Question

If the radius of a wire is decreased by n times then, what will be the new resistance?

Answer 1

$$\begin{gathered} Resis\tan ceofthewire=\frac{\rho L}{A} \\ initiallength=L \\ finallength{L}^{\text{'}}=\frac{L}{n} \\ Asvolumremainssame \\ initialvolume=finalvolume \\ AL=A^{\text{'}}{L}^{\text{'}} \\ AL=A^{\text{'}}\frac{L}{n} \\ {A}^{\text{'}}=An \\ \frac{R^{\text{'}}}{R}=\frac{{\displaystyle \frac{\rho L^{\text{'}}}{A^{\text{'}}}}}{\frac{\rho L}{A}}=\frac{{\displaystyle \frac{L}{n\times nA}}}{\frac{L}{A}}=\frac{1}{n^2} \\ {R}^{\text{'}}=\frac{R}{n^2} \\ Resis\tan cewilldecreaseby{n}^{2}times. \end{gathered}$$