A) \[{{\text{(}{{\text{R}}_{\text{1}}}\text{/}{{\text{R}}_{\text{2}}}\text{)}}^{\text{2}}}\]
B) \[{{\text{(}{{\text{R}}_{\text{1}}}\text{/2}{{\text{R}}_{\text{2}}}\text{)}}^{\text{2}}}\]
C) \[\text{R}\,_{1}^{2}\text{/2R}_{2}^{2}\]
D) \[\text{2}{{\text{R}}_{\text{1}}}\text{/}{{\text{R}}_{\text{2}}}\]
Correct Answer: C
Solution :
Given, \[{{a}_{y}}=2{{q}_{x}}\] 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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