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Question

$\mathrm{KE}=\frac{\mathrm{q}^{2} \mathrm{B}^{2} \mathrm{r}^{2}}{2 \mathrm{M}}$

$\mathrm{qV}=\frac{\mathrm{q}^{2} \mathrm{B}^{2} \mathrm{r}^{2}}{2

Vedclass · Accepted Answer

$\sqrt 2 \,:\,1\,:\,\sqrt 2 $

Vedclass · Answer

$\mathrm{KE}=\frac{\mathrm{q}^{2} \mathrm{B}^{2} \mathrm{r}^{2}}{2 \mathrm{M}}$

$\mathrm{qV}=\frac{\mathrm{q}^{2} \mathrm{B}^{2} \mathrm{r}^{2}}{2 \mathrm{M}}$

$\frac{2 m V}{B^{2} q}=r^{2}$

$\mathrm{r}=\sqrt{\frac{2 \mathrm{mV}}{\mathrm{B}^{2} \mathrm{q}}}$

$r \propto \sqrt {\frac{m}{q}} $

$r_{\alpha}: r_{p} \cdot r_{d}=\sqrt{\frac{M_{\alpha}}{q_{\alpha}}}: \sqrt{\frac{M_{p}}{q_{p}}}: \sqrt{\frac{M_{d}}{q_{d}}}$

$=\sqrt{\frac{4 M_{p}}{2 q_{p}}}: \sqrt{\frac{M_{p}}{q_{p}}}: \sqrt{\frac{2 M_{p}}{q_{p}}}$

$=\sqrt{2}: 1: \sqrt{2}$

$\alpha $ particle, proton and duetron enters in a uniform (transverse) magnetic field $'B'$ with same acceleration potential find ratio of radius of path followed by these particles.

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