📐 Dimensional Analysis
Every engineering equation must speak the same dimensional language. Learn to read the M–L–T code behind every quantity, check any formula's validity in seconds, and even derive equations from scratch — the way Rayleigh and Taylor did.
4Topics
20Worked Examples
30+Dimensions
12Practice Qs
📏Units & DimensionsThe M–L–T code
📜History & RelevanceFourier to Mars
⚖️HomogeneityComparing equations
🧪Deriving EquationsMethod of dimensions
⚙️ApplicationsLive calculators
📝PracticeGraded problems
🎯What you will learn
Express any engineering quantity in base dimensions (M, L, T, I, Θ, N, J); test whether equations are dimensionally homogeneous; find the units of unknown constants; and derive relationships between physical quantities using the method of dimensions.
💡 The golden idea: you can only add, subtract or equate quantities that share the same dimensions. Metres plus seconds is meaningless — and every valid equation obeys this.
🧭How to use this resource
1. Read the rules banner at the top of each tab
2. Experiment with the interactive tool
3. Check the key vocabulary panel
4. Follow all 5 worked examples, then practise
📏
Units & Dimensions
Every mechanical quantity is a combination of Mass, Length and Time — its dimensional "recipe".
🔧Dimension Explorer — decode any quantityPick a quantity, then press Explore
Choose a quantity from the list and press Explore to see its dimensions derived step by step from its defining equation…
📖 Key Vocabulary
Base unit — one of the seven SI units everything else is built from (kg, m, s, A, K, mol, cd).
Derived unit — a combination of base units, e.g. the newton (kg m s⁻²).
Dimension — the physical nature of a quantity, independent of the unit chosen: [speed] = LT⁻¹ whether measured in m/s or mph.
Dimensionless — a pure number with dimensions M⁰L⁰T⁰, e.g. strain, Reynolds number, π.
👁️ Visual aid: every quantity on the right is just M, L and T blocks multiplied together with different powers.
Units are a human choice; dimensions are physics. Speed can be m/s, km/h or knots, but its dimensional recipe is always L T⁻¹. To find any quantity's dimensions, write its defining equation and substitute the dimensions of everything on the right side.
📋The Full Engineering Dimension Table
| Quantity | Defining relation | Dimensions | SI unit |
|---|---|---|---|
| Area | length × length | L² | m² |
| Volume | length³ | L³ | m³ |
| Velocity | distance ÷ time | L T⁻¹ | m s⁻¹ |
| Acceleration | velocity ÷ time | L T⁻² | m s⁻² |
| Force | mass × acceleration | M L T⁻² | N = kg m s⁻² |
| Weight | mass × g | M L T⁻² | N |
| Energy / Work | force × distance | M L² T⁻² | J = kg m² s⁻² |
| Power | energy ÷ time | M L² T⁻³ | W = kg m² s⁻³ |
| Pressure / Stress | force ÷ area | M L⁻¹ T⁻² | Pa = kg m⁻¹ s⁻² |
| Density | mass ÷ volume | M L⁻³ | kg m⁻³ |
| Momentum | mass × velocity | M L T⁻¹ | kg m s⁻¹ |
| Impulse | force × time | M L T⁻¹ | N s |
| Torque / Moment | force × distance | M L² T⁻² | N m |
| Angular velocity | angle ÷ time | T⁻¹ | rad s⁻¹ |
| Angular acceleration | angular velocity ÷ time | T⁻² | rad s⁻² |
| Frequency | cycles ÷ time | T⁻¹ | Hz = s⁻¹ |
| Moment of inertia | mass × length² | M L² | kg m² |
| Spring stiffness | force ÷ extension | M T⁻² | N m⁻¹ |
| Dynamic viscosity μ | stress ÷ velocity gradient | M L⁻¹ T⁻¹ | Pa s |
| Kinematic viscosity ν | μ ÷ density | L² T⁻¹ | m² s⁻¹ |
| Surface tension | force ÷ length | M T⁻² | N m⁻¹ |
| Flow rate (volume) | volume ÷ time | L³ T⁻¹ | m³ s⁻¹ |
| Strain | extension ÷ length | 1 (dimensionless) | — |
| Young's modulus | stress ÷ strain | M L⁻¹ T⁻² | Pa |
| Specific heat capacity | energy ÷ (mass × temp.) | L² T⁻² Θ⁻¹ | J kg⁻¹ K⁻¹ |
| Thermal conductivity | power ÷ (length × temp.) | M L T⁻³ Θ⁻¹ | W m⁻¹ K⁻¹ |
| Electric charge | current × time | I T | C = A s |
| Voltage (p.d.) | power ÷ current | M L² T⁻³ I⁻¹ | V |
| Resistance | voltage ÷ current | M L² T⁻³ I⁻² | Ω |
| Gravitational constant G | from F = Gm₁m₂/r² | M⁻¹ L³ T⁻² | N m² kg⁻² |
✏️Worked Example 1 — Straight from the definitionType: Definition
Question Find the dimensions of acceleration.
