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Master Study Module: Units, Dimensions, and Error Analysis
Target: JEE Main & Advanced (AIR < 500) Physics Core Faculty
- 1 Unit Systems and Foundations Contents
- 1.1 First Principles: What is a Physical Quantity?
- 1.2 The SI System (International System of Units)
- 1.3 Supplementary Units
- 2 Theory of Dimensions & Derivations
- 2.1 Step-by-Step Derivations of Advanced Quantities
- 3 Dimensional Analysis & Homogeneity
- 3.1 Transcendental Functions Rule
- 3.2 System Transformations & Conversions
- 3.3 Limitations of Dimensional Analysis
- 4 Premium High-Frequency Identity Groupings
- 5 Phase A: 20 Basic Practice Questions
- 6 Phase B: 20 JEE Main Level Questions
- 7 Phase C: 20 JEE Advanced Level Questions
- 8 Advanced Multi-Concept Question Formats
- 8.1 Assertion-Reason Type
- 8.2 Integer-Type Questions
- 8.3 Multiple Correct Type
- 9 Comprehensive Formula Sheet
- 10 One Final Challenge Test
1 Unit Systems and Foundations
1.1 First Principles: What is a Physical Quantity?
Any quantity that can be measured and by which the laws of physics can be described is called a Physical Quantity (PQ). Every measurement consists of two parts: a numerical value (n) and a unit (u). P Q = n × u (1)
Because the actual magnitude of a physical quantity is invariant regardless of the system of units chosen:
n 1 u 1 = n 2 u 2 =⇒ n ∝
u
Visualizing Invariance
If you measure the length of a rod, it remains the same whether you express it as 1 meter or 100 centimeters. As the unit size decreases (m → cm), the numerical value increases (1 → 100).
1.2 The SI System (International System of Units)
The International System of Units (SI) defines seven fundamental (base) quantities and two supple- mentary quantities.
Table 1: SI Base Quantities Base Physical Quantity SI Unit Symbol Technical Grounding Reference Mass kilogram kg Fixed via Planck constant h = 6. 62607015 × 10 −^34 J · s Length meter m Distance light travels in vacuum in 1/ 299 , 792 , 458 s Time second s Hyperfine transition periods of Cesium-133 atom Electric Current ampere A Elementary charge e = 1. 602176634 × 10 −^19 C Thermodynamic Temp. kelvin K Boltzmann constant kB = 1. 380649 × 10 −^23 J/K Amount of Substance mole mol Avogadro constant NA = 6. 02214076 × 1023 mol−^1 Luminous Intensity candela cd Monochromatic source at 540 × 1012 Hz
1.3 Supplementary Units
- Plane Angle (θ): Measured in radian (rad). Defined as arc length divided by radius: dθ = dsr.
- Solid Angle (Ω): Measured in steradian (sr). Defined as area divided by radius squared: dΩ = dAr 2.
CRITICAL CONCEPTUAL TRAP (JEE Advanced)
Supplementary quantities are dimensionless, but they do possess units. Therefore, having no dimensions does not mean a quantity is unitless. However, a completely unitless quantity must always be dimensionless (e.g., refractive index).
2 Theory of Dimensions & Derivations
The dimensions of a physical quantity are the powers to which the base quantities must be raised to represent that quantity. We use mechanical formulas as [M aLbT c].
4 Premium High-Frequency Identity Groupings
- Category 1 ([T −^1 ]): Frequency (ν), Angular Velocity (ω), Velocity Gradient (dv/dx), Decay Constant (λ).
- Category 2 ([M L^2 T −^2 ]): Energy, Work, Torque, Heat, P ∆V , I^2 Rt, q 2 2 C ,^
1 2 LI
- Category 3 ([M L−^1 T −^2 ]): Pressure, Stress, Young’s/Bulk/Shear Modulus, Energy Density.
- Category 4 ([T 1 ]): LR , RC,
LC.
