Physics - Current Electricity Question with Solution | TestHub

Given, $I=12\mathrm{m},d=0.8\mathrm{cm}$, $R=3.5\times 10^{-3}\Omega$, thickness $=1.25\mathrm{mm}$, Resistivity of the material of the rod $\begin{array}{rl}\rho & =\frac{RA}{I}=\frac{3.5\times 10^{-3}\times \pi {\left(0.4\times 10^{-2}\right)}^2}{1.2}\\ & =46.6\times 10^{-9}\pi \Omega \mathrm{m}\end{array}$ $\begin{array}{rl}& \text{ Resistance of disc, }R=\frac{\rho \text{ (Thickness) }}{\text{ Area of cross-section }}\\ & =\frac{46.6\times 10^{-9}\pi \times \left(1.25\times 10^{-3}\right)}{\pi {\left(1\times 10^{-2}\right)}^2}\\ & =5.82\times 10^{-7}\Omega \end{array}$