Verified Solution • 2025–26 Session moderate

QJEEM19Ph1753: JEE Main 2019 Physics - Class 11 JEEM19Ph1753 Solution: Which of the following combinations has the dimension of…

Which of the following combinations has the dimension of electrical resistance ( ∈\in 0 is the permittivity of vacuum and μ\mu 0 is the permeability of vacuum)?

Multiple Choice Options

  • A ∈0μ0\sqrt {{{{ \in _0}} \over {{\mu _0}}}}
  • B ∈0μ0{{{{ \in _0}} \over {{\mu _0}}}}
  • C μ0∈0\sqrt {{{{\mu _0}} \over {{ \in _0}}}}
  • D μ0∈0{{{{\mu _0}} \over {{ \in _0}}}}

Step-by-Step Verified Solution

According to Coulomb's law

F = 14π∈0q2r2{1 \over {4\pi { \in _0}}}{{{q^2}} \over {{r^2}}}

∴\therefore ∈0=14πq2Fr2{ \in _0} = {1 \over {4\pi }}{{{q^2}} \over {F{r^2}}}

[∈0]=[AT]2[MLT−2][L2]\left[ {{ \in _0}} \right] = {{{{\left[ {AT} \right]}^2}} \over {\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]}} = [M−1L−3T4A2]\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]

Force between two parallel current carrying wires,

FL=μ02πi2r{F \over L} = {{{\mu _0}} \over {2\pi }}{{{i^2}} \over r}

∴\therefore μ0{{\mu _0}} = 2πrFi2L{{2\pi rF} \over {{i^2}L}}

[μ0]=[L][MLT−2][A2][L]\left[ {{\mu _0}} \right] = {{\left[ L \right]\left[ {ML{T^{ - 2}}} \right]} \over {\left[ {{A^2}} \right]\left[ L \right]}} = [MLT−2A−2]\left[ {ML{T^{ - 2}}{A^{ - 2}}} \right]

From Ohm's law,

V = IR

∴\therefore R=VIR = {V \over I} =UIt×1I= {U \over {It}} \times {1 \over I}

[R] = [U][I]2[t]{{\left[ U \right]} \over {{{\left[ I \right]}^2}\left[ t \right]}} = [ML2T−2][A2][T]{{\left[ {M{L^2}{T^{ - 2}}} \right]} \over {\left[ {{A^2}} \right]\left[ T \right]}} = [ML2T−3A−2]\left[ {M{L^2}{T^{ - 3}}{A^{ - 2}}} \right]

Let, R ∝\propto [∈0]a[μ0]b{\left[ {{ \in _0}} \right]^a}{\left[ {{\mu _0}} \right]^b}

⇒\Rightarrow [ML2T−3A−2]\left[ {M{L^2}{T^{ - 3}}{A^{ - 2}}} \right] =

[M−1L−3T4A2]\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right] aa [MLT−2A−2]\left[ {ML{T^{ - 2}}{A^{ - 2}}} \right] b

⇒\Rightarrow [ML2T−3A−2]\left[ {M{L^2}{T^{ - 3}}{A^{ - 2}}} \right] =

[ M- aa + b L-3 aa + b T4 aa - 2b A2 aa - 2b ]

By comparing both sides we get,

- aa + b = 1

- 3 aa + b = 2

By solving we get,

aa = −12- {1 \over 2} and b = 12{1 \over 2}

∴\therefore R = [∈0]−12[μ0]12{\left[ {{ \in _0}} \right]^{ - {1 \over 2}}}{\left[ {{\mu _0}} \right]^{{1 \over 2}}} = μ0∈0\sqrt {{{{\mu _0}} \over {{ \in _0}}}}

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