T^2$ . By the master rule, with exponents 1 on $L$ and 2 on $T$ :

\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\,\frac{\Delta T}{T} \qquad (2)

where: $\Delta L$ — the length error, one least count of the metre scale; $\Delta T$ — the error in one period. The 2 is the exam's favourite bait: timing counts double because $T$ is squared.

Step 2. The procedure: you time $n$ oscillations in total time $t$ , so $T = t/n$ . The stopwatch contributes a fixed absolute error $\Delta t$ — least count plus start/stop reflexes — once per timing, not once per swing; dividing by the exact integer $n$ gives $\Delta T = \Delta t / n$ . Why $n$ is exact: counting swings is not a measurement — 20 is an integer with no uncertainty.

Step 3. See the gain concretely. One oscillation of a metre pendulum: $t \approx 2.0$ s, $\Delta t = 0.2$ s → $\Delta T/T = 10\%$ , which equation (2) makes a hopeless 20% in $g$ . Twenty oscillations: $t = 40.0$ s, the same $\Delta t$ → $\Delta T/T = 0.5\%$ ; the timing contribution falls from 20% to 1%, a twenty-fold improvement that cost nothing but patience.

Step 4. Assemble the standard numbers: $L = 98.0 \pm 0.1$ cm and $t = 40.0 \pm 0.2$ s for 20 oscillations. Then $T = 2.00$ s, $\Delta g/g = 0.1/98.0 + 2(0.2/40.0) = 0.00102 + 0.0100 \approx 0.011$ , i.e. 1.1% — and the timing term still dominates the budget by a factor of ten, telling you where the next improvement should go. Verified: $g = 4\pi^2 (0.980)/(2.00)^2 = 9.67\ \mathrm{m/s^2}$ , sensible for a real pendulum; $0.1/98.0 = 0.102\%$ ; $2 \times 0.5\% = 1.0\%$ ; sum $1.1\%$ , so $\Delta g \approx 0.11\ \mathrm{m/s^2}$ .

Limiting cases

Case	Result	Meaning
$n = 1$ vs $n = 20$ oscillations	timing % error 20× larger for $n=1$	procedure, not instrument, sets the precision
$a \to b$ in a difference	$\%$ error of $(a-b) \to \infty$	never design an experiment around a small difference of large measurements
exponent $0$	quantity drops out of the budget	anything absent from the formula cannot contribute error
$\mathrm{LC} \to 0$	single-reading error $\to$ 0	the T3 line is entirely an instrument statement
errors of tens of percent	rule underestimates ( $\varepsilon^2$ terms bite)	the linear expansion behind (1) assumes small errors
all error in one factor	budget collapses to $\\|n\\|\,\Delta x/x$	T1 contains T3 as the one-variable special case

Common trap“Percentage errors always just add”

Half true, and the half-truth costs marks in both directions. For products, quotients and powers, percentage errors add with exponent weights — forgetting the 2 on a squared radius is the single most common wrong option, deliberately planted. For sums and differences, percentages do not add at all: there it is absolute errors that add, and the percentage of the result must be recomputed against the (possibly tiny) result. A formula like $Z = a + b^2$ therefore needs two currencies in sequence: convert $b^2$ 's relative error to an absolute one, $\Delta(b^2) = b^2 \times 2\Delta b/b$ , then add absolutes: $\Delta Z = \Delta a + \Delta(b^2)$ . Mixing the rules, or applying them in the wrong order, is exactly what the planted option is waiting for.

Worked examples — the drill ladder

Each drill is fought twice — slow honest way, then template way — so you can feel the seconds being bought.

Drill 1 · Board level — the reporting ritual (T4)

Five measurements of a pendulum's period: 2.63 s, 2.56 s, 2.42 s, 2.71 s, 2.80 s. Report the period with its error, and give the percentage error.

The slow way is the only way — T4 is a ritual, not a shortcut — but knowing the ritual cold makes it mechanical.

Step 1. Mean: $\bar{T} = (2.63 + 2.56 + 2.42 + 2.71 + 2.80)/5 = 13.12/5 = 2.624$ s.

Step 2. Absolute deviations from the mean: $0.006,\ 0.064,\ 0.204,\ 0.086,\ 0.176$ s.

Step 3. Mean absolute error: $\overline{\Delta T} = (0.006+0.064+0.204+0.086+0.176)/5 = 0.536/5 = 0.107$ s.

