A) \[\frac{ml}{n{{e}^{2}}\tau A}\]
B) \[\frac{m{{\tau }^{2}}A}{n{{e}^{2}}l}\]
C) \[\frac{n{{e}^{2}}\tau A}{2ml}\]
D) \[\frac{n{{e}^{2}}A}{2m\tau l}\]
Correct Answer: A
Solution :
\[R=\rho \frac{l}{A}=\frac{n}{n{{e}^{2}}\tau }.\frac{l}{A}\]You need to login to perform this action.
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