A) 2.5 pF
B) 5.0 pF
C) 25 pF
D) 50 pF
Correct Answer: A
Solution :
[a] \[L=10\,\,mHz={{10}^{-2}}\,Hz\] \[f=1\,MHz={{10}^{6}}\,Hz\] \[f=\frac{1}{2\pi \,\sqrt{LC}}\Rightarrow \,\,\,\,\,\,\,{{f}^{2}}=\frac{1}{4{{\pi }^{2}}LC}\] \[\Rightarrow \,\,C=\frac{1}{4{{\pi }^{2}}{{f}^{2}}L}=\frac{1}{4\times 10\times {{10}^{-2}}\times {{10}^{12}}}\]
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