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Dimensional Analysis
Fundamental Dimensions and Base Units
Every physical quantity is expressed as a combination of seven fundamental quantities, representing the qualitative nature of the measurement.
$$[Q] = M^a L^b T^c I^d K^e mol^f cd^g$$
The general form of a dimensional formula using the 7 base SI dimensions.
$M$=Mass(kg)
$L$=Length(m)
$T$=Time(s)
$I$=Electric Current(A)
Mechanics
→Only M, L, and T are typically required.
The Big Three: Mass $[M]$, Length $[L]$, and Time $[T]$ form the core of mechanical dimensional analysis.
Qualitative Nature: Dimensions describe what a quantity is (e.g., a distance), regardless of its unit magnitude (meters vs. miles).
Some physical quantities have no dimensions — their dimensional formula is $M^0 L^0 T^0$. These dimensionless quantities arise as ratios of quantities with identical dimensions.
Common Examples: Strain ($\Delta L / L$), refractive index ($c/v$), relative density, coefficient of friction, and all trigonometric ratios.
Angles: Radian and steradian are dimensionless — they are ratios of lengths ($s/r$) and areas ($A/r^2$) respectively.
Shortcut: If a quantity is defined as a ratio of two quantities with the same dimensions, it is always dimensionless.
Deriving Dimensions for Derived Quantities
Dimensions of derived quantities are obtained by substituting the dimensions of fundamental quantities into their defining algebraic equations.
$$[Force] = [Mass] \times [Acceleration] = M L T^{-2}$$
Force is derived from Newton's Second Law, showing how derived units combine base units.
$F$=Force(newtons (N))
$m$=Mass(kilograms (kg))
$a$=Acceleration(metres per second squared (m/s²))
deriving Pressure units
→$[P] = [F]/[A] = [MLT^{-2}]/[L^2] = [ML^{-1}T^{-2}] = \text{Pa}$
deriving Energy units
→$[E] = [F][d] = [MLT^{-2}][L] = [ML^2T^{-2}] = \text{J}$
Square Brackets: Used to denote 'the dimension of' a specific physical variable.
Algebraic Manipulation: Dimensions can be treated as variables in algebraic operations: multiplied, divided, or raised to powers.
Shortcut: Build complex dimensions step by step — velocity is $L/T$, acceleration is $v/T = L/T^2$, force is $m \times a = ML/T^2$.
Common Dimensional Formulas
These are the building blocks for checking and deriving formulas
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Velocity: $[LT^{-1}]$
•
Acceleration: $[LT^{-2}]$
•
Force: $[MLT^{-2}]$
•
Energy / Work / Torque: $[ML^2T^{-2}]$
•
Power: $[ML^2T^{-3}]$
•
Pressure / Stress: $[ML^{-1}T^{-2}]$
•
Density: $[ML^{-3}]$
•
Momentum: $[MLT^{-1}]$
•
Viscosity (η): $[ML^{-1}T^{-1}]$
•
Gravitational constant (G): $[M^{-1}L^3T^{-2}]$
The Principle of Homogeneity
The Principle of Homogeneity states that only physical quantities with identical dimensions can be added, subtracted, or compared.
$$[LHS] = [RHS]$$
Ensuring equations are physically meaningful by matching dimensions on both sides.
$LHS$=Left Hand Side dimensions(N/A)
$RHS$=Right Hand Side dimensions(N/A)
$A + B = C$
→Must satisfy $[A] = [B] = [C]$.
Addition Rule: You cannot add 5 meters to 2 seconds; the operation is dimensionally invalid.
Consistency Check: Homogeneity is a necessary, but not sufficient, condition for an equation to be physically correct (it ignores dimensionless constants).
Checking Method: Compute dimensions of every term separately. If any term has different dimensions from the others, the equation is wrong.
Limiting Case Test: Does the formula make sense when $t=0$, $v=0$, or $r \to \infty$?
