Chapter 1 of 14

Karnataka (KSEEB)Class 11 Physics★ Unit I · part of the 23-mark block covering Units I and II in CBSE Class 11 Physics (70 marks total); dimensional analysis and significant figures are frequent in JEE and NEET

Units and Measurements

The seven SI base units are no longer defined by objects — since 2018 every one is pinned to a constant of nature. This chapter covers that, plus significant figures and dimensional analysis, worked through all five of the NCERT chapter's own solved examples. CBSE Class 11 Physics Chapter 1.

⏱ 20 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Name the seven SI base units and what each is now pinned to
2Count significant figures, and know why changing units never changes them
3Use the right rule: significant figures for ×÷, decimal places for +−
4Find the uncertainty in a result — percentage errors add for a product
5Write any dimensional formula, and test an equation for homogeneity
6Derive a relation by dimensional analysis, and say what it cannot give you

💡

Why this chapter matters

Every number you write for the rest of physics depends on this chapter. Since 2018 the SI units are pinned to constants of nature rather than objects — the kilogram no longer lives in a vault in Paris. And dimensional analysis will kill a wrong formula in fifteen seconds, though it can only ever prove one wrong, never right.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

🔗

Class 10 Science — measurement and SI units

Familiarity with metre, kilogram and second as standard units, and with writing numbers in scientific notation

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Class 10 Mathematics — exponents and powers

Manipulating powers of ten fluently is needed throughout scientific notation and unit conversion

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Class 10 Mathematics — algebraic manipulation

Equating powers on both sides of an equation, as used to solve for exponents in dimensional analysis

Units and Measurements

1. What This Chapter Covers

NCERT Physics Part I, Chapter 1 (Reprint 2026-27) has six sections. Note section 1.3.3 — it is where uncertainty now lives.

Textbook section	Topic	What it answers
1.1	Introduction	Why physics needs an agreed system of units
1.2	The International System of Units	What SI is, and how its base units are fixed today
1.3	Significant figures	How many digits of a measurement you can trust
1.3.3	Uncertainty in a result	How error carries through a calculation
1.4	Dimensions of physical quantities	What a quantity's "dimensions" mean
1.5	Dimensional formulae and equations	How to write them down
1.6	Dimensional analysis	What the method can and cannot do

What changed — and what did NOT

The old standalone "Errors in measurement" section is gone. Uncertainty is not. It survives as section 1.3.3, folded inside significant figures — and the CBSE 2026-27 syllabus lists it explicitly:

"significant figures, Determining the uncertainty in result. Dimensions of physical quantities, dimensional analysis and its applications." — CBSE Class XI Physics (Code 042), Unit I, session 2026-27

Topic	Status in the 2026-27 chapter
Uncertainty in a calculated result	In — textbook section 1.3.3, and examinable
Relative error, and the `±` notation	In — worked in textbook section 1.3.3
Combination-of-errors rule for a product	In — worked in textbook section 1.3.3
Systematic / random / gross error classification	Not in this chapter
The terms "absolute error", "percentage error"	Not used
Measuring instruments as a taught section	Removed from the body
Vernier callipers, screw gauge	Only in Exercises 1.6 and 1.8 — never taught

Two traps follow from this table.

Notes that tell you error analysis is off-syllabus are wrong — CBSE lists it as examinable. Section 6 below teaches section 1.3.3 in full.
Notes that drill you on systematic vs random errors are teaching an older edition. That classification is genuinely absent here.

Exercises 1.6 and 1.8 assume instruments the chapter never explains. Section 8 covers them.

2. Units and the SI System (Textbook 1.1 to 1.2)

Why units exist at all

A measurement is a comparison against an agreed standard. "The length is 5" means nothing until you say 5 of what.

Every quantity in physics can be built from a small independent set. So you only need units for that set.

Term	Meaning	Examples
Base quantity	Independent, not built from others	length, mass, time
Base unit	The unit of a base quantity	metre, kilogram, second
Derived unit	Combination of base units	m/s, kg·m/s², N, J
System of units	The complete set, base + derived	SI, CGS, FPS, MKS

Before SI

Three systems were in common use side by side:

System	Length	Mass	Time
CGS	centimetre	gram	second
FPS (British)	foot	pound	second
MKS	metre	kilogram	second

SI — Système International d'Unités — replaced them internationally. It was last revised in November 2018 by the General Conference on Weights and Measures.

