Correct Answer - C
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`