Class 11 · Physics · Chapter 1

Units and Measurements

This chapter builds the language of Physics: how we choose units, measure length, mass and time, report a measurement honestly through errors and significant figures, and use dimensions to check and derive relations. It has no board exam this year, but every later chapter, from motion to electromagnetism, quietly assumes you know this one.

10 sections11 NCERT questions23 board questions17-question quiz

Before you startWhat the board asks from this chapter

Class 11 has no board exam, so this chapter is not weighed in marks the way a Class 12 chapter is. It is tested fully in the school's own unit tests and final exam, and its ideas (SI units, significant figures, errors, dimensional analysis) resurface as 1-mark and 2-mark questions inside almost every Class 12 board paper and in JEE/NEET.

Question typeMarksUsually asked
MCQ1Dimensional formula of a quantity, significant figures in a given number, error propagation in a product or quotient, SI base unit identification.
Assertion–Reason1Statements testing whether a dimensionally correct equation must be numerically correct, or whether precision and accuracy mean the same thing.
Very short answer1Define a term (accuracy, systematic error, significant figure), state a dimensional formula, or name a method (parallax, echo).
Short answer2Small numerical on percentage error or significant figures; distinguish two close terms with one example each.
Short answer3Derive an equation's dimensional formula and check its correctness, or work a combined-error numerical with full steps.
Long answer5Full derivation of a relation using dimensional analysis (e.g. time period of a pendulum), or a multi-part numerical on errors and significant figures.

Hot topics, in order

  1. Rules for significant figures and rounding off
  2. Combination of errors (sum, product, power) in a measured quantity
  3. Dimensional formulae of common physical quantities
  4. Checking the dimensional correctness of an equation
  5. Deriving a relation by the method of dimensions (e.g. simple pendulum, orbital speed)
  6. Parallax method for large distances
  7. Difference between accuracy and precision, and between systematic and random error
  8. SI base units and the rules for writing symbols

The whole chapter, in orderNotes

01 1.1 Introduction

Physics is a science of measurement. Every law you will study this year, from Newton's second law to Coulomb's law, is a relation between measured quantities, and it only means something once you have agreed on what a metre, a second and a kilogram actually are. Physical quantities are split into two kinds. A fundamental (base) quantity is one that is measured directly and independently of others: length, mass, time, electric current, temperature, amount of substance and luminous intensity. A derived quantity is built from these by multiplication or division: speed = length/time, force = mass × acceleration, and so on. This chapter fixes the units for the base quantities, tells you how those units are realised in a laboratory, and gives you the tools, significant figures, error analysis and dimensional analysis, to handle any measurement honestly and to check whether an equation can possibly be correct before you even test it.

Exam tip Read every question in this chapter for the word being tested: 'accuracy' and 'precision' are different words with different meanings; examiners exploit that.

02 1.2 The International System of Units

Before SI, different countries used different systems: CGS (centimetre, gram, second), FPS (foot, pound, second) and MKS (metre, kilogram, second). To remove this confusion, the General Conference on Weights and Measures adopted the SI (Système International d'Unités), built on seven base units: metre (m) for length, kilogram (kg) for mass, second (s) for time, ampere (A) for electric current, kelvin (K) for temperature, mole (mol) for amount of substance and candela (cd) for luminous intensity. Two supplementary units are added for angles: radian (rad) for plane angle and steradian (sr) for solid angle. Every other physical unit (newton, joule, volt, ohm...) is derived from these seven by combining them according to the quantity's defining equation. SI also fixes rules for writing units correctly: unit symbols are never pluralised (5 kg, not 5 kgs), no full stop follows a symbol unless it ends a sentence, and a space separates the number from the unit (10 m, not 10m). Since 2019, all seven base units are defined by fixing the numerical value of a fundamental constant (e.g. the kilogram is now defined by fixing the Planck constant h), rather than by a physical artefact.

Exam tip A one-mark question often gives a derived unit (like the pascal or the watt) and asks for its base-unit expression — practise converting a few by heart.

03 1.3 Measurement of Length

For everyday lengths we use metre scales, vernier callipers and screw gauges. For very large distances, such as the distance to the moon or a nearby star, direct measurement is impossible, so we use the parallax method. Parallax is the apparent shift in the position of an object against a distant background when viewed from two different points. To find the distance D of a distant object, observe it from two points separated by a known baseline b, and measure the parallax angle θ (in radians) it subtends. Since θ is small, the arc length b ≈ Dθ, giving D = b/θ. The same idea, using the angular size of the object as seen from Earth, gives the diameter of a distant star once its distance is known: if α is the angular diameter, the star's diameter d = Dα. For distances within the solar system, the echo (radar) method is used: a radio pulse is sent, its reflection is timed, and distance = (speed of pulse × time)/2. For very small lengths (atomic and molecular size, of the order 10−10 m), electron microscopes and X-ray diffraction are used; a rough estimate can also be made from Avogadro's number and the molar volume of a substance.

