Correct Answer - A::C
When a charged particle is subjected to a perpendicular magnetic field, it will describe a circular path of radius, `r=mv//Bq` i.e. `rprop1//(q//m)`. Thus option (c) is true. As time period of revolution, `T=(2pim)/(Bq)` is independent of v and r, so the particle will return to its initial position after one revolution, hence the two particles of different charges will return to their initial positions after one revolution and their paths will touch each other there.