A wire of resistance 160   Ω is melted and drawn in a wire of one-fourth of its length. The new…

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

$R=\rho \frac{l}{A} ...\left(1\right)$

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

$R\text{'}=\rho \frac{l\text{'}}{4A} ...\left(2\right)$

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

$\begin{array}{rcl}\frac{R\text{'}}{R}& =& \frac{\rho \frac{l\text{'}}{4A}}{\rho \frac{l}{A}}\\ & \Rightarrow & R\text{'}=R\frac{l\text{'}}{4l} ...\left(3\right)\end{array}$

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

$\begin{array}{rcl}R\text{'}& =& \frac{{\displaystyle \frac{l}{4}}}{4l}\times 160 \mathrm{\Omega }\\ & =& 10 \mathrm{\Omega }\end{array}$