A) \[1.57\times {{10}^{-6}}\,\Omega m\]
B) \[5.25\times {{10}^{-7}}\,\Omega m\]
C) \[7.12\times {{10}^{-5}}\,\Omega m\]
D) \[2.55\times {{10}^{-7}}\,\Omega m\]
Correct Answer: D
Solution :
(d) \[2.55\times {{10}^{-7}}\,\Omega m\] \[l=1.0\,m;D=0.4\,mm=4\times {{10}^{-4}}m\] \[R=2\,\Omega \] \[A=\frac{\pi {{D}^{2}}}{4}=\frac{\pi \times {{(4\times {{10}^{-4}})}^{2}}}{4}=4\pi \times {{10}^{-8}}{{m}^{2}}\] Now, \[\rho =\frac{RA}{l}=\frac{2\times 4\pi \times {{10}^{-8}}}{1}=2.55\times {{10}^{-7}}\Omega m\]
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