A uniform wire of resistance 100 Ω is melted and recast in a wire of length double that of the original. What would be the resistance of the wire?

400 Ω

Given,


Initial Resistance, R1 of the wire= 100 Ω


Initial length of wire=l1, Final length of wire=l2= 2 l1


Formula used


The expression for the resistance, R, of a wire is



Where ρ is the resistivity, A is the area of cross-section and l is the length of the wire.


Hence, by knowing the final length and area we can calculate the final resistance by comparing it with that of the initial case. Also, the information that the volume of the wire would not change on recast, should be used.


Solution,


We know that the volume remains same after the recast.


If we represent the volume as a function of area and length, the above information can be expressed as,



Where subscripts ‘1’ and ‘2’ denotes the initial and final cases.


It is given that, l2= 2 l1, so the final area A2 can be represented as,



We use this to compare the two resistance,


The initial resistance, R1 is



Similarly, the final resistance, R2 is



Since the material is the same in both cases, ρ1= ρ2. By dividing eqn.3 by eqn.2, we get,



By substituting the relation between length and area at the initial and final cases, the above expression can be re-written as,



Or,



Or,



the resistance of the wire is 400 Ω


1