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NEET 2027 · Physics · Class 11 · Ch. 1 Units and Measurement

Errors in Measurement

Almost every recent NEET paper has carried an error-propagation item. Most of this topic sits outside the rationalised NCERT text but firmly inside the NTA syllabus — do not skip it.

PRIORITY 1TOPIC 3 OF 1520 QUESTIONS2 PYQ-BASEDSYLLABUS GAP TOPIC

01Syllabus warning — read this first

This topic was cut from the textbook but not from the exam

The rationalised NCERT removed the detailed treatment of accuracy, precision, systematic and random errors, and combination of errors. What survives in the printed chapter is the compressed §1.3.3, 'Rules for determining the uncertainty in the results of arithmetic calculations'.

The NTA syllabus for Unit I, Physics and Measurement, still lists least count, significant figures and errors in measurement. And NEET has kept asking: an error-propagation item on voltage and current appeared in 2025, error questions appeared in 2023, 2021, 2019 and 2017, and a screw-gauge item appeared in the 2024 re-examination.

Conclusion: everything on this page is examinable. Treat the gap between the textbook and the syllabus as a warning, not as permission to skip.

02Concept in plain language

No measurement gives the true value. The honest thing to report is not a single number but a range: a best estimate, plus a statement of how far off it might be.

Write a length as 2.62 ± 0.11 cm and you have said two things. The 2.62 is your best estimate, usually the mean of several readings. The 0.11 is your admission of how much room for doubt there is. Both halves are part of the answer; a measurement quoted without an uncertainty is incomplete.

Absolute, relative, percentage

The same doubt can be expressed three ways, and NEET switches between them freely:

Absolute error tells you about the instrument. Relative error tells you about the quality of the measurement. A 1 mm error is fine on a room but hopeless on a wire.

The one idea behind every propagation question

When a result is computed from several measured quantities, the doubts feed through. The rule is always the same and always pessimistic: assume the worst case, in which every error pushes the answer in the same direction. That is why errors are always added and never subtracted, no matter what operation the quantities themselves are undergoing.

Decide which of two rules applies before touching numbers

Sums and differences: add the absolute errors.
Products, quotients and powers: add the relative errors, each multiplied by the magnitude of its exponent.

Every propagation question in NEET is one of these two. Mixing them up — adding absolute errors for a product, say — is the single most common failure in this topic.

03Accuracy, precision and propagation

ACCURACY IS CLOSENESS TO THE TRUE VALUE · PRECISION IS CLOSENESS TO EACH OTHERaccurate & preciseprecise, not accurateaccurate, not preciseneitherSystematic error moves the whole cluster off centre (kills accuracy). Random error scatters the cluster (kills precision).Averaging many readings fights random error only. It never removes a zero error.
Fig. 1 — The accuracy–precision matrix. Read the position of the cluster for accuracy and its tightness for precision. The two are independent, which is why there are four boxes and not two.
REPEATED READINGS OF ONE LENGTH · THE MEAN IS THE BEST ESTIMATE2.632.562.422.712.802.442.532.61mean = 2.587 cmMean absolute error Δā = average of |ai − ā| · report the result as ā ± Δā, rounded to the same decimal place.
Fig. 2 — Repeated readings scatter about the mean. The mean is the best estimate; the average size of the deviations is the mean absolute error.
FOR PRODUCTS AND QUOTIENTS, RELATIVE ERRORS ADD — NEVER ABSOLUTE ONES V = 200 ± 4 V ΔV/V = 2% I = 20 ± 0.2 A ΔI/I = 1% R = V / I 2% + 1% = 3% R = 10 Ω ΔR = 0.3 Ω Powers multiply the relative error by the exponent: if Z = A²B³/C, then ΔZ/Z = 2(ΔA/A) + 3(ΔB/B) + (ΔC/C). Errors always add, even when the quantity is divided. There is no cancellation — you must assume the worst case.
Fig. 3 — The NEET 2025 item as a flow diagram. For a quotient, the percentages add, and only at the last step is the total converted back into an absolute error.

04The four kinds of error

the first three are all systematic — that is the grouping NEET tests
KindCauseBehaviourHow to reduce it
Systematic — instrumentalZero error, worn screw, wrongly graduated scaleSame size and same sign in every readingFind the zero error and subtract it; recalibrate or replace the instrument
Systematic — imperfect techniqueIgnoring buoyancy of air, heat lost to surroundings, neglecting the mass of a threadConsistently shifts the result one wayImprove the method or apply a correction term
Systematic — personalAlways holding the eye to one side, habitual bias in judging a coincidenceConsistent for a given observerChange the observer, or use a fixed viewing arrangement
RandomUnpredictable fluctuations in temperature, supply voltage, vibrationVaries in size and sign from reading to readingTake many readings and average — the only case where averaging helps
Least countThe finite resolution of the instrument itselfSets a floor on the uncertainty of a single readingUse a finer instrument; averaging cannot go below this floor
GrossHuman blunder — misreading a scale, recording the wrong digit, wrong formulaErratic and usually largeRepeat the measurement carefully; there is no statistical treatment for a mistake