θ = b/D ⇒ D = b/θ
Ddistance to the distant objectbbasis (separation between the two observation points)θparallax angle in radians
d = Dα
ddiameter of the distant object (e.g. a star)αangular diameter as seen from Earth
Exam tip In a parallax numerical, always convert the angle to radians first if it is given in degrees, arcseconds or arcminutes — this is the most common slip.

04 1.4 Measurement of Mass

Mass is measured against the SI kilogram using a common balance for ordinary masses and a sensitive spring or inertial balance in the absence of gravity (as on a spacecraft). For atomic and subatomic masses, a common unit is the unified atomic mass unit (u), defined as 1/12 of the mass of a carbon-12 atom: 1 u = 1.66 × 10−27 kg. Atomic and molecular masses are measured very precisely using a mass spectrograph, in which charged particles are deflected by electric and magnetic fields and their radius of curvature is used to find their mass-to-charge ratio. The range of masses in nature is enormous: from about 10−30 kg for an electron to about 1055 kg for the known universe, which is why NCERT wants you to appreciate the scale as much as the method.

1 u = 1.66 × 10−27 kg
uunified atomic mass unit, 1/12 the mass of a ¹²C atom
Exam tip If asked to name the instrument for atomic mass, the answer is 'mass spectrograph', not just 'spectrometer' — NCERT uses this exact term.

05 1.5 Measurement of Time

Time is measured using a clock; the SI second was earlier defined from Earth's rotation, but that rotation is not perfectly uniform, so the modern definition uses atomic vibrations instead. One SI second is defined as the time taken by 9,192,631,770 periods of the radiation emitted during a specific transition of the caesium-133 atom. Devices built on this principle, atomic clocks, are accurate to about 1 second in many thousands of years, and India's National Physical Laboratory (NPL), New Delhi maintains the atomic clock that provides Indian Standard Time.

Exam tip A common one-mark question just asks which atom's transition defines the second — the answer is caesium-133.

06 1.6 Accuracy, Precision of Instruments and Errors in Measurement

Accuracy is how close a measured value is to the true value; precision is how close repeated measurements are to each other, regardless of the true value — a clock can be precise (giving the same wrong time every day) without being accurate. Every measurement has some uncertainty, called an error. Errors are of three kinds. Systematic errors are one-sided (always positive or always negative) and arise from instrumental defects (zero error), imperfect technique, or an external condition; they can be reduced by correcting the instrument or the method. Random errors arise from unpredictable, uncontrollable causes and cause a scatter of readings around the true value; they are reduced by taking many readings and averaging. Gross errors are human mistakes in reading or recording.

If a1, a2, ..., an are n measurements of a quantity, the best estimate is their arithmetic mean amean. The absolute error in each reading is Δai = |amean − ai|, and the mean absolute error is the average of these. The relative (fractional) error is Δamean/amean, and multiplying it by 100 gives the percentage error. When quantities are combined, their errors combine by definite rules: for a sum or difference the absolute errors add; for a product or quotient the relative errors add; and for a quantity raised to a power, the relative error is multiplied by that power.

Δamean = (Δa1 + Δa2 + ... + Δan)/n
Δameanmean absolute error
Relative error = Δamean/amean; % error = (Δamean/amean) × 100
ameanmean value of the measured quantity
Z = A ± B ⇒ ΔZ = ΔA + ΔB
ΔZabsolute error in the sum/difference
Z = ApBq/Cr ⇒ ΔZ/Z = p(ΔA/A) + q(ΔB/B) + r(ΔC/C)
ΔZ/Zrelative error in Zp, q, rpowers of the respective quantities
Exam tip In the product/quotient rule, errors of quantities in the denominator still ADD, they never subtract — this is the single most common mistake.

07 1.7 Significant Figures

The significant figures in a measured number are all the digits that are reliably known plus one estimated (uncertain) digit; they tell you how precisely a quantity was actually measured. The rules: all non-zero digits are significant; zeros between two non-zero digits are significant (1005 has 4 sig. figs.); leading zeros (before the first non-zero digit) are never significant, they only fix the decimal point (0.0025 has 2 sig. figs.); trailing zeros after a decimal point are significant (2.500 has 4 sig. figs.); trailing zeros in a number without a decimal point are ambiguous unless written in scientific notation (2500 could have 2 to 4 sig. figs., but 2.500 × 103 has exactly 4). Scientific (power-of-ten) notation is preferred exactly because it removes this ambiguity.

When numbers are combined, the result cannot be more precise than the least precise input. In addition or subtraction, the result is rounded to the same number of decimal places as the term with the fewest decimal places. In multiplication or division, the result is rounded to the same number of significant figures as the factor with the fewest significant figures. Rounding off follows the standard rule: if the digit to be dropped is more than 5, the preceding digit is raised by 1; if less than 5, it is left unchanged; if exactly 5, the preceding digit is raised by 1 if it is odd and left unchanged if it is even (this keeps rounding unbiased on average).

Exam tip For addition/subtraction match decimal places; for multiplication/division match sig. figs. — mixing these two rules is the examiner's favourite trap.

