Physics · Units, Dimensions & Error Analysis

JEE Main 2026 — 22 January, Morning Shift — Question 42

Match the LIST-I with LIST-II

LIST-ILIST-II
A. Spring constantI. ML2 T−2 K−1\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~K}^{-1}
B. Thermal conductivityII. ML0 T−2\mathrm{ML}^{0} \mathrm{~T}^{-2}
C. Boltzmann constanIII. ML2 T−3 A−2\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}
D. Inductive reactanceIV. MLT−3 K−1\mathrm{MLT}^{-3} \mathrm{~K}^{-1}

Choose the correct answer from the options given below:

  1. Option A:

    A−II,B−I,C−IV,D−IIIA-II, B-I, C-IV, D-III

  2. Option B:

    A−I,B−IV,C−II,D−IIIA-I, B-IV, C-II, D-III

  3. Option C:

    A−III,B−II,C−IV,D−IA-III, B-II, C-IV, D-I

  4. Option D:

    A−II,B−IV,C−I,D−IIIA-II, B-IV, C-I, D-III

    Correct

Answer: D

Step-by-step solution

(A) F=KxF=K x [MLT−2]=[K][L]\left[\mathrm{MLT}^{-2}\right]=[\mathrm{K}][\mathrm{L}] [K]=ML0 T−2[\mathrm{K}]=\mathrm{ML}^{0} \mathrm{~T}^{-2} (B) Thermal conductivity dQdt=kAℓΔT\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{kA}}{\ell} \Delta \mathrm{T} ML2 T−3=[k]L2 K L\mathrm{ML}^{2} \mathrm{~T}^{-3}=\frac{[\mathrm{k}] \mathrm{L}^{2} \mathrm{~K}}{\mathrm{~L}} [K1]=MLT−3 K−1\left[\mathrm{K}^{1}\right]=\mathrm{MLT}^{-3} \mathrm{~K}^{-1} (C) Boltzman constant [K]=ML2 T−2 K−1[\mathrm{K}]=\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~K}^{-1} (D) Inductive reactance [V][I]=ML2 T−3 A−1 A\frac{[\mathrm{V}]}{[\mathrm{I}]}=\frac{\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-1}}{\mathrm{~A}} =ML2 T−3 A−2=\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Units and Dimensions Analysis
Match the LIST-I with LIST-II LIST-I LIST-II --- --- A. Spring… | JEE Main 2026 PYQ with Solution · DhiX AI