05Formula sheet

Errors in a single quantity
Best estimate (mean)
ā = (a₁ + a₂ + … + an) / n
the arithmetic mean of n readings
Absolute error of one reading
Δai = |ā − ai|
the size of the deviation; sign discarded
Mean absolute error
Δā = (1/n) ∑ |Δai|
Reported result
a = ā ± Δā
quote both to the same decimal place
Relative error
δa = Δā / ā
a pure number
Percentage error
δa % = (Δā / ā) × 100
Least count error
Δa = least count
for a single reading on any instrument
Zero-error correction
true = observed − zero error
positive zero error → subtract; negative → add
Propagation of errors
Sum  Z = A + B
ΔZ = ΔA + ΔB
absolute errors add
Difference  Z = A − B
ΔZ = ΔA + ΔB
still add — never subtract
Product  Z = AB
ΔZ/Z = ΔA/A + ΔB/B
relative errors add
Quotient  Z = A/B
ΔZ/Z = ΔA/A + ΔB/B
same rule as the product
Power  Z = ApBq/Cr
ΔZ/Z = |p|(ΔA/A) + |q|(ΔB/B) + |r|(ΔC/C)
each exponent's magnitude multiplies its own relative error
Exact numerical factor
no contribution
½, 2π, 4/3 and counted integers have infinite significant figures

06The propagation rules in full

Why absolute errors add for a sum

If Z = A + B, the largest Z can be is (A + ΔA) + (B + ΔB), which exceeds A + B by ΔA + ΔB. The smallest is short by the same amount. So ΔZ = ΔA + ΔB.

Why they still add for a difference

If Z = A − B, the largest Z occurs when A is at its highest and B at its lowest: (A + ΔA) − (B − ΔB) = (A − B) + (ΔA + ΔB). The errors add again. This is why a difference of two close measurements is so dangerous — the central value shrinks while the error does not.

Why relative errors add for a product

If Z = AB then Z + ΔZ = (A + ΔA)(B + ΔB) = AB + AΔB + BΔA + ΔAΔB. The last term is a product of two small quantities, so it is dropped. Dividing through by Z = AB leaves ΔZ/Z = ΔA/A + ΔB/B.

Why the exponent multiplies

A³ is just A × A × A, so its relative error is counted three times. The same argument extends to fractional powers: √A contributes half, and A1/3 contributes a third.

The two-second method for exam conditions

Write the exponents in a row above the variables, ignoring all minus signs. Multiply each by its percentage error. Add. Done.

P = a³b²/(√c · d) → 3(1) + 2(2) + ½(3) + 1(4) = 3 + 4 + 1.5 + 4 = 12.5%

07Exceptions and traps

Averaging never removes a systematic error

The single most-tested distinction in this topic. More readings shrink random error towards zero and leave systematic error exactly where it was. A tight cluster in the wrong place is the signature.

Differences destroy precision

Subtracting two nearly equal measurements keeps the absolute error and shrinks the value, so the relative error can reach 100% or more. NCERT's own example: 0.307 m − 0.304 m = 0.003 m, where three significant figures become one.

The error rules give the maximum, not the likely, error

Adding all the contributions assumes every error conspires in the same direction. In real statistics you would add in quadrature. NEET always wants the simple sum — the maximum possible error — unless a question explicitly says otherwise.

Exact numbers are error-free

½ in ½mv², 2π in the pendulum formula, 4/3 in a sphere's volume, and the count n in T = t/n are all exact. They contribute nothing to the error budget and have infinitely many significant figures.

Read whether the question wants absolute or percentage

NEET frequently gives the data as absolute errors (200 ± 4 V), asks you to work in percentages, and then wants the answer back as an absolute error (± 0.3 Ω). Losing track of which form you are in costs the mark even when the physics is right.

