In terms of potential difference V , electic current I , permittivity ε _ 0 , permeability μ _ 0 and speed of light c , the dimensionally correct equation(s) is (are)

Correct answer: A. μ _ 0 I ^ 2 = ε _ 0 V ^ 2 Using C = 1 μ _ 0 ϵ _ 0 & R = μ _ 0 ϵ _ 0 we can check corrections (i) μ _ 0 I ^ 2 = ϵ _ 0 V ^ 2 μ _ 0 ϵ _ 0 = V ^ 2 I ^ 2 = R ^ 2 ∴ R ^ 2 = R ^ 2 Correct (ii) ϵ _ 0 I = μ _ 0 V ϵ _ 0 μ _ 0 = (V)/(I) 1 R ^ 2 = R Not correct (iii) I = ϵ _ 0 cV (1)/(V) = ϵ _ 0 c = ϵ _ 0 μ _ 0 ϵ _ 0 (1)/(R) = ϵ _ 0 μ _ 0 = (1)/(R) Correct (iv) μ _ 0 CI = ϵ _ 0 V μ _ 0 C ϵ _ 0 = (V)/(I) μ _ 0 ϵ _ 0 1 μ _ 0 ϵ _ 0 = R ∴ 1 ϵ _ 0 μ _ 0 ϵ _ 0 = R ⇒ R ϵ _ 0 = R Incorrect

JEE Advanced 2015Paper 2

Exam: JEE Advanced
Year: 2015
Paper: Paper 2
Subject: Physics
Chapter: Units and Dimensions

JEE Advanced 2015 · Paper 2 · Physics · Units and Dimensions

In terms of potential difference V , electic current I ,...

In terms of potential difference

V

, electic current

I

, permittivity

{ε}_{0}

, permeability

{\mu}_{0}

and speed of light

c

, the dimensionally correct equation(s) is (are)

Options

A ${\mu}_{0}{I}^{2} = {ε}_{0}{V}^{2}$
B ${ε}_{0}I = {\mu}_{0}V$
C $I = {ε}_{0}cV$
D ${\mu}_{0}cI = {ε}_{0}V$

Correct answer

A. ${\mu}_{0}{I}^{2} = {ε}_{0}{V}^{2}$

Step-by-step solution

Using

C = \dfrac{1}{\sqrt{{\mu}_{0}{ϵ}_{0}}}

R = \sqrt{\dfrac{{\mu}_{0}}{{ϵ}_{0}}}

we can check corrections
(i)

{\mu}_{0}{I}^{2} = {ϵ}_{0}{V}^{2}

\dfrac{{\mu}_{0}}{{ϵ}_{0}} = \dfrac{{V}^{2}}{{I}^{2}} = {R}^{2}

∴ {R}^{2} = {R}^{2}

Correct
(ii)

{ϵ}_{0}I = {\mu}_{0}V

\dfrac{{ϵ}_{0}}{{\mu}_{0}} = \dfrac{V}{I}

\dfrac{1}{{R}^{2}} = R

Not correct
(iii)

I = {ϵ}_{0}cV

\dfrac{1}{V} = {ϵ}_{0}c = \dfrac{{ϵ}_{0}}{\sqrt{{\mu}_{0}{ϵ}_{0}}}

\dfrac{1}{R} = \sqrt{\dfrac{{ϵ}_{0}}{{\mu}_{0}}} = \dfrac{1}{R}

Correct
(iv)

{\mu}_{0}CI = {ϵ}_{0}V

\dfrac{{\mu}_{0}C}{{ϵ}_{0}} = \dfrac{V}{I}

\dfrac{{\mu}_{0}}{{ϵ}_{0}}\dfrac{1}{\sqrt{{\mu}_{0}{ϵ}_{0}}} = R

∴ \dfrac{1}{{ϵ}_{0}}\sqrt{\dfrac{{\mu}_{0}}{{ϵ}_{0}}} = R ⇒ \dfrac{R}{{ϵ}_{0}} = R

Incorrect

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