A) \[4.4\times {{10}^{-6}}ohm\times m\]
B) \[2.2\times {{10}^{-6}}ohm\times m\]
C) \[1.1\times {{10}^{-6}}ohm\times m\]
D) \[0.22\times {{10}^{-6}}ohm\times m\]
Correct Answer: B
Solution :
\[R=\frac{\rho L}{A}\Rightarrow 0.7=\frac{\rho \times 1}{\frac{22}{7}{{(1\times {{10}^{-3}})}^{2}}}\] \[\rho =2.2\times {{10}^{-6}}\]ohm-m.
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