A wire of length L is drawn such that its diameter is reduced to...

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}$L=4 L Now, we know that the resistance is given by$R=\rho \frac{L}{A}$.$$\begin{aligned} \therefore \quad& R^{\prime}=\rho \frac{L^{\prime}}{A^{\prime}}=\rho \frac{4 L}{A / 4}=16 R . \\ & {\left[\becauseA^{\prime}=\frac{\pi D^2}{16}=\frac{A}{4}\right]} \\ \therefore \quad & R^{\prime}=16 \times10=160 \Omega . \end{aligned}$$