If a capillary tube is tilted to $45^o$ and $60^o$ from the vertical,then the ratio of the lengths $l_1$ and $l_2$ of the liquid columns in it will be:

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

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