08Numbers to remember

each of these has appeared as a question in its own right
SituationResult to carry into the exam
Radius measured to 1%circumference 1%, area 2%, volume 3%
Side of a cube measured to 1%surface area 2%, volume 3%, density (with exact mass) 3%
Square root of a quantitycontributes half of that quantity's percentage error
Cube root of a quantitycontributes one third
Kinetic energy from m and vΔm/m + 2(Δv/v)
g from a pendulum (l and T)Δl/l + 2(ΔT/T)
Resistance from V and IΔV/V + ΔI/I
Timing n oscillations instead of onedivides the percentage error in T by n
Vernier callipers, 20 divisions on 1 mm scaleleast count 0.05 mm = 0.005 cm
Screw gauge, pitch 1 mm, 100 divisionsleast count 0.01 mm = 10 μm
Metre scaleleast count 1 mm
Common relative-error benchmark (NCERT)0.01 g on 1.02 g is 1%; on 9.89 g it is 0.1%

09Scientists to remember

lightly examined, but they appear in matching and statement items
NameWhat to attach to the name
Carl Friedrich GaussThe normal (Gaussian) distribution, which describes how random errors scatter about the mean, and the method of least squares for fitting data. The bell curve behind 'take many readings and average'.
Pierre VernierThe vernier scale (1631), which made sub-division readings possible and is the origin of the term 'vernier constant' for least count.
William Gascoigne / Jesse RamsdenEarly development of the micrometer screw gauge, the instrument behind pitch and circular-scale least count.
Lord Kelvin"When you cannot measure it, your knowledge is of a meagre and unsatisfactory kind" — the standard epigraph for this chapter.

10Twenty worked questions

About the PYQ labels

Items tagged PYQ follow the wording and structure of questions that have appeared in NEET papers. Year labels are indicative and worth cross-checking against the official NTA paper. Every number on this page was recomputed before printing.

Q01PYQ · NEET 2025Propagation

In an electrical circuit the voltage is measured as V = (200 ± 4) V and the current as I = (20 ± 0.2) A. The value of the resistance is:

  1. (10 ± 0.3) Ω
  2. (10 ± 0.2) Ω
  3. (10 ± 0.4) Ω
  4. (10 ± 0.1) Ω
Show full solution
Given
V = 200 ± 4 V, I = 20 ± 0.2 A.
Asked
R together with its absolute error.
Concept
R = V/I is a quotient, so the relative errors add. Convert back to an absolute error only at the very end.
Formula
ΔR/R = ΔV/V + ΔI/I
Baby steps
  1. Central value: R = 200/20 = 10 Ω.
  2. ΔV/V = 4/200 = 0.02 = 2%.
  3. ΔI/I = 0.2/20 = 0.01 = 1%.
  4. ΔR/R = 2% + 1% = 3%.
  5. ΔR = 0.03 × 10 = 0.3 Ω, so R = (10 ± 0.3) Ω.
Answer
(a) (10 ± 0.3) Ω
Why not others
(a)Correct.
(b)This uses only the current's error.
(c)This comes from adding the absolute errors 4 and 0.2 in some way — not permitted for a quotient.
(d)This uses only the voltage's error, misread.
Shortcut
For a quotient, percentages simply add: 2 + 1 = 3. Then 3% of 10 is 0.3. Two lines, no algebra.
Where it goes wrong
Errors add even though the quantities are divided. Division never cancels error — you must assume the worst case in which both errors push the answer the same way.
Q02Propagation

The mass of a cube is measured with a percentage error of 2% and the length of its side with a percentage error of 1%. The maximum percentage error in the calculated density is:

  1. 3%
  2. 4%
  3. 6%
  4. 5%
Show full solution
Given
Δm/m = 2%, ΔL/L = 1%, cube of side L.
Asked
Maximum percentage error in density.
Concept
Density = mass/volume, and volume = L³. A cubed quantity contributes three times its percentage error.
Formula
ρ = m/L³ → Δρ/ρ = Δm/m + 3(ΔL/L)
Baby steps
  1. Volume V = L³, so ΔV/V = 3 × 1% = 3%.
  2. ρ = m/V → Δρ/ρ = Δm/m + ΔV/V.
  3. = 2% + 3% = 5%.
  4. Note that the length was measured once but is used three times, so its error counts thrice.
Answer
(d) 5%
Why not others
(a)This treats the volume as if it were a single length.
(b)This uses 2 × 1% for the volume instead of 3 × 1%.
(c)This doubles the mass error as well.
(d)Correct.
Shortcut
Any exponent, whole or fractional, multiplies the percentage error. Cube → ×3, square → ×2, square root → ×½, cube root → ×⅓.
Where it goes wrong
The side is a single measurement, so students often add 1% only once. Because L appears three times in L³, its error appears three times too.
Q03Propagation

A physical quantity is given by P = a³b² / (√c × d). The percentage errors in a, b, c and d are 1%, 2%, 3% and 4% respectively. The maximum percentage error in P is:

