Question

If a wire of resistance `R` is melted and recast into a wire of length `2/5`th that of the original wire, what is the resistance of the new wire?

Answer 1

Let

`"initial length of the wire be"=L_i`

`"initial cross section of the wire be"=A_i`

`"initial resistance of the wire be"=R_i`

After melting and recast let

`"final length of the wire be"=L_f`

`"final cross section of the wire be"=A_f`

`"final resistance of the wire be"=R_f`

The volume of the material of the wire in both state will be same. So `L_iA_i=L_fA_f....(1)`

Again it is given that `L_f=2/5L_i=>L_f/L_i=2/5`

Now from (1)

`L_iA_i=L_fA_f`

`=>A_i/A_f=L_f/L_i.....(2)`

Let `rho` be the resustivity of the material of the wire.
Then for initial wire

`R_i=rhoL_i/A_i.....(3)`

And for initial wire

`R_f=rhoL_f/A_f.....(4)`

Dividing (4) by (3) we get

`R_f/R_i=L_f/L_ixxA_i/A_f=(L_f/L_i)^2=(2/5)^2=4/25`

`=>R_f/R_i =0.16`

`=>R_f =0.16R_i`

So the resistance of the new recast wire will be `0.16` times the resistance of the original wire.

Answer 2

`R_2 = 4/25 R`
Or
`R_2 = 0.16 R`

Explanation:

During the recasting process the same volume of metal that was melted was used to form the new wire. For that reason we know that the volume (`V`) of both wires is equal. The resisitivity (`ρ`) is also the same for both wires because exactly the same material is used for both wires.

Volume is related to the length and cross-sectional area of the wire by the relationship:
`V=AL`
I will define the initial dimensions of the wire as `V`, `A` and `L`. The dimensions of the second wire will be `V`, `A_2` and `L_2`.

Initial resistance of the wire is: `R = (ρ L)/A`

The question specifies that the length is reduced to `(2L)/5`. So `L_2 = (2L)/5`. Now use the fact that the volumes are equal to get an expression for `A_2` in terms of `A`:
`A_2 = V/L_2 = V/((2L)//5) = (5V)/(2L)`

Also `V=AL` so then the equation becomes:
`A_2 = (5AL)/(2L) = 5A/2`

Now that we have expresssions for both `L_2` and `A_2` we can use them in the equation for resistance to find `R_2` in terms of `R`:
`R_2 = (ρ L_2)/A_2 = (ρ (2L)//5)/((5A)/2) = (ρ 4L)/(25A) = 4/25 ((ρ L)/A)`
`⇒ R_2 = 4/25 R`

Or
`R_2 = 0.16 R`