PhysicsCBSE Class 11 · Chapter 1

Units and Measurement Class 11: Notes, Dimensions and Solved Numericals

Updated 14 October 2026 · Fact-checked

Units and Measurement is Chapter 1 of CBSE Class 11 Physics. It covers why we need units, the SI system with its seven base units and derived units, significant figures and rounding off, how uncertainty carries into a calculated result, and dimensions of physical quantities with dimensional analysis and its applications.

What Units and Measurement is about

Every number in physics comes with a unit. Saying a rod is '3' long means nothing until you say 3 m or 3 cm. This chapter sets up the language the rest of Class 11 Physics is written in. You meet the International System of Units (SI), built on seven base quantities (length, mass, time, electric current, temperature, amount of substance and luminous intensity). Every other unit, from the newton to the pascal, is a derived unit made from these.

The second idea is honesty about precision. A measured value is reported with all the digits you are sure of plus one doubtful digit: these are its significant figures. When you multiply, divide, add or subtract measured values, simple rules tell you how many digits the answer deserves and how the uncertainty in the inputs carries into the result.

The third idea is dimensions. Any mechanical quantity can be written as a combination of mass [M], length [L] and time [T]; force, for example, is $[M L T^{- 2}]$ . This lets you check whether an equation could be right, convert units between systems and even guess the form of a formula. The rationalised 2026-27 book has dropped the old 'Physical World' chapter and the sections on measuring length, mass and time and on errors, so this chapter is now short but basic: dimensional analysis comes back in almost every chapter of Class 11 and 12 Physics and in JEE and NEET.

Weightage: This chapter is part of Unit I, Physical World and Measurement. The syllabus groups Unit I with Unit II, Kinematics and Unit III, Laws of Motion: together they carry 23 of the 70 theory marks in the school annual exam.Source: CBSE Physics (042) syllabus 2026-27, Class XI course structure, Unit I Physical World and Measurement

Key concepts

Units, base quantities and the SI system

NCERT section 1.1 and 1.2

To measure a quantity we compare it with a fixed reference standard called a unit. A measurement is always a number times a unit: in 5.2 m, the number is 5.2 and the unit is the metre.

Base (fundamental) units are chosen for a small set of independent quantities. Derived units are built from them; for example, speed has unit m s $^{- 1}$ and force has unit kg m s $^{- 2}$ , given the special name newton (N).

Older systems were the CGS (centimetre, gram, second), FPS (foot, pound, second) and MKS (metre, kilogram, second) systems. Today the SI (Système Internationale d'Unités) is used worldwide. It has seven base quantities:

Quantity	SI unit	Symbol
Length	metre	m
Mass	kilogram	kg
Time	second	s
Electric current	ampere	A
Thermodynamic temperature	kelvin	K
Amount of substance	mole	mol
Luminous intensity	candela	cd

Two supplementary units are used for angles: the radian (rad) for a plane angle, $d θ = \frac{d s}{r}$ , and the steradian (sr) for a solid angle, $d Ω = \frac{d A}{r ^{2}}$ . Both are ratios of like quantities, so they have no dimensions.

Writing units well: use the symbol without a full stop (5 kg, not 5 kgs or 5 Kg); units named after people are lower case when written in full (newton) but the symbol takes a capital (N).

Worked example

The density of mercury is 13.6 g cm $^{- 3}$ . Express it in SI units. [1 mark]

Try it first, then show the solutionSolution

1 g = $1 0^{- 3}$ kg and 1 cm $^{3}$ = $1 0^{- 6}$ m $^{3}$
$13.6 g c m^{- 3} = 13.6 \times \frac{1 0 ^{- 3} k g}{1 0 ^{- 6} m ^{3}} = 13.6 \times 1 0^{3} k g m^{- 3}$

Answer: $1.36 \times 1 0^{4} k g m^{- 3}$

Significant figures and how to count them

NCERT section 1.3

A measured value should show how precise the measurement is. We report every digit we are sure of plus the first uncertain digit. Together these are the significant figures. A pendulum period written as 2.47 s says that 2 and 4 are reliable and 7 is the doubtful digit: three significant figures. Precision depends on the least count of the instrument, and changing the unit never changes the number of significant figures.

Counting rules:
1. Every non-zero digit is significant.
2. Zeros between non-zero digits are significant (4005 has four).
3. In a number less than 1, zeros after the decimal point but before the first non-zero digit are not significant (0.0034 has two). The zero written before the decimal point is never significant.
4. Trailing zeros in a number without a decimal point are not significant (4500 m has two).
5. Trailing zeros in a number with a decimal point are significant (2.500 has four; 0.0700 has three).
6. Exact numbers such as 2 in $2 π r$ or the count of oscillations have unlimited significant figures.

To remove any doubt about trailing zeros, write the value in scientific notation $a \times 1 0^{b}$ with $1 \leq a < 10$ : all digits of $a$ are significant and the power of ten does not count. So 4.500 m = $4.500 \times 1 0^{3}$ mm still has four significant figures. The power $b$ (after rounding $a$ to 1 or 10) gives the order of magnitude of the quantity.

