A material B has twice the specific resistance of A. A circular wire made of B has twice the diameter of a wire made of A. Then for the two wires to have the same resistance, the ratio of lB / lA of their respective lengths must be

Let the specific resistances, lengths, radii and areas of wires A and B be ρA,lA,rA,AA and ρB,lB,rB,AB respectively.Resistance of A =  RA=ρAlAπrA2Resistance of B= RB=ρBlBπrB2According to given question, ρB=2ρA,rB=2rA and RA=RB ThereforeρAlAπrA2=ρBlBπrB2Further, ρAlAπrA2=2ρAlBπ2rA2∴ lBlA=21=2