Question #222af

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Answer 1 · 2016-12-18T08:04:14

Given

#d->"Diameter of the wire"=0.46mm=46xx10^-5m#

#l->"Length of the wire"=730cm=7.3m#

#rho->*"Resistivity of the wire"=110xx10^-8ohmm#

So #A->"Area of cross section"=(pid^2)/4#

We know resustance #R# of the wire is rlated with above noted physical quantities as follows.

#R=rhol/A=(4*rho*l)/(pi*d^2)#

# =(4*110xx10^-8xx7.3)/(3.14*(46xx10^-5)^2)~~48.3Omega #