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

In terms of potential difference V, electric current I, permittivity $${\varepsilon _0}$$, permeability $${\mu _0}$$ and speed of light c, the dimensionally correct equation(s) is(are) :
$${\mu _0}{I^2} = {\varepsilon _0}{V^2}$$
$${\varepsilon _0}I = {\mu _0}V$$
$$I = {\varepsilon _0}cV$$
$${\mu _0}cI = {\varepsilon _0}V$$

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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