08 1.8 Dimensions of Physical Quantities

The dimensions of a physical quantity are the powers to which the base quantities must be raised to represent it. Physics uses seven dimensional symbols: [M] for mass, [L] for length, [T] for time, [A] for electric current, [K] for temperature, [mol] for amount of substance and [cd] for luminous intensity. For example, speed = distance/time has the dimensional formula [M0L1T−1], because it involves length to the first power and time to the power −1, and no mass at all. Writing a quantity's dimensional formula this way, in square brackets with the powers of M, L, T (and other base quantities where needed), is the language the rest of the chapter is built on.

[Speed] = [M0L1T−1]
[M], [L], [T]dimensions of mass, length and time
Exam tip Angle, being a ratio of two lengths (arc/radius), is dimensionless: [M0L0T0]. So are strain, refractive index and relative density.

09 1.9 Dimensional Formulae and Dimensional Equations

A dimensional equation is the equation obtained when a physical quantity is equated to its dimensional formula, for example [F] = [MLT−2] for force. To write the dimensional formula of any derived quantity, start from its defining relation in terms of quantities you already know the dimensions of, and substitute. For instance, force = mass × acceleration = mass × (velocity/time) = mass × (length/time)/time, giving [F] = [MLT−2]; and since pressure = force/area, [P] = [MLT−2]/[L2] = [ML−1T−2]. Building up dimensional formulae step by step from familiar definitions, rather than memorising a long list, is both faster and less error-prone in the exam.

[F] = [MLT−2]
Fforce
[P] = [ML−1T−2]
Ppressure
[E] = [ML2T−2]
Eenergy/work
Exam tip Derive, don't memorise: write the defining formula, substitute known dimensions, simplify. It also protects you if the question uses an unfamiliar quantity.

10 1.10 Dimensional Analysis and its Applications

Dimensional analysis rests on the principle of homogeneity: in a dimensionally correct equation, every term on both sides must have the same dimensions, because you can only add, subtract or equate quantities of the same kind. This gives three uses. First, checking correctness: if the dimensions on the two sides of an equation do not match, the equation must be wrong (though matching dimensions do not guarantee the equation is numerically correct — a dimensionless constant could still be wrong or missing). Second, deriving a relation: if you can guess which quantities a result depends on, you can assume a power relation between them, match dimensions on both sides, solve for the unknown powers, and arrive at the form of the formula (up to an undetermined dimensionless constant). For example, assuming the time period T of a simple pendulum depends on its length l, the mass m of the bob and g, writing T = k lagbmc and matching dimensions gives a = ½, b = −½, c = 0, so T ∝ √(l/g) — mass drops out entirely, which the method itself reveals. Third, converting units from one system to another using n1u1 = n2u2. The method has real limits: it cannot fix the dimensionless constant k, it fails if a quantity depends on more than three unknown powers (only three equations are available, one each for M, L, T), and it cannot be applied if the relation involves a sum of terms like s = ut + ½at2 or a trigonometric/exponential/logarithmic function, since dimensional analysis cannot tell you the numerical coefficients or that such functions must be dimensionless in their argument.

T ∝ √(l/g) ⇒ T = 2π√(l/g)
Ttime period of a simple pendulumllength of the pendulumgacceleration due to gravity
n1u1 = n2u2
n1, u1numerical value and unit in system 1n2, u2numerical value and unit in system 2
Exam tip State all three limitations when asked 'limitations of dimensional analysis' — most students give only one and lose marks.

VocabularyKey terms

Fundamental (base) quantityA physical quantity, such as length, mass or time, that is measured independently and not defined in terms of any other quantity.
Derived quantityA physical quantity, such as speed or force, whose unit is obtained by combining the units of base quantities.
SI systemThe International System of Units, built on seven base units (m, kg, s, A, K, mol, cd) and two supplementary units (rad, sr), used worldwide for scientific measurement.
ParallaxThe apparent change in the position of an object when viewed from two different points, used to measure very large distances.
Unified atomic mass unit (u)A unit of mass equal to 1/12 the mass of a carbon-12 atom, used for atomic and molecular masses; 1 u = 1.66 × 10−27 kg.
AccuracyHow close a measured value is to the true value of the quantity being measured.
PrecisionHow close repeated measurements of the same quantity are to one another, independent of how close they are to the true value.
Systematic errorAn error that is consistently biased in one direction, arising from an instrument, technique or external condition, and can be corrected once identified.
Random errorAn error arising from unpredictable, small variations that scatter repeated readings on both sides of the true value; reduced by averaging many readings.
Absolute errorThe magnitude of the difference between the mean value of a quantity and one individual measurement of it.
Percentage errorThe relative (fractional) error in a measurement expressed as a percentage: (mean absolute error/mean value) × 100.
Significant figuresAll the digits in a measured number that are known reliably plus the one digit that is estimated or uncertain.
DimensionsThe powers to which the base quantities (M, L, T, ...) must be raised to express a given physical quantity.
Dimensional formulaAn expression showing how, and to what powers, the base quantities enter into a physical quantity, e.g. [MLT−2] for force.
Principle of homogeneity of dimensionsThe rule that every term on both sides of a correct physical equation must have the same dimensions.
Order of magnitudeThe power of 10 nearest to the value of a quantity, used to make quick, approximate comparisons of size.

