A section of fixed smooth circular track of radius `R` in vertical plane is shown in the figure. A block is released from position `A` and leaves the track at `B` The radius of curvature of its trajectory just after it leaves the track `B` is ?

Options

A. R
B. `(R )/(4)`
C. `(R )/(2)`
D. none of these

Answer

C

Explanation

By energy conservation between A and B
`rArr Mg.(2R)/5+0=(MgR)/5+1/2 MV^(2)`
`V=sqrt((2gR)/5)`
Now, radius of curvature =`(V_(bot)^(2))/(a_(r))=(2gR//5)/(g cos 37)=R/2`