A copper wire of length l and r 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.
Text Solution
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Let `R_(c)` and `R_(n)` be the resistance of copper wire and nickel plated portion of wire respectively. Then `R_(c) = (rho_(c) l)/(pi r^(2)) , and R_(n) = ( rho_(n) l)/(pi (2 r)^(2) - pi r^(2))` Here, `R_(C)` and `R_(n)` are in parallel, the equivalent resistance of the wire is `R = (R_(c) R_(n))/(R_(c) +R_(n)) = (rho_(c) l //(pi r^(2)) xx rho_(n)l//[pi(2r)^(2) - pi r^(2)])/((rho_(c) l)/(pi r^(2)) + (rho_(n)l)/([pi(2r)^(2) - pir^(2)]))` `(rho_(c ) rho_(n) l)/(pi r^(2) (rho_(n) + 3 rho_(c )))`
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