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Physics · Units, Dimensions & Error Analysis

JEE Main 2024 — 31 January, Shift 2 — Question 47

Consider two physical quantities A and B related to each other as E=B−x2AtE=\frac{B-x^{2}}{A t} where E,xE, x and tt have dimensions of energy, length and time respectively. The dimension of AB is

  1. Option A:

    L−2M1 T0\mathrm{L}^{-2} \mathrm{M}^{1} \mathrm{~T}^{0}

  2. Option B:

    L2M−1 T1\mathrm{L}^{2} \mathrm{M}^{-1} \mathrm{~T}^{1}

    Correct
  3. Option C:

    L−2M−1 T1\mathrm{L}^{-2} \mathrm{M}^{-1} \mathrm{~T}^{1}

  4. Option D:

    L0M−1 T1\mathrm{L}^{0} \mathrm{M}^{-1} \mathrm{~T}^{1}

Answer: B

Step-by-step solution

[B]=L2[B]=L^{2}

A=x2tE=L2TML2 T−2=1MT−1\mathrm{A}=\frac{\mathrm{x}^{2}}{\mathrm{tE}}=\frac{\mathrm{L}^{2}}{\mathrm{TML}^{2} \mathrm{~T}^{-2}}=\frac{1}{\mathrm{MT}^{-1}} [A]=M−1 T[A]=\mathrm{M}^{-1} \mathrm{~T}

[AB]=[L2M−1 T1][\mathrm{AB}]=\left[\mathrm{L}^{2} \mathrm{M}^{-1} \mathrm{~T}^{1}\right]

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
31 January, Shift 2
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Units and Dimensions Analysis

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