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Q.

A smooth ring is kept on a smooth horizontal surface. From a point P of the ring a particle is projected at an angle α  to the radius vector at P. If e is the coefficient of restitution between the ring and the particle and the particle will return to the point of projection after two reflections then 1e+1e2+1e3  is equal to__

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a

sin2α

b

cos2α

c

tan2α

d

cot2α

answer is D.

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Detailed Solution

Let u be the velocity of projection at P. We can find the velocity of rebound at point Q fromtanβ=usinαeucosα=tanαeSimilarly, tanγ=tanβe=tanαe2 2α+2β+2γ=π α+β+γ=π2 α+β=π2-γ tanα+tanβ1-tanαtanβ=cotγ=e2tanα tanα+tanαe1-tanαtanαe=e2tanα on  solving we get  cot2α=e+e2+1e3=1e+1e2+1e3
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