A wire $50\, cm$ 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
AIPMT 1994, Medium
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$R = \frac{V}{i} = \rho \frac{l}{A}$ $ \Rightarrow $ $\frac{2}{4} = \rho \frac{{50 \times {{10}^{ - 2}}}}{{{{(1 \times {{10}^{ - 3}})}^2}}}$
$ \Rightarrow \rho = 1 \times {10^{ - 6}}\,\Omega m$.
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