3. The Seven Base Units, and the 2018 Redefinition (Textbook 1.2)

This is the most quietly remarkable idea in the chapter.

Every SI base unit is now defined by declaring a constant of nature to be exact. No unit depends on a physical object any more.

Base quantity	Unit	Symbol	Fixed by declaring
Length	metre	m	speed of light `c = 299792458 m/s`
Mass	kilogram	kg	Planck constant `h = 6.62607015 × 10⁻³⁴ J s`
Time	second	s	caesium-133 frequency `Δν = 9192631770 Hz`
Electric current	ampere	A	elementary charge `e = 1.602176634 × 10⁻¹⁹ C`
Temperature	kelvin	K	Boltzmann constant `k = 1.380649 × 10⁻²³ J/K`
Amount of substance	mole	mol	Avogadro constant `Nₐ = 6.02214076 × 10²³ /mol`
Luminous intensity	candela	cd	luminous efficacy `K_cd = 683 lm/W`

You do not need to memorise these numbers. They are quoted only to show how precisely each constant is now known.

Why the kilogram changed

Until 2019 the kilogram was an actual metal cylinder — platinum-iridium, kept in a vault outside Paris.

That is a poor foundation. A physical object can pick up contamination or be damaged. If it drifts, every kilogram on Earth drifts with it.

The fix reverses the logic:

Old approach	New approach
Object defines the unit	Constant defines the unit
Realising a kg = compare to the object	Realising a kg = an experiment
Only one master copy exists	Any equipped lab can reproduce it

The metre now depends on the second

Light speed is fixed by definition. So once the caesium clock fixes the second, the metre follows automatically.

The metre is defined in terms of the second — not the other way round.

Two supplementary units

Unit	For	Defined as	Dimensions
radian (rad)	plane angle	arc ÷ radius	dimensionless
steradian (sr)	solid angle	area ÷ radius²	dimensionless

Both are ratios of like quantities, which is why neither has dimensions.

4. Significant Figures (Textbook 1.3)

What the term means

Every measurement carries uncertainty. The way you write a number should show how well you actually know it.

A significant figure count = all reliably known digits, plus the first uncertain one.

Example. A pendulum period written as 1.62 s:

Digit	Status
1	certain
6	certain
2	first uncertain digit

That is three significant figures.

The counting rules

Start from one clean case. 2.308 cm has four significant figures. Now change units:

2.308 cm = 0.02308 m = 23.08 mm = 23080 μm

All four still have four significant figures. Changing units cannot make an instrument sharper.

Rule	Example	Count
All non-zero digits count	`2.38`	3
Zeros between non-zeros count	`6.032`	4
Leading zeros never count	`0.002308`	4
Trailing zeros with a decimal point count	`3.500`	4
Trailing zeros without a decimal point are ambiguous	`12300`	3

The trailing-zero trap

Take a length reported as 4.700 m. Those zeros were written deliberately — otherwise the measurer would have written 4.7 m. So it has four significant figures.

Now convert it:

4.700 m = 470.0 cm = 4700 mm = 0.004700 km

Read 4700 mm on its own and the no-decimal rule gives you two figures. That is wrong. A unit change cannot alter measured precision.

The problem is that a plain integer with trailing zeros is genuinely ambiguous.

The fix: always use scientific notation

Write every measurement as a × 10ᵇ, with 1 ≤ a < 10.

Form	Significant figures
`4.700 × 10² cm`	4
`4.700 × 10³ mm`	4
`4.700 × 10⁻³ km`	4

The power of ten carries no significance. Every digit in a is significant. No judgement calls remain.

Order of magnitude

Round a to 1 or 10, and the leftover 10ᵇ is the order of magnitude.

Quantity	Value	Order
Earth's diameter	`1.28 × 10⁷ m`	10⁷
Hydrogen atom diameter	`1.06 × 10⁻¹⁰ m`	10⁻¹⁰

So Earth is 17 orders of magnitude larger than a hydrogen atom — one checkable sentence from two unwieldy numbers.

Exact numbers

Some numbers are exact by definition or by counting. They have unlimited significant figures and never limit a result.

