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?

2 Answers
Nov 24, 2016

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.

Nov 25, 2016

Answer:

#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#.

enter image source here

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#