QJEEAdv15Ph91ec: JEE Advanced 2015 Physics - Class 12 JEEAdv15Ph91ec Solution: In terms of potential difference V, electric current I,…

The speed of light in vacuum is given by, $c = {1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}$ and The impedance (resistance) of free space is defined a $R = \sqrt {{{{\mu _0}} \over {{\varepsilon _0}}}}$ . Using this check the dimensional correctness of equations.

(a) ${\mu _0}{I^2} = {\varepsilon _0}{V^2}$

${{{\mu _0}} \over {{\varepsilon _0}}} = {{{V^2}} \over {{I^2}}} = {R^2}$

$\therefore$ ${R^2} = {R^2}$ which is dimensionally correct.

(b) ${\varepsilon _0}I = {\mu _0}V \Rightarrow {{{\varepsilon _0}} \over {{\mu _0}}} = {V \over I} \Rightarrow {1 \over {{R^2}}} = R$

which is dimensionally incorrect.

(c) $I = {\varepsilon _0}cV \Rightarrow {I \over V} = {\varepsilon _0}C = {{{\varepsilon _0}} \over {\sqrt {{\varepsilon _0}{\mu _0}} }}$ = $\sqrt {{{{\varepsilon _0}} \over {{\mu _0}}}}$ = ${1 \over R}$

${1 \over R} = \sqrt {{{{\varepsilon _0}} \over {{\mu _0}}}} \Rightarrow {1 \over R} = {1 \over R}$

which is dimensionally correct.

(d) ${\mu _0}cI = {\varepsilon _0}V \Rightarrow {{{\mu _0}c} \over {{\varepsilon _0}}} = {V \over I}$

${{{\mu _0}} \over {{\varepsilon _0}}}{1 \over {\sqrt {{\varepsilon _0}{\mu _0}} }} = R \Rightarrow {1 \over {{\varepsilon _0}}}\sqrt {{{{\mu _0}} \over {{\varepsilon _0}}}} = R$

${R \over {{\varepsilon _0}}} = R$ , which is dimensionally incorrect.