Question

A vertical glass capillary tube, open at both ends, contains some water. Which of the following shapes can not be taken by the water in the tube?

• A.
• B.
• C.
• D.

Answer 1

Answer: (A), (B), (D)

Since the capillary tube is vertical, the weight of liquid column $W$ must be balanced by force due to surface tension acting in vertically upward direction. Only then, the liquid can remain inside the tube.

Surface tension force acts along the perimeter of the liquid meniscus, and the contact angle is $\theta$.
$F_{\text{up}}=(2\pi r\times T)\times \cos \theta =2\pi rT\cos \theta$
For the same liquid, the contact angle $\theta$ will be the same at each meniscus.

Option $\left(a\right)$:


Net surface tension force in upward direction will be,
$F_{\text{up}}=2\pi rT\cos \theta -2\pi rT\cos \theta =0$
Due to the shape of the meniscus at the upper and lower surfaces of the liquid, the vertical component of surface tension force will be in oppposite directions.
$\Rightarrow$ Weight of liquid $W$ won't be balanced.

Option $\left(b\right)$:


Net surface tension force in upward direction is zero.
$F_{\text{up}}=0$
Due to the shape of the meniscus at upper and lower surface of the liquid, the vertical component of surface tension force and weight of liquid$\left(W\right)$ will be in downward direction.
$F_{\text{down}}=W+2\pi rT\cos \theta +2\pi rT\cos \theta$
$\Rightarrow$ Weight of liquid $W$ won't be balanced.

Option $\left(c\right)$:


Net surface tension force in upward direction will be,
$F_{\text{up}}=2\pi rT\cos \theta +2\pi rT\cos \theta =4\pi rT\cos \theta$
Due to the shape of the meniscus at the upper and lower surfaces of the liquid, the vertical components of surface tension force will be added along vertical direction.

For equilibrium of liquid:
$4\pi rT\cos \theta =W$
$\Rightarrow$ Weight of liquid $W$ can be balanced by surface tension force.

Option $\left(d\right)$:


Due to flat meniscus, the surface tension force on the liquid will be acting along horizontal direction as shown in figure.

Hence, weight of liquid $W$ will not be balanced.
$\therefore$ Shape of liquid given in options $\left(a\right),\left(b\right),\left(d\right)$ are not possible.