$A$ wire of length $L$ and resistance $R$ is stretched so that the length is doubled and area of cross-section is halved. How will its
$(a)$ resistance change?
$(b)$ resistivity change?

(A) $R = \rho \frac{L}{A}$
Where $R$ is the resistance and $\rho$ is the resistivity of the material.
Initial length of the wire $= L$. The new length is $L' = 2L$ and the new area of cross-section is $A' = A/2$.
$(a)$ The new resistance $R'$ is given by:
$R' = \rho \frac{L'}{A'} = \rho \frac{2L}{A/2} = 4 \left( \rho \frac{L}{A} \right) = 4R$.
Thus,the resistance becomes $4$ times the original resistance.
$(b)$ Resistivity $(\rho)$ is an intrinsic property of the material and depends only on the nature of the material and temperature. Therefore,it remains unchanged.