Dimensional Analysis Flashcards: Equation Checks & Coefficient Units
Practice 48 short physics prompts: check equation dimensions, derive coefficient units, judge function inputs, and recognize what a dimensional pass can establish.
આ ડેક વિશે
A missing power of time can survive several lines of algebra. These 48 dimensional analysis flashcards give you short scalar physics proposals to check before you trust an equation.
The cards practice six notation and scope decisions, ten complete-term equation checks, fourteen unknown coefficient calculations, eight function-input tasks, six constrained repairs, and four limits of dimensional evidence. Each equation-check prompt maps a fully defined proposal to a dimensional verdict and a short ledger. Each coefficient prompt asks for one coefficient’s mass, length, and time powers together with a coherent SI unit. Function prompts check exp, ln, sine, and cosine inputs under the usual dimensionless-argument and radian convention; logarithm prompts also state the real-domain conditions.
Repairs ask you to change one specified exponent, unit, ledger entry, or function argument. They repair dimensional consistency without claiming to recover a unique physical law. The limit cards ask what dimensions can establish about numerical factors, signs, real models, and undefined symbols.
You should already know the usual mechanics quantities and basic algebra. The deck leaves out unit-name lookup tables, prefixes, numerical conversions, medical calculations, full-course or exam coverage, calculus derivations, Buckingham Pi, and similarity analysis. It does not automatically reverse every expression: deriving a coefficient and explaining a mismatch are the useful retrieval tasks here.
The sequence establishes notation and complete-term checks first, then mixes coefficient and function tasks before constrained repairs and evidence limits. Closely related prompts stay at least three unrelated cards apart. Long-term spacing belongs to the review scheduler.
The underlying facts were checked against NIST SP 330 §2, Durham’s dimensional-analysis guidance, and Oxford’s calculations-with-units guidance. All trial expressions, calculations, wording, and arrangement were independently authored; no source exercises, competitor cards, or examination material were copied.
AI tools drafted and sequenced the cards and generated the decorative cover. Every final front and back was manually read and solved during author review; the cover was checked at full size and across anticipated crops. The original questions, answers, organization, metadata, and generated cover are shared under CC0 1.0 to the extent applicable rights exist. The scientific facts remain common knowledge, and the cited sources retain their own rights. This is independent general physics practice, with no curriculum or examination alignment claimed.
આ ડેકનાં કાર્ડ
કાર્ડ 1
પ્રશ્ન
In mechanics dimensional notation, what do M, L, T and the brackets [q] mean for a quantity q?
જવાબ
M denotes mass, L length, and T time; [q] denotes the dimension of q. These symbols describe dimensions, not chosen measurement units.
કાર્ડ 2
પ્રશ્ન
A displacement x is measured in metres or centimetres. What stays the same in its dimensional description when the unit changes?
જવાબ
Its dimension stays L (length). Metres and centimetres are different units for the same dimension; the numerical value changes with the unit.
કાર્ડ 3
પ્રશ્ન
For nonzero scalar quantities q and r, how do you obtain the dimensions of qr and q/r from [q] and [r]?
જવાબ
Multiply dimensions for qr and divide them for q/r. Add base-dimension exponents in a product and subtract them in a quotient.
કાર્ડ 4
પ્રશ્ન
For a scalar equation y = a + b, where y, a, and b are physical quantities, what dimensional condition must hold for all three complete terms?
જવાબ
[y] = [a] = [b]. Check each complete addend, not just a product formed from the whole right-hand side. Matching dimensions is necessary, but does not establish physical validity.
કાર્ડ 5
પ્રશ્ન
Does the scalar proposal x = uv/a + 0.3at² pass a dimensional check? Here x is displacement, u and v velocities along one axis, a nonzero acceleration, and t elapsed time.
જવાબ
Passes. [uv/a] = (LT⁻¹)²/(LT⁻²) = L and [at²] = (LT⁻²)T² = L, matching [x]. The numerical factor 0.3 is dimensionless; this does not establish a motion law.
કાર્ડ 6
પ્રશ્ન
A coefficient k stays constant during an experiment. Does that fact alone tell you whether k is dimensionless?
જવાબ
No. Being constant means its value does not change in that setting. Its dimensions must come from its definition or the equation in which it appears.
કાર્ડ 7
પ્રશ્ન
Does the scalar proposal F = ρAv² pass a dimensional check? Here F is force magnitude, ρ mass density, A area, and v speed.
