Question

Q.6. A copper wire has a 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

$$\begin{gathered} DearStudent, \\ \rho =\frac{RA}{l} \\ \rho =resistivity \\ R=resis\tan ce \\ A=area \\ l=length \\ 1.6\times {10}^{-8}=\frac{10\times {\mathrm{\pi r}}^2}{l} \\ 1.6\times {10}^{-8}=\frac{10\times \pi \times 0.25\times 0.25\times {10}^{-3}\times {10}^{-3}}{l} \\ l=\frac{10\times \pi \times 0.25\times 0.25\times {10}^{-3}\times {10}^{-3}}{1.6\times {10}^{-8}}=1.22\times 100=122m \\ ifdiameterisdoubleditwilldoubletheradiussothelengthwillincreasetofourtimes. \\ Regards \end{gathered}$$