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, \[r=4,\,g=9.8\,\,m/{{s}^{2}}\] \[\therefore \] \[{{T}_{\max }}=2\,\pi \sqrt{\frac{4}{9.8}}=2\times 3.14\times 0.6389=4s\] \[\Rightarrow \] \[{{T}_{\max }}=4\,s.\]You need to login to perform this action.
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