arthham

JEE Advanced 2015 · Units and Measurements

In terms of potential difference VV, electric current II, permittivity ε0\varepsilon_0, permeability μ0\mu_0 and speed of light cc, the dimensionally correct equation(s) is(are)

Answers: A. μ0I2=ε0V2\mu_0 I^{2} = \varepsilon_0 V^{2} · C. I=ε0cVI = \varepsilon_0 c V

Watch this solved on the board The full walkthrough, written out a line at a time with narration explaining each step as it appears. Pause anywhere and ask why a line is there. Free to watch

Or read the worked solution below. Jump to the steps ↓

Written solution

Step 1: Find the dimensions of the given physical quantities.

The dimensions of potential difference VV, electric current II, permittivity ε0\varepsilon_0, permeability μ0\mu_0, and speed of light cc are determined from their physical definitions.

[V]=[W][q]=ML2T−2AT=ML2T−3A−1[V] = \frac{[W]}{[q]} = \frac{\text{ML}^2\text{T}^{-2}}{\text{AT}} = \text{ML}^2\text{T}^{-3}\text{A}^{-1} [I]=A[I] = \text{A} [ε0]=M−1L−3T4A2[\varepsilon_0] = \text{M}^{-1}\text{L}^{-3}\text{T}^{4}\text{A}^{2} [μ0]=MLT−2A−2[\mu_0] = \text{MLT}^{-2}\text{A}^{-2} [c]=LT−1[c] = \text{LT}^{-1}

Step 2: Analyze the first candidate equation.

We compare the dimensions of both sides for the equation μ0I2=ε0V2\mu_0 I^2 = \varepsilon_0 V^2.

[LHS]=[μ0I2]=(MLT−2A−2)(A2)=MLT−2[\text{LHS}] = [\mu_0 I^2] = (\text{MLT}^{-2}\text{A}^{-2})(\text{A}^2) = \text{MLT}^{-2} [RHS]=[ε0V2]=(M−1L−3T4A2)(ML2T−3A−1)2=MLT−2[\text{RHS}] = [\varepsilon_0 V^2] = (\text{M}^{-1}\text{L}^{-3}\text{T}^{4}\text{A}^{2})(\text{ML}^2\text{T}^{-3}\text{A}^{-1})^2 = \text{MLT}^{-2}

Since [LHS]=[RHS][\text{LHS}] = [\text{RHS}], this equation is dimensionally correct.

Step 3: Analyze the second candidate equation.

We compare the dimensions of both sides for the equation ε0I=μ0V\varepsilon_0 I = \mu_0 V.

[LHS]=[ε0I]=(M−1L−3T4A2)(A)=M−1L−3T4A3[\text{LHS}] = [\varepsilon_0 I] = (\text{M}^{-1}\text{L}^{-3}\text{T}^{4}\text{A}^{2})(\text{A}) = \text{M}^{-1}\text{L}^{-3}\text{T}^{4}\text{A}^{3} [RHS]=[μ0V]=(MLT−2A−2)(ML2T−3A−1)=M2L3T−5A−3[\text{RHS}] = [\mu_0 V] = (\text{MLT}^{-2}\text{A}^{-2})(\text{ML}^2\text{T}^{-3}\text{A}^{-1}) = \text{M}^{2}\text{L}^{3}\text{T}^{-5}\text{A}^{-3}

Since [LHS]≠[RHS][\text{LHS}] \neq [\text{RHS}], this equation is dimensionally incorrect.

Step 4: Analyze the third candidate equation.

We compare the dimensions of both sides for the equation I=ε0cVI = \varepsilon_0 c V.

[LHS]=[I]=A[\text{LHS}] = [I] = \text{A} [RHS]=[ε0cV]=(M−1L−3T4A2)(LT−1)(ML2T−3A−1)=A[\text{RHS}] = [\varepsilon_0 c V] = (\text{M}^{-1}\text{L}^{-3}\text{T}^{4}\text{A}^{2})(\text{LT}^{-1})(\text{ML}^2\text{T}^{-3}\text{A}^{-1}) = \text{A}

Since [LHS]=[RHS][\text{LHS}] = [\text{RHS}], this equation is dimensionally correct.

Step 5: Analyze the fourth candidate equation.

We compare the dimensions of both sides for the equation μ0cI=ε0V\mu_0 c I = \varepsilon_0 V.

[LHS]=[μ0cI]=(MLT−2A−2)(LT−1)(A)=ML2T−3A−1[\text{LHS}] = [\mu_0 c I] = (\text{MLT}^{-2}\text{A}^{-2})(\text{LT}^{-1})(\text{A}) = \text{ML}^2\text{T}^{-3}\text{A}^{-1} [RHS]=[ε0V]=(M−1L−3T4A2)(ML2T−3A−1)=L−1TA[\text{RHS}] = [\varepsilon_0 V] = (\text{M}^{-1}\text{L}^{-3}\text{T}^{4}\text{A}^{2})(\text{ML}^2\text{T}^{-3}\text{A}^{-1}) = \text{L}^{-1}\text{TA}

Since [LHS]≠[RHS][\text{LHS}] \neq [\text{RHS}], this equation is dimensionally incorrect.

Answer: The dimensionally correct equations are μ0I2=ε0V2\mu_0 I^2 = \varepsilon_0 V^2 and I=ε0cVI = \varepsilon_0 c V.

More from Units and Measurements

All 37 Units and Measurements questions →

Units and Measurements: formulas and what the chapter covers →