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. $R=\frac{2l}{\pi \alpha a^3}$

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 $dA=2\pi rdr$
$\therefore$ Electric resistance between ends of conductor of considered element is $dR=\rho \frac{l}{dA}=\rho \frac{l}{2\pi rdr}=\frac{\alpha l}{2\pi r^{3}dr}[$ as resistivity is $\rho =\frac{\alpha }{r^2}]$
The system may be assumed as a parallel combination of a number of such types of elementary resistance.
$\therefore \frac{1}{R}=\underset{0}{\int }\frac{1}{dR}=\frac{2\pi }{\alpha l}\underset{0}{\overset{a}{\int }}(r{)}^{3}dr$
or $\frac{1}{R}=\frac{2\pi }{\alpha l}\left(\frac{a^4}{4}\right)$
$\therefore R=\frac{2\alpha l}{\pi a^4}$