A wire 250 cm long and $1 mm^2$ in cross-section carries a current of 4 A when connected to a 2 V battery. The resistivity of the wire is

$0.2 \times 10^{-6}\Omega-m$ $0.2 \times 10^{-7}\Omega-m$ $5\times 10^{-6}\Omega-m$ $4 \times 10^{-6}\Omega-m$

$0.2 \times 10^{-6}\Omega-m$

$ R = \frac{V}{I} = 0.5 \Omega$ $ R = \frac{\rho l}{A} $ $\Rightarrow \rho = \frac{RA}{l} = 0.2\times 10^{-6} \Omega-m$