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Units and Measurements question

2025 · 3 Apr · Shift 2 · Q64
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Units and Measurements question

2025 · 3 Apr · Shift 2 · Q64

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1

 Match the LIST-I with LIST-II \text { Match the LIST-I with LIST-II } Match the LIST-I with LIST-II 

LIST-I

LIST-II
A.
 Boltzmann constant \text { Boltzmann constant } Boltzmann constant 

I
ML2 T−1\mathrm{ML}^2 \mathrm{~T}^{-1}ML2 T−1

B
 Coefficient of viscosity \text { Coefficient of viscosity } Coefficient of viscosity 

II
MLT−3 K−1\mathrm{MLT}^{-3} \mathrm{~K}^{-1}MLT−3 K−1

C
 Planck’s constant \text { Planck's constant } Planck’s constant 

III
ML2 T−2 K−1\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}ML2 T−2 K−1

D
 Thermal conductivity \text { Thermal conductivity } Thermal conductivity 

IV
ML−1 T−1\mathrm{ML}^{-1} \mathrm{~T}^{-1}ML−1 T−1

Choose the correct answer from the options given below:
  1. A
    A - III, B - IV, C - I, D - II
  2. B
    A - III, B - IV, C - II, D - I
  3. C
    A - III, B - II, C - I, D - IV
  4. D
    A - II, B - III, C - IV, D - I
View written solutionFree

Correct answer: A

  1. Find dimensions of each quantity in LIST-I

We match each physical quantity with its dimensional formula.


  1. A. Boltzmann constant

Boltzmann constant kBk_BkB​ has unit: energytemperature\frac{\text{energy}}{\text{temperature}}temperatureenergy​

Since energy has dimensions: [E]=ML2T−2[E] = ML^2T^{-2}[E]=ML2T−2

So, [kB]=ML2T−2K=ML2T−2K−1[k_B] = \frac{ML^2T^{-2}}{K} = ML^2T^{-2}K^{-1}[kB​]=KML2T−2​=ML2T−2K−1

This matches III.

So, A→IIIA \to IIIA→III


  1. B. Coefficient of viscosity

Coefficient of viscosity η\etaη has dimensions: [η]=ML−1T−1[\eta] = ML^{-1}T^{-1}[η]=ML−1T−1

This matches IV.

So, B→IVB \to IVB→IV


  1. C. Planck's constant

Planck's constant hhh has unit: energy×time\text{energy} \times \text{time}energy×time

Thus, [h]=(ML2T−2)(T)=ML2T−1[h] = (ML^2T^{-2})(T) = ML^2T^{-1}[h]=(ML2T−2)(T)=ML2T−1

This matches I.

So, C→IC \to IC→I


  1. D. Thermal conductivity

Thermal conductivity kkk comes from Fourier’s law: Qt=kAΔTL\frac{Q}{t} = kA\frac{\Delta T}{L}tQ​=kALΔT​

So, k=Q/t⋅LAΔTk = \frac{Q/t \cdot L}{A\Delta T}k=AΔTQ/t⋅L​

Now,

  • Q/tQ/tQ/t is power ⇒ML2T−3\Rightarrow ML^2T^{-3}⇒ML2T−3
  • Multiply by LLL ⇒ML3T−3\Rightarrow ML^3T^{-3}⇒ML3T−3
  • Divide by area L2L^2L2 ⇒MLT−3\Rightarrow MLT^{-3}⇒MLT−3
  • Divide by temperature KKK ⇒MLT−3K−1\Rightarrow MLT^{-3}K^{-1}⇒MLT−3K−1

Thus, [k]=MLT−3K−1[k] = MLT^{-3}K^{-1}[k]=MLT−3K−1

This matches II.

So, D→IID \to IID→II


  1. Final matching

A→III,B→IV,C→I,D→IIA \to III, \quad B \to IV, \quad C \to I, \quad D \to IIA→III,B→IV,C→I,D→II

This corresponds to Option A.

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