A) \[22.5\,\Omega \]
B) \[45\,\Omega \]
C) \[9\,\Omega \]
D) \[11.25\,\Omega \]
Correct Answer: D
Solution :
For single wire cable \[R=\frac{\rho l}{\pi {{r}^{2}}}=\frac{\rho l}{\pi {{(9\times {{10}^{-3}})}^{2}}}=5\,\Omega \] \[\Rightarrow \] \[\frac{\rho l}{\pi }=81\times 5\times {{10}^{-6}}\] \[=4.05\times {{10}^{-4}}\] For other wire of cable \[R=\frac{\rho l}{\pi r_{1}^{2}}=4.05\times {{10}^{-4}}\times \frac{1}{{{(3\times {{10}^{-3}})}^{2}}}\] \[=\frac{405}{9}\times \frac{{{10}^{-6}}}{{{10}^{-6}}}\] \[R=45\,\Omega \] When 4 were each of resistance\[R\]are connected in parallel then effective resistance Will be \[{{R}_{p}}=\frac{R}{4}=\frac{45}{4}=11.25\Omega \]
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