Q. A copper wire of length $L$ and radius $r$ is nickel plated till its final radius become $R$ but length remains $L$. If the resistivity of nickel and copper be $\rho_{n}$ and $\rho_{c}$ respectively, the conductance of the nickelled wire is
A) $\frac{\pi r^{2}}{L.\rho_{c}}$
B) $\frac{\pi(R^{2} - r^{2})}{L{.\rho}_{n}}$
C) $\frac{\pi}{L}\left\lbrack \frac{r^{2}}{\rho_{c}} + \frac{(R^{2} - r^{2})}{\rho_{n}} \right\rbrack$
D) $\frac{L\rho_{c}}{\pi r^{2}} + \frac{L.\rho_{n}}{\pi(R^{2} - r^{2})}$
Explanation
(c) Resistance of copper part of wire $R_{c} = \frac{\rho_{c}.L}{A_{c}} = \frac{\rho_{c}.L}{\pi r^{2}}$ and
Resistance of nickel portion of wire $R_{n} = \frac{\rho_{c}.L}{A_{n}} = \frac{\rho_{c}.L}{\pi(R^{2} - r^{2})}$
As these two resistances are in parallel, hence conductance
of the nickelled wire $C = \frac{1}{R} = \frac{1}{R_{c}} + \frac{1}{R_{n}} = \frac{\pi r^{2}}{\rho_{c}.L} + \frac{\pi(R^{2} - r^{2})}{\rho_{n}.L}$
$\mathbf{=}\frac{\pi}{L}\left\lbrack \frac{r^{2}}{\rho_{c}} + \frac{R^{2} - r^{2}}{\rho_{n}} \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.