A) \[2\pi \sqrt{\frac{2{{a}^{2}}}{G\lambda }}\]
B) \[\frac{\sqrt{2}}{\pi }\times \frac{1}{\sqrt{15LC}}\]
C) \[\frac{1}{\sqrt{2}}\pi \times \frac{1}{\sqrt{15LC}}\]
D) \[\frac{2\sqrt{2}}{\pi }\times \frac{1}{\sqrt{15LC}}\]
Correct Answer: D
Solution :
We know that \[{{I}_{\max }}={{(\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}})}^{2}}\] In this case \[{{I}_{1}}=-{{I}_{P}}={{I}_{0/2}}\] and \[{{I}_{2}}={{I}_{Q}}={{I}_{0}}/2\] \[\therefore \] \[{{I}_{\max }}={{\left( \sqrt{\frac{{{I}_{0}}}{2}}+\sqrt{\frac{{{I}_{0}}}{2}} \right)}^{2}}=2{{I}_{0}}\]
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