Keep this on one pageFormula sheet

NameExpressionNotes
Parallax distanceD = b/θθ in radians; b is the baseline between the two observation points.
Angular diameterd = Dαα is the angular size (radians) of a distant object seen from Earth, D its distance.
Unified atomic mass unit1 u = 1.66 × 10−27 kgDefined as 1/12 the mass of a ¹²C atom.
Mean absolute errorΔamean = (Σ|amean − ai|)/nAverage of the absolute errors of n readings.
Percentage error(Δamean/amean) × 100Relative error expressed as a percentage.
Error in sum/differenceΔZ = ΔA + ΔB, for Z = A ± BAbsolute errors always add, even for a difference.
Error in product/quotientΔZ/Z = ΔA/A + ΔB/B, for Z = AB or A/BRelative errors add.
Error in a powerΔZ/Z = n(ΔA/A), for Z = AnRelative error is multiplied by the power.
Dimensional formula of force[F] = [MLT−2]From F = ma.
Dimensional formula of pressure/stress[P] = [ML−1T−2]From pressure = force/area.
Dimensional formula of energy/work[E] = [ML2T−2]From work = force × distance.
Time period of a simple pendulumT = 2π√(l/g)Derived by dimensional analysis; the constant 2π cannot be found by this method.
Unit conversion rulen1u1 = n2u2Used to convert a physical quantity from one system of units to another.

Practise with a pencilDiagrams that earn marks

Parallax method for the distance of a star/planet

Draw the Earth's orbit as a large circle around the Sun; mark two positions of the Earth six months apart as the baseline (2 AU). From both positions draw sight lines to a distant star, meeting at the star with a small angle θ between them (the parallax angle). Label the baseline b, the distance D, and mark θ at the star. Write D = b/θ beside it.

Systematic vs random error on a dartboard

Draw four small target/dartboard circles in a 2×2 grid. Top-left: hits tightly clustered but off-centre (precise, not accurate → systematic error). Top-right: hits scattered evenly around the bullseye (accurate on average, not precise → random error). Bottom-left: hits scattered and off-centre (neither accurate nor precise). Bottom-right: hits tightly clustered on the bullseye (both accurate and precise, ideal measurement). Label each panel clearly.

Building up the dimensional formula of force step by step

Draw three boxes in a row connected by arrows: 'mass, velocity, time → acceleration = velocity/time = [LT⁻²]', then 'force = mass × acceleration → [M][LT⁻²] = [MLT⁻²]'. This visual chain shows the derive-don't-memorise method for any derived quantity.

Vernier callipers and screw gauge, labelled

Draw a vernier callipers with main scale, vernier scale, jaws for external/internal measurement, and the depth strip; mark least count = 1 MSD − 1 VSD. Separately draw a screw gauge with the sleeve (main scale), thimble (circular scale), ratchet and the object held between the spindle and anvil; mark least count = pitch/number of circular scale divisions.

TextbookNCERT questions the board keeps asking

2.1 Fill in the blanks: (a) The volume of a cube of side 1 cm is equal to ___ m3. (b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ___ (mm)2.2 m

(a) Side = 1 cm = 10−2 m, so volume = (10−2)3 = 10−6 m3.

(b) Surface area of a solid cylinder = 2πr2 + 2πrh = 2πr(r + h). Here r = 2.0 cm = 20 mm, h = 10.0 cm = 100 mm. Area = 2π(20)(20 + 100) = 2π(20)(120) = 4800π ≈ 1.5 × 104 mm2.

2.2 State the number of significant figures in the following: (a) 0.007 m2 (b) 2.64 × 1024 kg (c) 0.2370 g cm−3 (d) 6.320 J (e) 6.032 N m−2 (f) 0.0006032 m2.2 m

(a) 1 sig. fig. (only the 7 is significant; leading zeros are not). (b) 3 sig. figs. (2, 6, 4). (c) 4 sig. figs. (2, 3, 7, 0 — trailing zero after decimal counts). (d) 4 sig. figs. (6, 3, 2, 0). (e) 4 sig. figs. (6, 0, 3, 2 — the zero is between non-zero digits). (f) 4 sig. figs. (6, 0, 3, 2 — leading zeros before 6 are not significant).

2.7 The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide. The slide is projected onto a screen, and the area of the house on the screen is 1.55 m2. What is the linear magnification of the projector-screen arrangement?3 m

Area magnification = area on screen/area on slide = 1.55 m2/1.75 cm2. Convert: 1.55 m2 = 1.55 × 104 cm2. Area magnification = (1.55 × 104)/1.75 ≈ 8857.

Linear magnification = √(area magnification) = √8857 ≈ 94.1.

