Question

A wire is stretched so that its length becomes 6/5 times its original length. If its original resistance is 25 ohm find its new resistance and resistivity. 

Answer 1

Resistance becomes n² times of length is made n times its original length then answer would be (6/5)²25=36Ω

Answer 2

Resistance of a conductor of length L and an area of uniform cross section is 
                     R = ρ L / A ,              ρ = resistivity of the material
                                given     R = 25 Ω

Volume of the material in the conductor = L A.
Let us assume that the density of the material remains the same.  Hence the volume Vol of the conductor remains the same.
 
    Vol = A L = A' L'        => A' = A L / L'

After stretching the wire :    L' = 6 L / 5 

             A' = 5 A L / 6 L = 5 A / 6

Resistance of stretched wire =  R' = ρ L' / A' = ρ  ( 6 L /5) *  [ 6 / (5 A) ]

             R' = 36 ρ L / (25 A)  = 36 R / 25

New resistance = 36 * 25 / 25 Ω = 36 Ω