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

Dear Student

r1 = R
r2 = 2R

​Specific resistance, $\rho$ = $\frac{RA}{L}$
where, R is resistance, A is cross-sectional area and L is length

So, $\rho$
Or, $\frac{{\rho }_{1}}{{\rho }_{2}}$ = $\left(\frac{{r}_{1}}{{r}_{2}}{\right)}^{2}$        $⇒$    ​$\frac{{\rho }_{1}}{{\rho }_{2}}$ = ​$\left(\frac{R}{2R}\right)$2         $⇒$     ​$\frac{{\rho }_{1}}{{\rho }_{2}}$ = $\frac{1}{4}$

So, the ratio is 1:4

Regards

• -4
1/4 will the ratio
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