Question

A long cylindrical conductor of radius $a$ and length $l$ is made of a material whose resistivity depends only on the distance $r$ from the axis of the conductor as $\rho =\frac{\alpha }{r^{2}}$ , where $\alpha $ is a constant. The resistance of the conductor across the two ends will be

• A. $R=\frac{4 \alpha l}{\pi a^{3}}$
• B. $R=\frac{2 \alpha l}{\pi a^{4}}$
• C. $R=\frac{4 \alpha }{\pi l a^{3}}$
• D. $R=\frac{2 l}{\pi \alpha a^{3}}$

Answer 1

Answer: (B)

We consider a long co-axial hollow cylinder of radius $r$ and thickness $dr$.
The cross sectional area of considered element is $d A=2 \pi r d r$
$\therefore $ Electric resistance between ends of conductor of considered element is $dR =\rho \frac{l}{ dA }=\rho\frac{l}{2\pi r\,dr} = \frac{\alpha l}{2\pi r^3 dr}\left[\right.$ as resistivity is $\left.\rho=\frac{\alpha}{r^{2}}\right]$
The system may be assumed as a parallel combination of a number of such types of elementary resistance.
$\therefore \frac{1}{ R }=\int\limits_{0} \frac{1}{ dR }=\frac{2\pi }{\alpha l} \int\limits_{0}^{a}( r )^{3} dr$
or $\frac{1}{ R }=\frac{2 \pi }{\alpha l}\left(\frac{a^{4}}{4}\right) $
$\therefore R=\frac{2 \alpha l}{\pi a^{4}}$