Exact number	Where it appears
the 2	`d = 2r`
the 2π	`T = 2π√(l/g)`
the n	dividing by n readings when averaging

5. Arithmetic With Significant Figures (Textbook 1.3)

Two different rules

This is the most commonly confused pair in the chapter.

Operation	What you count	Example
Multiply / divide	fewest significant figures	`5.74 ÷ 1.2 = 4.8`
Add / subtract	fewest decimal places	`436.32 + 227.2 = 663.8`

Why they differ

Multiplication — each factor's fractional uncertainty carries into the product, so the weakest factor caps the result.
Addition — what matters is the last trustworthy decimal position. Adding 227.2 g cannot give you a trustworthy hundredths digit, because that input has none.

Worked case. 436.32 g + 227.2 g + 0.301 g = 663.821 g by raw arithmetic.

227.2 has only one decimal place → answer is 663.8 g.

Using the multiplication rule here would give 664 g, which misstates the precision in the other direction.

Subtraction loses figures

0.307 m − 0.304 m = 0.003 m

Both inputs had three significant figures. The answer has one. You cannot write 3.00 × 10⁻³ m — that invents two digits the measurement never had.

Example 1.1 — cube

A cube's side is measured as 7.203 m → 4 significant figures, so both answers round to 4.

Quantity	Raw value	Reported
Surface area `6a²`	`311.299254 m²`	`311.3 m²`
Volume `a³`	`373.714754 m³`	`373.7 m³`

Example 1.2 — density

5.74 g occupies 1.2 cm³.

Input	Significant figures
mass `5.74 g`	3
volume `1.2 cm³`	2 — limiting

5.74 ÷ 1.2 = 4.8666… → reported as 4.8 g/cm³.

The extra digits on your calculator are false precision, not extra accuracy.

Rounding when the dropped digit is exactly 5

The convention is round to the nearest even digit.

Number	Rounds to	Why
`2.745`	`2.74`	4 is already even
`2.735`	`2.74`	3 is odd, bumps to even 4

Both land on 2.74, from opposite sides. That is deliberate, not coincidence.

Never round mid-calculation

Carry one extra digit through intermediate steps.

Approach	Result
Round `1/9.58` to 3 figures → `0.104`, then invert	`9.62` — drifted
Keep `0.1044`, then invert	`9.58` — correct

Round twice and the error compounds.

6. Uncertainty in a Calculated Result (Textbook 1.3.3)

This section is examinable. CBSE lists "Determining the uncertainty in result" under Unit I.

Writing a measurement with its uncertainty

A metre scale reading of 16.2 cm carries an uncertainty of one least division, ± 0.1 cm.

That can be written two ways — absolute, or as a percentage:

Measurement	Absolute form	Percentage form
length `l`	`16.2 ± 0.1 cm`	`16.2 cm ± 0.6%`
breadth `b`	`10.1 ± 0.1 cm`	`10.1 cm ± 1%`

The percentage form is what makes combining errors easy.

Combining errors in a product

Rule: for a product or quotient, the percentage errors add.

Worked, straight from the chapter:

Step	Working
Area	`l × b = 16.2 × 10.1 = 163.62 cm²`
Percentage errors add	`0.6% + 1% = 1.6%`
Convert back to absolute	`1.6% of 163.62 = 2.6 cm²`
Raw result	`163.62 ± 2.6 cm²`
Quoted result	`164 ± 3 cm²`

The final rounding matters. An uncertainty of 2.6 is itself only known roughly, so it is quoted as 3, and the value is rounded to match — 164, not 163.62.

Relative error depends on the number, not just the digit count

Two masses measured on the same balance, both accurate to ± 0.01 g:

Measurement	Relative error	Working
`1.02 g`	± 1%	`0.01 / 1.02 × 100`
`9.89 g`	± 0.1%	`0.01 / 9.89 × 100`

Same instrument, same absolute uncertainty — but ten times the relative error on the smaller mass. This is why weighing a small sample on a coarse balance is a bad idea.

Subtraction can destroy significant figures

12.9 g − 7.06 g, both given to three significant figures.

You cannot write 5.84 g. The correct answer is 5.8 g.

Addition and subtraction combine uncertainties by decimal places, not by significant figures — so a result can end up with fewer significant figures than either input started with.

