Question

A copper wire of length $l$ and radius $r$ is nickel plated till its final radius is $R$ and length $l$. If the specific resistance of nickel and copper be ${\rho }_n$ and ${\rho }_c$, then the conductance of wire with nickel plating is

• A. $\frac{\pi r^2}{l{\rho }_c}$
• B. $\frac{\pi (R^2-r^2)}{l{\rho }_n}$
• C. $\frac{\pi }{l}[\frac{r^2}{{\rho }_c}+\frac{R^2-r^2}{{\rho }_n}]$
• D. $\frac{l{\rho }_c}{\pi r^2}+\frac{l{\rho }_n}{\pi (R^2-r^2)}$

Answer 1

The correct option is C $\frac{\pi }{l}[\frac{r^2}{{\rho }_c}+\frac{R^2-r^2}{{\rho }_n}]$

We can treat the whole wire as a copper wire and a nickel wire(plating) connected in parallel.
copper wire:-
area of copper wire($a$) $=\pi r^2$
length $=l$
resistivity $={\rho }_c$
$\Rightarrow$ conductance $=\frac{a}{l{\rho }_c}$
$=\frac{\pi r^2}{l{\rho }_c}$
nickel wire:-
area($a$) $=\pi (R^2-r^2)$
length $=l$
resistivity $={\rho }_n$
$\Rightarrow$ conductance $=\frac{a}{l{\rho }_n}$
$=\frac{\pi (R^2-r^2)}{l{\rho }_c}$
The conductance of the nickel plated copper wire is the sum of these two conductances
$=\frac{\pi }{l}\times [\frac{r^2}{{\rho }_c}+\frac{R^2-r^2}{{\rho }_n}]$