JEE Advance - Physics (2015 - Paper 2 Offline - No. 5)
Explanation
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.
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