The length of a metallic wire is increased by $20 \%$ and its area of cross-section is reduced by $4 \%$. The percentage change in resistance of the metallic wire is $.......$ (in $\%$)

In order to quadruple the resistance of a uniform wire,a part of its length $x$ was uniformly stretched till the final length of the entire wire was $1.5$ times the original length $l$. The fraction of the wire that was stretched,$x/l$,is equal to:

$A$ $1\,m$ long wire is broken into two unequal parts $X$ and $Y$. The $X$ part of the wire is stretched into another wire $W$. The length of $W$ is twice the length of $X$ and the resistance of $W$ is twice that of $Y$. Find the ratio of the length of $X$ to the length of $Y$.

The resistance of a conductor at $20\,^{\circ}C$ and $500\,^{\circ}C$ is $20\,\Omega$ and $60\,\Omega$ respectively. At what temperature will the resistance be $25\,\Omega$?

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

The reciprocal of resistance is