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The length of a germanium rod is `0.58 cm` and its area of cross-section is `10cm^(2)`. If for germanium `n_(i)=2.5xx10^(19)m^(-3),mu_(h)=0.19 m^(2)//V-s,mu_(e)=0.39 m^(2)//V-s`, then the resistance of the rod will be-A. `2.5k Omega`B. `4.0kOmega`C. `5.0kOmega`D. `10.0kOmega` |
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Answer» Correct Answer - B We have, `R=(rhol)/(A)=(L)/(n_(i)e(mu_(e)+mu_(h))A)` `=(0.928xx10^(-2))/(2.5xx10^(19)xx1.6xx10^(-19)(0.19)xx10^(-6))` `=4000Omega=4.0kOmega` |
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