A) \[2.4\times {{10}^{-6}}\Omega m\]
B) \[1.2\times {{10}^{-6}}\Omega m\]
C) \[1.2\times {{10}^{-8}}\Omega m\]
D) \[2.4\times {{10}^{-8}}\Omega m\]
Correct Answer: D
Solution :
The resistance R of a particular conductor is related to the resistivity \[\rho \]of its material by \[R=\frac{\rho l}{A}\] or \[\rho =\text{resistivity}=\frac{RA}{l}\] Given, \[R=0.072\,\Omega ,\] \[A=2\,mm\times 2\,mm=4\times {{10}^{-6}}{{m}^{2}},l=12\,mm\] \[\therefore \] \[\rho =\frac{0.072\times 4\times {{10}^{-6}}}{12}\] \[=0.024\times {{10}^{-6}}\Omega \,m\] \[=2.4\times {{10}^{-8}}\Omega \,m\]
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