(2)

As we know, $I=nAe{v}_d$

where, $I=100\text{mA,}$ diameter $=d$ and $v_d=v$

$100$ $\text{mA}=nAev$                                                                ...(i)

Now, $I=200$$\text{mA}$, diameter $=\frac{d}{2}$ and $v_d=v\text{'}$

$200$$\text{mA}=nA\text{'}ev\text{'}$                                                            ...(ii)

From equations (i) and (ii),

$\frac{100}{200}=\frac{nAev}{nA\text{'}ev\text{'}} ; \frac{1}{2}=\frac{\pi \times d^2/4\times v}{\pi \times d^2/16\times v\text{'}}$$;$      $\frac{1}{2}=\frac{4v}{v\text{'}} \Rightarrow v\text{'}=8v$

(2)

The relation between current and drift velocity is given as  $I=n_{e}A{v}_d$

Rearranging and substituting the values,

$v_d=\frac{I}{neA}=\frac{10}{{10}^{22}\times 1.6\times {10}^{-19}\times \pi \times {10}^{-6}} ; v_d=\frac{6.25}{\pi }\times {10}^3$ $\mathrm{m}/\mathrm{s}$

(4)

length, $l=10$$\text{m}$

radius, $r=\frac{{10}^{-2}}{\sqrt{\pi }}$$\text{m}$

Resistance, $R=10$$\Omega$

Electric field, $E=10$ $\text{V/m}$

The current density, $j=\sigma E=\frac{E}{\rho }$                              ...(i)

So, $\rho =\frac{RA}{l}=\frac{10\times \pi \times \frac{{10}^{-4}}{\pi }}{10}={10}^{-4}$ $\Omega \text{m}$

From equation (i),                                                       ...(ii)

$V=\frac{10}{{10}^{-4}}={10}^5$ ${\text{A/m}}^2$

(2)

The formula of drift velocity is $v_d=\frac{eE}{m}\tau$

Current density $J=\frac{I}{A}=\frac{neA{v}_d}{A}=ne{v}_d$

Resistivity is $\rho =\frac{m}{n{e}^2\tau }\Rightarrow \tau =\frac{m}{n{e}^2\rho }$

Resistance is $R=\frac{V}{I} ; \rho \frac{l}{A}=\frac{El}{I}\Rightarrow \rho =\frac{EA}{I}=\frac{E}{J}$

where, $E=$electric field, $A=$area of cross section

$e=$electronic charge, $n=$number of density of electrons,

$\tau =$relaxation time.

(2)

Here, $v_d=7.5\times {10}^{-4}$$\text{m/s}, E=3\times {10}^{-10}$ $\text{V/m}$

$\text{Mobility, }\mu =\frac{v_d}{E}=\frac{7.5\times {10}^{-4}}{3\times {10}^{-10}} ; \mu =2.5\times {10}^6$ ${\text{m}}^2{\text{V}}^{-1}{\text{s}}^{-1}$