Class 11Physics · MeasurementFull chapter

Units and Measurements

Q: A calorie is a unit of heat energy and it equals 4.2 J, where 1 J = 1 kg m² s⁻². Suppose a new system of units is used in which the unit of mass equals α kg, the unit of length equals β m and the unit of time equals γ s. Show that in this new system, 1 calorie equals 4.2 α⁻¹β⁻²γ² new units of energy.

Energy has the dimensional formula [ML²T -2 ]. Using the general relation for converting a physical quantity between two systems of units, n 1 u 1 = n 2 u 2 , with the powers a = 1, b = 2, c = -2 (from the dimensions of energy): Given: n 1 = 4.2 (value in SI, where M 1 = 1 kg, L 1 = 1 m, T 1 = 1 s); new system has M 2 = α kg, L 2 = β m, T 2 = γ s Formula: n 2 = n 1 (M 1 /M 2 ) 1 (L 1 /L 2 ) 2 (T 1 /T 2 ) -2 Substitution: n 2 = 4.2 × (1/α) 1 × (1/β) 2 × (1/γ) -2 = 4.2 × (1/α) × (1/β²) × γ² Result: n 2 = 4.2 α -1 β -2 γ², so 1 calorie = 4.2 α -1 β -2 γ² new units of energy, as required.

Q: The mass and volume of a body are measured to be (5.00 ± 0.05) kg and (2.00 ± 0.02) m³ respectively. Find the density of the body along with its maximum percentage error.

Given: m = 5.00 kg (Δm = 0.05 kg), V = 2.00 m³ (ΔV = 0.02 m³) Formula: ρ = m/V; for a quotient, Δρ/ρ = Δm/m + ΔV/V Substitution: ρ = 5.00/2.00 = 2.50 kg m -3 ; Δρ/ρ = (0.05/5.00) + (0.02/2.00) = 0.01 + 0.01 = 0.02, i.e., 2%; Δρ = 0.02 × 2.50 = 0.05 kg m -3 Result: ρ = (2.50 ± 0.05) kg m -3 , with a maximum percentage error of 2% .

Q: Using dimensional analysis, check the correctness of Einstein's mass-energy relation E = mc², where E is energy, m is mass and c is the speed of light.

Given: [E] = [ML²T -2 ] (energy), [m] = [M], [c] = [LT -1 ] (speed) Formula: compare [E] with [mc²] Substitution: [mc²] = [M][LT -1 ]² = [M][L²T -2 ] = [ML²T -2 ] Result: [E] = [mc²] = [ML²T -2 ] on both sides, so the equation is dimensionally correct .

Q: Distinguish between accuracy and precision of a measurement, with a suitable example.

Accuracy is the closeness of a measured value to the true (actual) value of the quantity being measured. Precision is the closeness among a set of repeated measurements of the same quantity, regardless of how close they are to the true value. For example, suppose the true length of a rod is 10.00 cm. If five readings taken with one instrument are 9.99, 10.01, 9.98, 10.02 and 10.00 cm, the measurements are both accurate and precise, since they cluster tightly around the true value. If a second, poorly calibrated instrument gives readings of 9.20, 9.22, 9.21, 9.23 and 9.20 cm, these readings are precise (they closely agree with one another) but not accurate, since they are all far from the true value of 10.00 cm — indicating a systematic (zero) error in that instrument.

Q: Define absolute error, relative error and percentage error for a set of repeated measurements. How is percentage error related to relative error?

For n repeated measurements a 1 , a 2 , ..., a n of a quantity, with mean value a mean : Absolute error of the i-th reading is Δa i = |a mean − a i |; it carries the same unit as the measured quantity itself. Relative (fractional) error is the mean absolute error divided by the mean value, Δa mean /a mean ; it is a pure (dimensionless) number. Percentage error is the relative error expressed as a percentage: percentage error = (Δa mean /a mean ) × 100%. Percentage error is simply the relative error multiplied by 100; expressing it this way allows the precision of measurements of different physical quantities, in different units, to be compared on the same scale.

Q: State the number of significant figures in 4.700 × 10⁵.

In scientific notation N × 10 x , every digit of N is significant. Here N = 4.700 has the digits 4, 7, 0, 0, all significant (the trailing zeros after the decimal point are significant). Therefore, 4.700 × 10⁵ has 4 significant figures .

Q: Using the method of dimensions, check whether the relation v² = u² + 2as is dimensionally correct, where v and u are velocities, a is acceleration and s is displacement.

