A) \[{{(2{{R}_{1}}/{{R}_{2}})}^{2}}\]
B) \[{{({{R}_{1}}/2{{R}_{2}})}^{2}}\]
C) \[R_{1}^{2}/2R_{2}^{2}\]
D) \[2{{R}_{1}}/{{R}_{2}}\]
Correct Answer: C
Solution :
Given, \[{{q}_{y}}=2{{q}_{x}}\] Radius of circular path in a magnetic field is given by \[r=\frac{mv}{Bq}\] \[\therefore \] \[v=\frac{Brq}{m}\] \[{{v}^{2}}=\frac{{{B}^{2}}{{r}^{2}}{{q}^{2}}}{{{m}^{2}}}\] or \[m{{v}^{2}}=\frac{{{B}^{2}}{{r}^{2}}{{q}^{2}}}{m}\] \[\therefore \] \[KE=\frac{1}{2}m{{v}^{2}}=\frac{{{B}^{2}}{{r}^{2}}{{q}^{2}}}{2m}\] ? (i) When charge particle is accelerated by potential\[V\], then its kinetic energy \[KE=Vq\] ... (ii) From Eqs. (i) and (ii) \[Vq=\frac{{{B}^{2}}{{r}^{2}}{{q}^{2}}}{2m}\] \[m=\frac{{{B}^{2}}{{r}^{2}}q}{2V}\] \[\therefore \] \[m\propto {{r}^{2}}q\] \[\therefore \] \[\frac{{{m}_{1}}}{{{m}_{2}}}=\frac{r_{1}^{2}{{q}_{1}}}{r_{2}^{2}{{q}_{2}}}\] \[=\frac{R_{1}^{2}}{R_{2}^{2}}\times \frac{q}{2q}=\frac{R_{1}^{2}}{2R_{2}^{2}}\]
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