Question

A conductor of length $I$ has a nonuniform cross-section. The radius of cros-ssection varies linearly from $a$ to $b$. Find the resistance of the material is $p$. Find the resistance of the conductor across its ends.

• A. $\frac{2\rho l}{\pi ab}$
• B. $\frac{\rho l}{\pi ab}$
• C. $\frac{2\rho l}{9\pi ab}$
• D. $\frac{4\rho l}{5\pi ab}$

Answer 1

The correct option is B $\frac{\rho l}{\pi ab}$

Let the elementary ring of width $dx$ and radius $r$.

From figure, $y=mx+c$ (straight line equation)

Using $y=0$ at $x=0$ $⟹c=0$

Also $y=b-a$ at $x=l$ $⟹$ $m=\frac{b-a}{l}$

$⟹$ $y=\frac{b-a}{l}$

$\therefore$ Radius of elementary ring $r=a+mx=a+\frac{b-a}{l}x$

Now resistance of elementary ring $dR=\frac{\rho dx}{\pi r^2}=\frac{\rho l^2}{\pi }\frac{dx}{\left(\right(b-a)x+al{)}^2}$

$\therefore$ Total resistance $R=\frac{\rho l^2}{\pi }{\int }_o^l\frac{dx}{\left(\right(b-a)x+al{)}^2}=\frac{-\rho l^2}{(b-a)\pi }\times \frac{1}{(b-a)x+al}{|}_o^l$

$⟹$ $R=\frac{-\rho l^2}{\pi (b-a)}[\frac{1}{bl}-\frac{1}{al}]=\frac{\rho l}{\pi ab}$