Define Acceleration = change in velocity ÷ time
Substitute
[a] = [v] ÷ [t] = (L T⁻¹) ÷ TAnswer [a] = L T⁻² — length per time, per time.
🔗Worked Example 2 — Building through a chain of formulasType: Chained
Question Starting from F = ma, find the dimensions of work and then of power.
Force
[F] = [m][a] = M × L T⁻² = M L T⁻²Work
[W] = [F][d] = M L T⁻² × L = M L² T⁻²Power
[P] = [W] ÷ [t] = M L² T⁻² ÷ TAnswer [P] = M L² T⁻³ — each new quantity is built from the last.
➗Worked Example 3 — Quantities defined by divisionType: Ratio
Question Find the dimensions of pressure and show the pascal in base units.
Define Pressure = force ÷ area
Substitute
[p] = M L T⁻² ÷ L² = M L⁻¹ T⁻²Units
1 Pa = 1 kg m⁻¹ s⁻²Answer [p] = M L⁻¹ T⁻² — note the negative index on L: dividing pushes indices down.
🌊Worked Example 4 — Extracting dimensions from a physical lawType: From a law
Question Newton's law of viscosity is τ = μ (dv/dy), where τ is shear stress and dv/dy is the velocity gradient. Find the dimensions of dynamic viscosity μ.
Rearrange
μ = τ ÷ (dv/dy)Pieces
[τ] = M L⁻¹ T⁻² (a stress); [dv/dy] = (L T⁻¹) ÷ L = T⁻¹Combine
[μ] = M L⁻¹ T⁻² ÷ T⁻¹ = M L⁻¹ T⁻¹Answer [μ] = M L⁻¹ T⁻¹, so the SI unit Pa s = kg m⁻¹ s⁻¹.
⭕Worked Example 5 — Proving a quantity is dimensionlessType: Dimensionless
Question The Reynolds number is Re = ρvd/μ. Show that Re is dimensionless.
Numerator
[ρvd] = (M L⁻³)(L T⁻¹)(L) = M L⁻¹ T⁻¹Denominator
[μ] = M L⁻¹ T⁻¹Divide
[Re] = (M L⁻¹ T⁻¹) ÷ (M L⁻¹ T⁻¹) = M⁰L⁰T⁰Answer Re = 1 (dimensionless) — which is exactly why the same Re value means the same flow behaviour at any scale.
🧠 Common slip: confusing units with dimensions. km/h and m/s are different units but identical dimensions (L T⁻¹). Dimensional analysis ignores the numbers and the unit system entirely.
📜
History & Relevance in Engineering
Two centuries of an idea that still checks every equation you will ever write.
🕰️Interactive Timeline — click an era to exploreSix milestones, 1822 → today
Click any milestone above to reveal its story and why it still matters to engineers…
📖 Key Vocabulary
Dimensional homogeneity — Fourier's principle that all terms in a valid equation share the same dimensions.
Rayleigh's method — assuming a power-law relationship and equating indices of M, L, T.
Buckingham π theorem — the formal rule for how many independent dimensionless groups a problem contains.
