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))`

At which angle from vertical from bottom most point `F`. The normal reaction on ball due to pipe will change its direction `(` in terms of radially outwards and inwards `) :`

Options

A. `theta=180^(@)`
B. `theta = cos^(-1)(-(2)/(3))`
C. `theta = cos^(-1)(-(5)/(6))`
D. None of these

Answer

C

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)`

`mg cos (180- theta )=(mv^(2))/(l) ....(1)`
Applying `W-E` theoroen between points `F & P:`
`(1)/(2) m u^(2)=.(1)/(2) mv^(2)+mgl(1-cos theta )`
`v^(2)=u^(2)-2gl(1-cos theta ) ......(2)`
on putting the value of `v^(2)` from `(2)` in `(1)`
`mg cos (180-theta )=(m)/(l)(u^(2)-2gl(1-cos theta )]`
`-g l cos theta =u^(2)-2gl+2glcos theta `
`-3gl cos theta =900-2xx10xx20`
`cos theta -(500)/(3gl)=-(500)/(600)`
`cos theta =-5//6`