Applications: Checking and Deriving Relations
Dimensional analysis is used to verify consistency of existing formulas and derive proportionality relations between variables.
$$X \propto a^p b^q c^r$$
The method of undetermined powers for deriving physical laws.
$X$=Dependent variable(varies)
$p, q, r$=Undetermined exponents(dimensionless)
Constants
→Dimensionless constants (like $2\pi$ or $1/2$) cannot be found using this method.
Checking Correctness: Compare dimensions of all terms in an equation to catch algebraic errors.
Power Method Step-by-Step: (1) Write $X = k \cdot a^p b^q c^r$. (2) Substitute dimensions on both sides. (3) Equate exponents of M, L, T to form simultaneous equations. (4) Solve for $p, q, r$.
Proportionality Shortcut: Once you derive $T \propto \sqrt{l/g}$, doubling $l$ increases $T$ by $\sqrt{2}$; quadrupling $g$ halves $T$ — no need to recalculate from scratch.
Dimensionless Constants Are Invisible: The method yields only the functional form. Constants like $2\pi$, $1/2$, $4/3$ must be found experimentally.
Use Unit Cancellation to quickly eliminate incorrect formulas without full dimensional analysis.
$$\text{Check: Do the units on LHS match RHS?}$$
Quick elimination technique - if units don't match, the formula MUST be wrong.
$\text{LHS units}$=Units of the quantity being calculated(variable)
$\text{RHS units}$=Combined units of all terms on right side(must equal LHS)
checking $v = \sqrt{2gh}$
→$[v] = \text{m/s}$, RHS: $\sqrt{(\text{m/s}^2)(\text{m})} = \sqrt{\text{m}^2/\text{s}^2} = \text{m/s}$ ✓
checking $F = mv$ (wrong)
→$[F] = \text{kg m/s}^2$, RHS: $(\text{kg})(\text{m/s}) = \text{kg m/s}$ ✗ Different!
checking $E = mc^2$
→$[E] = \text{J} = \text{kg m}^2/\text{s}^2$, RHS: $(\text{kg})(\text{m/s})^2 = \text{kg m}^2/\text{s}^2$ ✓
Quick Elimination: In problems with multiple candidate formulas, eliminate 2-3 options just by checking units first.
Dimensionless Quantities: Angle, strain, refractive index, relative density, coefficient of friction have no units.
Limitations of Dimensional Analysis
While powerful, dimensional analysis has inherent limitations that prevent it from being a complete tool for deriving physical laws.
Dimensionless Constants: Cannot determine pure numbers like $2\pi$, $1/2$, or $4/3$ — e.g., the pendulum formula $T = 2\pi\sqrt{l/g}$ gives only $T \propto \sqrt{l/g}$.
Same Dimensions, Different Quantities: Work ($F \cdot d$) and torque ($F \times r$) both have dimensions $[ML^2T^{-2}]$ but are physically different.
Transcendental Functions: Cannot derive formulas involving $\sin$, $\cos$, $\log$, or $e^x$ — their arguments must be dimensionless, but the method cannot predict their presence.
Multiple Variables: If more than 3 unknowns exist (in mechanics), there are not enough equations (M, L, T) to solve for all powers uniquely.
Uncertainty Propagation
When performing calculations with measured values, uncertainties propagate through the result. The rules depend on the mathematical operation involved.
$$\delta R = \sqrt{(\frac{\partial R}{\partial x} \delta x)^2 + (\frac{\partial R}{\partial y} \delta y)^2}$$
General formula for uncertainty propagation (simplified rules below are used in practice).
$\delta R$=Uncertainty in the result(same as R)
$\delta x, \delta y$=Uncertainties in measured quantities(same as x, y)
Addition/Subtraction: Add absolute uncertainties. If $x = a \pm \delta a$ and $y = b \pm \delta b$, then $\delta(x \pm y) = \delta a + \delta b$.
Multiplication/Division: Add percentage uncertainties. For $R = xy$: %uncertainty in $R$ = %uncertainty in $x$ + %uncertainty in $y$.