Dimensionless group — a combination like Re or the Froude number that lets scale models predict full-size behaviour.
👁️ Visual aid: click the matching buttons in the tool above to read each milestone's story.
Dimensional analysis began as a philosophical remark in Joseph Fourier's 1822 treatise on heat and grew into one of engineering's sharpest everyday tools. Its modern relevance is fourfold: error-catching (an inhomogeneous equation is wrong before you compute anything), scale modelling (wind tunnels, ship tanks), reducing experiments (fewer dimensionless groups than raw variables), and estimation (order-of-magnitude answers when full theory is impossible).
🕰️Case Example 1 — Rayleigh and the pendulumType: Classic derivation
Story Rayleigh showed you can find a pendulum's period without solving any differential equation.
Assume
t = k ma lb gcEquate M: a = 0; L: b + c = 0; T: −2c = 1 → c = −½, b = ½
Result
t = k √(l/g) — mass drops out entirely, a real physical discovery made purely from dimensions. Experiment gives k = 2π.🌪️Case Example 2 — Reynolds and scale modelsType: Scale modelling
Story Osborne Reynolds (1883) found that flow switches from smooth to turbulent at the same value of the dimensionless group ρvd/μ, whatever the pipe.
Consequence A 1:20 model aircraft in a wind tunnel behaves like the full-size aircraft if the Reynolds numbers match.
Engineering use Match Re by raising the tunnel speed or pressurising the air — this is how every airframe and F1 car is still developed.
Takeaway Dimensionless groups make small, cheap experiments stand in for huge, expensive ones.
🚢Case Example 3 — Froude and ship hullsType: Competing groups
Story William Froude towed model hulls and found wave resistance depends on the group
Fr = v/√(gL).Check
[v/√(gL)] = (L T⁻¹)/√(L T⁻² × L) = (L T⁻¹)/(L T⁻¹) = 1 ✓ dimensionlessTension A model cannot match Fr and Re simultaneously — engineers match Fr for wave drag and correct viscous drag by calculation.
Takeaway Dimensional analysis also tells you the limits of model testing, not just the possibilities.
💥Case Example 4 — G. I. Taylor and the 1945 blast photographsType: Estimation
Story From declassified photographs showing only the fireball radius R at time t, Taylor estimated the Trinity test's energy — then still secret.
Assume R depends on energy E, air density ρ and time t:
R = k Ea ρb tcSolve Equating M, L, T indices gives a = 1/5, b = −1/5, c = 2/5, so
R = k (E t²/ρ)1/5Takeaway Rearranged for E, one photograph gave ≈ 20 kilotons — embarrassingly close to the classified value. Dimensions alone can crack a state secret.
🛰️Case Example 5 — Mars Climate Orbiter, 1999Type: Cautionary
Story NASA's $327 million orbiter burned up in the Martian atmosphere because one software team supplied thrust data in pound-force seconds while another expected newton seconds.
Analysis The two quantities share dimensions (M L T⁻¹, an impulse) but differ by a factor of 4.45 — a unit error that dimensional bookkeeping disciplines are designed to catch.
Lesson Homogeneity checks catch wrong dimensions; rigorous unit tracking catches wrong numbers. Professional engineering requires both.
Takeaway Every calculation you submit should state its units at every line — this module is where that habit starts.
🧠 Why this matters at Level 4: in coursework and industry alike, the first question a reviewing engineer asks of any new formula is "is it dimensionally consistent?" — it is the fastest possible error filter.
⚖️
Dimensional Homogeneity — Comparing Equations
Every term in a valid equation must carry identical dimensions. This is your equation lie-detector.
🔧Homogeneity Checker — put equations on trialPick an equation, then press Check
Choose an equation and press Check — the tool tests every term against the left side and delivers a verdict. Two of the equations hide deliberate faults…
📖 Key Vocabulary
Homogeneous — all terms share the same dimensions; the equation could be physically valid.
Term — a chunk separated by + or −; each one is tested individually.
Necessary, not sufficient — homogeneity can prove an equation wrong, but never fully prove it right (constants may still be wrong).
