Answer:
Option A
Explanation:
According to question, force acting on the particle inside the magnetic field is given by
F_{B}=qvB\sin \theta
[where \theta= angle between v and B]
This force FB provides necessary centripetal force for circular motion of the charged particle.
So, \frac{mv^{2}}{r}=qvB\sin \theta
Now, for particles x and y and \theta = 90°
\frac{m_{x}v_{x}^{2}}{r_{1}}=qv_{x}B...................(i)
\frac{m_{y}v_{y}^{2}}{r_{2}}=qv_{y}B...........(ii)
From Eqs.(i) and (ii) , we get
\frac{m_{x}v_{x}^{}}{m_{y}v_{y}^{}}=\frac{r_{1}}{r_{2}}
as, \frac{v_{x}^{}}{v_{y}^{}}=1
So, \frac{m_{x}^{}}{m_{y}^{}}=\frac{r_{1}}{r_{2}}