Power Rule: Multiply percentage uncertainty by the power. For $R = x^n$: %uncertainty in $R$ = $n \times$ %uncertainty in $x$.
Special rules apply for uncertainties in timing experiments and in the average of repeated measurements.
Timing Experiments: Uncertainty in a period is $\delta T = \frac{\text{least count of timer}}{\text{number of vibrations}}$. Counting more vibrations reduces uncertainty.
Example: 30 pendulum swings timed at 54.6 s with a stopwatch (least count 0.1 s) gives $T = 1.82 \pm 0.003$ s, since $\delta T = 0.1/30 = 0.003$ s.
Mean of Repeated Measurements: (1) Calculate the average. (2) Find the absolute deviation of each reading from the average. (3) The mean deviation is the uncertainty in the average.
Example: Six diameter readings 1.20, 1.22, 1.23, 1.19, 1.22, 1.21 mm give mean = 1.21 mm, deviations = 0.01, 0.01, 0.02, 0.02, 0.01, 0, mean deviation = 0.01 mm. Result: $1.21 \pm 0.01$ mm.
Applying uncertainty rules to real measurements requires careful attention to which rule applies at each step of the calculation.
Step 1: Identify whether the operation is addition/subtraction or multiplication/division.
Step 2: For addition/subtraction, compute absolute uncertainties. For multiplication/division, convert to percentage uncertainties first.
Step 3: Combine the uncertainties using the appropriate rule, then convert back to absolute uncertainty for the final answer.
Significant Figures: Uncertainty is quoted to 1 significant figure, and the result is rounded to match the decimal place of the uncertainty.
Combined Operations: When a formula involves both powers and products (e.g., $V = \frac{\pi d^2 l}{4}$), apply the power rule first ($2 \times \%\delta d$), then add percentage uncertainties of all factors.
Application Strategy
Worked Examples
Problem Statement
Derive the dimensional formula for Pressure ($P$).
$P = F/A$Newton's Second Law
Strategic Toolkit
$P = \frac{F}{A}$
Step 1
Define Formula
Pressure is defined as force per unit area.
$P = \frac{Force}{Area}$
Step 2
Dimensions of Components
Force is $MLT^{-2}$ and Area is $L^2$.
$[F] = MLT^{-2}, [A] = L^2$
Step 3
Substitute and Simplify
Combine the dimensions and simplify the exponent of L.
$[P] = \frac{MLT^{-2}}{L^2} = ML^{1-2}T^{-2} = ML^{-1}T^{-2}$
Problem Statement
Check the dimensional correctness of the equation: $\frac{1}{2} mv^2 = mgh$.
Kinetic EnergyPotential EnergyHomogeneity
Strategic Toolkit
$\text{LHS} = \text{RHS}$
Step 1
Analyze LHS
Analyze Kinetic Energy. Ignore the dimensionless constant $1/2$.
$[LHS] = [M][LT^{-1}]^2 = ML^2T^{-2}$
Step 2
Analyze RHS
Analyze Gravitational Potential Energy ($m$ is mass, $g$ is acceleration, $h$ is height).
$[RHS] = [M][LT^{-2}][L] = ML^2T^{-2}$
Step 3
Conclusion
Since both sides have the same dimensions, the equation is dimensionally consistent.
$[LHS] = [RHS] = ML^2T^{-2}$
Problem Statement
Using dimensional analysis, derive the relation for the time period $T$ of a simple pendulum, given it depends on length $l$, mass $m$, and acceleration due to gravity $g$.
Method of undetermined powersDimensional analysisSimple pendulum
Strategic Toolkit
$T = k \cdot m^a \cdot l^b \cdot g^c$
Step 1
Write the Assumed Relation
Assume $T$ depends on $m$, $l$, and $g$ raised to unknown powers.
$[T] = [M]^a [L]^b [LT^{-2}]^c = [M]^a [L]^{b+c} [T]^{-2c}$
Step 2
Equate Exponents
Compare powers of M, L, and T on both sides.
