JEE PYQ: Units & Measurements - Question ID e4e5e231d709 (JEE Main 2016)

Electrical conductivity ($\sigma$) is defined as the reciprocal of electrical resistivity ($\rho$). From Ohm’s law in microscopic form, we have:

• $\vec{J}$ is the current density (current per unit area),
• $\vec{E}$ is the electric field (potential gradient),
• $\sigma$ is the electrical conductivity.

Our goal is to express $\sigma$ in terms of the fundamental dimensions: mass ($M$), length ($L$), time ($T$), and current ($I$).

Step 1: Dimensions of current density $\vec{J}$

Current density is current ($I$) divided by area ($L^2$). Hence:

Step 2: Dimensions of electric field $\vec{E}$

Electric field is potential difference ($V$) divided by distance ($L$). Potential difference has dimensions of work per unit charge:

Therefore:

Step 3: Relate $\sigma$ to $\vec{J}$ and $\vec{E}$

From $\vec{J} = \sigma \vec{E}$, we solve for $\sigma$:

Substitute the dimensions of $J$ and $E$:

Step 4: Simplify the expression

Multiply numerator and denominator:

Combine like terms:

Step 5: Match with given options

The derived dimension $M^{-1}\,L^{-3}\,T^{3}\,I^{2}$ corresponds exactly to option C.

Students often confuse the dimensions of conductivity with those of resistivity or conductance. A frequent mistake is to forget that conductivity is the reciprocal of resistivity, leading to sign errors in the exponents of $M$, $L$, and $T$. Another pitfall is miscounting the exponent of current ($I$), especially when dealing with the electric field’s dependence on charge. Always double-check the algebraic manipulation of exponents to avoid such errors.