7. Dimensions and Dimensional Analysis (Textbook 1.4 to 1.6)

What dimensions are

Every mechanical quantity reduces to a combination of [M], [L] and [T] raised to powers.

Quantity	Built from	Dimensional formula
Volume	length³	`[M⁰L³T⁰]`
Speed	length ÷ time	`[M⁰LT⁻¹]`
Acceleration	length ÷ time²	`[M⁰LT⁻²]`
Force	mass × acceleration	`[MLT⁻²]`
Energy / Work	force × length	`[ML²T⁻²]`
Mass density	mass ÷ volume	`[ML⁻³T⁰]`

Dimensions deliberately discard magnitude. Initial velocity, final velocity and average speed are physically different, but all are [LT⁻¹].

Formula vs equation

Term	Meaning	Example
Dimensional formula	the powers themselves	`[M⁰LT⁻¹]`
Dimensional equation	quantity equated to its formula	`[v] = [M⁰LT⁻¹]`

The principle of homogeneity

Only quantities with the same dimensions can be added or subtracted.

Adding a velocity to a force is as meaningless as adding three apples to two hours. So every term on both sides of a correct equation must match dimensionally.

Test case: x = x₀ + v₀t + ½at²

Term	Dimensions
`x`	`[L]`
`x₀`	`[L]`
`v₀t`	`[LT⁻¹][T]` = `[L]`
`½at²`	`[LT⁻²][T²]` = `[L]`

All four match. The equation passes.

The test runs one way only

Failing it proves an equation wrong. Passing it never proves an equation right.

A bare numerical factor has no dimensions, so a dimensional check simply cannot see it.

Example 1.3 — checking an equation

Check ½mv² = mgh.

Side	Working	Result
Left	`[M][LT⁻¹]²`	`[ML²T⁻²]`
Right	`[M][LT⁻²][L]`	`[ML²T⁻²]`

They match — dimensionally correct.

Example 1.4 — what dimensions can rule out

Five candidate formulas for kinetic energy. True dimension is [ML²T⁻²].

Candidate	Dimensions	Verdict
`K = m²v³`	`[M²L³T⁻³]`	eliminated
`K = ½mv²`	`[ML²T⁻²]`	survives
`K = ma`	`[MLT⁻²]`	eliminated — that is force
`K = (3/16)mv²`	`[ML²T⁻²]`	survives
`K = ½mv² + ma`	mixed	eliminated — illegal sum

Two survive, and this is the point of the example.

½mv² and (3/16)mv² are dimensionally identical. The method cannot separate them, because it cannot see the numbers ½ and 3/16.

Only the derivation of kinetic energy from the work-energy theorem (Chapter 5) settles it. The answer is ½mv².

Example 1.5 — deriving a relationship

A pendulum's period T may depend on length l, bob mass m, and g.

Step 1. Assume a product form: T = k·lˣ·gʸ·mᶻ, with k dimensionless.

Step 2. Substitute dimensions:

[T¹] = [L]ˣ [LT⁻²]ʸ [M]ᶻ = L⁽ˣ⁺ʸ⁾ T⁻²ʸ Mᶻ

Step 3. Equate powers separately:

Base	Equation	Solution
M	`z = 0`	`z = 0`
L	`x + y = 0`	`x = ½`
T	`−2y = 1`	`y = −½`

Step 4. Result: T = k√(l/g)

Two things fall out of this:

Mass drops out entirely (z = 0). A pendulum's period does not depend on the bob's weight — a real, testable physics result.
k cannot be found this way. The full Newtonian derivation gives k = 2π, so T = 2π√(l/g).

Always state that limitation. The derivation is incomplete without it — you have found the form of the relation, not the constant.

What the method can and cannot do

Good for	Cannot do
Sanity-checking a formula fast	Find dimensionless constants (2π, ½)
Converting between unit systems	Handle sin, log, eˣ relationships
Finding the form of a relation	Separate torque from energy (both `[ML²T⁻²]`)
Up to three independent variables	Four or more unknowns

8. What Exercises 1.6 and 1.8 Still Assume

These three exercises lean on the removed instrument material. Here is the minimum you need.

