JEE Main 2014 · 11 Apr · Shift 1 · Physics Q1

PhysicsSingle correct+4 / -1Question 1 of 90

In terms of resistance R and time T, the dimensions of ratio $\dfrac\mu\epsilon$ of the permeability $\mu$ and permittivity $\epsilon$ is :

[RT^{-2}]

[R^2T^{-1}]

[R^2]

[R^2T^2]

Answer and worked solution

Official answer: C

$\sqrt{\mu/\epsilon}$ has the dimensions of resistance (it is the impedance of free space), so $\dfrac\mu\epsilon$ has dimensions $[R^2]$ .

JEE Main 2014 · 11 Apr · Shift 1, Physics question 1