2.9 The Sun is a hot plasma with mass density in the range of 103 kg m−3. Estimate the ratio of density of the Sun to the density of water. Given that mass of the Sun = 2.0 × 1030 kg and its radius = 7.0 × 108 m, calculate the density of the Sun in the wake of these numbers and check.3 m

Volume of Sun = (4/3)πR3 = (4/3)π(7.0 × 108)3 ≈ (4/3)π(3.43 × 1026) ≈ 1.44 × 1027 m3.

Density = mass/volume = (2.0 × 1030)/(1.44 × 1027) ≈ 1.39 × 103 kg m−3.

This is close to, and even a little more than, the density of water (103 kg m−3) — a useful reminder that the Sun's average density is comparable to water despite being a plasma, because its huge gravity compresses the core far more than the outer layers are rarefied.

2.16 The unit of length convenient on the atomic scale is known as an angstrom, 1 Å = 10−10 m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m3 of a mole of hydrogen atoms?3 m

Radius r = 0.5 Å = 0.5 × 10−10 m. Volume of one atom = (4/3)πr3 = (4/3)π(0.5 × 10−10)3 ≈ 5.24 × 10−31 m3.

One mole contains NA = 6.023 × 1023 atoms. Total atomic volume = 5.24 × 10−31 × 6.023 × 1023 ≈ 3.16 × 10−7 m3.

2.17 One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen (use the answer to the previous question)? Why is this ratio so large?3 m

Molar volume = 22.4 L = 22.4 × 10−3 m3. Atomic volume (from Q2.16) ≈ 3.16 × 10−7 m3.

Ratio = (22.4 × 10−3)/(3.16 × 10−7) ≈ 7.08 × 104, of the order 104 to 105.

The ratio is this large because in a gas the average separation between molecules is far greater than the size of a molecule itself — most of the gas's volume is empty space, unlike in a liquid or solid where molecules are packed close together.

2.20 The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun?3 m

1 light year = 9.46 × 1015 m and 1 parsec = 3.08 × 1016 m.

Distance = 4.29 ly = 4.29 × 9.46 × 1015 m ≈ 4.06 × 1016 m.

In parsecs: 4.06 × 1016/3.08 × 1016 ≈ 1.32 parsec.

Parallax angle θ = 1/(distance in parsec) arcsecond = 1/1.32 ≈ 0.76 arcsecond.

2.23 The Sun's angular diameter is measured to be 1920 arcseconds. The distance of the Sun from the Earth is 1.496 × 1011 m. What is the diameter of the Sun?3 m

Convert 1920 arcseconds to radians: 1 arcsecond = 4.85 × 10−6 rad, so α = 1920 × 4.85 × 10−6 ≈ 9.31 × 10−3 rad.

Diameter d = Dα = 1.496 × 1011 × 9.31 × 10−3 ≈ 1.39 × 109 m.

2.27 A famous relation in physics relates the 'moving mass' m of a particle to its 'rest mass' m0 as m = m0/√(1 − v2), where v is the speed of the particle relative to the observer, and c is the speed of light. Guess where the factor c has to be inserted from dimensional considerations.3 m

For (1 − v2) inside the square root to be dimensionless (since it must be, being subtracted from the pure number 1), v2 must itself be dimensionless. But v has dimensions of speed [LT−1], so v2 must be divided by another speed squared. The only constant speed available is c, the speed of light.

So the correct relation is m = m0/√(1 − v2/c2) — c must appear as v2/c2 inside the root.

additional-1 A physical quantity X is given by X = (2A3B2)/(√C D). If the percentage errors in A, B, C and D are 4%, 2%, 3% and 1% respectively, find the percentage error in X.3 m

For a product/quotient of powers, relative (percentage) errors add, each multiplied by the magnitude of its power.

%error in X = 3(%error in A) + 2(%error in B) + ½(%error in C) + 1(%error in D)

= 3(4%) + 2(2%) + ½(3%) + 1(1%) = 12% + 4% + 1.5% + 1% = 18.5%.

MCQ-1 Derive, using dimensional analysis, the relation for the time period T of a simple pendulum of length l, oscillating under gravity g (the mass of the bob does not appear), and state the value found for the powers of l and g.5 m

Assume T = k la gb, where k is a dimensionless constant. Writing dimensions: [T] = [L]a[LT−2]b = [La+b T−2b].

Comparing powers: for L, a + b = 0; for T, −2b = 1 ⇒ b = −1/2, so a = 1/2.

Hence T = k l1/2g−1/2 = k√(l/g); experiment (or a fuller derivation) fixes k = 2π, giving the familiar T = 2π√(l/g).