Worked example

State the number of significant figures in: (a) 0.00450 kg (b) 3.0800 m (c) 4500 s (d) $1.20 \times 1 0^{3}$ N (e) 0.10010 A. [2 marks]

Try it first, then show the solutionSolution

(a) Leading zeros do not count; 4, 5 and the trailing zero after the decimal do: 3
(b) All digits count, including the trailing zeros after the decimal point: 5
(c) No decimal point, so the trailing zeros do not count: 2
(d) In scientific notation all digits of 1.20 count: 3
(e) Ignore the first zero; 1, 0, 0, 1, 0 all count: 5

Answer: (a) 3 (b) 5 (c) 2 (d) 3 (e) 5

Arithmetic with significant figures and rounding off

NCERT section 1.3.1 and 1.3.2

A calculated answer cannot be more precise than the data it comes from.

Multiplication and division: keep as many significant figures as the input with the fewest significant figures.
Addition and subtraction: keep as many decimal places as the input with the fewest decimal places.

Rounding off: drop the extra digits. If the first dropped digit is more than 5, raise the digit before it by 1; if less than 5, leave it. If the dropped digit is exactly 5 (with nothing after it), the convention is: leave the preceding digit unchanged if it is even, raise it by 1 if it is odd. So 4.365 becomes 4.36, while 4.375 becomes 4.38.

In a long calculation keep one extra digit in the intermediate steps and round only at the end, so that rounding errors do not pile up.

Worked example

(a) A block of mass 3.513 kg has a volume of 1.25 m $^{3}$ . Find its density to the correct significant figures. (b) Add 12.11 m, 18.0 m and 1.013 m. [2 marks]

Try it first, then show the solutionSolution

(a) $ρ = \frac{3.513}{1.25} = 2.8104$ ; the volume has 3 significant figures, so $ρ = 2.81 k g m^{- 3}$
(b) Sum $= 31.123$ m; 18.0 has only one decimal place, so the answer is 31.1 m

Answer: (a) 2.81 kg m $^{- 3}$ (b) 31.1 m

Uncertainty in the result of a calculation

NCERT section 1.3.3

A reading like 16.2 cm taken with a metre scale (least count 0.1 cm) really means $16.2 \pm 0.1$ cm. The relative (fractional) uncertainty is $\frac{0.1}{16.2} \approx 0.6%$ .

How uncertainty carries into a result:

Product or quotient ( $Z = A B$ or $Z = A / B$ ): the relative uncertainties add: $\frac{Δ Z}{Z} = \frac{Δ A}{A} + \frac{Δ B}{B}$ .
A power is a repeated product, so for $Z = A^{n}$ the relative uncertainty is $n$ times that of $A$ .
Sum or difference ( $Z = A \pm B$ ): the absolute uncertainties add, $Δ Z = Δ A + Δ B$ . Subtracting two nearly equal values can therefore leave a result with a large relative uncertainty and fewer significant figures.

The same absolute uncertainty matters more for a smaller number: ±0.01 g on 1.02 g is about 1%, but on 9.89 g it is only about 0.1%.

Worked example

The length and breadth of a card are $5.0 \pm 0.1$ cm and $2.0 \pm 0.1$ cm. Find its area with the uncertainty. [2 marks]

Try it first, then show the solutionSolution

Relative uncertainties: $\frac{0.1}{5.0} = 2%$ and $\frac{0.1}{2.0} = 5%$ ; for a product they add: 7%
Area $= 5.0 \times 2.0 = 10 c m^{2}$ ; uncertainty $= 7%$ of 10 = 0.7 cm $^{2}$

Answer: Area $= 10 c m^{2}$ with an uncertainty of about $0.7 c m^{2}$ (7%)

Dimensions and dimensional formulae

NCERT section 1.4 and 1.5

The dimensions of a quantity are the powers to which the base quantities must be raised to represent it. In mechanics we need only [M], [L] and [T]. Square brackets mean 'the dimensions of'.

The dimensional formula shows these powers; writing quantity = formula gives a dimensional equation. Build each one from a defining equation:

Velocity = displacement/time: $[M^{0} L T^{- 1}]$
Acceleration: $[M^{0} L T^{- 2}]$
Force = mass × acceleration: $[M L T^{- 2}]$
Work, energy = force × distance: $[M L^{2} T^{- 2}]$
Power = work/time: $[M L^{2} T^{- 3}]$
Pressure, stress = force/area: $[M L^{- 1} T^{- 2}]$
Momentum and impulse: $[M L T^{- 1}]$
Density: $[M L^{- 3}]$
Frequency: $[T^{- 1}]$
Gravitational constant, from $F = \frac{G m _{1} m _{2}}{r ^{2}}$ : $[M^{- 1} L^{3} T^{- 2}]$

Only the kind of quantity matters, not its size: initial velocity, average velocity and speed all have $[L T^{- 1}]$ . Pure numbers, angles, strain and refractive index are dimensionless.

Worked example

Find the dimensional formula of (a) pressure and (b) the gravitational constant G. [2 marks]

Try it first, then show the solutionSolution

(a) Pressure = force/area = $\frac{[ M L T ^{- 2} ]}{[ L ^{2} ]} = [M L^{- 1} T^{- 2}]$
(b) $G = \frac{F r ^{2}}{m _{1} m _{2}} = \frac{[ M L T ^{- 2} ] [ L ^{2} ]}{[ M ^{2} ]} = [M^{- 1} L^{3} T^{- 2}]$

Answer: (a) $[M L^{- 1} T^{- 2}]$ (b) $[M^{- 1} L^{3} T^{- 2}]$

Checking equations: the principle of homogeneity

NCERT section 1.6 and 1.6.1

Only quantities with the same dimensions can be added, subtracted or equated. So every term on both sides of a correct equation must have the same dimensions: the principle of homogeneity of dimensions.

