(D) The resistance $R$ of a cylindrical conductor depends on the following factors:
$(i)$ Length $(l)$: Resistance is directly proportional to the length $(R \propto l)$.
$(ii)$ Area of cross-section $(A)$: Resistance is inversely proportional to the area of cross-section $(R \propto 1/A)$.
$(iii)$ Nature of the material (resistivity,$\rho$).
$(iv)$ Temperature.
When a conductor is stretched to double its length,its volume remains constant. If the original length is $l$ and the new length is $l' = 2l$,then the new area of cross-section $A'$ will be $A/2$ because $V = A \times l = A' \times l'$.
The new resistance $R'$ is given by:
$R' = \rho \times (l' / A') = \rho \times (2l / (A/2)) = 4 \times (\rho \times l / A) = 4R$.
Thus,the resistance becomes $4$ times the original resistance.