Instrument	Least count	Typical resolution
Vernier calliper	main-scale division ÷ vernier divisions	0.05–0.02 mm
Screw gauge	pitch ÷ circular-scale divisions	0.01 mm

Least count = the smallest change an instrument can actually resolve.

Worked: a screw gauge of pitch 1.0 mm with 100 circular divisions has least count 1.0 ÷ 100 = 0.01 mm.

More divisions do not mean unlimited precision

Past a point, extra markings outrun what the screw itself can resolve. Backlash — slack in the thread — becomes the real limit, not the dial.

Why more readings help

Any single reading carries random scatter in both directions.

Average enough readings and that scatter increasingly cancels. A mean of 100 readings sits closer to the true value than a mean of 5.

That is the whole reasoning behind Exercise 1.8(c).

9. Summary

A measurement is a comparison against a standard; a number without a unit means nothing.
Seven SI base units: metre, kilogram, second, ampere, kelvin, mole, candela.
Since 2018, every base unit is fixed by declaring a constant of nature exact.
The kilogram left its platinum-iridium cylinder behind; the metre now depends on the second.
Significant figures = reliable digits + the first uncertain one.
A change of units never changes the significant-figure count.
Scientific notation removes all trailing-zero ambiguity.
Multiply/divide → fewest significant figures. Add/subtract → fewest decimal places.
Round a dropped 5 to the nearest even digit; never round mid-calculation.
Dimensions are the powers of [M], [L], [T] that build a quantity.
Homogeneity: every term in a correct equation shares dimensions.
A dimensional check can disprove an equation, never prove one.
Dimensional analysis gives the form of a relation, never its dimensionless constant.
Still examinable: uncertainty in a result (Textbook 1.3.3) — relative error, and percentage errors adding for a product.
Not in this chapter: systematic/random error classification, and instruments as a taught topic.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Significant figures

all reliable digits + the first uncertain one

1.62 s → 3 significant figures

Scientific notation

a × 10^b, with 1 ≤ a < 10

The only unambiguous way to write trailing zeros

Multiply or divide

keep the FEWEST significant figures

5.74 ÷ 1.2 = 4.8, not 4.87

Add or subtract

keep the FEWEST decimal places

A different rule — decimal places, not sig figs

Rounding a dropped 5

round to the EVEN neighbour

2.745 → 2.74 and 2.735 → 2.74

Measurement with uncertainty

l = 16.2 ± 0.1 cm = 16.2 cm ± 0.6%

section 1.3.3 — absolute and percentage form

Combining errors in a product

percentage errors ADD

0.6% + 1% = 1.6% → 164 ± 3 cm²

Relative error

(uncertainty ÷ value) × 100%

±0.01 g is ±1% on 1.02 g but ±0.1% on 9.89 g

Dimensional formulae

force [MLT⁻²] · energy [ML²T⁻²] · density [ML⁻³T⁰]

Torque is ALSO [ML²T⁻²] — dimensions can't separate them

Homogeneity

every term in a correct equation shares dimensions

Can disprove an equation, never prove one

⚠️

Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Believing uncertainty is off-syllabus for this chapter

✓ It is examinable. CBSE 2026-27 Unit I lists 'Determining the uncertainty in result', and NCERT keeps it as section 1.3.3 inside significant figures.

WATCH OUT

✗ Revising systematic vs random vs gross errors

✓ That classification returns zero hits in the current chapter — it is older-edition material. Revise section 1.3.3 instead.

WATCH OUT

✗ Treating 'dimensionally correct' as 'correct'

✓ ½mv² and (3/16)mv² pass identically; a dimensional check cannot see a bare number. Passing is necessary, never sufficient.

WATCH OUT

✗ Thinking a unit change alters the significant-figure count

✓ 2.308 cm = 23.08 mm = 0.02308 m — four figures throughout. Converting units cannot sharpen an instrument.

WATCH OUT

✗ Using the multiplication rule for an addition

✓ Multiply/divide counts significant figures; add/subtract counts decimal places. 436.32 + 227.2 = 663.8 g, not 664 g.

WATCH OUT

✗ Quoting an area as 163.62 ± 2.6 cm²

✓ Round the uncertainty to one figure and match the value to it: 164 ± 3 cm². Carrying 163.62 implies precision the readings never had.

