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Q.
A wire of resistance $R$ is stretched till its radius is half of the original value. Then, the resistance of the stretched wire is :-
Solution:
Resistance $R$ of a wire of length $1$ and radius $r$ with specific resistance $K$ of its material is given by:
$R =\frac{ K l}{\pi r ^{2}} \ldots$(i)
As the wire is stretched, mass $M$ of the wire remains unchanged.
But $M =\pi r ^{2} ld$ or $1=\frac{ M }{\pi r ^{2} d }\ldots$(ii)
Hence, $R =\frac{ KM }{\pi^{2} r ^{4} d } \ldots . . .$(iii)
Let the resistance of teh stretched wire be $R'$;
then
$R' =\frac{ KM }{\pi^{2} d ( r / 2)^{4}}$
$=(2)^{4}\left[\frac{ KM }{\pi^{2} dr ^{4}}\right]=(2)^{4} R =16 R$