A particle of mass `m` is released from a height `H` on a smooth curved surface which ends into a vertical loop of radius `R`, as shown. Choose the correct alernative(s) if `H=2R`.

Options

A. The particles reaches the top of the loop with zero velocity
B. The particle cannot reach the top of the loop.
C. The particle breaks off at a height `h=R` from the base of the loop.
D. The particle breaks off at a height `h` from the base of the loop such that `Rlthlt2R`

Answer

A, B, D

Explanation

Velocity on reaching the base of circle is
`u=sqrt(2gH)=sqrt(2g2R)=sqrt(4gR)`
This is less than `sqrt(5gR)`. Hence, the particle cannot complete the circle. But this velocity is more than `sqrt(2gR)` so the particle will break off the loop at a height above the centre before reaching the top.