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Units and Measurements
Dimensional Analysis
17 questions from 16 papers, each with its answer.
- questions
- 17
- papers
- 16
- years asked, 2017–2026
- 2 of 10
- of Units and Measurements
- 32.08%
Appears in
Concepts tested
- Change of fundamental units 2
- Dimensions of physical quantities 2
- Comparing dimensions
- Deriving a relation by dimensions
- Dimensional analysis
- Dimensionless quantities: solid angle vs strain/angle
- Dimensions of energy density
- Dimensions of h and I
- Dimensions of physical constants
- Dimensions of Planck's constant
- Dimensions of resistance
- energy in terms of F, A and T
- Finding dimensionless combination of constants
- Principle of homogeneity
- Speed of light from constants
All Questions
2024 2 questions
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The quantities which have the same dimensions as those of solid angle are;Solid angle $= \frac{area}{r^2}$ is dimensionless. Strain (ratio of two lengths) and plane angle ($\frac{arc}{radius}$) are also dimensionless. So strain and angle have the same dimensions as solid angle.
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A force defined by $F=\alpha t^2+\beta t$ acts on a particle at a given time $t$. The factor which is dimensionless, if $\alpha$ and $\beta$ are constants, is:$F = \alpha t^2 + \beta t$ $[\alpha t^2] = [M^1L^1T^{-2}] \Rightarrow [\alpha] = [M^1L^1T^{-4}]$ $[\beta t] = [M^1L^1T^{-2}] \Rightarrow [\beta] = [M^1L^1T^{-3}]$ $\left[\frac{\alpha t}{\beta}\right] = \frac{[M^1L^1T^{-4}][T]}{[M^1L^1T^{-3}]} = [M^0L^0T^0]$ So $\frac{\alpha t}{\beta}$ is dimensionless.
2024 Re-NEET 1 question
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The potential energy of a particle moving along x-direction varies as $V=\dfrac{Ax^2}{\sqrt{x+B}}$. The dimensions of $\dfrac{A^2}{B}$ are:By homogeneity, $B=[L]$. Since $V=[ML^2T^{-2}]=\dfrac{A[L^2]}{[L^{1/2}]}$, $A=[ML^{1/2}T^{-2}]$. So $\dfrac{A^2}{B}=\dfrac{M^2LT^{-4}}{L^{1/2}}=[M^2L^{1/2}T^{-4}]$.
2021 1 question
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If force $[F]$, acceleration $[A]$ and time $[T]$ are chosen as the fundamental physical quantities. Find the dimensions of energy.Let $E \propto F^aA^bT^c$ $[ML^2T^{-2}] = [MLT^{-2}]^a[LT^{-2}]^b[T]^c$ M: a = 1 ; L: a + b = 2 ⇒ b = 1 ; T: $-2a - 2b + c = -2$ ⇒ c = 2 $E = [F][A][T^2]$
2015 AIPMT-I 1 question
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If energy (E), velocity (V) and time (T) are chosen as fundamental quantities, the dimensional formula of surface tension will beLet $\sigma = E^aV^bT^c$.
$[ML^0T^{-2}] = [ML^2T^{-2}]^a\,[LT^{-1}]^b\,[T]^c = [M^a L^{2a+b} T^{-2a-b+c}]$
Comparing powers: $a = 1$; $2a + b = 0 \Rightarrow b = -2$; $-2a - b + c = -2 \Rightarrow c = -2$.
So the dimensional formula of surface tension is $[EV^{-2}T^{-2}]$.
2015 AIPMT-II 1 question
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If dimensions of critical velocity $v_c$ of liquid flowing through a tube are expressed as $[\eta^x\rho^yr^z]$, where $\eta$, $\rho$ and r are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of x, y and z are given by:$[LT^{-1}] = [ML^{-1}T^{-1}]^x[ML^{-3}]^y[L]^z = [M^{x+y}L^{-x-3y+z}T^{-x}]$
$-x = -1 \Rightarrow x = 1$; $x + y = 0 \Rightarrow y = -1$; $-x - 3y + z = 1 \Rightarrow z = -1$
2014 AIPMT 1 question
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If force (F), velocity (V) and time (T) are taken as fundamental units, then the dimensions of mass are:$[\text{mass}] = \frac{\text{Force}}{\text{Acceleration}} = \frac{\text{Force}}{\text{Velocity/time}}$
$[\text{mass}] = [FV^{-1}T]$
2012 AIPMT-MAINS 1 question
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The dimensions of $(\mu_0\varepsilon_0)^{-1/2}$ areThe speed of light in vacuum is $c = \frac{1}{\sqrt{\mu_0\varepsilon_0}} = (\mu_0\varepsilon_0)^{-1/2}$
So its dimensions are those of velocity: $[LT^{-1}]$
2011 AIPMT-PRE 1 question
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The dimensions of $(\mu_0\varepsilon_0)^{-\frac{1}{2}}$ areThe speed of light in vacuum is related to the permeability and permittivity of free space by $c = \frac{1}{\sqrt{\mu_0\varepsilon_0}}$
So $(\mu_0\varepsilon_0)^{-\frac{1}{2}} = \frac{1}{\sqrt{\mu_0\varepsilon_0}} = c$
The given expression is therefore a speed, and speed = distance/time.
Dimensions of $(\mu_0\varepsilon_0)^{-\frac{1}{2}}$ = $[LT^{-1}]$
2010 AIPMT-PRE 1 question
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The dimension of $\frac{1}{2}\varepsilon_0E^2$, where $\varepsilon_0$ is permittivity of free space and E is electric field, is$\frac{1}{2}\varepsilon_0E^2$ is the energy stored per unit volume in an electric field, i.e., the energy density.
