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A particle initially at rest starts moving from point A on the surface of a fixed smooth hemisphere ofradius r as shown. The particle looses its contact with hemisphere at point B. C is centre of the hemisphere.The equaion relating $α and β$ is

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3 \sin α = 2 \cos β

3 \sin β = 2 \cos α

2 \sin α = 3 \cos β

2 \sin β = 3 \cos α

answer is C.

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Detailed Solution

According to LCE

${(K + U)}_{A} = {(K + U)}_{B}$

$mgr \cos α = \frac{1}{2} {mv}_{B}^{2} + mgR \sin β$

$v_{B}^{2} = 2 gR (\cos α - \sin β)$

At ‘B’ particle leaves the contact than N = 0

$mg \sin β = \frac{{mv}_{B}^{2}}{r}$

$g \sin β = \frac{2 gr (\cos α - \sin β)}{r}$

$\sin β = 2 \cos α - 2 \sin β$

$2 \cos α = 3 \sin β$

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