જવાબ
Passes. [ρAv²] = (ML⁻³)L²(L²T⁻²) = MLT⁻², matching force. The check does not determine a numerical coefficient or establish a force model.
કાર્ડ 8
પ્રશ્ન
Two quantities have the same dimensions. Is that enough to establish that adding them is physically meaningful?
જવાબ
No. Their physical meaning and scalar or vector character must also fit. Matching dimensions is a necessary condition for an allowed sum, not permission to add arbitrary quantities.
કાર્ડ 9
પ્રશ્ન
Does the scalar proposal E = mv + Fx pass a dimensional check? Here E is energy, m mass, v speed, F force magnitude, and x distance.
જવાબ
Fails. [mv] = MLT⁻¹, while [Fx] = ML²T⁻² and [E] = ML²T⁻². The mv addend has the wrong dimensions for this sum.
કાર્ડ 10
પ્રશ્ન
In the scalar proposal F = bv, F is force magnitude and v speed. What dimensions and coherent SI unit must b have? Use M, L, T for mass, length, time.
જવાબ
[b] = MT⁻¹; coherent SI unit kg·s⁻¹. Divide force by speed: (MLT⁻²)/(LT⁻¹) = MT⁻¹. Substitution gives (MT⁻¹)(LT⁻¹) = MLT⁻².
કાર્ડ 11
પ્રશ્ન
Does the scalar proposal P = Fv + E/t pass a dimensional check? Here P is power, F force magnitude, v speed, E energy, and t nonzero elapsed time.
જવાબ
Passes. [Fv] = (MLT⁻²)(LT⁻¹) = ML²T⁻³ and [E/t] = ML²T⁻³, matching power. Each complete addend passes.
કાર્ડ 12
પ્રશ્ન
In the scalar proposal E = cx³, E is energy and x displacement. What dimensions and coherent SI unit must c have? Use M, L, T for mass, length, time.
જવાબ
[c] = ML⁻¹T⁻²; coherent SI unit kg·m⁻¹·s⁻². Divide energy by length cubed: ML²T⁻²/L³ = ML⁻¹T⁻². Substitution gives (ML⁻¹T⁻²)L³ = ML²T⁻².
કાર્ડ 13
પ્રશ્ન
Is the argument in exp(t/τ) dimensionally valid under the usual dimensionless-argument convention? Here t is elapsed time and τ a positive time scale.
જવાબ
Yes. [t/τ] = T/T = 1, so the argument is dimensionless. The exponential itself is also dimensionless; τ > 0 keeps the denominator nonzero.
કાર્ડ 14
પ્રશ્ન
Does the scalar proposal v² = u² + ar² pass a dimensional check? Here v and u are speeds, a acceleration, and r distance.
જવાબ
Fails. [v²] = [u²] = L²T⁻², but [ar²] = (LT⁻²)L² = L³T⁻². The ar² addend has an extra power of length.
કાર્ડ 15
પ્રશ્ન
In the scalar proposal p = dv², p is pressure and v speed. What dimensions and coherent SI unit must d have? Use M, L, T for mass, length, time.
જવાબ
[d] = ML⁻³; coherent SI unit kg·m⁻³. Divide pressure by speed squared: (ML⁻¹T⁻²)/(L²T⁻²) = ML⁻³. Substitution gives (ML⁻³)(L²T⁻²) = ML⁻¹T⁻².
કાર્ડ 16
પ્રશ્ન
In the scalar proposal x = A + Bt + Ct², x is displacement and t elapsed time. What dimensions and coherent SI unit must B have? A and C have the dimensions needed for their terms. Use M, L, T for mass, length, time.
જવાબ
[B] = LT⁻¹; coherent SI unit m·s⁻¹. The complete term Bt must have dimension L, so [B] = L/T. Substitution gives (LT⁻¹)T = L.
કાર્ડ 17
પ્રશ્ન
Is ln(r/r₀) dimensionally valid, and is it defined as a real number? Here r and r₀ are positive lengths; use the usual dimensionless-argument convention.
જવાબ
Yes to both. [r/r₀] = L/L = 1, and r/r₀ > 0. The ratio is dimensionless and meets the real logarithm’s positive-argument requirement.
કાર્ડ 18
પ્રશ્ન
Does the scalar proposal a = (v − u)/t pass a dimensional check? Here a is acceleration, v and u velocities along one axis, and t nonzero elapsed time.