Energy density = $\frac{\text{energy}}{\text{volume}}$
Dimensions of energy = $[ML^2T^{-2}]$; dimensions of volume = $[L^3]$
Dimensions of energy density = $\frac{[ML^2T^{-2}]}{[L^3]}$
$= [ML^{-1}T^{-2}]$
2009 AIPMT 1 question
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If the dimensions of a physical quantity are given by $[M^aL^bT^c]$, then the physical quantity will beForce = mass × acceleration: $[F] = [MLT^{-2}]$, i.e., a = 1, b = 1, c = –2. So the first statement is wrong.
Pressure = force/area: $[P] = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$, i.e., a = 1, b = –1, c = –2. This matches.
Velocity: $[v] = [LT^{-1}]$, i.e., a = 0, b = 1, c = –1. So the third statement is wrong.
Acceleration: $[LT^{-2}]$, i.e., a = 0, b = 1, c = –2. So the fourth statement is wrong.
Hence the quantity is pressure if a = 1, b = –1, c = –2.
2008 AIPMT 1 question
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Which two of the following five physical parameters have the same dimensions ?
(a) Energy density
(b) Refractive index
(c) Dielectric constant
(d) Young's modulus
(e) Magnetic fieldEnergy density $= \dfrac{\text{energy}}{\text{volume}} = \dfrac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}]$
Young's modulus $= \dfrac{\text{stress}}{\text{strain}} = \dfrac{\text{force/area}}{\text{dimensionless}} = \dfrac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$
So energy density and Young's modulus have the same dimensions, $[ML^{-1}T^{-2}]$.
Refractive index is a ratio of two speeds and the dielectric constant is a ratio of two permittivities, so both are dimensionless.
Magnetic field $B = \dfrac{F}{qv}$ has dimensions $[MT^{-2}A^{-1}]$, which matches none of the others.
Among the given pairs, the pair with the same dimensions is (a) energy density and (d) Young's modulus.
2007 AIPMT-PRE 1 question
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Dimensions of resistance in an electrical circuit, in terms of dimension of mass M, of length L, of time T and of current I, would beResistance: $R = \dfrac{V}{I}$
Potential difference $V = \dfrac{\text{work}}{\text{charge}} = \dfrac{[ML^2T^{-2}]}{[IT]} = [ML^2T^{-3}I^{-1}]$
$[R] = \dfrac{[V]}{[I]} = \dfrac{[ML^2T^{-3}I^{-1}]}{[I]}$
$[R] = [ML^2T^{-3}I^{-2}]$
Check using power: $P = I^2R \Rightarrow [R] = \dfrac{[ML^2T^{-3}]}{[I^2]} = [ML^2T^{-3}I^{-2}]$
2006 AIPMT-PRE 1 question
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The velocity $v$ of a particle at time $t$ is given by $v = at + \dfrac{b}{t + c}$, where $a$, $b$ and $c$ are constants. The dimensions of $a$, $b$ and $c$ are respectively :-By the principle of homogeneity, only quantities of the same dimensions can be added, and every term of an equation must have the same dimensions.
In $(t + c)$, $c$ is added to time, so $[c] = [t] = T$.
$[v] = [at] \Rightarrow [a] = \dfrac{[v]}{[t]} = \dfrac{LT^{-1}}{T} = LT^{-2}$
$[v] = \dfrac{[b]}{[t + c]} \Rightarrow [b] = [v][t] = (LT^{-1})(T) = L$
So the dimensions of $a$, $b$ and $c$ are $LT^{-2}$, $L$ and $T$ respectively.
2005 AIPMT 1 question
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The ratio of the dimension of Planck's constant and that of the moment of inertia is the dimension of :Planck's constant $h = \dfrac{E}{\nu}$, so its unit is J s: $[h] = [ML^2T^{-2}][T] = [ML^2T^{-1}]$
Moment of inertia $I = mr^2$, with unit kg m$^2$: $[I] = [ML^2]$
$\dfrac{[h]}{[I]} = \dfrac{ML^2T^{-1}}{ML^2}$
$= [T^{-1}]$
$[T^{-1}]$ is the dimension of frequency (also of angular velocity).
So the ratio has the dimension of frequency.
2001 AIPMT 1 question
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The dimension of Planck constant equals to that of :From $E = h\nu$: $h = \dfrac{E}{\nu}$
$[h] = \dfrac{ML^2T^{-2}}{T^{-1}} = ML^2T^{-1}$
Angular momentum $L = mvr$: $[L] = M \cdot LT^{-1} \cdot L = ML^2T^{-1}$
So Planck's constant has the same dimensions as angular momentum (this is why Bohr's condition $mvr = \dfrac{nh}{2\pi}$ is dimensionally correct).
Energy is $ML^2T^{-2}$, momentum is $MLT^{-1}$ and power is $ML^2T^{-3}$.
2000 AIPMT 1 question
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Which pair does not have equal dimensions :Energy and torque: both $ML^2T^{-2}$; equal.
Force: $MLT^{-2}$; impulse = force $\times$ time: $MLT^{-1}$; not equal.
Angular momentum and Planck's constant: both $ML^2T^{-1}$; equal.
Elastic modulus and pressure: both $ML^{-1}T^{-2}$; equal.
So force and impulse do not have equal dimensions.