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An $\alpha$-particle (mass 4 amu) and a singly ....

An $\alpha$-particle (mass 4 amu) and a singly charged sulfur ion (mass 32 amu) are initially at rest. They are accelerated through a potential V and then allowed to pass into a region of uniform magnetic field which is normal to the velocities of the particles. Within this region, the $\alpha$-particle and the sulfur ion move in circular orbits of radii $r_{\alpha}$ and $r_S$, respectively. The ratio ($r_S/r_\alpha$) is _____.

Solution

$r = \sqrt {\frac{{2mV}}{{q{B^2}}}} $

So, $r \propto \sqrt {\frac{m}{q}} $

$\frac{{{r_S}}}{{{r_\alpha }}} = \sqrt {\frac{{{m_S}}}{{{q_S}}} \times \frac{{{q_\alpha }}}{{{m_\alpha }}}}  = \sqrt {\frac{{{m_S}}}{{{m_\alpha }}} \times \frac{{{q_\alpha }}}{{{q_S}}}}  = \sqrt {\frac{{32}}{4} \times \frac{2}{1}}  = 4$

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A man starts walking from the point P (-3, 4) ....

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR} $ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1 $ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$
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