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Units and Measurements question

2021 · 31 Aug · Shift 1 · Q59
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Units and Measurements question

2021 · 31 Aug · Shift 1 · Q59

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
Which of the following equations is dimensionally incorrect? Where t = time, h = height, s = surface tension, θ\thetaθ= angle, ρ\rhoρ= density, a, r = radius, g = acceleration due to gravity, v = volume, p = pressure, W = work done, T = torque, ∈\in∈ = permittivity, E = electric field, J = current density, L = length.
  1. A
    v=πpa48ηLv = {{\pi p{a^4}} \over {8\eta L}}v=8ηLπpa4​
  2. B
    h=2scos⁡θρrgh = {{2s\cos \theta } \over {\rho rg}}h=ρrg2scosθ​
  3. C
    J=∈∂E∂tJ = \in {{\partial E} \over {\partial t}}J=∈∂t∂E​
  4. D
    W=ΓθW = \Gamma \thetaW=Γθ
View written solutionFree

Correct answer: NO OPTION IS DIMENSIONALLY INCORRECT; A, B, C AND D ARE ALL DIMENSIONALLY CORRECT.

  1. We check dimensions of each equation.

Useful dimensions:

  • Volume flow rate / volume per unit time in Poiseuille-type formula: if written as vvv, likely volumetric flow rate, so [v]=L3T−1[v] = L^3 T^{-1}[v]=L3T−1
  • Pressure: [p]=ML−1T−2[p] = M L^{-1} T^{-2}[p]=ML−1T−2
  • Radius: [a]=[r]=L[a]=[r]=L[a]=[r]=L
  • Viscosity: [η]=ML−1T−1[\eta] = M L^{-1} T^{-1}[η]=ML−1T−1
  • Length: [L]=L[L]=L[L]=L
  • Surface tension: [s]=MT−2[s] = M T^{-2}[s]=MT−2
  • Density: [ρ]=ML−3[\rho] = M L^{-3}[ρ]=ML−3
  • Acceleration due to gravity: [g]=LT−2[g]=L T^{-2}[g]=LT−2
  • Current density: [J]=IL−2[J]=I L^{-2}[J]=IL−2
  • Permittivity: from ∇⋅E=ρ/ε\nabla\cdot E = \rho/\varepsilon∇⋅E=ρ/ε, we get [ε]=M−1L−3T4I2[\varepsilon]=M^{-1}L^{-3}T^4I^2[ε]=M−1L−3T4I2
  • Electric field: [E]=MLT−3I−1[E]=M L T^{-3} I^{-1}[E]=MLT−3I−1
  • Work: [W]=ML2T−2[W]=M L^2 T^{-2}[W]=ML2T−2
  • Torque: [Γ]=ML2T−2[\Gamma]=M L^2 T^{-2}[Γ]=ML2T−2
  • Angle θ\thetaθ is dimensionless.
  1. Check option A: v=πpa48ηLv=\frac{\pi p a^4}{8\eta L}v=8ηLπpa4​ Ignoring numerical constants, [pa4ηL]=(ML−1T−2)(L4)(ML−1T−1)(L)\left[\frac{p a^4}{\eta L}\right]=\frac{(M L^{-1}T^{-2})(L^4)}{(M L^{-1}T^{-1})(L)}[ηLpa4​]=(ML−1T−1)(L)(ML−1T−2)(L4)​ =ML3T−2MT−1=L3T−1=\frac{M L^3 T^{-2}}{M T^{-1}}=L^3 T^{-1}=MT−1ML3T−2​=L3T−1 This matches [v]=L3T−1[v]=L^3T^{-1}[v]=L3T−1. So A is dimensionally correct.
  1. Check option B: h=2scos⁡θρrgh=\frac{2s\cos\theta}{\rho r g}h=ρrg2scosθ​ Since cos⁡θ\cos\thetacosθ is dimensionless, [sρrg]=MT−2(ML−3)(L)(LT−2)\left[\frac{s}{\rho r g}\right]=\frac{M T^{-2}}{(M L^{-3})(L)(L T^{-2})}[ρrgs​]=(ML−3)(L)(LT−2)MT−2​ =MT−2ML−1T−2=L=\frac{M T^{-2}}{M L^{-1} T^{-2}}=L=ML−1T−2MT−2​=L This matches [h]=L[h]=L[h]=L. So B is dimensionally correct.
  1. Check option C: J=ε∂E∂tJ=\varepsilon \frac{\partial E}{\partial t}J=ε∂t∂E​ Now, [ε∂E∂t]=[ε][E]T−1\left[\varepsilon \frac{\partial E}{\partial t}\right]=[\varepsilon][E]T^{-1}[ε∂t∂E​]=[ε][E]T−1 =(M−1L−3T4I2)(MLT−3I−1)T−1=(M^{-1}L^{-3}T^4I^2)(M L T^{-3}I^{-1})T^{-1}=(M−1L−3T4I2)(MLT−3I−1)T−1 =IL−2=I L^{-2}=IL−2 This matches [J]=IL−2[J]=I L^{-2}[J]=IL−2. So C is dimensionally correct.
  1. Check option D: W=ΓθW=\Gamma\thetaW=Γθ Since angle is dimensionless, [Γθ]=[Γ]=ML2T−2[\Gamma\theta]=[\Gamma]=M L^2 T^{-2}[Γθ]=[Γ]=ML2T−2 which matches work. So D is dimensionally correct.
  1. Conclusion All four equations are dimensionally correct. Therefore no option is dimensionally incorrect.

Hence, the stored answer A is not correct.

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