A $4\, \Omega$ resistance wire is doubled on itself. Calculate the new resistance of the wire.

Given $R=4 \Omega$

When a wire is doubled on it, its length would become half and area of cross-section would double. That is, a wire of length $L$ and area of cross-section A becomes of length $L / 2$ and area of cross- section $2 A$ i.e.

$L_{n}=\frac{L}{2}$ and $A_{n}=2 A$

Using equation $R =\frac{\rho L }{ A },$ we have

$R _{n}=\frac{\rho L _{n}}{ A _{n}}$

Dividing, we have

$\frac{ R _{n}}{ R }=\frac{ L _{n}}{ A _{n}} \times \frac{ A }{ L }=\frac{ L / 2}{2 A } \times \frac{ A }{ L }=\frac{1}{4}$

Since the old resistance is $4 \Omega$, therefore, the new resistance becomes one-fourth of the previous value. Hence, $R _{n}=1 ohm$