Two beads $A$ and $B$ of masses $m_1$ and $m_2$ respectively are placed on a smooth vertical circular wire of radius $R$ as shown in the figure. Bead $A$ is given a very gentle push so that it slides down and collides with bead $B$ and comes to rest. After the collision,bead $B$ reaches the height of the center of the circle. Then $m_1 : m_2$ is equal to:

  • A
    $1 : \sqrt{2}$
  • B
    $1 : 2$
  • C
    $1 : 4$
  • D
    $\sqrt{2} : 1$

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