A copper wire of length l and radius r is nickel plated till its final radius is 2r. If the resistivity of the copper and nickel are `rho_(c)` and `rho_(n)`, then find the equivalent resistance of the wire.

Options

A. `(l)/(pir^2[(1)/(rho_(c))+(3)/(rho_(n))])`
B. `(l)/(pir^2[(1)/(rho_(c))-(3)/(rho_(n))])`
C. `(2l)/(pir^2[(1)/(rho_(c))-(3)/(rho_(n))])`
D. `(2l)/(pir^2[(1)/(rho_(c))+(3)/(rho_(n))])`

Answer

A

Explanation

Copper and nickel are connected in parallel. Now , area of copper wire is `A_1=pi r^2`
and area of nickel wire is `A_2=pi(2r)^2-pi r^2=3 pi r^2`
`(1)/(R_p)=(1)/(R_1)+(1)/(R_2)`
As, `R=(rhol)/(pi r^2)`
`(1)/(R_p)=(pi r^2)/(rho_cl)+(3 pi r^2)/(rho_nl)=(pi r^2rho_nl+3pi r^2rho_cl)/(rho_clxxrho_nl)`
or ,` (1)/(R_p)=(pi r^2)/(rho_crho_nl)[rho_n+3rho_c]`
or `(1)/(R_p)=(pi r^2)/(l)[(rho_n)/(rho_c.rho_n)+(3rho_c)/(rho_crho_n)]`
or `(1)/(R_p)=(pir^2)/(l)[(1)/(rho_c)+(3)/(rho_n)]`
or `R_p=(l)/(pi r^2[(1)/(rho_c)+(3)/(rho_n)])`