$M: a = 0, \quad L: b + c = 0, \quad T: -2c = 1$
Step 3
Solve
From $T$: $c = -1/2$. From $L$: $b = 1/2$. Mass has no effect ($a = 0$).
$T = k \sqrt{\frac{l}{g}}$
Step 4
Note the Limitation
Dimensional analysis gives the proportionality but not the constant $k = 2\pi$.
$T = 2\pi\sqrt{\frac{l}{g}}$
Problem Statement
A metal cylinder has diameter $d = 1.22 \pm 0.01$ cm and length $l = 5.35 \pm 0.01$ cm. Find the volume and its uncertainty.
Power ruleMultiplication ruleCombined uncertainty
Strategic Toolkit
$V = \frac{\pi d^2 l}{4}$
Step 1
Calculate Percentage Uncertainties
Convert absolute uncertainties to percentages for each measured quantity.
$\%\delta d = \frac{0.01}{1.22} \times 100 = 0.8\%, \quad \%\delta l = \frac{0.01}{5.35} \times 100 = 0.2\%$
Step 2
Apply Power Rule to Diameter
Since $V \propto d^2$, the power rule doubles the percentage uncertainty in $d$.
$\%\delta(d^2) = 2 \times 0.8\% = 1.6\%$
Step 3
Add Percentage Uncertainties (Multiplication)
For multiplication ($d^2 \times l$), add the percentage uncertainties.
$\%\delta V = 1.6\% + 0.2\% = 1.8\%$
Step 4
Compute Final Result
Calculate volume and convert percentage uncertainty back to absolute.
$V = \frac{\pi (1.22)^2 (5.35)}{4} = 6.25 \text{ cm}^3, \quad \delta V = 1.8\% \times 6.25 \approx 0.1 \text{ cm}^3$. Result: $V = 6.2 \pm 0.1 \text{ cm}^3$
Problem Statement
Find the dimensions of viscosity $\eta$ from Stokes' law: $F = 6\pi\eta r v$.
Stokes' lawIsolating variablesDimensional analysis
Strategic Toolkit
$\eta = \frac{F}{6\pi r v}$
Step 1
Isolate η
Rearrange Stokes' law. Ignore the dimensionless constant $6\pi$.
$[\eta] = \frac{[F]}{[r][v]}$
Step 2
Substitute Known Dimensions
Force is $MLT^{-2}$, radius is $L$, velocity is $LT^{-1}$.
$[\eta] = \frac{MLT^{-2}}{L \cdot LT^{-1}} = \frac{MLT^{-2}}{L^2T^{-1}}$
Step 3
Simplify
Combine the exponents of each base dimension.
$[\eta] = ML^{1-2}T^{-2-(-1)} = ML^{-1}T^{-1}$
Problem Statement
Check the dimensional correctness of the equation $E = mc^2$.
Dimensions of EnergyDimensions of Mass and SpeedHomogeneity Principle
Strategic Toolkit
$E = mc^2$
Step 1
LHS Dimensions
Energy [E] is Work = Force × Distance.
$[E] = [Force] \times [Length] = [MLT^{-2}][L] = [ML^2T^{-2}]$
Step 2
RHS Dimensions
Calculate dimensions for mass [m] and speed [c] squared.
$[m] = [M], [c] = [LT^{-1}] \implies [c^2] = [L^2T^{-2}]$
Step 3
Multiply RHS
Combine the dimensions.
$[mc^2] = [M][L^2T^{-2}] = [ML^2T^{-2}]$
Step 4
Compare
LHS and RHS match, verifying correctness.
$[ML^2T^{-2}] = [ML^2T^{-2}] \implies \text{Correct}$
Memorize formulas
Dimensions of Force
The fundamental 'building block' for most mechanical quantities.
Formula
$[F] = MLT^{-2}$
Dimensions of Energy/Work
Applies to Kinetic, Potential, and Heat energy.
