CBSE Class 12 · 2005 · Current ElectricityShort Answer

\textbf{Resistivity} $\rho$ is defined by $R=\rho \frac{\ell }{A}$ — numerically the resistance of a unit cube; SI unit $\Omega \text{m}$. Derivation: $I=neA{v}_d$ with $v_d=\frac{eE}{m}\tau$ and $E=\frac{V}{\ell }$ gives $I=\frac{n{e}^{2}A\tau }{m}\cdot \frac{V}{\ell }$. By Ohm's law $R=\frac{V}{I}=\frac{m\ell }{n{e}^2\tau A}$; comparing with $R=\rho \frac{\ell }{A}$, \textbf{OR:} the fall of potential along a uniform current-carrying wire is proportional to its length ($V=k\ell$). Balancing each cell in turn gives $E_1=k{\ell }_1$, $E_2=k{\ell }_2$, so $\frac{E_1}{E_2}=\frac{{\ell }_1}{{\ell }_2}$. Sensitivity rises when the potential gradient $k$ is made small (use a longer wire or a smaller driving current).