# The radius of conducting wire is doubled. What will be the ratio of new specific resistance to the old one

r

_{1}= R

r

_{2}= 2R

Specific resistance, $\rho $ = $\frac{RA}{L}$

where, R is resistance, A is cross-sectional area and L is length

So, $\rho $ = $\frac{R\left(\pi {r}^{2}\right)}{L}$

Or, $\frac{{\rho}_{1}}{{\rho}_{2}}$ = $(\frac{{r}_{1}}{{r}_{2}}{)}^{2}$ $\Rightarrow $ $\frac{{\rho}_{1}}{{\rho}_{2}}$ = $\left(\frac{R}{2R}\right)$

^{2 }$\Rightarrow $ $\frac{{\rho}_{1}}{{\rho}_{2}}$ = $\frac{1}{4}$

So, the ratio is 1:4

Regards

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