JEE PHYSICS MCQ: UNITS & MEASUREMENTS

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Analyzing +4 | -1

From the following combinations of physical constants (expressed through their usual symbols) the only combination, that would have the same value in different systems of units, is:

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Correct option: B

\frac{\mathrm{e}^2}{2 \pi \varepsilon_0 \mathrm{Gm}_{\mathrm{e}}^2}\left(\mathrm{~m}_{\mathrm{e}}=\right.

mass of electron)

Option B is correct because the combination e² / (2π ε0 Gm_e²) is dimensionless and therefore has the same value in all systems of units.

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