  1. 10%
  2. 12%
  3. 12.5%
  4. 14%
Show full solution
Given
P = a³b²c−1/2d−1; errors 1%, 2%, 3%, 4%.
Asked
Maximum percentage error in P.
Concept
Take the magnitude of every exponent, multiply by that variable's percentage error, and add. Signs are ignored because we want the worst case.
Formula
ΔP/P = 3(Δa/a) + 2(Δb/b) + ½(Δc/c) + 1(Δd/d)
Baby steps
  1. From a: 3 × 1% = 3%.
  2. From b: 2 × 2% = 4%.
  3. From c: ½ × 3% = 1.5%.
  4. From d: 1 × 4% = 4%.
  5. Total = 3 + 4 + 1.5 + 4 = 12.5%.
Answer
(c) 12.5%
Why not others
(a)Obtained by dropping the c term entirely.
(b)Obtained by taking √c as contributing nothing.
(c)Correct.
(d)Obtained by giving c its full 3% instead of half.
Shortcut
Write the exponents in a row above the variables, multiply down the column, add across. Never mind the plus and minus signs — only the size of the exponent matters.
Where it goes wrong
Forgetting that a square root is a power of ½. It contributes half the error, not none and not all of it.
Q04Types of error

Which of the following is a systematic error?

  1. Random fluctuation of the supply voltage during a reading
  2. A screw gauge that reads 0.03 mm when its jaws are fully closed
  3. Parallax that shifts unpredictably from reading to reading
  4. Irregular changes in room temperature during a long experiment
Show full solution
Given
Four sources of error.
Asked
Which one is systematic.
Concept
A systematic error has a definite cause and pushes every reading the same way by the same amount. A random error changes size and sign unpredictably from reading to reading.
Formula
corrected reading = observed reading − zero error
Baby steps
  1. A closed-jaw reading of 0.03 mm is a zero error.
  2. It adds exactly +0.03 mm to every single reading, in the same direction.
  3. That is the definition of a systematic (instrumental) error.
  4. The other three vary unpredictably, so they are random.
Answer
(b) A screw gauge that reads 0.03 mm when its jaws are fully closed
Why not others
(a)Fluctuating supply voltage is unpredictable — random.
(b)Correct. It is an instrumental systematic error, removable by subtracting the zero error.
(c)Unpredictable parallax is random; a consistently tilted eye position would be systematic.
(d)Irregular temperature drift is random; a steady drift in one direction would be systematic.
Shortcut
Ask one question: does it push every reading the same way? Yes → systematic. Unpredictable → random.
Where it goes wrong
Averaging many readings reduces random error but does absolutely nothing to a zero error. The only cure for a systematic error is to find its cause and correct for it.
Q05PYQ · NEET 2023Types of error

Errors in measurement which arise due to unpredictable fluctuations in temperature, voltage supply, or mechanical vibrations of experimental setups are called:

  1. random errors
  2. systematic errors
  3. instrumental errors
  4. gross errors
Show full solution
Given
Errors caused by unpredictable fluctuations.
Asked
The name of this class of error.
Concept
The word unpredictable is the whole answer. Systematic errors are predictable in sign and size; random errors are not.
Formula
best estimate = arithmetic mean of many readings
Baby steps
  1. Fluctuations that cannot be predicted give readings that scatter about the true value.
  2. They are equally likely to be positive or negative.
  3. These are random errors.
  4. They are reduced (though never eliminated) by taking a large number of readings and averaging.
Answer
(a) random errors
Why not others
(a)Correct.
(b)Systematic errors have a known cause and a consistent direction.
(c)Instrumental errors are a type of systematic error, such as a zero error or a faulty scale.
(d)Gross errors are outright mistakes by the observer — misreading a scale, recording the wrong digit — and are not part of the statistical treatment.
Shortcut
Four-way split worth memorising: systematic (instrumental, imperfect technique, personal bias), random (unpredictable), least-count (resolution limit), gross (human blunder).
Where it goes wrong
Least-count error sits slightly apart: it is the smallest value the instrument can resolve, so it sets a floor below which no amount of averaging can help.
Q06Mean and mean absolute error

Five measurements of a length give 2.63, 2.56, 2.42, 2.71 and 2.80 cm. The mean absolute error is closest to:

  1. 0.05 cm
  2. 0.09 cm
  3. 0.13 cm
  4. 0.11 cm
Show full solution
Given
Readings 2.63, 2.56, 2.42, 2.71, 2.80 cm.
Asked
Mean absolute error.
Concept
Find the mean, take the size of each deviation from it (ignore signs), and average those sizes.
Formula
Δā = (1/n) ∑ |ai − ā|
Baby steps
  1. Mean ā = (2.63+2.56+2.42+2.71+2.80)/5 = 13.12/5 = 2.624 cm.
  2. Deviations: +0.006, −0.064, −0.204, +0.086, +0.176.
  3. Magnitudes: 0.006, 0.064, 0.204, 0.086, 0.176. Sum = 0.536.
  4. Δā = 0.536/5 = 0.1072 ≈ 0.11 cm.
  5. Result quoted as 2.62 ± 0.11 cm; relative error = 0.1072/2.624 ≈ 4%.
Answer
(d) 0.11 cm
Why not others
(a)Too small — this ignores the two largest deviations.
(b)Comes from dropping the 2.42 reading.
(c)Comes from dividing by 4 instead of 5.
(d)Correct.
Shortcut
Signs are discarded before averaging. If you average the signed deviations you will always get zero, which is the standard check that your mean is right.
Where it goes wrong
Discarding the reading that looks 'off' (2.42 here). Unless there is a recorded reason to reject it, every reading stays in.
Q07Sum and difference