Step 4. The ritual's grammar: error to one significant figure, $\overline{\Delta T} \approx 0.1$ s; the value ends at the same decimal place. Report $\boxed{T = 2.6 \pm 0.1\ \mathrm{s}}$ , percentage error $0.1/2.6 \approx \boxed{4\%}$ . Writing $2.624 \pm 0.107$ s — three claims of precision the scatter cannot support — is the classic marked error the ritual exists to prevent. Timing note: 90 seconds, all arithmetic. MCQ options usually differ only in the rounding grammar of Step 4, so Steps 1–3 can be done coarsely.

Drill 2 · Competitive level — the exponent harvest (T1)

The density of a wire is computed from $\rho = \dfrac{m}{\pi r^2 l}$ . The percentage errors are 2% in $m$ , 1% in $r$ and 3% in $l$ . Find the maximum percentage error in $\rho$ .

The slow way. Write $\rho(1+\varepsilon)$ , expand the quotient, track four terms, hope no sign slips — two minutes and an algebra risk.

The shortcut way. Step 1. "Maximum % error in a formula" — T1. Step 2. Harvest exponents: $m^{1}, r^{-2}, l^{-1}$ → moduli $1, 2, 1$ . Step 3. $\%\rho = 1(2\%) + 2(1\%) + 1(3\%) = \boxed{7\%}$ . The bait is 6%, for those who forgot the amplifier on $r$ ; $\pi$ contributes nothing. Timing note: 25 seconds. Say the moduli aloud in your head — "one, two, one" — before touching the percentages; that habit is what protects the $r^2$ .

Drill 3 · Competitive level — the dominant culprit (T2)

A quantity is computed as $Z = \dfrac{a^2\, b^{3/2}}{c^3\, d}$ . Each of $a, b, c, d$ is measured with 1% error. Which measurement limits the experiment, and what is the maximum % error in $Z$ ?

The slow way. Compute the full budget, then compare — T2 is T1 plus a comparison, so the shortcut only reorders the work.

The shortcut way. Step 1. Weights: $|2|,\ |{\tfrac32}|,\ |{-3}|,\ |{-1}|$ . Step 2. With equal 1% inputs the weights are the contributions: $2\%,\ 1.5\%,\ 3\%,\ 1\%$ — the culprit is $\boxed{c}$ . Step 3. Total: $2 + 1.5 + 3 + 1 = \boxed{7.5\%}$ . When the inputs' errors are not equal, compute each $|n| \times \%$ product — a small quantity on a coarse instrument routinely beats a bigger exponent. "Which needs the most careful measurement" always means "largest term in the budget", never "largest exponent" by itself. Timing note: 35 seconds with equal errors, under a minute otherwise. If the options name a quantity rather than a number, you may not even need the total.

Drill 4 · Competitive level — the difference ambush (T5)

Two rods measure $25.2 \pm 0.1$ cm and $24.8 \pm 0.1$ cm. What is the percentage error in the difference of their lengths?

The slow way — and the planted trap: "each is accurate to $0.1/25 \approx 0.4\%$ , so the difference is good to about 0.8%." Wrong currency, wrong by two orders of magnitude.

The shortcut way. Step 1. A difference: absolute errors add, then convert. Step 2. $\Delta = 0.1 + 0.1 = 0.2$ cm on $25.2 - 24.8 = 0.4$ cm. Step 3. $0.2/0.4 = \boxed{50\%}$ . Two beautifully precise measurements, one nearly meaningless difference. Timing note: 20 seconds. Whenever close values are subtracted, expect a shockingly large answer and distrust any option under 10%.

Figure

The difference ambush — precise inputs, worthless output

Drill 5 · Competitive level — the oscillation divider (T6)

In a simple-pendulum determination of $g$ , the length is $98.0 \pm 0.1$ cm and the time for 20 oscillations is $40.0 \pm 0.2$ s. Find the maximum percentage error in $g$ .

The slow way. Re-derive $g = 4\pi^2 L/T^2$ , differentiate mentally, wonder whether the 20 divides the error — the exact confusion the question is engineered to exploit.