Given: [v] = [u] = [LT -1 ], [a] = [LT -2 ], [s] = [L] Formula: compare the dimensions of each term: [v²], [u²], [2as] Substitution: [v²] = [L²T -2 ]; [u²] = [L²T -2 ]; [2as] = [LT -2 ][L] = [L²T -2 ] (the number 2 is dimensionless) Result: LHS = [L²T -2 ] and RHS = [L²T -2 ] + [L²T -2 ] = [L²T -2 ]; since LHS = RHS, the equation is dimensionally correct .

Q: Using dimensional analysis, derive an expression for the time period T of a simple pendulum, given that it may depend on the mass m of the bob, the length l of the pendulum, and the acceleration due to gravity g.

Given: Let T = k·m a l b g c , where k is a dimensionless constant, [T] = [T], [m] = [M], [l] = [L], [g] = [LT -2 ] Formula: [M 0 L 0 T 1 ] = [M] a [L] b [LT -2 ] c = [M a L b+c T -2c ] Substitution: equating powers: for M, a = 0; for T, -2c = 1 ⇒ c = -½; for L, b + c = 0 ⇒ b = ½ Result: T = k·m 0 l 1/2 g -1/2 = k√(l/g). The mass exponent works out to zero automatically, showing that the period does not depend on the mass of the bob; a full derivation gives k = 2π, so T = 2π√(l/g) .

Q: (a) Distinguish between systematic errors and random errors, giving one example of each. (b) The diameter of a wire, measured with a screw gauge in five trials, was found to be 1.34 mm, 1.38 mm, 1.35 mm, 1.32 mm and 1.36 mm. Calculate the mean absolute error and the percentage error in the measurement.

(a) Systematic errors occur in a definite direction (always positive or always negative) and arise from causes such as instrumental defects (e.g., a zero error in a screw gauge), imperfect experimental technique, or personal bias of the observer; they can be reduced by proper calibration and correction. Random errors occur irregularly, sometimes positive and sometimes negative, due to unpredictable fluctuations in experimental conditions (e.g., small variations in an observer's reaction time while using a stopwatch); they are reduced, though never fully eliminated, by taking a large number of readings and computing the mean. (b) Given: readings 1.34, 1.38, 1.35, 1.32, 1.36 mm Formula: a mean = Σa i /n; Δa mean = Σ|a mean − a i |/n; % error = (Δa mean /a mean ) × 100% Substitution: a mean = (1.34+1.38+1.35+1.32+1.36)/5 = 6.75/5 = 1.35 mm. Absolute errors: |1.35−1.34| = 0.01, |1.35−1.38| = 0.03, |1.35−1.35| = 0.00, |1.35−1.32| = 0.03, |1.35−1.36| = 0.01 mm; sum = 0.08 mm, so Δa mean = 0.08/5 = 0.016 mm ≈ 0.02 mm Result: mean diameter = (1.35 ± 0.02) mm; percentage error = (0.016/1.35) × 100% ≈ 1.2%

The whole chapter in one place — read it, then test yourself. Clear notes, formula sheet, a practice quiz, and worked NCERT solutions & PYQs.

In this chapter

The International System of Units (SI) and Fundamental Quantities Dimensions and Dimensional Formulae Dimensional Analysis: Checking Equations and Deriving Relations Significant Figures and Rules for Arithmetic Errors in Measurement: Accuracy, Precision and Percentage Error Propagation of Errors in Sums, Products and Powers

The International System of Units (SI) and Fundamental Quantities

Quick answer The SI system is built on seven independently-defined base units (metre, kilogram, second, ampere, kelvin, mole, candela); every other unit is derived by combining these through the defining equation of the quantity.

Physics is a science of measurement. Every measurement of a physical quantity needs two parts: a numerical value and a unit — for instance, a length of 5 m means 5 times the reference unit called the metre. A unit is an internationally accepted reference standard for a physical quantity.

To avoid the confusion of many older unit systems (CGS, FPS, MKS), scientists across the world adopted the International System of Units (SI). The SI is built on seven base (fundamental) units, each defined independently, from which every other physical quantity's unit is built.

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)

Since 2019, all seven base units have been redefined by fixing the exact numerical value of a fundamental constant of nature (such as the speed of light c, the Planck constant h, or the Avogadro constant N_A), instead of relying on a physical prototype like a metal bar or cylinder. This makes the units exactly reproducible in any well-equipped laboratory, anywhere.

Besides the seven base units, the SI also defines two supplementary units, both dimensionless: the radian (rad) for plane angle and the steradian (sr) for solid angle. Every other unit — for speed, force, energy, pressure, and so on — is called a derived unit, since it is built from the base units through the defining relation of that quantity (for example, 1 newton = 1 kg·m·s^-2, from Newton's second law).

