Specific conductance (Κ) is related to resistivity as

The correct answer is Option (4) → \(\kappa=\frac{1}{\rho}\)

The resistance of a conductor is proportional to its length l and inversely proportional to its area of cross section A; the constant is the resistivity \( \rho \). Conductance G is the inverse of resistance, and conductivity (specific conductance) \( \kappa \) is the inverse of resistivity:

    \( \displaystyle R = \rho\,\frac{l}{A} \)

    \( \displaystyle G = \frac{1}{R} = \frac{1}{\rho}\cdot\frac{A}{l} = \kappa\,\frac{A}{l} \)

    \( \displaystyle \kappa = \frac{1}{\rho} \)

So specific conductance is \( \kappa = 1/\rho \), with the SI unit S m⁻¹ (often S cm⁻¹).

Shortcut: Check units: \( \rho \) is in Ω m and \( \kappa \) in S m⁻¹ = Ω⁻¹ m⁻¹, which only \( 1/\rho \) gives. \( R\rho \) gives Ω² m, \( \rho/R \) gives m, and \( R/\rho \) gives m⁻¹.

NCERT (2026-27 edition):
Conductance of Electrolytic Solutions – …