The shortcut way. Step 1. Name it: time for $n$ oscillations → T6, line pre-assembled: $\Delta g/g = \Delta L/L + 2\Delta t/t$ (the $n$ cancels in the relative timing error: $\Delta T/T = (\Delta t/n)/(t/n) = \Delta t/t$ ). Step 2. Numbers: $0.1/98.0 + 2(0.2/40.0) = 0.1\% + 1.0\% = \boxed{1.1\%}$ . Step 3. Sanity: the timing term dominates ten-to-one — which is why the experiment times twenty swings, not one; with $n = 1$ the same stopwatch poisons $g$ with a 20% error. Timing note: 40 seconds. The distractor options split the students who forgot the 2 (0.6%) from those who mangled the $n$ (avoid both by writing the line before the numbers).

Exam traps

Each of these failure modes appears in the archives as a deliberately planted wrong option.

The unweighted harvest. Adding raw percentages while ignoring exponents (Drill 2's 6% bait). Defence: read the formula as moduli first — "one, two, one" — before any arithmetic.

The dropped modulus. A negative exponent does not subtract error: $Z = a/b$ with both 1% wrong gives 2%, never 0%. The question says maximum, so every term enters positively.

Currency mixing. $\Delta x$ is absolute (units); $\Delta x/x$ is relative (pure). Sums add the first; products add the second. A formula with both — $Z = (a - b)^2 c$ — is processed inside-out: absolute rule on $(a-b)$ , convert to relative, then harvest the exponent.

The mid-formula sum. In $Z = a + b^2$ the power rule may NOT be applied to the whole expression — there is no "exponent of $a$ ". Convert $b^2$ 's error to absolute form and add absolutes: any plus sign inside the formula switches you out of T1.

Mean when max is asked (and vice versa). JEE and NEET overwhelmingly want the maximum (worst-case) error — all terms added. If a paper explicitly says "mean absolute error" on a data table, you are in T4's ritual instead. Read which error the stem names before computing anything.

The error's own digits. Quoting $\Delta g = 0.1064\ \mathrm{m/s^2}$ flunks the reporting grammar: uncertainties get one significant figure, and the value ends where the error ends. On integer-answer JEE questions this grammar is the question.

Unit mismatch inside a ratio. $\Delta x/x$ demands both in the same unit — $0.1$ cm against $98.0$ cm, never against $0.98$ m. Percentages are unit-free only after the division is honest.

Origins · 1 minCarl Friedrich Gauss and how this was discovered

Carl Friedrich Gauss

1777–1855 · Göttingen

The six templates are the pocket edition of a theory Gauss built for the highest-stakes measurements of his age. In Theoria Motus (1809) he asked what a stack of imperfect observations can honestly assert, and answered with the normal law of errors and least squares. Then he lived the theory: directing the triangulation of the Kingdom of Hanover from 1818, he ran error budgets across thousands of sightings — inventing the heliotrope to shrink the dominant term, exactly the T2 move. The exam's "maximum error" rules are the classroom-safe cousin of his machinery: cruder, but built on his insight that errors must be propagated, not ignored — and that knowing which measurement limits you is worth more than averaging the rest.

History

1809

— Gauss's Theoria Motus states the normal error law and least squares, founding formal error analysis.

1816

— Gauss's paper on the precision of observations gives the first systematic treatment of an observer's error budget.

1818

— The Hanover survey begins: two decades of triangulation run as a managed error budget, heliotrope included.

1960

— The SI is established under the BIPM, standardising the units in which uncertainties are stated.

1993

— The ISO Guide to the Expression of Uncertainty in Measurement (GUM) fixes the modern international rules that the exam versions approximate.

Going further

Texts. NCERT Physics XI, Ch. 1 (Units and Measurement) — source of the T4 ritual and the exam's error rules · H.C. Verma, Concepts of Physics Vol. 1, Ch. 2 · Irodov-class collections for harder T1/T5 hybrids · BIPM, GUM (2008) for the professional version these rules approximate.

Provenance. Fresh module (exam-technique class); templates distilled from the JEE-Main/NEET archive patterns; drill numbers original and recomputed. A verifier should recompute: Drill 1 mean $13.12/5 = 2.624$ , deviations summing to $0.536$ , mean error $0.107 \to 0.1$ , report $2.6 \pm 0.1$ s (≈4%); Drill 2: $2 + 2(1) + 3 = 7\%$ ; Drill 3 weights $2, 1.5, 3, 1 \to c$ , total $7.5\%$ ; Drill 4: $0.2/0.4 = 50\%$ ; Drill 5: $0.1/98.0 + 2(0.2/40.0) = 1.1\%$ , $g = 4\pi^2(0.980)/4.00 = 9.67\ \mathrm{m/s^2}$ ; anchors $102^2 = 10404$ , $1/0.98 = 1.0204$ , $\sqrt{1.02} = 1.00995$ ; quiz values $4\%$ , $c$ , $100\%$ ( $0.2/0.2$ ), fixed- $\Delta t$ argument, $5\% + 2\% = 7\%$ .

