Verified Solution • 2025–26 Session moderate

QJEEM19Phddbb: JEE Main 2019 Physics - Class 11 JEEM19Phddbb Solution: In SI units, the dimensions of is :

In SI units, the dimensions of ∈0μ0\sqrt {{{{ \in _0}} \over {{\mu _0}}}} is :

Multiple Choice Options

  • A A–1 TML3
  • B A2T3M–1L–2
  • C AT–3ML3/2
  • D AT2M–1L–1

Step-by-Step Verified Solution

∈0μ0\sqrt {{{{ \in _0}} \over {{\mu _0}}}} = ∈0μ0∈0{{{ \in _0}} \over {\sqrt {{\mu _0}{ \in _0}} }} = c ×\times ∈0{{ \in _0}}

∴\therefore [∈0μ0]\left[ {\sqrt {{{{ \in _0}} \over {{\mu _0}}}} } \right] = [LT−1]×[∈0]\left[ {L{T^{ - 1}}} \right] \times \left[ {{ \in _0}} \right]

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

∴\therefore ∈0=q24πr2F{ \in _0} = {{{q^2}} \over {4\pi {r^2}F}}

⇒\Rightarrow [∈0]=[AT]2[MLT−2]×[L2]\left[ {{ \in _0}} \right] = {{{{\left[ {AT} \right]}^2}} \over {\left[ {ML{T^{ - 2}}} \right] \times \left[ {{L^2}} \right]}} = [A2M−1L−3T4]\left[ {{A^2}{M^{ - 1}}{L^{ - 3}}{T^4}} \right]

∴\therefore [∈0μ0]\left[ {\sqrt {{{{ \in _0}} \over {{\mu _0}}}} } \right] = [LT−1]\left[ {L{T^{ - 1}}} \right] ×\times [A2M−1L−3T4]\left[ {{A^2}{M^{ - 1}}{L^{ - 3}}{T^4}} \right]

                 = [A2T3M–1L–2]

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