1.

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`

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`


Discussion

No Comment Found

Related InterviewSolutions