Question

A wire of resistance $R$ is stretched till its radius is half of the original value. Then, the resistance of the stretched wire is :-

• A. 2R
• B. 4R
• C. 8R
• D. 16R

Answer 1

Answer: (D)

Resistance $R$ of a wire of length $1$ and radius $r$ with specific resistance $K$ of its material is given by:
$R=\frac{Kl}{\pi r^2}\dots$(i)
As the wire is stretched, mass $M$ of the wire remains unchanged.
But $M=\pi r^{2}ld$ or $1=\frac{M}{\pi r^{2}d}\dots$(ii)
Hence, $R=\frac{KM}{{\pi }^2{r}^{4}d}\dots ...$(iii)
Let the resistance of teh stretched wire be $R^{\prime }$;
then
$R^{\prime }=\frac{KM}{{\pi }^{2}d(r/2{)}^4}$
$=(2{)}^4\left[\frac{KM}{{\pi }^{2}d{r}^4}\right]=(2{)}^{4}R=16R$