Question

A capillary tube with inner cross-section in the form of a square of side ${}^{\prime }{a}^{\prime }$ is dipped vertically in a liquid of density $\rho$ and surface tension $\sigma$ which wets the surface of the capillary tube with an angle of contact $\theta$. The approximate height to which the liquid will rise in the tube is: (Neglect the effect of surface tension at the corners of the capillary tube)

• A. $\frac{2\sigma \text{cos}\theta }{a\rho g}$
• B. $\frac{4\sigma \text{cos}\theta }{a\rho g}$
• C. $\frac{8\sigma \text{cos}\theta }{a\rho g}$
• D. $\frac{4\sigma \text{sec}\theta }{a\rho g}$

Answer 1

The correct option is B $\frac{4\sigma \text{cos}\theta }{a\rho g}$

Upward force due to surface tension on the top surface of the liquid is,
$F=4a\sigma \cos \theta$
The force due to surface tension will be acting on the liquid surface along the perimeter of the square$\left(4a\right)$.

The component of surface tension force along vertical direction $\left(4a\sigma \cos \theta \right)$ will balance the weight of the liquid column.

If liquid is raised to a height $h$, then
$F=mg$
$4a\sigma \cos \theta =(A\times h\times \rho )g$
$4a\sigma \cos \theta =a^{2}h\rho g$
$\therefore h=\frac{4\sigma \cos \theta }{a\rho g}$