Formula
$[E] = ML^2T^{-2}$
Dimensions of Power
Work done per unit time ($E/T$).
Formula
$[P] = ML^2T^{-3}$
Dimensions of Gravitational Constant
From Newton's law: $F = Gm_1m_2/r^2$.
Formula
$[G] = M^{-1}L^3T^{-2}$
Dimensions of Pressure
Force per unit area ($F/A$). Same as stress.
Formula
$[P] = ML^{-1}T^{-2}$
Dimensions of Viscosity
From Stokes' law: $F = 6\pi\eta rv$.
Formula
$[\eta] = ML^{-1}T^{-1}$
Uncertainty: Power Rule
Percentage uncertainty multiplied by the power factor.
Formula
$\% \delta(x^n) = n \times \% \delta x$
Uncertainty: Addition/Subtraction
Absolute uncertainties add for sums and differences.
Formula
$\delta(x \pm y) = \delta x + \delta y$
Uncertainty: Multiplication/Division
Percentage uncertainties add for products and quotients.
Formula
$\% \delta(xy) = \% \delta x + \% \delta y$
Dimensional Formula Method
General form for deriving unknown relationships via dimensions.
Formula
$[Q] = [M^a L^b T^c] \implies \text{compare powers of M, L, T}$
Review
1
Find the dimensional formula for the Universal Gravitational Constant ($G$) given the formula $F = G \frac{m_1 m_2}{r^2}$.
Ans
$M^{-1} L^3 T^{-2}$
Review
Tap to reveal
Rearrange the gravitational law for $G$
Start from $F = G\frac{m_1m_2}{r^2}$ and isolate the constant so its dimensions can be read directly from the other quantities.
$$G = \frac{Fr^2}{m_1m_2}$$
Substitute the base dimensions
Use $[F] = [MLT^{-2}]$, $[r^2] = [L^2]$, and $[m_1m_2] = [M^2]$ to combine powers of each base quantity.
$$[G] = \frac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1}L^3T^{-2}]$$
2
In the equation $v = a + bt + ct^2$, where $v$ is velocity and $t$ is time, what are the dimensions of $c$?
Ans
$L T^{-3}$
Review
Tap to reveal
Match the $ct^2$ term to velocity
Every term in the sum must have the same dimensions as $v$, so the $ct^2$ term must also reduce to velocity.
$$[c][t^2] = [v]$$
Divide by time squared
Since $[v] = LT^{-1}$ and $[t^2] = T^2$, the dimensions of $c$ follow directly.
$$[c] = \frac{LT^{-1}}{T^2} = LT^{-3}$$
3
Which pair of physical quantities has the same dimensional formula?
Ans
Impulse and Momentum
Review
Tap to reveal
Compute the impulse dimensions
Impulse is force multiplied by time, so its dimensions are the dimensions of force with one power of time added back.
$$[I] = [F][t] = [MLT^{-2}][T] = [MLT^{-1}]$$
Compare with momentum
Momentum is $mv$, which gives the same base dimensions, so this is the matching pair.
$$[p] = [M][LT^{-1}] = [MLT^{-1}]$$
4
If Force ($F$), Velocity ($V$), and Time ($T$) are taken as fundamental units, what is the dimensional formula for Energy?
Ans
$F V T$
Review
Tap to reveal
Rewrite distance using the new base quantities
In the $F, V, T$ system, distance is not fundamental, but it can be expressed as velocity multiplied by time.
$$d = VT$$
Use work as force times distance
Energy has the same dimensions as work, so multiply force by the distance expression to get the final form.
$$E = F \cdot d = FVT$$
5
Work and torque both have the same dimensional formula $[ML^2T^{-2}]$. This demonstrates which limitation of dimensional analysis?
Ans
It cannot distinguish physically different quantities with the same dimensions
Review
Tap to reveal
Notice the shared dimensional form
Both quantities reduce to force times length, so dimensional analysis treats them as identical at the level of units.
$$[W] = [\tau] = [ML^2T^{-2}]$$
Identify the limitation
Because dimensions only track units, they cannot separate two physically different quantities that happen to share the same formula.
