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**Hint:**The relaxation time of the free electrons drifting in a conductor is given by the formula, $T=v\dfrac{m}{e}E$

The expression of electrical resistivity of the material is obtained by modifying the expression,

Electric field is equal to E and voltage is equal to V

$\begin{align}

& E=\dfrac{V}{L} \\

& V=IR \\

\end{align}$

These all are helpful in finding the answer.

**Complete step-by-step solution:**First of all, let us check what relaxation time actually means.

Relaxation time is the time needed for the exponential decrease of the variable from its initial value to the $\dfrac{1}{{e}^{th}}$ value or 0.368 of that value. The relaxation time of an electron is the time gap between the successive collision of electrons in a conductor.

Now let us check the question,

Let I is the current that flows through the conductor, n is the number of electrons in that conductor, A is the area of the conductor, v is the drift velocity of the electron, e is the charge of the electron in that conductor, E is the electric field in that region, m is the mass of electron and T is the relaxation time period.

Then, $V=IR$

Rearranging this equation will give, $T=v\dfrac{m}{e}E$

The current flowing in the conductor I is

$I=-neA{{v}_{d}}=neA\left( e\dfrac{E}{m} \right)TI=\dfrac{n{{e}^{2}}EA}{m}T$

Also we know that electric field is given by the equation,

$E=\dfrac{V}{L}$

Substituting this in equation of current will give,

$I=\dfrac{n{{e}^{2}}VA}{mL}T\dfrac{V}{I}=\dfrac{mL}{n{{e}^{2}}TA}$

According to ohm's law,

$V=IR$

$R=\dfrac{V}{I}$

Therefore the equation becomes

\[R=\left( \dfrac{m}{n{{e}^{2}}T} \right)\dfrac{L}{A}; R=\rho \dfrac{L}{A}\]

\[R=\rho \dfrac{L}{A}\]

**is the required solution.**

**Note:**As the temperature of a conductor decreases, the number of collisions of electrons in the conductor also decreases. Therefore the free travel time increases or relaxation time increases. This is the temperature dependence of relaxation time.

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