Two resistances R₁ = (100 ± 3) Ω and R₂ = (200 ± 4) Ω are connected in series. The equivalent resistance is:

  1. (300 ± 1) Ω
  2. (300 ± 3.5) Ω
  3. (300 ± 7) Ω
  4. (300 ± 5) Ω
Show full solution
Given
R₁ = 100 ± 3 Ω, R₂ = 200 ± 4 Ω, in series.
Asked
Equivalent resistance with its error.
Concept
For a sum or a difference, the absolute errors add. Relative errors are irrelevant here.
Formula
R = R₁ + R₂ ; ΔR = ΔR₁ + ΔR₂
Baby steps
  1. Central value: 100 + 200 = 300 Ω.
  2. ΔR = 3 + 4 = 7 Ω.
  3. So R = (300 ± 7) Ω.
  4. Relative error = 7/300 ≈ 2.3%, smaller than either input's — a genuine advantage of adding quantities.
Answer
(c) (300 ± 7) Ω
Why not others
(a)This subtracts the errors, which is never allowed.
(b)This averages the errors.
(c)Correct.
(d)This uses some other combination; only the plain sum is permitted.
Shortcut
Two rules, and no third: sums and differences → add absolute errors; products and quotients → add relative errors. Decide which one you are looking at before you touch a number.
Where it goes wrong
For a difference, the absolute errors still add. They never subtract, even though the quantities do.
Q08Sum and difference

Two lengths are measured as A = (5.00 ± 0.05) cm and B = (4.90 ± 0.05) cm. The percentage error in (A − B) is:

  1. 2%
  2. 100%
  3. 10%
  4. 50%
Show full solution
Given
A = 5.00 ± 0.05 cm, B = 4.90 ± 0.05 cm.
Asked
Percentage error in A − B.
Concept
Subtracting two nearly equal numbers keeps the absolute error but destroys the central value, so the relative error explodes.
Formula
Δ(A−B) = ΔA + ΔB
Baby steps
  1. A − B = 5.00 − 4.90 = 0.10 cm.
  2. Δ(A−B) = 0.05 + 0.05 = 0.10 cm.
  3. Percentage error = (0.10/0.10) × 100 = 100%.
  4. The result is 0.10 ± 0.10 cm — effectively worthless.
Answer
(b) 100%
Why not others
(a)1% is the relative error of A alone.
(b)Correct — and a warning: the difference carries no useful information at all.
(c)This halves the total absolute error.
(d)This uses only one of the two absolute errors.
Shortcut
If two measurements are close together, never design an experiment around their difference. Measure the difference directly instead.
Where it goes wrong
NCERT makes the same point with 0.307 m − 0.304 m = 0.003 m, where three significant figures collapse to one. Subtraction is the only operation that can reduce the number of significant figures.
Q09Propagation

In a simple pendulum experiment, the length is measured with a 1% error and the time period with a 2% error. The maximum percentage error in the value of g obtained is:

  1. 5%
  2. 3%
  3. 4%
  4. 7%
Show full solution
Given
Δl/l = 1%, ΔT/T = 2%.
Asked
Maximum percentage error in g.
Concept
Rearrange the pendulum formula so that g is the subject before applying the error rules. T sits in the denominator squared, so it contributes twice its percentage error.
Formula
T = 2π√(l/g) → g = 4π²l/T² → Δg/g = Δl/l + 2(ΔT/T)
Baby steps
  1. Square both sides: T² = 4π²l/g.
  2. Rearrange: g = 4π²l / T².
  3. 4π² is an exact number and contributes no error.
  4. Δg/g = 1% + 2(2%) = 5%.
Answer
(a) 5%
Why not others
(a)Correct.
(b)This counts the time error only once.
(c)This drops the length error.
(d)This adds an extra term that is not there.
Shortcut
Rearranging first is not optional. If you apply the error rules to T = 2π√(l/g) directly you will get the error in T, which is not what was asked.
Where it goes wrong
If instead the question gives the error in the total time for n oscillations, that same percentage carries straight over to T, because T = t/n and n is an exact number.
Q10Accuracy vs precision

A set of readings of a quantity whose true value is 25.0 comes out as 27.1, 27.2, 27.0 and 27.1. These readings are:

  1. accurate and precise
  2. accurate but not precise
  3. neither accurate nor precise
  4. precise but not accurate
Show full solution
Given
True value 25.0; readings 27.1, 27.2, 27.0, 27.1.
Asked
Classify the set.
Concept
Precision is agreement among the readings. Accuracy is agreement with the true value. They are independent properties.
Formula
accuracy ↔ closeness to true value ; precision ↔ closeness to each other
Baby steps
  1. The readings spread over only 0.2 — tightly clustered, so highly precise.
  2. Their mean, 27.1, is 2.1 away from the true value 25.0 — not accurate.
  3. So the set is precise but not accurate.
  4. The pattern points to a systematic error of about +2.1, such as a zero error.
Answer
(d) precise but not accurate
Why not others
(a)The mean is well away from 25.0, so accuracy fails.
(b)The reverse of the truth — the scatter is very small.
(c)Precision is clearly good here.
(d)Correct. A tight cluster in the wrong place is the signature of a systematic error.
Shortcut
Tight cluster, wrong place → systematic error. Loose cluster, right place on average → random error. The dartboard picture in Fig. 1 above encodes both cases.
Where it goes wrong
Averaging more readings will make this set more precise and no more accurate. Precision without accuracy is the more dangerous failure, because it looks convincing.
Q11Zero error

A screw gauge shows a reading of +0.05 mm when its jaws are closed with nothing between them. While measuring a wire it reads 3.35 mm. The correct diameter of the wire is:

  1. 3.40 mm
  2. 3.35 mm
  3. 3.30 mm
  4. 3.25 mm
Show full solution
Given
Zero error = +0.05 mm; observed reading = 3.35 mm.
Asked
Corrected diameter.
Concept
A zero error is added to every reading by the instrument, so it must be subtracted from every reading by you. The zero correction is the negative of the zero error.
Formula
true value = observed reading − zero error
Baby steps
  1. Zero error = +0.05 mm (a positive zero error).
  2. Zero correction = −0.05 mm.
  3. True diameter = 3.35 − 0.05.
  4. = 3.30 mm.
Answer
(c) 3.30 mm
Why not others
(a)This adds the zero error instead of subtracting it.
(b)This ignores the zero error entirely.
(c)Correct.
(d)This subtracts twice.
Shortcut
Positive zero error → subtract. Negative zero error → add. Say it as: 'the instrument was already reading high, so take that away.'
Where it goes wrong
A negative zero error (the zero mark sitting above the reference line) means you add its magnitude. Read the sign of the zero error before deciding the operation, every time.
Q12Relative error

A mass of 1.02 g and a mass of 9.89 g are each measured to ± 0.01 g. Which measurement has the smaller relative error, and what is it?

  1. 1.02 g, about 1%
  2. 9.89 g, about 0.1%
  3. both are the same
  4. 1.02 g, about 0.1%
Show full solution
Given
Both measurements carry the same absolute error, ± 0.01 g.
Asked
Which has the smaller relative error, and its value.
Concept
Relative error depends on the size of the quantity as well as on the absolute error. The same absolute error is a smaller fraction of a larger measurement.
Formula
relative error = Δa / a
Baby steps
  1. For 1.02 g: 0.01/1.02 = 0.0098 ≈ 1%.
  2. For 9.89 g: 0.01/9.89 = 0.00101 ≈ 0.1%.
  3. The larger measurement carries the smaller relative error.
  4. So 9.89 g, about 0.1%.
Answer
(b) 9.89 g, about 0.1%
Why not others
(a)1.02 g has the larger relative error, about 1%.
(b)Correct — this is the worked example in NCERT §1.3.3(3).
(c)They differ by a factor of ten.
(d)Right measurement paired with the wrong figure.
Shortcut
Same instrument, bigger object → better relative accuracy. This is why you weigh a hundred identical washers together instead of one at a time.
Where it goes wrong
Absolute error and relative error move independently. A tiny absolute error can still be a huge relative error if the quantity itself is tiny — see the subtraction question above.
Q13Propagation

The mass of a body is measured with a 2% error and its speed with a 3% error. The maximum percentage error in the calculated kinetic energy is:

  1. 8%
  2. 5%
  3. 6%
  4. 11%
Show full solution
Given
Δm/m = 2%, Δv/v = 3%.
Asked
Maximum percentage error in kinetic energy.
Concept
K = ½mv². The ½ is an exact number and contributes nothing; the square on v doubles its error contribution.
Formula
ΔK/K = Δm/m + 2(Δv/v)
Baby steps
  1. K = ½ m v².
  2. The factor ½ is exact → no error contribution.
  3. ΔK/K = 2% + 2(3%).
  4. = 2% + 6% = 8%.
Answer
(a) 8%
Why not others
(a)Correct.
(b)This adds 2% and 3% as if v were not squared.
(c)This uses only the speed term.
(d)This adds an extra 3%.
Shortcut
Exact numerical factors — ½, 2π, 4/3, 6 — never contribute error. They have infinitely many significant figures.
Where it goes wrong
Momentum p = mv would give 2 + 3 = 5%. Kinetic energy gives 8%. The only difference is the square, and it is worth 3 percentage points.
Q14Types of error

Which of the following errors cannot be reduced by repeating the measurement many times and averaging?