- Category 5 ([LT −^1 ]): Speed of light √μ^10 ε 0 , EB ,
q T μ ,
q Y ρ.
5 Phase A: 20 Basic Practice Questions
- Find the dimensional formula of linear momentum. Solution: p = m · v =⇒ [p] = [M ][LT −^1 ] = [M LT −^1 ]
- Find the dimensions of Kinetic Energy. Solution: KE = 12 mv^2 =⇒ [KE] = [M ][LT −^1 ]^2 = [M L^2 T −^2 ]
- Deduce the dimensional formula of Surface Tension (T = F/l). Solution: [T ] = [[Fl]^ ] = [M LT^
− (^2) ] [L] = [M L
0 T − 2 ]
- State the dimensions of refractive index. Solution: μ = cv =⇒ [μ] = [LT^
− (^1) ] [LT −^1 ] = [M^
(^0) L (^0) T 0 ] (Dimensionless)
- Find the dimensions of power (P ). Solution: P = WorkTime = [M L
(^2) T − (^2) ] [T ] = [M L
2 T − 3 ]
- What are the dimensions of frequency? Solution: f = (^) T^1 =⇒ [f ] = [T −^1 ]
- Determine the dimensional formula for impulse. Solution: I = F · ∆t =⇒ [I] = [M LT −^2 ][T ] = [M LT −^1 ]
- Find the dimensions of Angle (θ). Solution: θ = (^) RadiusArc =⇒ [θ] = [[LL]] = [M 0 L^0 T 0 ]
- Determine the dimensional formula of density. Solution: ρ = (^) VolumeMass =⇒ [ρ] = (^) [[ML 3 ]] = [M L−^3 T 0 ]
- Find the dimensions of specific heat capacity (c) where Q = mc∆θ.
Solution: [c] = (^) [m[][∆Q]θ] = [M L
(^2) T − (^2) ] [M ][θ] = [M^
(^0) L (^2) T − (^2) θ− (^1) ]
- Find the dimensions of Planck’s constant (h) using E = hν.
Solution: [h] = [[Eν]] = [M L
(^2) T − (^2) ] [T −^1 ] = [M L
2 T − 1 ]
- Find the dimensions of Torque (τ =r ⃗× F⃗ ). Solution: [τ ] = [L][M LT −^2 ] = [M L^2 T −^2 ]
- Deduce the dimension of angular momentum (L = mvr). Solution: [L] = [M ][LT −^1 ][L] = [M L^2 T −^1 ]
- Determine the dimensions of electrical resistance (R).
Solution: R = VI = WqI = [M L
(^2) T − (^2) ] [AT ][A] = [M L
2 T − 3 A− 2 ]
- Find the dimensional formula of Electric Potential (V ).
Solution: [V ] = (^) [Charge][Work] = [M L
(^2) T − (^2) ] [AT ] = [M L
2 T − 3 A− 1 ]
- Determine the dimensions of capacitance (C). Solution: C = (^) Vq =⇒ [C] = (^) [M L 2 [TAT −^3 ]A− (^1) ] = [M −^1 L−^2 T 4 A^2 ]
- Check if the formula v^2 = u^2 + 2as is dimensionally correct. Solution: [v^2 ] = [L^2 T −^2 ], [u^2 ] = [L^2 T −^2 ], [2as] = [LT −^2 ][L] = [L^2 T −^2 ]. All match.
- Convert a force of 1 Newton into Dynes.
Solution: n 2 = 1 ×
1 kg 1 g
×
(^) 1 m 1 cm
×
(^) 1 s 1 s
= 10^3 × 102 = 10^5 Dynes.
- Find the dimensions of coefficient of thermal conductivity (K) from H = KAl∆ θt.