Worked Example: Find the SI unit of kinetic energy in terms of base units, and calculate the kinetic energy of a 2 kg object moving at 3 m s^-1.

Given: mass m = 2 kg, speed v = 3 m s^-1
Formula: KE = ½mv²
Substitution: KE = ½ × 2 kg × (3 m s^-1)² = ½ × 2 × 9 kg·m²·s^-2
Result: KE = 9 kg·m²·s^-2 = 9 J (the joule is thus a derived unit equal to kg·m²·s^-2)

Kinetic energy KE = ½mv² J · m in kg, v in m·s⁻¹, gives KE in kg·m²·s⁻² = joule (J)

Newton in base units 1 N = 1 kg·m·s⁻² from F = ma

Seven SI base units m, kg, s, A, K, mol, cd length, mass, time, electric current, thermodynamic temperature, amount of substance, luminous intensity

Remember

The SI system has exactly 7 base units and 2 supplementary (dimensionless) units — radian and steradian.
Since 2019, every SI base unit is fixed by a defined value of a fundamental constant, making the system universal and reproducible.
A derived unit is always built from base units through the physical relation that defines that quantity, e.g. N = kg·m·s⁻².
Every term in a calculation must use the same, consistent unit system to avoid errors.

Dimensions and Dimensional Formulae

Quick answer The dimensions of a quantity are the powers of M, L, T (and other base quantities) needed to express it; writing a quantity in this power form gives its dimensional formula.

The dimensions of a physical quantity are the powers to which the base quantities — mass (M), length (L), time (T), and, where needed, current (A), temperature (K), etc. — must be raised to represent that quantity. Expressing a quantity in this power form is called its dimensional formula; the equation obtained by equating a physical quantity to its dimensional formula is called a dimensional equation.

For example, since velocity = displacement / time, its dimensional formula is [M⁰L¹T^-1], usually written simply as [LT^-1]. Dimensions carry no information about the magnitude or the exact unit chosen — only about how a quantity is built from the base quantities.

Dimensional formulae of some common derived quantities:

Area = length × length → [L²]
Volume = length × length × length → [L³]
Density = mass/volume → [ML^-3]
Velocity = displacement/time → [LT^-1]
Acceleration = velocity/time → [LT^-2]
Momentum = mass × velocity → [MLT^-1]
Force = mass × acceleration → [MLT^-2]
Work / Energy = force × distance → [ML²T^-2]
Power = work/time → [ML²T^-3]
Pressure / Stress = force/area → [ML^-1T^-2]

Worked Example: Newton's law of gravitation states F = Gm₁m₂/r². Find the dimensional formula of the gravitational constant G.

Given: F = Gm₁m₂/r², so G = Fr²/(m₁m₂)
Formula (dimensions): [G] = [F][r²] / ([m₁][m₂])
Substitution: [G] = [MLT^-2][L²] / [M][M] = [MLT^-2L²] / [M²]
Result: [G] = [M^-1L³T^-2] — consistent with the known SI value G = 6.674×10^-11 N·m²·kg^-2 = 6.674×10^-11 m³·kg^-1·s^-2.

Velocity [v] = [LT⁻¹]

Acceleration [a] = [LT⁻²]

Force [F] = [MLT⁻²]

Work / Energy [W] = [ML²T⁻²]

Power [P] = [ML²T⁻³]

Pressure / Stress [P] = [ML⁻¹T⁻²]

Gravitational constant [G] = [M⁻¹L³T⁻²] derived from F = Gm₁m₂/r²

Remember

The dimensional formula shows how a quantity is built from M, L, T (and others); it says nothing about the quantity's numerical magnitude.
Quantities with identical dimensional formulae (e.g., work and torque, both [ML²T⁻²]) need not be physically the same kind of quantity.
Pure ratio quantities (angle, strain, refractive index, relative density) are dimensionless, [M⁰L⁰T⁰].
The dimensional formula of a given physical quantity is the same in every system of units.

Dimensional Analysis: Checking Equations and Deriving Relations

Quick answer By the principle of homogeneity — that both sides of a correct equation must have identical dimensions — dimensional analysis can check an equation's correctness, help derive the form of a relation, and convert units between systems.

Dimensional analysis rests on the principle of homogeneity of dimensions: in any correct physical equation, the dimensions of every term on the left-hand side must equal the dimensions of every term on the right-hand side. This single idea gives dimensional analysis three major uses.

1. Checking the correctness of an equation. If the two sides of an equation do not have matching dimensions, the equation must be wrong. (Matching dimensions do not guarantee an equation is fully correct, since purely numerical/dimensionless constants cannot be checked this way.)