WATCH OUT

✗ Rounding at every intermediate step

✓ Round 1/9.58 to three figures then invert and you get 9.62, not 9.58. Carry one spare digit and round once at the end.

WATCH OUT

✗ Expecting dimensional analysis to give the constant

✓ It gives T = k√(l/g) and stops. k = 2π comes only from the full derivation — say so explicitly, or the derivation is incomplete.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex s

Exercises

Unit conversions, dimensional analysis, significant figures, and precision reasoning (Q1.1-1.17)

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Units and Measurements?

10 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

10 questions~7 min worth ~23 marks in Karnataka (KSEEB) exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•A number without a unit means nothing.
•Seven SI base units, each now fixed by a constant of nature.
•The metre is defined via the second, not the reverse.
•Significant figures = reliable digits + the first uncertain one.
•Changing units never changes the significant-figure count.
•×÷ → fewest significant figures. +− → fewest decimal places.
•Round a dropped 5 to the even neighbour.
•Uncertainty is examinable, and lives in section 1.3.3.
•For a product, percentage errors add.
•Quote 164 ± 3 cm², not 163.62 ± 2.6 cm².
•Homogeneity disproves an equation; it never proves one.
•Dimensional analysis gives the form, never the constant.

Karnataka (KSEEB) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: Unit I sits inside the 23-mark block covering Units I and II (CBSE Class 11 Physics, 70 marks)

Question type	Marks each	Typical count	What it tests
SI units and significant figures	1-2	1-2	Naming an SI base unit or its defining constant; counting significant figures; applying the correct rule (multiply/divide vs add/subtract) to a calculation
Uncertainty and error combination	2-3	1	Writing a measurement with its absolute or percentage uncertainty; combining percentage errors in a product or quotient; rounding a result to match its uncertainty
Dimensional formulae and dimensional analysis	3-5	1	Writing dimensional formulae; checking homogeneity term by term; deriving the form of a relation and stating what the method cannot give

Prep strategy

Learn the chapter as three blocks: what a unit IS (1.1-1.2), how honestly to WRITE a number (1.3), and what dimensions can DO for you (1.4-1.6)
Memorise dimensional formulae of the common quantities — velocity, acceleration, force, energy, density — and be able to derive any other from its defining equation
Practise the pendulum-style derivation until the 'equate powers of M, L, T separately' step is automatic; it is the single most examined derivation here
Always state the limitation alongside a dimensional result: k cannot be found this way
Do not revise error propagation for this chapter — it is off-syllabus for 2026-27 and no marks are available for it
For every significant-figure question, first ask whether the operation is multiply/divide (count figures) or add/subtract (count decimal places) before touching the arithmetic

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Catching a wrong formula before it costs anything

A dimensional check takes fifteen seconds and rules out an inconsistent equation before it is ever used in a design, an experiment, or an exam answer.

The Mars Climate Orbiter

A 1999 NASA mission was lost because one team supplied thrust data in pound-force seconds while the software expected newton-seconds — a units mismatch, not a physics error.

Calibration laboratories worldwide

Because the base units are now pinned to constants of nature, any suitably equipped lab can realise a kilogram or a metre independently, without shipping an artefact to Paris.

Reporting a scientific result honestly

Significant figures are how a published measurement communicates the limits of the instrument that produced it, rather than the limits of the calculator that processed it.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Write the dimensional formula of every term explicitly before comparing them — the working is the answer, not the verdict

Check homogeneity term by term; show each term reduces to the same dimension

End every dimensional derivation by stating that the dimensionless constant cannot be found this way — without it the derivation is incomplete

Decide first whether the operation is multiply/divide (significant figures) or add/subtract (decimal places) before any arithmetic

Carry one extra digit through intermediate steps and round once at the end

Uncertainty is examinable — CBSE 2026-27 Unit I lists 'Determining the uncertainty in result'

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

Use the Buckingham Pi theorem to generalise the product-form method beyond three variables, and identify the dimensionless groups governing a given problem

STRETCH

Construct the Planck units — Planck length, mass and time — from c, h and G alone, and show that each is uniquely determined by dimensional reasoning

STRETCH

Show that the Reynolds number is dimensionless, and explain why that single fact allows a scale model in a wind tunnel to predict full-size behaviour

STRETCH

Given that the drag force on a sphere may depend on its radius, the fluid density and the flow speed, derive the form of the drag equation by dimensions and state exactly what remains undetermined

🚀

JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainError propagation in kinetic energyFormula application

The mass and speed of a body are measured with percentage errors of 2% and 3% respectively. Find the maximum percentage error in the kinetic energy $K = \frac{1}{2} m v^{2}$ .

