Physics · Units, Dimensions & Error Analysis

JEE Main 2024 — 30 January, Shift 1 — Question 31

Match List -I with List -II

List-IList-II
A.Coefficient of viscosityI.[ML2  T−2]\left[ {{\rm{M}}{{\rm{L}}^2}{\rm{\;}}{{\rm{T}}^{ - 2}}} \right]
B.Surface TensionII.[ML2  T−1]\left[ {{\rm{M}}{{\rm{L}}^2}{\rm{\;}}{{\rm{T}}^{ - 1}}} \right]
C.Angular momentumIII.[ML−1  T−1]\left[ {{\rm{M}}{{\rm{L}}^{ - 1}}{\rm{\;}}{{\rm{T}}^{ - 1}}} \right]
D.Rotational kinetic energyIV.[ML0  T−2]\left[ {{\rm{M}}{{\rm{L}}^0}{\rm{\;}}{{\rm{T}}^{ - 2}}} \right]
  1. Option A:

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

  2. Option B:

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

  3. Option C:

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

    Correct
  4. Option D:

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

Answer: C

Step-by-step solution

F=ηAdvdyF=\eta A \frac{d v}{d y}

[MLT−2]=η[L2][T−1]\left[M L T^{-2}\right]=\eta\left[L^{2}\right]\left[T^{-1}\right]

η=[ML−1T−1]\eta=\left[M L^{-1} T^{-1}\right]

S.T =Fℓ=[MLT−2][L]=[ML0T−2]=\frac{F}{\ell}=\frac{\left[M L T^{-2}\right]}{[L]}=\left[M L^{0} T^{-2}\right]

L=mvr=[ML2T−1]L=m v r=\left[M L^{2} T^{-1}\right]

K⋅E=12Iω2=[ML2T−2]K \cdot E=\frac{1}{2} I \omega^{2}=\left[M L^{2} T^{-2}\right]

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Units and Dimensions Analysis
Match List -I with List -II List-I List-II --- --- --- --- A.… | JEE Main 2024 PYQ with Solution · DhiX AI