Question

A cylindrical tube of length $l$ has inner radius $a$ while outer radius $b$. The resistance of the tube between its inner and outer surfaces will be
$($resistivity of material is $\rho )$

• A. $\ln \left(\frac{b^2}{a}\right)$
• B. $\frac{\rho }{2\pi l}\ln \frac{b}{a}$
• C. $\frac{\rho }{2\pi b}\ln a$
• D. $\frac{{\ln }^2\left(a/b\right)}{2\pi l}$

Answer 1

The correct option is B $\frac{\rho }{2\pi l}\ln \frac{b}{a}$

The resistance needs to calculated between inner and outer surface, hence the cross-sectional area for current will vary along radial direction.


Let us take a section at a distance of $x$ from center of tube with thickness $dx$ as shown in cross-sectional view of the element.


By using the formula of resistance,

$R=\frac{\rho l}{A}$

So, the resistance of the element will be

$dR=\frac{\rho dx}{2\pi xl}$

Here, $A=2\pi xl$ as the surface area (curved surface) for current flow.

After integrating both side we get,

$\Rightarrow \int dR=\int \frac{\rho dx}{2\pi xl}$

Substituting limits of $x$ from $a$ to $b$

$\Rightarrow R=\underset{x=a}{\overset{x=b}{\int }}\frac{\rho dx}{2\pi xl}$

Thus,

$R=\frac{\rho }{2\pi l}\underset{a}{\overset{b}{\int }}\frac{dx}{x}=\frac{\rho }{2\pi l}(\ln x{)}_a^b$

$\Rightarrow R=\frac{\rho }{2\pi l}[\ln b-\ln a]$

$\therefore R=\frac{\rho }{2\pi l}\left[\ln \left(\frac{b}{a}\right)\right]$

$\begin{array}{c}\text{Why this question?}\\ \text{Tip: When resistance for hollow cylindrical}\\ \text{tube is asked between inner \& outer surfaces,}\\ \text{then curved surface area i.e.}A=\pi Dl\\ \text{will be taken into account, and it is varying}\\ \text{with distance from center of tube.}\end{array}$