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  • The resistivity is a function of the radial position for a cylindrical conductor of radius <img alt="\mathrm{r_{0}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Br_%7B0%7D%7D" /

The resistivity is a function of the radial position for a cylindrical conductor of radius \mathrm{r_{0}} and length \mathrm{L, \rho=\rho_{0}\left(1+\beta r^{2}\right)}. Then what is the inverse of the total resistance?

Option: 1

\mathrm{\frac{\pi}{\rho_{0} L \beta} \ln \left(1+\beta r_{0}^{2}\right)}


Option: 2

\mathrm{\frac{\pi}{2 \rho_{0} L \beta} \ln \left(1+\beta r_{0}^{2}\right)}


Option: 3

\mathrm{\frac{2 \pi}{\rho_{0} L \beta} \ln \left(1+\beta r_{0}^{2}\right)}


Option: 4

None of the above


Answers (1)

best_answer

The resistance of a small cylindrical portion of the rod between radii \mathrm{r} and \mathrm{(r+\delta r)} is given by :

\mathrm{\mathrm{R}^{\prime}=\frac{\rho_{0}\left(1+\beta \mathrm{r}^{2}\right) \mathrm{L}}{2 \pi r \delta r}}
Since all these are connected between the same two points, the resistances are in parallel,

\mathrm{Y =\int \frac{1}{R^{\prime}}=\int_{0}^{r_{0}} \frac{2 \pi r d r}{\rho_{0} L\left(1+\beta r^{2}\right)} }
\mathrm{=\frac{\pi}{\rho_{0} L} \int_{r}^{r_{0}} \frac{2 r d r}{1+\beta r^{2}}=\frac{\pi}{\rho_{0} L \beta} \ln \left(1+\beta r_{0}^{2}\right) }

Posted by

Ajit Kumar Dubey

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