The table given below provides the resistivity of three samples in $\Omega \cdot m$:

Sample	Resistivity $(\Omega \cdot m)$
$A$	$1.6 \times 10^{-8}$
$B$	$7.5 \times 10^{17}$
$C$	$44 \times 10^{-6}$

$(a)$ Which of them is a good conductor? Which is an insulator? Why?
$(b)$ $A$ wire of resistance $4 \ \Omega$ is folded on itself to double its thickness. Calculate the new resistance of the wire.

$(A)$ Sample $A$ is a good conductor, and sample $B$ is an insulator.
Metals and alloys typically have low resistivity in the range of $10^{-8} \ \Omega \cdot m$ to $10^{-6} \ \Omega \cdot m$, making them good conductors of electricity. In contrast, insulators possess very high resistivity, typically in the order of $10^{12} \ \Omega \cdot m$ to $10^{17} \ \Omega \cdot m$.
$(b)$ When a wire of resistance $R = 4 \ \Omega$ is folded on itself, its length $l$ becomes $l' = l/2$ and its cross-sectional area $A$ becomes $A' = 2A$.
The resistance formula is $R = \rho (l/A)$.
The new resistance $R'$ is given by:
$R' = \rho (l'/A') = \rho ((l/2) / (2A)) = (1/4) \times \rho (l/A) = R/4$.
Substituting $R = 4 \ \Omega$:
$R' = 4 \ \Omega / 4 = 1 \ \Omega$.
Therefore, the new resistance is $1 \ \Omega$.