Question

There is a hole of radius r in a cylindrical glass pot. To what depth in the sea can it be immersed so that water may not enter it? (Surface tension of water is T)
A. $\dfrac{2T}{r}$
B. $\dfrac{2T}{rgd}$
C. $\dfrac{T}{rgd}$
D. $\dfrac{rgd}{2}$

Answer 1

As a first step, you could recall the expression for force due to surface tension and also of the pressure at depth h. You could hence find the force exerted on the hole in terms of the pressure at depth h. We know that, at maximum depth at which the glass pot can be immersed without seawater entering, both these forces are equal. And hence rearrange the terms to get the answer.

Complete answer:
In the question, we are given a cylindrical glass pot that has a hole of radius r. We are asked to find the depth at which we can immerse the glass pot with a hole such that water may not enter it. The surface tension of water is given as T.
We know that the force due to surface tension T is given by,
${{F}_{s}}=T2\pi r\cos \theta$
Where, ${{F}_{s}}$ = force due to surface tension,
$r$ = radius of the surface at which this force is acting
$\theta$ = angle of contact
Angle of contact between the liquid and the solid surface is a measure of the degree of wetting.
Here, in the given case, $r$ is the radius of the hole in glass pot, T is the surface tension of water and let $\theta$ be zero, so,
$\Rightarrow {{F}_{s}}=T2\pi r\cos 0$
$\Rightarrow {{F}_{s}}=T2\pi r$ …………………………. (1)
Let $d$ be the density of seawater and$g$be the acceleration due to gravity, then, we know that pressure due to sea water at depth h is given by,
$P=dgh$
But, $P=\dfrac{F}{A}$
Where $F$ is the force acting on the circular hole by water that results in this pressure and $A$ is the cross-sectional area of the hole given by,
$A=\pi {{r}^{2}}$
$\Rightarrow {{F}_{w}}=dgh\times \pi {{r}^{2}}$ ……………………… (2)
When the force due to pressure exerted by water exceeds that due to surface tension, then the water will enter through the hole. So, we could equate (1) and (2) to get the depth,
$\Rightarrow T2\pi r=dgh\times \pi {{r}^{2}}$
$\Rightarrow h=\dfrac{2T}{rdg}$
Therefore, the pot can be kept at a depth of $h=\dfrac{2T}{rdg}$ without letting the water enter through the hole.

So, the correct answer is “Option B”.

Note:
If the water had entered the pot through hole, we could say that the adhesion to the walls is stronger than the cohesion between the liquid molecules, thus resulting in capillary action. The adhesion together with surface tension produces this action. Adhesion of water with the walls of the glass pot results in an upward force on the liquid at edges and the surface tension keeping the surface intact will drag the whole surface upward.