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In the diagram shown , the charge +Q is fixed another charge +2q and mass M is projected from a distance R from the fixed charge . Minimum separation between the charges if the velocity becomes $\;\large\frac{1}{\sqrt{3}}\;$ times of the projected velocity , at this moment

$(a)\;\large\frac{\sqrt{3}R}{2}\qquad(b)\;\large\frac{1}{\sqrt{3}R}\qquad(c)\;\large\frac{1}{2R}\qquad(d)\;None\;of\;these$

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Answer : (a) $\;\large\frac{\sqrt{3}R}{2}$
Explanation :
Angular momentum of 2q charge about Q is constant . Therefore
$m\;v\;R\;sin 30^{0}=\large\frac{m\;v}{\sqrt{3}}\;r_{min} \;sin \theta$
When the two charges will be at minimum distance their relative velocity along the line joining them is zero . So $\theta=90^{0}$
$\large\frac{m\;v\;R}{2}=\large\frac{m\;v\;r_{min}\times1}{\sqrt{3}}$
$r_{min}=\large\frac{\sqrt{3} R}{2}\;.$
answered Feb 12, 2014 by yamini.v
 

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