question_answer

A cone filled with water is revolved in a vertical circle of radius 4 m and the water does not fall down. What must be the maximum period of revolution?                                                     [BHU M-2003]

A)  \[2\,\,s\]                                            

B)  \[4\,\,s\]

C)   \[1\,\,s\]                                           

D)  \[6\,\,s\]

Correct Answer: B

Solution :

                 When cone filled with water is revolved in a vertical circle of radius 4m, then speed of body is different at different points of the circular path. Let \[{{v}_{c}}\] be the minimum speed at highest point. When \[{{T}_{A}}=0\,\,and\,\,{{v}_{A}}={{v}_{C}},\] then \[mg=\frac{mv_{c}^{2}}{r}\] \[\Rightarrow \]                                               \[{{v}_{c}}=\sqrt{gr}\] This is critical speed of body should be greater than the critical speed for body to continue its circular path.                 \[\therefore \]                                  \[{{v}_{c}}\ge \sqrt{gr}\] Also, \[v=r\omega \] where, \[\omega \] is angular speed and \[\omega =\frac{2\pi }{T},\]                 \[\therefore \]                  \[v=r\frac{2\pi }{T}\]                 \[\Rightarrow \]                               \[\frac{2\pi \,r}{T}\ge \sqrt{rg}\]                 \[\Rightarrow \]                               \[T\le \frac{2\pi \,r}{\sqrt{rg}}=2\,\pi \sqrt{\frac{r}{g}}\] Given, \[r=4,\,g=9.8\,\,m/{{s}^{2}}\]                 \[\therefore \] \[{{T}_{\max }}=2\,\pi \sqrt{\frac{4}{9.8}}=2\times 3.14\times 0.6389=4s\]                 \[\Rightarrow \]                               \[{{T}_{\max }}=4\,s.\]