Solution: [K] = (^) [A[][∆H][θl]][t] = [M L
(^2) T − (^2) ][L] [L^2 ][θ][T ] = [M LT^
− (^3) θ− (^1) ]
6 Phase B: 20 JEE Main Level Questions
- The velocity of a particle depends on time as v = at + (^) t+bc. Find the dimensions of a, b, and c. Solution: By homogeneity, [c] = [t] = [T ]. Next, [at] = [v] =⇒ [a] = [LT −^2 ]. Finally,
(^) b t+c
[v] =⇒ (^) [[Tb] ] = [LT −^1 ] =⇒ [b] = [L].
- Find the dimensions of a/b in the van der Waals equation:
P + (^) Va 2
(V − b) = RT. Solution:
(^) a V 2
= [P ] =⇒ [a] = [M L−^1 T −^2 ][L^6 ] = [M L^5 T −^2 ]. Since [b] = [V ] = [L^3 ], we get [a/b] = [M L^2 T −^2 ].
- If Force (F ), Velocity (V ), and Time (T ) are chosen as fundamental units, find the dimensional formula of mass. Solution: [M ] = [F ]a[V ]b[T ]c^ =⇒ [M 1 L^0 T 0 ] = [M LT −^2 ]a[LT −^1 ]b[T ]c. Solving gives a = 1, b = − 1 , c = 1 =⇒ [F 1 V −^1 T 1 ].
- Find the dimensions of 12 ε 0 E^2 (Energy Density).
Solution: Energy Density = (^) VolumeEnergy = [M L
(^2) T − (^2) ] [L^3 ] = [M L
− 1 T − 2 ].
- Find the dimensions of B 2 2 μ 0 (Magnetic Energy Density). Solution: Magnetic energy density shares the exact same mechanical dimension as electrical energy density: [M L−^1 T −^2 ].
- Find the dimensions of Electrical Conductivity (σ). Solution: σ = (^1) ρ = (^) EJ. Doing the division fields: [M −^1 L−^3 T 3 A^2 ].
- If the density of a cube is calculated by measuring its mass and side length, and the errors are 1 .5% and 1%, find the maximum error in density. Solution: ∆ρρ = ∆MM + 3 ∆LL = 1.5% + 3(1%) = 4.5%.
- Find the dimensions of RL where R is resistance and L is inductance. Solution: The inductive time constant is τ = LR. Thus, [ RL ] = [T −^1 ].
- The position of a particle at time t is given by x(t) = v α^0 (1 − e−αt). Find the dimensions of α and v 0. Solution: Exponent [αt] = [1] =⇒ [α] = [T −^1 ]. Then
(^) v 0 α
= [L] =⇒ [v 0 ] = [LT −^1 ].
- Find the dimensions of the coefficient of mutual inductance (M ). Solution: E = −M dIdt =⇒ [M ] = [M L
(^2) T − (^3) A− (^1) ] [AT −^1 ] = [M L
2 T − 2 A− 2 ].
- In a new system of units, the unit of mass is 10 kg, length is 100 m, and time is 1 minute. What is the value of 1 Joule in this system? Solution: n 2 = 1 ×
10
×
100
×
60
- Find the dimensions of Solar Constant (energy received per unit area per unit time). Solution: S = (^) AreaEnergy×Time = [M L
(^2) T − (^2) ] [L^2 ][T ] = [M T^
− 3 ].
- If X = 3Y Z^2 , find the dimensions of Y if X is capacitance and Z is magnetic field induction. Solution: [Y ] = (^) [[BC]] 2 = [M^
− (^1) L− (^2) T 4 A (^2) ] [M T −^2 A−^1 ]^2 = [M^
− 3 L− 2 T 8 A 4 ].
- Find the dimensions of
q g λ where^ g^ is acceleration due to gravity and^ λ^ is wavelength. Solution:
hq g λ
i
q [LT −^2 ] [L] = [T^
− 1 ].
- Express impedance (Z) in terms of fundamental SI parameters. Solution: Impedance has the identical dimension as resistance: [M L^2 T −^3 A−^2 ].
