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Physics · Units, Dimensions & Error Analysis

JEE Main 2026 — 5 April, Morning Shift — Question 2

L, C and R represents physical quantities inductance, capacitance and resistance respectively. The dimensional formula ML2T−4A−2M L^2 T^{-4} A^{-2} corresponds to ________.

  1. Option A:

    RLC\frac{R}{LC}

    Correct
  2. Option B:

    RLCR LC

  3. Option C:

    CLR\frac{C}{LR}

  4. Option D:

    1LRC\frac{1}{L R C}

Answer: A

Step-by-step solution

[R]=ML2T−3A−2[R] = ML^2T^{-3}A^{-2}, [L]=ML2T−2A−2[L] = ML^2T^{-2}A^{-2}, [C]=M−1L−2T4A2[C] = M^{-1}L^{-2}T^4A^2. Then [R/LC]=(ML2T−3A−2)/(ML2T−2A−2⋅M−1L−2T4A2)=ML2T−4A−2[R/LC] = (ML^2T^{-3}A^{-2}) / (ML^2T^{-2}A^{-2} \cdot M^{-1}L^{-2}T^4A^2) = ML^2T^{-4}A^{-2}.

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
5 April, Morning Shift
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Units and Dimensions Analysis

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