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 :
Resistance of a conductor is given by \[R=\frac{\rho l}{A}\] where\[l\]is the length of conductor. A its area of cross-section and p its resistivity. So, \[\rho =\frac{RA}{l}\] ...(i) Given, \[R=0.072\,\Omega ,\] \[A=(2\times 2)m{{m}^{2}}=4\times {{10}^{-6}}{{m}^{2}},l=12\,m\] Substituting the given values is Eq.(i), we get \[\rho =\frac{0.072\times 4\times {{10}^{-6}}}{12}=2.4\times 1000\,\Omega \,m\]
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