A wire of resistance 160 Ω is melted and drawn in a wire of...

The formula to calculate the resistance of the original wire is given by

$R = ρ\dfrac{l}{A} . . . \left(1\right)$

When the new wire is formed, its resistance is given by

$R ' = ρ\dfrac{l '}{4A} . . . \left(2\right)$

Divide equation (2) by equation (1) and simplify to obtain the expression for the new resistance.

$\begin{cases}\dfrac{R '}{R} & = & \dfrac{ρ\dfrac{l '}{4A}}{ρ\dfrac{l}{A}} \\ & ⇒ & R ' = R\dfrac{l '}{4l} . . . \left(3\right)\end{cases}$

Substitute the values of the known parameters into equation (3) to calculate the required new resistance.

$\begin{cases}R ' & = & \dfrac{\dfrac{l}{4}}{4l} \times 160Ω \\ & = & 10Ω\end{cases}$