જવાબ
Passes. The difference v − u has dimension LT⁻¹; dividing by t gives LT⁻², matching acceleration. Both terms inside the difference have the same dimensions.
કાર્ડ 19
પ્રશ્ન
In the scalar proposal F = k/r², F is force magnitude and r a positive distance. What dimensions and coherent SI unit must k have? Use M, L, T for mass, length, time.
જવાબ
[k] = ML³T⁻²; coherent SI unit kg·m³·s⁻². Multiply force by r²: (MLT⁻²)L² = ML³T⁻². Substitution gives ML³T⁻²/L² = MLT⁻².
કાર્ડ 20
પ્રશ્ન
What dimensions and coherent SI unit must k have in exp(−kt²)? Here t is elapsed time and k a scalar coefficient; use the usual dimensionless-argument convention and M, L, T for mass, length, time.
જવાબ
[k] = T⁻²; coherent SI unit s⁻². The argument requires [k]T² = 1. The minus sign changes neither dimensions nor the real exponential’s domain.
કાર્ડ 21
પ્રશ્ન
Does the scalar proposal p = F/A + ρv pass a dimensional check? Here p is pressure, F force magnitude, A positive area, ρ mass density, and v speed.
જવાબ
Fails. [F/A] = ML⁻¹T⁻², matching pressure, but [ρv] = (ML⁻³)(LT⁻¹) = ML⁻²T⁻¹. The second addend is incompatible.
કાર્ડ 22
પ્રશ્ન
In the scalar proposal v = h√r, v is speed and r a positive distance. What dimensions and coherent SI unit must h have? Use M, L, T for mass, length, time.
જવાબ
[h] = L¹ᐟ²T⁻¹; coherent SI unit m¹ᐟ²·s⁻¹. Divide speed by √r: LT⁻¹/L¹ᐟ² = L¹ᐟ²T⁻¹. Substitution gives (L¹ᐟ²T⁻¹)L¹ᐟ² = LT⁻¹.
કાર્ડ 23
પ્રશ્ન
Is sin(vt) dimensionally valid under the usual dimensionless-argument and radian convention? Here v is speed and t elapsed time.
જવાબ
No. [vt] = (LT⁻¹)T = L, so the argument is a length rather than dimensionless. A sine output cannot repair an invalid input.
કાર્ડ 24
પ્રશ્ન
In the scalar proposal E = jm², E is energy and m mass. What dimensions and coherent SI unit must j have? Use M, L, T for mass, length, time.
જવાબ
[j] = M⁻¹L²T⁻²; coherent SI unit kg⁻¹·m²·s⁻². Divide energy by mass squared: ML²T⁻²/M² = M⁻¹L²T⁻². Substitution gives (M⁻¹L²T⁻²)M² = ML²T⁻².
કાર્ડ 25
પ્રશ્ન
Does the scalar proposal E = F/t pass a dimensional check? Here E is energy, F force magnitude, and t nonzero elapsed time.
જવાબ
Fails. [F/t] = MLT⁻³, while energy has dimension ML²T⁻². Both the length and time exponents differ.
કાર્ડ 26
પ્રશ્ન
In the scalar proposal F = qat, F is force magnitude, a acceleration, and t elapsed time. What dimensions and coherent SI unit must q have? Use M, L, T for mass, length, time.
જવાબ
[q] = MT⁻¹; coherent SI unit kg·s⁻¹. [at] = LT⁻¹, so [q] = MLT⁻²/(LT⁻¹) = MT⁻¹. Substitution gives (MT⁻¹)(LT⁻¹) = MLT⁻².
કાર્ડ 27
પ્રશ્ન
What dimensions and coherent SI unit must h have in ln(hp)? Here p is positive pressure and h a positive scalar coefficient; use the usual dimensionless-argument convention and M, L, T for mass, length, time.
જવાબ
[h] = M⁻¹LT²; coherent SI unit kg⁻¹·m·s². [h][p] = 1 requires the inverse of ML⁻¹T⁻². The positive values also ensure hp > 0 for a real logarithm.
કાર્ડ 28
પ્રશ્ન
In the scalar proposal p = s/r, p is pressure and r a positive distance. What dimensions and coherent SI unit must s have? Use M, L, T for mass, length, time.
જવાબ
[s] = MT⁻²; coherent SI unit kg·s⁻². Multiply pressure by length: (ML⁻¹T⁻²)L = MT⁻². Substitution gives MT⁻²/L = ML⁻¹T⁻².