Dimensional analysis cannot distinguish physically different quantities with the same dimensions.
6
The mass, length, and breadth of a rectangular block are measured as $m = 250.0 \pm 0.5$ g, $l = 8.0 \pm 0.1$ cm, $b = 4.0 \pm 0.1$ cm, and height $h = 2.0 \pm 0.1$ cm. What is the percentage uncertainty in its density $\rho = \frac{m}{lbh}$?
Ans
8.95%
Review
Tap to reveal
Find the percentage uncertainty of each measurement
For a quotient, each fractional uncertainty contributes positively, so convert every absolute uncertainty into a percentage first.
$$\%\delta m = 0.2\%,\ \%\delta l = 1.25\%,\ \%\delta b = 2.5\%,\ \%\delta h = 5.0\%$$
Add the fractional uncertainties for density
Because $\rho = m/(lbh)$, the percentage uncertainty is the sum of the percentage uncertainties of all four measured quantities.
$$\%\delta \rho = 0.2 + 1.25 + 2.5 + 5.0 = 8.95\%$$
7
Which of the following expressions for energy is dimensionally incorrect? ($m$ = mass, $v$ = velocity, $g$ = gravitational acceleration, $h$ = height, $k$ = spring constant with dimensions $MT^{-2}$)
Ans
$E = \frac{1}{2}kv^2$
Review
Tap to reveal
Check the dimensions of the suspicious term
The only way to test each formula is to replace every symbol by its base dimensions and compare with energy.
$$[kv^2] = [MT^{-2}][L^2T^{-2}] = [ML^2T^{-4}]$$
Compare with energy
Energy must have dimensions $[ML^2T^{-2}]$, so the extra factor of $T^{-2}$ makes this expression inconsistent.
Option C is dimensionally incorrect.
8
The dimensional formula of Planck's constant $h$ (where $E = hf$, $E$ is energy and $f$ is frequency) is:
Ans
$ML^2T^{-1}$
Review
Tap to reveal
Solve the relation for $h$
Since $E = hf$, divide energy by frequency to isolate Planck's constant.
$$[h] = \frac{[E]}{[f]}$$
Substitute energy and frequency dimensions
Energy has dimensions $[ML^2T^{-2}]$ and frequency has dimensions $[T^{-1}]$, so the time power shifts by one.
$$[h] = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$$
9
Using dimensional analysis to derive the speed of sound $v$ in a medium depending on density $\rho$ and modulus of elasticity $E$ (stress/strain), the correct relation is:
Ans
$v \propto \sqrt{E/\rho}$
Review
Tap to reveal
Assume a power-law form
Write the speed as $v = kE^a\rho^b$ and match the dimensions of each factor to those of velocity.
$$[LT^{-1}] = [ML^{-1}T^{-2}]^a[ML^{-3}]^b$$
Match the exponents
Equating powers gives $a = \tfrac12$ from time and $b = -\tfrac12$ from mass, which leaves the length power consistent.
$$v = kE^{1/2}\rho^{-1/2} = k\sqrt{\frac{E}{\rho}}$$
10
A stopwatch with least count 0.1 s is used to time 50 vibrations of a pendulum, recording 102.5 s. What is the uncertainty in a single time period?
Ans
0.002 s
Review
Tap to reveal
Use the stopwatch least count for total-time uncertainty
The timing error applies to the full 50-vibration measurement, so the total uncertainty is taken as the least count of the stopwatch.
$$\delta t = 0.1\ \text{s}$$
Spread the uncertainty over one period
Since one period is the total time divided by 50, the uncertainty per vibration is also divided by 50.
$$\delta T = \frac{0.1}{50} = 0.002\ \text{s}$$
11
The centripetal acceleration $a$ of a particle moving in a circle depends on radius $r$ and speed $v$. Using dimensional analysis, if $a \propto r^n v^m$, what are $n$ and $m$?