Stuck? Show the approach

For a product of powers, percentage errors multiply by the exponent and add. Here $K \propto m^{1} v^{2}$ , so the error in $m$ counts once and the error in $v$ counts twice.

Show the full solution

$\frac{Δ K}{K} = \frac{Δ m}{m} + 2 \frac{Δ v}{v} = 2% + 2 (3%) = 2% + 6% = 8%$

Answer: 8%

The trap

A common slip is adding the errors as 2% + 3% = 5%, forgetting that v is squared in the formula — each power multiplies its variable's percentage error before the errors are summed.

JEE MainError in g from a pendulum experimentFormula application

In an experiment to find g using $T = 2 π l / g$ , the length l is measured with 1% error and the time period T with 2% error. Find the maximum percentage error in the computed value of g.

Stuck? Show the approach

Rearrange for g first: $g = 4 π^{2} l / T^{2}$ , so $g \propto l^{1} T^{- 2}$ . The sign of the exponent does not matter for a maximum-error estimate — only its magnitude does.

Show the full solution

$\frac{Δ g}{g} = \frac{Δ l}{l} + 2 \frac{Δ T}{T} = 1% + 2 (2%) = 1% + 4% = 5%$

Answer: 5%

The trap

Students sometimes divide by 2 instead of multiplying by 2 for T, mistaking the square root in the original formula for T's own exponent — but once g is isolated, T appears as T^-2, not T^(1/2).

JEE MainDimensions of viscosity from Stokes' lawDimensional analysis

Stokes' law gives the viscous drag on a sphere as $F = 6 π η r v$ , where r is radius and v is speed. Find the dimensional formula of the coefficient of viscosity $η$ .

Stuck? Show the approach

Rearrange for the unknown and substitute the known dimensions of each remaining quantity: force, radius, and speed.

Show the full solution

$η = \frac{F}{6 π r v} ⟹ [η] = \frac{[ M L T ^{- 2} ]}{[ L ] [ L T ^{- 1} ]} = [M L^{- 1} T^{- 1}]$

Answer: [M L^-1 T^-1]

The trap

The numerical factor 6π is dimensionless and must be dropped before comparing dimensions — carrying it through as if it had units is a frequent error when rearranging a formula for an unknown constant.

JEE AdvancedCombined error in a four-variable derived quantityMulti-step error propagation

A quantity is calculated as $P = \frac{a ^{2} b ^{3}}{c d}$ , where a, b, c, d are measured with percentage errors of 1%, 2%, 3% and 4% respectively. Find the maximum percentage error in P.

Stuck? Show the approach

Treat P as a product of powers, including the negative power for d in the denominator and the fractional power (1/2) for the square root of c. Every exponent multiplies its variable's own percentage error, and a division still adds its error — it never subtracts.

Show the full solution

$\frac{Δ P}{P} = 2 \frac{Δ a}{a} + 3 \frac{Δ b}{b} + \frac{1}{2} \frac{Δ c}{c} + \frac{Δ d}{d}$ $= 2 (1%) + 3 (2%) + \frac{1}{2} (3%) + 4% = 2% + 6% + 1.5% + 4% = 13.5%$

Answer: 13.5%

The trap

Because d sits in the denominator, some students subtract its error instead of adding it. Errors always add for both multiplication and division — a quantity in the denominator does not get its error sign flipped, since an error can push the true value either way regardless of where the variable sits in the formula.

JEE AdvancedDimensional derivation with an unreachable constantDerivation with conceptual trap

The fundamental frequency $ν$ of a stretched string may depend on its length l, the tension T in it, and its mass per unit length $μ$ . Use dimensional analysis to find how $ν$ depends on l, T and $μ$ , and explain why the method alone cannot fix the exact numerical relation used in practice, $ν = \frac{1}{2 l} T / μ$ .

