A) \[{{10}^{-6}}\,ohm\]
B) \[2.5\times {{10}^{-5}}\,ohm\]
C) \[{{10}^{-8}}\,ohm\]
D) \[5\times {{10}^{-4}}\,ohm\]
Correct Answer: A
Solution :
\[R=\frac{\rho l}{A}=50\times {{10}^{-8}}\times \frac{50\times {{10}^{-2}}}{{{(50\times {{10}^{-2}})}^{2}}}={{10}^{-6}}\Omega \]
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