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JEE Main 202622 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Morning Shift
Subject
Physics
Chapter
Units and Dimensions

JEE Main 2026 · 22 January · Morning Shift · Physics · Units and Dimensions

Match the List-I with List-II ( List-I List-II A. Spring constant...

Match the List-I with List-II
(
List-IList-II
A.Spring constantI.ML2T−2K−1\mathrm{M L^{2} T^{-2} K^{-1}}
B.Thermal conductivityII.ML0T−2\mathrm{M L^{0} T^{-2}}
C.Boltzmann constantIII.ML2T−3A−2\mathrm{M L^{2} T^{-3} A^{-2}}
D.Inductive reactanceIV.MLT−3K−1\mathrm{M L T^{-3} K^{-1}}
)
Choose the correct answer from the options given below:

Options

  1. AA-II, B-IV, C-I, D-III
  2. BA-I, B-IV, C-II, D-III
  3. CA-II, B-I, C-IV, D-III
  4. DA-III, B-II, C-IV, D-I

Correct answer

A. A-II, B-IV, C-I, D-III

Step-by-step solution

A. Spring constant: F=kx⇒[k]=MLT−2/L=ML0T−2F = kx \Rightarrow [k] = MLT^{-2}/L = ML^0T^{-2} →\rightarrow II
B. Thermal conductivity: Q/t=KA(dT/dx)⇒[K]=ML2T−3/(L2⋅K/L)=MLT−3K−1Q/t = KA(dT/dx) \Rightarrow [K] = ML^2T^{-3}/(L^2 \cdot K/L) = MLT^{-3}K^{-1} →\rightarrow IV
C. Boltzmann constant: E=kBT⇒[kB]=ML2T−2K−1E = k_BT \Rightarrow [k_B] = ML^2T^{-2}K^{-1} →\rightarrow I
D. Inductive reactance: same dimension as resistance =ML2T−3A−2= ML^2T^{-3}A^{-2} →\rightarrow III
Matching: A-II, B-IV, C-I, D-III.
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