$A$ piece of wire having resistance $R$ is cut into four equal parts. $(a)$ How will the resistance of each part compare with the original resistance? $(b)$ If the four parts are placed in parallel,how will the joint resistance compare with the resistance of the original wire?

(N/A) Since the wire is divided into four equal parts,the new length of each part becomes one-fourth of the original length. Since resistance is directly proportional to length $(R \propto l)$,the resistance of each part becomes one-fourth of the original resistance,$i.e., R' = R/4$.
$(b)$ When these four parts are connected in parallel,the equivalent resistance $R_p$ is given by the formula:
$\frac{1}{R_p} = \frac{1}{R'} + \frac{1}{R'} + \frac{1}{R'} + \frac{1}{R'}$
Substituting $R' = R/4$:
$\frac{1}{R_p} = \frac{4}{R} + \frac{4}{R} + \frac{4}{R} + \frac{4}{R} = \frac{16}{R}$
Therefore,$R_p = R/16$.
Thus,the joint resistance is $1/16$ times the resistance of the original wire.