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, \[{{a}_{t}}=-g\sin \theta =-g\theta \] \[{{a}_{t}}=-g\frac{x}{(R-r)}\] Motion is S.H.M., with time period \[T=2\pi \sqrt{\frac{\text{displacement}}{\text{acceleration}}}\]\[=2\pi \sqrt{\frac{x}{\frac{gx}{(R-t)}}}=2\pi \sqrt{\frac{R-r}{g}}\]You need to login to perform this action.
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