A) \[\tilde{\ }{{10}^{10}}/{{m}^{-3}}\]
B) \[\tilde{\ }{{10}^{11}}{{m}^{-3}}\]
C) \[\tilde{\ }{{10}^{12}}{{m}^{-3}}\]
D) \[\tilde{\ }{{10}^{13}}{{m}^{-3}}\]
Correct Answer: C
Solution :
We know that \[fC=q\sqrt{{{N}_{\max }}}\] \[\Rightarrow \]\[{{N}_{\max }}=\frac{f{{C}^{2}}}{81}=\frac{{{(10\times {{10}^{6}})}^{2}}}{81}\] \[=\frac{{{10}^{14}}}{81}=1.24\times {{10}^{12}}/{{m}^{3}}\] \[={{10}^{12}}{{m}^{-3}}\]
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