Physics · Units, Dimensions & Error Analysis

JEE Main 2026 — 28 January, Evening Shift — Question 45

Match List-I with List-II.

List IList II
A. Coefficient of viscosityI. [ML−1 T−2]\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]
B. Surface tensionII. [ML2 T−2]\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]
C. PressureIII. [ML0 T−2]\left[\mathrm{ML}^{0} \mathrm{~T}^{-2}\right]
D. Surface energyIV. [ML−1 T−1]\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]

Choose the correct answer from the options given below:

  1. Option A:

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

  2. Option B:

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

    Correct
  3. Option C:

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

  4. Option D:

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

Answer: B

Step-by-step solution

(A) η=FdrAdv=[MLT−2][L][L2][LT−1]=[ML−1 T−1]\eta=\frac{\mathrm{Fdr}}{\mathrm{Adv}}=\frac{\left[\mathrm{MLT}^{-2}\right][\mathrm{L}]}{\left[\mathrm{L}^{2}\right]\left[\mathrm{LT}^{-1}\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right] (B) S=FL=[MLT−2][L]=[MT−2]\mathrm{S}=\frac{\mathrm{F}}{\mathrm{L}}=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{L}]}=\left[\mathrm{MT}^{-2}\right] (C) P=FA=[MLT−2][L2]=[ML−1 T−2]\mathrm{P}=\frac{\mathrm{F}}{\mathrm{A}}=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^{2}\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] (D) E=S×A=[MT−2][L2]=[ML2 T−2]\mathrm{E}=\mathrm{S} \times \mathrm{A}=\left[\mathrm{MT}^{-2}\right]\left[\mathrm{L}^{2}\right]=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]

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 List-I with List-II. List I List II --- --- A. Coefficient of… | JEE Main 2026 PYQ with Solution · DhiX AI