A) 5.6 ´ 10?5 W
B) 2 ´ 10?5 W
C) 4 ´ 10?5 W
D) None of these
Correct Answer: A
Solution :
By using \[R=\rho .\frac{l}{A};\] here \[A=\pi (r_{2}^{2}-r_{1}^{2})\] Outer radius r2 = 5cm Inner radius r1 = 5 ? 0.5 = 4.5 cm So \[R=1.7\times {{10}^{-8}}\times \frac{5}{\pi \{{{(5\times {{10}^{-2}})}^{2}}-{{(4.5\times {{10}^{-2}})}^{2}}\}}\] \[=5.6\times {{10}^{-5}}\Omega \]You need to login to perform this action.
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