A wire of length L is drawn such that its diameter is reduced to half of its original diameter. If the initial resistance of the wire were$10\Omega$, its new resistance would be _________.

$$\begin{gathered} \mathrm{For} \mathrm{a} \mathrm{wire} \mathrm{of} \mathrm{length} \mathrm{l}, \mathrm{area} \mathrm{A}, \mathrm{specific} \mathrm{resistance} \mathrm{\sigma } , \mathrm{the} \mathrm{resistance} \mathrm{is} \mathrm{given} \mathrm{by} \\ \text{R}=\mathrm{\rho }\frac{\mathrm{l}}{\text{A}}\to \left(1\right) \\ \mathrm{If} \mathrm{original} \mathrm{diameter} \mathrm{of} \mathrm{wire} \mathrm{be} \mathrm{d}, \mathrm{then} \mathrm{new} \mathrm{diameter} \mathrm{is} \frac{\mathrm{d}}{2}, \\ \mathrm{Original} \mathrm{area} \mathrm{of} \mathrm{cross} – \mathrm{section} \mathrm{is} \frac{{\mathrm{\pi d}}^2}{4} \mathrm{and} \mathrm{final} \mathrm{area} \mathrm{of} \mathrm{cross} – \mathrm{section} \mathrm{is}\frac{{\mathrm{\pi d}}^2}{16} \\ \mathrm{Since} \mathrm{volume} \mathrm{remains} \mathrm{constant}, \mathrm{on} \mathrm{pulling} \mathrm{the} \mathrm{wire} \mathrm{we} \mathrm{have} \\ \frac{{\mathrm{\pi d}}^2}{4}\times \mathrm{l}=\frac{{\mathrm{\pi d}}^2}{16}\times {\mathrm{l}}^{|} \\ \Rightarrow {\mathrm{l}}^{|}=\frac{16}{4}\mathrm{l}=4\mathrm{l} \\ {\text{R}}^{|}=\mathrm{\rho }\frac{{\mathrm{l}}^{|}}{{\text{A}}^{\text{|}}}\to \left(2\right) \\ \therefore {\text{R}}^{\text{|}}=\mathrm{\rho }.\frac{4\mathrm{l}}{\text{A}\mathrm{l}4}=16\text{R} \\ \mathrm{Hence}, \mathrm{new} \mathrm{resistance} 16\times 10=160\mathrm{\Omega } \end{gathered}$$