Previous years and frequently askedBoard questions with model answers

MCQ Which of the following pairs of physical quantities has the same dimensional formula?
  1. Work and power
  2. Force and pressure
  3. Work and torque
  4. Energy and momentum
Frequently asked1 m
Answer: (c)
Work = force × distance and torque = force × perpendicular distance, so both have dimensional formula [ML2T−2]. Work and power differ by a factor of [T]; force and pressure differ by [L2]; energy [ML2T−2] and momentum [MLT−1] also differ — so only work and torque match.
MCQ The number of significant figures in 0.06900 is:
  1. 2
  2. 3
  3. 4
  4. 5
Frequently asked1 m
Answer: (c)
Leading zeros (0.0) are not significant. 6, 9, 0, 0 (the trailing zeros after the decimal point) are all significant, giving 4 significant figures.
MCQ The dimensional formula of surface tension is:
  1. [MLT−2]
  2. [ML0T−2]
  3. [ML2T−2]
  4. [ML−1T−2]
Frequently asked1 m
Answer: (b)
Surface tension = force/length = [MLT−2]/[L] = [ML0T−2].
MCQ A physical quantity P is related to four observables a, b, c and d as P = a2b3/(√c d). The percentage errors in a, b, c and d are 1%, 2%, 3% and 4% respectively. What is the percentage error in P?
  1. 12.5%
  2. 13.5%
  3. 14.5%
  4. 16%
Frequently asked1 m
Answer: (b)
%error in P = 2(%error in a) + 3(%error in b) + ½(%error in c) + 1(%error in d) = 2(1%) + 3(2%) + ½(3%) + 1(4%) = 2 + 6 + 1.5 + 4 = 13.5%.
MCQ Which of these is a dimensionless quantity?
  1. Angular velocity
  2. Strain
  3. Torque
  4. Angular momentum
Frequently asked1 m
Answer: (b)
Strain is the ratio of change in dimension to original dimension (both lengths), so it is dimensionless [M0L0T0].
MCQ Two resistors of resistances R1 = (4 ± 0.5) Ω and R2 = (16 ± 0.5) Ω are connected in series. What is the percentage error in the equivalent resistance?
  1. 2%
  2. 5%
  3. 8%
  4. 10%
Frequently asked1 m
Answer: (b)
For series, R = R1 + R2 = 20 Ω and ΔR = ΔR1 + ΔR2 = 0.5 + 0.5 = 1 Ω. %error = (1/20) × 100 = 5%.
MCQ Which pair of physical quantities has different dimensions?
  1. Impulse and momentum
  2. Work and energy
  3. Pressure and stress
  4. Force and pressure
Frequently asked1 m
Answer: (d)
Force has dimensional formula [MLT−2] while pressure has [ML−1T−2] — they differ by [L2], so this pair has different dimensions; all the other pairs listed are dimensionally identical.
Assertion–Reason Assertion (A): A dimensionally correct equation need not be numerically (physically) correct. Reason (R): A dimensionless proportionality constant cannot be found using dimensional analysis.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
Frequently asked1 m
Answer: (a)
A is true: dimensional correctness only checks that the powers of M, L, T match, not the value of any dimensionless constant, so an equation can look right dimensionally while being numerically wrong. R correctly explains why: since the method cannot fix that constant, an error hidden inside it goes undetected, which is exactly why (A) is true.
Assertion–Reason Assertion (A): Precision of a measurement is different from its accuracy. Reason (R): Accuracy depends on the number of significant figures, while precision depends only on the least count of the instrument.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
Frequently asked1 m
Answer: (c)
A is true: precision is the closeness of repeated readings to each other, while accuracy is the closeness to the true value; the two are distinct ideas. R is false, because accuracy is not simply a matter of significant figures — an instrument can give many significant figures that are all precisely wrong (a systematic error), so accuracy depends on freedom from systematic error, not on digit count alone.
Assertion–Reason Assertion (A): The equation s = ut + ½at2 cannot be derived purely by the method of dimensions. Reason (R): Dimensional analysis cannot determine dimensionless numerical constants that appear in a sum of terms.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
Frequently asked1 m
Answer: (a)
A is true: this relation is a sum of two terms of the same dimensions ([L]), and dimensional analysis has no way of finding the relative numerical coefficients (1 and ½) of the two terms if the equation itself were unknown. R gives exactly this reason, so R correctly explains A.
Very short answer Define 'systematic error' and name one source of it.Frequently asked1 m

A systematic error is an error that consistently occurs in the same direction (either always positive or always negative) in a set of measurements, rather than scattering randomly. One common source: instrumental error, such as zero error in a vernier callipers or a screw gauge, where the instrument itself is not correctly calibrated.

Very short answer Round off the following to three significant figures: (a) 6.658 (b) 15.75 (c) 3245.Frequently asked2 m

(a) 6.658 → 6.66 (the dropped digit 8 > 5, so the preceding digit rounds up from 5 to 6).

(b) 15.75 → 15.8 (the dropped digit is exactly 5 and the preceding digit 7 is odd, so it is raised to 8).

(c) 3245 → 3.24 × 103 (the dropped digit is exactly 5 and the preceding digit 4 is even, so it stays unchanged).

Very short answer Write the dimensional formulae of (a) angular momentum and (b) universal gravitational constant G.Frequently asked2 m

(a) Angular momentum L = mvr, dimensions [M][LT−1][L] = [ML2T−1].

(b) From F = Gm1m2/r2, G = Fr2/(m1m2), so [G] = [MLT−2][L2]/[M2] = [M−1L3T−2].

