Question

Question 6
A copper wire has diameter 0.5 mm and resistivity of $1.6\times {10}^{-8}\Omega m$. What will be the length of this wire to make its resistance $10\Omega$ ? How much does the resistance change if the diameter is doubled?

Answer 1

Area of cross-section of the wire, $A=\pi {\left(\frac{d}{2}\right)}^2$
Diameter = 0.5 mm = 0.0005 m
Resistance, $R=10\Omega$
We know that
$R=\rho \frac{l}{A}$
$l=\frac{RA}{\rho }$
$=\frac{10\times 3.14\times (\frac{0.0005}{2}{)}^2}{1.6\times {10}^{-8}}$ $=\frac{10\times 3.14\times 25}{4\times 1.6}=122.7$
$\therefore$ length of the wire = 122.7 m
If the diameter of the wire is doubled, new diameter = $2\times 0.5=1mm=0.001m$
Let new resistance be R’
$R^{\prime }=\rho \frac{l}{A}$ $=\frac{1.6\times {10}^{-8}\times 122.7}{\pi (\frac{1}{2}\times {10}^{-3}{)}^2}$ $=\frac{1.6\times {10}^{-8}\times 122.7\times 4}{3.14\times {10}^{-6}}$
$=250.09\times {10}^{-2}=2.5\Omega$
Therefore, the length of the wire is 122.7 m and the new resistance is $2.5\Omega$.