A smooth horizontal fixed pipe is bent in the form of a vertical circle of radius `20m` as shown in figure. A small glass ball is shown in horizontal portion of pipe at speed `30m//s` as shown from end `A`. `(` Take `g=10m//s^(2))`

With what speed ball willcome out from point `B :`

Options

A. `30 m//s`
B. `20sqrt(2) m//s`
C. `10sqrt(5) m//s`
D. None of these

Answer

A

Explanation

In the given situation if the speed becomes zero at the highest point then also the particle can complete the circle as there is no chance for the particle to loose constact in this case.
`u_(m i n)=` minimum speed required to complete vertical circle
`=sqrt(4gR)=sqrt(4xx10xx20)=sqrt(800)m//s`
`30 m//s gt sqrt(800)`
so it can easily complete the vertical circel
Now, for point `C`
`k_(f)+p_(f)=p_(i)+k_(i)`
`(1)/(2) mv_(c)^(2)+mgh_(c)=0+(1)/(2) m (30)^(2)`
`v_(c)^(2)=(30)^(2)-2gh_(c)`
As `h_(c)=h_(E)=R,` heights of points `C & E` from reference so `V_(E)=V_(C)`
As there will be no energy dissipation, it will come out at the same speed at which it enters.