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Question

(A) We know that resistance $R$ is given by $R = \rho \frac{l}{S}$.Conductivity $\sigma$ is the reciprocal of resistivity $\rho$,so $\sigma = \frac{1}

Vedclass · Accepted Answer

$\frac{GR}{\rho}$

Vedclass · Answer

(A) We know that resistance $R$ is given by $R = ho \frac{l}{S}$.Conductivity $\sigma$ is the reciprocal of resistivity $ho$,so $\sigma = \frac{1}{ho}$.Conductance $G$ is the reciprocal of resistance $R$,so $G = \frac{1}{R}$,which implies $R = \frac{1}{G}$.Substituting $R = \frac{1}{G}$ into the resistance formula: $\frac{1}{G} = ho \frac{l}{S}$.Rearranging for $ho$: $ho = \frac{S}{Gl}$.Since $\sigma = \frac{1}{ho}$,we have $\sigma = \frac{Gl}{S}$.Alternatively,using the given options,we check the expression $\sigma = \frac{Gl}{S}$.Wait,let us re-evaluate the expression $\sigma = \frac{1}{ho}$.From $R = ho \frac{l}{S}$,we have $ho = \frac{RS}{l}$.Thus,$\sigma = \frac{1}{ho} = \frac{l}{RS}$.Since $G = \frac{1}{R}$,we can write $\sigma = \frac{Gl}{S}$.Looking at the options provided,if we consider $\sigma = \frac{1}{ho}$ and $G = \frac{1}{R}$,then $GR = 1$. Therefore,$\sigma = \frac{GR}{ho}$ is mathematically consistent as $\sigma = \frac{1}{ho} 	imes 1 = \frac{1}{ho} 	imes (GR) = \frac{GR}{ho}$.

$A$ cylindrical metal wire of length $l$ and cross-sectional area $S$ has resistance $R$,conductance $G$,conductivity $\sigma$,and resistivity $\rho$. Which one of the following expressions for $\sigma$ is valid?

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