Question

A copper wire has diameter 2 mm and resistivity of $3.14\times {10}^{-7}\Omega m$. What will be the length of this wire to make its resistance $10\Omega$ ?

Answer 1

Area of cross-section of the wire, $A=\pi {\left(\frac{d}{2}\right)}^2$
Diameter = 2 mm = 0.002 m
Resistance, $R=10\Omega$
We know that
$R=\rho \frac{l}{A}$
$l=\frac{RA}{\rho }$
$=\frac{10\times 3.14\times (\frac{0.002}{2}{)}^2}{3.14\times {10}^{-7}}$

$=\frac{10\times 3.14\times (0.001{)}^2}{3.14\times {10}^{-7}}$

$=\frac{10\times 3.14\times (10{)}^{-6}}{3.14\times {10}^{-7}}$

$=\frac{10\times 3.14\times 10}{3.14}=100m$

$\therefore$ length of the wire = 100 m