Transcendental argument — whatever sits inside sin, cos, ln or e^( ) must be dimensionless.
👁️ Visual aid: when the dimensions on each side don't match, the equation "tips over" — it cannot be physically valid.
To compare or verify equations: write the dimensions of the left side, then of every term on the right side, and demand that they all match. One mismatched term condemns the whole equation — no arithmetic required.
✏️Worked Example 1 — Verifying a two-term equationType: Verify
Question Show that v = u + at is dimensionally homogeneous.
Left side
[v] = L T⁻¹Term u
[u] = L T⁻¹ ✓Term at
[at] = (L T⁻²)(T) = L T⁻¹ ✓Answer All terms are L T⁻¹ — the equation is homogeneous.
🧩Worked Example 2 — Multi-term with a coefficientType: Multi-term
Question Verify s = ut + ½at².
Left side
[s] = LTerm ut
(L T⁻¹)(T) = L ✓Term ½at² The ½ is dimensionless:
(L T⁻²)(T²) = L ✓Answer Homogeneous. Note that the check would pass equally with ⅓ or 7 in place of ½ — dimensions cannot see pure numbers.
🕵️Worked Example 3 — Catching a fraudType: Detect invalid
Question A student writes v² = u² + 2as². Compare the terms and decide if it can be correct.
Left side
[v²] = L² T⁻²Term u²
L² T⁻² ✓Term 2as²
(L T⁻²)(L²) = L³ T⁻² ✗Answer The final term fails — the equation is invalid. (The correct form is 2as, dimensions (L T⁻²)(L) = L² T⁻² ✓.)
🔎Worked Example 4 — Finding the units of an unknown constantType: Unknown constant
Question Aerodynamic drag is modelled as F = kv². What are the dimensions and SI units of k?
Rearrange
k = F ÷ v²Substitute
[k] = (M L T⁻²) ÷ (L² T⁻²) = M L⁻¹Answer [k] = M L⁻¹, units kg m⁻¹. (In the full model F = ½CdρAv², that M L⁻¹ is exactly what ρA supplies — leaving Cd dimensionless.)
⚙️Worked Example 5 — Engineering comparison across an equation (Bernoulli)Type: Engineering
Question Show that all three terms of p + ½ρv² + ρgh share the same dimensions, and state what they represent.
Term p
M L⁻¹ T⁻² (pressure)Term ½ρv²
(M L⁻³)(L² T⁻²) = M L⁻¹ T⁻² ✓Term ρgh
(M L⁻³)(L T⁻²)(L) = M L⁻¹ T⁻² ✓Answer Every term is a pressure — equivalently an energy per unit volume (M L² T⁻² ÷ L³). Bernoulli's equation is energy bookkeeping in disguise.
🧠 Common slip: checking only the two sides of the = sign and skipping the middle terms. Homogeneity is a term-by-term test — one bad term anywhere sinks the equation.
🧪
Deriving Equations & Checking Validity
The method of dimensions: propose a power law, balance the indices, and the shape of the physics appears.
🔧Derivation Builder — watch the indices solve themselvesPick a problem, then press Derive
Choose a problem and press Derive to see the power-law assumption, the three index equations and the solved formula appear step by step…
📖 Key Vocabulary
Index (plural: indices) — the unknown powers a, b, c in the assumed relationship.
Method of dimensions — Rayleigh's technique of equating indices of M, L and T on both sides.
Empirical constant — the dimensionless k that dimensions cannot supply; found by experiment.
Limitation — the method fails for sums of terms, transcendental functions, and more than three unknowns per M–L–T system.
👁️ Visual aid: the same three steps drive every derivation in the tool above.
✏️Worked Example 1 — Full derivation (pendulum)Type: Full derivation
Question The period t of a simple pendulum may depend on its mass m, length l and gravity g. Derive the form of the equation.
