A) \[\sqrt{\frac{2L}{(g+a)\sin \theta }}\]
B) \[\sqrt{\frac{2L}{(g-a)\sin \theta }}\]
C) \[\sqrt{\frac{2L}{a\sin \theta }}\]
D) \[\sqrt{\frac{2L}{g\sin \theta }}\]
Correct Answer: B
Solution :
Key Idea: Net acceleration in downward motion decreases. The free body diagram of the set up is shown. Net acceleration is \[(g-a)\sin \theta \]You need to login to perform this action.
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