Using it to test an equation: find the dimensions of each term separately. If any term differs, the equation is certainly wrong. If all terms match, the equation may be right but is not proven right, because dimensionless factors (like $\frac{1}{2}$ or $2 π$ ) cannot be checked.

Arguments of functions such as $sin θ$ , $e^{x}$ and $lo g x$ must be dimensionless, and so must the functions themselves.

Worked example

Check whether $s = u t + \frac{1}{2} a t^{2}$ is dimensionally correct. [2 marks]

Try it first, then show the solutionSolution

[s] = [L]
$[u t] = [L T^{- 1}] [T] = [L]$ and $[\frac{1}{2} a t^{2}] = [L T^{- 2}] [T^{2}] = [L]$ (½ has no dimensions)
Every term has dimension [L], so the equation is dimensionally correct.

Answer: Dimensionally correct (all terms are [L])

Deducing relations and converting units

NCERT section 1.6.2

Deducing a relation: if you know which quantities a result depends on, assume it is proportional to a product of their powers, equate the powers of M, L and T on both sides and solve. The constant of proportionality must come from experiment or theory.

Converting a value between systems: a quantity with dimensions $[M^{a} L^{b} T^{c}]$ has numerical values in two systems related by
$n_{2} = n_{1} [\frac{M _{1}}{M _{2}}]^{a} [\frac{L _{1}}{L _{2}}]^{b} [\frac{T _{1}}{T _{2}}]^{c}$

Limitations of dimensional analysis:

It cannot find dimensionless constants (the $2 π$ in a pendulum formula).
It fails if the quantity depends on more than three quantities (only three equations from M, L and T), or on sums of terms.
It cannot handle exponential, logarithmic or trigonometric relations.
A dimensionally correct equation is not necessarily correct.

Worked example

The centripetal force F on a body moving in a circle depends on its mass m, speed v and radius r. Use dimensions to find the form of the relation. [3 marks]

Try it first, then show the solutionSolution

Let $F = k m^{a} v^{b} r^{c}$ , so $[M L T^{- 2}] = [M]^{a} [L T^{- 1}]^{b} [L]^{c} = [M^{a} L^{b + c} T^{- b}]$
Compare powers: M: a = 1; T: −b = −2 so b = 2; L: b + c = 1 so c = −1
$F = k \frac{m v ^{2}}{r}$ (theory gives k = 1)

Answer: $F \propto \frac{m v ^{2}}{r}$

Physics in real life: Units and Measurement

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Significant figures in a recipe for medicine

Suppose a pharmacist weighs 2.45 g of a powder on a balance and dissolves it in water to make 1.5 L of solution.

Find the concentration in g L $^{- 1}$ and report it to the correct number of significant figures.

Try it first, then show the solutionSolution

Concentration $= \frac{2.45}{1.5} = 1.6333... g L^{- 1}$
The volume has only 2 significant figures, so keep 2
Concentration $= 1.6 g L^{- 1}$

Answer: 1.6 g L $^{- 1}$

Memory hook: A chain is as strong as its weakest link: the answer is as precise as the least precise reading.

Uncertainty in a volume

Imagine you measure the side of a sugar cube with a ruler as $1.5 \pm 0.1$ cm and want to know how well you know its volume.

Suppose the side is $1.5 \pm 0.1$ cm. Find the volume and its percentage uncertainty.

Try it first, then show the solutionSolution

Volume $= 1. 5^{3} = 3.375 \approx 3.4 c m^{3}$
Relative uncertainty of the side $= \frac{0.1}{1.5} \approx 6.7%$
For a cube, $V = a^{3}$ , so the relative uncertainty is $3 \times 6.7% = 20%$ , about $\pm 0.7 c m^{3}$

Answer: About $3.4 \pm 0.7 c m^{3}$ (20% uncertainty)

Memory hook: Cube the side, triple the error.

Dimensional check of a formula

Suppose a friend, revising late at night, writes the kinetic energy of a scooter as $K = \frac{1}{2} m v$ .

Use dimensions to show that the formula cannot be right, and suggest the correct power of v.

Try it first, then show the solutionSolution

Energy has dimensions $[M L^{2} T^{- 2}]$
$[m v] = [M] [L T^{- 1}] = [M L T^{- 1}]$ , which is momentum, not energy
To get $L^{2} T^{- 2}$ , v must be squared: $K \propto m v^{2}$

Answer: Wrong; K must be proportional to $m v^{2}$ (in fact $K = \frac{1}{2} m v^{2}$ ).

Memory hook: If the dimensions do not match, the formula is wrong for sure.

