Chapter 1: Units & Measurement | Class 11 Physics | AAI ATC Exam Prep

📌 Chapter Overview

🌐 SI System & 7 Base Units

🔢 Significant Figures & Rules

➕ Arithmetic with Sig. Figs.

📐 Dimensions of Physical Quantities

✅ Dimensional Consistency Check

🔬 Deriving Relations by Dimensions

The International System of Units (SI)

Every measurement involves comparison with a reference standard called a unit. The internationally accepted system is the Système Internationale d'Unites (SI), adopted in 1971 and revised in 2018 by BIPM.

Earlier Systems of Units

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

7 SI Base Units

Base Quantity	Unit Name	Symbol	Key Definition Basis
Length	metre	m	Speed of light in vacuum (c)
Mass	kilogram	kg	Planck constant (h)
Time	second	s	Caesium-133 atom frequency
Electric current	ampere	A	Elementary charge (e)
Temperature	kelvin	K	Boltzmann constant (k)
Amount of substance	mole	mol	Avogadro constant (Nₐ)
Luminous intensity	candela	cd	Luminous efficacy of 540 THz radiation

      📌 Two supplementary units — radian (rad) for plane angle and steradian (sr) for solid angle — are dimensionless quantities.

      Plane angle: dθ = ds/r | Solid angle: dΩ = dA/r²

🎯 1 mole = 6.02214076 × 10²³ elementary entities (Avogadro number). The type of entity (atom, molecule, ion, electron) must always be specified when using mole.

🎯 Practice MCQs — SI Units

Q1 Which SI base unit is defined using the Planck constant h = 6.626 × 10⁻³⁴ J·s?

A Second

B Metre

C Kilogram

D Ampere

✅ The kilogram is defined by fixing the numerical value of Planck's constant h = 6.62607015 × 10⁻³⁴ J·s (= kg·m²·s⁻¹).

Q2 The solid angle dΩ is defined as dA/r². If an area of 4 m² is intercepted at radius 2 m from the apex, the solid angle is:

A 2 sr

B 1 sr

C 8 sr

D 0.5 sr

✅ dΩ = dA/r² = 4/(2²) = 4/4 = 1 steradian

Q3 A vehicle moves at 18 km/h. What is its speed in m/s?

A 18 m/s

B 10 m/s

C 5 m/s

D 0.3 m/s

✅ 18 km/h = 18 × (1000/3600) m/s = 18/3.6 = 5 m/s
Distance covered in 1 s = 5 m.

Significant Figures

Significant figures = all certain digits + the first uncertain digit in a measurement. They indicate the precision of measurement — NOT affected by change of units.

📋 Rules for Counting Significant Figures

All non-zero digits are significant (e.g., 2308 → 4 s.f.)
Zeros between two non-zero digits are always significant (e.g., 3007 → 4 s.f.)
Leading zeros (zeros before first non-zero digit) are NOT significant (e.g., 0.00230 → 3 s.f.)
Trailing zeros without decimal point are NOT significant (e.g., 1230 → 3 s.f.)
Trailing zeros with decimal point ARE significant (e.g., 3.500 → 4 s.f.)
In scientific notation a × 10ᵇ, all digits in 'a' are significant

Order of Magnitude

Express number as a × 10ᵇ (1 ≤ a < 10). Round a → 1 if a ≤ 5, or → 10 if a > 5. Then the number ≈ 10ᵇ, where b is the order of magnitude.

Earth diameter = 1.28 × 10⁷ m → Order of magnitude = 7
H atom diameter = 1.06 × 10⁻¹⁰ m → Order of magnitude = −10
Earth is 17 orders of magnitude larger than H atom

Rounding Off Rules

      🔢 If digit to be dropped > 5 → raise preceding digit by 1

      🔢 If digit to be dropped < 5 → leave preceding digit unchanged

      🔢 If digit to be dropped = 5 → if preceding digit is even, drop it; if odd, raise by 1

🎯 Practice MCQs — Significant Figures

Q4 How many significant figures are in the measurement 0.0006032 m²?

A 7

B 5

C 4

D 2

✅ Leading zeros (0.000) are not significant. Digits 6, 0, 3, 2 → 4 significant figures.
(The zero between 6 and 3 is significant as it lies between two non-zero digits.)

