JEE PYQ: Units & Measurements - Question ID 3d4b016f2159 (JEE Main 2019)

Core Formula & Concept:

In electromagnetism, the two fundamental constants are:

• $\epsilon_0$ – the permittivity of free space, and
• $\mu_0$ – the permeability of free space.

Their SI units are:

• $\epsilon_0$ has units of farad per metre (F/m), which in base SI units is $A^2\,s^4\,kg^{-1}\,m^{-3}$.
• $\mu_0$ has units of henry per metre (H/m), which in base SI units is $kg\,m\,s^{-2}\,A^{-2}$.

The quantity $\sqrt{\epsilon_0/\mu_0}$ appears naturally in the expression for the speed of light in vacuum: $c = \frac{1}{\sqrt{\epsilon_0\,\mu_0}}.$ Hence, $\sqrt{\epsilon_0/\mu_0}$ must have the dimensions of velocity⁻¹, i.e. $L^{-1}T$.

Step-by-Step Derivation:

Step 1 – Write the dimensions of $\epsilon_0$ and $\mu_0$

$[\epsilon_0] = A^2\,T^4\,M^{-1}\,L^{-3} \quad\text{and}\quad [\mu_0] = M\,L\,T^{-2}\,A^{-2}.$

Step 2 – Form the ratio $\epsilon_0/\mu_0$

$\left[\frac{\epsilon_0}{\mu_0}\right] = \frac{A^2\,T^4\,M^{-1}\,L^{-3}}{M\,L\,T^{-2}\,A^{-2}} = A^{2-(-2)}\,T^{4-(-2)}\,M^{-1-1}\,L^{-3-1} = A^4\,T^6\,M^{-2}\,L^{-4}.$

Step 3 – Take the square root

$\left[\sqrt{\frac{\epsilon_0}{\mu_0}}\right] = \bigl(A^4\,T^6\,M^{-2}\,L^{-4}\bigr)^{1/2} = A^{2}\,T^{3}\,M^{-1}\,L^{-2}.$

Step 4 – Match with the given options

The derived dimension $A^2\,T^3\,M^{-1}\,L^{-2}$ exactly matches option B. Common Traps & Exam Tip:

1. Sign errors in exponents: Students often mis-count the negative exponents when dividing $[\epsilon_0]$ by $[\mu_0]$. Double-check each exponent algebraically. 2. Square-root confusion: Forgetting to halve each exponent after taking the square root leads to incorrect dimensions. 3. Option misreading: Option A has $A^{-1}$ while B has $A^2$. A quick dimensional check shows $A^2$ is the only plausible choice.

Exam Tip: Always express both constants in base SI units first, then perform the algebra step by step. This avoids unit-conversion errors and ensures full marks.