In the figure `ADB` & `BEF` are two fixed circular paths. A block of mass m enters in the tube `ADB` through point `A` with minimum velocity to reach point `B`. From there it moves on another circular path of radius `R`'. There it is just able to complete the circle.

Options

A. velocity at `A` must be `sqrt(4Rg)`
B. velocity at `A` must be `sqrt(2Rg)`
C. `(R')/(R ) = (2)/(3)`
D. the normal reaction ast point `E` is `6 mg`

Answer

B, C, D

Explanation

In the figure…………….

For minimum velocity at `A` :
`(1)/(2)mV_(A)^(2)=mgR implies V_(A) = sqrt(2gR)`
Now , `(1)/(2)mV_(B)^(2)+mgR' = (1)/(2)m V_(E )^(2)`
As , `V_(B) = sqrt(2gR)`
For looping the loop ,
`V_(E ) = sqrt(5gR')`
`:. (1)/(2)m2gR+mgR' = (1)/(2)m 5gR'`
`:. (R')/(R ) = (2)/(3)`
And also, `N - mg = (mV_(E )^(2))/(R')`
`N-mg= (m5gR')/(R')`
`N = 6mg`