a given wire of resistance 1ohm is stretched to double its length. what will be its new resiastance ?how to solve

Here,

R = 1 Ω

Length = L

CSA = A

R = ρL/A

=> ρL/A = 1  ---1.

The new length L’ = 2L

And new CSA = A’

So we have

Volume of the wire = AL

Now after the act of stretching the volume remains constant.

AL = A’L’

=> AL = 2LA’

=> A’ = ½A

Now the resistance of the wire becomes

R’ = ρL’/A’

  = ρ(2L)/(A/2)

   = 4(ρL/A)

By eqn.1

ρL/A = 1

  R’ = 4×1 =  4Ω

So new resistance becomes = 4Ω