Question

A conductor of length l has a resistance of 4 ohm.if this resistance wire is melted and drawn to half of its original length

(a) what will be the resistance and resistivity of the new wire

Answer 1

Let r be radius of the original wire,
Resistance of original wire,R = 4 ohm
R =ρl/A
where l is the original length of the wire,
A be the area of the wire = πr²
So,
ρ = 4×πr² /l

Volume of wire remains same as we decrease the length of wire.

Now, new length after melting and redrawn, l' = l/2
new resistance,r'
As volume remains constant so,
Volume before melting = Volume after melting
πr²l = πr'² l/2
r' = √2r
So,new resistance be,
R' = $$\begin{gathered} \frac{4\times \pi r^2}{l}\times \frac{l}{2}\times \frac{1}{\pi (\sqrt{2}r{)}^2} \\ =1ohm \end{gathered}$$
Resistivity of new wire remains same.
Resistivity does not change with length and area of cross section , it changes with temperature.