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Physics · Units, Dimensions & Error Analysis

JEE Main 2025 — 7 April, Evening Shift — Question 52

The dimension of μ0ε0\sqrt{\frac{\mu_{0}}{\varepsilon_{0}}} is equal to that of : ( μ0=\mu_{0}= Vacuum permeability and ε0=\varepsilon_{0}= Vacuum permittivity)

  1. Option A:

    Inductance

  2. Option B:

    Resistance

    Correct
  3. Option C:

    Capacitance

  4. Option D:

    Voltage

Answer: B

Step-by-step solution

[μ0]=[MLT−2 A−2]\left[\mu_{0}\right]=\left[\mathrm{MLT}^{-2} \mathrm{~A}^{-2}\right] [ε0]=[M−1L−3T4A2]\left[\varepsilon_{0}\right]=\left[M^{-1} L^{-3} T^{4} A^{2}\right]

[μ0ε0]=[ML2 T−3 A−2]≃[R]\left[\sqrt{\frac{\mu_{0}}{\varepsilon_{0}}}\right]=\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right] \simeq[R]

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
7 April, Evening Shift
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Units and Dimensions Analysis

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