Worked Example: Check whether v² = u² + 2as is dimensionally correct, where v, u are velocities, a is acceleration and s is displacement.

Given: [v] = [u] = [LT^-1], [a] = [LT^-2], [s] = [L]
Formula (dimensions of each term): [v²] = [LT^-1]² = [L²T^-2]; [u²] = [L²T^-2]; [2as] = [LT^-2][L] = [L²T^-2] (2 is a pure number, dimensionless)
Substitution: LHS = [L²T^-2]; RHS = [L²T^-2] + [L²T^-2] = [L²T^-2]
Result: LHS = RHS = [L²T^-2], so the equation is dimensionally correct.

2. Deriving the form of a relation between physical quantities, when it is known (or assumed) which quantities a given quantity depends on, but the exact formula is not known.

Worked Example: The time period T of a simple pendulum is expected to depend on its length l and the acceleration due to gravity g. Derive the form of T using dimensional analysis.

Given: Let T = k·l^a·g^b, where k is a dimensionless constant
Formula (dimensions): [T] = [L]^a[LT^-2]^b = [L^a+bT^-2b]
Substitution: comparing powers with [M⁰L⁰T¹]: for T, -2b = 1 ⇒ b = -½; for L, a + b = 0 ⇒ a = ½
Result: T = k·l^1/2g^-1/2 = k√(l/g). A fuller derivation shows k = 2π, giving the familiar T = 2π√(l/g); the dimensionless constant itself can never be found by dimensional analysis alone.

3. Converting a physical quantity from one system of units to another, using n₁u₁ = n₂u₂, since the actual magnitude of a physical quantity does not change no matter which unit system is used to express it.

Principle of homogeneity [LHS] = [RHS] dimensions on both sides of a correct equation must match, term by term

Unit conversion relation n₁u₁ = n₂u₂ used to convert the numerical value of a quantity between two systems of units

Simple pendulum period T = k·l^(1/2)·g^(-1/2) = 2π√(l/g) exponents ½, -½ found by dimensional analysis; constant 2π cannot be found this way

Remember

Principle of homogeneity: the dimensions of every term on both sides of a valid physical equation must be identical.
Dimensional analysis can check equations and derive the form of a relation, but it can never fix dimensionless constants (like 2π or ½).
It cannot distinguish between two physically different quantities that share the same dimensions (e.g., work vs torque).
It fails for relations containing trigonometric, exponential or logarithmic functions of a dimensional quantity, and when a quantity depends on more than three independent unknowns (since only M, L, T give three equations).

Significant Figures and Rules for Arithmetic

Quick answer Significant figures are the reliably known digits of a measurement plus one estimated digit; addition/subtraction is limited by decimal places, while multiplication/division is limited by the fewest significant figures.

The significant figures in a measured (or calculated) number are all the digits that are reliably known plus one estimated (uncertain) digit. They indicate how precisely a quantity has been measured; writing extra digits beyond the significant ones falsely suggests more precision than the instrument actually provides.

Rules for counting significant figures:

All non-zero digits are significant (1234 has 4 significant figures).
Zeros between two non-zero digits are significant (1005 has 4 significant figures).
Leading zeros (to the left of the first non-zero digit) are never significant — they only fix the position of the decimal point (0.0025 has 2 significant figures).
Trailing zeros after a decimal point are significant (1.200 has 4 significant figures).
Trailing zeros in a number with no decimal point are ambiguous; scientific notation removes the ambiguity (1500 could have 2, 3 or 4 significant figures; writing it as 1.5×10³ fixes it at 2, while 1.500×10³ fixes it at 4).
In scientific notation N×10^x, every digit of N is significant.

Rules for arithmetic with significant figures:

Addition/subtraction: the result is rounded to the same number of decimal places as the term with the fewest decimal places.
Multiplication/division: the result is rounded to the same number of significant figures as the term with the fewest significant figures.
Rounding off: if the digit to be dropped is more than 5, the preceding digit is raised by 1; if it is less than 5, the preceding digit is unchanged; if it is exactly 5, the preceding digit is rounded to the nearest even digit.

Worked Example (addition): Add three lengths 4.237 m, 2.1 m and 0.55 m, and express the sum to the correct number of significant figures.

Given: 4.237 m (3 decimal places), 2.1 m (1 decimal place), 0.55 m (2 decimal places)
Formula: add all terms, then round to the fewest decimal places present (here, 1, from 2.1 m)
Substitution: 4.237 + 2.1 + 0.55 = 6.887 m
Result: rounded to 1 decimal place → 6.9 m

Worked Example (multiplication): Multiply 4.2 (2 significant figures) by 1.70 (3 significant figures).

