${\mu _0}$ and ${\varepsilon _0}$ denote the permeability and permittivity of free space, the dimensions of ${\mu _0}{\varepsilon _0}$ are
${\mu _0}$ and ${\varepsilon _0}$ denote the permeability and permittivity of free space, the dimensions of ${\mu _0}{\varepsilon _0}$ are
- A
$L{T^{ - 1}}$
- B
${L^{ - 2}}{T^2}$
- C
${M^{ - 1}}{L^{ - 3}}{Q^2}{T^2}$
- D
${M^{ - 1}}{L^{ - 3}}{I^2}{T^2}$
Similar Questions
The quantities $\quad x=\frac{1}{\sqrt{\mu_{0} \epsilon_{0}}}, y=\frac{E}{B}$ and $z=\frac{l}{C R}$ are defined where $C-$ capacitance $R-$Resistance, $l-$length, $E-$Electric field, $B-$magnetic field and $\varepsilon_{0}, \mu_{0},$ -free space permittivity and permeability respectively. Then....
The quantities $\quad x=\frac{1}{\sqrt{\mu_{0} \epsilon_{0}}}, y=\frac{E}{B}$ and $z=\frac{l}{C R}$ are defined where $C-$ capacitance $R-$Resistance, $l-$length, $E-$Electric field, $B-$magnetic field and $\varepsilon_{0}, \mu_{0},$ -free space permittivity and permeability respectively. Then....
- [JEE MAIN 2020]
The physical quantity which has dimensional formula as that of $\frac{{{\rm{Energy}}}}{{{\rm{Mass}} \times {\rm{Length}}}}$ is
The physical quantity which has dimensional formula as that of $\frac{{{\rm{Energy}}}}{{{\rm{Mass}} \times {\rm{Length}}}}$ is
The focal power of a lens has the dimensions
The focal power of a lens has the dimensions
The dimension of the ratio of magnetic flux and the resistance is equal to that of :
The dimension of the ratio of magnetic flux and the resistance is equal to that of :
The dimension of $\frac{1}{{\sqrt {{\varepsilon _0}{\mu _0}} }}$ is that of
The dimension of $\frac{1}{{\sqrt {{\varepsilon _0}{\mu _0}} }}$ is that of