Verified Solution • 2025–26 Session moderate

QJEEAdv16Ph7c5b: JEE Advanced 2016 Physics - Class 11 JEEAdv16Ph7c5b Solution: A length-scale (l) depends on the permittivity ( ) of a…

A length-scale (l) depends on the permittivity ( ε\varepsilon ) of a dielectric material, Boltzmann constant (kB), the absolute temperature (T), the number per unit volume (n) of certain charged particles, and the charge (q) carried by each of the particles. Which of the following expression(s) for I is(are) dimensionally correct?

Multiple Choice Options

  • A l=(nq2εkbT)l = \sqrt {\left( {{{n{q^2}} \over {\varepsilon {k_b}T}}} \right)}
  • B l=(εkbTnq2)l = \sqrt {\left( {{{\varepsilon {k_b}T} \over {n{q^2}}}} \right)}
  • C l=(q2εn2/3kBT)l = \sqrt {\left( {{{{q^2}} \over {\varepsilon {n^{2/3}}{k_B}T}}} \right)}
  • D l=(q2εn1/3kBT)l = \sqrt {\left( {{{{q^2}} \over {\varepsilon {n^{1/3}}{k_B}T}}} \right)}

Step-by-Step Verified Solution

[n]=[L−3];[q]=[AT][n] = [{L^{ - 3}}];[q] = [AT]

[ε]=[M−1L−3A2T4][\varepsilon ] = [{M^{ - 1}}{L^{ - 3}}{A^2}{T^4}]

[T]=[L][T] = [L]

[l]=[L][l] = [L]

[kB]=[M1L2T−2K−1][{k_B}] = [{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}]

(a) RHS

=[L−3A2T2][M−1L−3T4A2][M1L2T−2K−1][K]= \sqrt {{{[{L^{ - 3}}{A^2}{T^2}]} \over {[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}][{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][K]}}}

=[L−3A2T2][L−1A2T2]=[L−2]=[L−1]= \sqrt {{{[{L^{ - 3}}{A^2}{T^2}]} \over {[{L^{ - 1}}{A^2}{T^2}]}}} = \sqrt {[{L^{ - 2}}]} = [{L^{ - 1}}] Wrong

(b) RHS

=[M−1L−3T4A2][M1L2T−2K−1][K][L−3A2T2]= \sqrt {{{[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}][{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][K]} \over {[{L^{ - 3}}{A^2}{T^2}]}}}

=[L−1A2T2][L−3A2T2]=L= \sqrt {{{[{L^{ - 1}}{A^2}{T^2}]} \over {[{L^{ - 3}}{A^2}{T^2}]}}} = L Correct

(c) RHS

=[A2T2][M−1L−3T4A2][L−2][M1L2T−2K−1][K]= \sqrt {{{[{A^2}{T^2}]} \over {[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}][{L^{ - 2}}][{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][K]}}}

=[L3]= \sqrt {[{L^3}]} Wrong

(d) RHS

=[A2T2][M−1L−3T4A2][L−1][M1L2T−2K−1][K]= \sqrt {{{[{A^2}{T^2}]} \over {[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}][{L^{ - 1}}][{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][K]}}}

=[A2T2][L−2T2A2]=[L]= \sqrt {{{[{A^2}{T^2}]} \over {[{L^{ - 2}}{T^2}{A^2}]}}} = [L] Correct

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