Given: 4.2 has 2 significant figures; 1.70 has 3 significant figures
Formula: multiply, then round to the fewest significant figures present (here, 2)
Substitution: 4.2 × 1.70 = 7.14
Result: rounded to 2 significant figures → 7.1

Addition/subtraction rule Round result to the fewest decimal places among the terms

Multiplication/division rule Round result to the fewest significant figures among the terms

Rounding a dropped '5' Round to the nearest even preceding digit applies only when the dropped digit is exactly 5

Remember

Significant figures communicate how precisely a value is known; never write more digits than the measurement justifies.
Leading zeros are never significant; zeros between non-zero digits, and trailing zeros after a decimal point, are always significant.
Addition/subtraction is limited by the least precise decimal place; multiplication/division is limited by the fewest significant figures.
Multiplying or dividing by an exact (counted) number does not reduce the number of significant figures in the result.

Errors in Measurement: Accuracy, Precision and Percentage Error

Quick answer Every measurement carries some uncertainty; systematic errors bias readings in one direction while random errors scatter them, and the mean absolute, relative and percentage error quantify the overall uncertainty.

Every measurement has some uncertainty compared with the quantity's true value — this uncertainty is called an error. Accuracy tells us how close a measured value is to the true value; precision tells us how close repeated measurements are to one another (a set of readings can be precise yet inaccurate, if there is a constant instrumental error).

Systematic errors: occur consistently in the same direction (always positive or always negative). They arise from instrumental defects (e.g., zero error), imperfect experimental technique, or personal bias, and can usually be reduced or corrected by careful calibration and technique.
Random errors: occur irregularly, in both directions, from unpredictable fluctuations in conditions or observation. They are reduced (never fully eliminated) by taking many readings and averaging.

The smallest change a measuring instrument can detect is its least count; the uncertainty arising from it is called the least count error, minimised by using an instrument of finer resolution.

For n repeated readings a₁, a₂, ..., a_n of a quantity, the best estimate of the true value is their mean, a_mean. The following quantities describe the overall uncertainty:

Absolute error of the i-th reading: Δa_i = |a_mean − a_i|
Mean absolute error: the average of all the individual absolute errors
Relative (fractional) error: mean absolute error divided by the mean value
Percentage error: the relative error expressed as a percentage

Worked Example: Five measurements of the time period of a pendulum are 2.63 s, 2.56 s, 2.42 s, 2.71 s and 2.80 s. Find the mean absolute error and the percentage error.

Given: readings 2.63, 2.56, 2.42, 2.71, 2.80 s
Formula: a_mean = Σa_i/n; Δa_mean = Σ|a_mean − a_i|/n; % error = (Δa_mean/a_mean) × 100%
Substitution: a_mean = (2.63+2.56+2.42+2.71+2.80)/5 = 13.12/5 = 2.62 s. Absolute errors: 0.01, 0.06, 0.20, 0.09, 0.18 s; sum = 0.54 s, so Δa_mean = 0.54/5 = 0.11 s
Result: T = (2.62 ± 0.11) s; percentage error = (0.11/2.62) × 100% ≈ 4.2%

Mean value a_mean = (a₁ + a₂ + ... + aₙ) / n

Absolute error Δaᵢ = |a_mean − aᵢ|

Mean absolute error Δa_mean = (Σ|Δaᵢ|) / n

Relative error Δa_mean / a_mean

Percentage error (Δa_mean / a_mean) × 100%

Remember

Accuracy = closeness to the true value; precision = closeness among repeated readings — a measurement can be precise without being accurate.
Systematic errors bias all readings the same way and can be corrected; random errors scatter readings and are reduced by averaging over many trials.
Absolute error carries the same unit as the measured quantity; relative and percentage error are dimensionless ratios.
Percentage error is the standard way to quote experimental uncertainty since it allows comparison across different quantities and units.

Propagation of Errors in Sums, Products and Powers

Quick answer When a result is calculated from several measured quantities, absolute errors add for sums/differences while relative errors add (each multiplied by its power) for products, quotients and power-law relations.

When a final result is calculated from two or more measured quantities, each carrying its own error, the individual errors combine according to definite rules that depend on the mathematical operation used.

Sum or difference (Z = A ± B): the absolute errors add, ΔZ = ΔA + ΔB — errors never cancel, even when the quantities themselves are subtracted.
Product (Z = AB): the relative errors add, ΔZ/Z = ΔA/A + ΔB/B.
Quotient (Z = A/B): the relative errors again add, ΔZ/Z = ΔA/A + ΔB/B.
Power law (Z = A^pB^q/C^r): each relative error is multiplied by the magnitude of its power before being added, ΔZ/Z = p(ΔA/A) + q(ΔB/B) + r(ΔC/C). A quantity raised to a higher power (such as radius r in V = (4/3)πr³, where p = 3) therefore contributes a proportionally larger error to the result.