Topics in this chapter

Units and Measurement formula sheet

Plane and solid angle: $d θ = \frac{d s}{r} (r a d), d Ω = \frac{d A}{r ^{2}} (sr)$
Scientific notation and order of magnitude: $x = a \times 1 0^{b}, 1 \leq a < 10$; Round $a$ to 1 (if $a \leq 5$ ) or 10 (if $a > 5$ ) to get the order of magnitude.
Relative uncertainty of a product or quotient: $Z = A B or \frac{A}{B} : \frac{Δ Z}{Z} = \frac{Δ A}{A} + \frac{Δ B}{B}$
Uncertainty of a power: $Z = A^{n} : \frac{Δ Z}{Z} = n \frac{Δ A}{A}$
Uncertainty of a sum or difference: $Z = A \pm B : Δ Z = Δ A + Δ B$
Common dimensional formulae: $[F] = [M L T^{- 2}], [W] = [M L^{2} T^{- 2}], [P] = [M L^{2} T^{- 3}], [p] = [M L^{- 1} T^{- 2}]$; Force, work/energy, power, pressure.
Gravitational constant: $[G] = [M^{- 1} L^{3} T^{- 2}]$
Converting a numerical value between unit systems: $n_{2} = n_{1} [\frac{M _{1}}{M _{2}}]^{a} [\frac{L _{1}}{L _{2}}]^{b} [\frac{T _{1}}{T _{2}}]^{c}$

How the school exam asks Units and Measurement

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
SI units and conversions	Name the base quantities and units; express a value in SI; spot wrongly written units.	Write the conversion factor for each unit (1 g = $1 0^{- 3}$ kg, 1 cm = $1 0^{- 2}$ m) and raise it to the right power.
Counting significant figures	MCQ or one-liners: how many significant figures in a given number; effect of changing units.	Apply the six rules; when in doubt rewrite in scientific notation.
Calculations with significant figures	Report an area, volume, density or sum to the right number of figures; round off a number.	× and ÷: fewest significant figures; + and −: fewest decimal places; round only at the end, using the even-odd rule for a 5.
Uncertainty in a result	Given readings with ± values, find the uncertainty or percentage uncertainty in an area, volume or density.	Convert each to a relative uncertainty, multiply by the power, add, then convert back to an absolute value.
Dimensional formula and checking equations	Find dimensions of a constant; check whether an equation is correct; find dimensions of a and b in a given relation.	Use homogeneity: each added term has the same dimensions as the left side.
Deriving a formula or converting units by dimensions	3-5 mark questions: time period, wave speed or force in terms of given quantities; value of a unit in another system; limitations.	Assume a power law, compare powers of M, L, T, solve three equations; for conversions use $n_{2} = n_{1} [M_{1} / M_{2}]^{a} [L_{1} / L_{2}]^{b} [T_{1} / T_{2}]^{c}$ .

Tricks to remember Units and Measurement

Short memory aids. Say each one aloud once, then use it on a question the same day.

Seven SI base units: 'King Mohan Saw A Monkey Carrying Kites'
kilogram, metre, second, ampere, mole, candela, kelvin.
Use it when: Listing base units.
× ÷ count figures; + − count decimals
Multiply/divide: fewest significant figures wins. Add/subtract: fewest decimal places wins.
Use it when: Every significant-figure calculation.
Five rounds to even
Dropping exactly a 5? Make the previous digit even: 4.365 → 4.36, 4.375 → 4.38.
Use it when: Rounding questions.
F = MLT⁻², then multiply
Learn force; work = force × L gives $M L^{2} T^{- 2}$ ; power = work ÷ T gives $M L^{2} T^{- 3}$ ; pressure = force ÷ $L^{2}$ gives $M L^{- 1} T^{- 2}$ .
Use it when: Building dimensional formulae quickly.
Powers multiply errors
In $Z = A^{p} B^{q} / C^{r}$ , the percentage uncertainty is $p % A + q % B + r % C$ : always add, never subtract.
Use it when: Uncertainty in a result.

Units and Measurement: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Counting significant figures · 1 mark · MCQExam level

The number of significant figures in 0.05020 is:
(a) 2
(b) 3
(c) 4
(d) 5

Try it first, then show the solutionSolution

Leading zeros (0.0) are not significant; 5, 0, 2 and the trailing 0 after the decimal point are.

Answer: (c) 4

Example 2: Correct dimensions, wrong equation · 1 mark · Assertion-reasonExam level

Assertion (A): A dimensionally correct equation may still be physically wrong.
Reason (R): Dimensional analysis gives no information about dimensionless constants.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

$s = u t + a t^{2}$ is dimensionally correct but wrong: the missing ½ is a pure number that dimensions cannot detect. R explains A.

Answer: (a)

Example 3: Dimensions of constants in a gas equation · 2 marks · Very short answerHarder exam level

For a real gas, $(P + \frac{a}{V ^{2}}) (V - b) = R T$ , where P is pressure and V volume. Find the dimensions of a and b. [2 marks]

Try it first, then show the solutionSolution

Only like quantities can be added: $[b] = [V] = [L^{3}]$
$[\frac{a}{V ^{2}}] = [P]$ , so $[a] = [M L^{- 1} T^{- 2}] [L^{6}] = [M L^{5} T^{- 2}]$

Answer: $[a] = [M L^{5} T^{- 2}]$ , $[b] = [L^{3}]$

Example 4: Newton to dyne · 2 marks · Very short answerHarder exam level

Using dimensions, convert 1 N into the CGS unit of force (dyne). [2 marks]

