Question

A material of resistivity $\rho$ is formed in the shape of a truncated cone of altitude h. The top end has a radius "a" while bottom end has a radius "b". Assuming a uniform current density through any circular cross-section of the cone, find the resistance between the two ends.

• A. $\frac{\rho h}{2\pi ab}$
• B. $\frac{2\rho h}{\pi ab}$
• C. $\frac{\rho h}{\pi ab}$
• D. $\frac{\rho h}{\pi (a+b)}$

Answer 1

The correct option is C

$\frac{\rho h}{\pi ab}$

Here, $\frac{r-a}{y}=\frac{b-a}{h}\Rightarrow y=\frac{h(r-a)}{b-a}\Rightarrow dy=\frac{h}{(b-a)}dr$
Resistance of the thin elementary disc is $dR=\frac{\rho dy}{\pi r^2}=\frac{\rho }{\pi }\frac{h}{(b-a)}\frac{dr}{r^2}$
$R_{total}=\frac{\rho h}{\pi (b-a)}{\int }_a^b\frac{1}{r^2}dr=\frac{\rho h}{\pi (b-a)}{\left[\frac{-1}{r}\right]}_a^b=\frac{\rho h}{\pi \left(ab\right)}$