Q5 The mass of a body is measured as 5.74 g and volume as 1.2 cm³. The density expressed to correct significant figures is:

A 4.783 g/cm³

B 4.78 g/cm³

C 4.8 g/cm³

D 5.0 g/cm³

✅ In division, result has as many s.f. as the least precise input.
Mass: 3 s.f. | Volume: 2 s.f. → result must have 2 s.f.
5.74/1.2 = 4.7833... → rounded to 2 s.f. = 4.8 g/cm³

Q6 The sum of 436.32 g + 227.2 g + 0.301 g rounded to correct significant figures is:

A 663.821 g

B 664 g

C 663.8 g

D 663.82 g

✅ In addition/subtraction → retain as many decimal places as the least precise number.
227.2 g has 1 decimal place (least). Sum = 663.821 → rounded to 1 decimal = 663.8 g

Dimensions of Physical Quantities

The dimensions of a physical quantity are the powers to which the base quantities (M, L, T, A, K, mol, cd) must be raised to represent it. Dimensions are written in square brackets [ ].

Key Dimensional Formulae

Physical Quantity	Dimensional Formula	SI Unit
Length	[M⁰ L¹ T⁰]	m
Area	[M⁰ L² T⁰]	m²
Volume	[M⁰ L³ T⁰]	m³
Velocity / Speed	[M⁰ L T⁻¹]	m/s
Acceleration	[M⁰ L T⁻²]	m/s²
Force	[M L T⁻²]	N (newton)
Energy / Work	[M L² T⁻²]	J (joule)
Power	[M L² T⁻³]	W (watt)
Pressure	[M L⁻¹ T⁻²]	Pa (pascal)
Mass Density	[M L⁻³ T⁰]	kg/m³
Momentum	[M L T⁻¹]	kg·m/s
Frequency	[M⁰ L⁰ T⁻¹]	Hz

      💡 Magnitudes are NOT considered in dimensional analysis — only the quality (type) of the quantity matters. Velocity, speed, change in velocity — all have dimension [L T⁻¹].
    

⚠️ Dimensionless quantities: angles (radian), refractive index, relative density, strain. Arguments of trig, log, and exponential functions must ALWAYS be dimensionless.

🎯 Practice MCQs — Dimensions

Q7 The dimensional formula of pressure is:

A [M L² T⁻²]

B [M L T⁻²]

C [M L⁻¹ T⁻²]

D [M⁰ L T⁻¹]

✅ Pressure = Force/Area = [M L T⁻²] / [L²] = [M L⁻¹ T⁻²]

Q8 Which of the following pairs has the same dimensions?

A Force and Energy

B Power and Force

C Work and Energy

D Pressure and Force

✅ Work = Force × displacement = [M L T⁻²][L] = [M L² T⁻²]
Energy is also [M L² T⁻²]. Both Work and Energy share the same dimensions.

Q9 If G = 6.67 × 10⁻¹¹ N·m²/kg², what is its value in CGS units (cm³·s⁻²·g⁻¹)?

A 6.67 × 10⁻¹¹ cm³·s⁻²·g⁻¹

B 6.67 × 10⁻⁸ cm³·s⁻²·g⁻¹

C 6.67 × 10⁻⁸ cm³·s⁻²·g⁻¹

D 6.67 × 10⁻⁵ cm³·s⁻²·g⁻¹

✅ [G] = [M⁻¹ L³ T⁻²]
1 N·m²/kg² = 1 kg·m·s⁻² × m² × kg⁻² = m³·kg⁻¹·s⁻²
= (100 cm)³ × (1000 g)⁻¹ × s⁻² = 10⁶/10³ cm³·g⁻¹·s⁻² = 10³
G = 6.67×10⁻¹¹ × 10³ = 6.67 × 10⁻⁸ cm³·s⁻²·g⁻¹

Dimensional Consistency of Equations

The Principle of Homogeneity of Dimensions: Every term in a physical equation must have the same dimensions. Only quantities with same dimensions can be added or subtracted.

How to Check an Equation

Write the dimensional formula of every term → if all terms have identical dimensions → equation is dimensionally consistent.