Try it first, then show the solutionSolution

Force: $[M L T^{- 2}]$ , so $n_{2} = n_{1} [\frac{1 k g}{1 g}]^{1} [\frac{1 m}{1 c m}]^{1} [\frac{1 s}{1 s}]^{- 2}$
$n_{2} = 1 \times 1 0^{3} \times 1 0^{2} \times 1 = 1 0^{5}$

Answer: 1 N = $1 0^{5}$ dyne

Example 5: Volume of a sphere with uncertainty · 3 marks · Short answerHarder exam level

The radius of a ball bearing is measured as $2.50 \pm 0.01$ cm. Find its volume to the correct significant figures and the uncertainty in the volume. [3 marks]

Try it first, then show the solutionSolution

$V = \frac{4}{3} π r^{3} = \frac{4}{3} π (2.50)^{3} = 65.45 c m^{3}$ ; r has 3 significant figures, so V = 65.4 cm $^{3}$ ( $\frac{4}{3}$ and π are exact or taken to more figures)
Relative uncertainty in r $= \frac{0.01}{2.50} = 0.4%$ ; for $r^{3}$ it is $3 \times 0.4% = 1.2%$
$Δ V = 1.2% \times 65.4 \approx 0.8 c m^{3}$

Answer: $V = 65.4 \pm 0.8 c m^{3}$

Example 6: Deriving the speed of sound in a gas · 5 marks · Long answerHarder exam level

(a) Assuming that the speed v of sound in a gas depends on the pressure P and the density ρ of the gas, derive the relation by dimensions. (b) Find v for air with $P = 1.0 \times 1 0^{5}$ Pa and $ρ = 1.3 k g m^{- 3}$ (constant = 1). (c) If the density of the gas is made four times at the same pressure, what happens to v? (d) State two limitations of this method. [5 marks]

Try it first, then show the solutionSolution

(a) $v = k P^{a} ρ^{b}$ ; $[M^{0} L T^{- 1}] = [M L^{- 1} T^{- 2}]^{a} [M L^{- 3}]^{b} = [M^{a + b} L^{- a - 3 b} T^{- 2 a}]$
M: a + b = 0; T: −2a = −1, so $a = \frac{1}{2}$ and $b = - \frac{1}{2}$ ; check L: $- a - 3 b = - \frac{1}{2} + \frac{3}{2} = 1$ ✓. So $v = k P / ρ$
(b) $v = \frac{1.0 \times 1 0 ^{5}}{1.3} = 7.7 \times 1 0^{4} \approx 2.8 \times 1 0^{2} m s^{- 1}$
(c) $v \propto \frac{1}{ρ}$ : four times the density halves v
(d) It cannot find k; it cannot be used if v depends on more than three quantities or on a sum of terms.

Answer: $v = k P / ρ$ ; about $2.8 \times 1 0^{2}$ m s $^{- 1}$ ; v is halved

Example 7: Finding an unknown unit of mass · 3 marks · Short answerHOTS / competency

In a new system the unit of force is 100 N, the unit of length is 10 m and the unit of time is 100 s. What is the unit of mass in this system, in kg? [3 marks]

Try it first, then show the solutionSolution

$[F] = [M L T^{- 2}]$ , so $M = \frac{F T ^{2}}{L}$
Unit of mass $= \frac{100 N \times ( 100 s ) ^{2}}{10 m}$
$= \frac{100 \times 1 0 ^{4}}{10} = 1 0^{5} k g$

Answer: $1 0^{5}$ kg

Example 8: Oscillating drop · 3 marks · Short answerHOTS / competency

A small liquid drop oscillates under surface tension. Its period T is thought to depend on the density ρ, the radius r and the surface tension S (force per unit length). Find the form of the relation. [3 marks]

Try it first, then show the solutionSolution

$[S] = \frac{[ M L T ^{- 2} ]}{[ L ]} = [M T^{- 2}]$ ; let $T = k ρ^{a} r^{b} S^{c}$
$[T] = [M L^{- 3}]^{a} [L]^{b} [M T^{- 2}]^{c}$ : M: a + c = 0; L: −3a + b = 0; T: −2c = 1
$c = - \frac{1}{2}$ , $a = \frac{1}{2}$ , $b = \frac{3}{2}$ , so $T = k \frac{ρ r ^{3}}{S}$

Answer: $T = k ρ r^{3} / S$

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

A pendulum experiment in the lab

Suppose a student measures the length of a simple pendulum as 1.00 m (to 0.001 m) and the time period as 2.01 s (to 0.01 s), then uses $g = \frac{4 π ^{2} l}{T ^{2}}$ .

(i) · 1 mark
How many significant figures does the period have?
Answer
Answer: 3
2, 0 and 1 all count.
(ii) · 1 mark
Find g to the correct significant figures.
Answer
Answer: 9.77 m s $^{- 2}$
$g = \frac{4 π ^{2} \times 1.00}{2.0 1 ^{2}} = 9.7716...$ ; three significant figures: 9.77
(iii) · 1 mark
Find the percentage uncertainty in g.
Answer
Answer: About 1.1%
$\frac{Δ g}{g} = \frac{Δ l}{l} + 2 \frac{Δ T}{T} = 0.1% + 2 \times 0.50% \approx 1.1%$
(iv) · 1 mark
Which measurement should be improved first, and how?
Answer
Answer: The time period: time 20 oscillations and divide.
T enters squared and has the larger relative uncertainty; timing many oscillations reduces it.

