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)/(pi r^(2)[(1)/(rho_(c))+(3)/(rho_(n))])`
B. `(l)/(pi r^(2)[(1)/(rho_(c))-(3)/(rho_(n))])`
C. `(2l)/(pi r^(2)[(1)/(rho_(c))-(3)/(rho_(n))])`
D. `(2l)/(pi r^(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_(1)^(2)= pi r^(2)`
and area of nickel wire is
`A_(2)= pi ( 2 r )^(2) - pi r^(2)= 3 pi r^(2)`
`:.(1)/(R_(p))=(1)/(R_(1))+(1)/(R_(2)) " " ( :'R=(rho l)/(pi r^(2)))`
`:.(1)/(R_(p))=(pi r^(2))/( rho_(c)l)+(3 pi r^(2))/( rho_(n)l)=(pi r^(2)rho_(n)l + 3 pi r^(2) rho_(c)l)/(rho_(c) l xx rho_(n) l)`
`(1)/(R_(p))=(pi r^(2))/(rho_(c)rho_(n)l) [ rho_(n)+3 rho_(c)]`
`(1)/(R_(p))=( pi r^(2))/(1) [ (rho_(n))/(rho_(c).rho_(n))+(3 rho_(c))/(rho_(c)rho_(n))]`
`(1)/(R_(p))=(pi r^(2))/(l) [ (1)/(rho_(c))+(3)/( rho_(n))]`
`R_(p)=(1)/(pi r^(2)[ (1)/( rho_(c))+(3)/( rho_(n))])`