Answer:
Given: Resistance of a wire, \[R=10\,\Omega \] Radius, \[r=0.01\,cm=0.01\times {{10}^{-2}}m\] Resistivity, \[\rho =50\times {{10}^{-8}}\,\Omega m\] Area of cross-section A, \[=\pi {{r}^{2}}=3.14\times {{(0.01\,\times {{10}^{-2}})}^{2}}\] \[=3.14\times 0.01\times 0.01\times {{10}^{-4}}\] \[=3.14\times {{10}^{8}}\,{{m}^{2}}\] We know, \[R=\rho \frac{l}{A}\] or \[l=\frac{R\mathrm{ }\times \mathrm{ }A}{\rho }=\frac{10\times 3.14\times \mathrm{ }{{10}^{-8}}}{50\mathrm{ }\times \mathrm{ }{{10}^{-8}}}=0.628m\]
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