Question

A uniform capillary tube of inner radius $r$ is dipped vertically into a beaker filled with water. The water rises to a height $h$ in the capillary tube above the water surface in the beaker. The surface tension of water is $\sigma$. The angle of contact between water and the wall of the capillary tube is $\theta$. Ignore the mass of water in the meniscus. Which of the following statements is (are) true?

• A. For a given material of the capillary tube, $h$ decreases with increase in $r$
• B. For a given material of the capillary tube, $h$ is independent of $\sigma$
• C. If this experiment is performed in a lift going up with a constant acceleration, then $h$ decreases
• D. $h$ is proportional to contact angle $\theta$

Answer 1

Answer: (A), (C)

If this experiment is performed in a lift going up with a constant acceleration, then $h$ decreases

relation between surface tension to water density is given by,
$\frac{2\sigma }{R}=\rho gh$
$\Rightarrow h=\frac{2\sigma }{R\rho g}$ .....(1)
where,
$R$= Radius of meniscus
$r$= Radius of capillary
$h$=height of tube

Relation between radius of capillary and contact angle $\theta$
$R=\frac{r}{\cos \theta }$ .....(2)
$\theta$ = contact angle
putting eq. (2) value into eq. (1)
$h=\frac{2\sigma \cos \theta }{r\rho g}$ .....(3)

from the above expression a relationship between $h$ and $r$ is,
$h\propto \frac{1}{r}$
so as $h$ decreases $r$ increase simultaneously.
Hence, option A is correct.

For option B,
$\Rightarrow h=\frac{2\sigma }{R\rho g}$
$h\propto \sigma$
So, for a given material of the capillary tube, $h$ is dependent on $\sigma$. Hence option B is wrong.

For option C,
If lift is going upwards with the constant acceleration then the acceleration due to the gravity will be
$g_{lift}=g+a$
So, from equation 3,
$h=\frac{2\sigma \cos \theta }{r\rho (g+a)}$
where, $h\propto \frac{1}{(g+a)}$
hence, option C is correct.

For option D,
from eq. (3)
$h=\frac{2\sigma \cos \theta }{r\rho g}$
where, $h\propto cos\theta$ not $h\propto \theta$
Hence, option D is wrong.