  1. Random error caused by mechanical vibration
  2. Parallax error that varies from reading to reading
  3. Statistical fluctuation in a counting experiment
  4. Zero error of the measuring instrument
Show full solution
Given
Four error sources.
Asked
Which one averaging cannot remove.
Concept
Averaging works only when errors are equally likely to be positive or negative, so they cancel in the long run. A systematic error always has the same sign and simply survives the averaging.
Formula
ā → true value only if errors are random
Baby steps
  1. Random errors scatter both ways about the true value.
  2. Their mean tends to zero as the number of readings grows.
  3. A zero error adds the same amount every time.
  4. Its mean is that same amount, so averaging leaves it untouched.
Answer
(d) Zero error of the measuring instrument
Why not others
(a)Vibration produces scatter in both directions — averaging helps.
(b)Varying parallax is random — averaging helps.
(c)Counting fluctuations are random — averaging helps, which is why long counts are used.
(d)Correct. A systematic error must be found and corrected for, not averaged away.
Shortcut
Averaging fights scatter, not bias. If all your readings are wrong in the same direction, more readings just give you a more confident wrong answer.
Where it goes wrong
NCERT Exercise 1.8(c) asks why 100 measurements beat 5. The answer is about random error only — a full answer should say that systematic error is unaffected.
Q15Propagation

A quantity is given by Y = A²√B / (C1/3 D³). If the percentage errors in A, B, C and D are 1%, 2%, 3% and 4% respectively, the maximum percentage error in Y is:

  1. 10%
  2. 12%
  3. 16%
  4. 14%
Show full solution
Given
Y = A²B1/2C−1/3D−3; errors 1%, 2%, 3%, 4%.
Asked
Maximum percentage error in Y.
Concept
Multiply each percentage error by the magnitude of its exponent, then add. Fractional and negative exponents are handled by size alone.
Formula
ΔY/Y = 2(ΔA/A) + ½(ΔB/B) + ⅓(ΔC/C) + 3(ΔD/D)
Baby steps
  1. From A: 2 × 1% = 2%.
  2. From B: ½ × 2% = 1%.
  3. From C: ⅓ × 3% = 1%.
  4. From D: 3 × 4% = 12%.
  5. Total = 2 + 1 + 1 + 12 = 16%.
Answer
(c) 16%
Why not others
(a)Comes from dropping the C term.
(b)Comes from taking D's exponent as 2 rather than 3.
(c)Correct. Notice D dominates: a large exponent on a poorly measured quantity swamps everything else.
(d)Comes from halving the D contribution.
Shortcut
Scan for the biggest product of exponent × error first. If one term dominates as badly as D does here, you can often eliminate wrong options before finishing the sum.
Where it goes wrong
Cube root means an exponent of ⅓, not 3. Reading 'cube root' as 'cube' turns a 1% contribution into a 9% one.
Q16Least count error

A balance has a least count of 1 mg. When it is used to weigh an object of mass 10 g, the relative error in the measurement is:

  1. 0.1%
  2. 0.01%
  3. 1%
  4. 0.001%
Show full solution
Given
Least count = 1 mg; measured mass = 10 g.
Asked
Relative error, as a percentage.
Concept
The least count sets the smallest absolute uncertainty an instrument can give. Convert both quantities to the same unit before dividing.
Formula
relative error = least count / measured value
Baby steps
  1. 1 mg = 10⁻³ g.
  2. Relative error = 10⁻³ / 10 = 10⁻⁴.
  3. As a percentage: 10⁻⁴ × 100 = 0.01%.
  4. Weighing a 1 g object on the same balance would give 0.1% — ten times worse.
Answer
(b) 0.01%
Why not others
(a)This is the value for a 1 g object.
(b)Correct.
(c)This is the value for a 0.1 g object.
(d)One order of magnitude too small — the conversion to percent was applied twice.
Shortcut
Always convert to a common unit before dividing. Mixing mg with g is the only real difficulty in this question type.
Where it goes wrong
Forgetting the × 100 that turns a fraction into a percentage — or applying it twice. State clearly which form you are working in.
Q17Propagation

The radius of a sphere is measured with a percentage error of 1%. The maximum percentage error in its calculated volume is:

