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(C) The specific conductivity $(\kappa)$ is given by the formula: $\kappa = G 	imes \frac{l}{A}$, where $G$ is conductance, $l$ is thickness, and $A$

Vedclass · Accepted Answer

$40 \ mS \ cm^{-1}$

Vedclass · Answer

(C) The specific conductivity $(\kappa)$ is given by the formula: $\kappa = G 	imes \frac{l}{A}$, where $G$ is conductance, $l$ is thickness, and $A$ is the area of the electrode. Given: Conductance $G = \frac{314}{5} \ S = 62.8 \ S$. Diameter $= 20 \ mm = 2 \ cm$, so radius $r = 1 \ cm$. Area $A = \pi r^2 = 3.14 	imes (1 \ cm)^2 = 3.14 \ cm^2$. Thickness $l = 20 \ \mu m = 20 	imes 10^{-4} \ cm$. Substituting these values: $\kappa = 62.8 	imes \frac{20 	imes 10^{-4}}{3.14} \ S \ cm^{-1}$. $\kappa = 20 	imes 20 	imes 10^{-4} \ S \ cm^{-1} = 400 	imes 10^{-4} \ S \ cm^{-1} = 0.04 \ S \ cm^{-1}$. Converting to $mS \ cm^{-1}$: $0.04 \ S \ cm^{-1} 	imes 1000 \ mS/S = 40 \ mS \ cm^{-1}$.

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