JEE PYQ: Units & Measurements - Question ID 55ce6153f2cb (JEE Main 2019)

Electrical resistance ($R$) is defined by Ohm’s law as the ratio of voltage ($V$) to current ($I$): $R = \frac{V}{I}$ To find the dimensions of resistance, we must express voltage and current in terms of fundamental mechanical units (mass $M$, length $L$, time $T$, and current $A$).

Key formulas and dimensional relations:

• Voltage (potential difference): $V = \frac{\text{Work}}{\text{Charge}} = \frac{ML^2T^{-2}}{AT} = ML^2T^{-3}A^{-1}$
• Current: $I = A$ (ampere)
• Resistance: $R = \frac{V}{I} = ML^2T^{-3}A^{-2}$
• Permittivity of vacuum ($\epsilon_0$): From Coulomb’s law, $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$, so $\epsilon_0 = \frac{A^2T^4}{ML^3}$ (dimensions: $M^{-1}L^{-3}T^4A^2$)
• Permeability of vacuum ($\mu_0$): From the Biot-Savart law, $dB = \frac{\mu_0}{4\pi} \frac{I \, dl \sin\theta}{r^2}$, so $\mu_0 = \frac{ML}{A^2T^2}$ (dimensions: $MLT^{-2}A^{-2}$)

We need to find which option has the same dimensions as resistance $R = ML^2T^{-3}A^{-2}$.

Step 1: Compute dimensions of $\epsilon_0$ and $\mu_0$

From above: $[\epsilon_0] = M^{-1}L^{-3}T^4A^2$ $[\mu_0] = MLT^{-2}A^{-2}$

Step 2: Compute dimensions of each option

Option A: $\sqrt{\frac{\epsilon_0}{\mu_0}}$

Compute $\frac{\epsilon_0}{\mu_0}$: $\frac{[\epsilon_0]}{[\mu_0]} = \frac{M^{-1}L^{-3}T^4A^2}{MLT^{-2}A^{-2}} = M^{-2}L^{-4}T^6A^4$ Take square root: $\sqrt{M^{-2}L^{-4}T^6A^4} = M^{-1}L^{-2}T^3A^2$ This does not match $ML^2T^{-3}A^{-2}$. So Option A is incorrect.

Option B: $\frac{\epsilon_0}{\mu_0}$

From above: $\frac{[\epsilon_0]}{[\mu_0]} = M^{-2}L^{-4}T^6A^4$ This does not match $ML^2T^{-3}A^{-2}$. So Option B is incorrect.

Option C: $\sqrt{\frac{\mu_0}{\epsilon_0}}$

Compute $\frac{\mu_0}{\epsilon_0}$: $\frac{[\mu_0]}{[\epsilon_0]} = \frac{MLT^{-2}A^{-2}}{M^{-1}L^{-3}T^4A^2} = M^2L^4T^{-6}A^{-4}$ Take square root: $\sqrt{M^2L^4T^{-6}A^{-4}} = ML^2T^{-3}A^{-2}$ This matches the dimensions of resistance. So Option C is correct.

Option D: $\frac{\mu_0}{\epsilon_0}$

From above: $\frac{[\mu_0]}{[\epsilon_0]} = M^2L^4T^{-6}A^{-4}$ This does not match $ML^2T^{-3}A^{-2}$. So Option D is incorrect.

Students often confuse the dimensions of $\epsilon_0$ and $\mu_0$, especially their exponents. A frequent mistake is misapplying the square root or inverting the ratio incorrectly. Always write down the full dimensional formula for each constant and perform the algebra step-by-step. Remember that resistance has dimensions $ML^2T^{-3}A^{-2}$, and verify each option against this. Option C is the only one that matches.