Worked Example: In an experiment, a voltmeter reads V = 5.0 ± 0.1 V and an ammeter reads I = 2.0 ± 0.2 A. Using R = V/I, find the resistance along with its percentage error.

Given: V = 5.0 V (ΔV = 0.1 V), I = 2.0 A (ΔI = 0.2 A)
Formula: R = V/I; for a quotient, ΔR/R = ΔV/V + ΔI/I
Substitution: R = 5.0/2.0 = 2.5 Ω; ΔR/R = (0.1/5.0) + (0.2/2.0) = 0.02 + 0.10 = 0.12, i.e., 12%; ΔR = 0.12 × 2.5 Ω = 0.3 Ω
Result: R = (2.5 ± 0.3) Ω, i.e., R = 2.5 Ω with a percentage error of 12%

Error in sum/difference Z = A ± B ⇒ ΔZ = ΔA + ΔB

Error in product Z = AB ⇒ ΔZ/Z = ΔA/A + ΔB/B

Error in quotient Z = A/B ⇒ ΔZ/Z = ΔA/A + ΔB/B

Error in power-law relation Z = AᵖBᵠ/Cʳ ⇒ ΔZ/Z = p(ΔA/A) + q(ΔB/B) + r(ΔC/C)

Remember

For sums and differences, absolute errors always add — they never partially cancel, even when the quantities are subtracted.
For products and quotients, it is the relative (fractional) errors that add.
In a power-law relation, the exponent multiplies the relative error of that quantity, so quantities raised to higher powers must be measured with the greatest care.
Error propagation gives the maximum possible (worst-case) error in the final calculated result.

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

KE = ½mv²

Kinetic energyJ

1 N = 1 kg·m·s⁻²

Newton in base units

m, kg, s, A, K, mol, cd

Seven SI base units

[v] = [LT⁻¹]

Velocity

[a] = [LT⁻²]

Acceleration

[F] = [MLT⁻²]

Force

[W] = [ML²T⁻²]

Work / Energy

[P] = [ML²T⁻³]

Power

[P] = [ML⁻¹T⁻²]

Pressure / Stress

[G] = [M⁻¹L³T⁻²]

Gravitational constant

[LHS] = [RHS]

Principle of homogeneity

n₁u₁ = n₂u₂

Unit conversion relation

T = k·l^(1/2)·g^(-1/2) = 2π√(l/g)

Simple pendulum period

Round result to the fewest decimal places among the terms

Addition/subtraction rule

Round result to the fewest significant figures among the terms

Multiplication/division rule

Round to the nearest even preceding digit

Rounding a dropped '5'

a_mean = (a₁ + a₂ + ... + aₙ) / n

Mean value

Δaᵢ = |a_mean − aᵢ|

Absolute error

Δa_mean = (Σ|Δaᵢ|) / n

Mean absolute error

Δa_mean / a_mean

Relative error

(Δa_mean / a_mean) × 100%

Percentage error

Z = A ± B ⇒ ΔZ = ΔA + ΔB

Error in sum/difference

Z = AB ⇒ ΔZ/Z = ΔA/A + ΔB/B

Error in product

Z = A/B ⇒ ΔZ/Z = ΔA/A + ΔB/B

Error in quotient

Z = AᵖBᵠ/Cʳ ⇒ ΔZ/Z = p(ΔA/A) + q(ΔB/B) + r(ΔC/C)

Error in power-law relation

Test yourself

Tap an answer to check it instantly — you'll see why it's right, and what to revise if it isn't.

0 correct · 0/12 answered

Q1 SI Units easy

Which of the following is a fundamental (SI base) unit?

Q2 Dimensional Formulae easy

What is the dimensional formula of force?

Q3 Dimensional Analysis medium

From F = Gm₁m₂/r², the dimensional formula of the gravitational constant G is:

Q4 Significant Figures medium

The number of significant figures in the measurement 0.00620 is:

Q5 Arithmetic with Significant Figures medium

The sum of the lengths 4.237 m, 2.1 m and 0.55 m, expressed to the correct number of significant figures, is:

Q6 Arithmetic with Significant Figures medium

The product 4.2 × 1.70, rounded to the correct number of significant figures, is:

Q7 Dimensional Analysis medium

Dimensional analysis can be used to check the correctness of a physical equation because, in a valid equation:

Q8 Dimensional Analysis hard

For a simple pendulum, assuming T ∝ lᵃgᵇ, dimensional analysis gives:

Q9 Errors in Measurement medium

In five measurements, the mean time period of a pendulum was 2.62 s with a mean absolute error of 0.11 s. The percentage error in the measurement is closest to:

Q10 Error Propagation medium

A resistance R = V/I is calculated from a voltage measured with 2% error and a current measured with 10% error. The percentage error in R is:

Q11 Accuracy and Precision easy

A set of repeated measurements is said to be precise if:

Q12 Dimensions easy

Which of the following physical quantities is dimensionless?

