A) 7.42
B) 10.56
C) 14.38
D) 16.48
E) 18.29
Correct Answer: A
Solution :
Given: \[l=5\,cm=5\times {{10}^{-2}}m,\] \[\rho =3.5\times {{10}^{-5}}\Omega m\] \[A=\pi (1\times {{10}^{-4}}-0.25\times {{10}^{-4}})\] \[=0.75\pi \times {{10}^{-4}}m\] \[=7.5\pi \times {{10}^{-5}}m\] \[\therefore \] \[R=\frac{\rho l}{A}=\frac{3.5\times {{10}^{-5}}\times 5\times {{10}^{-2}}}{7.5\pi \times {{10}^{-5}}}\] \[=7.42\times {{10}^{-3}}\Omega \]You need to login to perform this action.
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