Very short answer Name the method used to measure (a) very large distances such as that of a star, and (b) very small distances such as the size of a molecule.Frequently asked2 m

(a) The parallax method, using the apparent angular shift of the object against a distant background when observed from two points of a known baseline.

(b) Electron microscopy or X-ray diffraction; a rough order-of-magnitude estimate can also be made from the molar volume of a substance and Avogadro's number.

Short answer The period of oscillation of a simple pendulum is T = 2π√(L/g). The measured value of L is 20.0 cm known to 1 mm accuracy, and the time for 100 oscillations is found to be 90 s using a clock of 1 s resolution. What is the accuracy in the determination of g?Frequently asked3 m

Time for one oscillation T = 90/100 = 0.9 s. Resolution of the clock is 1 s over 100 oscillations, so ΔT/T = 1/90 (fractional error in T using the total time). Actually the fractional error is calculated on the measured total time: Δ(time)/time = 1/90.

Since T ∝ time, the fractional error in T equals 1/90, and since g ∝ L/T2, Δg/g = ΔL/L + 2ΔT/T.

ΔL/L = 0.1 cm/20.0 cm = 0.005. 2ΔT/T = 2 × (1/90) = 0.0222.

Δg/g = 0.005 + 0.0222 = 0.0272 ≈ 2.7%, so g is accurate to about 3%.

Short answer The distance of the Moon from the Earth is about 3.84 × 105 km. Using the parallax method with a baseline of 1276 km (the Earth's diameter), calculate the parallax angle in radians and in degrees.Frequently asked3 m

θ = b/D = 1276/(3.84 × 105) ≈ 3.32 × 10−3 rad.

Converting to degrees: θ(°) = θ(rad) × (180/π) = 3.32 × 10−3 × 57.3 ≈ 0.19°.

Short answer Explain, with one example each, the difference between accuracy and precision.Frequently asked3 m

Accuracy is how close a measurement is to the true value; precision is how close repeated measurements are to each other, irrespective of the true value.

Example: suppose the true length of a rod is 10.00 cm. If four measurements give 9.98, 9.99, 10.01, 10.02 cm, they are both accurate (close to 10.00) and precise (close to each other). If instead they give 9.50, 9.51, 9.49, 9.50 cm, they are precise (tightly clustered) but not accurate (far from 10.00, likely due to a systematic zero error). A set like 9.7, 10.3, 9.9, 10.5 cm is accurate on average but not precise, showing scatter (random error).

Short answer State the principle of homogeneity of dimensions. Use it to check whether the equation v = u + at is dimensionally correct.Frequently asked3 m

The principle of homogeneity states that in a physically (dimensionally) correct equation, every term on both sides must have the same dimensions, since only quantities of the same dimension can be added, subtracted or equated.

Checking v = u + at: [v] = [LT−1]; [u] = [LT−1]; [at] = [LT−2][T] = [LT−1]. All three terms have the dimension [LT−1], so the equation is dimensionally correct (and homogeneous).

Long answer (a) Explain the terms absolute error, relative error and percentage error with their formulae. (b) Derive the rule for combination of errors in a product of two quantities Z = AB.Frequently asked5 m

(a) For n measurements a1, a2, ..., an of a quantity with mean amean: the absolute error in a reading ai is Δai = |amean − ai|. The mean absolute error is Δamean = (Σ|Δai|)/n, The relative (fractional) error is Δamean/amean, and the percentage error is this relative error × 100.

(b) Let Z = AB, with measured values A ± ΔA and B ± ΔB. Then Z ± ΔZ = (A ± ΔA)(B ± ΔB) = AB ± AΔB ± BΔA ± ΔAΔB. Dividing by Z = AB and dropping the doubly-small term ΔAΔB: ΔZ/Z = ± (ΔA/A ± ΔB/B). Taking the maximum possible error (worst case, all errors adding): ΔZ/Z = ΔA/A + ΔB/B — the relative error in a product is the sum of the relative errors of the factors.

Long answer (a) State the rules for determining significant figures with two examples each for (i) addition/subtraction and (ii) multiplication/division. (b) Add 4.237 g and 2.51 g, and give the answer to the correct number of significant figures/decimal places.Frequently asked5 m

(a)(i) In addition/subtraction, the result is rounded to the number of decimal places of the term with the FEWEST decimal places. Example 1: 12.3 + 4.56 = 16.86 → rounded to one decimal place (fewest, from 12.3) → 16.9.

(a)(ii) In multiplication/division, the result is rounded to the number of significant figures of the factor with the FEWEST significant figures. Example 1: 4.11 × 3.6 = 14.796 → 3.6 has 2 sig. figs., so the answer is rounded to 2 sig. figs. → 15.

(b) 4.237 + 2.51 = 6.747. The term 2.51 has only 2 decimal places (the fewest), so the answer is rounded to 2 decimal places: 6.75 g.

Long answer Using the method of dimensions, derive an expression for the viscous force F on a small sphere of radius r moving with velocity v through a fluid of viscosity η, given F depends on η, r and v. (This is the basis of Stokes' law.)Frequently asked5 m

Let F = k ηa rb vc, where k is a dimensionless constant.

