Question

A vessel, whose bottom has round holes with diameter $0.1\,{\text{mm}}$, is filled with water. The maximum height up to which water can be filled without leakage is
A. $100\,{\text{cm}}$
B. $75\,{\text{cm}}$
C. $50\,{\text{cm}}$
D. $30\,{\text{cm}}$

Answer 1

If we want to fill the vessel up to a height h without leakage the pressure on the top layer of the water at a height h must be equal to the pressure on the liquid drops formed at the bottom holes of the vessel. The pressure on the liquid drop also includes the pressure due to the surface tension of the water.

Formulae used:
The rise in the height $h$ of water in a capillary tube is given by
$h = \dfrac{{2T}}{{R\rho g}}$ …… (1)
Here, $T$ is the surface tension of water, $R$ is radius of the capillary tube, $\rho$ is density of the water and $g$ is acceleration due to gravity.

Complete step by step answer:
We have given that the diameter of the holes at the bottom of the vessel is $0.1\,{\text{mm}}$.
$D = 0.1\,{\text{mm}}$
The surface tension of water is $75\,{\text{dyne/cm}}$.
$T = 75\,{\text{dyne/cm}}$

Let us determine the radius of the holes at the bottom of the vessel.
$R = \dfrac{{0.1\,{\text{mm}}}}{2}$
$\Rightarrow R = 0.05\,{\text{mm}}$
$\Rightarrow R = \left( {0.05\,{\text{mm}}} \right)\left( {\dfrac{{{{10}^{ - 1}}\,{\text{m}}}}{{1\,{\text{cm}}}}} \right)$
$\Rightarrow R = 5 \times {10^{ - 3}}\,{\text{cm}}$
Hence, the radius of the holes is $5 \times {10^{ - 3}}\,{\text{cm}}$. The radius of the holes at the bottom of the vessel and the radius of the water drops formed at the bottom of the vessel is the same.

The pressure ${P_{top}}$ on the water at the top of the water at the height $h$ is the sum of the atmospheric pressure ${P_0}$ and pressure at height.
${P_{top}} = {P_0} + h\rho g$

The pressure ${P_{drop}}$ on the water drops at the bottom of the vessel is equal to the sum of the atmospheric pressure ${P_0}$ and pressure on drop due to surface tension.
${P_{drop}} = {P_0} + \dfrac{{2T}}{R}$

The pressure on the water at the top of the water at the height $h$ is equal to the pressure on the liquid drops at the holes of the vessel in order to fill the vessel without leakage.
${P_{top}} = {P_{drop}}$
Substitute ${P_0} + h\rho g$ for ${P_{top}}$ and ${P_0} + \dfrac{{2T}}{R}$ for ${P_{drop}}$ in the above equation.
${P_0} + h\rho g = {P_0} + \dfrac{{2T}}{R}$
$\Rightarrow h\rho g = \dfrac{{2T}}{R}$
$\Rightarrow h = \dfrac{{2T}}{{R\rho g}}$

Substitute $75\,{\text{dyne/cm}}$ for $T$, $5 \times {10^{ - 3}}\,{\text{cm}}$ for $R$, $1\,{\text{g/c}}{{\text{m}}^{\text{3}}}$ for $\rho$ and $1000\,{\text{cm/}}{{\text{s}}^2}$ for $g$ in the above equation.
$\Rightarrow h = \dfrac{{2\left( {75\,{\text{dyne/cm}}} \right)}}{{\left( {5 \times {{10}^{ - 3}}\,{\text{cm}}} \right)\left( {1000\,{\text{cm/}}{{\text{s}}^2}} \right)\left( {1\,{\text{g/c}}{{\text{m}}^{\text{3}}}} \right)}}$
$\therefore h = 30\,{\text{cm}}$
Therefore, the maximum height up to which the water can be filled without leakage is $30\,{\text{cm}}$.

Hence, the correct option is D.

Note: One can also solve the same question by another method. One can equate the weight of the drop with the vertical component of the surface tension due to water as the weight of the water drop in vertically downward direction is balanced by the vertical component of the surface tension due to water acting in vertically upward direction.