કાર્ડ 29
પ્રશ્ન
Does the scalar proposal ρ = m/r² pass a dimensional check? Here ρ is mass density, m mass, and r a positive distance.
જવાબ
Fails. [m/r²] = ML⁻², while mass density has dimension ML⁻³. The denominator supplies area dimensions instead of volume dimensions.
કાર્ડ 30
પ્રશ્ન
In the scalar proposal x = A + Bt + Ct², x is displacement and t elapsed time. What dimensions and coherent SI unit must C have? A and B have the dimensions needed for their terms. Use M, L, T for mass, length, time.
જવાબ
[C] = LT⁻²; coherent SI unit m·s⁻². The complete term Ct² must have dimension L, so [C] = L/T². Substitution gives (LT⁻²)T² = L.
કાર્ડ 31
પ્રશ્ન
Is cos(at/v) dimensionally valid under the usual dimensionless-argument and radian convention? Here a is acceleration, t elapsed time, and v nonzero speed.
જવાબ
Yes. [at/v] = (LT⁻²)T/(LT⁻¹) = 1. The argument is dimensionless, and v ≠ 0 keeps the quotient defined.
કાર્ડ 32
પ્રશ્ન
In the scalar proposal E = wv³, E is energy and v speed. What dimensions and coherent SI unit must w have? Use M, L, T for mass, length, time.
જવાબ
[w] = ML⁻¹T; coherent SI unit kg·m⁻¹·s. Divide energy by speed cubed: (ML²T⁻²)/(L³T⁻³) = ML⁻¹T. Substitution gives (ML⁻¹T)(L³T⁻³) = ML²T⁻².
કાર્ડ 33
પ્રશ્ન
Does the scalar proposal t = r/v + v/a pass a dimensional check? Here t is elapsed time, r distance, v nonzero speed, and a nonzero acceleration.
જવાબ
Passes. [r/v] = L/(LT⁻¹) = T and [v/a] = (LT⁻¹)/(LT⁻²) = T, matching elapsed time. This does not establish the proposal as a motion law.
કાર્ડ 34
પ્રશ્ન
What dimensions and coherent SI unit must q have in sin(qr²)? Here r is distance and q a scalar coefficient; use the usual dimensionless-argument and radian convention and M, L, T for mass, length, time.
જવાબ
[q] = L⁻²; coherent SI unit m⁻². The argument requires [q]L² = 1. With this choice qr² can serve as a dimensionless radian argument.
કાર્ડ 35
પ્રશ્ન
In the scalar proposal P = zr², P is power and r distance. What dimensions and coherent SI unit must z have? Use M, L, T for mass, length, time.
જવાબ
[z] = MT⁻³; coherent SI unit kg·s⁻³. Divide power by r²: ML²T⁻³/L² = MT⁻³. Substitution gives (MT⁻³)L² = ML²T⁻³.
કાર્ડ 36
પ્રશ્ન
The scalar proposal x = ut + a²t/u fails a dimensional check. Here x is displacement, u nonzero velocity, a acceleration, and t elapsed time. Change only the exponent of t in the second addend to make that term match x. What exponent is required?
જવાબ
The exponent must be 3: use a²t³/u. [a²tⁿ/u] = (L²T⁻⁴)Tⁿ/(LT⁻¹) = LTⁿ⁻³; matching L requires n − 3 = 0. This repairs dimensions only, not the motion model.
કાર્ડ 37
પ્રશ્ન
In the scalar proposal v = n/(ρr²), v is speed, ρ positive mass density, and r a positive distance. What dimensions and coherent SI unit must n have? Use M, L, T for mass, length, time.
જવાબ
[n] = MT⁻¹; coherent SI unit kg·s⁻¹. [ρr²] = ML⁻¹, so [n] = (LT⁻¹)(ML⁻¹) = MT⁻¹. Substitution gives MT⁻¹/(ML⁻¹) = LT⁻¹.
કાર્ડ 38
પ્રશ્ન
In the scalar proposal E = αmv², E is energy, m mass, v speed, and α a dimensionless number. Can a dimensional check determine α’s numerical value?
જવાબ
No. [mv²] = ML²T⁻² for every dimensionless α. A numerical value needs additional physical or experimental information.
કાર્ડ 39
પ્રશ્ન
What dimensions and coherent SI unit must j have in cos(jmv)? Here m is mass, v speed, and j a scalar coefficient; use the usual dimensionless-argument and radian convention and M, L, T for mass, length, time.