- A quantity z is given by z = A (^4) B 1 / 3 CD^3 /^2. Find the relative error in^ z. Solution: ∆zz = 4 ∆AA + 13 ∆BB + ∆CC + 32 ∆DD.
- Find the dimension of 2 hπ. Solution: Constants don’t shift dimensions. [h] = [M L^2 T −^1 ].
- The force operating on a body moving in a fluid is given by F = Av^2 + Be−kt. Find the dimensions of A. Solution: [Av^2 ] = [F ] =⇒ [A][L^2 T −^2 ] = [M LT −^2 ] =⇒ [A] = [M L−^1 ].
- Find the dimensions of Poynting vector ( S⃗ = (^) μ^10 E⃗ × B⃗ ).
Solution: It scales as energy flux: [S] = [M T −^3 ].
- Which two have identical dimensions? (i) Force Constant, (ii) Surface Tension, (iii) Spring Constant. Solution: All three are force per unit length, yielding identical dimensions: [M T −^2 ].
- What is the dimension of magnetic susceptibility (χm)? Solution: χm = MH = A/mA/m = [M 0 L^0 T 0 ] (Dimensionless).
- Find the dimension of (^) ∂B/∂tE where E is electric field and B is magnetic field.
Solution: From Faraday’s law, [[EL]] =
∂B
∂t
=⇒ the ratio yields length [L].
- Calculate the dimensions of displacement current density (Jd). Solution: It matches basic current density: [AL−^2 ].
- If critical velocity vc depends on viscosity η, density ρ, and radius r, derive the grouping. Solution: Proportionality matching yields vc ∝ (^) ρrη , leading directly to the Reynolds Number formu- lation.
8 Advanced Multi-Concept Question Formats
8.1 Assertion-Reason Type
Q51.
- Assertion: A supplementary quantity like an angle has a unit but no dimensions.
- Reason: Quantities with units must always have dimensions.
Solution: Choice C. The assertion is accurate (radians have no dimension), but the reason is false since supplementary units provide a clear counter-demonstration.
8.2 Integer-Type Questions
Q52. In the formula X = 3αβ^2 , X has dimensions of energy density and β has dimensions of magnetic field intensity. If the dimension of α is [M xLyT z^ Aw], find the integer value of x + y + z + w. Solution: [M L−^1 T −^2 ] = [α][M T −^2 A−^1 ]^2 =⇒ [α] = [M −^1 L−^1 T 2 A^2 ]. Summing the components: − 1 − 1 + 2 + 2 = 2. Answer: 2
8.3 Multiple Correct Type
Q53. Identify pairs that share exactly identical dimensions:
- (A) Torque and Work
- (B) Stress and Young’s Modulus
- (C) Light year and Wavelength
Solution: (A, B, C) All choices correctly specify matching internal pairs.
9 Comprehensive Formula Sheet
Physical Quantity Standard Formula Dimensional Formula Universal Gravitation (G) F = G m^1 rm 2 2 [M −^1 L^3 T −^2 ] Planck’s Constant (h) E = hν [M L^2 T −^1 ] Coefficient of Viscosity (η) F = 6πηrv [M L−^1 T −^1 ] Permittivity (ε 0 ) F = (^4) πε^10 q^1 rq 22 [M −^1 L−^3 T 4 A^2 ] Boltzmann Constant (kB ) P V = N kB T [M L^2 T −^2 θ−^1 ]
10 One Final Challenge Test
- Find dimensions of a · b in P = b−x 2 at where^ P^ is power. Solution: [b] = [L^2 ]. Then
L 2
aT
= [M L^2 T −^3 ] =⇒ [a] = [M −^1 T 2 ]. Consequently, [a · b] = [M −^1 L^2 T 2 ].
- Determine dimensions of P = EJ 2 M 5 G^2 where^ J^ is angular momentum. Solution: Substituting terms shows that all elements cancel perfectly out: [M 0 L^0 T 0 ].