  1. 3%
  2. 1%
  3. 2%
  4. 0.33%
Show full solution
Given
Δr/r = 1%; V = (4/3)πr³.
Asked
Maximum percentage error in volume.
Concept
Only the exponent matters. The constant 4π/3 is exact and carries no error.
Formula
ΔV/V = 3(Δr/r)
Baby steps
  1. V = (4/3)πr³.
  2. 4π/3 is exact → no contribution.
  3. ΔV/V = 3 × 1%.
  4. = 3%. For the surface area 4πr² it would be 2%.
Answer
(a) 3%
Why not others
(a)Correct.
(b)This ignores the cube.
(c)This is the surface-area answer.
(d)This divides by 3 instead of multiplying.
Shortcut
Learn the triple as one line: radius 1% → circumference 1%, area 2%, volume 3%.
Where it goes wrong
If the diameter is measured with 1% error, the radius also has 1% error, since r = d/2 and 2 is exact. The answer is unchanged.
Q18Experimental design

A student measures the period of a pendulum by timing 20 oscillations with a stopwatch of least count 0.1 s. The total time recorded is 40 s. The percentage error in the period T is:

  1. 2.5%
  2. 0.5%
  3. 1%
  4. 0.25%
Show full solution
Given
Least count 0.1 s; total time t = 40 s for n = 20 oscillations.
Asked
Percentage error in T.
Concept
T = t/n, and n is an exact counted number with no error. So the relative error in T equals the relative error in the total time — which is exactly why we time many oscillations.
Formula
T = t/n → ΔT/T = Δt/t
Baby steps
  1. Δt = 0.1 s (the least count).
  2. Δt/t = 0.1/40 = 0.0025.
  3. Since n is exact, ΔT/T = Δt/t = 0.0025.
  4. As a percentage: 0.25%. Timing a single swing would have given 5%.
Answer
(d) 0.25%
Why not others
(a)This uses t = 4 s.
(b)This uses n = 10 rather than 20.
(c)This uses t = 10 s.
(d)Correct.
Shortcut
Timing n oscillations instead of one divides the percentage error by n. This is the whole reason the experiment is done that way, and it is a favourite one-line reasoning question.
Where it goes wrong
Do not also divide Δt by 20. The stopwatch's least count applies to the single total reading you took, not to each oscillation.
Q19Propagation

In an experiment, the percentage errors in measuring M, L and T are 1%, 2% and 3% respectively. The maximum percentage error in a quantity X = M L² / T³ is:

  1. 6%
  2. 12%
  3. 14%
  4. 9%
Show full solution
Given
Errors 1%, 2%, 3% in M, L, T; X = M L² T⁻³.
Asked
Maximum percentage error in X.
Concept
Exponent × error, summed, with all signs ignored.
Formula
ΔX/X = ΔM/M + 2(ΔL/L) + 3(ΔT/T)
Baby steps
  1. From M: 1 × 1% = 1%.
  2. From L: 2 × 2% = 4%.
  3. From T: 3 × 3% = 9%.
  4. Total = 1 + 4 + 9 = 14%.
Answer
(c) 14%
Why not others
(a)This adds 1 + 2 + 3 without applying the exponents.
(b)This drops the M term.
(c)Correct.
(d)This keeps only the T term.
Shortcut
Because T is both the worst-measured quantity and the one with the largest exponent, it supplies nine of the fourteen points. In a real experiment, that is where you would spend your effort improving.
Where it goes wrong
T is in the denominator, but its exponent is still counted as 3, not −3. Errors always add; there is no cancellation.
Q20Precision

Which of the following measurements is the most precise?

  1. 5.00 km
  2. 5.00 mm
  3. 5.00 m
  4. 5.00 cm
Show full solution
Given
Four measurements, each written to three significant figures.
Asked
The most precise measurement.
Concept
All four have the same number of significant figures, so they are equally accurate in relative terms. Precision here is decided by the smallest absolute uncertainty, which means the smallest unit.
Formula
uncertainty ≈ ±1 in the last written digit
Baby steps
  1. 5.00 km implies ±0.01 km = ±10 m.
  2. 5.00 m implies ±0.01 m = ±1 cm.
  3. 5.00 cm implies ±0.01 cm = ±0.1 mm.
  4. 5.00 mm implies ±0.01 mm — by far the smallest.
  5. So 5.00 mm is the most precise.
Answer
(b) 5.00 mm
Why not others
(a)The coarsest of the four: uncertain to ten metres.
(b)Correct — smallest unit, therefore smallest absolute uncertainty.
(c)Uncertain to a centimetre.
(d)Uncertain to a tenth of a millimetre.
Shortcut
Same significant figures → smallest unit wins. Precision follows the size of the unit; the count of significant figures is what stays fixed under a change of unit.
Where it goes wrong
This question is deliberately built so that the significant-figure rule gives no answer at all. Changing units never changes the number of significant figures — NCERT states this explicitly — so you must compare absolute uncertainties instead.