Assume a power-law relationship:
t = k ma lb gcDimensions
[t] = T, [m] = M, [l] = L, [g] = L T⁻²Substitute
T¹ = Ma · Lb · (L T⁻²)cExpand bracket
(L T⁻²)c = Lc T−2c, so the right side is Ma Lb+c T−2cEquate M left has M⁰, right has Ma:
0 = a → a = 0 (mass drops out completely)Equate T left has T¹, right has T−2c:
1 = −2c → c = −½Equate L left has L⁰, right has Lb+c:
0 = b + cBack-substitute put c = −½ into the L-equation:
0 = b + (−½) → b = ½Rebuild with a = 0, b = ½, c = −½:
t = k m⁰ l½ g−½ = k √l ÷ √gAnswer
t = k √(l/g); experiment shows k = 2π, giving the familiar t = 2π√(l/g). Mass genuinely does not matter.🎸Worked Example 2 — Two-variable derivation (wave on a string)Type: Two variables
Question The speed v of a transverse wave on a string depends on tension F and mass per unit length μ. Find the relationship.
Assume
v = k Fa μb with [F] = M L T⁻², [μ] = M L⁻¹Dimensions
L T⁻¹ = Ma+b La−b T−2aEquate T: −1 = −2a → a = ½; M: 0 = a + b → b = −½; L check: a − b = 1 ✓
Answer
v = k √(F/μ) — experiment gives k = 1 exactly.🌬️Worked Example 3 — Three-variable derivation (drag)Type: Three variables
Question Aerodynamic drag F depends on air density ρ, frontal area A and speed v. Derive the drag equation's form.
Assume
F = k ρa Ab vc → M L T⁻² = (M L⁻³)a(L²)b(L T⁻¹)cEquate M: a = 1; T: −2 = −c → c = 2; L: 1 = −3a + 2b + c = −3 + 2b + 2 → b = 1
Answer
F = k ρAv² — the familiar F = ½CdρAv² with k = ½Cd. Doubling speed quadruples drag: dimensions predicted it.✅Worked Example 4 — Validity check of a proposed formulaType: Validity check
Question A textbook claims the velocity of liquid leaving a tank is v = √(2gh). Check its validity dimensionally.
Right side
[√(2gh)] = √((L T⁻²)(L)) = √(L² T⁻²) = L T⁻¹Left side
[v] = L T⁻¹ ✓Answer Dimensionally valid (Torricelli's theorem). But remember: the check confirms the form, not the 2 — homogeneity is necessary, not sufficient.
🚧Worked Example 5 — Where the method reaches its limitsType: Limitations
Question Try to derive the range R of a projectile launched at speed u and angle θ, and explain what the method can and cannot deliver.
Assume
R = k ua gb × f(θ) — θ is an angle, hence dimensionless, so it can only appear inside an unknown function f.Equate
L = (L T⁻¹)a(L T⁻²)b → L: 1 = a + b; T: 0 = −a − 2b → a = 2, b = −1Result
R = f(θ) u²/g — correct shape, but the method cannot reveal f(θ) = sin 2θ.Answer Dimensions find the skeleton of an equation; dimensionless quantities (angles, ratios, pure numbers) stay invisible and need theory or experiment.
🧠 Common slip: forgetting the third index equation as a consistency check. If your solved a, b, c fail the leftover equation, the assumed variable list is wrong or incomplete.
⚙️
Engineering Applications
Four live scenarios — each with its own calculator. Enter your own numbers.
🌊 Reynolds Number — Flow Regime Predictor
The dimensionless group Re = ρvd/μ decides whether pipe flow is smooth (laminar, Re < 2300) or chaotic (turbulent, Re > 4000) — at any scale, in any fluid.
Re = ρ v d ÷ μ (dimensionless)
📟 Try it — Reynolds Number
Enter the four fluid values, then press Calculate…
🌬️ Drag Force — the Equation We Derived
Using F = ½CdρAv² from the Deriving tab: dimensional analysis gave the shape, wind tunnels supply Cd (≈ 0.3 for a modern car, ≈ 1.1 for a flat plate).
