A cylindrical wire of length L and radius r has a resistance R. The resistance of another wire of the same material but half its length and half its radius will be :

Correct option is C

$$\rho\ (\text{rho}) = \frac{RA}{L}\\[8pt]R = \rho \frac{L}{A} = \rho \frac{L}{\pi r^{2}}\\[14pt]\text{For second wire:}\\[8pt]L = \frac{L}{2}, \quad r = \frac{r}{2}\\[14pt]R_1 = \rho \frac{\frac{L}{2}}{\pi \left(\frac{r}{2}\right)^{2}}= \rho \frac{\frac{L}{2}}{\frac{\pi r^{2}}{4}}= \rho \frac{2L}{\pi r^{2}}\\[14pt]\boxed{R_1 = 2R}$$​​