Inhomogeneous conductor

E=V\L. HERE L=1 SO E=V J(r)=(r^2\k).V dI=J(r).d(r) consider a circular ring at a distance of r from axis.its area dr will be 2pi r dr. so dI=(2 pi r^3 dr).V\k integrating we get I=(pi r^4).V\2k so V=pi 2 k.I(pi^2.r^4) also given pi^2 r^4=A^2=10^-8 putting these values we get V=753600I comparing it with V=IR we get R=753600 ohms

Since the cross-sectional area element is $dA=2\pi rdr$ , the total current is given by $I=\int\limits_0^{r_0}jdA$ , where ${r_0}$ is the radius of the cross-section (and therefore $A=\pi r_0^2$ ). Because $E=\frac{V}{L}$ , we have $I=\int\limits_0^{r_0}{\frac{r^2}{k}\frac{V}{L}2\pi rdr} = \frac{2\pi V}{kL} \int\limits_0^{r_0} r^3dr = \frac{2\pi V}{kL} \frac{r_0^4}{4}$ . Replacing $r_0^4=\frac{A^2}{\pi^2}$ and knowing that $R=\frac{V}{I}$ , we get $R=\frac{2\pi kL}{A^2}$ . Numerical value $R=753982,236$ $\Omega$

For a homogeneous conductor of length L and cross sectional area A we know that its resistance is given by $R=\rho \frac{L}{A}.$ Let us mentally divide the conductor into thin wires with cross sectional area $\Delta A$ . Since they are thin, they can be treated as homogeneous conductors and the resistance of each thin wire is then
$\rho_{thin}=\rho(r) \frac{L}{\Delta A}.$ We notice that all the thin wires are connected in parallel and therefore $\frac{1}{R_{eq}}=\sum{\frac{1}{\rho_{thin}}}= \sum{\frac{1}{\rho(r)}\frac{\Delta {A}}{L} }=\frac{1}{kL} \int r^{2} dA= \frac{2\pi}{kL} \int_{0}^{a} r^{3} dr= \frac{\pi a^{4}}{2k L}$ Hence $R_{eq}= \frac{2 k L}{\pi a^{4}}=2 \pi k \frac{L}{A^{2}}=753982 \Omega.$

Given a cylinder , Take an elemental cylinder of thickness dr (lies between r and r+dr) which has length L . The resistance of this element can be found and such cylinders arranged in a parallel way gives the total cylinder. Hence integrating our elemental resistance ...... gives total resistance