Question

A long cylindrical glass vessel has a small hole of radius t at its bottom. The depth to which the vessel can be lowered vertically in the deep water bath (surface tension T) without any water entering inside is
A . $\dfrac{4T}{\rho rg} \\ $
B . $\dfrac{3T}{\rho rg} \\ $
C . $\dfrac{2T}{\rho rg} \\ $
D . $\rho rg$

Answer 1

Hint: We are given a cylindrical glass vessel with a small hole. To find the depth to which the vessel can be lowered vertically in deep water, we remember the formula of hydrostatic pressure and the excess pressure. We know water just comes with the force, these two pressures become equal. By equating the two formulas, we get our desirable answer.

Formula used:
The formula of hydrostatic pressure = $\rho hg$
Where $\rho $= density of water in $kg/{{m}^{3}}$
G = gravitational force
H = height in m
And P = water pressure in Pa
The formula of excess pressure = $\dfrac{2T}{R}$
Where R is the radius of curvature and T is the surface tension of the liquid.

Complete step by step solution:
Given a cylindrical vessel which has a small hole of radius t.
We have to find the depth to which the vessel can be lowered vertically in the deep water. Water cannot penetrate till the hydrostatic pressure is greater than the excess of pressure that comes into play due to surface tension.

We know the formula of hydrostatic pressure = $\rho hg$
And we know the formula of excess pressure = $\dfrac{2T}{R}$
At the limiting condition, when water is just about to penetrate, these two pressures should be equal.
That means $\rho hg$= $\dfrac{2T}{R}$
Solving the above equation, we get
$\therefore h=\dfrac{2T}{\rho gr}$

Thus, option C is the correct answer.

Note: Surface tension is the film of a liquid which occurs on the surface because of the attraction of the surface particles by the liquid that tries to minimise the surface area of liquid drop .This phenomenon is called surface tension. We must remember the formulas to find out the answer otherwise we have to derive the whole equation.