A) \[\omega =\sqrt{rg\,\sin \theta }\]
B) \[\omega =\sqrt{g/r\,\,\cos \theta }\]
C) \[\omega =\sqrt{\frac{gr}{\cos \theta }\,\,}\]
D) \[\omega =\sqrt{\frac{gr}{\tan \theta }\,\,}\]
Correct Answer: A
Solution :
The overall conductivity of a semiconductor is \[{{C}_{2}}{{H}_{5}}CHBr-C{{H}_{3}}\,\,and\,\,{{C}_{2}}{{H}_{5}}-C{{H}_{2}}-C{{H}_{2}}Br\] \[PC{{l}_{5}}\] Also \[PC{{l}_{5}}(g)PC{{l}_{3}}(g)+C{{l}_{2}}(g)\] (For an intrinsic semiconductor) \[PC{{l}_{5}}\] Now. \[\frac{X}{a}={{\left( \frac{{{K}_{p}}}{P} \right)}^{1/2}}\] On differentiating, \[\frac{X}{a}=\frac{{{K}_{p}}}{{{K}_{p}}+p}\] \[\frac{X}{a}={{\left( \frac{{{K}_{p}}}{{{K}_{p}}+p} \right)}^{1/2}}\] \[\frac{X}{a}={{\left( \frac{{{K}_{p}}+p}{{{K}_{p}}} \right)}^{1/2}}\]
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