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**Hint:**Resistivity: The resistivity or specific resistance of a material of a conductor is the resistance offered by a wire of this material of unit length and unit area of cross section. The $S.I$ unit of resistivity is ohm-meter. It is donated by $'\rho '$ .The resistivity of a conductor depends upon its temperature and it increases with the increases in temperature of the conductor.

Conductivity: The Electrical conductivity is the measure of the ease at which an electric charge or heat can pass through a material. It is the inverse of electrical resistivity. It is denoted by $'\sigma '$ .

**Complete step by step answer:**

Consider a conductor having length $'\ell '$ and an area of cross-section $'A'$. Let $'n'$ be the number of electrons per unit volume in the conductor. If an electric field $E$ is applied across the two ends of the conductor, then the drift velocity (in magnitude) of electrons is given by.

${\vartheta _d} = \dfrac{{eE}}{m}\tau {\text{ - - - - - - }}\left( 1 \right)$

Here $\tau $ is the average relaxation time $'m'$ is the mass of an electron.

Now the current flowing through the conductor due to drift electrons is given by.

$I = neA{\vartheta _d}{\text{ - - - - - - - }}\left( 2 \right)$

Put the value of ${\vartheta _d}$ in $\left( 2 \right)$ from $\left( 1 \right)$

$\implies$ $I = neA\dfrac{{eE\tau }}{m} = \dfrac{{n{e^2}AE\tau }}{m}{\text{ - - - - - }}\left( 3 \right)$

If $V$ is the potential difference applied across the end of the conductor

$\implies$ $E = \dfrac{V}{\ell }$

Now put value of E in $\left( 3 \right)$

$\implies$ $I = \dfrac{{nA{e^2}V\tau }}{{m\rho }}$

Now

$\implies$ $\dfrac{V}{I} = \dfrac{{m\ell }}{{n{e^2}\tau A}}$

But according to ohm's law, $\dfrac{V}{I} = R$, the resistance of the conductor

$\implies$ $R = \dfrac{{m\ell }}{{n{e^2}\tau A}} = \dfrac{m}{{n{e^2}\tau }}.\dfrac{\ell }{A}$

Now comparing the above result with the expression of resistance

$\implies$ $R = \rho \dfrac{\ell }{A}$

Now here we have the resistivity of the material.

$\rho = \dfrac{m}{{n{e^2}\tau }}$

So from above result it is cleared that

$\rho \propto \dfrac{1}{\tau }{\text{ - - - - - - - - }}\left( 4 \right)$

Also the conductivity is the reciprocal of resistivity i.e.

$\implies$ $\sigma = \dfrac{1}{\rho }{\text{ - - - - - - - - }}\left( 5 \right)$

$\implies$ $\sigma = $ is conductivity here.

Now from $\left( 4 \right){\text{ & }}\left( 5 \right)$ it is cleared that

$\sigma \propto \tau $

Now according the question, the ratio of resistivity to conductivity is

$\implies$ $\dfrac{\rho }{\sigma } \propto \dfrac{1}{{{\tau ^2}}}$

So when temperature increases relaxat time will decrease and hence the ratio of resistivity and conductivity will increase.

**So option ‘A’ is correct.**

**Note:**

1. Temperature dependence of the resistivity of a semiconductor and insulator:

The variation of resistivity with temperature in case of semiconductor with temperature is different from that in case of conductor. In case of semiconductor and an insulator, the resistivity at temperature $T$ is given by

$\rho = \rho \times exp({\dfrac{E_g}{KT}})$

Here $K$ is the Boltzman Constant $Eg$ is the energy band gap between valency and conduction bonds in the atoms of such materials.

2. For pure conductors, such as copper, the resistivity increases linearly with the temperature in the temperature range around and above the room temperature.

If ${\rho _o}$ and $\rho $ is the resistivity of the material of a conductor at $0^\circ C$ and $\theta ^\circ C$ respectively, then it is found that

$\rho = {\rho _o}\left( {1 + \alpha \theta } \right)$

Here $\alpha $ is the temperature coefficient of resistivity.

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