A wire of length L is drawn such that its diameter is reduced to half of its original diameter. If the initial resistance of the wire were 10  Ω, its new resistance would be

For a wire of length l, area A, specific resistance ρ, the resistance is given byR= ρ lA                                      ....1If original diameter of wire be d, then new diameter is d/2Original area of cross-section is πd2/4 and final area of cross-section is πd2/16Since volume remains constant, on pulling the wire we haveπd24  ×  l  =  πd216  ×  l'⇒    l' =164  l  =  4 lAlso  R'= ρ l'A'                                        ....2∴      R'= ρ. 4 lA/4  =  16  RHence, new resistance 16  ×  10=160   Ω