F = ½ C_d ρ A v² → [F] = (M L⁻³)(L²)(L² T⁻²) = M L T⁻² ✓
📟 Try it — Drag Force
Enter the four values, then press Calculate…
⏱️ Pendulum Period — Rayleigh's Result, Live
t = 2π√(l/g): derived from dimensions in the History tab, constant supplied by physics. Change g to see the same pendulum on the Moon (1.62 m/s²) or Mars (3.71 m/s²).
t = 2π √(l ÷ g) → [√(l/g)] = √(L ÷ L T⁻²) = T ✓
📟 Try it — Pendulum Period
Enter length and gravity, then press Calculate…
🏗️ Cantilever Deflection — Homogeneity as Quality Control
End deflection of a loaded cantilever: δ = FL³/(3EI). Before trusting any beam formula, an engineer confirms it returns a length — this one does: (M L T⁻²)(L³) ÷ (M L⁻¹ T⁻²)(L⁴) = L ✓
δ = F L³ ÷ (3 E I)
📟 Try it — Cantilever Deflection
Enter the four beam values, then press Calculate…
🎮
Quick-Fire Dimensions Game
12 questions — dimensions, homogeneity and derivations mixed. How many can you get?
⚡ Dimensions Challenge
Score: 0Question: 1/12Streak: 0🔥
Press Start to begin!
📝
Practice Problems
Graded from warm-up to challenge. Attempt fully before revealing the working.
Question 1 🥉2 marks
Write down the dimensions of (a) momentum, (b) power.
(a) [mv] = M × L T⁻¹ = M L T⁻¹
(b) [W/t] = M L² T⁻² ÷ T = M L² T⁻³
(b) [W/t] = M L² T⁻² ÷ T = M L² T⁻³
Question 2 🥉2 marks
Express the joule in SI base units using dimensional reasoning.
[E] = [F][d] = (M L T⁻²)(L) = M L² T⁻²
1 J = 1 kg m² s⁻²
1 J = 1 kg m² s⁻²
Question 3 🥉2 marks
Is the equation v = u + 2as dimensionally homogeneous? Justify.
[v] = L T⁻¹, [u] = L T⁻¹ ✓
[2as] = (L T⁻²)(L) = L² T⁻² ✗
Not homogeneous — invalid. (2as belongs in v² = u² + 2as.)
[2as] = (L T⁻²)(L) = L² T⁻² ✗
Not homogeneous — invalid. (2as belongs in v² = u² + 2as.)
Question 4 🥉2 marks
Show that strain and the coefficient of friction are both dimensionless.
Strain = extension ÷ length = L ÷ L = 1 ✓
μ = friction force ÷ normal force = (M L T⁻²) ÷ (M L T⁻²) = 1 ✓
μ = friction force ÷ normal force = (M L T⁻²) ÷ (M L T⁻²) = 1 ✓
Question 5 🥈3 marks
Verify that E = ½mv² + mgh is dimensionally homogeneous.
[E] = M L² T⁻²
[½mv²] = M (L T⁻¹)² = M L² T⁻² ✓
[mgh] = M (L T⁻²)(L) = M L² T⁻² ✓
Homogeneous — all terms are energies.
[½mv²] = M (L T⁻¹)² = M L² T⁻² ✓
[mgh] = M (L T⁻²)(L) = M L² T⁻² ✓
Homogeneous — all terms are energies.
Question 6 🥈3 marks
The stiffness of a spring is defined by F = kx. Find the dimensions of k and hence its SI units.
k = F ÷ x → [k] = (M L T⁻²) ÷ L = M T⁻²
Units: kg s⁻² = N m⁻¹
Units: kg s⁻² = N m⁻¹
Question 7 🥈4 marks
In the gas equation p = ρRT̄ (T̄ = absolute temperature), find the dimensions of the specific gas constant R.
R = p ÷ (ρT̄)
[R] = (M L⁻¹ T⁻²) ÷ (M L⁻³ × Θ) = L² T⁻² Θ⁻¹
Units: m² s⁻² K⁻¹ = J kg⁻¹ K⁻¹ (for air, R ≈ 287 J kg⁻¹ K⁻¹)
[R] = (M L⁻¹ T⁻²) ÷ (M L⁻³ × Θ) = L² T⁻² Θ⁻¹
Units: m² s⁻² K⁻¹ = J kg⁻¹ K⁻¹ (for air, R ≈ 287 J kg⁻¹ K⁻¹)
Question 8 🥈4 marks
Check the validity of the orbital period claim t² = 4π²r³/(GM), given [G] = M⁻¹L³T⁻².