NCERT solutions & previous-year questions

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NCERT questions 6

1 A calorie is a unit of heat energy and it equals 4.2 J, where 1 J = 1 kg m² s⁻². Suppose a new system of units is used in which the unit of mass equals α kg, the unit of length equals β m and the unit of time equals γ s. Show that in this new system, 1 calorie equals 4.2 α⁻¹β⁻²γ² new units of energy.Dimensional Analysis (Unit Conversion)

Energy has the dimensional formula [ML²T^-2]. Using the general relation for converting a physical quantity between two systems of units, n₁u₁ = n₂u₂, with the powers a = 1, b = 2, c = -2 (from the dimensions of energy):

Given: n₁ = 4.2 (value in SI, where M₁ = 1 kg, L₁ = 1 m, T₁ = 1 s); new system has M₂ = α kg, L₂ = β m, T₂ = γ s
Formula: n₂ = n₁ (M₁/M₂)¹(L₁/L₂)²(T₁/T₂)^-2
Substitution: n₂ = 4.2 × (1/α)¹ × (1/β)² × (1/γ)^-2 = 4.2 × (1/α) × (1/β²) × γ²
Result: n₂ = 4.2 α^-1β^-2γ², so 1 calorie = 4.2 α^-1β^-2γ² new units of energy, as required.

2 Round off the following measurements to three significant figures: (i) 7.364 cm (ii) 0.02898 kg (iii) 6.755 × 10² sSignificant Figures

To round to 3 significant figures, keep the first three significant digits and adjust the third based on the digit that follows (round up if it is 5 or more; keep unchanged if it is less than 5; if the dropped digit is exactly 5 with nothing after it, round the preceding digit to the nearest even digit).

(i) 7.364 cm: significant digits are 7, 3, 6, 4. The digit after the third significant figure is 4 (less than 5), so it is dropped unchanged → 7.36 cm
(ii) 0.02898 kg: significant digits are 2, 8, 9, 8 (leading zeros are not significant). The digit after the third significant figure is 8 (5 or more), so the third digit 9 is rounded up to 10, carrying over → 0.0290 kg
(iii) 6.755 × 10² s = 675.5 s: significant digits are 6, 7, 5, 5. The dropped digit is exactly 5, so the preceding digit (5, odd) is rounded up to the nearest even digit, 6 → 676 s, i.e., 6.76 × 10² s

3 The mass and volume of a body are measured to be (5.00 ± 0.05) kg and (2.00 ± 0.02) m³ respectively. Find the density of the body along with its maximum percentage error.Error Propagation

Given: m = 5.00 kg (Δm = 0.05 kg), V = 2.00 m³ (ΔV = 0.02 m³)
Formula: ρ = m/V; for a quotient, Δρ/ρ = Δm/m + ΔV/V
Substitution: ρ = 5.00/2.00 = 2.50 kg m^-3; Δρ/ρ = (0.05/5.00) + (0.02/2.00) = 0.01 + 0.01 = 0.02, i.e., 2%; Δρ = 0.02 × 2.50 = 0.05 kg m^-3
Result: ρ = (2.50 ± 0.05) kg m^-3, with a maximum percentage error of 2%.

4 Using dimensional analysis, check the correctness of Einstein's mass-energy relation E = mc², where E is energy, m is mass and c is the speed of light.Dimensional Analysis

Given: [E] = [ML²T^-2] (energy), [m] = [M], [c] = [LT^-1] (speed)
Formula: compare [E] with [mc²]
Substitution: [mc²] = [M][LT^-1]² = [M][L²T^-2] = [ML²T^-2]
Result: [E] = [mc²] = [ML²T^-2] on both sides, so the equation is dimensionally correct.

5 Distinguish between accuracy and precision of a measurement, with a suitable example.Accuracy and Precision

Accuracy is the closeness of a measured value to the true (actual) value of the quantity being measured. Precision is the closeness among a set of repeated measurements of the same quantity, regardless of how close they are to the true value.

