Physics · Units, Dimensions & Error Analysis

JEE Main 2025 — 28 January, Evening Shift — Question 49

Match List-I with List-II

List-IList-II
(A) Angular Impulse(I) [M0 L2 T−2]\quad\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right]
(B) Latent Heat(II)[ML2 T−3 A−1]\quad\left[\mathrm{M}\mathrm{L}^{2}\mathrm{~T}^{-3}\mathrm{~A}^{-1}\right]
(C) Electrical resistivity(III) [ML2 T−1]\left[\mathrm{M} \mathrm{L}^{2} \mathrm{~T}^{-1}\right]
(D) Electromotive force(IV) [ML3 T−3 A−2]\left[\mathrm{M} \mathrm{L}^{3} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]

Choose the correct one from the options given below:

  1. Option A:

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

    Correct
  2. Option B:

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

  3. Option C:

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

  4. Option D:

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

Answer: A

Step-by-step solution

. Angular impulse =[ML2 T−1]=\left[\mathrm{M} \mathrm{L}^{2} \mathrm{~T}^{-1}\right]

Latent Heat =[M0 L2 T−2]=\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right]

Electrical resistivity =[ML3 T−3 A−2]=\left[\mathrm{M} \mathrm{L}^{3} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]

Electromotive force =[ML2 T−3 A−1]=\left[\mathrm{M} \mathrm{L}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Units and Dimensions Analysis
Match List-I with List-II List-I List-II --- --- (A) Angular Impulse… | JEE Main 2025 PYQ with Solution · DhiX AI