A) \[2\pi \sqrt{\left( \frac{\left( R-r \right)1.4}{g} \right)}\]
B) \[2\pi \sqrt{\left( \frac{R-r}{g} \right)}\]
C) \[2\pi \sqrt{\left( \frac{rR}{a} \right)}\]
D) \[2\pi \sqrt{\left( \frac{R}{gr} \right)}\]
Correct Answer: B
Solution :
Tangential acceleration, \[{{a}_{t}}=-g\sin \theta =-g\theta \]You need to login to perform this action.
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