Prerequisites. Absolute, Relative & Percentage Errors · Propagating Errors: Sums & Differences · Propagating Errors: Products, Quotients & Powers · Significant Figures, Rounding & Reporting · Least Count & Instrument Errors (with The Vernier Calipers and The Screw Gauge for T3's readings). Leads to. Limiting & Extreme Cases · Option Elimination & Answer Checking — the track capstone, where the error audit joins the dimension audit in a full option-killing toolkit.

Why do we always add the errors? Couldn't they cancel each other?

They could — and often do. But the exam asks for the maximum error: the bound that holds even if every input conspires against you. Since you cannot know the true direction of any individual error, the worst case is the only defensible promise. Research labs use a gentler statistical combination — adding in quadrature, $\sqrt{\sum \varepsilon_i^2}$ , the GUM convention — which assumes independent random errors partially cancel. For JEE and NEET, add linearly; that is the syllabus default.

When do I use ΔZ and when ΔZ/Z? I keep mixing them.

Match the currency to the operation and to the ask. Sums and differences work in absolute errors ( $\Delta$ , with units); products, quotients and powers work in relative errors ( $\Delta/x$ , unit-free); a mixed formula is processed inside-out, converting at each boundary. Then read the question again: "error in Z" wants $\Delta Z$ (units!), "percentage error" wants $100 \times \Delta Z/Z$ . Half the wrong options in the archive are the right number in the wrong currency.

Why does the smallest measured quantity usually dominate the error?

Because relative error is $\Delta x / x$ , and small $x$ means a small denominator. A metre-long wire measured on a mm scale carries 0.1%; its half-millimetre diameter on a screw gauge of least count 0.01 mm carries $0.01/0.50 = 2\%$ — twenty times worse — and it usually enters squared (as $r^2$ ), doubling to 4%. That double whammy, small value with high exponent, is why the wire diameter is the perpetual villain of T2 questions, and why experiments spend their best instrument on their smallest dimension.

Does π (or any constant) add to the error?

No. Constants of mathematics ( $\pi$ , 2, $\tfrac43$ ) and supplied constants ("take $g = 9.8\ \mathrm{m/s^2}$ ") are exact for error purposes: you did not measure them, so they contribute nothing — $\rho = m/(\pi r^2 l)$ has exactly three error sources. One nuance: a constant given with an explicit uncertainty (" $G = 6.67 \pm 0.01$ ") has been promoted to a measured quantity — include it like any other factor, exponent weight and all.

Maximum error, mean absolute error, standard error — which one does my exam want?

JEE-Main, JEE-Advanced and NEET propagation questions want the maximum error — linear addition with exponent weights — unless the stem says otherwise. Mean absolute error appears only in the T4 ritual: repeated readings, deviations from the mean, average of the moduli. Standard error and quadrature belong to college labs, not the HS syllabus — though they explain why your textbook calls the exam rule an upper bound. Reliable tell: a data table means T4; a formula with given percentages means T1/T2.

Why does a square root HALVE the error? Squaring doubles it, so shouldn't a root too?

Run the inversion: if $w = \sqrt{a}$ , then $a = w^2$ , so $a$ 's relative error is twice $w$ 's — read backwards, $w$ 's error is half of $a$ 's. Amplification follows the exponent as written: $a^{1/2}$ contributes $\tfrac12 (\Delta a/a)$ . Concrete check: $\sqrt{1.02} \approx 1.00995$ — a 2% error under the root emerges as 1%. Anchor pair: square → ×2, root → ×½.

Can a percentage error be more than 100%? My answer of 50% on a difference felt illegal.

Perfectly legal, and the exam counts on you flinching. Percentage error compares the uncertainty to the result — and if the result is a small difference of large numbers, the ratio can be 50%, 100%, or beyond (past 100%, your measurement cannot even fix the difference's sign). Nothing is broken: the inputs were precise; the design of subtracting near-equal values destroyed the information. That is why good experiments measure differences directly, and why T5 answers that look scandalously large are usually the correct options.