Physics - Current Electricity Question with Solution | TestHub

Potential difference$=5\mathrm{V}$, length $=\text{0}.\text{1 m}=l$ and the electron speed $=$ drift velocity ${\text{v}}_{\text{d}}=\text{2}.\text{5}\times 10^{-4}{\text{m s}}^{-1}$, electron density $\left(\text{n}\right)=8\times 10^{28}{\text{m}}^{-3}$ and the charge on each electron(e) $=\text{1}.\text{6}\times 10^{-19}\mathrm{C}$. We know $\text{i}=\text{nAe}{v}_{\text{d}}...(i)$. And $V=iR...(ii)$

Resistance $R$ is also equal to, $\frac{\rho l}{\text{A}}$ $\Rightarrow \text{R}=\frac{\rho l}{\text{A}}$

$\rho =\frac{\text{AR}}{l}$ $\left[\rho =\text{Resistivity}\right]$

$=\frac{\text{A}}{l}\times \frac{V}{\text{i}}$ [from (ii)]$=\frac{\text{Av}}{l\times \text{nA}\text{ e}{\text{v}}_{\text{d}}}$ & [from (i)]$=\frac{\text{V}}{l\times \text{n}\times \text{e}\times {\text{v}}_{\text{d}}}$

$=\frac{5}{\text{0}.\text{1}\times 8\times 10^{28}\times \text{1}.\text{6}\times 10^{-19}\times \text{2}.\text{5}\times 10^{-4}}$

$=\text{0}.\text{156}\times 10^{-4}\Omega \text{m}=\text{1}.\text{56}\times 10^{-5}\Omega \text{m}$