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 \[R=\frac{\rho l}{A}\] When\[l\]is the length of conductor. A its area of cross-section and p its resistivity \[\rho =\frac{RA}{l}\] ?.(i) Given \[R=0.072\,\Omega \] \[A=(2\times 2)m{{m}^{2}}\] \[=4\times {{10}^{-6}}{{m}^{2}}\]. \[l=12m\] Substituting the given values is Eq. (i) we get \[\rho =\frac{0.072\times 4\times {{10}^{-6}}}{12}=2.4\times {{10}^{-8}}\,\Omega \,m\]
You need to login to perform this action.
You will be redirected in
3 sec
