A) \[2\sqrt{\frac{\pi gL}{{{\mu }_{0}}}\tan \theta }\]
B) \[\sqrt{\frac{\pi \lambda gL}{{{\mu }_{0}}}\tan \theta }\]
C) \[\sin \theta \sqrt{\frac{\pi \lambda gL}{{{\mu }_{0}}\cos \theta }}\]
D) \[2\sin \theta \sqrt{\frac{\pi \lambda gL}{{{\mu }_{0}}\cos \theta }}\]
Correct Answer: D
Solution :
\[T\sin \theta =\frac{{{\mu }_{0}}{{I}^{2}}x}{4\pi L\sin \theta }\] \[T\cos \theta =\lambda xg\] \[T\cos \theta =\frac{{{\mu }_{0}}{{I}^{2}}}{4\pi L\sin \lambda g}\Rightarrow I=2\sin \theta \sqrt{\frac{\pi \lambda gL}{{{\mu }_{0}}\cos \theta }}\]You need to login to perform this action.
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