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Physics · Units, Dimensions & Error Analysis

JEE Advanced 2024 — Paper 1 — Question 18

A dimensionless quantity is constructed in terms of electronic charge e, permittivity of free space ε0\varepsilon_{0}, Planck's constant hh, and speed of light cc. If the dimensionless quantity is written as eαε0βhγcδe^{\alpha} \varepsilon_{0}^{\beta} h^{\gamma} c^{\delta} and n is a non-zero integer, then (α,β,γ,δ)(\alpha, \beta, \gamma, \delta) is given by

  1. Option A:

    (2n,−n,−n,−n)(2 n,-n,-n,-n)

    Correct
  2. Option B:

    (n,−n,−2n,−n)(\mathrm{n},-\mathrm{n},-2 \mathrm{n},-\mathrm{n})

  3. Option C:

    (n,−n,−n,−2n)(\mathrm{n},-\mathrm{n},-\mathrm{n},-2 \mathrm{n})

  4. Option D:

    (2n,−n,−2n,−2n)(2 n,-n,-2 n,-2 n)

Answer: A

Step-by-step solution

eαε0βhγcδ=k e^{\alpha} \varepsilon_{0}^{\beta} h^{\gamma} c^{\delta}=k eαhγcδ=ε0n\mathrm{e}^{\alpha} \mathrm{h}^{\gamma} \mathrm{c}^{\delta}=\varepsilon_{0}^{\mathrm{n}} [AαTα][ML2 T−1]γ[LT−1]δ=[A2nT4nM−nL−3n]\left[A^{\alpha} T^{\alpha}\right]\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]^{\gamma}\left[\mathrm{LT}^{-1}\right]^{\delta}=\left[\mathrm{A}^{2 \mathrm{n}} \mathrm{T}^{4 \mathrm{n}} \mathrm{M}^{-\mathrm{n}} \mathrm{L}^{-3 n}\right] So, α=2n,γ=−n\alpha=2 n, \gamma=-n

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2024
Paper
Paper 1
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Units and Dimensions Analysis
A dimensionless quantity is constructed in terms of electronic charge… | JEE Advanced 2024 PYQ with Solution · DhiX AI