Right side: [r³/(GM)] = L³ ÷ (M⁻¹L³T⁻² × M) = L³ ÷ (L³T⁻²) = T²
Left side: [t²] = T² ✓
Dimensionally valid — Kepler's third law passes.
Left side: [t²] = T² ✓
Dimensionally valid — Kepler's third law passes.
Question 9 🥇4 marks
The frequency f of a stretched wire depends on its length l, tension F and mass per unit length μ. Derive the relationship.
f = k lᵃ Fᵇ μᶜ → T⁻¹ = Lᵃ(M L T⁻²)ᵇ(M L⁻¹)ᶜ
M: 0 = b + c; T: −1 = −2b → b = ½, c = −½
L: 0 = a + b − c = a + 1 → a = −1
f = (k/l)√(F/μ) — the string equation, k = ½ for the fundamental.
M: 0 = b + c; T: −1 = −2b → b = ½, c = −½
L: 0 = a + b − c = a + 1 → a = −1
f = (k/l)√(F/μ) — the string equation, k = ½ for the fundamental.
Question 10 🥇4 marks
The velocity of sound c in a gas depends on its pressure p and density ρ. Derive the form of the equation and check it with air at p = 101 kPa, ρ = 1.29 kg/m³ (take k = √1.4).
c = k pᵃ ρᵇ → L T⁻¹ = (M L⁻¹T⁻²)ᵃ(M L⁻³)ᵇ
M: 0 = a + b; T: −1 = −2a → a = ½, b = −½
c = k√(p/ρ)
Check: √(1.4 × 101000 ÷ 1.29) = √109612 ≈ 331 m/s ✓
M: 0 = a + b; T: −1 = −2a → a = ½, b = −½
c = k√(p/ρ)
Check: √(1.4 × 101000 ÷ 1.29) = √109612 ≈ 331 m/s ✓
Question 11 🥇5 marks
Poiseuille's law states Q = πΔp r⁴/(8μL) for volume flow through a pipe. Verify it is dimensionally homogeneous.
Left: [Q] = L³ T⁻¹
Right: [Δp r⁴ ÷ (μL)] = (M L⁻¹T⁻²)(L⁴) ÷ ((M L⁻¹T⁻¹)(L))
= (M L³ T⁻²) ÷ (M T⁻¹) = L³ T⁻¹ ✓ homogeneous
Right: [Δp r⁴ ÷ (μL)] = (M L⁻¹T⁻²)(L⁴) ÷ ((M L⁻¹T⁻¹)(L))
= (M L³ T⁻²) ÷ (M T⁻¹) = L³ T⁻¹ ✓ homogeneous
Question 12 🥇5 marks
Following G. I. Taylor, a blast wave's radius obeys R = k(Et²/ρ)^(1/5). (a) Verify this is dimensionally correct. (b) With k ≈ 1, ρ = 1.2 kg/m³, estimate E if R = 140 m at t = 0.025 s.
(a) [Et²/ρ] = (M L²T⁻²)(T²) ÷ (M L⁻³) = L⁵ → fifth root gives L ✓
(b) E = ρR⁵/t² = 1.2 × 140⁵ ÷ 0.025²
140⁵ = 5.37824 × 10¹⁰ → E = 1.2 × 5.37824e10 ÷ 6.25e−4
≈ 1.03 × 10¹⁴ J ≈ 25 kilotons of TNT — Taylor's famous estimate.
(b) E = ρR⁵/t² = 1.2 × 140⁵ ÷ 0.025²
140⁵ = 5.37824 × 10¹⁰ → E = 1.2 × 5.37824e10 ÷ 6.25e−4
≈ 1.03 × 10¹⁴ J ≈ 25 kilotons of TNT — Taylor's famous estimate.