$A$ capillary tube of radius $0.1 \,mm$ is dipped in water. The water rises to a height of $2 \,cm$ in the tube. If the surface tension of water is $0.072 \,N/m$, the contact angle between water and the wall of the tube is:

  • A
    $\theta = \cos^{-1}\left(\frac{1}{3.6}\right)$
  • B
    $\theta = \cos^{-1}\left(\frac{1}{7.2}\right)$
  • C
    $\theta = \cos^{-1}\left(\frac{1}{1.8}\right)$
  • D
    $\theta = \cos^{-1}\left(\frac{1}{6.2}\right)$

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