Checking a stranger's formula

Suppose a website claims that the height h reached by a ball thrown up with speed v is $h = \frac{v ^{2}}{2 g}$ , while a second site says $h = \frac{v}{2 g}$ . Take g as the acceleration due to gravity.

(i) · 1 mark
Find the dimensions of $\frac{v ^{2}}{2 g}$ .
Answer
Answer: [L]
$\frac{[ L ^{2} T ^{- 2} ]}{[ L T ^{- 2} ]} = [L]$
(ii) · 1 mark
Find the dimensions of $\frac{v}{2 g}$ .
Answer
Answer: [T]
$\frac{[ L T ^{- 1} ]}{[ L T ^{- 2} ]} = [T]$
(iii) · 1 mark
Which formula can be ruled out, and why?
Answer
Answer: $h = v /2 g$
Its right side has dimensions of time, not length.
(iv) · 1 mark
Does the dimensional check prove the first formula correct?
Answer
Answer: No
It only shows it is consistent; the factor ½ comes from the equations of motion.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Check the dimensional correctness of $v^{2} = u^{2} + 2 a s$ . [2 marks] (2 marks)

Step	Marks
$[v^{2}] = [u^{2}] = [L^{2} T^{- 2}]$	1
$[2 a s] = [L T^{- 2}] [L] = [L^{2} T^{- 2}]$ ; all terms match, so the equation is dimensionally correct	1

Each edge of a cube is measured as 4.05 cm. Find its total surface area and volume to the correct significant figures, and explain your rounding. [3 marks] (3 marks)

Step	Marks
The edge has 3 significant figures, so the results keep 3	1
Surface area $= 6 \times 4.0 5^{2} = 98.415 \approx 98.4 c m^{2}$	1
Volume $= 4.0 5^{3} = 66.430 \approx 66.4 c m^{3}$	1

(a) State the principle of homogeneity of dimensions. (b) The speed v of a wave on a stretched string depends on the tension T and the mass per unit length μ. Derive the form of the relation using dimensions. (c) Find v when T = 100 N and μ = 0.01 kg m $^{- 1}$ , taking the constant as 1. (d) Give one limitation of this method. [5 marks] (5 marks)

Step	Marks
(a) In a correct equation every term on both sides has the same dimensions	1
(b) $v = k T^{a} μ^{b}$ ; $[L T^{- 1}] = [M L T^{- 2}]^{a} [M L^{- 1}]^{b}$	1
M: a + b = 0; L: a − b = 1; T: −2a = −1, so $a = \frac{1}{2}$ , $b = - \frac{1}{2}$ , giving $v = k T / μ$	1
(c) $v = 100/0.01 = 1 0^{4} = 100 m s^{- 1}$	1
(d) It cannot give the dimensionless constant k (or: fails for more than three variables)	1

Where students lose marks in Units and Measurement

The slips examiners see most often, and what to write instead.

Mistake: Counting trailing zeros wrongly
Mixing the rules for numbers with and without a decimal point.
Wrong working
2.500 has 2 significant figures
Right working
2.500 has 4 (trailing zeros after a decimal point count)
Do this: Rewrite in scientific notation; every digit of the base number counts.
Usually costs: 1 mark.
Mistake: Using the significant-figure rule for addition
Applying the multiplication rule everywhere.
Wrong working
312.45 + 125.3 + 0.208 reported as 438 (3 significant figures)
Right working
438.0 (one decimal place, like 125.3; the exact sum is 437.958)
Do this: For + and −, count decimal places, not significant figures.
Usually costs: 1 mark.
Mistake: Subtracting relative uncertainties for a quotient
Thinking division should 'cancel' errors.
Do this: Relative uncertainties always add, for products and quotients alike: the worst case is what we report.
Usually costs: 1-2 marks.
Mistake: Claiming that a dimensionally correct equation is correct
Forgetting that pure numbers have no dimensions.
Do this: Write: 'the equation is dimensionally consistent, so it may be correct'. $s = u t + a t^{2}$ is consistent but wrong.
Usually costs: 1 mark in assertion-reason items.
Mistake: Writing units carelessly
Habit from everyday writing.
Wrong working
5 Kgs, 10 Sec, 3 mtrs
Right working
5 kg, 10 s, 3 m
Do this: Use standard symbols, no plurals, no full stops.
Usually costs: Half a mark per answer in some school marking.
Mistake: Rounding at every step
Trying to keep the working neat.
Do this: Keep one extra digit in intermediate results and round only the final answer.
Usually costs: The last significant figure, and sometimes the mark.

NCERT exercise guide: Units and Measurement

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercises 1.1-1.5 (units and conversions)

What it tests: Filling in units, converting between unit systems, and the meaning of 'large' and 'small' for a quantity.

How to approach it: Write each unit in terms of SI base units, then substitute the conversion factors raised to the right power; for a new system use $n_{2} = n_{1} [M_{1} / M_{2}]^{a} [L_{1} / L_{2}]^{b} [T_{1} / T_{2}]^{c}$ .

Same kind as Exercise 1.3

In a new system the unit of mass is 10 kg, the unit of length is 10 m and the unit of time is 10 s. What is the numerical value of 1 J in this system?

