The electric resistance of a certain wire of iron is R. If its length and radius are both doubled, then

Solution:

The formula for resistance of wire is

$R=\frac{\rho l}{A}$

where $\rho$ = specific resistance of the wire

$\Rightarrow R\propto \frac{l}{A}$

$\Rightarrow R\propto \frac{1}{r^2} \left(\because A=\pi r^2\right)$

$\begin{array}{l}\therefore \frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}=\frac{{\mathrm{l}}_1}{{\mathrm{l}}_2}\times \frac{{\mathrm{r}}_2^2}{{\mathrm{r}}_1^2}....\left(\mathrm{i}\right)\\ \mathrm{Given},{\mathrm{l}}_1=\mathrm{l},{\mathrm{l}}_2=2\mathrm{l},{\mathrm{r}}_1=\mathrm{r},{\mathrm{r}}_2=2\mathrm{r},{\mathrm{R}}_1=\mathrm{R}\end{array}$

Substituting these values in Eq. (i), we have

$\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}=\frac{\mathrm{l}}{2\mathrm{l}}\times \frac{{\left(2\mathrm{r}\right)}^2}{{\mathrm{r}}_{ }^2}$

$\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}=2\mathrm{and}{\mathrm{R}}_2=\frac{\mathrm{R}}{2}$

Therefore, resistance will be halved.
Now the specific resistance of the wire does not depend on the geometry of the wire, hence, it remains unchanged.

Aliter: $\mathrm{R}\propto \frac{\mathrm{l}}{{\mathrm{r}}^2}\Rightarrow \frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}=\frac{{\mathrm{l}}_2^{ }}{{\mathrm{l}}_1}\times \frac{{\mathrm{r}}_1^2}{{\mathrm{r}}_2^2}=\left(\frac{2}{1}\right)\times {\left(\frac{2}{1}\right)}^2$

$=\frac{1}{2}$

$\Rightarrow {\mathrm{R}}_2=\frac{{\mathrm{R}}_1}{2},$  specific resistance doesn't depend upon length and radius.