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
$$\eqalign{ & \Rightarrow R \propto \frac{l}{A} \cr & \Rightarrow R \propto \frac{l}{{{r^2}}}\,\,\left( {\because A = \pi {r^2}} \right) \cr & \therefore \frac{{{R_1}}}{{{R_2}}} = \frac{{{l_1}}}{{{l_2}}} \times \frac{{r_2^2}}{{r_1^2}}\,......\left( {\text{i}} \right) \cr} $$
Given, $${l_1} = l,{l_2} = 2l,{r_1} = r,{r_2} = 2r,{R_1} = R.$$
Substituting these values in Eq. (i), we have
$$\eqalign{ & \frac{{{R_1}}}{{{R_2}}} = \frac{l}{{2l}} \times \frac{{{{\left( {2r} \right)}^2}}}{{{r^2}}} \cr & \frac{{{R_1}}}{{{R_2}}} = \frac{l}{{2l}} \times \frac{{{{\left( {2r} \right)}^2}}}{{{r^2}}} \cr & \frac{{{R_1}}}{{{R_2}}} = 2 \cr & \Rightarrow {R_2} = \frac{R}{2} \cr} $$
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.