JEE PHYSICS MCQ: UNITS & MEASUREMENTS

1 Medium

1 of 1

Remembering +4 | -1

\mu_0

and

\varepsilon_0

are the permeability and permittivity of free space, respectively, then the dimension of

\left(\frac{1}{\mu_0 \varepsilon_0}\right)

is:

View answer and explanation

Correct option: B

\mathrm{L}^2 / \mathrm{T}^2

Option B (

\mathrm{L}^2 / \mathrm{T}^2

) is correct because

\frac{1}{\mu_0 \varepsilon_0}

represents the square of the speed of light, which has dimensions of length squared divided by time squared.

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\sqrt{\frac{\text { Modulus of elasticity }}{\text { Density }}}

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\frac{1}{\left(\sqrt{\varepsilon_0 \mu_0}\right) t} \text {. }

\left[ {{\mathrm{M}^{ - 1}}{\mathrm{L}^{ - 3}}{\mathrm{A}^2}{T^3}} \right]

{\varepsilon _0}L\frac{{\Delta V}}{{\Delta t}},\;

{\varepsilon _0}

{\Delta V}

{\Delta t}