JEE PHYSICS MCQ: UNITS & MEASUREMENTS

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The unit of permittivity of free space

{\varepsilon _0}

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Correct option: D

Coulom{b^2}/newton - metr{e^2}

Option D (

Coulom{b^2}/newton - metr{e^2}

) is correct because it is derived from Coulomb's law, where permittivity is inversely proportional to force and directly proportional to the square of the charge and inversely proportional to the square of the distance.

More Questions from UNITS & MEASUREMENTS

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Assertion :

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

has the unit

\mathrm{ms}^{-1} \text {. }

Reason: Acceleration has the dimensions of

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

The method of dimensional analysis can be used to derive which of the following relations? Statement-I: Dimensions of constant of proportionality of constants can be derived from dimensional method
Statement-II: Numerical value of constant of proportionality can be found from experiments only The dimensions of torque are The physical quantity having the dimensions

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

is A quantity X is given by

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

, where

{\varepsilon _0}

is the permittivity of free space, L is a length,

{\Delta V}

is a potential difference and

{\Delta t}

is a time interval. The dimensional formula for X is the same as that of