Show the solutionSolution

Energy: $[M L^{2} T^{- 2}]$ , so a = 1, b = 2, c = −2
$n_{2} = 1 \times (\frac{1}{10})^{1} (\frac{1}{10})^{2} (\frac{1}{10})^{- 2}$
$n_{2} = \frac{1}{10} \times \frac{1}{100} \times 100 = 0.1$

Answer: 1 J = 0.1 new unit of energy

Exercises 1.10-1.12 (significant figures)

What it tests: Counting significant figures, and reporting areas, volumes, densities and sums correctly.

How to approach it: Count with the six rules; × ÷ use fewest significant figures; + − use fewest decimal places.

Same kind as Exercise 1.12

A bag has mass 3.40 kg. Two packets of 15.21 g and 15.23 g are added. What is (a) the total mass, (b) the difference in mass of the packets, each to the correct precision?

Show the solutionSolution

(a) $3.40 k g + 0.01521 k g + 0.01523 k g = 3.43044 k g$ ; 3.40 has two decimal places, so 3.43 kg
(b) $15.23 - 15.21 = 0.02 g$ (two decimal places)

Answer: (a) 3.43 kg (b) 0.02 g

Exercises 1.13-1.17 (dimensions)

What it tests: Restoring missing constants by dimensions, checking formulae, and estimates using orders of magnitude.

How to approach it: Use homogeneity: each term must have the same dimensions; insert powers of c, g or other constants to fix a mismatch.

Same kind as Exercise 1.14

A student writes the displacement of a particle in simple harmonic motion as $y = a sin (2 π t)$ , where a is a length and t a time. Is this dimensionally correct? If not, correct it using the period T.

Show the solutionSolution

The argument of a sine must be dimensionless.
$[2 π t] = [T]$ , which is not dimensionless, so the formula is wrong.
Dividing by the period gives a pure number: $y = a sin \frac{2 π t}{T}$

Answer: Wrong; $y = a sin (2 π t / T)$ is correct dimensionally.

Practice questions on Units and Measurement

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
Which of these is not an SI base unit?
(a) kelvin
(b) candela
(c) newton
(d) mole
Hint
One of them is made from kg, m and s.
Answer
(c) newton (it is derived: kg m s $^{- 2}$ )
Q2 · 1 mark · MCQExam level
The dimensional formula of power is:
(a) $[M L^{2} T^{- 2}]$
(b) $[M L^{2} T^{- 3}]$
(c) $[M L T^{- 3}]$
(d) $[M L^{- 1} T^{- 2}]$
Hint
Work divided by time.
Answer
(b) $[M L^{2} T^{- 3}]$
Q3 · 1 mark · Very short answerExam level
Round off each to three significant figures: (a) 2.746 (b) 6.235 (c) 6.245.
Hint
Exactly 5: make the preceding digit even.
Answer
(a) 2.75 (b) 6.24 (3 is odd, so raise it) (c) 6.24 (4 is even, so keep it)
Q4 · 1 mark · Assertion-reasonExam level
Assertion (A): Angle is a dimensionless quantity, but it has a unit.
Reason (R): Angle is the ratio of arc length to radius.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Ratio of two lengths.
Answer
(a): a ratio of two lengths has no dimensions, yet we still name its unit, the radian.
Q5 · 2 marks · Very short answerHarder exam level
A rectangular plate has length 12.4 cm and breadth 3.6 cm. Find its area and perimeter to the correct precision.
Hint
Area: significant figures. Perimeter: decimal places.
Answer
Area $= 44.64 \approx 45 c m^{2}$ (2 significant figures, as 3.6 has two). Perimeter $= 2 (12.4 + 3.6) = 32.0 c m$ (one decimal place).
Q6 · 2 marks · Very short answerHarder exam level
Find the dimensional formulae of (a) impulse (b) surface tension (force per unit length) (c) angular velocity.
Hint
Impulse = force × time.
Answer
(a) $[M L T^{- 1}]$ (b) $[M T^{- 2}]$ (c) $[T^{- 1}]$
Q7 · 3 marks · Short answerHarder exam level
The value of G in SI units is $6.67 \times 1 0^{- 11} N m^{2} k g^{- 2}$ . Find its value in CGS units using dimensions.
Hint
$[G] = [M^{- 1} L^{3} T^{- 2}]$ .
Answer
$n_{2} = 6.67 \times 1 0^{- 11} \times (1 0^{3})^{- 1} \times (1 0^{2})^{3} \times 1 = 6.67 \times 1 0^{- 11} \times 1 0^{- 3} \times 1 0^{6} = 6.67 \times 1 0^{- 8}$ dyne cm $^{2}$ g $^{- 2}$ .
Q8 · 3 marks · Short answerHarder exam level
The mass of a cube is $5.25 \pm 0.05$ g and its side is $1.20 \pm 0.01$ cm. Find its density and the percentage uncertainty in the density.
Hint
$ρ = m / a^{3}$ .
Answer
$ρ = \frac{5.25}{1.2 0 ^{3}} = \frac{5.25}{1.728} = 3.04 g c m^{- 3}$ . Uncertainty $= \frac{0.05}{5.25} + 3 \times \frac{0.01}{1.20} = 0.95% + 2.5% \approx 3.5%$ , so $ρ = 3.04 \pm 0.11 g c m^{- 3}$ .
Q9 · 2 marks · Very short answerHarder exam level
Check by dimensions whether $T = 2 π \frac{m}{k}$ is correct for a mass m on a spring of spring constant k (force per unit extension).
Hint
$[k] = [M T^{- 2}]$ .
Answer
$[k] = \frac{[ M L T ^{- 2} ]}{[ L ]} = [M T^{- 2}]$ , so $\frac{[ M ]}{[ M T ^{- 2} ]} = [T]$ . Both sides are time: dimensionally correct.
Q10 · 3 marks · Short answerHOTS / competency
The speed v of water flowing out of a small hole at depth h below the surface of a tank may depend on h, g and the density ρ of the water. Derive the form of the relation by dimensions and explain what the result says about the density.
Hint
Let $v = k h^{a} g^{b} ρ^{c}$ .
Answer
$[L T^{- 1}] = [L]^{a} [L T^{- 2}]^{b} [M L^{- 3}]^{c}$ : M: c = 0; T: −2b = −1 so $b = \frac{1}{2}$ ; L: $a + b - 3 c = 1$ so $a = \frac{1}{2}$ . $v = k g h$ . Since c = 0, the speed does not depend on the density of the liquid (theory gives $k = 2$ ).
Q11 · 3 marks · Short answerHOTS / competency
A student measures two lengths as $25.3 \pm 0.1$ cm and $24.9 \pm 0.1$ cm and finds their difference. Find the difference with its uncertainty, and the percentage uncertainty. What lesson does this give about subtracting nearly equal measurements?
Hint
Absolute uncertainties add in a difference.
Answer
Difference $= 0.4 \pm 0.2$ cm, a 50% uncertainty, although each length was known to about 0.4%. Subtracting nearly equal values wipes out most of the precision, so such measurements should be avoided or made much more precisely.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Name the seven SI base quantities.Show answer
Length, mass, time, electric current, thermodynamic temperature, amount of substance and luminous intensity.
2.What are the two supplementary SI units?Show answer
The radian (plane angle) and the steradian (solid angle).
3.How many significant figures in 0.02040?Show answer
Four (2, 0, 4, 0).
4.State the rule for significant figures in multiplication.Show answer
The result keeps as many significant figures as the factor with the fewest.
5.What is the dimensional formula of energy?Show answer
$[M L^{2} T^{- 2}]$ .
6.State the principle of homogeneity of dimensions.Show answer
In a correct equation every term has the same dimensions.
7.Give one limitation of dimensional analysis.Show answer
It cannot find dimensionless constants such as $2 π$ .
8.Round 7.465 to three significant figures.Show answer
7.46 (the dropped digit is 5 and the preceding 6 is even).