Stuck? Show the approach

Assume a product form $ν = k l^{x} T^{y} μ^{z}$ with k dimensionless, substitute the dimensions of each quantity, and equate powers of M, L, T separately.

Show the full solution

$[ν] = [T^{- 1}]$ , $[l]^{x} = [L]^{x}$ , $[T]^{y} = [M L T^{- 2}]^{y}$ , $[μ]^{z} = [M L^{- 1}]^{z}$ .

Equating M: $y + z = 0$ . Equating T: $- 2 y = - 1 \Rightarrow y = 1/2$ , so $z = - 1/2$ . Equating L: $x + y - z = 0 \Rightarrow x = - y + z = - 1$ .

$ν = k l^{- 1} T^{1/2} μ^{- 1/2} = \frac{k}{l} \frac{T}{μ}$

Answer: nu = (k/l) sqrt(T/mu), with k a dimensionless constant the method cannot determine

The trap

Dimensional analysis correctly gives the l^-1 sqrt(T/mu) form but can never produce the factor of 1/2 in front, since that factor is a pure number with no dimensions to track — it comes only from the full wave-mechanics derivation (Chapter 14), not from this method.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 Physics examMedium

JEE Main and Advanced (Units & Dimensions)Medium

NEET PhysicsMedium

NSEP and Indian National Physics OlympiadMedium

KVPY-style aptitude and reasoning papersLow

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Partly removed, and the distinction matters for your marks. Uncertainty itself is very much in: NCERT keeps it as section 1.3.3, 'Rules for Determining the Uncertainty in the Results of Arithmetic Calculations', tucked inside the significant-figures section rather than standing alone. The CBSE 2026-27 syllabus for Unit I names it directly — 'significant figures, Determining the uncertainty in result'. So you must know the ± notation, relative error, and the rule that percentage errors add for a product. What genuinely has gone is the older edition's classification of errors into systematic, random and gross — a text search of the current chapter returns zero hits for any of those terms — along with the standalone sections on measuring instruments. So: revise 1.3.3 properly, and skip the systematic-versus-random tables your coaching notes may still carry.

Because it only ever checks that both sides of an equation balance dimensionally, and a bare number has no dimensions to balance. K = ½mv² and K = (3/16)mv² are dimensionally identical, so the method passes both equally and has no way to prefer one. It also cannot touch relationships built on sin, log or exponential functions, since those functions require dimensionless arguments before you even begin, and it cannot separate two quantities that happen to share a dimension — torque and energy are both [ML²T⁻²] despite being physically unrelated. What it gives you is the shape of a relation, which is often exactly what you need, plus a very fast way to rule wrong formulas out.

Until 2019 the kilogram was defined by an actual platinum-iridium cylinder stored outside Paris. That is an uncomfortable foundation for a system of units — a physical object can pick up contamination, or be damaged, and if it drifts then by definition every kilogram in the world drifts with it. The 2018 revision instead declares the Planck constant to be exactly 6.62607015 × 10⁻³⁴ J s, so a kilogram is now realised through experiment rather than compared against one irreplaceable object. For your calculations nothing changes — the new definitions were chosen precisely so that a kilogram stays the same size as before. What changes is that the standard is now reproducible in any suitably equipped lab, rather than existing in exactly one vault.

That is exactly the point — they must have the same count, because converting between units cannot make your instrument more or less precise. Both carry four significant figures (2, 3, 0, 8), and so do 0.02308 m and 23080 μm. The location of the decimal point is irrelevant to the count. The one form that causes trouble is a plain integer with trailing zeros, like 4700 mm, where you genuinely cannot tell from the digits whether those zeros were measured or are just placeholders. That ambiguity is the whole reason the book recommends writing every measurement in scientific notation, where the question never arises.

Because they are limited by different things. In a multiplication, the fractional uncertainty of each factor carries through to the product, so the factor known to the fewest significant figures caps the whole result. In an addition, what matters is the absolute position of the last trustworthy digit: adding 436.32 g to 227.2 g cannot give you a trustworthy hundredths digit, because one of the inputs simply does not have one — so the sum is capped at one decimal place, 663.8 g, regardless of how many significant figures each input had. Applying the multiplication rule here would give 664 g, which misrepresents the precision in the other direction.

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