Physics - Current Electricity Question with Solution | TestHub

Let the original diameter of the wire be $D$. Therefore the new diameter is $D/2$. Original area of cross-section is $\frac{\pi D^2}{4}$ and the final area of cross-section is $\frac{\pi D^2}{16}$. The new length of the wire is given by $$L\times \frac{\pi D^2}{4}=L^{\prime }\times \frac{\pi D^2}{16}\Rightarrow L^{\prime }=\frac{16}{4}L=4L$$ Now, we know that the resistance is given by $R=\rho \frac{L}{A}$. $$\begin{array}{rl}\therefore & R^{\prime }=\rho \frac{L^{\prime }}{A^{\prime }}=\rho \frac{4L}{A/4}=16R.\\ & \left[\because A^{\prime }=\frac{\pi D^2}{16}=\frac{A}{4}\right]\\ \therefore & R^{\prime }=16\times 10=160\Omega .\end{array}$$