A) \[7.43\]
B) \[10.56\]
C) \[14.38\]
D) \[16.48\]
Correct Answer: A
Solution :
The resistance \[(R)\] of wire of length (and area\[A\]is \[R=\rho \frac{1}{A}\] Given,\[l=5\,\,cm=5\times {{10}^{-2}}m,\,\,\rho =3.5\times {{10}^{-5}}\Omega \text{-}m\] \[A=\pi (r_{1}^{2}-r_{2}^{2})=\pi (1\times {{10}^{-4}}-0.25\times {{10}^{-4}})\] \[\Rightarrow \] \[A=0.75\pi \times {{10}^{-4}}m\] \[\Rightarrow \] \[A=7.5\pi \times {{10}^{-5}}m\] \[\therefore \] \[R=\frac{3.5\times {{10}^{-5}}\times 5\times {{10}^{-2}}}{7.5\pi \times {{10}^{-5}}}\] \[=7.43\times {{10}^{-3}}\Omega \]
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