Example: x = x₀ + v₀t + ½at² [x] = [L] [x₀] = [L] [v₀t] = [LT⁻¹][T] = [L] [½at²] = [LT⁻²][T²] = [L] → All terms have [L] — dimensionally consistent ✓

      ⚠️ Important limitations of dimensional analysis:

      ① Cannot determine dimensionless constants (e.g., π, ½, 2π)

      ② Cannot distinguish between quantities with same dimensions

      ③ A dimensionally correct equation may still be wrong, but a dimensionally incorrect equation is always wrong

🎯 Velocity cannot be added to force. Electric current cannot be subtracted from temperature. Only same-dimension quantities can be combined by addition or subtraction.

🎯 Practice MCQs — Dimensional Consistency

Q10 The equation ½mv² = mgh is dimensionally checked. The dimensions of LHS [½mv²] are:

A [M L T⁻¹]

B [M L T⁻²]

C [M L² T⁻²]

D [M² L² T⁻²]

✅ [½mv²] = [M][L T⁻¹]² = [M][L² T⁻²] = [M L² T⁻²]
RHS: [mgh] = [M][L T⁻²][L] = [M L² T⁻²] ✓ — dimensionally consistent.

Q11 Which formula for kinetic energy (K) can be ruled out on dimensional grounds? (K has dimensions [M L² T⁻²])

A K = m²v³ (dimensions [M² L³ T⁻³])

B K = ½mv²

C K = (3/16)mv²

D Both B and C

✅ K must have [M L² T⁻²].
K = m²v³ → [M²][L³ T⁻³] = [M² L³ T⁻³] ≠ [M L² T⁻²] → ruled out
½mv² and (3/16)mv² both give [M L² T⁻²] ✓ (correct dimensionally).

Q12 A formula K = ½mv² + ma is proposed. Why is this dimensionally incorrect?

A Both terms have different units only

B ½mv² has [ML²T⁻²] and ma has [MLT⁻²] — different dimensions cannot be added

C ma is not a form of energy

D The formula has no constant

✅ [½mv²] = [M L² T⁻²] (energy) vs [ma] = [M][L T⁻²] = [M L T⁻²] (force).
Adding energy and force is dimensionally forbidden — they have different dimensions.

Deriving Relations by Dimensional Analysis

If a physical quantity depends on other quantities, we can derive the relationship using dimensional analysis by assuming a product form.

Method — Step by Step

Example: Time period T of simple pendulum depends on l, m, g Assume: T = k · lˣ · mʸ · gᶻ (k = dimensionless constant) [T¹] = [L]ˣ [M]ʸ [L T⁻²]ᶻ [M⁰ L⁰ T¹] = Lˣ⁺ᶻ · T⁻²ᶻ · Mʸ Equating: y = 0; −2z = 1 → z = −½; x + z = 0 → x = ½ T = k · l^(½) · g^(−½) = k√(l/g) Actual value: k = 2π → T = 2π√(l/g) ✓

      📌 Dimensional analysis can determine ONLY up to 3 unknown exponents (corresponding to M, L, T). It cannot find dimensionless constants (k). It also cannot distinguish between quantities with same dimensions.
    

Unit Conversion Using Dimensional Analysis

To convert a quantity from one system to another: write dimensional formula, express each base unit as a ratio, and compute the numerical factor.

1 kg·m²·s⁻² = ? g·cm²·s⁻² = 1 × (1000 g) × (100 cm)² × (s)⁻² = 1000 × 10000 g·cm²·s⁻² = 10⁷ g·cm²·s⁻² (i.e., 1 J = 10⁷ erg)

🎯 Practice MCQs — Deriving Relations

Q13 The time period T of oscillation of a mass m on a spring depends on m and spring constant k (dimensions [M T⁻²]). By dimensional analysis, T ∝:

A √(k/m)

B √(m/k)

C m/k

D mk

✅ T = C · mˣ · kʸ → [T] = [M]ˣ [M T⁻²]ʸ = M^(x+y) · T^(−2y)
−2y = 1 → y = −½; x + y = 0 → x = ½
T ∝ m^(½) · k^(−½) = √(m/k)

Q14 A calorie = 4.2 J. In a system where unit of mass = α kg, length = β m, time = γ s, the magnitude of 1 calorie is:

A 4.2 α β γ

B 4.2 α⁻¹ β⁻² γ²

C 4.2 α β⁻² γ

D 4.2 α β² γ⁻²

✅ Energy [E] = [M L² T⁻²]. In new system: n = n₁(M₁/M₂)¹(L₁/L₂)²(T₁/T₂)⁻²
= 4.2 × (1 kg/α kg) × (1 m/β m)² × (1 s/γ s)⁻²
= 4.2 × α⁻¹ × β⁻² × γ² = 4.2 α⁻¹ β⁻² γ²

Q15 The relativistic mass formula is m = m₀/√(1 − v²/c²). Dimensionally, why must it be v²/c² and not v²?

A c is a fundamental constant

B v is always small

C The argument of √ must be dimensionless — v²/c² makes it [L²T⁻²]/[L²T⁻²] = dimensionless

D √v² would not make physical sense

✅ Arguments of all mathematical functions (√, sin, log, exp) must be dimensionless.
v² alone has [L² T⁻²] — not dimensionless. v²/c² = [L²T⁻²]/[L²T⁻²] = dimensionless ✓
So the correct formula must have v/c → confirmed by dimensional analysis.

Measurement Applications & Error Analysis

Arithmetic Rules with Significant Figures

Multiplication/Division → result has s.f. equal to the LEAST s.f. in inputs Addition/Subtraction → result has decimal places equal to the LEAST decimal places in inputs

Error Propagation

For a product or quotient z = x × y: the relative error in z is the sum of relative errors in x and y.

Δz/z = Δx/x + Δy/y (for multiplication/division) Δz = Δx + Δy (for addition/subtraction)

Precision of Measuring Devices

Instrument	Least Count / Precision
Vernier Callipers (20 div)	1/20 mm = 0.05 mm
Screw Gauge (pitch 1mm, 100 div)	1/100 mm = 0.01 mm
Optical instrument (wavelength of light)	~10⁻⁷ m (most precise)

      📌 Linear magnification of projector: M = √(A_screen / A_slide) = √(1.55 m² / 1.75 cm²)

      📌 Thickness of hair: average width / magnification = 3.5 mm / 100 = 0.035 mm

      📌 Scientific notation avoids all ambiguity about trailing zeros and significant figures.

🌟 Density of Sun = Mass/Volume = 2×10³⁰ / (4π/3 × (7×10⁸)³) ≈ 1.4 × 10³ kg/m³ — in the range of densities of liquids/solids (not gases), despite being a plasma!

🎯 Practice MCQs — Measurement Applications

Q16 A rectangular sheet has length 4.234 m, breadth 1.005 m, and thickness 0.0201 m. Its volume to correct significant figures is:

A 0.0855 m³

B 0.0855 m³ (3 s.f.)

C 0.08551 m³

D 0.086 m³

✅ Volume = 4.234 × 1.005 × 0.0201 = 0.085528... m³
Least s.f. in inputs: thickness 0.0201 has 3 s.f. → result: 3 s.f.
V = 0.0855 m³

Q17 A student measures human hair through a 100× microscope. Average width seen = 3.5 mm. The actual thickness of the hair is:

A 3.5 mm

B 350 mm

C 0.035 mm

D 0.35 mm

✅ Actual thickness = observed width / magnification = 3.5 mm / 100 = 0.035 mm

Q18 The total atomic volume (m³) of 1 mole of hydrogen atoms, given radius of H atom = 0.5 Å = 0.5 × 10⁻¹⁰ m, is approximately:

A 7.9 × 10⁻⁷ m³

B 3.15 × 10⁻⁷ m³

C 6.02 × 10²³ m³

D 1.0 × 10⁻³⁰ m³

✅ Volume of 1 H atom = (4/3)π r³ = (4/3)π(0.5×10⁻¹⁰)³
= (4/3)π × 1.25×10⁻³¹ = 5.24×10⁻³¹ m³
Total volume = 6.023×10²³ × 5.24×10⁻³¹ ≈ 3.15 × 10⁻⁷ m³

🚀 Master AAI ATC Physics with Aviate Learnings!

Watch full video lectures, join live classes & submit your doubts directly to Kanishk Sir

▶ Subscribe on YouTube 📋 Submit Your Query

📏 Units & Measurement