Question

A capillary tube of length $l = 50\,{\text{cm}}$ and radius $r = 1/4\,{\text{mm}}$ is immersed vertically into water. Find the capillary rise in the tube in cm, to the nearest integer. Angle of contact=$0^\circ$. Take coefficient of surface tension as $72\,{\text{dyne/cm}}$, $g = 1000\,{\text{cm}} \cdot {{\text{s}}^{ - 2}}$. Round of to the nearest integer.

Answer 1

Use the formula for the surface tension of a liquid. This equation gives the relation between the surface tension of the liquid, rise in the level of a capillary tube, density of the liquid, acceleration due to gravity and angle of contact.

Formula Used: The surface tension of the liquid is given by
$T = \dfrac{{hr\rho g}}{{2\cos \theta }}$ …… (1)

Here, $T$ is the surface tension of the liquid, $h$ is the rise in the capillary tube, $r$ is the radius of the capillary tube, $\rho$ is the density of the liquid, $g$ is the acceleration due to gravity and $\theta$ is the angle of contact.

Complete step by step answer:
The capillary tube of radius $1/4\,{\text{mm}}$ and length $50\,{\text{cm}}$ is immersed vertically in the water.
$l = 50\,{\text{cm}}$
$r = \dfrac{1}{4}\,{\text{mm}}$

The surface tension is $72\,{\text{dyne/cm}}$.
$T = 72\,{\text{dyne/cm}}$

Convert the unit of the radius of the capillary tube in centimeters.
$r = \dfrac{1}{4}\,{\text{mm}}$
$\Rightarrow r = 0.25\,{\text{mm}}$
$\Rightarrow r = \left( {0.25\,{\text{mm}}} \right)\left( {\dfrac{{{{10}^{ - 1}}\,{\text{cm}}}}{{1\,{\text{mm}}}}} \right)$
$\Rightarrow r = 0.025\,{\text{cm}}$

Hence, the radius of the capillary tube is $0.025\,{\text{cm}}$.

The density of the water is $1\,{\text{g}} \cdot {\text{c}}{{\text{m}}^{ - 3}}$.
$\rho = 1\,{\text{g}} \cdot {\text{c}}{{\text{m}}^{ - 3}}$

Determine the rise in the level of the capillary tube.

Rearrange equation (1) for the rise $h$ in the level of the capillary tube.
$h = \dfrac{{2T\cos \theta }}{{r\rho g}}$

Substitute $72\,{\text{dyne/cm}}$ for $T$, $0^\circ$ for $\theta$, $0.025\,{\text{cm}}$ for $r$, $1\,{\text{g}} \cdot {\text{c}}{{\text{m}}^{ - 3}}$ for $\rho$ and $1000\,{\text{cm}} \cdot {{\text{s}}^{ - 2}}$ for $g$ in the above equation.
$h = \dfrac{{2\left( {72\,{\text{dyne/cm}}} \right)\cos 0^\circ }}{{\left( {0.025\,{\text{cm}}} \right)\left( {1\,{\text{g}} \cdot {\text{c}}{{\text{m}}^{ - 3}}} \right)\left( {1000\,{\text{cm}} \cdot {{\text{s}}^{ - 2}}} \right)}}$
$\Rightarrow h = \dfrac{{2\left( {72\,{\text{dyne/cm}}} \right)\left( 1 \right)}}{{\left( {0.025\,{\text{cm}}} \right)\left( {1\,{\text{g}} \cdot {\text{c}}{{\text{m}}^{ - 3}}} \right)\left( {1000\,{\text{cm}} \cdot {{\text{s}}^{ - 2}}} \right)}}$
$\Rightarrow h = 5.76\,{\text{cm}}$

Hence, the rise in the capillary tube is $5.76\,{\text{cm}}$.

Additional information:
If the length of the capillary tube is less than the rise in the level of the capillary tube then the liquid stops rising after reaching the top horizontal end of the tube. This happens because the surface tension becomes horizontal and the vertical force to pull the liquid upward becomes zero.

Note: Since all the physical quantities in the present case are in the CGS system of units, the unit of radius of the capillary tube is converted to the centimeter in the CGS system of units.