A) \[t=2\pi \sqrt{\frac{l}{g}}\]
B) \[t=2\pi \sqrt{\frac{l\sin \theta }{g}}\]
C) \[t=2\pi \sqrt{\frac{l\cos \theta }{g}}\]
D) \[t=2\pi \sqrt{\frac{l}{g\cos \theta }}\]
Correct Answer: C
Solution :
Radius of circular path in the horizontal plane\[r=l\sin \theta \]You need to login to perform this action.
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