For example, suppose the true length of a rod is 10.00 cm. If five readings taken with one instrument are 9.99, 10.01, 9.98, 10.02 and 10.00 cm, the measurements are both accurate and precise, since they cluster tightly around the true value. If a second, poorly calibrated instrument gives readings of 9.20, 9.22, 9.21, 9.23 and 9.20 cm, these readings are precise (they closely agree with one another) but not accurate, since they are all far from the true value of 10.00 cm — indicating a systematic (zero) error in that instrument.

6 Define absolute error, relative error and percentage error for a set of repeated measurements. How is percentage error related to relative error?Errors in Measurement

For n repeated measurements a₁, a₂, ..., a_n of a quantity, with mean value a_mean:

Absolute error of the i-th reading is Δa_i = |a_mean − a_i|; it carries the same unit as the measured quantity itself.
Relative (fractional) error is the mean absolute error divided by the mean value, Δa_mean/a_mean; it is a pure (dimensionless) number.
Percentage error is the relative error expressed as a percentage: percentage error = (Δa_mean/a_mean) × 100%.

Percentage error is simply the relative error multiplied by 100; expressing it this way allows the precision of measurements of different physical quantities, in different units, to be compared on the same scale.

Previous-year board questions 4

Q1 State the number of significant figures in 4.700 × 10⁵. CBSE 2020 1 mark

In scientific notation N × 10^x, every digit of N is significant. Here N = 4.700 has the digits 4, 7, 0, 0, all significant (the trailing zeros after the decimal point are significant). Therefore, 4.700 × 10⁵ has 4 significant figures.

Q2 Using the method of dimensions, check whether the relation v² = u² + 2as is dimensionally correct, where v and u are velocities, a is acceleration and s is displacement. CBSE 2022 2 marks

Given: [v] = [u] = [LT^-1], [a] = [LT^-2], [s] = [L]
Formula: compare the dimensions of each term: [v²], [u²], [2as]
Substitution: [v²] = [L²T^-2]; [u²] = [L²T^-2]; [2as] = [LT^-2][L] = [L²T^-2] (the number 2 is dimensionless)
Result: LHS = [L²T^-2] and RHS = [L²T^-2] + [L²T^-2] = [L²T^-2]; since LHS = RHS, the equation is dimensionally correct.

Q3 Using dimensional analysis, derive an expression for the time period T of a simple pendulum, given that it may depend on the mass m of the bob, the length l of the pendulum, and the acceleration due to gravity g. CBSE 2023 3 marks

Given: Let T = k·m^al^bg^c, where k is a dimensionless constant, [T] = [T], [m] = [M], [l] = [L], [g] = [LT^-2]
Formula: [M⁰L⁰T¹] = [M]^a[L]^b[LT^-2]^c = [M^aL^b+cT^-2c]
Substitution: equating powers: for M, a = 0; for T, -2c = 1 ⇒ c = -½; for L, b + c = 0 ⇒ b = ½
Result: T = k·m⁰l^1/2g^-1/2 = k√(l/g). The mass exponent works out to zero automatically, showing that the period does not depend on the mass of the bob; a full derivation gives k = 2π, so T = 2π√(l/g).

Q4 (a) Distinguish between systematic errors and random errors, giving one example of each. (b) The diameter of a wire, measured with a screw gauge in five trials, was found to be 1.34 mm, 1.38 mm, 1.35 mm, 1.32 mm and 1.36 mm. Calculate the mean absolute error and the percentage error in the measurement. CBSE 2024 5 marks

(a) Systematic errors occur in a definite direction (always positive or always negative) and arise from causes such as instrumental defects (e.g., a zero error in a screw gauge), imperfect experimental technique, or personal bias of the observer; they can be reduced by proper calibration and correction. Random errors occur irregularly, sometimes positive and sometimes negative, due to unpredictable fluctuations in experimental conditions (e.g., small variations in an observer's reaction time while using a stopwatch); they are reduced, though never fully eliminated, by taking a large number of readings and computing the mean.

(b)

Given: readings 1.34, 1.38, 1.35, 1.32, 1.36 mm
Formula: a_mean = Σa_i/n; Δa_mean = Σ|a_mean − a_i|/n; % error = (Δa_mean/a_mean) × 100%
Substitution: a_mean = (1.34+1.38+1.35+1.32+1.36)/5 = 6.75/5 = 1.35 mm. Absolute errors: |1.35−1.34| = 0.01, |1.35−1.38| = 0.03, |1.35−1.35| = 0.00, |1.35−1.32| = 0.03, |1.35−1.36| = 0.01 mm; sum = 0.08 mm, so Δa_mean = 0.08/5 = 0.016 mm ≈ 0.02 mm
Result: mean diameter = (1.35 ± 0.02) mm; percentage error = (0.016/1.35) × 100% ≈ 1.2%

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