Ans
$n = -1, m = 2$
Review
Tap to reveal
Write the dimensional form of the assumed law
Replace $r$ and $v$ by their base dimensions and force the result to match acceleration.
$$[r^nv^m] = L^{n+m}T^{-m} = [LT^{-2}]$$
Equate powers of $L$ and $T$
Matching time gives $m = 2$, and then matching length gives $n = -1$.
$$a \propto r^{-1}v^2 = \frac{v^2}{r}$$
12
Which of the following pairs have the same dimensions?
Ans
Work and Torque
Review
Tap to reveal
Check the force-length combinations
Work and torque each combine force with a length, so both collapse to the same dimensional formula.
$$[W] = [F][d] = [ML^2T^{-2}],\qquad [\tau] = [F][r] = [ML^2T^{-2}]$$
Reject the other options
The other pairs involve extra division by area, time, or time again, so their dimensions are not the same.
Work and Torque are the matching pair.
13
Surface tension is defined as force per unit length. What are the dimensions of surface tension?
Ans
$[MT^{-2}]$
Review
Tap to reveal
Use the definition force per unit length
Surface tension is force divided by length, so its dimensions come from dividing the force dimensions by one power of length.
$$[\gamma] = \frac{[F]}{[L]} = \frac{[MLT^{-2}]}{[L]}$$
Cancel the length dimension
After dividing by length, the remaining base dimensions are mass and time only.
$$[\gamma] = [MT^{-2}]$$
14
The frequency $f$ of a stretched string depends on tension $T$, length $L$, and mass per unit length $\mu$. By dimensional analysis, $f$ is proportional to:
Ans
$\sqrt{T/\mu}/L$
Review
Tap to reveal
Test the candidate combination
For a frequency, the expression must reduce to $T^{-1}$, so check the dimensions of the proposed form directly.
$$\frac{\sqrt{T/\mu}}{L} \sim \frac{\sqrt{[MLT^{-2}]/[ML^{-1}]}}{[L]}$$
Simplify the dimensions
The square root becomes $[LT^{-1}]$, and dividing by $L$ leaves exactly the dimensions of frequency.
$$\frac{\sqrt{[L^2T^{-2}]}}{[L]} = [T^{-1}]$$
15
Which pair does NOT have the same dimensions?
Ans
Torque and Power
Review
Tap to reveal
Compare torque and power directly
Torque is force times distance, while power is work per time, so the time power is different immediately.
$$[\tau] = [ML^2T^{-2}],\qquad [P] = [ML^2T^{-3}]$$
Select the non-matching pair
Since the dimensions differ by one power of time, this is the only option that does not match.
Torque and Power do not have the same dimensions.
16
What is the dimensional formula of angular momentum?
Ans
$[ML^2T^{-1}]$
Review
Tap to reveal
Use the definition of angular momentum
For a particle in circular motion, angular momentum is mass times velocity times radius.
$$L = mvr$$
Replace each factor by its dimension
Multiply the base dimensions of mass, speed, and length to get the final dimensional formula.
$$[L] = [M][LT^{-1}][L] = [ML^2T^{-1}]$$
17
If $[A] = [ML^2T^{-2}]$ and $[B] = [MT^{-2}]$, what are the dimensions of $\frac{A}{B}$?
Ans
$[L^2]$
Review
Tap to reveal
Divide the two dimensional formulas
Subtract the powers of matching base quantities when taking the ratio $A/B$.
$$\frac{[A]}{[B]} = \frac{[ML^2T^{-2}]}{[MT^{-2}]}$$
Cancel common factors
Mass and time cancel exactly, leaving only the length term.
$$\frac{[A]}{[B]} = [L^2]$$
Question 1 of 17
0:00
1
Find the dimensional formula for the Universal Gravitational Constant ($G$) given the formula $F = G \frac{m_1 m_2}{r^2}$.