Dimensions: [F] = [MLT−2]; [η] = [ML−1T−1] (coefficient of viscosity); [r] = [L]; [v] = [LT−1].

So [MLT−2] = [ML−1T−1]a[L]b[LT−1]c = [Ma L−a+b+c T−a−c].

Comparing powers: For M: a = 1. For T: −a − c = −2 ⇒ c = 2 − a = 1. For L: −a + b + c = 1 ⇒ −1 + b + 1 = 1 ⇒ b = 1.

So F = k η1 r1 v1 = kηrv. Experiment shows k = 6π, giving Stokes' law: F = 6πηrv.

Case-based A student measures the length of a metal rod five times with a metre scale and obtains the readings (in cm): 25.4, 25.6, 25.3, 25.5, 25.2.
(i) Find the mean length of the rod.
(ii) Find the absolute error in each reading and the mean absolute error.
(iii) Express the result as (mean ± mean absolute error).
(iv) State one way the student could reduce the random error in this measurement.
Frequently asked4 m

(i) Mean = (25.4 + 25.6 + 25.3 + 25.5 + 25.2)/5 = 127.0/5 = 25.4 cm.

(ii) Absolute errors: |25.4−25.4| = 0.0, |25.6−25.4| = 0.2, |25.3−25.4| = 0.1, |25.5−25.4| = 0.1, |25.2−25.4| = 0.2 cm. Mean absolute error = (0.0+0.2+0.1+0.1+0.2)/5 = 0.6/5 = 0.12 cm.

(iii) Result = (25.4 ± 0.12) cm, usually reported to the precision of the instrument, e.g. (25.40 ± 0.12) cm.

(iv) Take a larger number of readings and average them, or use a more precise instrument (e.g. vernier callipers instead of a plain metre scale) to reduce the scatter caused by random error.

Case-based In an experiment, a physical quantity Z is calculated from Z = A2B3/C, where A = (2.0 ± 0.1), B = (1.0 ± 0.1) and C = (4.0 ± 0.2) (all in appropriate SI units).
(i) Find the value of Z.
(ii) Find the percentage error in A, B and C individually.
(iii) Find the total percentage error in Z.
(iv) Find the absolute error in Z, and express the result as Z ± ΔZ.
Frequently asked4 m

(i) Z = A2B3/C = (2.0)2(1.0)3/4.0 = 4.0 × 1.0/4.0 = 1.0.

(ii) %error in A = (0.1/2.0) × 100 = 5%. %error in B = (0.1/1.0) × 100 = 10%. %error in C = (0.2/4.0) × 100 = 5%.

(iii) %error in Z = 2(%error in A) + 3(%error in B) + 1(%error in C) = 2(5%) + 3(10%) + 1(5%) = 10% + 30% + 5% = 45%.

(iv) ΔZ = 45% of 1.0 = 0.45. Result: Z = 1.0 ± 0.45 (in the appropriate SI unit for Z).

Test yourselfPractice quiz

Marks lost every yearCommon mistakes

  • Treating a dimensionally correct equation as automatically numerically correct — dimensional analysis cannot verify or find a dimensionless constant like 2π or ½.
  • Subtracting errors for a difference Z = A − B instead of adding them; absolute errors ALWAYS add for both sum and difference.
  • Confusing the rule for addition/subtraction (match decimal places) with the rule for multiplication/division (match significant figures), and applying the wrong one.
  • Counting leading zeros as significant (e.g. treating 0.0025 as having 5 significant figures instead of 2).
  • Using accuracy and precision as synonyms in an answer instead of defining them separately with an example each.
  • Forgetting to convert an angle to radians before using θ = b/D in a parallax numerical, especially when the angle is given in degrees or arcseconds.
  • Assuming dimensional analysis can derive any formula — forgetting its three limitations (cannot find the constant, cannot handle more than three unknown powers, cannot handle sums, trigonometric or exponential relations).

How to prepareStrategy for this chapter

This chapter carries no board weightage in Class 11, but treat it as the toolkit you will use in every numerical for the next two years, so master it now rather than revisiting it under exam pressure in Class 12. Split your revision into three blocks. First, learn the SI base units and the parallax/echo methods well enough to state them in one line each — these are quick 1-mark recall questions. Second, drill the error-combination rules (sum/difference adds absolute error, product/quotient adds relative error, a power multiplies relative error) and the significant-figure rules until you can apply them without hesitating; work at least ten numericals until the arithmetic is automatic. Third, practise deriving 3–4 standard relations by dimensional analysis (time period of a pendulum, Stokes' law, orbital speed) so that a 5-mark 'derive using dimensional analysis' question never surprises you, and be ready to state the method's three limitations from memory.

Do not try to memorise a long table of dimensional formulae; instead memorise the defining equation of each quantity (force = ma, pressure = force/area, energy = force × distance) and build up the dimensions from there in the exam — it is faster and self-checking. Keep a small personal formula sheet with the seven base dimensions and ten common derived ones, and revise it the night before any test on this chapter.

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