Best order to study it

Learn what a unit is, base and derived units, and the table of seven SI base units.
Practise counting significant figures until the six rules are automatic.
Learn the multiplication and addition rules and rounding off, with the even-odd rule.
Learn how uncertainties combine in products, powers and sums.
Build the dimensional formulae of 15-20 common quantities from their definitions.
Practise checking equations, deducing relations and converting units with dimensions; finish with the limitations.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you list the seven SI base quantities with units and symbols?
Can you explain why the radian has no dimensions?
Can you count significant figures in 0.00300, 30.0 and 300?
Can you report a product and a sum to the correct precision?
Can you round a number ending in 5 correctly?
Can you find the percentage uncertainty in the volume of a sphere from that in its radius?
Can you write dimensional formulae for force, work, power, pressure, momentum and G?
Can you check an equation using homogeneity of dimensions?
Can you derive the form of a formula from the quantities it depends on?
Can you state three limitations of dimensional analysis?

Units and Measurement: frequently asked questions

Is the 'Physical World' chapter still in Class 11 Physics?

No. The rationalised NCERT book (Reprint 2026-27) has removed it, so Units and Measurement is now Chapter 1. Measurement of length, mass and time and the old section on errors in measurement are also gone.

Are the rules for errors still needed?

The detailed error theory (mean absolute error and so on) was removed, but the syllabus still asks you to determine the uncertainty in a result. You should know how relative uncertainties add in products, quotients and powers, and how uncertainties combine in sums.

How is the Class 11 Physics paper set?

It is a school annual exam of 70 theory marks (3 hours) plus 30 practical marks. CBSE gives unit weightage and a typology table but no sample paper, so many schools follow the Class 12 pattern of MCQs, assertion-reason, short and long answers and case-based questions.

Why does changing units not change significant figures?

Changing units only moves the decimal point or changes the power of ten; the precision of the measurement, which the significant figures show, stays the same.

Can dimensional analysis derive the full formula for the period of a pendulum?

It gives T \propto \sqrt{l/g} but not the factor 2\pi, which is dimensionless. That has to come from theory (you derive it in Oscillations) or experiment.

Why is this chapter important later?

Dimensions are used to check answers in every chapter, from Laws of Motion to Waves, and in Class 12 Physics (for example the dimensions of \varepsilon_0 or \mu_0). JEE and NEET also set direct questions on dimensions and significant figures.

Based on: CBSE Physics (042) syllabus 2026-27 (Classes XI-XII); NCERT Physics Part I, Textbook for Class XI, Chapter 1 (Reprint 2026-27)