A) \[\frac{4}{9}\]
B) \[\frac{9}{5}\]
C) \[\frac{1}{2}\]
D) \[\frac{1}{3}\]
Correct Answer: A
Solution :
\[\operatorname{r}=\frac{2mE}{eB}~~~~\therefore E = eV\] \[\operatorname{r}=\frac{2mE}{eB}~=\sqrt{\frac{2mV}{e{{B}^{2}}}}\] Two particles each of charge 6C under same magnetic field \[\operatorname{r}\propto \sqrt{m}\] \[\frac{{{\operatorname{r}}_{1}}}{{{\operatorname{r}}_{2}}}=\sqrt{\frac{{{m}_{1}}}{{{m}_{2}}}}\] squaring both sides \[\frac{{{m}_{1}}}{{{m}_{2}}}={{\left( \frac{{{\operatorname{r}}_{1}}}{{{\operatorname{r}}_{2}}} \right)}^{2}}=\frac{4}{9}\]You need to login to perform this action.
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