જવાબ
[j] = M⁻¹L⁻¹T; coherent SI unit kg⁻¹·m⁻¹·s. [mv] = MLT⁻¹, so multiplying by its inverse makes jmv dimensionless.
કાર્ડ 40
પ્રશ્ન
For the scalar proposal F = cv², a unit ledger assigns c the unit kg·s⁻¹. Here F is force magnitude and v speed. Replace only c’s unit with the coherent SI unit required by this proposal.
જવાબ
Use kg·m⁻¹. Force divided by speed squared gives (kg·m·s⁻²)/(m²·s⁻²) = kg·m⁻¹; the stated kg·s⁻¹ would instead give kg·m²·s⁻³ for cv².
કાર્ડ 41
પ્રશ્ન
In the scalar proposal t = g√(m/r), t is elapsed time, m positive mass, and r a positive distance. What dimensions and coherent SI unit must g have? Use M, L, T for mass, length, time.
જવાબ
[g] = M⁻¹ᐟ²L¹ᐟ²T; coherent SI unit kg⁻¹ᐟ²·m¹ᐟ²·s. [√(m/r)] = M¹ᐟ²L⁻¹ᐟ². Multiplying by the stated [g] leaves T.
કાર્ડ 42
પ્રશ્ન
Two scalar proposals differ only by a plus sign versus a minus sign between complete terms that all have the same dimensions. Can dimensional analysis select the physically correct sign?
જવાબ
No. Changing a plus sign to a minus sign does not change the dimensions. The sign needs information about the physical model and the definitions or directions of the quantities.
કાર્ડ 43
પ્રશ્ન
The proposed real logarithm ln(r) uses a positive length r. With a fixed positive reference length r₀, repair only its argument using one quotient of the two lengths. The corrected logarithm must increase as r increases.
જવાબ
Use ln(r/r₀). The argument has dimension L/L = 1 and is positive. The reference length makes the logarithm meaningful without choosing a unit-dependent raw length value. With r₀ fixed, it increases with r; ln(r₀/r) would decrease.
કાર્ડ 44
પ્રશ્ન
A ledger claims [Fr²] = ML²T⁻², with F force magnitude, r distance, and M, L, T mass, length, time. Keep Fr² unchanged and correct only the dimensional ledger.
જવાબ
[Fr²] = ML³T⁻². Force contributes MLT⁻² and r² contributes L², so the length exponents add to 3. The expression therefore does not have energy dimensions.
કાર્ડ 45
પ્રશ્ન
A proposed scalar physics equation passes all term and function-input dimensional checks. What can you conclude about whether it describes a real system?
જવાબ
Only that these dimensional checks found no inconsistency. A pass does not establish the model, assumptions, dependence on variables, numerical factors, or agreement with measurements.
કાર્ડ 46
પ્રશ્ન
In the scalar proposal p = dm/r³, a ledger calls d dimensionless. Here p is pressure, m mass, and r a positive distance. Replace only that entry with the dimensions and coherent SI unit required for d; use M, L, T for mass, length, time.
જવાબ
[d] = L²T⁻²; coherent SI unit m²·s⁻². Since [m/r³] = ML⁻³, [d] = (ML⁻¹T⁻²)/(ML⁻³) = L²T⁻². Substitution gives (L²T⁻²)(ML⁻³) = ML⁻¹T⁻².
કાર્ડ 47
પ્રશ્ન
In the scalar proposal t = r/v + u, t is elapsed time, r distance, and v and u nonzero speeds. A nonzero acceleration a is available. Replace only the addend u with u/aⁿ. Which exponent n makes that addend match time?
જવાબ
The exponent must be 1: use u/a. [u/aⁿ] = L¹⁻ⁿT⁻¹⁺²ⁿ; n = 1 removes length and leaves T. This repairs dimensions only; it does not establish a motion law.
કાર્ડ 48
પ્રશ્ન
You see the proposed equation Q = ab, but the author supplies no definitions or dimensions for Q, a, or b. What information is missing before you can judge dimensional consistency?
જવાબ
The meanings or dimensions of Q, a, and b. Letters alone do not determine dimensions; once supplied, the check compares [Q] with [a][b].
48 કાર્ડ
Dimensional Analysis Flashcards: Equation Checks & Coefficient Units
અભ્યાસ શરૂ કરવા Nibomo ખૂલશે.