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$SI$ इकाई में, $\sqrt {\frac{{{ \varepsilon _0}}}{{{\mu _0}}}} $ की विमा है?
A$A{T^{ - 3}}M{L^{3/2}}$
B${A^{ - 1}}TM{L^3}$
C${A^2}{T^3}{M^{ - 1}}{L^{ - 2}}$
D$A{T^2}{M^{ - 1}}{L^{ - 1}}$
(JEE MAIN-2019)
Solution
$\begin{array}{l}
Dimension\,of\,\sqrt {\frac{{{\varepsilon _0}}}{{{\mu _0}}}} \\
\left[ {{\varepsilon _0}} \right] = \left[ {{M^{ – 1}}{L^{ – 3}}{T^4}{A^2}} \right]\\
\left[ {{\mu _0}} \right] = \left[ {ML{T^{ – 2}}{A^{ – 2}}} \right]\\
Dimension\,of\,\,\sqrt {\frac{{{\varepsilon _0}}}{{{\mu _0}}}} = {\left[ {\frac{{{M^{ – 1}}{L^{ – 3}}{T^4}{A^2}}}{{ML{T^{ – 2}}{A^{ – 2}}}}} \right]^{\frac{1}{2}}}\\
= {\left[ {{M^{ – 2}}{L^{ – 4}}{T^6}{A^4}} \right]^{1/2}}\\
= \left[ {{M^{ – 1}}{L^{ – 2}}{T^3}{A^2}} \right]
\end{array}$
Dimension\,of\,\sqrt {\frac{{{\varepsilon _0}}}{{{\mu _0}}}} \\
\left[ {{\varepsilon _0}} \right] = \left[ {{M^{ – 1}}{L^{ – 3}}{T^4}{A^2}} \right]\\
\left[ {{\mu _0}} \right] = \left[ {ML{T^{ – 2}}{A^{ – 2}}} \right]\\
Dimension\,of\,\,\sqrt {\frac{{{\varepsilon _0}}}{{{\mu _0}}}} = {\left[ {\frac{{{M^{ – 1}}{L^{ – 3}}{T^4}{A^2}}}{{ML{T^{ – 2}}{A^{ – 2}}}}} \right]^{\frac{1}{2}}}\\
= {\left[ {{M^{ – 2}}{L^{ – 4}}{T^6}{A^4}} \right]^{1/2}}\\
= \left[ {{M^{ – 1}}{L^{ – 2}}{T^3}{A^2}} \right]
\end{array}$
Standard 11
Physics
