A) \[1.28\times {{10}^{-8}}\Omega m\]
B) \[1.57\times {{10}^{-8}}\Omega m\]
C) \[1.75\times {{10}^{-9}}\Omega m\]
D) \[1.88\times {{10}^{-8}}\Omega m\]
Correct Answer: B
Solution :
\[R=40\,\Omega ,d=1mm,I=2m\] \[\rho =\frac{RA}{I},\rho =\frac{R\pi {{d}^{2}}}{4I},\frac{40\times 2\times 1}{4\times 7\times 2\times {{10}^{8}}}=\frac{800}{56\times {{10}^{8}}}\] \[=\frac{880\times {{10}^{-8}}}{560}\]\[\therefore \]\[\rho =1.57\times {{10}^{-8}}\Omega m\]
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