A) \[1.1\times {{10}^{3}}rad\text{ }{{s}^{-1}}\]
B) \[2.1\times {{10}^{3}}\text{ }rad\text{ }{{s}^{-1}}\]
C) \[3.1\times {{10}^{3}}rad\text{ }{{s}^{-1}}\]
D) \[4.1\times {{10}^{3}}rad\text{ }{{s}^{-1}}\]
Correct Answer: A
Solution :
(a) \[1.1\times {{10}^{3}}rad\,{{s}^{-1}}\] Here, \[C=30\mu F=30\times {{10}^{-6}}F\] \[L=27\,mH=27\times {{10}^{-3}}H\] \[\therefore \omega =\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{27\times {{10}^{-3}}\times 30\times {{10}^{-6}}}}\] \[=\frac{1}{\sqrt{81\times {{10}^{-8}}}}\] \[=\frac{{{10}^{4}}}{9}=1.1\times {{10}^{3}}rad\,{{s}^{-1}}\]
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