If the masses of three wires of same material are in the ratio of $1: 2: 3$ and their lengths are in the ratio of $3: 2: 1$,then electrical resistances of these wires are in the ratio

Equal potentials are applied on an iron and copper wire of the same length. In order to have the same current flow in the two wires, the ratio $r_{\text{iron}} / r_{\text{copper}}$ of their radii must be (Given that specific resistance of iron = $1.0 \times 10^{-7} \, \Omega \cdot \text{m}$ and specific resistance of copper = $1.7 \times 10^{-8} \, \Omega \cdot \text{m}$)

If a conducting wire of length $L$ is uniformly stretched to double its length, then its conductivity becomes . . . . . . .

Consider a block of conducting material of resistivity $\rho$ shown in the figure. Current $I$ enters at $A$ and leaves from $D$. We apply the superposition principle to find the voltage $\Delta V$ developed between $B$ and $C$. The calculation is done in the following steps:
$(i)$ Take current $I$ entering from $A$ and assume it to spread over a hemispherical surface in the block.
$(ii)$ Calculate field $E(r)$ at distance $r$ from $A$ by using Ohm's law $E = \rho j$,where $j$ is the current per unit area at $r$.
$(iii)$ From the $r$ dependence of $E(r)$,obtain the potential $V(r)$ at $r$.
$(iv)$ Repeat $(i)$,$(ii)$ and $(iii)$ for current $I$ leaving $D$ and superpose results for $A$ and $D$.
For current entering at $A$,the electric field at a distance $r$ from $A$ is

$A$ wire of resistance $3 \Omega$ is stretched to twice its original length. The resistance of the new wire will be (in $\Omega$)

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