A) 1 × \[{{10}^{-6}}\]
B) 4 × \[{{10}^{-7}}\]
C) 3 × \[{{10}^{-7}}\]
D) 2 × \[{{10}^{-7}}\]
Correct Answer: A
Solution :
\[R=\frac{V}{i}=\rho \frac{l}{A}\]Þ \[\frac{2}{4}=\rho \frac{50\times {{10}^{-2}}}{{{(1\times {{10}^{-3}})}^{2}}}\]Þ \[\rho =1\times {{10}^{-6}}\Omega m\].
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