Courses
Courses for Kids
Free study material
Offline Centres
More
Store

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) $\dfrac{{\pi {r^2}}}{{l{\rho _c}}}$B) $\dfrac{{\pi ({R^2} - {r^2})}}{{l{\rho _n}}}$C) $\dfrac{\pi }{l}\left[ {\dfrac{{{r^2}}}{{{\rho _c}}} + \dfrac{{{R^2} - {r^2}}}{{{\rho _n}}}} \right]$D) $\dfrac{{l{\rho _c}}}{{\pi {r^2}}} + \dfrac{{l{\rho _n}}}{{\pi \left( {{R^2} - {r^2}} \right)}}$

Last updated date: 13th Jun 2024
Total views: 51.6k
Views today: 0.51k
Verified
51.6k+ views
Hint: In this problem, we can treat the whole wire as a copper wire and having nickel wire (plating) connected in parallel. Calculate the conductance of copper wire and nickel wire separately. The effective conductance will be the same as the sum of two conductances connected in parallel.

Complete step by step solution:
In this problem, as we have discussed in the hint above, the copper wire after nickel plating will act as two conductances in parallel. Let us find the conductance for the copper wire.
As per the given information,
$l$ is the length of the copper wire
$r$ is the initial radius of the copper wire
${\rho _c}$ is the specific resistance of the copper wire
The conductance of copper wire ${G_c}$ will be given as
${G_c} = \dfrac{{area}}{{length \times specific\,resistance}}$
The copper wire is circular in shape, the area of the copper wire will be given as
$area = \pi {r^2}$
Thus, the conductance of the copper wire will be
${G_c} = \dfrac{{\pi {r^2}}}{{l \times {\rho _c}}}$--equation $1$
After the nickel plating is done, the new dimensions will be
$l$ is the length of the nickel plating
$R - r$ is the radius of the nickel plating
${\rho _n}$ is the specific resistance of nickel
The conductance of nickel wire ${G_n}$ will be given as
${G_n} = \dfrac{{area}}{{length \times specific\,resistance}}$
The nickel-plated wire will be circular in shape, the area of this nickel wire will be given as
$area = \pi ({R^2} - {r^2})$
Thus, the conductance of the nickel wire will be
${G_n} = \dfrac{{\pi \left( {{R^2} - {r^2}} \right)}}{{l \times {\rho _n}}}$--equation $2$
The effective conductance of two systems in parallel is the arithmetic sum of their individual conductance. Therefore, the effective conductance ${G_{effective}}$ will be given as
${G_{effective}} = {G_c} + {G_n}$
Substituting the value from equation $1$ and equation $2$ , we can have
${G_{effective}} = \dfrac{{\pi {r^2}}}{{l \times {\rho _c}}} + \dfrac{{\pi \left( {{R^2} - {r^2}} \right)}}{{l \times {\rho _n}}}$
Taking $\dfrac{\pi }{l}$ common, we can have
${G_{effective}} = \dfrac{\pi }{l}\left[ {\dfrac{{{r^2}}}{{{\rho _c}}} + \dfrac{{\left( {{R^2} - {r^2}} \right)}}{{{\rho _n}}}} \right]$
Thus, the conductance of wire with nickel plating is $\dfrac{\pi }{l}\left[ {\dfrac{{{r^2}}}{{{\rho _c}}} + \dfrac{{\left( {{R^2} - {r^2}} \right)}}{{{\rho _n}}}} \right]$ siemens.

Therefore, option C is the correct option.

Note: Conductance is the inverse of resistance. Similarly, conductivity is the inverse of resistivity. Remember that the effective conductance for two conductors is the arithmetic sum of their individual conductance. The SI unit of electrical conductance is siemens, it is inverse of ohm.