$A$ 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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$A$ glass capillary tube is in the shape of a truncated cone with an apex angle $\alpha$ so that its two ends have cross sections of different radii. When dipped in water vertically,water rises in it to a height $h$,where the radius of its cross section is $b$. If the surface tension of water is $S$,its density is $\rho$,and its contact angle with glass is $\theta$,the value of $h$ will be ($g$ is the acceleration due to gravity).

Surface tension of two liquids (having same densities), $T_1$ and $T_2$, are measured using the capillary rise method utilizing two tubes with inner radii of $r_1$ and $r_2$ where $r_1 > r_2$. The measured liquid heights in these tubes are $h_1$ and $h_2$ respectively. [Ignore the weight of the liquid above the lowest point of the meniscus]. If $T_1 = T_2$, which of the following relations is satisfied?

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Water rises in a capillary tube of radius $r$ up to a height $h$. The mass of water in the capillary is $m$. The mass of water that will rise in a capillary of radius $r/3$ will be:

What is capillary action? Derive the formula for the rise of liquid in a capillary tube immersed vertically in a liquid.

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