Units & Dimensions

JEE Main level, single correct, medium

135 of 149

Question 135 of 149

In terms of resistance RR and time TT, the dimensions of ratio με\frac{\mu}{\varepsilon} of the permeability μ\mu and permittivity ε\varepsilon is:

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Answer: (C) [R2]\left[\mathrm{R}^2\right]

Dimensions of μ=[\mu=\left[\right. MLT −2 A−2]\left.^{-2} \mathrm{~A}^{-2}\right]
Dimensions of ∈=[M−1 L−3 T4 A2]\in=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right]
Dimensions of R=[ML2 T−3 A−2]\mathrm{R}=\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]
∴ Dimensions of μ Dimensions of e=[MLT−2 A−2][M−1 L−3 T4 A2]=[M2 L4 T−6 A−4]=[R2]\begin{aligned} &\therefore \frac{\text { Dimensions of } \mu}{\text { Dimensions of } e}=\frac{\left[\mathrm{MLT}^{-2} \mathrm{~A}^{-2}\right]}{\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right]} \\ &=\left[\mathrm{M}^2 \mathrm{~L}^4 \mathrm{~T}^{-6} \mathrm{~A}^{-4}\right]=\left[\mathrm{R}^2\right] \end{aligned}

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