question_answer

A bucket, full of water is revolved in a vertical circle of radius 2m. What should be the maximum time-period of revolution so that the water does not fall out of the bucket?

A) 1 s                                          

B) 2 s

C) 3 s                                          

D) 4 s

Correct Answer: A

Solution :

Key Idea: The minimum velocity which the bucket should have to complete the full circle is\[v\ge \sqrt{5gR}.\] The water contained in the revolving bucket experiences a centrifugal force which is always equal and opposite to the centripetal force. The minimum velocity required to complete the circle with radius R and gravity g is  \[v\ge \sqrt{5gR}.\] Also\[v=R\omega \] where, \[\omega \]is angular velocity \[\therefore \]                        \[R\omega \ge \sqrt{5gR}\] \[\Rightarrow \]                      \[\omega \ge \sqrt{\frac{5g}{R}}\] Also \[\omega =\frac{2\pi }{T}\] where, T is time period \[\therefore \]                      \[T\le 2\pi \sqrt{\frac{R}{5g}}\] Given, \[~R=2\text{ }m,\text{ }g=10\text{ }m/{{s}^{2}}\] \[\therefore \]    \[{{T}_{\max }}=2\pi \sqrt{\frac{2}{5\times 10}}=\frac{2\pi }{5}=1.256\,s\approx 1s\]