Physics · Units, Dimensions & Error Analysis

JEE Main 2025 — 3 April, Evening Shift — Question 58

Match the LIST-I with LIST-II. Choose the correct answer from the options given below.

LIST-ILIST-II
A.Boltzmann constantI.ML2  ⁣ ⁣  ⁣ ⁣ T−1\text{M}{{\text{L}}^{2}}\text{ }\!\!~\!\!\text{ }{{\text{T}}^{-1}}
B.Coefficient of viscosityII.MLT−3  ⁣ ⁣  ⁣ ⁣ K−1\text{ML}{{\text{T}}^{-3}}\text{ }\!\!~\!\!\text{ }{{\text{K}}^{-1}}
C.Planck's constantIII.ML2  ⁣ ⁣  ⁣ ⁣ T−2  ⁣ ⁣  ⁣ ⁣ K−1\text{M}{{\text{L}}^{2}}\text{ }\!\!~\!\!\text{ }{{\text{T}}^{-2}}\text{ }\!\!~\!\!\text{ }{{\text{K}}^{-1}}
D.Thermal conductivityIV.ML−1  ⁣ ⁣  ⁣ ⁣ T−1\text{M}{{\text{L}}^{-1}}\text{ }\!\!~\!\!\text{ }{{\text{T}}^{-1}}
  1. Option A:

    A−III,B−IV,C−I,D−IIA-I I I, B-I V, C-I, D-I I

    Correct
  2. Option B:

    A−III,B−II,C−I,D−IVA-I I I, B-I I, C-I, D-I V

  3. Option C:

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

  4. Option D:

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

Answer: A

Step-by-step solution

kt=k t= Energy ⇒k=ML2 T−2 K−1\Rightarrow k=\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~K}^{-1}

A→IIIA \rightarrow \mathrm{III}

F=6πηru⇒η=MLT−2LLT−1F=6 \pi \eta r u \Rightarrow \eta=\frac{\mathrm{MLT}^{-2}}{\mathrm{LLT}^{-1}}

η=ML−1 T−1\eta=\mathrm{ML}^{-1} \mathrm{~T}^{-1}

B →\rightarrow IV

E=hν=h=ML2 T−1E=h \nu=h=\mathrm{ML}^{2} \mathrm{~T}^{-1}

C→IC \rightarrow I

H=kAΔTl=k=ML2 T−3LK=MLT−3 K−1H=\frac{k A \Delta T}{l}=k=\frac{\mathrm{ML}^{2} \mathrm{~T}^{-3}}{\mathrm{LK}}=\mathrm{MLT}^{-3} \mathrm{~K}^{-1} D →\rightarrow II

Answer key and solution verified before publishing.

Practise Units, Dimensions & Error Analysis

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Exam
JEE Main 2025
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Units and Dimensions Analysis
Match the LIST-I with LIST-II. Choose the correct answer from the… | JEE Main 2025 PYQ with Solution · DhiX AI