A) \[2\pi \sqrt{\frac{L}{g\cos \alpha }}\]
B) \[2\pi \sqrt{\frac{L}{g\sin \alpha }}\]
C) \[2\pi \sqrt{\frac{L}{g}}\]
D) \[2\pi \sqrt{\frac{L}{g\tan \alpha }}\]
Correct Answer: A
Solution :
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