Answer
64.8k+ 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.
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.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)