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A proton and an $\alpha$-particle, accelerated through the same potential difference, enter a region of uniform magnetic field normally. If the radius of the proton orbit is $10 \mathrm{~cm}$, then radius of $\alpha$-particle is
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$10 \sqrt{2} \mathrm{~cm}$
Radius of path $r_{\text {time }}=\frac{1}{\beta} \sqrt{\frac{2 \mathrm{mv}}{\mathrm{q}}}$
$\therefore \quad \frac{\mathrm{r}_{\alpha}}{\mathrm{r}_{\mathrm{p}}}=\sqrt{\frac{\mathrm{m}_{\alpha}}{\mathrm{m}_{\mathrm{p}}}} \sqrt{\frac{\mathrm{q}_{\mathrm{p}}}{\mathrm{q}_{\alpha}}}$
or, $\quad \frac{\mathrm{r}_{\alpha}}{10}=\sqrt{\frac{4}{2}} \Rightarrow \mathrm{r}_{\alpha}=10 \sqrt{2} \mathrm{~cm}$
$\therefore \quad \frac{\mathrm{r}_{\alpha}}{\mathrm{r}_{\mathrm{p}}}=\sqrt{\frac{\mathrm{m}_{\alpha}}{\mathrm{m}_{\mathrm{p}}}} \sqrt{\frac{\mathrm{q}_{\mathrm{p}}}{\mathrm{q}_{\alpha}}}$
or, $\quad \frac{\mathrm{r}_{\alpha}}{10}=\sqrt{\frac{4}{2}} \Rightarrow \mathrm{r}_{\alpha}=10 \sqrt{2} \mathrm{~cm}$
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