The diameter of one of the bases of a truncated cone is $100 \, mm$. If the diameter of this base is increased by $21 \%$ such that it still remains a truncated cone with the height and the other base unchanged,the volume also increases by $21 \%$. The radius of the other base (in $mm$) is

  • A
    $65$
  • B
    $55$
  • C
    $45$
  • D
    $35$

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